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1

THE EUCLIDE ANSPACES

Unit Structure :

1.0 Introduction

1.1 The Vector Spacen.

1.2 The Inner Product ofn.

1.3 The Metric Topology ofn.

1.4 Orientingn.

1.5 Exercises

1.0 INTRODUCTIO N

Differential geometry makes use of a lot of linear algebra and

multi -variable calculus. We utilize this unit consisting of Chapters

1,2,3 ,of the study material to recollect basic concepts and

elementary results o f both, linear algebra and multi -variable

calculus.

To begin with, in this chapter, we will recapitulate elementary

algebra and geometry of the Euclidean Spaces2,3,4......nn.

We discuss their basic features ab initio in three parts; (i) the real

vector space structure ofn, (ii) the inner product and the resulting

metric topology ofnand (iii) its standard orientation.

InChapter 2 we recall the algebra of linear endomorphisms

ofn, reaching finally the groupnSOof its orientation

preserving linear automorphisms and discuss some of its properties.

Actually we introduce the whole groupnGLand them

concentrate m ore on its sub -groupnOconsisting of all

orthogonal automorphisms ofnand their matrix representations.

We explain here, the total derivativeDf pof a vector valued

functionfxof a multi -variable12, ,....,nxx x x as a linear

transformation elaborating its role as a local linear approximation tofin neighborhoods of the point p(in the domain of)

Chapter 3 is a mix -bag o f some more linear algebra and a rather

long recap of basic concepts and elementary and yet fundamental

results of differential calculus (such as the inverse function theorem,munotes.in

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implicit function theorem the rank theorem….). Throughout we are

emphasizing the role ofDf pas a linear transformation

approximatingfaround p.

In what is to follow, we make use of both -the linear algebra

apparatus and the multi -variable calculus machinery in a crucial

way.For ex ample ,we differentiate a curve at a point to get the

tangent line -a linear (and hence a more amicable) curve

approximating the bending and twisting thecurve.

Similarly we approximate a (continuously bending) surface

by the tangent plane to the surf ace at a point of it.

Approximating the non-linear real world by linear objects is

indeed a fruitful, common practice .Differential geometry

emphasizes this practice.

Actually smooth curves in23and smooth surfaces in3are the main geometric objects of our interest but the analysis of

their geometry often leades us to higher dimensional Euclidean

geometry. Therefore we are treating their generality, emphasizing

particular cases of2and3.

For further details regarding the portion of this unit, the

reader should consult (1) Linear Algen, (2) Undergraduate Analysis,

both books authored by Serg Lang; and of course, the text books

recommended by th e University.

1.1 THE VECTOR SPACEnThroughout this set of notes,denotes the real number

system (aka the “real line” ). Following subsets of it appear here and

there in the text :

1,2,3,..... 00,1, 2,...., ,.....U

n

...., 2, 1,0,1,2,.....Letnbe any integer2.munotes.in

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3nstands for the set of all order ed12, ,...., ,xx xn-tuples of

real numbers. For the sake of notational economy, we denote it byx; thus;12: , ,... ,....knxx x xx the real numberkxoccupying thethkplace in thent u p l e xabove is thethkcoordinate ofx.

For any12, ,....,nxx x x n-tuples and12, ,....,nyy y y and

for any a, we put :11 2 2 : , ,....nnxy x y x y x y

and12 : , ,.......,na x ax ax ax

(again, for the notational simplicity, we will often writeaxin place

ofax)

The declarativesandgive rise to the algebraic operations:

a)addition o f vectors :

:

,nn nxy x yand b) multiplication of vectors by real numbers :

:

,nnax axThe resulting algebraic system,,nis areal vector space

(and therefore we call its elements vectors .I nstead of the complete

triple,,nwe will indicate onlyn, the underlying vector space

operations,being understood.

The dimension of this vector space isn. For, the elements

12 ,, ....,nee eofngiven by

1

21,0,....,0

0,1,....,00,....,1,....,00,.......,0k

th

ne

e

ek placee

munotes.in

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enable us to write every12, ....,nnxx x x uniquely in the

form :11 2 2 .....nnxx e x e x eand therefore, the set12, ,...,nee econsisting of thenvectors is a vector basis ofn.

We call12, ,...,nee ethestandard vector basis ofn.

It turns out that anyndimensional real vector space can be

identified withn(the identification being by meanse of an

isomorphism of vector spaces). Thus, the Euclidean spaces2,3....nnare prototypes of allfinite dim ensional real vector

spaces.

Let us note at this stage a slight deviation from the classical

vector notations in case of2and3:

In the 2 -dimensional coordinate geometry we identified a

plane with2by meanse of a Cartesian coordinate frameXOYand

then we dealt with the points of the plane in terms of the coordinate

pairs,xyw.r.t. our choice frameXOY.Similarly we used to

identify the physical space with3by meanse of an orthogonal

coordinate fromeOX Y Zand the resulting Cartesian coordinates of

a point were,,xyz. In the present con text, we use the notations12,xxin place of,xyof the planar coordinate geometry and the

triples123,,xxxin place of,,xyz. Also instead of the unit vectors,,ij k(along the axes of theOX Y Zframe) we will bring123,,eeeof

the standard basis.

Also, the arrows,,uvwover the vectors,,uvware banished,

we simply wri te,,uvweven though they are vectors.

One more point : We often consider a lower dimensionalmimbedded in a higher dimensionalnby meanse of the natural

imbedding map :

taking a point12, ,......,mxx xofmto the point

12, ,......, , 0.......0m

nmxx x

ofn. Thus occasionally we consider the

vector spacemas asubspace of a higher dimensionaln.munotes.in

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1.2 THE I NNER PRODUCT OFn:

For any12, ....,nxx x x and12, ....,nyy y y innwe

consider the sum :11 2 2 ........nnxy xy xy. Denoting it by,xywe

get the map :

,:,,nnxy xy

Note the following properties of the map,:

a)The map,isbilinear iei.e.for any,,xyzinnand for any

a, b, c in, we have

i),,,ax by z a x z b y z

ii),, ,x by cz b x y c x z

b),issymmetric,,,xy yxfor all,xyinnand

c),ispositive definite i.e.,0xxfor allnxand move over

xxwhen and only when0 0 0....0x

The map ,:nnis called the standard inner

product ofn.

In what is to follow, we consider the vector spacenequipped with the inner product,i.ewe consider the

quadruple,, ,,n ; it is thendimensional Euclidean space .

For the usual reason, we adopt and use the shorter notationnfor

the quadruple.

Thus, the Euclidean spacenis not just a barren set, it is a

mathematical space carrying two distinct structures, namely itsndimensional real vector space structure together with the

standard inner product of it. Of course, thes e two structures are

compatible with each other. One manifestation of this compatibility

is the bilineanity of the inner product : the inner product respects the

vector space operations ofn. Several other forms of themunotes.in

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compatibili ty between the algebraic and geometric features ofnwill be witnessed while studying these notes.

We proceed to explain that the inner product ofnis

geometric in nature; it gives rise to a metric i.e . a distance function

onn:

For eachnxwe writexfor,xx.

This gives rise to the function ::0 ,n

xxWe interpretxas the length of the vector x and call the map:0 ,ntheEuclidean norm onn.

The normand the inner product < , > are related by the

following inequality :

Prop osition 1 : For every,xyinnwe have ,xy x yand the

equality holds when and only whenya xfor some a(i.e. when

x and y are parallel vectors).

The above inequality is variously called the Schwarz

inequality , the Cauchy -Schwarz inequality or the CBS inequality

(CBS being the acronym for Cauchy -Buniyakowski -Schwarz, the

mathematicians who invented this inequality independently.)

Proof : The inequality is a trivial equality in case when either of x, y

is a zero vector, say y = 0. For, in that case, we have

,0 ,0 0,0 ,0xxxx

Thus,0 2 ,0xxwhich implies,0 0x. We therefore

proceed to consider0y(and therefore0y.)Now, for any

a, we have,0x ay x aythat is,

2,2 , 0xx a x y a yy .

i.e.22 22, 0xa x y a yfor any a.munotes.in

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In particular, for2,xyaythe above inequality reduces to

22 2

2

24,,20xy xy yx

yy .

Thus,2

2

2,0xyx

yor equivalenty put, we get22 2,xy x ywhich gives the desired inequalit y.

Next if,xyare parallel, sayya xfor some athen we getya xand then2,x y x ax a yxy

Thus when x and y are parallel vector, the Schwa rz inequality

becomes equality.

Finally supposexy x ywithyoand therefore,0y. Consider2,xyayand then we have

22 222 22

24

2

2

2,2 ,

,,2

,x ay x ay x a x y a yxy xy yx

yy

xyx

y

0by the assumed equality.

Thus, we have0x ay x ayand thereforexa y(with2,xyay)

The CBS inequality leades us to a geometric interpretation of

the inner product : Already we have treated:,xx xas themunotes.in

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length of the vectornx. Note that this interpretation is consist ant

with the usual length (Pythagorean) of a vector in32.

Secondly consider any pair x, y of non zero vector inn.W e

rewrite the CBS inequality in the form,11xy

xy.

This suggest s that we interprete the quantity,xyxyas the

cosine -cos-of the angle -between thevectors x, y.

This consideration inspires us to declare the perpendicularity

relation between vectors inn;xyif,0xy.

Also note that the classical Pythagoream property (about the

lengths of sides of a right angled triangle) continues to hold in the

present (higher dimensional) context : If x, y a re any elements ofnwithxythen222xy x y.

To see this, consider,

2

22,

2

20xy xy xyxx xy yyxy

Thus222xy x yholds for all x, y innwithxy.

Note that the vectors12, ....,nee ein the standard basis12, ,...,nEe e e are pairwise orthogonal and each of them has unit

length. We express this property by saying that the standard basis ofnisorthonormal . More generally a subsetofnisorthonormal

if its elements satisfy the following two conditions :

i)1xfor each xii)If x, y are any two distinct elements ofthenxy(i.e.,0xy).

Note that an orthonormal subset12.....nvv vofnis

linearly independ ent.For ifmunotes.in

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11 2 20mm av av a v…………………………… (*)

holds for some real numbers12, ,....., ,maa athen we deduce that

120m aa a. To get this take the inner product of the equality(*)

with each1vto get.

11 2 2... ... 0ii i i i m m i av v av v av v a v v i.e.200 1 0 0ii maa a awhich gives0iafor each1ii m. This justifies our claim that the orthonormal set1mvvis linearly independent. On t he other hand any linearly

independent subset ofngives rise to an orthonormal subset having

as many elements as those of the linearly independent subset. We

prove this fact in the following proposition :

Proposition 2 : Any line arly independent subset12,.....,mvv v

ofnngives rise to an orthonormal subset12...mff fofnin

which each'1fis a linear c ombination of12,.....,ivv v m.

Proof :1vbeing an element of linearly independent set is non zero.

Therefore10vand therefore111:vfvis a well defined unit

vector.

Next , we consider21 122 1 1 2 2

1vv vvv f f v

v. This vector

also is non -zero. (For, otherwise we would get21 12 2

1vv vv

vwhich

contradicts the linear independence of the elements of the set A. We

put22 1 1222 1 1vv v ffvv v f

Clearly121ffand12ff.

In the neat step, we consider3vand obtain the vector33 1 13 2 2vv f fv f ffrom it. Invokingthe linear independence of

the set A, we again get that this vector is non -zero. Using this last

observation, we construct :munotes.in

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1033 1 13 2 2333 1 13 2 2vv f fv f ffvv f fv f f

Imitating this procedure successively, we obtain the desired

ortho-normal set12, ....mff fwhere thekffor 2kmis given

inductively by11 2 2 1 111 2 2 1 1...

, ...kk k k k k

k

kk k k k kvv f fv f f v fffvv f fv f f v ff

This method of obtaining an orthonormal set12, ....mff ffrom

a linearly independent set12, ....mvv vof vectors is called the

Gram -Schmidt orthonormalization process. Application of this

process to an arbitrary basis ofnenables us to get a new vector

basis which is orthonormal.

1.3 THE METRIC TOPOLOGY OFn:

The inner product;ofngives rise to a complete

separable metric topology on it in the following way :

For any,xyinn, we put :,,dx y x y x y x yor equivalently 21,n

jj

jdx y x y . This assignment gives rise

to the map ::0 ,

,,nndxy d xy x y This map isin fact a metric onn:

We readily have :

i),0dx yfor all,xyinnand,0dx yif and only ifxy.

ii),,dx y dy xfor,xyinn.

More over, for any,,xyzinn, we h avemunotes.in

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2 2

2

22

22,

,

,2 , ,

2,

2dx z x z

xyyz

xyyz xyyzxy xy xy yz yz yzxy xy yz yz

xy xyyz yz

By the CBS inequality. 22;,xy yzdx y dy z

Thus 22,,dx z dx y dy z for all,,xyzinn.

Thereby we get the triangle inequality ,

,,dx z dx y dy zfor all,,xyzinn.

Thus, the Euclidean spacenis actually a metric base but we

will not indicate its metric. All the topological considerations will be

in referenc e to this Euclidean metric topology. Among all the

properties of the metric spacen, we mention only the following

two :

i)nis a complete metric space ;

ii)nis separable.

Property (ii) can be seen here itself : Letnbe the set of all

ordered n -types12, ,...naa aof rational numbers. Then the setnis a

countable, and dense subjectnand he ncenis separable.

We pro ve property in the following proposition :

Proposition 3 : The metric spacenis complete.

Proof : We consider a Cauchy sequence:kvkinn. Writing

each termkvin terms of its coordinates12, ;...nkk k kvv v v .

We split the sequence:kvkinto n sequences of real

numbers:12: , : ...... :n

kk kvk vk vk .munotes.in

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Note that for each.kinand for each1ii nwe have :

jjke k ke kvvvv ,……………………….. (*)

The inequalities (*) imply that the Cauchy property of:kvkinduces Cauchy pr operty in each of the coordinate

sequences :12: , : ...... :n

kk kvk vk vk .

By the completeness of the real line, we get real numbers

12, ,....nww wwhich are limits of the (Cauchy) coordinate sequences :

12

12lim , lim ,,.... limnkk n kkk kwv wv wv .

We form the vector12, ,...., .nnww w w Finally note that

1,n

kkdv w v wwhich (together with

the above deduction thatkvwaskfor1n)implies that

kv.

Thus, each Cauchy sequence:kvkinnconverges to anwand thereforenis complete.

We observe one more property of the metric topology ofn.

Let A be any subset of amand let a be any point of A. Let

12, ,... :n ff f A be funct ions, all being continuous at a.

Let :nfAbe the map given by12 , ,......nnfy fy f y f y for each y.

Proposition 4 : The map :nfAis continuous ata.

Proof : Letobe given. Then foron, continuity of each1ifi nat a implies that there existiosuch that

iify fanfor all,ya i.munotes.in

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Consider12min , ....n . Thenoand,ya

implies,1ya ii n and therefore iify fanfor

1in. This set of inequalities implies :fy fafor all,yaproving continuity offata.

1.4 ORIE NTINGn:

Orientation ofnand its orientability in two different ways is

yet another aspect of its geometry. Here, we give a brief, heuristic

introduction to the main ideas related to the orientations ofn.W e

use only elemen tary geometric concepts. A precise algebraic

formulation of it (in terms of orthonormal transformations ofn)

will be given in the ne xtchapter.

The term orientation applies primarily to orthogonal frames inn. W try to reach the vast expense ofnby means of an

orthogonal frame12....ss nff f FF associated with an orthonormal

vector basis12....nff f.

Recall, an orthogonal frame12....snff fF is obtained by

laying its axes12 .....nOX OX OX along the vectors

12,....nff frespectively.

Schematic depition of an orthogonal frame12....snff fF

Fig. 1

Clearly the frame12....snff fF andthe ordered orthonormal vector

basis12....nff fspecify each other and therefore, we often talk of

them interchangeably.

We consider various12....snff fFmunotes.in

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Of course, there is the standard orthornormal frame12....nee eF introduced earlier. But this particular frame may not

be the best choice to study a specific geometric / physical problem.

For example in studying the rotational motion of a spinning top

(with its nail tip remaining stationary on the ground ) we need

consider besides the stationary frame the rotating body framesFwhich is an orthogonal frame fixed in the top and therefore it is a

moving orthogonal frame. And we study the rotational motion of the

top by studying how th e body framesFchanges its orientation with

respect to the stationary framesF.Thus, we need two distinct

orthogonal frames to study the dynamics of a spinning top.

We therefore consider allorthogonal frames12....snff fF and

compare them with the standard frame12....nsee eFF

Fig. 2 : The stationary framesFand the body framesFHow do we compare two frames?

It is in tuitively clear that we can rotatesFabout the common

origin and make it coincide withsF. This corresponds to a change12 3 1 2 3.... ....ff f e e eof the orthogonal bases associated with the

two framessFand0sF.

Recall now the elementary facts of linear algebra. (We will

discuss more about there in the next chapter.)

Each change of orthonormal basis (and therefore that of the

associated orthonormal fra mes)12 1 2.... ....nnee e f f fgives rise

to a unique orthogonal linear transformation :nnT given byiTe f ifor1in.munotes.in

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Each such T has the property det (T)=+1 or -1. We use these

properties t ocompare two frames. We say that two orthogonal

framessFand0sFhave the same orientation if det (T) = +1

and they have the opposite orientation if det T = -1.

We regard this classification of orthogon al frames into two

disjoint families as two orientation sof the spacen; we call them

the “standard orientation” and the “opposite orientation” ofn.

Thus we have the following :The standard orientation ofnpertains to the orthogonal

frames12....snff fF with the property that the associated

:nnT (with1iiTe f i n) has det T = +1.

The opposite orien tation ofnpertains to any orthogonal frame12....snff fF with det T = -1.

Thus each Euclidean spacencarries two distinct orientations,

namely (a) the standard orientation as described i nand (b) the opposite

orientation described in.

Applying all this consideration to3, our physical space ;we

have an equivalent, but rather tangible description in the po pular

language : Orthogonal frames being left handed andright handed :

12....snff fF is right handed if the frame can be grabed by right

hand so that the thumb points in the direction of3f.

On the other hand12....snff fF is left handed if it can be grabed

by the left hand so that the thumb (again) points in the direction

of3f.

Left and Right Handedness of Orthogonal Frames

Figure 3munotes.in

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Of course the right handed frames dete rmine the standard

orientation of3while the left handed frames determine the opposite

orientation of it.

1.5 EXERCISES :

1)Prove that...nnis a dense subset ofn.

2)Apply Grah m-Schmidt process to obtain orthonormal sets from

the given (linearly independent) subsets :

a)21, 3 ; 2, 4b)31, 3,1 1, 4,1 0, 2,1 c)31, 2, 3 , 2, 3,1 , 3,1, 2 3)Prove that any n -dimensional real vector space is isomorphic

withn.

4)Prove that any two vector bases innhave equal number of

elements.

5)Give all the details regarding the proof that:0 ,nndgiven by,,dx y x y x yinn, is a metric.

6)Describe a real vector space which is not isomorphic with anyn. (Justify your claim s)

7)Let :, :nm mfg be maps such that fis continuous at

anpand g is continuous offp q. Prove continuity of

:ngof at p.

8)Recall :(i) a multi -index is an ordered n -tuple.12, ,....,n where eachi.

ii)

1niiiii) 121, , ,...,nininixx x x x x

iv)A polynomial in the multi -variable12, ,...,nn xx x xis finite

linear combination.:

mpx ax

Prove (a) each monomial :nxis continuous onn.

b)andtherefore each :np is continuous onn.

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2

ORTHOGO NAL TRA NSFORMATIO NS

Unit Structure :

2.0 Introduction

2.1 Linear Transformation

2.2 Algebra of Matrices

2.3 Determinant of a Linear Endomorphism o fn2.4 Trace of an Operator

2.5 Orthogonal Linear Transformations

2.6 The Total Derivative

2.7 Exercises

2.0INTRODUCTIO N

In the preceding chapter, an Euclidean space was introduced

as a mathematical system consisting of the setncarrying three

mutually compatible structures, namely (i) the n -dimensional real

vector space structure, (ii) the inner product giving rise to the metric

topology ofnand (iii) the standard orien tation of it.

In this chapter we will discuss linear transformations between

Euclidean spaces and their properties. In particular, we will come

across the groupnGL(,,GL n GL n are other notations for the

same )consisting of bijective linear self maps of the vector spacen(aself-map is a map of the type :fi.e. a map of a setto

the same set.) Actually we are moving towards a sub -groupnSO(or,,SO n SO netc.) ofnGL; it is the group of

symmetries or the automorphisms of the Eudidean spacen. These

transformations -being symmetries ofn-help us understand the

shapes of geometric objects residing inn: smooth curves, smooth

surfaces, higher dimensional s mooth manifolds…. Also, being

automorphisms of the vector space, they play an important role in

the derivation of many results of differential geometry.

We begin with a recall of basic concepts of linear algebra. (:

linear transformations between Euclidean spaces, their matrix

representation, the algebra of linear transformations and its reflectinmunotes.in

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in the albegra of matrices and so on, reaching finally the groupsSO n). We will say a little more about the forms of the matrices in2SO and3SO .

We will also recall a bit of diffe rential calculus of vector

valued functions:nmfof a multi -variable12, ,....,n xx x x . Recalling the definition of the total derivativeDf pof such afat a pas a linear transformation fromntomwe take the view -point that differentiation of a function at a point

is a process employed to approximate a general (differentiable) map

locally by a linear transformation. This is an important

interpretation, because we can now use all the machinery of linear

algebra to get information about the local behaviour of such afaround a point p of its domain of definition.

Basic result s of differential calculus mentioned in this chapter

and the next one are : the inverse mapping theorem, the implicit

mapping and the rank theorems Picard’s existence / uniqueness

theorem about the solution of an ODE and so on. We state these

results (they go without proof) here in this set of notes because they

are used here and there in differential geometry and therefore, a

student should know at least the precise enunciations of these results.

Detailed proofs of them are equally important and the reader can

consult a suitable analysis book (e.g. one of the text -books by Serg

Lang)

2.1 LI NAR TRA NSFORMATIO NS

Definition 1 :

a)A linear transformation fromntomis a map :

:nnT which satisfies the identity :T ax by aT x bT y

for all,xyinnand for all a, b in.

Occasionally we speak of a linear map instead of a linear

transformation.

b)A linear self map :nnT is called a linear endomorphis m(or

often merely an endomorphism) ofn. It is also said to be an

operato ronn.munotes.in

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c)A bijective linear endomorphism :nnT is said to be a

linear automorphis m(or only an automorphis m) ofnd)A linear map :nT is called a linear form onn.

We adopt the following notations :

,nmLdenotes the set of all linear maps : :nmT .

Endnis the set of all linear endomorphisions :nnT Autnis the set of all linear automorphison s ofn.

*ndenotes the set of all linear forms onnWe note here a few basic properties of linear linear

transformations and their spaces listed above. Most of these

properties are stated here without proof, because they are discussed

routinely in any linear algebra courses. The reason why these

properties are listed here is only to refresh readers memory about the

precise statements and the full import of these properties :

I)If S, T are linear transformations fromntomand if,are

an real numbers then they combine to give a map :

:nmST………………………… ………. (*)

which is given by :ST x S x T for allnx.

