Paper-III-Calculus-on-Manifolds-munotes

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Chapter 1
Multilinear Algebra
Unit Structure :
1.1O b j e c t i v e
1.2ktensor
1.3 Alternating Tensor
1.4W e d g eP r o d u c t
1.5B a s i sf o r ⇤k(V)
1.6V o l u m eE l e m e n to f V
1.7C h a p t e rE n dE x e r c i s e
1.1 Objectives
After going through this chapter you will be able to:
1. Define a multilinear function, ktensor, alternating tensor and
wedge product.
2. Learn algebraic properties of alternating tensor and wedge prod-
uct.
3. Identify basis and dimension of subspace of tensor.
4. Learn the concept of volume element.
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1.2ktensor
Multilinear Function: IfVis a vector space over R,w ew i l l
denote the kfold product V⇥V⇥...⇥VbyVk. A function T:Vk!R
is called multilinear if for each iwith 1 ikwe have
T(v1,v2,···,vi+v0
i..., v k)=T(v1,v2,···,vi,···,vk)+T(v1,v2,···,v0
i,···,vk),
T(v1,v2,···, avi,···,vk)=aT(v1,v2,···,vi,···,vk).
Example: Consider the function f:R3!Rdefined as, f(x, y, z )=
xyz.S h o wt h a t fis 3linear.
Solution: We begin by fixing xandzand treat fas a function of one
variable y.
Consider f(x,↵y1+y2,z)=x(↵y1+y2)z
=x(↵y1)z+x(y2)z
=↵xy1z+xy2z
=↵f(x, y 1,z)+f(x, y 2,z).
shows that fis linear in y.
Similarly we can show that fis linear in xandzvariables.
ktensor: A multilinear function T:Vk!Ris called a ktensor
onVand the set of all ktensors denoted by =k(V), becomes a vector
space over Rif for S, T2=k(V)a n d a2Rwe define
(S+T)(v1,v2,···,vi,···,vk)=S(v1,v2,···,vi,···,vk)+T(v1,v2,···,vi,···,vk),
(aS)(v1,v2,···,vi,···,vk)=aS(v1,v2,···,vi,···,vk).
Tensor Product: There is an operation connecting the various
spaces =k(V). If S2=k(V)a n d T2=l(V), we define the tensor
product S⌦T2=k+l(V)b y
S⌦T(v1,v2,···,vk,vk+1,···,vk+l)=S(v1,v2,···,vk)·T(vk+1,···,vk+l).
Note: The order of the factors SandTis crucial here since S⌦T
andT⌦Sare far from equal.
T⌦S(v1,v2,···,vl,vl+1,···,vl+k)=T(v1,v2,···,vl)·S(vl+1,···,vl+k).
Example: IfS1,S22=k(V),T2=l(V),U2=m(V)a n d a2R
then Show that
(1) ( S1+S2)⌦T=S1⌦T+S2⌦T,
(2) S⌦(T1+T2)=S⌦T1+S⌦T2,
(3) ( aS)⌦T=S⌦(aT)=a(S⌦T),
(4) ( S⌦T)⌦U=S⌦(T⌦U).
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Notes:
(1) Both ( S⌦T)⌦UandS⌦(T⌦U)a r eu s u a l l yd e n o t e ds i m p l y S⌦T⌦U.
(2) higher-order products T1⌦T2⌦,···⌦Trare defined similarly.
(3) The =1(V)i sj u s tt h ed u a ls p a c e V⇤.
Note: Any vector space has a corresponding dual vector space (or dual
space) consisting of all linear forms on. , together with the vector space
structure of pointwise addition and scalar multiplication by constants.
Theorem-01: Letv1,···,vnbe a basis for V, and let '1,'2,···'nbe
the dual basis, 'i(vj)=ij. Then the set of all kfold tensor products
'i1⌦'i2⌦···⌦'ik,1i1,···,ikn
is a basis for =k(V), which therefore has dimension nk.
Proof Note that
'i1⌦'i2⌦···⌦'ik(vj1,vj2,···,vjk)=i1,j1·i2,j2···ik,jk
=(
1i f j1=i1;···;jk=ik,
0o t h e r w i s e .
Step I: Claim: 'i1⌦'i2⌦···⌦'ikspan =k(V).
Ifw1,w2,···,wkarekvectors with wi=nP
j=1aijvjandTis in =k(V),
then
T(w1,w2,···,wk)=nX
j1,j2,···,jk=1a1,j1···· ak,jkT(vj1,vj2,···vjk)
and
'i1⌦'i2⌦···⌦'ik(w1,,w 2,···,wk)=a1,j1····ak,jk'i1⌦'i2⌦···⌦'ik(vj1,vj2,···vjk)
'i1⌦'i2⌦···⌦'ik(vj1,vj2,···vjk)=(
1i f j1=i1;···;jk=ik,
0o t h e r w i s e .
