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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
44
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
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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
munotes.in
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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,
munotes.in
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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
munotes.in
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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
munotes.in
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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
todD,N p N C tdtmunotes.in
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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
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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
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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
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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
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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
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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.
munotes.in
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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
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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
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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