This mapSTis also a linear transformation fromntom.

II)Let,nmLbe the set of all linear transformations fromntom. Then the operation (*) (described above) combining two linear

transformations S, T and two real number,producing the linear

transformationSTis an algebraic operation giving the set,nmLthe structure of a real vector space. Thus, the set,nmLtogether with the operation (*)is a real vecto r space. We

will prove that the dimension of this vector space is m.n.

III)In particular the set*nis a vector space and its dimension is.1 .nnmunotes.in

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We justify this claim by describing a bijective linear map*:nnas follows.

Let*nTbe arbitrary.

For each1ii nwe putiiyT e. We form the vector12, ,.....nyy y y . Now for eachnxwe ha ve :11 2 2

11 2 2

11 2 2, ....

....

......nnnnnnTx Txe x e x exT e xT e x T exy xy xy

xy

Thus, with each*nTis associated anysatisfying.,,nTx x y x .

Clearly this y (associated with the*nT) is unique. We

putOT y. Now we have the map*:nnO. It is easy to

prove that this map()His bijective and linear.

IV) a)If :, :nm mkSTthen0:nkTSis also

linear.

b)If :nnTis bijective linear then its inverse

1:nnTalso is linear .

V)Let12, ,...nff fbe any vector basis ofnand let12, ,.....nvv vbe

any vectors in am. Then there exists a unique linear

:nmThaving the property :1iiTf v i n.

The unique linear T is given as follows :

Let11 2 2 ...nnxx f x f x fbe any vector inn. Then by linearity

of T, we have11 2 2

12 2

11 2 2nnnnnnTx Tx f x f x fxT f x T f x T fxv xv xv

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VI) In particular we consider m = n and in place of12, ,...,nff fwe take the standard basis12, ,...,nee eofn.Next for

any pair,1 ,ij ij n, we consider the set12, ,.....,nvv vwhereijveand0ivfor all other1n. By property (V) above, we

get a unique linear :nnijT satisfying.ij k

jTe o i f kiei fk iThus,ij i jTx x efor allnx.

We consider the set:1 ,ijTi j n. It is easy to prove that

this set is linearly independent

1i j ij

ij nTo

only when allijo.

On the other hand we prove that any T is a linear combination

of:1 ,ijTi j n: In fact let1

jnT ei ij ej for i n

. Then

1i j ijij nTT

.

This shows that:1 ,ijTi j nis a a vector basis of

Endnand thus it is a vector space having dimension2n.

VII) Thus, the set Endncarries two kinds of algebraic

operations namely. (a) the vector space op erations and (b) the

composition:nn no End End End taking a pair S, T toSoT. Note that ‘o’distributes over the vector space operationsSTR S T S R .

Thus Endnis a real n -dimensional algebra.

VIII) LetnGLbe the set of all linear automorphisms ofn(We

often denote itby,GL nor byGL n.) The setnGLhas the

following properties :

i)nIG L, (I being the identity transformation onn.

ii)If,nT GLthen1nT also GLiii)If S, T are both innGLthennS T GL munotes.in

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In other words, the system,nGL ois a group. We call it

then-dimensional general linear group .

2.2 ALGEBRA OF MATRICES :

Matrices are computational counterparts of linear

transformations. With each o perator T ofn, we associate square

matrix and the neumerical calculations done on matrices given

information about their predessors.

Recall for any m, n in, a matrix of sizemnis an array

1

1imijjnAaof real numbers arranged in m rows and n columns (the

numbersijabeing placed at the cross -roads of ith row and jth

column:

11 1 1 1

11 1

1

1

12...... ......................... ..................... .......................... .............................. ............, ... .. .......zj n

iz j i n

im ij

jnmm m j m naa a a

aa a a

Aaaa a a

We often write onlyijainstead of1

1imijjnawhenever the

size of the matrix is understood.

We represent a linear :nmTby amnmatrix (T) and

use the latter as a computational device to get informati on of the

linear transformation T.

Recall two matricesijAaandijBbof the same size are

equal A =B if and only ifij ijabholds for all pairs (i, j) -,,Mm ndenot es the set of all real matrices of sizemn.

When m = n, we write,Mnfor the set,,Mn nand the

matrices in it are said to be square matrices (of sizenn.)

The set, Mm nhas the structure of a real vector space : IfijAaijBbare any two matrices and if,areany two realmunotes.in

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numbers, then we defineABas themnmatrixijcCgiven

by :

1, 1ij ij ijcab i mj nThusABis the matrixij ijab .

For any pair,1 1ij i m j n, letijAbe the matrix :

there being zeros at all places inijAexcept at theijth-place

where we have 1.

It now follows that any matrixijAa,,Mm ncan be

expressed uniquely as the linear combination.ij ijAa Athe sum above extending over all pairsijwith 1imand

1jnand consequently, the vector space, Mm mhas

dimension m.n.

We now recall the multiplication of matrices : for any m,n

and p in, let,,ij Aa M m n,,jk Bb M n p. Then

the matrix,,ik Dd M m pgiven by

1ik ij jkjmda bis defined as

the productDA B(the factors A, B of D in the indicated order).

Note that both, the productsABandBAare defined only

when m = n = p i.e. wh en both A, B are square matrices of the same

size. We pursue this case (i.e. of square matrices) by the following

hands -on account :

The set,Mncarries the following algebraic operations

(all explained above in the more general c ontext) :

Addition of matrices :munotes.in

## Page 24

24:, ,,Mn Mn MnAB A B

Multiplication of a matrix by a real number ::, ,,Mn Mn

A

Multiplication of matrices:, , ,,,Mn Mn Mn

AB AB

The set,Mntogether with the ab ove three algebraic

operations is an2ndimensional associative (real) algebra with

identity (i.e. it is a combination of a2n-dimensional real vector

space and aring with identity.)

Thus, on one hand we have the operator algebra Autnand

on the other hand, we have the algebra,Mnofnnreal

matrices. We proceed to explain below that an orthonomal vector

basis ofnestablishes an isomorphism (of algebras) between the

two.

Thus let12, ,....,nff fF be an orthonormal basis ofn.

Now for anT Endand for each1jfj nwe get the vectorjTfexpressing it as the linear combination :11j ij i

iTf af j n…………………............... (*)

We collect the coefficientsijain (*) above and form the matrixijawhich we den ote by [T] or more accurately byTF.

Thus the orthonormal basis ofngives rise to the map

End,s nMn AA AFWe note the followi ng properties of this map -

i)The map is a bijection betweennEndand,Mnmunotes.in

## Page 25

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ii)The map preserves the algebraic operations on the two sets, that

is,the following equalities hold :

,sij FII being the identity operator onnandijis the

identitynnmatrix :

ST S T

SSST S TThese properties -descri bed in (I) and (II) above -taken

together imply that the map (*) is an isomorphism between the two

algebraic systems.

The third proper -property (III) stated below -is about the

dependence of the matrix representationTFof an opera tor T on the

orthonormal basis:FLet12, ,....,nff fF and12, ,...ngg gG be two

orthonormal bases ofn. If T is any operator onn, then the two

bases asso ciate the matricesTFandTG. We seek a relation

between the two matrix representation. Towards this aim, we

consider the matrixijCCdescribing the change of the vector

basesFGthus for each ,1jj nwe have :

1nj ij iigc fApplying T to this equality, we get

1nj ij iiTg c T f

Now ifijTaFandijTbGthen we have

11nnik i k i k i kkkTf a f T g b g and therefore, we getmunotes.in

## Page 26

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11

11nn

ej e ij ij

i

nnij ki kikbg c T fca f

Therefore

11 11nn nnej ke k j ke kek kbc f ca f that is,kl ej k k ej kkkcb f a c f .

Equating coefficients of eachkf, we get

11nnke ej ke ejcb ac .

The abo ve equalities are obtained for each pair (k, j) with

1,kj nand therefore, we get the equality of the matrices :ke ej ke ejeecb acthat is, we haveCB AC.

Now note that C is invertible (it being the matrix onnecting

two vector bases,andg) and therefore the last equality implies

1..BCA C

that is :1TC T CGF.

We summarize it in the third property of matrices :

III) For any two orthonormal bases,FGofnand for any operator

T onn, we have :1TC T CGF

Cin above being the matrix of the change of vector bases

fromFtoG.

We use this property crucially in defining the determinant of

an operator T onn.munotes.in

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2.3DETERMI NANT OF A LI NEAR EUDOMORPHISM

OFn:

First we define the determinant of a square matrix and then

extend itto linear endomorphisms.

Recall, first, the permutation group S(n) of set {1,2,….,n}.

Also recall that eachSnhas its signature1, 1.

Definition 3 : For a square matrixijAaof sizenn, the

determinant det(A) is the number 1, 2, , det .......ii n nSnAa a a

,1..........iiinSna

………………….. (*)

Now we have a functi on :

det :, Mn detAAWe mention (without proof) following three properties of this

function.

1)det (I) =1, I being the identitynnmatrix :ijI2)det det detAB A Bfor any A, B in,Mnand

3)a matrixijAais invertible if an only ifdetAo.

Note that property ( 2) above has the following import ant

consequence : If C is any invertiblennmatrix, then for anynnmatrix A we have the equality :1det detA C AC

In fact we have11

1det det det

det det detC AC C ACCA C

= det Amunotes.in

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We also have 1111 det det detdetIC CCC and

therefore 11detdetCC. Applying this result we get :11det det det det

detCA C C A CA

We use the property of determinant of square matrices to

definedetTof an operator T onn.

For an orthonormal basisFwe consider the matrixTFof T

w.r.t.F. Invol ving the formula*above, we consider detTFand

then we observe that this number, thus arrived at, is actually

independent of the vector basisFused (and therefore, it is actually

an attribute of the operator T itself and not that of its matrix

representa tion.) For, if gis any other orthonormal basis ofn, then

we have :1TC T CGFand therefore,1det det

detTC T CT

GF

F

Thus,det detTTFGfor any orthonormal basesFandGofn. We definedetTto be this common value :det det detTT T FG.

Now we have the function :det :nEnd.

This map has the following prop erties

det 1I, I being the identity operator onndet det detST S T for all S T innEnd.

An operator T is invertible if and only ifdetTO.

If T is invert ible, then 11detdetTT.munotes.in

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2.4TRACE OF A NOPERATOR

There is yet another invariant associated with an operator T

onn, namely its trace . Like the determinant of an operator, we

define it first for a square matr ix and then extend it to an operator.

Definition 4 : The trace of a matrixijAa-denoted bytr A-is

given by

1niiitr A a.

Note that for anyijAa,ijBbwe have

1nik kjkAB a b

and therefore,

11

11:nnik kiik

nnki ikkitr AB a bbatr BA

Thustr AB tr BAfor any A, B in,Mn. We use this

property to definetr Tof an operator T : Choose a ny orthonormal

basisFand considerTr TFas defined above. We claim t hat this

number does not depend on the orthonormal basisF. For, letFandGbe two orthnormal bases with C as the matrix describing

the changeFtoG. Then for any operator T onn, we have1TC T CGF

and therefore,

11,tr T tr C T Ctr T C Ctr T

GF

F

F

This leades us to the definitiontr T tr T Tr TFG.

Now, we have the function:n tr End Two of the, properties of this map are

1)tr I nmunotes.in

## Page 30

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2)tr S T tr T Sfor any, S, T innEnd.

In what is to follow, we will be using only the standard

orthonormal basis12, ,.....,nee eas a convenient choice and

therefore the matrix representationTof T will be u nderstood to be

with respect to the standard orthornormal basis:TT.

2.5ORTHOGO NAL LI NEAR TRA NSFORMATIO NS:

We single out a sub -group of the groupnGL.

Definition : A linear :nnT isorthogonal if it preserves the

inner product.,,Tx Ty x yfor all,xyinn.nOdenotes the set of all orthorgonal :nnT .

Note the fo llowing elementary properties of orthogonal

transformations.

A linear :nnT is orthogonal if and only if it preserves the

Euclidean norm of the vectors.

As an immediate consequence of the above we get that an

orthogonal T is bijecti ve.

T is orthogonal if and only if,i j ijTe T efor all

,, 1 ,ij ij n.

i) The identity map :nI is orthogonal

ii) If T is orthogonal then so is1Tiii) If S, T are orthogonal, th enso isSoT.

Thus, the composition operation : ,ST S o Tbecomes a

binary operation on the setnOin such a way that,nOois a

sub-group ofnGL. We denote this sub -group by the underlying

setnOonly and call it the n-dimensional orthogonal group .

We characterize an orthogonal :nnT in terms of a

property of its matrix.munotes.in

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Proposition 1 : A lin ear :nnT is orthogonal if and only if its

matrixATsatisfiestAA I.

Proof : Suppose, T is orthogonal. Then for any pair (i, j) with

1,ij nwe have

,,ij i j i jTe Te ee

Now

11,nn

ik i k je j

kkTe a e Te a e and therefore :

11

1

,

,,, ,,nn

ij k i k j

k

ki j k

k

ki j k

k

ki kj

kTe Te a e a eaa e e

aa

aa

Note that

1nki kjkaais the thijentry in the matrix.tAA.Now

the equalities.

11,n

ki kj ij

kaa ij n

1nki kj ijkaa that istAAI.

The proof of the converse is left as an exercise.

We consider the determinant of an orthogonal T. On one

hand1tAAforATand therefore, we getdet 1tAA. But2det det det det det detttA A A product A A A A .Thus

2det 1Aholds for an orthogonal T with A = det (T). We consider

all orthogonal T withdet 1T.

Let: det 1nnSO T O T .munotes.in

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Note that because the mapdet :nO is multiplicative,

the setnSOis a sub -group of the groupnO.

Definition : The groupnSOis the n-dimensional special li near

group .

In the next chapter we will define orientations ofnusing

the groupnSO.

2.6THE TOTAL DERIVATIVE :

Letbe an open subset ofn,pa point ofand let

:mfbe any map.

We explain in few words the concept of total derivative of

such a vector valued function of a multi -variable12...n xx x xas a linear transformation ::nmDf p .

Definition 7 :fis differentiable at p if there exists a linear map

:nmT such thatlim

hofph fp T hOhNote that the quantityfph fp T hhis defined for

non-zero but smallnhand the limit being zero indicates thatfph fp T his a quantity of second order smallness in

comparison with the “increment” h. Thus differentiability of f at p is

about approximating the vari ationfph fpoffaround p by

the linear map :nmT .

Recall from analysis that any linear :nmT is continuous

at O (actually at every point ofn). Consequently for af;

differentiable at p;fph fp T h OandTh O

ashOimplies thatfph fp OashOi.e.fis

continuous at p. Thus, the classical result :differentiability of a

function at a point implies continuity of it at the same point -

continues to hold in the present context also.munotes.in

## Page 33

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Next, note that iffis differentiable at a pthen the linear

T appearing in the definition must be unique. To see this consider

two linear maps12,:nmTT satisfying :1

2fph fp ThOh

fph fp ThOh

ashO. Then we get12Th ThOhashO. But this implies12TT. To see this, consider any hon -zero2x. Then for k,

large enough use cons iderxhkso thathOask. Thus

12xxTTkkOx

kaskBut12

12xxTTTx Txkk

x x

k

Therefore,12

12lim

kxxTTTx Tx kkOx x

k .

This giv es12Tx Txwheneverxo. But12To To oby the linearity of1Tand2T.Therefore12Tx Txholds for allnxi.e.12TT.

We call the unique T the total derivative offat p and denote

it byDf p. Thus, whenfis differentiable at a point p of its

domain, its tot al derivativeDf pis a linear transformationnmDf p .munotes.in

## Page 34

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We consider the matrixDf pof the total derivative.

Suppose1

1imijjnDf p a. We ask what areija.To answer this we

have

limjtofpt e j fpDf p et .

The ithcomponent of this vector equation islimii

ijtofpt e j fpat

that is i

ij

jfapx. Thus we get i

jfDf p px. In classical

literature this matrix i

jfpxis called the Jacobean matrix of the

total derivativeDf p.

In particular if the mapfis differentiable at every pthen

we get the map:nmDf Lwith the partial derivatives

:i

jf

x.

We say that the map :mfiscontinuously differentiable

onif (i)fis differentiable at every pand if (ii) all the partial

derivatives :1 , 1i

jfim jnx are continuous on.

We will discuss more differential calculus in Chapter 3.

2.7 EXERCISES:

1)Let :nnT be linear. Prove that there exists a constant

Csuch thatTx C xholds for allnx.

Hence or otherwise deduce that any linear :nnT is

continuous at every point ofn.

2)Prove that any linear :nT satisfies,,Ts Ty x yfor

every,xyinnif and only ifTx xfor allnx.munotes.in

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3)LetATfor anT End.Suppose A satisfiestAA I.

Prove that T is ortrogonal.

4)Prove thatnOis a group andnSOis a normal sub -group of

it.

5)Prove thatnSOhas exactly two cosets innO.

6)Provetr ST tr TSholds for all linear S, :nnT .

7)Prove : If :nnT is linear, thenDT p Tfor everynp.

8)Let ,mnbe open sets and lotPbe arbitrary withaf p.

Let :, :fgbe maps such that (i)fis differentiable

at p, (ii) g is differentiable atqf p. Prove thatgofis

differentiable at p and derive :D gof p Dg q Df p.

9)Prove thatnGLis an open subset ofnEnd.

munotes.in

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3

ISOMETRIES OFn,SMOOTH

FUNCTIO NSONnUnit Structure :

3.0 Introduction

3.1 Isometries ofn3.2 Orientations ofn3.3 Smooth Functions

3.4 Basic Theorems of Differential Calculus

3.5 Exercises

3.0 INTRODUCTIO N

Having introduced the groupnOof orthogonal linear

transformation ofn, we discuss a large r group of transformations

of an, namely the group of isometries ofnwhere an isometry ofnis a bijective self map ofnwhich preserves the distance

between its points. First, we derive the basic result describing an

isometry as a rigid motion ofnie.amap which is a composition of

a rotation and a translation inn. We verify that such rigid motions

innform a group.

In the remaining part, we discuss some basic theorems of

differential calculus. We introduce the function spaceCof

smooth real valued functions of a multivariable ranging in an open

subsetofn.

3.1ISOMETRIES OFnDefinition 1 : An isometry ofnis a bijective map

:nnf which preserves distance between any two points ofn:fx fy x yfor all,xyinn.

Here are some simple facts about the isometries ofn:munotes.in

## Page 37

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Every orthogonal linear :nnT is an isometry ofn; for, let,xyinnbe arbitrary. Then we have :,Tx y Tx y x y x yfor all,xyinni.e.,Tx Ty Tx Ty x y x yand therefore :Tx Ty x yfor every,xyinnFor eachnalet :nnaT be the map given byaTx xafor everynx. The bijective mapaTiscalled the

translation map inndetermined by its element a. Clearly, eachaTis an isometry ofn.

If :nnf is an isometry ofnthem its inverse

1:nnfalso is an isometry ofn.

If ,:nnfg are isometries ofn, then so is their

composition :nngf LetnIsobe the set of all isometries ofn. It then follows

that the composition of self maps ofnwhen restricted tonIsobecomes a binary operation onnIsoand the resulting algebraio

system :,0nIso

is a group. It is the group of isometies ofn. It is easy to see that

orthogonal transformations ofnconstitute a sub group of,0nIso. Also the set of all translational maps i.e.:naTa is

also another subgroup of the isometry group.

Now, we obtain a result regarding the structure of an isometry

ofn.

Let :nnf be an isometry.

Let0af. Define :nnR by0Rx f x f for

eachnx. Thus, we have :munotes.in

## Page 38

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000

0n

afx fx f f

fx f a

TR x x

We prove below that :nnR is linear and preserves inner

product ofn(and therefore is an orthogonal transformation ofn).

Now, for any,xyinnwe havefx fy R x R y

and thereforeRx Ry f x f y x y(sincefis

isometry. Thus we have,,Rx Ry Rx Ry x yx yand therefore :,2 , ,Rx Rx Rx Rx Ry Ry 2, ,xx xy yy

i.e. 222,Rx Rx Ry Ry 222xx y yfor all,xyinn………………..(*)

Recall00Rand therefore00 0Rx Rx Rx R x xand similarlyRy y. Using these results, above yields. …………… (*),,Rx Ry xyfor all,xyinn.

We use the id entity (**) to deduce linearly of R as follows.

First, note that (**) implies that:1iTe i nis

orthonormal. Therefore, for any,nxwe have

1,niiiRx Rx Re Re

But again by (**) we have,,ii iRx Re x e xfor eachi1in. Therefore

1niizRx x Re…………………… (***)munotes.in

## Page 39

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for eachnx.

Now for any,xyinnand for any a, b inwe have

1

11n

ii i

i

nnii iiiiR ax by ax by R eax R e by R e

aR x bR yusing …………………… (***)

This prove linearly of R.Thus we have prove above both :

linearly in (***) a ndinner product preserving property (* *) and

therefore T is orthogonal.

Uniqueness of the decomposition0afTRis left as an

exercise for the reader.

We summarize this result in the following :

Proposition 1 : Every isometry :nnf is expressi ble uniquely

inthe form0afT Rwhere R is an orthogonal transformation ofnandaTis the translation with0af.

3.2ORIE NTATIO NS OFn:

The concept of orientation ofnwas introduced in Chapter 1

in terms of families of orthogonal frames ofn. It was shown thatnhas exactly two orientations. In this chapter we reformulate it

shightly differently so as to involve the groupSO n. We bring

orthonormal bases in place of the orthogonal frames and decompose

the set of all orthonormal bases into two classes, they are

equivalence classes of a c ertain equivalence relation, the later being

introduced in terms of the groupnSO.

To begin with, note that each orthogonal frameFinndetermines and is determined by an orde red orthornormal basis12, ...,nff f, the ithunit vectorifpointing along the ithaxisofF.

Thus there is a 1 -1 correspondance between orthogonal framesFinnand ordered orthonormal vector bases12, ...,nff f. Now we

consider ordered orthonormal bases instead of orthogonal frames to

specify orientations ofn. We make this choice because now we aremunotes.in

## Page 40

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acquainted with the groupnSO,the elements of the group

enabling us to change from one frame to another similarly oriented

frame.

Letdenote the set of all ordered orthonormal vector bases

ofn. We introduce a relationon the setas follows : Let12, ...,nff fand12, ...,ngg gbe any two ordered orthonormal bases.

Then there exists a unique orthogo nal linear :nnT satisfyingiiTf gfor1in. Moreover, we have :det 1Tor-1.

We set :12 12,...., ,....,nnff f g g g if an only ifdet 1T.

Clear ly the relationthus defined is an equivalence relation

on. Therefore it decomposes the setinto disjoint subsets of if

namely the equivalence classes of the relation:

Each equivalence class is said to determine an orientation ofn.

Finally becausedet 1Tor-1 for eachTwe see that

there are two distinct equivalence cl asses and hence two distinct

orientation sofn.

The equivalence class containing the standard basis12, ,.....,nee e.

To describe the other class consider the vector basis12, ,.....,nee e ofn. Let :nnT be the linear transformation

ofnassociated with the change12, ,.....,nee eto12, ,.....,nee e .

Clearly

and thereforedet 1T. Thu s12, ,.....,nee e belongs to the other

equivalence class ie the other orientation ofn. Therefore, we havemunotes.in

## Page 41

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The equivalence class containing12, ,.....,nee e is the

opposite orientation ofn.