)'i1⌦'i2⌦···⌦'ik(w1,,w 2,···,wk)=a1,j1····ak,jkifj1=i1;···;jk=ik
This gives us
T(w1,w2,···,wk)=nX
i1,i2,···,ik=1T(vi1,vi2,···vik)·'i1⌦'i2⌦···⌦'ik(w1,,w 2,···,wk).
Thus T=nP
i1,i2,···,ik=1T(vi1,vi2,···vik)·'i1⌦'i2⌦···⌦'ik.
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Consequently the 'i1⌦'i2⌦···⌦'ikspan =k(V).
Step II: Claim: 'i1⌦'i2⌦···⌦'ikis linearly independent
Suppose now that there are numbers ai1,i2···iksuch that
nX
i1,i2···ikai1,i2···ik'i1⌦'i2⌦···⌦'ik=0.
Applying both sides of this equation to ( vj1,vj2,···vjk)
nX
i1,i2···ikai1,i2···ik'i1⌦'i2⌦···⌦'ik(vj1,vj2,···vjk)=0 .
This yields ai1,i2···ik=0.Thus the 'i1⌦'i2⌦···⌦'ikare lineraly
independent.
hence by step I and II, we conclude
'i1⌦'i2⌦···⌦'ik,1i1,···,ikn
is a basis for =k(V), which therefore has dimension nk.
Example: Determine which of the following are tensors on R4and
express those in terms of elementary tensors.
f(x, y, z )=3 x1y2z3x3y1z4
g(x, y, z )=2 x1x2z3+x3y1z4
Solution:
(a)fis a 3-tensor since it is linear with respect to each variable x,y,
z. (Verify)
If!1,!2,!3,!4is the dual basis of the standard basis e1,..., e4in
R4, then
f=3!1⌦!2⌦!3!3⌦!1⌦!4.
(b)gis not a tensor since gis not linear as
g(ax, y, z )=2 ax1ax2z3+ax3y1z4=2a2x1x2z3+ax3y1z46=ag(x, y, z ).
Example: Consider the following tensors on R4,
f(x, y, z )=2 x1y2z2-x2y3z1
g(x, y)=!2⌦!1-2!3⌦!1
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where {!1,!2,!3,!4}is the dual basis of the standard basis {e1,. . . ,
e4}forR4.W r i t e f⌦gas a linear combination of elementary 5-tensors.
Solution: (b) Since f=2!1⌦!2⌦!2-!2⌦!3⌦!1.
f⌦g
=( 2!1⌦!2⌦!2-!2⌦!3⌦!1)⌦(!2⌦!1-2!3⌦!1)
=2!1⌦!2⌦!2⌦!2⌦!1-4!1⌦!2⌦!2⌦!3⌦!1+!2⌦!3
⌦!1⌦!2⌦!1-2!2⌦!3⌦!1⌦!3⌦!1.
Dual Transformation: Iff:V!Wis a linear transformation,
a linear transformation
f⇤:=k(W)!=k(V)i sd e fi n e db y
f⇤T(v1,v2,···,vk)=T(f(v1),f(v2),···,f(vk))
forT2=k(W)a n d v1,v2,···,vk2V.
Examples:
(1) Show that f⇤(S⌦T)=f⇤S⌦f⇤T.
(2) Show that an inner product on Vto be a 2-tensor or hi2=2(Rn).
Definition: We define an inner product on Vto be a 2-tensor Tsuch
that
Tis symmetric, that is T(v,w)=T(w,v)f o r v,w2Vand
Tis positive-definite, that is T(u, v)>0i fv6=0 .
We distinguish h,ias the usual inner product on Rn.
Theorem-02: IfTis an inner product on V, there is a basis v1,v2,··
·,vnforVsuch that T(vi,vj)=ij.(Such a basis is called orthonormal
with respect to T.) Consequently there is an isomorphism f:Rn!V
such that T(f(x),f(y)) = hx, yiforx, y2Rn. In other words f⇤T=
h,i.
Proof Letw1,w2,···,wnbe any basis of V. Define
w0
1=w1,
w0
2=w2T(w0
1,w2)
T(w0
1,w0
1)·w0
1,
w0
3=w3T(w0
1,w3)
T(w0
1,w0
1)·w0
1T(w0
2,w3)
T(w0
2,w0
2)·w0
2,
etc.
It is easy to check that T(w0
i,w0
j)=0i f i6=jand
w0
i6=0s ot h a t T(w0
i,w0
i)>0.
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Now define vi=w0
ip
T(w0
i,w0
i).
The isomorphism fmay be defined by f(ei)=vi.
Now Consider f⇤T(ei,ej)=T(f(ei),f(ei)) = T(vi,vj)=ij=hei,eji.
1.3 Alternating Tensor
Alternating Tensor: Aktensor !2=k(V)i sc a l l e da l t e r n a t i n g
if
!(v1,v2,···,vi,···,vj,···,vk)=!(v1,v2,···,vj,···,vi,···,vk)8v1,v2,···,vk2V.
(In this equation viandvjare interchanged and all other v’s are left
fixed.) The set of all alternating ktensors is clearly a subspace ⇤k(V)
of=k(V).
Note: Aktensor !2=k(V)i sc a l l e ds y m m e t r i ci f
!(v1,v2,···,vi,···,vj,···,vk)=!(v1,v2,···,vj,···,vi,···,vk)8v1,v2,···,vk2V.
Definition: IfT2=k(V), we define Alt( T)b y
Alt(T)(v1,v2,···,vk)=1
k!X
2Sksgn·T(v(1),v(2),···,v(k)),
where Skis the set of all permutations of the numbers 1 to k.
Note: Recall that the sign of a permutation denoted sgn ,i s
+1 if is even and 1i sis odd.
Theorem-03
(1) If T2=k(V), then Alt( T)2⇤k(V).
(2) If !2⇤k(V), then Alt( !)=!.
(3) If T2=k(V), then Alt(Alt( T)) = Alt( T).
Proof (1) Let ( i, j)b et h ep e r m u t a t i o nt h a ti n t e r c h a n g e s iandjand
leaves all other numbers fixed. If 2Sk, let 0=·(i, j). Then
Alt(T)(v1,v2,···,vj,···,vi,···,vk)
=1
k!X
2Sksgn·T(v(1),v(2),···,v(j),···,v(i),···,v(k)),
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=1
k!X
2Sksgn·T(v0(1),v0(2),···,v0(i),···,v0(j),···,v0(k)),
=1
k!X
02Sksgn0·T(v0(1),v0(2),···,v0(k)),
=Alt(T)(v1,v2,···,vk),
(2) If !2⇤k(V)a n d =(i, j), then
!(v(1),v(2),···,v(k))=s g n ·!(v1,v2,···,vk).
Since every is a product of permutations of the form ( i, j), this equa-
tion holds for all . Therefore
Alt!(v1,v2,···,vk)=1
k!X
2Sksgn·!(v(1),v(2),···,v(k))
=1
k!X
2Sksgn·sgn·!(v1,v2,···,vk)
=!(v1,v2,···,vk).
(3) follows immediately from (1) and (2).(Exercise)
1.4 Wedge product
Wedge product: If!2⇤k(V)a n d ⌘2⇤l(V), then !⌦⌘is
usually not in ⇤k+l(V). We will therefore define a new product, the
wedge product !^⌘2⇤k+l(V)b y
!^⌘=(k+l)!
k!l!Alt(!⌦⌘).
Example: Show that
(1) ( !1+!2)^⌘=!1^⌘+!2^⌘,
(2) !^(⌘1+⌘2)=!^⌘1+!^⌘2,
(3) a!^⌘=!^a⌘=a(!^⌘),
(4) !^⌘=(1)kl⌘^!,
(5) f⇤(!^⌘)=f⇤(!)^f⇤(⌘),
(6) ( !^⌘)^✓=!^(⌘^✓).
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Theorem-04
(1) If S2=k(V)a n d T2=l(V) and Alt( S)=0 ,t h e n
Alt(S⌦T) = Alt( T⌦S)=0 .
(2) Alt(Alt( !⌦⌘)⌦✓) = Alt( !⌦⌘⌦✓) = Alt( !⌦Alt(⌘⌦✓)).
(3) If !2⇤k(V),⌘2⇤l(V)a n d ✓2⇤m(V), then
(!^⌘)^✓=!^(⌘^✓)=(k+l+m)!
k!l!m!Alt(!⌦⌘⌦✓).
Proof: (1)Step I: Claim: Alt(S⌦T)=0
Alt(S⌦T)(v1,v2,···,vk+l)=1
(k+l)!X
2Sk+lsgn·(S⌦T)(v(1),v(2),···,v(k+l)).
(k+l)!Alt( S⌦T)(v1,v2,···,vk+l)
=X
2Sk+lsgn·S(v(1),v(2),···,v(k))·T(v(k+1),v(k+2),···,v(k+l)).
(1)
Case I: IfG⇢Sk+lconsists of all whcih leave k+1,k+2,···,k+l
fixed, then
X
2Gsgn·S(v(1),v(2),···,v(k))·T(v(k+1),v(k+2),···,v(k+l))
=X
02Sksgn0·S(v0(1),v0(2),···,v0(k))·T(v(k+1),v(k+2),···,v(k+l))
=0. (Since Alt( S)=0 )
Hence by equation (1), Alt( S⌦T)=0
Case II: Suppose 0/2G.