The Groups2SOand3SO:

We describe these groups in terms of their matrices.

First the matrices in2SO :

Let12,ffbe any orthonomal basis of2belonging to the

same orientation class of the standard basis12,eeof2.

Letbe the angle between1eand1fwhich is measured

counter -clockwise. Then the matrix of T iscos sinsin cos .

This shows that2SOconsists of all22:T having

matrix representations : (with respect to the standard basis12,ee:

22 cos , sin:, 2sin , cosSO T O T o

Next we describe the group3SOby means of the matrix

representationsTof its elements T with respect to the standard

basis123,,eee: We consider T obtained as the resultant21T T oTof

two rotations where (i)1Tis the rotation of the XOY -plane about the

Z-axis through an anglemeasured counter clockwise,2oand (ii)2Tis the rotation of the frame about the Y -axis through an

angle o:

The matrices are :

munotes.in

## Page 42

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33cos cos cos sin sin

:s i n c o s0 :

cos sin sin sin cosSO T O T o

3.3SMOOTH FU NCTIO NS:

We return to differential calculus and recall some more

terminology.

Letbe an open subset ofn.

For a :fand for a p, recall that the limits :limi

tofpt e fptare the partial derivatives

1

ifpi nxoffatp.

Suppose the functionfis such that

ifipxfor1inand

for all pexist. Then we get the functions :

:1

ifinx

from the functionf.

We say that the function :fiscontinuously

differentiable onif (i)

ifpxexists for each p, each1ii nand (ii) all the function :1

ifinx are

continuous on.munotes.in

## Page 43

431Cdenotes the set of a ll continuously differentiable

function on.

Next, we say that :fis twice continuously

differentiable onif (i)1fCand (ii) foreach 1ii n,

1

ifCx. Iffis twice continuously differentiable, then it

follows that for each,1ij ij nand for each p,

22;ij j iffppxx x x 1,ij n.2Cdenotes the space of all twice continuously

differentiable :f.

Higher order continuously differentiability of :fis

defined inductively : Suppose k times conti nuous differentiability offonis defined. Then we say thatfis1ktimes continuously

differentiable onif (i)

ifxxexists for each xand (ii) the

functions :

if

xare k times continuously differentiable on.

Iffis k times continuously differentiable onthen it

follows that for any multi -index12, ,...,n with

12 ......nk the mixed partial derivative

12

12:n

nDf p f pxx x

exists for all pand

the resulting funct ion ::;Df p Df p is continuous on.kCis the functions space of all k times continuous

differentiable functions :f. The functions spacekChas

the structure of a commutative ring with identity ;the ring operations

being addition and multiplication of function on.

Now, we have a decreasing sequence of functions spaces :12 kCC Cmunotes.in

## Page 44

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Weconsider the I ntersection ::kCC k .

This space contains non -trivial (= non constant) function : In

fact, we have the following :

a)If12,, ,nis any multi -ndex, then the monomial

12

12 :;nn xx x x x is inC.

Consequently any polynomial :11 mpx ax

is inC.

b)Given any compact k, openwith KU, there exists afCsatisfying :

i) 1fo n k,

ii) 0fo n U.

Above, we mentioned smooth functions defined on open

subsets ofn. In Chapter 8 we will extend the property of

smoothness to functions defined on open subsets of smooth surfaces.

Also, recall the smoothness of vector valued functions

defined on open subsets ofn. Let :mfbe any vector valued

functions. Let its components be12, ,..... :m ff f thus12, ,.....mfx fx f x f x for all x. Now we declare thatfis smooth if each of1...mffis inCin the above sease.

Moreover, for any multi -index12,, ,nwe define :12, ,......,mDfx Df x Df x Df x for each x.Dfxis the mixed partial derivative offatx.

3.4 BASIC T HEOREMS OF DIFFERE NTIAL

CALCULUS :

We recall here three of the basic theorems of differential

calculus, namely :

The inverse function theorem ,

The implicit function theorem ,

The rank theorem .munotes.in

## Page 45

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Of the three of them the first is an independent result of

fundamental nature and the other theorems are deduced from the

first. Other basic theorems of differential calculus such as Picard’s

theorem (regarding the existence and uniqueness of solution of an

ODE), the Frobaneous theorem and so on will be explai ned in the

latter chapters.

We begin with the first theorem of the above list :

Theorem 1 (The Inverse Function Theorem) :

Letbe an open subset ofn, p a point ofand let

:nfbe a smooth map.

Suppose, the derivative:nnDf p is a linear

isomorphism. Then there exist open subsetsUof,Vofnhaving

thefollowing properties :

i),pU fp Vii)fU Vand

iii) :fUVUis bijective with the inverse1

:fVUU

also

being smooth. (In other words, (iii ) meansfUis adiffcomo rphism

between U and V) .

ncFig. 1 (Inverse Function Theorem)

As explained earlier, the total derivativeDf pis a linear

map approximating the givenfin a neighborhood of p and

therefore, some of the properties of the approximating mapDf pshould reflect back on the local behaviour offaround the poin t p.

The theorem above asserts that indeed, the invertibility of the

approximating linear mapDf pensures local invertibility of the

functionf, the local inver seoffalso bei ng smooth.munotes.in

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A smooth bijective map :fU Vwith1:fVUalso

being smooth is said to be a smooth diffeomorphism between the

sets U and V. Thus, the inverse function theorem asserts that a

smooth map with its deriv ative at a point being invertible isa (local)

diffeomorphim in a neighborhood of that point.

Next, we discuss the implicit function theorem.

Theorem 2 (The Implicit Function Theorem) :

Let,a) ,mnUC VCbe open subsets .(Here we denote a

point of U by12, ,...,mxx x x and a point of V by12, ,...,nyy y y .)

b),:mff x y U V be a smooth map.

c)Suppose, for a point,pa b U Vthe matrix

1,i

ifab i j nyis invertible.

Then there exists an openUwithaUUand a smooth

map:gU Vwhich satisfies

i)ga band

ii),,fx g x fa bfor allxU.

Thus, the theorem asserts that when the condition ( c) is

satisfied, the equationfx y C fa bcan be solved to get the

variable y as a functions (smooth)yg xsatisfying the additional

proviso :bg a.

This result has applications everywhere in differential

geometry ;we explain here only a small aspect of it:

We are given a smooth function

:f,being an open subset of3. For a dwe consider the set,, : ,,Mx y z f x y z d . If not empty, then such aMM dis often called a level set of the function through the

value d.munotes.in

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We consider a non -empty level setMdof the functionfsatisfying the additional condition :

,, 0fxyzz

for all,,xyz M d. Then for any,,abc M dthe implicit

function theorem asse rts that there exists an open2Ucwith,ab Uand smooth map :gUwith,ga b csatisfying,, ,fx y g x y don U .

In other words a part of the level setMdcontaining a

given,,pa b cis the graph of a smooth function g and therefore it

looks like a surface. This observation is used very often in studying

local properties of smooth surfaces.

Finally, we describe the rank theorem.

First recall that a matrix (of sizemn) has rank k if the

matrix contains an invertible sub -matrix of sizekkand has no

invertible sub -matrix of size larger thankk.

Now, the theorem :

Theorem : (The Rank Theorem) :

Let :nfbe a smooth map -being an open subset ofm, the mapfhaving the property thatrk Df p kfor every

p.Then for every pthere exist :

i)an openmUcwith 0Uii)an opennVcwithfp Viii)a diffeomorphism:gU gUwith0gpiv)a diffeomorphism:nhV hV Such that the map

:nhfg U is given by

12 12...., ...., , .....mk

nkhfg x x x x x xo o for all12....,mxx x U.munotes.in

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3.5 EXERCISES :

1)Check if each of the following33:T is an isometry and then

express each of the T in the formaTRin case T is an isometry :

i) 23 32,, 2 , 4 , 313 13 13 13yz yzTx y z x ii) 33,, 5 , 2 , 422 2 2xzTx y z z y x iii),, 4 5 , 4 3 , 5Tx y z x y z x y z

iv) 27,, ,33xzTx y z y 2)Let1, fC p(being an open subset ofn)

Prove :

a)

21n

ifDf p u u pxi for allnub)Df p v w Df p v Df p w

3)If2fCthen prove that for every p,2Df pis a

symmetric bilinear form.

4)i)Construct a smooth map:0 1fsuch that1fxif1x0if2xii)Using the mapfof (i), construct a smooth2:0 , 1g such that10 1g on B200 , 4on Bmunotes.in

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4

SMOOTH CURVES

Unit Structure :

4.0 Introduction

4.1 Smooth Curves .

4.2 Curvature and Torsion of Frenet Curves .

4.3 Serret Frenet Formulae .

4.4 Signed Curvature of a Plain Curve .

4.5 Exercises

4.0 INTRODUCTIO N

In this unit (consisting of this chapter and the next, two) we

study the geometry of smoothly parametrized space curves. (After

discussing such curves, we will give indic ations of the geometry of

curves in higher dimensional E uclidean spaces also) .In this chapter

we begin with basic geometric features of ja smoothly prametrized

space curve, its reparametrization, its unit speed versio n, a moving

orthonormal frame along i t and so on.

Actually, we will consider a smaller class of curves consisting

of Frenet curves and explain how differentiation leades us to

geometric features of such curves. In particular. We introduce the

concepts of curvature and torsion of a curve whi ch are smooth

functions defined along a Frenet curve. Explaining their geometric

significance, we proceed to derive the basic equations -theSerret -

Frenet equations -associated with such curves. It is the central result

of the theory of Frenet curves t hat the two functions curvature and

torsion functions -of a curve determine the curve uniquely to within

an isometry of3. We derive this important result -the fundamental

theorem of curves -using Picard’s existence / uniquenes s theorem of

solutions of ODE .

Throughout, we are considering curves which are smooth (=

infinitely differentiable) This assumption (infinite differentaibility of

curves) is superfluous, for ,we are using only thrice continuous

differentiability of the p arametrized curves. We have adopted here

infinite differentiability as only a convenient set -up to derive themunotes.in

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basic theory. But on the other hand if a curve is not as much as thrice

continuously differentiable, then some of the tools of differential

calculus may not be applicable .

4.1 SMOOTH CURVES.

In this chapter, I, J, K denote intervals

Definition: A smooth curve is a smooth map3c:IThe curve is said to be parametrized by the independent

variable sof the map .sis the pa rameter of the curve and for a

0sIthe point c0(s )of the curve is said to have the parametric

value0s.

The set{c (s):s I}is called the trace of the curve c.

For e achsI ,writing the pointc( s )in terms of its Cartesian

coordinates:123cs ( x ( s ) , x ( s ) , x ( s ) )we get the real valued

function:

12 3x: I , x: I ,x: I Note that the curve3c:Iis smooth if and only if the

function1, 2 3x, x, x :I Rare smooth .

Let now:J Ibe a smooth, strictly monotonic increasing,

bijective function. The curve cand the functioncombine to get yet

anoth er curve:3cc 0: JDefinition 2: The curve3c:Jis said to be a reparamentrization

of the curve3c:IIfrJis the variable raging in J, then we speak ofras the

new parameter andthe reparametrizing map.

For the curve3c:I, we write cs

for2

2dc s d c s,c s for etc.dt dsmunotes.in

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Note that 123dx s dx s dx scs , ,ds ds ds 222

123

222dx s dx s dx sc( S ) , , etc.ds ds ds

Definition 3:3c:Iis said to be (i) regular ifcs 0for allsIand (ii) a Frenet curve ifcsandcsare linear ly

independent.

We will consider only smooth Frenet curves.

Definition 4: A Frenet curve is said to be a unit speed curve ifc s 1.Below, we show that a Frenet curve can be reparmetrized so

as to make it a unit speed curve .

Let3c:Ibe a Frenet curve .

For an arbitrary chosen0sIwe consider the integral :

0s

s( s ) c x dr s I sis the (signed) length of the segment of the curve clying

between the poi nts0c s and c sof it. Note the following

s0if0ssands0if0ss d( s )cs 0ds for allsI(by the regu larity assumption on c)

and therefore the functionssis a strictly monotonic

increasing function on its domain intervalI.

The mapssbeing continuous ,its range -we denote it byJis an interval .Now we have the function ::I Jwhich is strictly monotonic increasing and bijcetive function

betweenIandJ.munotes.in

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We consider the inverse function1:J I;We denote it by.Thus we get the strictly monotonic increasing, smooth map:J Iwhich is bijection between the indicted intervals .

We useto reparametrize the given curve :3cc : J Finally, we have: For anyrJdc (r)c( r )drdc sdsds dr

csdrdscscs

and thereforedc r c s1dr c s

for allrJ ,that is, the

reparametrized curve3c:J is a unit speed curve. Thus, a regular

curve when re -parametrized by its ar e-length becomes a unit speed

curve .

Note that we can regain the o riginal curve cfrom its unit

speed versionc:10cC .Therefore, we introduce many of the geometric aspects of the

given curve cin terms of those of its unit speed version. Also. note

thatcand its ( unit speed ) reparametrizationc, both have the same

trace.

Let us discuss a few simple examples of smooth curves, some

of which are Frenet curves while some of them are not.

The curve3c: given by234Cs s, s, s forsis

smooth but fails to be regula r atC 0 0,0,0.It is regularmunotes.in

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when restricted to any intervalInot containing0. In factc/Iis a Frenet curve for such aI.

Let3c:(0 , ) be the curve given by

2

3/2 s2 2cs s , , s s 023

Then we have 1/2cs 1 , s , 2 s ,

1c s 0,1,2s

Clearlycsandcsare linearly independent (i.e. non -

parallel) vectors for everycsand consequentlycis a Frenet curve.

Measuring are length from theC(0 )end, we get

0

0

2s C u du

(1 u )du

s2s s 02

Puttingsr(1sin the above nota tion) we get :

s2 r 1 1 r 0Therefore, the re -parametrization usingras the new

parameter gives the curve3c: 0 , 3/222cr 2 r 1 1 , r 1 2 r 1 , 2 r 1 13 forr0.

We consider a planar curve called the exponential spiral.Its

the curve2c: given byssc s e cos s,e sin s ,smunotes.in

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Fig. 1. The Exponential Spiral

Note that ssc s e cos s sin s , e cos s sin s givingscs 2e.Therefore cis a regular, but not a unit speed curve.

Moreover, we have

ssc s 2 e sin s,2 ecos s

Thus, in fact cis a Frenet curve.

To reparametrize itwith respect to its are length we consider

its signal arc length function. Taki ngc 0 1,0as the reference

point, we obtain the (signed) length function given by

0

s

r

0

sr s c dr2e dr2e 2

This givesrs log 12and therefore, the reparametrization

of the exponential spiral :

rs ssc( r ) 1 coslog 1 , 1 sin log 122 22

The Cycloide :

A wheel of radius a is rolling on the ( horizontalXa x i sof a

verticalXOY plane, moving with constant velocity. Then a

pointPheld fix o n the wheel rim traces a curve. This curve is called

a cycloide. Its parametric representation ( Parametrized by the timet) is

wtc( t ) wt a cos wt,a a sin ,ta munotes.in

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It is not a unit speed curve It is left as an exercise for the

reader to reparametrize it so as to get a unit speed curve.

Fig. Cycloide

The Elliptical Helix :

It is a curve which climbs up an elliptical cylinder with cross

section

2 2

2

22x x1a babChoosing yet another constantC0We ge t the curve.3c: c( t ) ( a cos t ,b sin t ,c.t ), t .The resulting curve is a Frenet curve The reader is invited to

verify this fact and to reparametrize it so as to get a unit speed curve.

4.2 CURVATURE A ND TORSIO NOF FRE NET CURVES

Let3c:I Rbe a Frenet curve, its parameter being denoted

bysI. As explained in the preceding section, we assume without

loss of generally that it a unit speed curve.

We use the notations:csfordc sdscsfor2

2dcsdscsfor3

3dcsdsand so on.munotes.in

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Puttingˆts cs,we get the tangential vector having unit

length. Moreoverts cs 0and is not parallel totsIn fact,

0i( s ) t ( s ) ;for, differentiating the identity.

=1We get 2 <ts, ts>=0and therefore, indeedts ts.Again usingts cs 0we introduce

i) csnscsand

ii)bs ts ns.

Now we get anorthonormal triadets, ns, bsof vectors

located at the pointcsofc.We call

tstheunit tangent tocatcsnstheprincipal normal tocatcsbsthebinormal tocatcsthe ordered triplets, ns, bsis called the Serret -Frenet

frame or the principal triade tocatcs(Often the Serret -

Frenet frame is referred to as the Frenet Frame. and the scalarks csis the curvature of the curve at the pointcs.

At a later stage we will associate one mo re scalar called the torsion

ofcatcsand denote itbys; it quanti fies the twisting of the

curvecat the pointcs.scsbstsc

Fig. 3 : The Principle Triadets, ns, bs

We proceed to explain how the scalarks 0describes the

bending of the curveCat its pointcs.munotes.in

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Fix arbitraril y a point say0cs pofc. Take two more

points1csand2cson the curve (without loss of generality,

assume that10 2sss.) We now prove the following

Proposition 2: suppose0k s 0.If12ssare near enough to0sthen012cs , cs , csare non -collinear (and therefore there is a

unique circle passing thought them.)

Proof: (By contradiction)

Under contrary assumption, suppose we can choose

parametric values12ssarbitrary near to0ssuch that the points102cs , cs , csare collinear. Now, because the (smoot h) curvecis

bending continuously, there exist parametric values12rrwith11 02 2srsrssuch that the tangent vectors12c( r ) c( r )are both

parallel to the line L.(The geometric situation is as in Fig.4 below. )

Fig. 4

Recall,1c( r )and2c( r )are both unit victors and therefore,

their being parallel to the lineLimplies their equality:

12c( r ) c( r )or equivalently put:21

21cr cr0rr…………………………… (*)

Recall,12ssare arbitrarily near to0s; we make12ssboth

approach0sindefinitely. T hen10 20rs , rsand therefore the

equation()in the limit becomes

10

2012 11

rs

21 rsc( r ) r c( r ) rlim 0rr

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But, the above limit is0cs ,Thus0cs 0Thus, we have

arrived at a con tradiction to the assumption0cs 0Therefore, indeed, when12s, sare near enough to0sthe three

points01 2cs , cs , cs are non -collinearWe c onsider the circle determined by the (Non -collinear) points01 2cs , cs , cs ; let it be denoted by0, 1 2Sss , sand its centre by0, 1 2Dss , s.

We prove below that the circle0, 1 2Sss , stakes a lim iting

position in the plane through0cscontaining0tsand0ns.

Clearly the limiting circle is the best curve reflecting the bending of

the curvecat its point0cs(The circles0, 1 2Sss , sapproximate c

around the point0csand the approximation improves as12 0ss s.

It turns out that radius of this limiting circle is01ks.We prove this

result in the following proposition

Proposition 3:

The circle0, 1 2Sss , s Stakes a limiting position in the plane

through0cscontaining0ts,0nsand its radius is01ks.

Proof: Let012DD s , s , s be the centre of the circleS.For a fixed

pair12s, sinI(near enough to0s) we consider the function:

f: Igiven byf(s) c s D ,c s DBecause the circleSpasses through0Cs,1C( s )and2Cs,

we geto12fs fs fsmunotes.in

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Applying mean value theorem of differential calculate toif s f s , i 0,1,2,we get12rrin I with11 02 2srsrssuch

that12f(r ) f(r )……………………………………….. (*)

Application of the same theorem tofwith()gives

3rIwith132rrrsuch that

3f(r ) 0……………………………………………. (**)

We take limit of()and()as12 0s, s sand consequently123 0r, r, r sThis gives00fs 0 fs But, we have:

0

00ss

ss

00xsfs l i m fs

lim 2 c s ; c s D2c s c s l i m D

This gives:00x0cs cs l i m D 0……………… ……… (***)

Next differentiation offstwice gives:fs 2 c s , c s D 2 c s , c s =2c s , c s D 2 Therefore:00f ( s )0sslim{ 2 c s ,c s D 2 }This gives00

01n( s ),c( s ) lim Dks…………………. ( ****)

Above, we have been writinglim Dfor the limitmunotes.in

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10

20012ss

sslim D( s ,s ,s )

Thus, from()we get:

(a) the pointlim Dlies on the line through0csand going

perpendicular to the vector0ts(equivalently putlim Dis a point

lying on the line through0csand extending in the direction of0nsFrom****we get:

(b) 001cs l i m Dks.

The observations (a), and (b) above give:

000nslim D c sks.

Therefore the circle012Ss , s , sindeed takes a limiting

position, lying i n the plane through0csparallel to00t s and n sin

such a way that its centre is 0o0nscsks. See Fig. 5 below:

Fig. 5

We call the limiting circle the osculating circle of the given curvecat its point0cs.munotes.in

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Here is some more terminology.

The plane throughcsspanned bytsandnsis the

osculating plane ofcat its pointcs.

The plane throughcsspanned bynsandbsis the

normal plane tocat its pointcs.

the plane throughcsspanned bytsandbsisthe plane

rectifying plane ofcatCs.

Thus, we have obtained above that to within s econd order of

approximation, the curve seems to live within its osculating plane atcsand is approximately a circle -the osculating circle atcs-and

having radius1ks.

Note one m ore geometric fact: the bino rmal maintains its

perpendicularity to the osculating plane as we move along the curve.

Therefore the mapsb sdescribes the movement of the binormal

as its foo ttraces the curvecwhile the foot moves forward, the

vectorbsrotates about the tangent line as its axis of rotation; in

other words it describes the twist in the curve We are interested in

the rate of twist -the rotation of the vectorbs.we denote the rate

of rotation ofbsbysand call it the torsion of the curvecat

the pointcs.

4.3THE SERRET -FRENET FORMULAE :

Inthe last section, we singled out a class of regular curves

which we called the Frenet curves and associated with such a3c:Ithe geometric objects namely

(i) a moving orthonormal framest s , n s , b s , s I and

ii) The two functions:k:I 0 ,c:Imunotes.in

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describing the bending and twisting of the curves .We now derive

differential equations in the vector fieldsst s , sn s ,sb s ,which rolate all the qu antities described in (i) and (ii)

above.

We already have:ts k ( s ) ns, s I Next, we havens 1i.e.s, ns 1. Differentiating this

identity we get.ns, ns 0 Therefore the vectornsis expressible as a linear

combination ofts, bs(Here we are using the facts thatts, ns, bs is an orthonormal vector basis and the vectorshas no component alongn( s )as derived above) We get.n( s )=sts sbs……………………… (*)

for some smooth functions,: I ;we find these functions .