LetG·0={·0:2G}and
letv0(1),v0(2),···,v0(k+l)=w1,w2·· ·,wk+l. Then
X
2G·0sgn·S(v(1),v(2),···,v(k))·T(v(k+1),v(k+2),···,v(k+l))
=2
4sgn0·X
02Gsgn0·S(w0(1),w0(2),···,w0(k))·3
5·T(wk+1,wk+2,···,wk+l)
=0. (Since Alt( S)=0 )
Hence by equation (1), Alt( S⌦T)=0
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Notice that G\G·0=.
In fact, if 2G\G·0, then =0·0for some 02Gand
0=·(0)12G, a contradiction.
We can then continue in this way, breaking Sk+lup into disjoint subsets;
the sum over each subset is 0, so that the sum over Sk+lis 0. Hence
Alt(S⌦T)=0 .
Step II: Claim: Alt(T⌦S) = 0 Show similarly as step I. Combining
step I and II, we obtain
Alt(S⌦T) = Alt( T⌦S)=0 .
(2)Step I: Claim: Alt(!⌦⌘⌦✓) = Alt( !⌦Alt(⌘⌦✓))
Consider Alt(Alt( ⌘⌦✓)⌘⌦✓) = Alt {Alt(⌘⌦✓)}Alt(⌘⌦✓).
By theorem (3(III)), we have Alt {Alt(⌘⌦✓)}= Alt( ⌘⌦✓),
hence we have
Alt(Alt( ⌘⌦✓)⌘⌦✓) = Alt( ⌘⌦✓)Alt(⌘⌦✓)=0 .
Hence by (1) we have
Alt(!⌦[Alt( ⌘⌦✓)⌘⌦✓]) = 0
Alt(!⌦Alt(⌘⌦✓))Alt(!⌦⌘⌦✓)=0
Alt(!⌦Alt(⌘⌦✓)) = Alt(!⌦⌘⌦✓)
Step II: Claim: Alt(Alt( !⌦⌘)⌦✓) = Alt( !⌦⌘⌦✓)
Similarly as per step I.
(3)Step I: Claim: (!^⌘)^✓=(k+l+m)!
k!l!m!Alt(!⌦⌘⌦✓).
By definition of wedge product have
(!^⌘)^✓=(k+l+m)!
(k+l)!m!Alt(( !^⌘)⌦✓)
again applying definition of wedge product have
(!^⌘)^✓=(k+l+m)!
(k+l)!m!Alt{((k+l)!
k!l!Alt(!⌦⌘))⌦✓}
(!^⌘)^✓=(k+l+m)!
(k+l)!m!(k+l)!
k!l!Alt{Alt(!⌦⌘)⌦✓}
By 2 above
(!^⌘)^✓=(k+l+m)!
k!l!m!Alt(!⌦⌘⌦✓)
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Step II: Claim: !^(⌘^✓)=(k+l+m)!
k!l!m!Alt(!⌦⌘⌦✓).
Similarly as per step I.
Note: (1)!^(⌘^✓)=( !^⌘)^✓=!^⌘^✓
and higher-order products !1^!2^···^!rare defined similarly.
(2) If an alternating tensor !and⌘are of odd order then !^⌘=⌘^!
(3) If an alternating tensor !is of odd order then !^!=0
Example: Consider the following tensors on R5
f(x, y, z )=3 x2y2z1-x1y5z4
g(x)=2 x1+x3
(a) Write Alt fas a linear combination of elementary alternating ten-
sors.
(b) Write (Alt f)^gas a linear combination of elementary alternating
tensors.
Solution:
(a) Recall that if I=(i1,...,ik)i sa nm u l t i - i n d e xa n d
!i1^···^!ik=!I:=k!Alt(!i1⌦···⌦!ik)( 1 . 1 )
Hence write fas a linear combination of elementary tensors,
f=3!2⌦!2⌦!1!1⌦!5⌦!4
Then by equation (2),
Altf= 3Alt( !2⌦!2⌦!1)Alt(!1⌦!5⌦!4)
=3
3!!2^!2^!11
3!!1^!5^!4
=1
3!!1^!5^!4
=1
3!!1^!4^!5
(b) Since g=2!1+!3so that
(Alt f)^g=1
3!!1^!4^!5^(2!1+!3)
=1
3!!1^!4^!5^!3
=-1
3!!1^!4^!3^!5
=1
3!!1^!3^!4^!5
Example 2: LetX1,X2,. . . , Xk2V and let '1,. . . , 'k2V⇤.
Show that '1^...^'k(X1,X2,. . ., Xk)=d e t [ 'i(Xj)]
Solution:
By definition,
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'1^...^'k(X1,X2,. . ., Xk)=(1+···+1)!
1!···1!Alt('1⌦···⌦'k)(X1,X2,
... ,Xk)
=k! Alt( '1⌦···⌦'k)(X1,X2,. . ., Xk)
=k!
k!P
2Sk(sign )'1(X(1))'2(X(2))···'k(X(k))
=d e t2
66664'1(X1)...'1(Xk)
.
.
.