Taking inner product of the identity()withts,we getns, ts s ts, ts s ts, bss . I s .0 i.e n s ,t s s ……………………………. (* *)

On the other hand, differentiating the identityns, ts 0givess, ts s, ts 0Therefores, ts s, ksns 0 i . ens, ts ksns, ns 0and thereforens, ts ks……………………….. (***)

Now,and()givens, ts ks.Finally combining

this identity withyields :ns ksts sbsmunotes.in

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Recall, we have introduced the functiontsas the function

describing the rotation of the unit vectorbsabout the vectortsas

its axis of rotation. Consequently we have the equation :bs tsns(the negative sign being introduced as a rotational convenience) .On

the other hand, differentiation of the identityns, bs 0givesnsbs ns, bs 0 i.e.nsbs ns, sns………………………. ( ****)

i.ensbs ts……………………………… ( ****)

Again taking inner product of the equationns ksts tsbswithbsgives,;ns bs psthus by (*** *) above we getss. This gives :ns ksts tsbsThus, we have obtained the triple of ODE

ts ksnsns ksts tsbsbs tsns

These equations are often written in the matrix form

ts 0 ks 0 tsdns ks 0 s nsdsbs 0 ts 0 bs These eq uations are called the Serret -Frenet equations of a

(Frenet) curve.

Thus associated with a Frenet curve is a pair of scalar valued

functions, defined along the curve namely the curvaturekand the

torsiont. In Chapter 6: we will prove that these two functions

together determine the curve uniquely to within a rigid motion of the

curve.munotes.in

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4.4SIGNED CURVATURE FOR A PLAI NCURVE :

For a curve living in a plain the binormalbsremains a

constant unit vector, namely one of the two unit vectors which are

perpendicular to the plain in which the curve is situated

Consequently the third of the Serret -Frenet equations givest0 .Onthe other hand we can make use of the standard (counter

clockwise) orientation of the plane to refine the (blunt) non -negative

curvature function and make it a function taking both non -negative/

negative values. We ascribe a signature toksas follows We

repla ce the principal normalnsby the vectorˆns(say) which is

obtained by rotatingts(about its footcsthrough2the rotation

being anticlockwise (It is here that we are using the standard

orientation of2) Now, we obtain the signed curvatureksofcatcsby using the defining equation:cs ksns

Fig. 6

For example, the curve2cs ( s , s) shas positive curvature

while the curve2ˆcs ( s , s) , shas negative curvature .

In passing, note the following simple fact :Identify the plane

with the complex plane.Then rotation of vectors anti -clockwise

through the angle2corresponds to multiplication of the vector (as a

complex number) by the imaginary unit i. This co nsideration leads to

the definition of the signed curvature:

dts ksi ts.dxmunotes.in

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4.5 EXERCISES :

1)Obtain principle triade of the curves given below at the indicated

points.

a)23,,cs s s s at1,1,1b)4cos , 4sin ,3cs s s s at4,0,0c)2,4 ,scs e s e at1, 0,12)Reparametrize the following curves soas to get unit speed

curves.

a)22 3,2 ,3 0cs s s s s at1,1,1b)os 2 , sin 2 ,0sscs e c s e s sc)4, 2 ,3 ,cs s s s3)Verify that the curve3:c given by

32 1cos 2 , cos 2 , sin 2 1 ,10 5 2cs s s s s is a unit speed curve

and obtain the curvature and torsion function of it.

4)Let33:L be an isometry of3and let3:cIbe a Frenet

curve.

Prove :

i)Locis also a Frenet curve

ii)both, c, Lochave the same curvature and torsion functio ns.

5)Suppose the curve3:c has non -vanishing curvature. Prove

that if all osculating planes of cpass through a fixed point, then c

is a plane curve.

6)Calculate the signed curvature function the curves :

i)2,,cs s s s O ii)2,,cs s s s O

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5

CURVATURE A ND TORSIO N

Unit Structure :

5.0 Introduction

5.1 Curvature and Torsion Functions

5.2 Signed Curvature of aPlane Curve

5.3 Elementary Properties of Curvature and Torsion

5.4 Exercises

5.0 INTRODUCTIO N

In the last chapter we co nsidered smooth Frenet curves and

defined the curvature and torsion functions of such curves. In

defining these terms, we used the unit speed kind of

paramentrization of the curves in an essential way (for example, we

were using the unit length property of the tangent vectorcsin

getting the perpendicularityts ns.) However, curves are

seldom in the unit speed parametrized form. We therefore need

develop equations to calculate these quantities applicable even when

the curves of our interest are arbitrarily parametrized (regular Frenet)

curves.

In this chapter, we develop the desired formulae for the

Frenet curves and then we proceed to study the geometry of such

curves in terms of the curvature and tersion f unctions.

5.1 CURVATURE A ND TORSIO NFUNCTIO NS:

Let3c:Ibe an arbitrarily parametrized regular Frenet

curve; its parameter being denoted byrI. Besides r, we need

consider the natural are -length (and hen ce unit speed)

parametrization for a while. Thus, we consider3c:J the are -

length parameter manifestation of c, the are -length parameter, as

usual, being denoted by sand the reparametrization map being:I J: r swhereis strictly monotonic increasing bijetive

map between the indicated intervals. For the sake of convenience wemunotes.in

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will denote differentiation w.r.t. r by a dot “.” e.g. dc rcrdr

while we will use the derivativ e notationdc sdsfor the are length

parametrization. Now the relations between the two parametrizations

are :cr c rcsr

Where we are writingSrforr

In terms of these notations, we have :cr csr

dcr csr srds

tr sr................................................ (*)

2 dcr tssr srtsds

2kssr ns srts

....................... (**)

33

3dcr kssr ns kssr nsds2s r s r k s n s s r t s s r s r k s n sdkssr ns kssr ksns 7sbsds2S r s r k s n s s r t s s r s r k s n s

This 33 3 dk scr sr ts sr ks sr 3 srs ksds

ns ks tsSrbs

............................................ (***)

Farming cross product of (*) and (**) we get :

32cr cr sr ksts ns sr ts ts

3sr ksbs 0

........................ (4*)

And therefore 3cr cr sr ksr bs

3sr ks 1 munotes.in

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Which gives

3cr crkr ksr

cr

Next, taking inner product of (***) with (4*) we get :

6 266cr cr, cr k rtssr

cr crts crcr

Using the above obtained expression for kr ksr . Thus we

get :

2cr cr, crtr

cr cr

(Above we have adapted the notationkrfor ksr .

Thus we have proved the following :

Proposition 1 : The curvature and torsion functions.k:I 0 , , t:I for a regular Frenet curve3c:Iare given by :

3cr crkr

cr

And

2cr cr, crtr

cr cr

=

2det c r ,c r ,c rcr cr

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(In the above determinant notation :det c t ,c t ,c tthe vectorscr, cr, c rare in the columns of the matrixcr, cr, c r .

An Illustrative Example : Obtain the curvature and torsion

functions for the circular helix :c r a cos r,a sin r,br ;raandbbeing both non -zero

constants.

Solution : We have :

c r a sin r,a cos r,bc r a cos r, a sin r,o

And c r a sin r, a cos r,o

Therefore 2c r c r ab sin r, abcos r,a , c r ,c r ,c r 22 2 2 2ab s i n r ab c o s r 0 . a 2ab

222cr cr a 1bba 1

And 22cr a b

. This gives :

23222ba 1kr

ab

and

2 2

22 22ab 1 atr

ba 1 ba 1

Here is another Illustrate Example :

Calculatek,tof the space curve :r2cr e, r , r , r .

Solution : We have :r2cr e, r , r

And therefore rc r e ,1,2rmunotes.in

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rc r e ,0,2

And rc r e ,0,0

This gives : 12 3r

reeec r c r det e 1 2re02 rr2,e 2r 1 , 2e

And therefore 2r 2cr e 4 r 1

2r 2cr cr e 4 r 8 r 5 4

And 2rcr, cr, cr e

These equalities give :

3

2r 2

32r 2 2cr cr

kr

cre4 r8 r 54e4 r 1

3

2r

2r 2cr cr, crtr

cr cr

e

e4 r 1

5.2 SIG NED CURVATURE OF A PLA NEC U R V E:

The concept of signed curvature of a plane curve was

introduced in Chapter 4. Here we tarry a while to explain a little

more about the u nderlying heuristics formula for the same of a

planar, regular but arbitrarily paramethrized Frenet curve.

Thus, let2c:Ibe a Frenet curve, its parameter being

denoted by r. we consider its unit speed parametrization also, the

associated unit speed parameter being (as usual)SS r. In ordermunotes.in

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to employ as few notations as possible, we write both the

parametrizations of the curve by the same symbolc:c r c rin

the sense cr cSr cS being the unit speed version ofcrandSS rbeing the unit speed parametrization map. With

this notational understanding in mind, we writetrfortsr nrfornsrand so on.

Now returning to the signed curvature we recall that we were

considering rotation of the unit tangenttrabout the pointcrthrough the angle2and thus getting tr. See the figure below:

Thus, at the pointcrof the curve c, we have the two unit

vectorsnrand tr. Clearly we have either tr nras indicated

in part (a) of the figure or tr nras shown in part (b).

Now, let us note the difference between the earlier (rather

blunt) case of the non -negative curvatureksand that of the

present signed curvatureks.

In definingkswe comparedcswith the principal normalns:

cs ksns

.................................... ........... (*)

While introducing the signed curvaturekswe are comparing cs

with ts:munotes.in

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cs

=ksts........ ....................................... (**)

thus arriving at the definition of signed curvatureksofcat the

pointcs. Consequently in view of the above observation

ns ts(as illustr ated in cases (a), (b) above) the equations (*)

and (**) give two possibilities :ks ks.

Now a few words about the notations : As mentioned above,

we are desirous of using as few notations as possible. Above, we

introduced the notat ionksfor the signed curvature besides the

earlierks. However, in a plane we will be considering the signed

curvature only and as such the two notations :ksandksare

superflows. We therefore abandonksand revert to the old notationksthrough we are dealing with the signed curvature. Thus from

now-onwardsksstands for the signed cu rvature of a planar curve

while in3it is the old non -negative curvature. (Also, we continue

with the practice of denoting by a dot : “.” differentiation with

respect to the given parameter, whileddsis the differentiation with

respect to the natural are length s of the curve). Now we write the

vector equation2

wdcskstsdsin terms of its components.

2

22

2

2

1 2

22

1dcsts

dsks

dc sts

ds

dcsdsksdcsdsEquivalently, put, we have the pair2

12

2

2

21

2d c s dc sksds ds.......d c s dc sksds ds

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The pairexpressed in terms of the given parameters takes

the form :

a) 2 2

11 22dr d r drcr cr k r cr ,ds ds ds

b) 2 2

22 22dr d r drcr cr k r crds ds ds

.

Multiplying (a) by 2cr

and (b) by 1cr

gives :

c) 2 2

2

12 1 2 22dr d r drcr cr cr cr k r crds ds ds

d) 2 2

2

21 1 2 12dr d r drcr cr cr cr k r crds ds ds

.

Subtract of (c) from (d) gives :

2

22

12 21 1 2dr drcr cr cr cr k rcr crds ds

and

therefore :

2212 21 1 2drcr cr cr cr k rcr crds

;which

in turn gives :

12

3det c r ,c rkrdr

ds

This is the desired formula for the signed curvaturekrof

the planer curve.

Note one more aspect of the signed curvature namely the

general non -negative curvature given by

3cr crkr

cr

involves differentiation of the curve only but the singed curvaturemunotes.in

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3det c r ,c rkr

cr

is not only terms of the derivatives cr, cr

of c, but it takes into consideration the anti -clockwise orientation of

the ambiant space2in a crucial way! For de fining the signed

curvature, we were considering the anticlockwise rotation of the unit

tangenttr.Had we chosen to rotate it in the clockwise manner,

the curvature could have changed its sign!

Let us consider two simple example s of curves and calculate

their signed curvature functions.

(I)2c: is given by the graph of the cosine curve :c r r,cos r.

Then we have (i) c r 1,sin r

(ii) 2cr 1 s i nr

(iii)c r 0, cos r

Now 10det c r ,c r det cos rsin r cos r

and

therefore the (signed) curvature of this curve is

322cos rkr1 sin r

.

II)2c: is given byrrc r e cos r,e sin r .

Then we have (i) rrc r e cos r sin r ,e sin r cos r

(ii) rcr 2 e

(iii) rrc r 2e sin r,2e cos r

Therefore

rr

rre cos r sin r 2e sin rdet c r ,c r dete cos r sin r 2e cos r 2r2e.

Therefore 2r3r r2e 1kr22 e 2 . e...............................munotes.in

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5.3 ELEME NTARY PROPERTIES OF CURVATURE

AND TORSIO N

Here is another description of the curvature of a space curve:

The curvature of a curve at a point of it measures angle variation of

the tangent vector per unit length of the are. To be more precise, we

have the following :

Proposition 2 : The curvaturekpofcat a point pof it is given bykp=qplimp,awhere qis a point on cwithqp ,p , qis

the length of the are of cbetween its points p,qandis the angle

between the tangents at pandq.

Proof : The angleis obtained by using the formula for the angle

in an isoscelese triangle :tq tpsin22

Therefore qp 0 02s i n22lim lim limp,a sinQ 2 p,q

0

0tp tp1l i m

cp cp

lim

cp

kp

Definition 5.1 :A regular curve having the property that the tangent

lines at all points of which make a constant a ngle with a fixed

direction is called a slope line .munotes.in

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Proposition 3 : (Lancert’s Theorem) aFrennet Curve isa slope line

if and only if the quotienttsksis constant.

Proof : First, suppose that there exists a constant unit vector e(the

direction) such thatts, eis the same for all s. Then we have

dts, e 0ds.

i.e.ks, ns, e 0andks 0for all simpliesns, e 0.

Differentiating this e quation we getks, ts tsbs, e 0 ;

which gives :bs, ets

ksks, e .......................................... (*)

Now, the above observation thatns efor all simplies that

the vector e remains in the rectifying planes. Combining this

observation with the assumption thatts, econstant implies thatbs, ealso is constant. Now (*) above gives constancy of the

functiontsrks.

Conversely suppose,tsksis independent of sand consider

the vector tsas bs tsks . Differentiation of the functionssa sgives.

dv s t stsns ksnsds k s0 and thusas afor a constant vector. The constancy of aand that

oftsksnow implies that the tangent vectors make constant angle

with the vector a and therefore, the curve is a slope line.

Proposition 4 : A Frennet curve3c:J lies on a sphere of radiusR0if and only if its curvature and torsion functionsk,t, satisfy the

identitymunotes.in

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222 2ks 1Rks ks ts

Proof : First, suppose that the curve clies on the sphere of radiuscentred at 0. Then we have :2cs 0 , cs 0 R.

Differentiating this identity w.r.t. s, we get2ts, cs 0.

The above identity implies thatcsliesinthe normal plane :cs s s sts and 22cs Rgives

222ss R.

Differentiating the identityts, cs 0givesks ns, cs ts, ts 0i.e.ks ns, s s sbs 1 0 which gives

1sks.................. ............................................. (*)

Next, differentiating the identityks ns, cs 1 0gives1ksn s , c s k s k s t s t s b s , c s 0 . Which gives1ks s k s t s ,s 0 and therefore we get

1

2kssks ts .................. ..................................... (**)

These values ofs, ssubstituted in the equation

222ssgives :

2122 2ks 1Rks t s ks .

Conversely, suppose the above equation is satisfied.

Differentiating it, we get :11 1

3 22xk s zk s k s d0k s s ds k s s ks .munotes.in

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This gives1

2ts k s dk s ds k s t s .

Next, consider the vector 1

2sk sas cs bsks k sts .

Differenting it, we get :

2

22ksts tsbsksnsas tsks k sks kstsnsdbsdsks ts ks ts0

Thus, dcs 0dsand thereforeasis a constant vector, sayas aand then we get :

2sk sac s b sks ks s

.

This gives :

22

2 2ks 1cs aks sks

.

But, by assumption, we have

2222ks 1Rks ks s

therefore, we get 22cs a Ri.e. the curve lies on the sphere

centred at aand having radiusR.

Proposition 5 : Letcbe a closed plane curve.

Then the integral

C1k s ds2is an integer.

Proof : We identify2with.

Also, we recall an elementary result of complex analysis. For

anyzx, e 1if and only ifz2 i mfor some m.

Definef:0 , L \0by pu ttingmunotes.in

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s

0f s exp k r dr ; s 0 L

(Here L is the length of the curve c)

Also, we consider the mapg: 0 , L given by

tsgs , s 0 , Lfs. Then we have,

2gs ts f s f sts f s

Then we get

2

2ts f s f stsgs

fsik s t s f s ik s f s t sfs

(Above we are dealing with the signed curvature of c and

therefore fs k s t s k s i t s i k s t s 0 .

Therefore g is a constant function. In particular,g0 gL

that is :t0 tLf0 fL. Now because c is a closed cu rve, we have0tt Lwhich in view of the last equality givesf0 fL.

Butf0 1and therefore we getfL 1i.e. 2

0exp i k r dr 1.

Therefore, by the above quoted r esult, we get L

Oi k r dr 2 imfor

some integer m. This gives

L

01krd r m2.

Lastly, we prove the following result (which is of

considerable technical importance in geometry / analysis).

Lemma : LetQ: ab be a continuously differentiable function.

Suppose the functionf:a , b given byfs e x p i Q s s a , b satisfies :fa 1andfb 1. Then

b

af r dr

and the inequality b ecomes equality if and only if it is

monotonic andQb Qa.munotes.in

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Proof : We have : fs i Q sfs

and therefore :

bb

aa

b

af s ds Q s f s dsQ s ds

Qb Qa

Becausefa 1 , fb 1, there exist integers,msuch

thatQa 2andQb 2m 2 m 1.

Therefore b

af r dr 1 2 l m.

The statement regarding the equality follows directly.

5.4 EXERCISES :

1)Compute the curvature and torsion functions of the following

curves.

a) ttaaac t t, e e ,0 t2 b)c t a t sint ,a 1 cost ,bt b c)23ct t , t, t t 02)Obtain the principal triadert r , n r , b rfor the

following curves :

i)2cr r , r, 2 rr 0 ii)rrcr 4 e, r , e iii)c r 2,10 cos r,5 sin r

iv)c r a cos r,a sin r,br r ,a,b being constants.

3)Prove : If all tangent vector (unit length) are drawn from the

origin of the curve23c t 3t,3t ,2t then their end points are on the

surface of a circular cone having axisxzy0 .munotes.in

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4)Let a plane curve be given in polar coordinatesr,byrr.

Using the notationdrrd1

prove that the are length of the curve

segment corresponding toverying ina,bis given by

b22

arr d

and the curvature functionkis :

2232222r rr rk

rr

.

5)Obtain the curvature functionkof the curve (called

Archimedean spiral) :ra ,a being a constant.

6)If a circle is rolled along a line (without slipping) then a fixed

point on the circle describe s a curve called the “cycloide”.

i)Obtain a parameterization of the cycloide generated by a

circle of radius.

ii)Obtain a unit speed parameterization of the same curve.

iii)Obtain expressions for the functionsst s , sn s , sb s forthe (unit speed) cycloide of

rad a>0.

iv)Obtain its curvature function.

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6

FUNDAME NTALTHEOREM OF CURVES

Unit Structure :

6.0 Introduction

6.1 The Fundamental Theorem of Curves

6.2 TheInitial ValueProblem of ODE

6.3 Proof of the Fundamental Theorem

6.4 Illustrative Examples

6.5 Smooth Curves In Higher Dimensional E uclidean Spaces

6.6 A Space -Filling Continuous Curve

6.7 Exercise

6.0INTRODUCTIO N

In the preceding chapter we studied that with each Frenet

curve3c:Iare associated two scalar functions, namely its

curvature function.k:I 0 ,and the torsion function t:I.

The fundamental theorem of curves, which we will study in this

chapter, deals with the converse : it asserts that the two functionsk:I 0 , , t:Idetermine the curve uni quely to within an

isometry of3.

The proof of this important theorem is based on a basic

existence / uniqueness theorem for the theory of ODE, namely the

Picard’s existence / uniqueness theorem on the solution of a first

order ODE. We therefore recall Picard’s theorem (statement only)

and then proceed to prove the fundamental theorem of curves.

After proving the main theorem, we discussed a few exercises

which illustrate various concepts related to space curves we have

come across.

A point regarding our differentiability assumptions need be

explained here : We are assuming throughout that all curves3c:Iare infinitely differentiable on I, we are also imposingmunotes.in

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regularity conditions on the derivativ es : ct o , ct o

for alltIand so on. Actually, we seldom differentiate curves more than

thrice in deriving any result or in calculating any quantity associated

with a curve. Infinite differentiability of curves is indeed superfluous

but it is used as a general set -up, it can be relaxed to just three times

continuous differentiability (but not any further because we are using

differentiation as a tool involvingct, ct, c tand their linear

independence and so on.)

For a long time, a curve was considered as a thin line in2or

in3which was mere a continuous image of an internal. Apart from

the fact that the tools of differential calculus are not appl icable to

such curves, there are space filling curves, which shatter the classical

expectation of a curve as a thin line. In 6.6 we discuss (rather

concisely) an example of a fat continuous curve filling a square.

6.1 THE FU NDAME NTAL THEOREM OF CU RVES

We begin here with the recall of some of the concepts

associated with a Frenet curve and then (only) state the enunciation

of the fundamental theorem. The proof of the theorem (as explained

above) makes use of Picard’s theorem in ODE and therefore we

discu ss Picard’s theorem in the next section (again only the

statement, no proof!) and then develop the proof of the main

theorem in 6.3. It is hoped that this approach will help the reader

develop the context to study the proof of the main theorem.

Recall t hat a smooth curve :3c:Iwith

cs 1 , cs 0

for allsIgives rise to the two functions :curvature k : I 0,and

torsion t : I .

These functions and the principal triad ets, ns, bsfor

eachsIassociated with the curve satisfy the ODE called the Serret

-Frenet formulae :

ts o ks o bsdns ks o ts nsdsbs o ts o bs Now we ask : Conversely given the following data :munotes.in

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two smooth functionsk:I o ,t:I

3a point p with a parametric value s I , and000an orthonormal triade of vectors t ,n ,b ,

is there a smooth curve3c:Ihaving the given smooth functionsk,as its curvature, and torsion; passing through the point p i.e.cs pand having the principal triade000t, n, bat its

pointpc s?

The fundamental theorem gives an affirmative answer.

Theorem 1 (The Fundamental Theorem of Space Curves) : Given :

i)smooth functions :k:I 0 , , t:Iii)3

00p, s Iand

iii)orthonormal vectors000t, n, bthere exists a unique Frenet curve3c:Iwhich has the

properties :0cs p000c has the principal triade t ,n ,s at c sandch a s ,k, ta si t sc u r v a t u r ea n dt o r s i o n f u n c t i o ns.We prove this theorem in 6.3.

6.2 THE I NITIAL VALUE PROBLEM OF ODE

We introduce here the initial value problem of ODE and state

without proof the existence / uniqueness theorem regarding the

solution of the initial value problem. The precise statement of the

theorem is to be used in proving the fundamental theorem (of

curves) in the next section.

Let I denote an open interval and let so be an arbitrary point

of it. Letn A:I M be any smooth matrix valued map and let0xbe any point ofn. We consider the ODE.munotes.in

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dxAs X ,ds(X being a variable ranging inn) and a

solutiontX tof the ODE is required to satisfy00Xs x. This

constitutes the initial value problem (I.V.P.) :

00dxAsX , X s xds......................................... (*)

Now, the theorem regarding the existence and uniqueness of

the solution of the I.V.P. (initial value problem) is the following :

Theorem 2 : The initial value problem (*) has a unique solutionnX:I defined on the whole of the internal I.

Consult Chapter 2 of this series of study materi al.

6.3 PROOF OF THE FU NDAME NTAL THEOREM

To begin with, we consider the principal triade mapst s , n s , b sthrough000t, n, bof a prospective curvecspassing through the given point0P.Putting

tsXs n s , s Ibs

We treatXsin two different but equivalent ways, namely :

Being an ordered triple of vectors in3it is a vector in9.

It is also a33matrix of which the top row consists of the

three components ofts, the middle row consists of those ofnsand the bottom row consisting of the components ofbs,

122

123

123ts ts tsXs n s n s n sbs bs bs We now consider the initial value problem :

0000tdxAsX s, X s ndsb ................................. (*)munotes.in

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Where

ok soAs ks o tsot s o .

Note that the ODE in (*) is nothing but the sys tem of the

Secret -Frenet equations.

By Theorem 2 above, we get a unique solution9X:I M 3 , of the initial value problem (*). Thus we get

the functions.333t:I

n:Ib:I

With00 0 00 0ts t , ns n , bs b . At this sta ge, we claim that the assumed orthonormility of000t, n, bimplies orthronomality ofts, ns, bsfor eachsI.

To get this result, we use the antisymmetry of the matrixAs,that

istAs As; (where tAsis the transpose ofAsWe have :

t tt

t

t

t t

tt tdd dXs Xs X s Xs X s Xsds ds ds

dXs Xs Xs A sXsds

AsX s X s X s AsX sXs A s Xs Xs A s Xs

tt

ttXs A s Xs Xs A s XsXs A s Xs Xs A s Xs0

This proves constancy of the matrix valued function

tsX s X s :munotes.in

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tt000

000 0

0Xs XS Xs XStt, n, b nb100010 I001

ThustXsX s I, that is, eachXsis an orthogonal

matrix; in other words for eachsI , t s , n s , b s is an

orthonormal triade of vectors.

Finally we get the desired curve3c:Iby putting :

0s

sc s p t r dr

Clearly the curve3c:Iis well -defined and satisfies; (a)0cs pand (b) cs ts

. Moreover, we have :

ts ksns, n ksts ts bs, b tsns

that

is, the curve3c:Isatisfied the Serret -Frenet equations havingk,tas its curvature and torsion functions . And then, the initial

conditions -that is,0cs p,000 0 0 0ts , ns , bs t , n , b impose

uniqueness on the solution curve c.

Thus, given smooth functionsk:I 0 ,and t:Ithe

theorem guarantees that there exist curves3c:Ihavingk,tas

their curvature and torsion functions. Next we claim that any two

such curves are related by an isometry of3i.e. one curve is the

isometric image of the other. To see this, consider any two such

curves say3c:Iand3c:Ichoose0sIarbitrarily.

Let0pc sand0pc s.

We putdp pthat is,pp d.

Also, let33A: be the unique orthogonal transformation

having the property :00 0 0At t An nand00Ab b.munotes.in

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The vector d and the orthogonal transformation A combine to

give the isometry33L: whereLx Ax dfor every3x.

We claimLoc c. To justify this claim, first denote the curve

Locby*3c: I . We have to verify*cs c s. To verify this

identi ty, it is enough to verify that both the curves*c,chave the

following properties :

i)they both satisfy the Serret -Frenet equations with the samek,t,

ii)they pass through the pointpand

iii)atp, both of them have the same principal triade.

We leave the verification of these facts as an exercise for the

reader.

6.4 ILLUSTRATIVE EXAMPLES

I)Determine all plane curves2c:Isatisfying

i)ks a ,a being a constant.

ii) 1kssforsOiii)

21ks 1 s 11s

Solution : Clearly because all curves are plain curves, we havetOandconsequently there is only one Serret -Frenet equation :dt sksnsds .

Nowtsbeing a unit vector, we can write it in the form :t s cos s , sin ssbeing the angle bet ween the vectortsand the X -axis.

Then we have :n s sin s ,cos smunotes.in

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Consequently, the equationdt sksnsds takes the form

dssin s , cos s k s sin s ,cos sds which givesdsksds.

Now i n case of (i) above we have :dsadsand therefores as b, for some constant b.

This givest s cos as b ,sin as b

Integrating this expression forts, we get

s

0c s p t r dr(becausedc stsds) being a fixed point

of3.12Pp p

s

12o

12p , p cos ar b ,sin ar b drsin as b cos as bp, paa

Therefore 12sin as b cos as bcs p , p , saa

.

ii)Now we haveds 1,s ods swhich givess2 s a, for

some constant and therefore, t s cos 2 s a ,sin 2 s a .

Integrating this equation, we get

s2

12o

ss

12oo

ss

12ooc s p t r dr, p p , pp , p cos 2 r a dr, sin 2 r a drp cos 2 r a dr,p sin 2 r a dr,

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The two defi nite integrals are left for the reader to evaluate

which he / she can, using methods of calculus (e.g. integration by

parts.)

iii)Now, we have :

2d1sds1sand therefore,1s sin s constant.

We choose a frame of reference such that the constant of

integration in above is zero :1s sin s i.e.s sin sand therefore 2cos s 1 sNowt s cos s ,sin s21s , s

This gives

ss2

12ooc s p p 1 r dr, rdr2s2

12orp 1 r dr,p2

(Again we leave the evaluation of the above definite integral

be completed by the reader.)

ii)For a plane unit speed curve2c:Ihaving curvature function

k:Iandthe Serret -Frenet framets, nsat a pointcsof

it, prove :

0

0s

s 0

s

0 0

so, k r drts ts1ns ns!k r dr,o

.

Proof : ForsI, letAsbe the22matrix :

0

0s

s

s

so , k r drAs

k r dr , o

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Also, let22X:I , Y:I be the functions given by :

0

0 0XstsAsns! Ys

.

Now we have :

0, k sdA s,ks, 0 ds and

0

0 0Xs tsAs dYs nsds !

0

0 0

0

0 00, k s t sAsks, 0 ns!0, k s t sAsks, 0 ns!

0, k s X s

ks, 0 Ys

Now we have :

1)The functionXssYssatisfies the ODEXs 0 , ks XsdYs ks, 0 Ysds and

2)00

00Xs tsYs nsIn other words the function onXssYssatisfies the ODE

(1) and the initial condition s (2). Therefore, the equation has the

solutionAs 00

0 00Xs ts tsAseYs ns ns!

0

0s

0

s

s

0

0 stsk r dr

1

!k r dr,ns

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6.5 SMOOTH CURVES I NHIGHER DIME NSIONAL

EUCLIDEA NSPACES

We describe, in very few words, some of the geometry of

smooth cu rvesnc:Iforn3. Our main intention is only to

indicate generalization to higher dimensions of the geometry of the

space curves which we have studied above. We only introduce

concepts and state some of the e lementary results, but every thing

going without proof! Interested reader can consult a standard

graduate level book such as : A course in Differential Geometry by

Withelm Klingenberg. (A Springer -Verlag publication).

Now, a smooth curve innis a smooth mapnc:I.

For arIwritingcrin terms of the Cartesian coordinates

:12 nc r c r ,c r ........c r we consider the derivatives ;

12 n

12 n

kk k k 12 m

rr r rc r c r ,c r ,.......,c rc r c r ,c r ,.......,c rc c ,c ,.......,c

the first one, namely cr

is the tangent vector to the curve at its

pointcr.

We have the straight -forward generalization of the notion of

reparametrization of c: Let:J Ibe a smooth, strictly

monotonic increasing and bijective map. Then the (smooth) curvenc. : Jis said to be obtained fromnc:Iby

reparametrization, the map:r J r sbeing the

parametrizatio n map.

Again for a fixed0rI(and thereby for a fixed point00pc rof c) and for a variablecr r I, the integralmunotes.in

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00j nrr 2

rrj1r c s ds c s ds;

ris the signed

length of the segment of the curve with0crandcras its end

points.

Note thatis a strictly monotonic increasing function

whenever cr o

for allrI. Also, note that the setJr : r I is an interval.

In the following we consider thosenc:Ifor which

cr o

for allrIholds. (which implies tha t the function:I Jis bijective). We use the strictly monotonic:I Jto

reparametrize c:

The reparametrized curvechas the property that cs 1

for allrJ, that is cis a unit speed curve.

Next, to get the n -dimensional analogue of the principal triadetp , np , bpof a space curve at a point p of it, we assume the

following property :

For eachrI, the set nc r ,c r ........c r

is linearly

independent. Applying the Gram -Schmidt orthogonalization to each

nc r ,c r ........c r

we get the orthonormal set12 ne r ,e r ........e rwith the property that for eachkkn , c r

is a linear combination of12 ke r ,e r ........e r. Now1ne r ........e rthus obtained, is the desired analogue of the

principal triade of a space curve. We call the set12 ne ,e ........eof unit

vector fields along c, the Frenet frame of the curve. Now, we have

the following two results :

Theorem 3 : Letnc:Ibe a smooth curve having its Frenet

frame1ne ........e. Then there are smooth functions

12 nk ,k ........k : I satisfying th e equations :munotes.in

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The functionsik: I are called the ith curvature sofcand

the above set of equations are the Frenet equations.

Theorem 4 : (Fundamental Theorem of Curves. )Let

12 nk ,k ........k : I be smooth cures12 n 2k ,k ........k 0on I. For a

fixedn00rI , pand for any orthonormal set10 20 n0e ,e ........ethere

exists a unique curvenc:Iparametrized by its are -length r

having the properties :

1)00cr p2)10 20 n0e ,e ........eis the Frenet frame of cat p. and

3)12 n 1k ,k ........k : Iare the curvature functions of c.

6.6 A SPACE -FILLI NGC O NTINUOUS CURVE

We conclude this chapter by discussing an example of a

continuous curve which is not a thin line but an area filling map

because it is continuous, lacking any differentiability properties.

This should convince the reader that a curve as a reasonable

geometric object it should be more than a merely continuous map, it

should ha ve, differentiability properties and the successive

derivatives having linear independence.

Theorem (Peano) : There exists a continuous surjective map (= a

curve)C : 0,1 0,1 0 1 RProof : We obtain the desired C as the uniform limit of a sequ encekC : 0,1 : k of continuous maps.munotes.in

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To construct1Cwe sub -divide (a)0,1into239sub-

intervals of equal length :i1 i1i999and (b) the rectangl eR

into nine Sub -rectangles of equal area as shown in the figure.

We construct1Cby mapping10,9linearly onto the diagonal1Dof the sub -rectangle1, then mapping1299linearly onto the

diagonal2Dof the rectangle2and so on.

Next we construct2Cin a similar manner : Sub -divide eachi1 i99into nine equal parts, the rectangle R, into nine sub

rectangles of it having equal areas, and mapping the intervali1 i99onto the nine diagonals ofiRin a similar manner.

Using the above proce dure we get the sequencekC : 0,1 : k .

Note the following properties of the sequencekC: k of

curves :

kk 11Ct C t k12.3 for allt 0,1

kk 1 k1 k1iiC C ....33 for all kandk1i3and

k1Ct Ct k123 for allk 0,1and for all k,inwith k.munotes.in

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The last of these properties imp lies that the sequencekC: k is uniformly Cauchy on0,1and as such it converges to

a continuous mapC : 0,1 .

Now, C is a continuous curve implies that its trace i.e. the setC r : s 0,1S a compact subset of. Moreover, this set

contains all the pointmmk,33formok , 3 , m and therefore,

the set is a dense subset of R. It then follows that this set is the whole

of R i.e. the continuous curve C maps0,1onto the rectangle R thus

C is a fat set and not a thin line.

6.7EXERCISE:

1)Let3c:Ibe a Frenet curve and let33L: be an

isometry. Pro ve that both the curves C, Lo chave the same curvature

and torsion functions.

2)Is it true that all curves3C: ab having common curvature

and torsion functions are isometric?

3)Letsbe a curve in its natu ral parametriz ation (=Unit

speed parametrization = are length parametrizational and letutbe

the same curve but with different parametrization the relation

between then beingut st. Prove :2 2 2

22dut ds d skt nt ttsdt dt dt 4)The Darbour vector of a curve with non -vanishing curvature is

the vectordt t k b. Prove that the Serret -Frenet formulae can be

written in the form :

dt dn dbdt , dn , dbds ds ds5)Consider the curves3c: 0Ldetermined by the unit tangent

of a regular curve3c: 0,L i.e.cs ts(unit tangent of c

at the pointcs). Assume thatksofcdoes not vanish anywhere

and prove thatcis a regular curve and obtain expressions for itsmunotes.in

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curvature and torsion functionsks, ts. Investigateks, tsin

case ofcs, the helix :c s a cos s,a sin s,bs 6)Let a plane curve be given in polar coordinates (r,) by r=f(),f:0 : 2 being a smooth function. Prove that the are -length s

between two points1122,f , ,fon the curve12is

given by 2

12 2sf f d

and the curvaturekof the

curves give by

2 12

32 2 22f 2f f fk

ff

7)Calculate the curvature of the cu rve given byrawhere ais a

positive constant.

8)Letnc:Ibe a Frenet curve inn, Prove :

n1niii1

ntc

c det k t

c

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7

REGULAR SURFACES

Unit Structure :

7.0 Introduction

7.1 Local Parametrization

7.2 Transition Functions and their Smoothness

7.3 Smooth Functions of Regular Surfaces

7.4 Exercises

7.0 INTRODUCTIO N

We think of a surface as a thin, smoothly bending sheet

having no creases, no corner s.......; a sheet spreading across a certain

region in the physical space3. Clearly we need two parameters -

its coordinates -to specify the points of such a thin sheet. Moreover

we need the coordinate systems which are adapted to the geometry

of such smooth surfaces.

Observing common surfaces such as a sphere, a two

dimensional torus, a cylinder, the M öbius band, a circular cone, etc

we find that indeed such coordinate systems are available a plenty

but only locally on a general surface, that is, each point of a surface

has a small enough neighborhood carrying a reasonable coordinate

system.munotes.in

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The above observation namely surfaces admitting coordinate

systems only locally -each point has a small enough neighborhood

carrying coordinates -leades as to the concept variously called a local

coordinate system a coordinate chart or a local parametrization.

Thus, mathematically, a smooth surface is a subset M of3admitting a nice set of coordinates in a neighborhood of each of its

points. These coordinate systems, being local, are not unique but

they are required to be smoothly related on the overlap of their

domains : one set of coordinates should be smooth functions of the

other coordinates! (This prop erty will be explained in detail at the

right stage.)

Using the local coordinates, we can differentiate functions

defined on a surface and this gives rise to a full -fledged differential

calculus on a surface. The resulting differential calculus is used as a

tool to study the highly sophisticated geometry of a surface -a

smoothly bending, thin portion of3. In particular, we study the

curvature properties of such a surface using the techniques of

differential calculus.

In this chapter we introduce the notion of a differential

structure of a surface and then proceed to explain differentiability of

functions, smooth (tangential and normal) vector fields, smooth

linear and bilinear forms on such smooth surfaces and so on. The

chapters next to this will explain the geometric features of smooth

surfaces.

Our discussion involves both the spaces2and3: we use

coordinates of2to (loally) parametize the su rface and3accommodates the surface. Although2is imbedded in3, we will

treat them as separate spaces, this is to avoide any notational

confusion (Higher dimensional Euclidean spaces also crop -up here

and there!)

The usual Cartesian coordinates in2will be denoted by12 12u, u ,v, vetc. In3we will use the triples such as123 12 3x, x, x , y, y, yetc. for the C artesian coordinates.r,will be the usual polar coordinates in

whiler, ,are the familiar spherical polar coordinates in

.munotes.in

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7.1 LOCAL PARAMETRIZATIO N

Let M be a non -empty subset of3. We will consider M

equipped with the subspace topology of3. Thus for eachpM,

the sets of the typeMB p ,for0form a fundamental

neighborhood system of p in the subspace topology of M. (Here, of

course,Bp ,is the open ball in3, centred at p and having radius0).

Definition 1 : A local parametrization of M around a poi nt p is a

tripleU, Q, Vconsisting of :

An open subsetUof2An open neighborhood. V of p (V being open with respect tothe

subspace topology ofM:V M W,W being an open

subset of3) and

A homeomorphismQ:U V,the tripleU, Q, Vhaving the

properties :

i)3Q:U is smooth and

ii)for eachqU ,the Jacobean matrix of Q at q:

11

12

22

Q

12

33

12QQqqQQJq q qQQqq has ran = 2.

Because1Q: V Uis well defined, for eachpV. We write1

12Qp u p , u p and regard12up , upas the coordinates

of p with respect to the local parametrizationU, Q, V. This

consideration lead susto the functions :12u, u : V and the resulting triple12V, u , uis called a local coordinate chart

on M around the point p; the functions12u, u : V being called the

coordinate functions of the coordinate chart.munotes.in

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Y

Here are some more e xplanations regarding the notion of a

loal parametrization :

Recall, in a local parametrizationU, Q, Vthe map1Q:U V Q V. As such we may mention either (U,Q) or1V, Qinstead of the whole triple. Using yet another

symbol, sayfor1Q, it is found that the pair

1V, V, Q is very useful. The mapassociates with

eachpVthe point ( say)qpand then we identify the

point p of M withqpof U and reard the coordinates12uq , uqas the coordinates of the pointpM. Thus we

are parametrizing the pat ch V on M by the coordinates on its

imageVU.

M, being a subset of3, a point p of M has its natural

Cartesian coordinates123xp , xp , xp. But it being a

thin sheet (a 2 dimensional geometric object so to speak ) the

coordinates -three of them -are not independent, one of them is

a function of the other two. Thus on the northern hemi -sphere

M give by32 2 2Mx , y , z x y z 1 , z 0 we have22z1 x y. Cartesian coordinates indeed are not

independent and therefore not very useful in calculations.

Secondly they do not reflect the spherical character of M.

(Indeed navigators do not mention the Cartesian coordinates,

the spherical polar coordinates,the (latitude ,

longitude) are their favourite choice!

All in all, the Cartesian coordinates of the ambient space3are not used to describe the geometry of M.munotes.in

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The main idea behind the new concept of a local

parametrizationU, Q, Vis to put the points p of the part V of

M in 1 -1 correspondance with the points q of2UCby

meanse of the homeomorphism Q so as to use the

independent coordinates12u, uof the associated point1qQ pas the coordinates12up , upof the point p of

our interest. And a careful choice of the coordinates12u, umay reflect better on the geometry of the portion V of M.

Thus, for example, on the nort hern hemi -sphere, we prefer the

independent coordinates,the latitude and longitude -

because they are better suited to the spherical geometry of the

hemi -sphere.

However, often a single parametrization fails to cover the

whole of M. a nd we need find a systemU, Q, V :of local parametrizations which together

cover the surface M, that is,MU V :. Such a

collection gives rise to the notion of a differential structure of

M; this notion is explain below.

We first define the simpler concept a surface covered by a

single coordinate chart.

Definition 2 : Aparametrized surface is a subset M of3which is

covered by a single parametrization i.e. there is a pairU, Qconsisting of (i) an open set U of2, (ii) a smooth map3Q:U such that the following conditions are satisfied.

a)QU Mb)Q:U Mis a homeo morphism and

c)QJqhas rank 2 at everyqUHere is an example of a parametrized surface; we consider the

graph of a smooth function of two real variables :

Let U be an open cubset of2and let f: Ube any

smooth function. We consider the graph of f i.e. the set3Mgiven by : 12 12 12Mu u f u u : u u U.

Now, letQ:U Mbe the smooth map given by12 12 12Quu uu, f uumunotes.in

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for all12uu U. Then indeed, Q is a homermorphism between U

and M ,moreover the Jacobean of Q at a12uu u Uis the matrix :

Q

12 12

1210

Ju 0 1

ffu, u u, uuu Clearly, this matrix has rank =2. Ther efore, the graph of such

a smooth f: Uis a parametrized surface. For example, take22 2

12 1 2Uu u : u u 1 and let f: Ube the smooth map

given by 22

12 1 2 12fuu 1u u uu U.

Clearly, the graph of this f is the northern hemisphere of unit

radius. Note that the parametrization of the hemi -sphere using this f

cannot be extended to any larger portion of the sphere. Thus on the

whole sphere, we need more than one local parametrizations to cover

it. This obs ervation motivates the following definition.

We are consisdering a subset M of3; it carrying the

subspace topology of3.

Definition 3 : A regular surface is a subset M of3having the

following property :

For eachpM, there exists a local parametrizationU, Q, Von M withpV.

A regular surface is often called a smooth surface.

As observed, we ha veQU Vand therefore we often write

onlyU, Qin place of the tripleU, Q, V.

A collectionDU Q :with the propertyMU U :is called a (smooth) c oordinate atlas on M.

Thus a parametrized surface is a special case of a regular

surface where a single coordinate chart iscovering the underlying

set. Of course we come across plenty of surfaces which are moremunotes.in

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general than parametrized surfaces. We d iscuss some examples of

them by describing the set M and then specifying a coordinate atlas

on it.

(I)A Sphere :

For a constanta0let 2332 2 2 2123 1Mx , x , x : x x x a

We consider the following open cover of the sphere M

consist ing of the six open hemi -spheres123456H, H, H, H, H, H,given

by :

11 2 3 1 21 2 3 1

31 2 3 2 41 2 3 2

51 2 3 3 61 2 3 3Hx , x , x M , x 0 , Hx , x , x M : x 0Hx , x , x M , x 0 , Hx , x , x M : x 0Hx , x , x M , x 0 , Hx , x , x M : x 0 Also let22 2 2

12 1 2Uu u : u u a ; U is an open subset

of2. We consider the following homeomorphismi :U Hi,1 i 6 :

12

12

12

12

12

12222

11 2 1 2 1 2

222

21 2 1 2 1 2

222

31 2 1 2 1 2

222

41 2 1 2 1 2

222

51 2 1 2 1 2

222

61 2 1 2 1 2uu a u u , uu ;uu Uuu a u u , uu ;uu Uuu u , a u u , u ;uu Uuu u , a u u , u ;uu Uuu uu, a u u ;uu UQu u u u , a u u ; u u U Then123456 DU , , U , , U , , U , , U , , U ,is a

coordinate atlas of the sphere M.

(II) The M öbius Band :

Let2Z 1,1 x, y : y 1,1 .

Define an equivalence relationon Z by declaringx,y x 2, yfor allx, y 1,1.munotes.in

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LetMZand:Z M, the natural projection .M is

given quotient topology of Z by the equivalence relation. Part (b) of

the figure below depicts t heMöbius band as a subset of3.

Now, let1Vx , y : 1 x 1 , 1 , y 1and2Vx , y : 0 x 2 ; 1 , y 1 .

Also, let1U 1,1 1,1and2U 0,2 1,1. And finally

let1212,UU .T hen it can be seen that11 22 DU, Uis a coordinate atlas on the set M.

The set M equipped with D is called the M öbius band.

Here is a geometric description of the M öbius band : We

consider the stripR 1,1 1,1. Twis ting the strip through180we bring the ends1 1,1and1 1,1together and glue

them in such a way that the end1 1,1comes upside down

and is glued to the other end.

An important property of regular surfaces is their

orientability. Orientability property of regular surfaces is explained

in the next chapter. M öbius band is a simple example of an

unoriented surface.