'k(X1)...'k(Xk)3
77775
1.5 Basis for ⇤k(V)
Theorem-05: The set of all
'i1^'i2^···^'ik,1i1,i2,···,ikn
is a basis for ⇤k(V), which therefore has dimension
✓n
k◆
=n!
k!(nk)!
Proof: Step I: Claim: 'i1^'i2^···^'ik,1i1,i2,···,ikn
spans ⇤k(V).
Letv1,v2,···vnbe a basis for Vand let '1,'2,···'nbe the dual basis.
If!2⇤k(V)⇢=k(V), then we can write
!=X
i1,i2,···ikai1,i2,···ik'i1⌦'i2⌦···⌦'ik.
Thus by theorem 3( II), we have
!= Alt( !)=X
i1,i2,···ikai1,i2,···ikAlt('i1⌦'i2⌦···⌦'ik).
Since by definition of wedge product, each Alt( 'i1⌦'i2⌦···⌦'ik)i sa
constant times one of the ( 'i1^'i2^···^'ik), these elements span ⇤k(V).
Step II: Claim: 'i1^'i2^···^'ik,1i1,i2,···,iknis linearly
independent.
Linear independence is proved as in Theorem-01.
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Step III: Claim: Dimension of ⇤k(V)i sn
k
=n!
k!(nk)!.
As⇤k(V)i ss e to fa l la l t e r n a t i n g ktensors which is subspace of =k(V),
clearly Dimension of ⇤k(V)i sn
k
=n!
k!(nk)!
Note: IfVhas dimension n, it follows from Theorem-05 that ⇤n(V)
has dimension 1.
Example: Let V be a vector space of dimension n=3 . T h es p a c eo f
alternating 2-tensors ⇤2(V⇤)h a st h ed i m e n s i o n
dim⇤2(V⇤)=3
2
=3!
2!(32)!=3
Theorem-06: Letv1,v2,···vnbe a basis for Vand let !2⇤n(V).
If!i=nP
j=1aijvjarenvectors in Vthen
!(w1,w2,···,wn)=d e t( aij)·!(v1,v2,···,vn).
Proof: Define ⌘2=n(Rn)b y
⌘((a11,a12,···,a1n),(a21,a22,···,a2n),···,(an1,an2,···,ann)
=!Pa1jvj,Pa2jvj,···,Panjvj
As!2⇤n(V)c l e a r l y ⌘2⇤n(Rn)
so⌘=·det (aij)f o rs o m e 2Rand
=⌘(e1,e2,···,en)=!(v1,v2,···,vn).
!(w1,w2,···,wn)=d e t( aij)·!(v1,v2,···,vn).
1.6 Volume Element of V
Orientation: Theorem-06 shows that a non zero !2⇤n(V)s p l i t s
the bases of Vinto two disjoint groups, those with !(v1,v2,···,vn)>0
and those for which !(v1,v2,···,vn)<0; ifv1,v2,···,vnandw1,w2,···,wn
are two bases and A=(aij)i sd e fi n e db y wi=Paijvjthen v1,v2,···,vn
andw1,w2,···,wnare in the same group if and only if det A>0.
This criterion is independent of !and can always be used to divide
the bases of Vinto two disjoint groups. Either of these two groups is
called an orientation for V. The orientation to which a basis v1,v2,···,vn
belongs is denoted by [ v1,v2,···,vn] and the other orientation is denoted
[v1,v2,···,vn].
Note: InRnwe define the usual orientation as [ e1,e2,···,en].
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Volume Element: The fact that dim ⇤n(Rn)=1i so b v i o u ss i n c e
det is often defined as the unique element !2⇤n(Rn)s u c ht h a t
!(e1,e2,···,en)=1 . B yt h e o r e m6
!(w1,w2,···,wn) = det ( aij)·!(e1,e2,···,en).
!(w1,w2,···,wn)=d e t( aij)
Suppose that an inner product TforVis given. If v1,v2,···,vnand
w1,w2,···,wnare two bases which are orthonormal with respect to T,
and the matrix A=(aij)i sd e fi n e db y wi=nP
j=1aijvj, then
ij=T(wi,wj)
=T(nX
k=1aikvk,nX
l=1ailvl)
=nX
k,l=1aikajlT(vk,vl)
=nX
k,l=1aikajlkl
=nX
k=1aikajk.
In other words, if ATdenotes the transpose of the matirix A, then we
have A·AT=I, so det( A)=±1.
It follows from Theorem-06 that if !2⇤n(V)s a t i s fi e s !(v1,v2,···,vn)=
±1, then !(w1,w2,···,wn)=±1. If an orientation µforVhas also
been given, it follows that there is a unique !2⇤n(V)s u c ht h a t
!(v1,v2,···,vn)=1w h e n e v e r v1,v2,···,vnis an orthonormal basis
such that [ v1,v2,···,vn]=µ.