A simplified description of orientability of a surface is that it

admits a continuous (actually a smooth) unit normal field. One can

see that the Möbius band does not admit such a unit normal field

because of the twist applied to the rectangle1,1 1,1in getting

theMöbius band o ut of it. Also note that the M öbius band has only

one side.munotes.in

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(III) Surfaces of Revolution :

We consider a smooth curve2c: a, b in the vertical

XOZ -plane2given by13ct x t, x t , in terms of th e two

smooth functions12x, x : a b .

LetUa , bo ,.

We consider3312:U , :U given by11 1 3s,t x t cos s,x t sin s,x t and21 1 3s,t x t cos s ,x t sin s ,x t .

Let312MU u U c.

Then it can be seen that11 22UQ, UQare local

parametrizations on M and11 22 DU , Q , U , Q is a coordinate

atlas on M; i tbeing the surface of revolution of the curve C about

theZ-axis of3.

Before discussing mo re illustrative examples, let us prove a

result. A variety of subsets of3-called level sets of smooth

functions -are regular surfaces. This claim is verified by applying

the result proved in Proposition 1 given below.

Letbe an open subset of3and let f:be a

smooth function. For a constant, the set :Mx ; f x(if non -empty) is called a level set of the function.

Proposition 1 : Letf, M,be as above .Suppose M is non -empty

and has the following property :

For eachxM, grad

123ffffx : x , x , xxxx is a

non-zero vector.

Then M is a regular surfa ce.munotes.in

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Proof : Let123pp , p , p be an arbitrary point of M. By assumption

gradf p 0.Assume, withought loss of generality that

3fp0x.

By the implicit function, there exists an open U2cand a

smooth function g: Uhaving the following properties :

a)12p, p Ub)12 3gp , p pc)for any12 12 12uu U , uu , g uuwith12 12fu u , g u u.

(In other words, the f unction g solves the equation123fx , x , xexpressing3xas a function12x, x.) The properties (a), (b), (c)

imply t hat putting12 12 12 12uu uu, guu uu Uthe tripleU, , Vis a local parametrization on M.

Therefore, M is a regular surface.

As an application of this result, we discuss the following

illustrative examples.

(IV) An ellipsoide 2 22

3 12

123 222x xxMx , x , x , 1abc wherea0 , b0 , c0are constants is a regular surface : Take

33,f : be the function 222312

123222xxxfx , x , xabc and

let1. Clearly gradf p 0,0,0for anypMand therefore,

M-the ellipsoide -is a regul ar surface.

V)The Parabolic Hyperboloid :

Let32 2123 3 1 2Mx , x , x ; x x x

Take33,f : given by22 3123 3 1 2 123fx , x , x x x x , x , x , x and0we see that

(grad f)23p 1, 2 p ,2 p 0,0 0 and therefore, the s et M isgiven

by32 2123 3 1 2Mx , x , x x x x is a regular surface.munotes.in

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VI) Example of a set which is not a regular surface :

Let32 2 2123 1 2 3Nx , x , x : x x x

Wecontend that this set is not a regular surface. Note thatO 0,0,0is a point of N. Now, if N were a regular surface, then

every point of N would have a local parametrization about that point.

We contend that the pointP 0,0,0of N has no local

parametrization about it.

We justify this claim by contradiction. Assume the point P

has a local parametrizationU, , V. Without loss of generally we

assume that U is the discDo ,with0,0 0,0,0 N. Now

consider any point12p, pas shown in the figure and let12q, qbe the

points in U=D0 ,with11 22qp , qp

Now the contradiction is :the points,12q, qinD0 ,can be

joined by a contin uous curve cnot passing through the pointqo obut the curvec. can not avoideq 0,0,0 !

consequently such a local parametrization aroundp 0,0,0of M

does not exist and therefore N is not a regular surface.

7.2 TRA NSITIO NFUNCTIO NSA ND THEIR

SMOOTH NESS

At this stage, we study on important aspect of local

coordinates on a regular surface M : Let,UandW,be local

parametrizations withQU Wnon-empty. Then any point p inQU W=N (say) has two sets of coordinates :1

12Qp u p , u p and1

12pw p , w p .munotes.in

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This gives rise to coordinate functions12u; u : N given

by1

12pu p , u p and12w; w :N given by1

12pw p , w p for allpN.

Now we can se ethat coordinates in one set are functions of

the coordinates in other set. In fact we have :112 1 1 2 2 1 2 1 2u, u u w, w , u w, w o w, w and112 11 2 21 2 1 2w, w w u, u , w u, u u, u .

It is an important (but tedius) result that these functions are

smooth functions (of the indicated variables). Here we give a

sketchy proof of this fact.

Proposition 1:The following functions are smooth :

1

11 2

1

21 2

1

11 2

1

21 2uw , w: N

uw , w: N

wu , u : N

wu , u : N

Proof : We prove smoothness of11 2 21 2wu , u , wu , u on the set1N. (Smoothness of the other two functions is obtained in a

similar proof.). We accompli sh this by verifying smoothness of

1oin a neighborhood of each1qN.

Thus choose arbitraity a1qN. Let1oq p.

Now recallJphas ra nk = 2 and therefore some22sub-

matrix of the matrix :

11

12

22

12

33

12ppwwJp p pwwppww is non -singular. Assume without loss of generality thatmunotes.in

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11

12

22

12ppwwppww is non -singular. Let32:be the projection map given by123 12x, x, x x, x . Then the non -singularity of the above sub -

matrix is the non -singularity ofoJp. Therefore, by the inverse

mapping theorem, we get whatois a local diff ermorphism in a

neighbourhood of p. This implies the invertibility ofin a

neighbourhood ofp. (Here, we are using local 1 -1 ness of bothoand.) No w, we have : 11 1

1o q o o oQHo o oQ

Thus, smoothness of both 1oando-implies

smoothness of the map1owhich is the map which gives the

change of coordinates12 1 2u, u w, w .

For the two parametrizations (U,),W,of M withQU W Nthe maps11 112 1 2o: N Nu, u w, w

and11 112 1 2o: N Nw, w u, u

both describing the change of coordi nates are alled the transition

maps between the sets11N and Q N . Transition maps describe

one set of coordinates as functions of the other set of coordinates.

And we have proved above that the transition maps are smooth

functions of the co ordinates equivalently put: the two sets of

coordinates -12u, uand12w, w-are smoothly related.

7.3 SMOOTH FU NCTIO NSONREGULAR SURFACES

Let M be a regular surface.

We will consider only two types of fun ctions and define their

being smooth :curves c : I Mand

functions f : M munotes.in

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Definition 4 : A curvec:I Mis smooth if3c:Iis smooth.

Itreadily follows that ifc:I Mis smooth in the sense of

this definition thanc:I Mis continuous.

Next, let (U,) be a local parametrization on M with

associated coordinate functions12u: u: U M . Then the

curve1oC : I Ucan be written in terms of its coordinates :1

12oC t u t ,u t for alltIwithct U. Thus we

get the functions12ut , utof the variable t. Now it ca n be seen that

the curve cis differentiable (= smooth) if both the real valued

functions12tu t , tu t of the real variabletIare smooth.

Finally we define smoothness of functions f: M.

Definitions 5 : f: Mis smooth if for every local

parametrization (U,) of M, the function1fo : Uis smooth.

Note that1fo Q : Uis a function of the two coordinate

variables12uuan U and therefore differentiability of1fois a

familiar concept.

We consider the setCMof all smooth functions

f: M. It is easy to see that th e operations of addition and

multiplication of functions f: Mgive the setCMthe

structure of a commutative and associative ring with identity.

Finally, letbe a non -empty open subset of a regular

surface M. Then it is easy to see thatalso is a regular surface. For

if (U,) is a local paramentrization of M, then putting1UUandUwe get a local parametrizationU,on. Such local parametrizationU,onobtained from

(U,) of M give a coordinate atlas forand thus,becomes a

regular surface in its own right. In particular, the function spacesCfor openMare well -defined.

In the next chapter, we w ill develop differential calculus on

M using these function spacesC.munotes.in

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7.4 EXERCISES

1)Let M be the subset of3obtained by rotating the parabola231x4 xabout the3xaxis. Describe smooth function f: Mwhich generates M.

2)The 2 -torus2Tis the surface generated by revolving the circle

22213xa x babout the3x-axis, a,bbeing constants with aExhibit a smooth coordinate atlas on2T.

3)Although the set32 2 2123 3 1 2Mx , x , x : x x x is not a

regular surface (as explained above) prove that its subsetMM 0is a regular surface.

4)Prove that a circular cylinder is a surface and describe a smooth

atlas on it.

5)Let M be a regular surface andan open subset of M. Let

f: Mbe a smooth map. Prove thatfis smooth on.

6)Let1Mand2Mbe regular surfaces, with12MMopen in both1Mand2M. Prove12MMis a regular surface.

7)Let M be a regular surface and let f, g:Mbe smooth

functions.

Prove :

a)fgis smooth on M.

b)fgis smooth on M.

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8

CALCULUS O NREGULAR SURFACES

Unit Structure :

8.0 Introduction

8.1 The Tangent Spaces3pT8.2 The Tangent SpacepTM8.3 Another Description of Tangent Vectors

8.4 Smooth Vector Fields

8.5 Smooth Forms on M

8.6 Exercises

8.0 INTRODUCTIO N

Having introduced regular surfaces M and the function spacesCfor various openMwe consider some more concepts

contributing to the calculus on a regular surface, namely : the

tangent spacespTMforpM, smooth vector fields on open

subsetsof M smooth linear and bilinear forms and their properties

and so on. The resulting calculus is then used as a too l to study the

geometry of M. The primary geometri features of a regular surface

M are two smooth symmetric bilinear forms the first fundamental

form I and the second fundamental forms II -they will be introduced

in the next chapter.

8.1 THE TA NGENT SPA CES3pTIn differential geometry, geometric object are highly

localized. In particular, we need consider the classical vectors -the

directed segments in3being located at various points of3. Thus

for a point3pand for a vector x in3, we consider the ordered

pairp,x ;it represents the vector x not emanating from the origin

of3but located at (or having its foot at) the point p.3pTdenotes the set of all such ordered pairs3px :xmunotes.in

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1143pClearly for a fixed point p in3, the set3pTis in 1 -1

correspondence with3:33pT

xp , x

Therefore the familiar inner product space structure of3induces an inner produce space structure onpT:p,x p,y p,x yap , x p , ax, ap,x , p,y x,y

8.2 THE TA NGENT SPACEpTMLet p be a point of the regular surface M.

Definition : A Vector3p p,x Tistangential to M at the point

p of M if there exists a smooth curvec: , Mfor some0with the properties : dcc0 P , c0 0 xdt

.pTMdenotes the set of all3p p,x Twhich are

tangential to M at p. we prove below thatpTMis a two -

dimensional subspace of the vector space3pT. Towards this aim,

consider a local paramentrizationU,of M withpU : 0 p. Recall that the derivative map3op *o: D o T U Tis an injective linear map. We

prove now that it maps22ooTT U ontopTM. To seemunotes.in

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this, consider app,x T M.By definition of a tangential vector

there exists a smooth curvec: , Mwithc0 pand

c0 x

. Assume without loss of generality thatct Ufor all t

in,. Now using bijective property of, considerc: , Usuch thatcc. Letco. ThisoTUand*ox. Thus proves thatop*:T U T M is surjective -

Consequentlyop*TU T M is a linear subspace of3pT.

Clearly, the above result implies that the mapop*O: T U T M is an isomorphism and thereforepTMis a

two dimensional subspace of3pT. We rest ate this fact in the

following :

Proposition 1 : For eachppM , TMis a two dimensional

subspace of3pT.

There is yet another noteworthy fact, namely the coordinate

chartU,around apMgives rise to a vector basis ofpTM:

Consider the curves12:n , n U , :n , n Ufor small

enoughn0;which are given by :12s s,0 s 0,s . s n,n. We have1201 , 0 ,00 , 1 which are vectors inoTUconstituting a

vector basis ofoTUconsequently the vectors.

12

11**00 0 , 00 0uu

form a vector

basis ofpTM.

Note that the maps :n,n M;s s,o andn,n M;s o,s aretwo smooth curves passing through p and giving the basic

tangent vectors

120,0 , 0,0uu respectively and therefore they

are vectors tangential to M at p.munotes.in

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Also, it is important to note thatpUwasarbitrary point

ofUand therefore the above discussion lead us to two vector

fields onU, both tangential to M at the points ofU: For each12uu , u U(and not only0,0as above) the vectors

12u, uuuaretangential to M at th e pointpu. Thus we get

two tangential vector fields12,uuonVUsuch that at each

pointpuof V,

12a, auuform a vector basis ofpTM.

There is one more point pertaining to the notation which we

explain right here : We adapt the notations

12p, puufor the

vectors

12u, uuurespectively at the pointpu V.

These notations -the pair

12p, puu representing

tangent vectors but partial differentiations in appearance are adapted

everywhere in mathematical literature because vectors operate on

functions by differentiation. We will explain more abou t this

notational convention below, but at this stage but we note that

because

12p, puuis a vector basis ofpTMfor any point p

ofUV, any vectorpp,x T Mis expressibl e as a linear

combination : 12

12ap apuu for a unique pair12aaof real

number.

Now, about the action of a tangent vector on a smooth

function :Letpp,x T Mand letfbe a smooth function defined

on an openWMwithpW. These two entities combine to

produce the real number (denoted in differential calculus by)xDf pthe derivative offat p along x. It is obtained as follows.

Choose a smooth curveC: , Wwithco pand co x

.

Then we lay :

xt0dDf p f ctdtmunotes.in

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Now, letU,be any ch art around p, its coordinate

functions being12uu. Using these coordinates, we write :12ct c t, c t,

Then we have :

12to to

12

12ddfc t fct , ct ,dt dt

ffc0 p c0 puu

But, we also have dcx0dt

12

12c0 p c0 puu

and therefore, we get;

x1 2

12Df p C o p C o o fuu

12

12ffc0 p c puu

.

To conclude, we have the following :

Given a pointpMand a pair of local coordinates12uuaround p (determined by a local parametrizationU,, we have the

following :

Anypp,x T Mcan be expressed uniquely in the form

12 1 212p,x a p a p a ,auu in.

Iff: Wis a smooth function, its domain of definition W

being an open subset of M withpWand ifU,is a local

parametrization around p, its coordinates bein g12uuthen

the real numberxDf p-the derivative offat p along x -is

given by x1 2

12ffDf p a p a puu where

12

12fxa p a puu .

The resulting mapxDp : CWhas the following

properties.munotes.in

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i)xx xD af bg p aD f p bD g p for allf, ginCW

and a, b inii)xx xDf gp Df pgp f p D gp for allf, ginCW

iii)Iff: W ; g : W are smooth functions, W andWbeing

open neighborhoods of p, thenfgonWWimplies :xxDf p D gp

For anyp,x , p,yinpTM, a, b in,

xy ax byDf p a D f p b D f p holds for allfC W(W being an open neighborhood of p).

The last property implies that any smooth f: Wgives

rise to a linear form onpTM; we will denote it bydf p. Thus the

linear formdf p:pTM is given by :x

12

12df p p,x D f p

ffxp xpuu

for all 12 p

12p,x x p x p T Muu .

In particular, the coordinate functions12uuof a local

parametrizationU,around p give rise to the linear forms12du p ,du ponpTM. Note that12du p ,du psatisfy

1 ijjdu p puand consequently we get :

12

12ffdf p p du p p du puu for any smooth

f: W.

Definition 2 : The linear formdf p:pTM is called the

differential offat the point p.

8.3 A NOTHER DESCRIPTIO NOF TA NGENT VECTORS

Above we have defined app,x T Mas a vector3xplaced at p for which there co rresponds a smooth curvemunotes.in

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119c: , Mwithc0 pand c0 u

. The vector thus defined

(tangential toM at p) operates on smooth functions f: Mproducing real numbersxDf pgiven by.

xtodDf p f C tdt

This action ofp,xon smoothfhas the following

properties (as we have noted them above) :

i)Iff, g inCMare such thatfgin some neighborhood of

p, thenxDf p=xDg p.

ii)xx xD af bg p a D f p b D g p for allf, ginCMand for all a, b in.

iii)xx xDf gp Df pgp f p D gp for allf, ginCW.

We prove below that conversely, properties (1) , (2) and (3)

above specify the vectorpp,x T Mcompletely. To be precise,

we prove the following.

Proposition 2 : LetL:C Mbe an operator satisfying the

following conditions :

1)Iff, g are such thatfgin some open nei ghborhood of

p, thenLf L g.

ii)LetL af bg a L f p b L g for allf, ginCMand

for all a, b iniii)LetLfg Lfgp f p L g for allf, ginCM

Then there exists a unique3x, tangential to M at p such

thatxLf D f p for allfCM

Next, to prove the existence of such ofp,x, note the

following two properties :

The result is alocal result in the sense that by property (1) of

L, the valueLffor anyfCM depends on the

variation offwithin (an arbitrarity chosen) neighborhood

of p.munotes.in

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Accordingly, we can chose a local parametrizationU,on

M with (i)UB 0 ,for someoand (ii)Q0 pand then forU; by property (1) of L, the behaviour offoutsideUdoes

not affectLf.If fconstant c(say), thenLf 0.

For, takingfg 1we have :2L1 L1 L1 1 1L1 2 L1, thusL1 2 L1and

thereforeL1 0.

NowLC C L1 0Thus,LC 0for any constant functionfC.

Now, for the above described choice ofU,consider the

finite Taylor expansion of afaround p

12 1 1 2 2

12fffuu f p u p p u p puu

2

iij j i j

i,j 1up u p g u for some smooth functions

ijg: U .

Applying the operator L to this identity, we get :

12 22

12

ii j j i j

ij

jj i i i j

ji

12 22

12

1

1ffLf Lf p L u p p L u p puu

up pp g p

up p p g p

ff0L u p p L u p p Ouu

fxp xu

2

2fpu

where we are putting11 1 22 2xL up ; xL up.W ef o rm the

vector 12 p

12xx p x p TMuu to getxLf D f p for

allfCM.munotes.in

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8.4 SMOOTH VECTOR FIELDS

A vector field on M is an assignment X, assigning a vectorXpto eachpM ; X p being tangential to M atpp:X p T Mfor eachpM.

A vector field X on M and a smooth function f: Mcombine to produce a function on M, -we denote it byXfwhere

for eachpMthe real numb erXf pis given by :xXf p D f p

wherepXp T Mis given byXp p , x.

Now letU,be any local parametrization on M is;its

coordinate functions bei ng12uu. Then for eachpUwe have:

12

12Xp X p p X p puu with12X, Xbeing smooth

functions onUVM. Therefore, for any smooth f: M,

weget 12

12Xf p X p p X p puu for everypU.

It now follows thatXfis smooth if the functionfissmooth. We

are interested in vector fields X on M which produce smooth

functions .

Definition 3 : A vector field X on M is smooth ifXf: M is

smooth whenever f: Mis smooth.XMdenotes the set of all smooth vector field X on M.

It now follows that a vector f ield X on M smooth (i.e.XX M) if it satisfied the follows condition : For any local

parametrizationU,, the representation :1211XX Xuuhas

the coefficient function12X, X : U be smooth. It can be verified

that the setXMhas the following algebric property :

If X, Y are smooth vector fields on M and if f, g are smooth

functions on M, then the vector fieldfX , gYis also a smoo th

vector field on M.munotes.in

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Thus the setXMof smooth vector fields on M is a module

over the ringCMof smooth functions.

Above, we considered smooth vector fields on aregular

surface M. Since every non -empty open subsetof M because a

regular surface, we have the well -defined concept of smooth vector

fields on an open subsetof M. We denote the resultingCmodule byX.

8.5 SMOOTH FORMS O NM

We consider now objects which are dual to the vector fields,

they are called smooth one -forms on M. First, (an arbitrary) one

form on M is an assignment of a liner formp wp: T M to

eachpM. We denote the collectionwp: p Mby w.

Now, note that a vector field X on M and a one -formww p : p M on M combine to give a function f: M: For

eachpM, we evaluate the one formp wp: T M on the

vectorpXp T Mto get the real numberwp X p; we

putfp w pXp . This gives the function :

f: M;pw p X pWe will be interested in those 1 -forms to which differential

calculus can be applied in a reasonable way. This motivates the

following definition :

Definition 4 : A1-form w on M is smooth if for every smooth

vector field X on M, the functionwX : M is smooth on M.

Now, we have the following list of simple facts related to

smooth 1 -forms and smooth vector fields on M :

1)If w, n are two smooth 1 -forms and if f, g:Mare any two

smooth f unctions then the combinationfw gngiven by

*f w gn p f p w p g p n p T p , p M

is a smooth 1 -

form on M.

Therefore, the set of all smooth 1 -forms on M is a module

over the ringCM.munotes.in

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2)For anypM, the forms12 pdu p ,du p :T M form a

vector basis of the dual space*

pTm ,(ofpM) and therefore, if w

is a 1 -form on M, then for eachpMwe everypM, we seet that

the 1 -form w gives rise to two functions f, g:Msuch that holds

for everypMand thus we have :12w fdu gdu.

3)Note further that

1wfuand

1wguconsequently, if w

is a smooth 1 -form on M then f,g (as above) must both be smooth

4)Now consider12w fdu gduand on arbitrary smooth vector

field

12Xh k , h , k : Muu being both smooth functions.

Then we havew X fh gk.

Consequently, we have : w is smooth if and only if both, f, g

are smooth functions. It also follows that the set of smooth 1 forms

is a module over the ringCMifw,nare smo oth 1 -forms and f, g

are smooth functions on M, thenfw gnis a smooth form on M.

5)If w is a smooth 1 -form on M and ifis an open subset of M,

then the restriction of w toisa smooth 1 -form on.

6)And a smooth function f: Mgives rise to a smooth 1 -form1212ffdu duuuon M, we denote it by df and call it the differential

of f, thus,

12ffdf X g hdu duwith12Xg hdu du.

We also consider smooth, symmetric 2 forms on M. first

recall a few algebraic terms.

Let E be a finite dimensional real vector space.

A bilinear form onis a map,: ; x , y x , ywhich is linear in each of the two vector variables x, y ranging on.

A bilinear form ,:is said to be

Symmetric ifx, y y,xholds for all x, y in.

Positive definite ifx,x 0forxandx,x 0only

whenx0.

Aninner product onif,is both symmetric and posit ive

definite.munotes.in

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Let12 ne ,e ,....,ebe a vector basis of, puttingij i jae , efor1i jn, we get the matrixijaof,with respect to the vector

basis12 ne ,e ,....,e. Note that -

i)ij i j1i , j nx, y a x ywhereniii1xx eandniii1yy e.

ii),is symmetri cif and only ifji jiaaholds for alli, j,1 i, j n.

Now, we introduce the notion of a smooth, symmetric bilinear

form on a regular surface.

Definition 5 : A bilinear form on a regular surface M is a rule -

denoted by B -which associates with eachpM, a bilinear form

B(p) on the tangent spacepTM:ppBp: T M T M ; u , v Bp u , vA bilinear form B on M and two tangent fields X, Y on M

combine to produce a functionBX Y: M :

For eachpMthe bilinear formpp Bp: T M T M evaluated overpXp , Yp T Mgives, the real numberBp X p, Yp. This specifies the functionBX , Y: M : BX , Y p Bp X p, Y p for everypM.