Note that det is the volume element of Rndetermined by the usual
inner product and usual orientation and that |det(v1,v2,···,vn)|is the
volume of the paralleopiped spanned by the line segments from 0 to
each of v1,v2,···,vn.
Volume Element of Rn:Ifv1,v2,···,vn12Rnand'is defined by
'(w) = det0
BBBBBBBB@v1
v2
·
·
·
vn1
w1
CCCCCCCCA,
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Then '2⇤1(V). Therefore there is a unique element z2Rnsuch
that
hw,zi='(w)=d e t0
BBBBBBBB@v1
v2
·
·
·
vn1
w1
CCCCCCCCA,
This zis the denoted v1⇥v2⇥···⇥vn1and called the cross product
ofv1,v2,···,vn1.
The following properties are immediate from the definition:
(1)v(1)⇥v(2)⇥···⇥v(n1)=s g n ·v1⇥v2⇥···⇥vn1,
(2)v1⇥v2⇥···⇥avi⇥···⇥vn1=a·(v1⇥v2⇥···⇥vn1),
(3)v1⇥v2⇥···⇥(vi+v0
i)⇥···⇥vn1=(v1⇥v2⇥···⇥vi⇥···⇥
vn1)+( v1⇥v2⇥···⇥v0
i⇥···⇥vn1).
1.7 Chapter End Exercise
1. Let T2=k(W)a n d S2=l(W). Show that f⇤(S⌦T)=f⇤S⌦
f⇤Twhere f⇤is a dual transformation of a linear transformation
f:V!W.
2. Let Vbe a vector space of dimension 5. Find the dimension of
the space of alternating 3 tensor ⇤3(V). Justify your answer.
3. Let !2⇤2(V),⌘2⇤3(V)a n d ✓2⇤4(V). Find the wedge
product ( !^⌘)^✓in terms of alternating tensor of tensor product
of!,⌘and✓.
4. Let S2⇤k(V)a n d T2⇤l(V) and Alt( T)=0t h e nc o m p u t e
T^S.
5. Let Vbe a vector space of dimension 3. Find the dimension of
the space of alternating 2 tensor ⇤2(V). Justify your answer.
6. Let !2⇤1(V),⌘2⇤2(V)a n d ✓2⇤3(V). Find the wedge
product ( !^⌘)^✓in terms of alternating tensor of tensor product
of!,⌘and✓.
7. Prove or disprove: An inner product on vector space Vto be a
2-tensor.
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8. If T2=k(V), then show that Alt(Alt( T)) = Alt( T).
9. If !2⇤k(V),⌘2⇤l(V)a n d ✓2⇤m(V), then show that
(!^⌘)^✓=(k+l+m)!
k!l!m!Alt(!⌦⌘⌦✓).
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Chapter 2
Di↵erential Forms
Unit Structure :
2.1O b j e c t i v e
2.2B a s i cP r e l i m i n a r i e s
2.3 Fields and Forms
2.4D i↵erential Forms
2.5P u l l b a c kF o r m s
2.6C h a p t e rE n dE x e r c i s e
2.1 Objectives
After going through this chapter you will be able to:
1. Learn the concept of tangent space.
2. Define Di ↵erential Forms and Pullback Forms.
3. Learn properties of Pullback Forms.
2.2 Basic Preliminaries
1. The Del operator:
r=@
@xˆi+@
@yˆj+@
@zˆk=✓@
@x,@
@y,@
@z◆
.
2. Gradient:
Suppose fis a function. rfis the gradient of f, sometimes denoted
grad f.
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grad f=rf=@f
@xˆi+@f
@yˆj+@f
@zˆk.
Example: Compute the gradient of f(x, y, z )=xyey2z
Solution: rf=@f
@xˆi+@f
@yˆj+@f
@zˆk=yey2zˆi+(xey2z+2 x y2ey2z)ˆj+ˆk(xy3ey2z).
3. Directional derivative
Definition: The directional derivative of fin the direction ~u, denoted
byD~uf, is defined to be,
D~uf=rf·~u
|~u|
Example: What is the directional derivative of f(x, y)=x2+xy,i n
the direction of ~i+2~jat the point (1, 1)?
Solution: Now we first find rf.
rf=(@
@x,@
@y)= ( 2 x+y,x)
=(3,1)
Let~u=~i+2~j
|~u|=p
12+22=p
5.
D~uf=rf·~u
|~u|=(3,1)·(1,2)p
5=p
5.
•Properties of the gradient deduced from the formula of Directional
derivatives
D~uf=rf·~u
|~u|=|rf||~u|cos(✓)
|~u|=|rf|cos(✓)
1. If ✓=0 ,i . e . ~upoints in the same direction as rf, then D~ufis
maximum. Therefore we may conclude that,
(i)rfpoints in the steepest direction.
(ii) The magnitude of rfgives the slope in the steepest direction.
2. At any point P,rf(P)i sp e r p e n d i u l a rt ol e v e ls e tt h r o u g ht h a t
point.
4. Divergence:
Definition: The Divergence is given by,
div~F=r·~F
where ~Fshould be vector field.
Example. Compute the divergence of ~F=(x2+y)ˆi+(y2-z)ˆj+
(z2+x)ˆk
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Solution: div~F=@
@xˆi+@
@yˆj+@
@zˆk·((x2+y)ˆi+(y2-z)ˆj+(z2+x)ˆk)
=2x+2y+2z.
5. Curl:
Definition: The curl is given by,
curl ~F=r⇥~F
More specifically, suppose ~F=(F1,F2,F3). Then
r⇥~F=ˆiˆjˆk
@
@x@
@y@
@z
F1F2F3
The cross product of two vectors is a vector, so curl takes a vector field
to another vector field.
Example. Compute the curl of ~F=(x2+y)ˆi+(y2-z)ˆj+(z2+x)ˆk
Solution: curl ~F=ˆiˆjˆk
@
@x@
@y@
@z
F1F2F3
=ˆi ˆj ˆk
@
@x@
@y@
@z
x2+yy2zz2+x
=ˆi-ˆj+ˆk=( 1 ,- 1 ,1 ) .
Example. Show that curl grad f=~0
Solution: curl grad f=r⇥r f
=ˆiˆjˆk
@
@x@
@y@
@z
@f
@x@f
@y@f
@z
=ˆiˆjˆk
@
@x@
@y@
@z
@
@x@
@y@
@z(f).
But the determinant of a matrix with two equal rows is 0, so the result
is~0.
Example. div(curl ~F)=0
Solution: div(curl ~F)=r·(r⇥f)
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=r·ˆiˆjˆk
@
@x@
@y@
@z
F1F2F3
=@
@x@
@y@
@z
@
@x@
@y@
@z
F1F2F3
=0 .
Example. Find Curl( rf)a n dD i v ( rf)
Solution: Curl( rf)=r⇥r f
=(fyz-fzy)ˆi+(fzx-fxz)ˆj+(fxy-fyx)ˆk
=0
Div(rf)=r·rf
=(@
@x,@
@y,@
@z)·(@f
@x,@f
@y,@f
@z)
=@2f
@x2+@2f
@y2+@2f
@z2.
2.3 Fields and Forms
Ifp2Rn, the set of all pairs ( p, v), for v2Rn, is denoted Rn
p,a n d
called the tangent space of Rnatp. This set is made into a vector space
in the most obvious way, by defining
(p, v)+( p, w)=( p, v+w),
a·(p, v)=( p, av).
Vector Field: A vector field is a function Fsuch that F(p)2Rn
p, for
each p2Rn. For each pthere are numbers F1(p),F2(p),···,Fn(p)s u c h
that
F(p)=F1(p)·(e1)p+F2(p)·(e2)p+···,Fn(p)·(en)p.
We thus obtain ncomponent functions Fi:Rn!R.
Note: (1) The vector field Fis called continuous, di ↵erentiable etc., if
the functions Fiare.
(2) A vector field defined only on an open subset of Rn.
(3) Operations on vectors yield operations on vector field when applied
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at each point separately. For example if FandGare vector fields and
fis a function, we define
(F+G)(p)=F(p)+G(p),
hF,Gi(p)=hF(p),G(p)i,
(f·F)(p)=f(p)F(p).
IfF1,F2,···,Fn1are vector fields on Rn, then we can similarly define
(F1⇥F2⇥···⇥Fn1)(p)=F1(p)⇥F2(p)⇥···⇥Fn1(p).
Gradient, Divergence and Curl: Introduce the formal symbol-
ism
O=nX
i=1Di·ei.
The gradient of a scalar field fis defined as Grad f=Of.
The divergence of a vector field Fis defined as Div F=nP
i=1DiFi.
we can write, symbolically, Div F=hO,Fi.
The curl of a vector field Fis defined as Curl F=O⇥F.
Ifn=3w ew r i t e ,i nc o n f o r m i t yw i t ht h i ss y m b o l i s m ,
(O⇥F)(p)=( D2F3D3F2)(e1)p+(D3F1D1F3)(e2)p+(D1F2D2F1)(e3)p.
2.4 Di ↵erential Forms
Di↵erential Forms or kForms: A function !with !(p)2
⇤k(Rn
p)i sc a l l e da kform on Rn, or simply a di ↵erential form where
⇤k(Rn
p)b et h es e to fa l la l t e r n a t i n g ktensors which is a subspace of
=k(Rn
p)a n d Rn
ptangent space of Rnatp.
If'1(p),'2(p),···,'n(p) is the dual basis to ( e1)p,(e2)p,···,(en)p, then
!(p)=X
i1for certain functions !i1,!i2,···,!ik.