It now f ollows that the following identities hold :

B fX ,gY f gB X ,Y for all functions f, g:Mand

for all vector fields X, Y on M.

i)12 1 2BX X, Y BX , Y BX, Y

ii)12 1 1 2BX , Y Y BXY BX , Y for all vector fields12 1 2X,X X , Y, Y, Y

Here is an example of an important bilinear form on M : Let,

for eachpM,pp Ip : T M T M be given by :p pIp v , w v , w v , wT M .

This gives rise to the following map

I:XM XM :munotes.in

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125IX , Y p Ip Xp , Yp for allpMand for all

smooth vector fields X,Y on M.

This biliner form is called the first fundamental form of the

surface. Read more about it in Chapter 9.

8.6 EXERCISE :

1)Forsmooth vector fields X, Y on M and for smooth functions

f, g , h:M, verify the following identies.

a)Xf . g Xf g fX g

b)IfXf 0for all smooth f: M, thenX0c)X af bg aX f bX g

2)Let X, Y be smooth vector fields on M giving the mapL:C M C M :Lf XYf YX f

Verify that L satisfies the properties (a), (b) (c) of exercise (1)

above, using Proposition 2 deduce that L gives rise to a smooth vetor

field on M. We denote this vector field byX, Yand call it the Lie-

prodct of X, Y in that order. It is also call the Lie -bracket of X, Y.

3)Prove that the operation of forming Lie -bracketX, Yof two

vector fields X,Y has the following properties :

i)X, Y=YXii)fX ,Y f X ,Y Y f X

iii)XY , Z Y , Z , X Z ,X , Y O

4)Prove that combing a smooth 1 -form w with a smooth vector

field X on M produces the functionswXwhich is smooth and the

operationw,X w Xis bilinear.

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9

PARAMETRIZED SURFACES

Unit Structure :

9.0 Introduction

9.1 An Oriented Parametrized Surface

9.2 The First Fundamental Form

9.3 The Shape Operator

9.4 Covariant Differentiation

9.5 Parallel Transport

9.6 Geodesics

9.7 Exercises

9.0 INTRODUCTIO N

In this chapter and the next, we will study some of the

elementary aspects of the geome try of an oriented regular surface M.

To begin with we wil discuss the geometry of such a M only at the

local level, that is, the geometric structure of a small enough pie ceo f

a surface in the form of an open neighbourhood of a point of it. After

getting familier with the local geometry, we will consider geometric

properties of M as a whole and prove some basic results about them.

Accordingly we begin with a surface element in the form

already introduced where it was termed a parametrized surface.

Reca lling the related concepts and explaining them again in the

present context, we introduce two basic geometric ingradients of a

parametrized surface namely the first and second fundamental forms

I and II on the tangent bundleTMofM. Both of them are

symmetric two forms onTM. These forms will lead us to a

number of geometric concepts on M : length of a smooth curve on

M, covariant differentiation of vector fields, parallel transport of

tangent vectors along smooth curves on M, geodesic curves on M,

principal curvature of M at a point of it, the Gaussian and mean

curvature tensor of M and so on. We introduce the intrinsic nature of

some of the geometric properties and conclude the next chapter with

the import ant theorem : Gauss’ theorema egragium.munotes.in

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9.1 A NORIE NTED PARAMETRIZED SURFACE

Let M be a parametrized surface, its parametrization beingU,F; thus U is an open subset of2, and3F: U is a smooth

map having the properties :

FU MandF: U Mis a homeomorphism, and

The JacobeanFJqhas rank = 2 at everyqUNow, using the homeomorphismF: U M, we identify

each point p of M with the point1Fpqof2Ucand regard

the native coordinates12uq , uqas the coordinates of p assigned

by the parametrizationU,F.

Thus, there are two sets of coordinates on M :

i)The Cartesian co -ordinates123xp , xp , xpgiven by the

(Cartersian) Co -ordinate system of the ambinat space3and

ii)the co -ordinates12up , updetermined by a parametrizationU,Fon M.

The co -ordinates12u, uare independent and are often better

adapted to the geometry of M while the Cartesian co -ordinates -

being coordinates of the ambient space3-are often used as

reference coordinates only. Thus, for aqUwe have :123Fq x q, x q, x q , Cartesian coordinates ofQq p M.

3 12

11 2 1x xx Fqq , q , quu u u 3 12

22 2 2x xx Fqq , q , quu u u and so on.

Note that we have adapted the notations 1

1p or pu

and 2

2p or pufor

12FFq, quurespectively and in view of

these notations, the rank condition -rank ofFJqbe 2 -ismunotes.in

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equivalent to the requirement that the vectors

12FFq, quube

independent elements of the tangent spacepTM. Also, keep in

mind that the pair

12q, quu has to pla y a double role (i) as a

vector basis ofpTMand (ii) as differential operators operating on

smooth functions f: Mgiving real numbers

12ffq, quu.

(In thse notations, the pointpQappears but it is considerd to be

identified withq:p F q).

Let us now consider vector fields on M, first those vector

fields which are tangential to M.

Recall, a vector field tangential to M (or a tangent field on M)

is a rule X associating with eachpMa vectorpXp T M.

Now since

12q, quu is a basis of the vector space.pTM,

such aXpcan be expressed uniqu ely as a linear combination :

12

12Xp X p X puu 12Xp , Xpbeing real numbers. This way the vector field

gives rise to the well -defined functions12X, X : M the vector

field then being expressible in the fo rm :

1212XX Xuu.

We regard the vector field X smooth if both the functions12XXare smooth on M. Now, for any smooth function f: M,

the vector field operates onfproducing a smooth functionXf: M given by :12

12Xf p Xp f

ffXp p Xp p ; pMuu

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On the other hand we have vector fields on M which are

perpendicular to M : A smooth map3Y:M considered as a

vector field on M (i e for eachpMthe vectorYpbeing

considered located at p) is normal to M ifpYp T Mfor eachpM. For example the vector field Y given by

12FFYp q quu(withFq p) for eachpMis such a

normal vector field on M. In particular the vector field N on M given

by 12

12pp

uuNp , p Mppuu

has the unit normal property.

Consequently for eachpM, the triple

12p, p, N puu forms a vector basis of3pTand the subset

12p, puu is a

vector basis of the subspacepTMof3pT. On account of t his

property the unit normal field N on M orients the parametrized

surface M. In what is follow, we will consider M to be oriented by

this normal field N.

9.2 THE FIRST FU NDAME NTAL FORM

Now we consider the standard inner product.,of3which

incudces the inner productp,on each3pT. We restrictp;to

the subspacepTMof3pTand d enote it byIp. Thus, for

eachpM, we have the symmetric, positive definite bilinear formpp Ip : T M T M given byIp p v , p w v , wfor

every pairpv , pwof vector s tangential to M at p.

Having introduced the inner productIponpTM, we will

write onlyv,win place of the full formIp p v , p w. This is

meant to simplify the notation whenever the point p of tangency of

the vectorsp,v , p,wis understood.

We consider the entire collectionIp : p Mas a single

entity and denote it by I.munotes.in

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Definition 1 : II p : p M is the first fundamental form of

the surface.

For eachpI, putting ij

ijgp p , puu

12FFp, puuforpM , 1i , j2, we get the matrix valued

function.pp 2 g: U T M T M :p M M .

It isthe matrix of the first fundamental form.

Now if

iiiXX i , Y Y iuu are two smooth vector fields

(tangential to M) then we get the mapIX , Y: M given by :

ij

ij ij

j

ij ij

ji j

ijIX , Y p Xp , YpXp p , Yp puu

Xi p Y p p , puu

Xi p Y p g p

We need consider the inverse of ea chijgp ;we denote the

resulting matrix byijgp, thus we have :kjik ijkgp gp.

Let us consider following examples of surfaces and obtain the

first fundamental forms for each of them :

(I)The(oriented) graph of a smooth functions : f: U;

U being an open subset of2Now, 12 12 12Mu , u , f u , u : u , u U .

The parametrization map3F: U is :munotes.in

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13112 12 12 12Fu , u u , u , f u , u p , u , u U .

Therefore,

12 1 2ff, ,1uu u u and then

12

22

12ff, ,1uuNp

ff1uu

the right hand side of the above equally being evaluated at the point12 12pu , u , f u , u for12u, u U.

Also, we have :2211 22

11ffg1 , g1uu and12 2112ffgg ,uuand

therefore the matrix of the first fundamental form of the surface :

2

11 2

ij212 2ff f1, ,uu ug

ff f,, 1uu u (II)A particular case of the above is the hemisphere of radiusao:

32 2 2 2 2 2123 1 2 3 1 2Mx , x , x : x x a ; x a x x

Now we have :22 2 212 1 2Uu , u ; u u a and the map

f: Uis22212 1 2fu , u a u u. Finding expressions for the

unit normal mappN p, the matrixijgpof the first

fundam ental form etc are left for the reader as an exercise.

(III) Let2Uand let23F: be the map given by22 312 11 2 1 2Fu , u u , u u , 4 u u,212uu. Now we have :munotes.in

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F1 2 121210J u u 2u 18u 3u forall212uu; the matrix clearly has

rank = 2 (because its submatrix

1102u 1is non singular) at every212uu. Consequently the set M :22 3 211 2 1 2 12 Mu , u u , 4 u u : u , u is a parametrized surface.

Now, we have

i)3F: U M is given by22 312 11 2 1 2Fu , u u , u u , 4 u u,

ii)The vectors 111Fu 1,2u ,8uuand 222F0,1,3uuspan the

tangent spacepTMwhere12pF u , u ;

iii)The unit normal field N on M is given by

22

12 2

222 2

12 22u 3u 4 , 3u ,1Np4u 3u 4 9u 1

and

iv)The matrix of the first fundamental forms is :22

11 2

24

12 21 68u , 2u 1 12u2u 1 12u , 1 9u (IV) We consider a unit speed curve3C:I and the associated

binormal field3b:Ialong it. Associated with the pair (a, b) is

the parametrized surface M :MC r s b r : r I , s Putting UI, let3F: U be given byF r,s C r sb r , r,s I .

Then we hav e: Fr,s t r s r n rr

Fr,s b rr

Clearly, FFr,s , r,srsare livearly independent vectors

and consequently (U,F) is a parametrization of the set M. Moreover,

the unit normal fi eld N on M is given by :munotes.in

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2 2sr t r rNp

sr 1

while the matrix of the first fundamental form is2 21s r ,00, 1 we resume our study of a parametrized surface Mhaving its

parametrization (U, F) :

LetS2be the unit sphere in3i.e.32 2 2

12 3 1 2 3S2 xx , x : x x x 1

Shifting the unit normalNpfrom the point p of M and

relocating it at3Owe get the map (denoted by the same letters) :N:M S 2

We call this ma pthe Gauss Map of the surface M. note that

the Gauss map on the unit32 2 2

12 3 1 2 3 3Mx x , x : x x x 1 , x o is the identity map

on M while that on the hemi sphere of radius32 2 2 2

12 3 1 2 3 3a0 : M x x , x : x x x a ; x 0 is : pNpaforpM.

Illustrative examples (I) ---(IV) above describe the Gauss

map of their surfaces.

9.3 THE SHAPE OPERATOR

We differentiate the Gauss map -defined above -at a point p of

M with respect to the vector spvTM. The resulting linear map -

the differential of the Gauss map at p -has important geometric

prosperties; we describe them below.

Let p be a point of M and letpvTM. We consider the

derivativevDN p. Thus, we choose a smoothC: , M

withC0 pand C0 v

. Then we have

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Note thatpD,N p T M. For, we have

NC t ,N C t 1

fort,and therefore

todNCt , NCp 0dti.e. tod2N C t , N C p 0dti.e.oDN p , N p Othe perpendicularity ofD,N pwithNpnow impliesD,N pis inpTM. Thus, the Gauss map, when differentiated at

apMgives the linear map :ppTM TMvD , N p

In what is to follow, we consider the mapvD , N p ,

the negative sign attached here is only to follow the standard practice

in mathematics literature. We denote the resulting (linear) map

bypL:pp pL: T M T M

Definition 2 :

The linear mappp pL: T M T Mis called the shape

operator of M at the point p.

The shape operatorpLis also called the Weingarten map of M

at p.

Considering the Weignarten mappLalong with the linear

productIpofpTM, we haave the important property of it :

Proposition 1 :pLis a self -adjoint linear endomorphism of the inner

product spacepTM , I p.munotes.in

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Proof : Since

12p, puu is a vector basis ofpTMit is

enough to verify the following equalities :

pp

ij i jLp , p p , L puu u u for1i , j2Proof : We have :

ippi u

iLp DNu

Npu

and therefore,

p

ij i j

2

ij i j

2

ijNFLp , p p puu u u

FFNp , p Np , puu u u

F0N p , puu

The first summond above is O, because

iFNq, q 0 ,u

(

iFqubeing tangential to M at qwhile N(q) is perpendicular to the

whole spaceqTM). Thus

2

p

ij i jFLp , p N p , puu u u .

Similarly we get :

2

p

ij i j

2

ijFp, L p N p, puu u u

FNp , puu

Combining these two equalities, we get

pp

ij i jLp , p p , L puu u u which leades

us to the self adjointness ofpL.munotes.in

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Next, we wi sh to find the matrix ofpLwith respect to the

vector basis

12p, puu ofpTM.

Suppose p

k ikLp , i k puu (of course, the

summation being overk1 , 2.) Taking inner product of the above

equality with

jpu, we get pk j

k ijL p , p ai gk puu .

But we already have

2

p

ij i jFLp , p N p , puu u u and therefore the

above equation gives 2

kj

k ijFN p , p i gk puu .

Letijgpbe the inverse of the matrixijgp. Using this

inverse matrix, we get

2

je jk

kj

jj k ij

e

ik k

k

ieFN p , p g p i gk p g puu

Thus 2

kj

ij

k ikFNp , p g puu ......................... (*)

This gives the matrixijof the Weingarten mappL.

Let us consider the following illustrative examples : M being

the graph of a smooth function f: U, (as usual U being an open

subset of2).

Now, we have the parametrisation map3F: U given by12 12 12 12Fu , u u , u , fu , u u , u U .

Writing2

12i j

12i jff fff , f i , j xuuu u etc.munotes.in

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i)22

ij21 2 2 1 2ij2222

12 2 12 1 121f , f f 1f , f f1ggff , 1 f ff , 1 f4f f ii)

11 2 21 2

22 22 22

12 12 12fu , u fu , u 1Np , ,

1f f 1f f 1f f and

iii) 2ijijFO,O, fuu

Substituting these expressions in the formulae (*) we get :

211 2 12 1 211 322 2

12

2

12 1 11 1 2

12 322 2

12

2

21 2 22 1 2

21 322 2

12f1f f ff1f f

f1 f f f f

1f f

f1 f f f f

1f f

and222 2 22 1 222 322 2

12f1 f f f f1f f

Taking2f: given by2212 1 1fu u u uwe get :

i)11 2 2f 2u , f 2 uii)11 22 12 21f2 f , f fOiii) 3322 2 22212 1 21f f 1 4 u 4 uand therefore

2

21 2

p 3222 212 2124 1 4u 16u u1L,16u u 4 1 4u14 u 4 u .

We combine the Weingartain mapspp pL: T M T M and

the first fundamental formpp p pI, : T M T Mto

get a bilinear mappp II p :T m T m for eachpMas

follows : If v, w are vectors inpTm, thenppII p v,w I p L v ,w L v ,w .munotes.in

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We denote the collectionII p : p Mby II and call it the

second fundamental form on M.

Note that the second fundamental form II comb ines two

smooth vector field X, Y on M and produces a smooth functionX, Y :M which is given by :pX, Y p L X p ; Y p p M .

Because I is bilinear and eachpLis self -adjoint, we get the

following identit ies :

II X ,Y II Y , X , X ,Y being smooth vector fields on M.

fX gY ,Z f X ,Z g Y ,Z

Second fundamental form is used to express curvature

properties of M, we will discuss this point in the next chapter.

9.4 CO VARIA NT DIFFERE NTIATIO N

Given a smooth tangent field X on M and apvTM

covariant differentiation is a process producing a vector -denoted byvXinpTM.

Recall :ToapvTM there correspo nds a smoothC: , Mhaving the properties C0 p , C0 v

. The two -

X, and C -Combine to give the smooth maptX C t ; t ,.

Differentiation of it gives 3vpdDX X C t Tdt . Note

that thoughvDXis a vector located at p, it is not (in general)

tangential to M at p.

To get a vector tangential to M at p, we project it down in the

subspacepTMof3pT; that is let3

ppp: T T M be

the desired projection thus,pw w w , N pN p for all3p wT. Now, we setmunotes.in

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v

to

vv vdX C tXpdtpD X DX DX , N pN p

Definition 3 :vXis the covariant derivative of X with respect topvTM.

The covariant derivative has the following properties

vw av bwXa X b X for all v,w inpTMa, b inand for all smooth vector fields X on M.

vv vXY X Yfor allpvTM and for all

smooth vector fields X and Y.

vv vfX D fXp fp X for allpvTMand

for all smooth vector fields X (RecallvDfis the usual

directional derivative of vtodf: D f f C tdt .

All these properties follow from (i) the properties of

vtodDf f C tdt and (ii) the linearity of the map3

ppp: T T M .

For apvTMand for a tangential vector field X on M we

intend to expressvpXT Musing the vector basis

12p, puu .

We adapt the notationsifor1uandipfor

ip, i 1 , 2uonly for a short while.

Let11 22 p 1 2vv p v p TM , v v and let11 22XX Xbe a vector field on M with12X, X : M being

smooth functions.munotes.in

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Now, we have :2 1v1 1 1 2 2 2 1 1 2 2p

11

12 1

12Xv X X v X X

XXvp vp puu

11

2222

12 2

12

11 1 12 2 pp

21 12 2 ppXXvp vp puu

vX p vX p

vX p vX p

Therefore, we need express eachijpas a linear

combination of1pand2p. Suppose :

i12

j ij 1 ij 2 ppp ppwhere1

ijp,2

ijp

are real numbers. (Indeed they depend oni, jand p). Also, we write

i2

j p

ijFpN puu.

Whereis some real number. Combining the above two equalities,

we get 2

12

ij 1 ij 2

ijFpp pp p N puu........... ( 1)

Note right here that 22

ij j iFFppuu u u implies11

ij jippand22

ij jipp.

Taking inner product of the equation (1) withkp, we get :

2

12

k ij 1k ij 2k

ijFp, p pg p pg puu ........... (2)munotes.in

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On the other hand, we have :

22

kk

ij i j j i k

2

jk j

ii k

12 ik

ji k 1 i k 2

iFF F Fp, p p p, p p, puu u u u uu

Fp, p , puu u

gpp , p p p pu

12 ik

ik 1 j ik 2 j

igpp g pp g pu

Combining (2) and (3) above, we get :

12 1 ik

ik 1k ij 2k ik 1 j

ig0p p g p p g p p g pu 2

ik 2 jpg p

Making cyclic permutations ini, j,kwe get two more

equalities :

1212 ki

jk 1i jk 2i ji 1 j ji 2k

jgpp g pp g p p g p p g pu

and

ij 12 12

ki 1 j ki kj 1k kj 2

kgpp g pp g j pp g pp g i pu

The operationyields :

kj ij ik 12

ij 1k ij 2k

ij k

2

ij k

1gp g gpp2 p gp2 gpuu u

2g p

This kj ij ik

ij k i

jkgp g gpp2 gpuu u

.

Multiplying the above equation bykmgand summing the

resulting equations fork1 , 2we getmunotes.in

## Page 142

142

kj ij ik km km

ij k

kk ij k

km

ij k k

kkgp g gp 1gp p g p gp2u u upg g pg p

ij em

m

ijp

p

This gives us the desired formula :

ij ki ik mk m

ij

k ij kg gp gp 1pg p p2u u u ............. (*)

Definition 4 :k

ijp1 i j k 2are called the Christoffel Symbols

of the surface M at the point p.

We thus get the function :k

ij:M.

Their defining property being : 2

k

i ij

k1jk

.

Now fro any 22ii ji21 j1vv p . X X

, we have

ivi j j ip

ij

j

ij i jj

ij i i

j k k

ik i i j k

ij k i j ii

k k

ii j i j k

ki i j iXv X

Xvp p v X p pu

X Xvp p vp p puu

Xvp v X p ppu

Thus k k

vi i j i j k

ki i j iXXv p v X p p pu .

The derivation (*) above gives a set of handy formulae to

calculate the Christoffel symbols. In particular, applying them to the

surfaces M which are graphs of functions f: U, we can obtainmunotes.in

## Page 143

143

these functionskij, for example, the formulae can be applied to

obtainkijon a hemisphere222 2123xxxa, or on a surface of

revolution such as22312xxxandso on. This is left as exercises for

the reader.

Also, above we were considering the covariant different

differentiatievXof X at a point. The concept generatizes

immediately : Given a pair of smooth vector fields say X, Y on M,

differentiate one of them say X with respect to the other, obtaining a

new vector field Z on M given byYpZp X. It can be

verified that the resulting Z is a smooth vector field. We denote Z byYXit is the covariant derivative of X with respect to Y.

9.5 PARALLEL TRA NSPORT

We new use covariant differentiation (the Christoffe symbolskij) to move tangent vectors along smooth curves on M the

movement preserving their tangentially, their length and the angle

between two of them.

To be more specific, letc:I Mbe a smooth curve,opc t, and 12

i2vv p v puu tangential to M at p. we want

totransfer v fromopc tto each pointctof the curve in such a

way that it is tangential to M atct, its length remaining unaltered.

This mode of transport of v then generates a vector field X along c

i.e. a map :cttI X tT M withXt vandoXt v.

We then say that the vector field X is obtained from v by parallel

transporting v along c. Such a vector field is obtained by solving a

pair of first order liner ODE (invo lving the Christoffe symbolskij.)

and using the vector v (which is to be parallel transported) as the

initial condition of the linear ODE.

Writing 12

12Xt X t c t X t c tuu , we get the

(unknown function12X, X : I . Now, we consider the initial

value problem :munotes.in

## Page 144

144

1 1

ij j 1 o 1

ij

2 2ij j 2 o 2ijdX tct X t 0 , X t vdt

dX tct ct X t 0 , X t vdt

............. (*)

(In above 1o 2 o

12vv c t v c tuu )

Note that this initial value problem (*) is equivalent to :

0CtXt o , Xt v ...................... ................................1*By Picards theorem, the above initial value problem (*) (or

equivalent version1*of it) has a unique solution :3X:I We say that the vectorCtXt T M is obtained from the

vector v by parallel transporting it toctalong c.

At this stage, we improve our notation slightly : Taking into

consideration the initial condtionoXt v,we writevXforX.

Thus eachoctvT M gives rise to the vector field3vX: I having the properties :

i)v ctXt T M for eachtI,

ii)

vctXt 0

iii)If v, w are inoctTMa, b in, thenav bw v wX aX bX

iv)For any v,octwT M, the associated vector fieldsvwX, Xsatisfy

the identityvwXt , X t v , wthat is, the parallel transport of

any two vectors v,octwT M, pressures the angle between them

(throughout the transport along c.)

To justify this last property, we have :

vw vwct

vwct

wvdXt , Xt Xt , X tdt

Xt , Xt0, X t X t 00

munotes.in

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Thereforevw v o w oXt , X t Xt, Xt v , w

This completes the verification of the claim that the parallel

transport preserves the inner product. In particular we have :

a)vXt vi.e. parallel transport preserves t he length of the

vectors and

b)Iftis the angle betweenvXtandwXtthen

vw

vw

vo wo

vo woXt , X tCos tXt XtXt, X tXt X t

oCos tfor all t and thereforeotti.e. parallel

transport of tangent vectors along a smooth curve preserves the

angle between them.

9.6 GEODESICS

Geoddesics are smoth curves on a surface which have parallel

tangent fields.

Definition 5 : A smooth curvec:I Msatisfying

Ctct O

is called a geodesic curve (or simply a geodesic)

Equivalently put, a smooth curvec:I Mthe second

derivative ct

of which is along the normal to the surface is a

geodesic.

Writing 12 1 2 2 2ct c t, c t u c t u c t we have

12

12

12ct c t, c tct ctct ctuu

and therefore we have

1

ij1i jCtij u1

2

ij2i j

ij u2ct c t ct c tc t ctct c tct ct c t

munotes.in

## Page 146

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Now

ctct O

yields.

1

ij1 ij

ijct c tct ct O

2

ij2i j

ijct c tct ct O

And then the existence and uniqueness theorem of solution of

the second order ODE with a prescribed initial conditions gives the

following result.

Theorem 1 : GivenpM, and thepvTMthere exists a unique

geodes ic curvep, vcc : I M(I being an open interval containing

0) having the following properties :

1)cis defined on the largest open interval I.

2)c0 pand c0 v

.

9.7 EXERCISES :

1)Let p, a, b be any vectors in3and let23F: be the map

given byF u,v p ua vbfor2u,v.

Prove :

i)2,Fgive rise to a parametrized surface if and only i fab 0.

ii)Puttingca b, prove that a3wis a point of the surface2MFif and only ifc; w p 0.

2)For each of the following surfaces obt ain the matrixijg, its

determinantijg det gthe inverse matrixijgand the unit

normal N :

a)F u,v R cos u cos v, R sinu,cos v, R sinv

b)F u,v u cos v;u sinv,bv

c)F u,v R r cosu cos v, R r cos u sin v,r sinu

R,r being constants.munotes.in

## Page 147

147

3)Calculatekijfor the surfaces M = Graph (f)2f: U being given by

a) 222uv

fu , v u , v2

b)32 3f u,v u 2uv 4uv v u,v 24)Let23F: be given by22Fu , v u , v , u v u , v .

Obtain

i)Expression forkijfor the surface2MFii)Derive equations for the geodesics on the above surface.

5)Obtain equations for the geodesics on the sphere (part of it)

parametrized by the usual longitude -lattitude anglesu,v:F u,v cos v cosu,cos v sinu,sin v

and prove that the great ci ucles are the geodesic curves on the

sphere.

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10

CURVATURE OF A REGULAR SURFACE

Unit Structure :

10.0 Introduction

10.1The Normal Curvature

10.2Principal Directions / Principal Curvatures :

10.3 The Riemannian Curvature Tensor

10.4 Locally Parametrized Smooth Surfaces

10.5 Exercises

10.0 I NTRODUCTIO N

We study now the main geometric feature of a regular surface

M, namely, its curvature. First, we introduce a number of scalar

quantities defined at each point p of M, namely.

i)the normal curvature of M along a tangential direction at p;

ii)the principal curvatures of M at p and

iii)the Gaussian and me an curvatures of M at a p.

And then we intro the Riemann curvature tensor which is a

biquadratic form on the tangent bundle of M. it is the carrier of

complete information about the curvature properties of the surface

M. Next, explaining the intrinsic / exterensic nature of geometric.

properties of M, we conclude the chapter by proving the important

result -the Theorema Egragium of C.F. Gauses -that the Gaussian

curvature function is an intrinsic property of a regular surface.

Throughout this chap ter, a regular surface is a subset M of3withF: U Mas its parametrization, its orientation being

specified by a given unti normal field3N:M .

10.1 THE NORMAL CURVATURE

Let p be a point of M and let v be a unit vector tangential to

M at p; it is to be treated as a direction vector (tangential to M) at p.munotes.in

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We choose a smooth unit speed curve :c: , Msatisfyingco pand co v

.

Assuming c0 o

we get the curvaturekpofcat p, which

given by : c0 k pn p

wherenpis the principal normal to c

atc0 p.

Now we have two unit vectors located at the point p, namely :

i)the principal normalnpofcat p and

ii)the unit normalNpto M at p.

In general, the two vectors are distinct.

We consider the decomposition of c0

into its components :

one along the normalNpand the other in the tangent planepTMof M :

c o c 0 tan c 0 normal

c 0 tan c 0 ,N p N pc 0 tan k p n p ,N p N p

This equality gives :kp np kp np (tangential)+kp np , Np Np.

Now, note the following :

tOt0tO

tO

pdc, Np c t, Nc tdt

ddct, Nct c0, Nctdt dt

d0c 0 , N c tdt

c0, L c0

II p c 0 ,c 0

munotes.in

## Page 150

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Where, of course,pp pL: T M T M is the shape operator

andpp II p : T M T M is the second fundamental form of

M; both at the point p.

Thus, the normal part of the curavaturekpdepends only on

the direction vc 0

ofcat p and not on the (whole) curve c: Ifc

andcare two curves on M withco p coand co v co,

thenkp(normal) =kp(normal). This is naturally so, because,

while passing through p in the direction v, the curve can wiggle on

the surface thus aff ecting the tangential component (in the surface

M) of its curvature but its normal bending being forced by the

bending of M in the direction v at p. As such it (the normal partkp(normal)) is attributed to the curvature property of M at p in the

direction vc 0

; we call it the normal curvature of M at p in the

direction v. We adapt the notationvkfor the normal curvature.

Above we have derived the equalityvkI I p v , v

This result is often called Musiner’s Theorem.

Consider the following simple cases :If M is a plane, then for anyc: , Mwe have co okco oandconsequ ently the normal part of it is zero

0k0for any unit vectorpvTM.Let M be a sphere of radius A>o and let p be a point of M.

Then for any unit vector v tangential to the sphere M at p, we

consider the great circlec: Mthrough p having tangent

vector v at p. Now, we know that OPnp Npaand

1kpa, consequently,v1kaLet M be the circular cylinder of radius a>o. We consider a

pointpMand a unit vector v tangential to the cylinder at

the point p. As usual,Npis the unit normal to the cylinder

at the point p. Thus, we have the two unit vectors, v andNpdetermining a planethrough the point p. Note that

the intersection Mis an ellipse E passing through p and

the given vector v is tangential to the el lipse at the point p.munotes.in

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Letbe the angle between the planeandNp. Clearly we

can take the ellipse E for the curvec: ,. Now note that

the curvature of E at the point p is coskpaand the angle

betweenNpand v isand consequently, the desired

normal curvaturevkof M at p in the direction v is given by :

2

vcoska.

We summarise the above discussion and formulate the

definition.

Let p be a point of a regular surface M and v, a unit vector

tangential to M at p. Choosing a smooth curvec: ,with

co p , co v

we consider its curvaturekpat p and the fractionkp v , Np. We find that it depends only on the bending property

of M at p in the direction v and not on the chosen curve :kp v , Np I Ip u , v . This leades us to the following

definition :

Definition 1 : GivenpvTMwithv1the numbervkp v , Np kis the normal curvature of M at p in the direction

v.

Here is another realization ofvk: We consider the plane P

through p containg the vector v andNp. It intersects the surface

M along a smooth curvec: , M. Obviously cpasses through

p and has unit tangent v at p.munotes.in

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Weconsider its curvaturekpand the associated quantity

kp v , Np I Ic 0 , c 0 u , v

giving us the normal

curvaturevkofMat p along v.

10.2 PRI NCIPAL DIRECTIO NS/PRINCIPAL

CURV ATURE S

Above we have obtained the expression :v

ppkI I u , vIL u , v L u , u

for the normal curvature of M at p along v; which in volves the shape

operatorpp pL: T M T M. We consider the eigenvalues and

eigen -vectors of it. RecallpLis self -adjoint and therefore its ei -

genvalues are real. We have the following two cases :pLhas a single (real) eigen -value sayand therefore,

pLI, I being the identity operator ofpTM. In this case,

every unit vectorpvTMis an eigen -vector ofppL: L v vpLhas two di stinct (real) eigen -values say,with.

Let u, v be the unit vectors inpTMcorresponding to the

eigen -values :pLu vandpLU v.

Inthe first case, that is whenpLhas a single eigenvaue, the

point p is said to be an umbitic point of M. For such an umbilic point

p of M, we have :

IfpvTMwithv1, thenpLu vand thereforemunotes.in

## Page 153

153v

pkI I u , vLv , Uv,U

U, U

This shows that the normal curvaturevkof M at an umbilic

point is the same in all directions at p.

Here are simple examples of umbili c points.

i)On a plane P in3, any pointpPis an umbilic point with

vk0for every unit vectorpvTP.

ii)Any point P on a sphere S of radius a > 0is an umb ilic point

withv1kafor every unit vectorpvTS.

iii)Let M be the surface of revolution generated by rotating the

parabola2zx , xabout the Z -axis. Then the pointp 0,0,0is n umbilic point. (In fact it is the only umbilic point

on the surface).

(Perhaps t he above claim is clear to the reader, but we advise

him / her to verify it mathematically in an exercise. )

In the other case, namely, whenpLhas two distinct

eigenvalues,with.let u, v be unit eigenvectors of,respectively (i.e.ppLu u , Lv v.) then as seen above we have

ukandvk. Moreoveruvand consequently, any unit vectorpwTPcan be expressed uniquenly in the form :wc o s us i n vwhereis the angle betw een u and w. Now, the

normal curvaturewkof M at p in the direction w is given by

wp

p

2

pp

2

pp

22

22

22kL w , wL cos u sin u , cos u sin vcos L u ,u sin cos L u,vcos sin L u ,u sin L u ,vcos u,u 0 0 sin v,v

cos u,u sin v,v

cos sin

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(In above the middle terms are zero each becausepLu , v=v,u 0becauseu, vand for the same reasonpLu , v 0).

Thus we get that the normal curvaturewkalong such aw cos u sin vis given by22

wk cos sin.

This formula forwkwhich expresswkas a linear combination

of the distinguished normal curvaturesuvk, kinvolving the angle,

is known as the Euler’s formula.

Now in view of Euler’s formula, it is clear that the eigen -

values,of the shape operatorpLare respectively the minimum

and maximum of the setwpk: w T M; w 1 .

Definition 2 : Suppose,pMis not an umbilic point. Then the

unit eigen -vectors u, v belonging to the minimum and maximum of

the normal curvatures,are called the principal curvature

directions of the surface M at the point p.

Definition 3 : Let,be the minimum and maximum values of the

normal curvature of M at p. Then the quantities;Kp

1Hp2are called respectively the Gaussian curvature and the mean

curvature of M at the point p.

Note that w hen p is not an umbilical point of M, then the

principal curvature directions u, v at p form an orthonormal basis ofpTM , I Pand the matrix of the shape operatorpLwith respect

to this orthonormal basisu,vis00and consequently we have

:

i)pK p det Land

ii) p1H p trace L2

We extend the above definition to an umbilic point also :

Now we havethe con stant valuewkfor all unit vectorsmunotes.in

## Page 155

155pwTM and we them have :22Kpand

1Hp2.

Thus, we have :

1)In case of a plane P in3, at any point p of P, we have :poand consequentlyKp 0 Hp.

ii)Let M be the sphere of radius a > O. Then for anypM, have

1aand therefore 21Kpaand 1Hpa.

iii)Let M be a circular cylinder of radius a > 0.

then at a point p of it, the principal directions are :

a)The line1Lthrough p, parallel to the axis of the cylinder and

b)The line2Ltangential to the cylinder at the point p and per

pendicular to1L.

The principal curvatures are0(the curvature of the line1L) and1a, the curvature of the cylinder) and therefore, we get :Kp 0and 1Hp2a.

iv)We consider upper half of the ellipsoide :

22 2

3

222xyzMx , y , z : 1 , z Oabc a, b, c being

constants withabc0. Let p be the pointp 0,0,c.

Note thatpTMis the plane through p which is parallel to the

XOY plane and the unit normal to M at p is the vector (0,0,1)

located at the point p.

Now recall, for each unit vectorpwTMwe consider the

planePwthrough p containingNpand w. The intersectionPw Nis the half part of an ellipse through the point p and the

curvature of t his are (of the ellipse) at p is the normal curvature of

the ellipsoide M at p in the direction w. in particular we consider the

unit vectorsu 1,0,0andv 0,1,0both located at the pointo,o,cofM.munotes.in

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ClearlyPu Mis the ellipse : 22

3

22xzx,0,z 1ac and

its curvature at the pointo,o,cis2ca.

SimilarlyPu Mis the ellipse : 22

3

22yzo,y,z 1bc

and its curvature at the point0,0,cis2cb.

Also note that the shape operatorpLhas eigen -vectors (1,0,0)

and (0,1,0) atp 0,0,c and the respective eigen -values2ca,2cb. Therefore, the vectorsu 1,0,0 v 0,1,0are the

principal directions of normal curvature and2ca,2cbare the

principal normal curvat ures of the ellipsoide M at the pointp 0,0,c. It now follows that the Gaussian and mean curvatures

are given by : 222cKpaband 22c1 1Hp2a b.

10.3 THE RIEMA NNIANCURVATURE TE NSOR

We introduce now the sophishicated curvature tensor on a

smooth, parametrized surface M. Being a smoothly verying field of

biquadratic forms on all the tangent spacespTMof M, i tencodes

all the curvature properties of the surface (and many more geo metric

properities of such a M. Naturaly it has very fine algetraic /

geometric / analytical features. A comparesnsive study of it therefore

leades one f ar beyond the scope of the syllabus; we cannot cover the

topic completely here. Instead, we introduce i t very briefly and

mention some of its properties and relate the tensor to the Gaussian

curvature of M. We then proceed to prove the grand “theorema

egregium ”of Gauss explaining the intrinsic nature of the geometry

of M.

To begin with recall the equat ions (already explained) :

a)2

ij ij

ij eFLNuu u

and

b)j

i

j ijNFLi , j , 2uu

where the functionsijL: M have the properties :munotes.in

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i)for eachj

ipM , Lpis the matrix of the shape operatorpp pL: T M T M with respect to the vector basis

12FFp, puu ofpTMand

ii)the functionsjii jL, Lare related as follows :jj k kii k i j i k jkkLg L , L g L .

Now, differentiating part (a) of ( *) we get :

32

ij ij

ij ijkki j k k k

m

ij m

ij k k m

mm km

ij m

ij k

m km

m

ij

ij k

kLFF F NLN Luu u u u uu u uFFLL Nuu u

L FNL Luu

u

mm

ij k

m m

ij

ij k

kFLLu

LLu

Similarly, we have :

m 3

kj mmkj i kj i mm ik j iFLL Fuu u u

kj

kj i

iLLL Nu

…………………… (***)

The subtraction (**) -(***) givesmunotes.in

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33

ki j ik j

mm

kj kj mm

kj k kj i

m ki

mm

ij k kj i

m

ij kj

ij ke kj ie

kiFF

uu u u uu

uu

FLL LLu

LLLL LL Nuu

Now we must have33

ki j ik jFFOuu u u uu and therefore

we get :mm

ij kj mm m m

ij k kj i ij k kj i

kiLL LL Ouu

along with

ij kj

ij ke kj ie

kiLLOuu .

We use the identity (***) written equivalently in the

following way

mm

ij kjmm m mij k kj i ij k kj ikiLL LLuu

.

Also reorganizing the indicesik,j ,mwer write :

ij mm ikijk ik mj ij mkm jk kRuu

all the indicesi, j,k, ,mtaking the values1,2.

Note that the functionsijkR: M satisfy :ijk ikjRR .

The collecti onijkR: 1i , j , k , 2are components of a

geometric object (related to M) called the curvature tensor of M.

We also introduce the functionsijkR: M 1 i , j , k , 2 bymijk ijkmRg m R.munotes.in

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Note that we can retrieveijkRfromijkRbymijk mijkmRg R .

This is indeed so, because the matricesijgandijgare the

inverses of each other.

Now the equality :

mmijk ik j ij kmRL L L L

multiplied bymgand then summed overm1 , 2gives :ijk ik jm ij kmRL L L L.

In particular, we have 21212 22 11 21RL L L ijdet LThus, for anypM, we have1212 ij

k

ik j

kR p det L pdet L p g p

j

i ij

ij

ijdet L p det g pdet L p det g pK p det g p Kpbeing, of course, the normal curvature of M at its point

p. Thus, we have obtaine d1212

ijRpKpdet g p………………… (G)

This is offen called Gauss’ formula for the normal curvature.

For the sake of convenience, we will refer to the Gauss

formula by the symbol (G).

Now, looking at the right hand side of (G) we notice that i t is

a complex expression involving the entriesijgof the first

fundamental form and their partial derivatives2ij ijkk igguu u. The

functionsijgare obtained by varying the parametrization maps12 3FF Fon the surface and all the partial derivatives too are

obtained by differentiating thei ijF, getc.munotes.in

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Consequently we infer that the Gaussian curvatureKpof M

at p is calculated by taki ng measurements on the surface and not

referring to the ambient space3.

On the other hand, there are geometric quantities pertaining to

M which involve the ambian space also : For example the unit

normal and its variation on the surface refer to the external space.

We call geometric quantities intrinsic to M if they are

obtained by taking measurements taken strictly on the surface M .

Thus, a geometric quantity is intrinsic if it is expressible in terms of

the first fundame ntal form of the surface.

Above we have explained the proof (!) of the following :

Theorema Egregium ofGauss : Gaussian curvature of a

surface is an intrinsic property of a surface.

(Here “Egregium” means “e xtraordinary”.)

And then let us note a property of surfaces which is not

“intrinsic”.

We consider the flat rectangleR x, y,0 : 0 x 1,0 y 2 in the XOY plane2.

We roll it up in the form of the circular cylinder :M x,cos y, sin y : 0 x 1,0 y 2

Note that we obtained M from R without crumpling the paper

(or without causing any kind of damage to the paper and

consequently any measurements taken on the surface either in its

rectangular form or in its cylindrical form are the same.munotes.in

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Geometrically, bot h the suffaces , R M have the same first

fundamental form.

But the mean curvature ofR0while that of the cylinder is

01 1022.

The above example shows that the mean curvature of a

surface is not an intrin sic property of surfaces. It dependes on the

way in which itis imbedded in the ambient space (i.e. the space3).

10.4 LOCALLY PARAMETRIZED SMOOTH SURFACES

In the preceeding part of this chapter, we considered smooth

surfaces M which were covered by single parametrizationsU,F :M F U. But we come across surfaces which are

parametrized only locally; such surfaces are over wheming in

mathematics. We introduce the concept here formally.

Let M be a non -empty su bset of3. We consider M give the

subspace topology of3.

By a smooth, local parametrization on M, we mean a pair

(U,F) consisting of an open subset U ofand a smooth map3F: U , the pair having the following properties :

i)FUis an open subset of M and theF: U F Uis a

homeomorphism.

ii)For eachqU, the Jacobean map23FJa : is injective

(equivalently put, it has rank 2)

A smooth atlas on M is a collectionDU , F : of

smooth local parametrizationsU, Fon M with the property :UFU : M.

A smooth, local ly parametrized surface is a set M on which is

specified a smooth atlas D. We indicate it by the notationM,D.

The collection D is called a smooth atlas of the surface and an

elementU, F Dis often called a coor dinate chart ofM,D.munotes.in

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Thus a (smooth) parametrized surface is a particular case of aM,Din which D has only one elementU,F. (We often speak ofM,Dbeing co vered by a single coordinate chart.) But, of course, a

set M may not be covered by a single coordinate chart. Moreover,

there are subsets of3which are so scattered in3that they do not

admit any smooth atlas.

We conclude this chapter by describing a smooth atlas on a

sphere of radius a > o and then generalizing this in the form of a

result which gives a large variety of locally parametrized surfaces :

Let22 22Mx , y , z : x y z a

We cons ider the open coverU, U, V, V, W, Wof M

where :

Ux , y , z M , z o , Ux , y , z M , z o

Vx , y , z M , y o , Vx , y , z M , y oW x, y,z M ,x o and W x, y,z M ,x o

Also let22 2Du , v , u v a ; it is an open subset of2.

Now defineF: D U, F: D U by

222 222Fu , v u , v , a u v Fu , v u , v , a u v andG: D V, G: D V by

222 222Gu , v u , a u v, u Gu , v u , a u v, u

and finally,H: D W, H: D W by

222 222Hu , v a u v , u , v , Hu , v a u v , u , v .

ThenU, F , V, G , W, H is a smooth atlas on the

sphere M.

Verification of this claim is left as an ex ercise for the reader.

Now, the following result.munotes.in

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Proposition 1 : Let W be an open subset of3and f: Wa

smooth function. For a, letMx , y , z W : f x , y , z a .

Suppose M sati sfies :fp o , o , ofor eachpM.

Then M carries a smooth atlas D.

We give a sketchy proof below :

Proof : Let123pp , p , p M .

Thenfp o. Assume without loss of general ity that

fpOx. Then by implicit function theorem there exists an open2pUwith12p, p Uand a smoothppg: U satisfyingpfx , y , g x , y afor allpx,y Uandp1 2 3gp p p.

Define3ppG: U by puttingppGx , y x , y , gx , y for

allpx,y U. ThenppU, Gis a local parametrization of M around

the point p. And thenppU, G : p M D=

is the desired smooth atlas on M.

10.5 EXERCISES :

1)Let M be the surface of revolution given byF b, r t cos ,r t sin t : t I ,O 2 for a given

r:I, Prove that the Gaussian curvature K and mean curvature

H fun ctions are given by

2

222

2rtKt ,

rt 1 rtrtrt 1 rt1Ht ,2rt 1 rt

2)Let S be the surface of reolution given bymunotes.in

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1F t, sint sin ,sint cos t,cos t log tan2t, 0, 0,22

Show that the surface has constant Gaussian curvatureK1.

3)Let M be the ellipsoid :

22

2yzMx , y , z : x 123 Prove that none of the points 12p 1,0,0 , p 0, 2 ,0 , 3p 0,0, 3 is an umbilic point.

4)Prove that (0,0,0) is the umbilic point of the surface22zx yand calculate the normal curvature of it at ( 0,0,0).

5)Find principal curvatures and principal directions of the

following surfaces at a point of them

i)a circular cylinder

ii)the saddle surfacezx y6)Let:I Mbe a smooth curve. Show that the norma l curvature

of M at a point ofin the direction(at that point) is given by :K k coswhere kis the curvature of(as a curve in3) andis the angle between the surface normal N and the

principal normal vector of the curve.

7)Find the normal, curvature of the surfacezf x , yat a point p

of it in the direction of the unit vector (a, b, c).

8)Let M be the hyperbolic paraboloid. 21zy2.Show that the

normal curvature of M at (0,0,0) along a unit vectorv cos ,sin ,0is :22

nk v cos sin cos 2munotes.in