Note:
1. The form !is continuous, di ↵erentiable, etc. if these functions
!i1,!i2,···,!ikare continuous, di ↵erentiable, etc.
2. Let !and⌘be two kforms then the sum ( !+⌘)(p)=!(p)+⌘(p).
3. The product ( f·!)(p)=f·!(p)a n d f·!is also written as f^!.
4. Let !bekform and and ⌘belforms then wedge product !^⌘
is (k+l)form given by ( !^⌘)(p)=!(p)^⌘(p).
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5. A arbitrary real valued function fis considered to be a 0 form.
Di↵erential Forms or kForms for a function f:Rn!R
:Iff:Rn!Ris di↵erentiable, then Df(p)2⇤1(Rn)i . e . Df(p)i s
1form. A 1 form df, defined by
df(p)(vp)=Df(p)(v)( 2 . 1 )
Let us consider in particular the 1 forms d⇡i.
Letxidenote the function ⇡i.
Since
dxi(p)(vp)=d⇡i(p)(vp)=D⇡i(p)(v)=vi(2.2)
We see that dx1(p),d x2(p),···,d xn(p)i sj u s tt h ed u a lb a s i st o( e1)p,(e2)p,··
·,(en)p.
Thus every kform !can be written
!=X
i1Note: Thus !=P
i1!i1dxi1is 1form.
!=P
i1!=P
i1Theorem-07: Iff:Rn!Ris di↵erentiable , then
df=D1f·dx1+D2f·dx2+···Dnf·dxn.
In classical notation, df=@f
@x1dx1+@f
@x2x2+···+@f
@xndxn
Proof:
df(p)(vp)=Df(p)(vp)=nX
i=1Dif(p)·viby equation 1
df(p)(vp)=nX
i=1Dif(p)·dxi(p)(vp)b y e q u a t i o n 2
This gives
df=D1f·dx1+D2f·dx2+···Dnf·dxn(2.4)
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2.5 Pullback Forms
Di↵erential Forms or kForms for a function f:Rn!Rm:
Pullback Forms : Consider a di ↵erentiable function f:Rn!Rm
we have a linear transformation Df(p):Rn!Rm. Another minor
modification therfore produces a linear transformation f⇤:Rn
p!Rm
f(p)
defined by
f⇤(vp)=( Df(p)(v))f(p) (2.5)
This linear transformation induces a linear transformation f⇤:⇤k(Rm
f(p))!
⇤k(Rn
p). If !is akform on Rmwe can therefore define a kform f⇤!
onRnby
(f⇤!)(p)=f⇤(!(f(p))) (2.6)
i.e. if v1,v2,···,vk2Rn
pthen
f⇤!(p)(v1,v2,···,vk)=!(f(p)(f⇤(v1),···,f⇤(vk)) (2.7)
Thus if !is akform on Rm, it can be pullback to Rnbyf⇤!then
f⇤!is an alternating ktensor on Rn
pand hence f⇤!iskform on Rn
and is known as pullback form of !byf
Theorem-08: Iff:Rn!Rmis di↵erentiable, then
(1) f⇤(dxi)=nP
j=1Djfi·dxj=nP
j=1@fi
@xjdxj.
(2) f⇤(!1+!2)=f⇤(!1)+f⇤(!2).
(3) f⇤(g·!)=( gf)·f⇤!.
(4) f⇤(!^⌘)=f⇤!^f⇤⌘.
Proof (1)
f⇤(dxi)(p)(vp)=( dxi)(f(p))(f⇤vp)b y e q u a t i o n 7
=(dxi)(f(p))(Df(p)(v))f(p)by equation 5
=(dxi)(f(p))"nX
j=1vj·Djf1(p),nX
j=1vj·Djf2(p),···,nX
j=1vj·Djfm(p)#
f(p)
=nX
j=1vj·Djfi(p)
=nX
j=1Djfi(p)·dxj(p)(vp)b y e q u a t i o n 2
Thus
f⇤(dxi)=nX
j=1Djfi·dxj=nX
j=1@fi
@xjdxj(2.8)
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(2) Let !1and!2bekforms. Consider
f⇤(!1+!2)(p)(v1,v2,···,vk)=( !1+!2)(f(p))(f⇤(v1),···,f⇤(vk)) by equation 7
=!1(f(p))(f⇤(v1),···,f⇤(vk)) + !2(f(p))(f⇤(v1),···,f⇤(vk))
=f⇤(!1)+f⇤(!2)
(3) Consider
f⇤(g·!)(p)(v1,v2,···,vk)=( g·!)(f(p))(f⇤(v1),···,f⇤(vk)) by equation 7
=,···,f⇤(vk)) since g is 0-form
=,···,f⇤(vk))
=(gf)·f⇤!
(4) Let !bekform and and ⌘belforms then wedge product !^⌘
is (k+l)form given by ( !^⌘)(p)=!(p)^⌘(p).
Consider
f⇤(!^⌘)(p)(v1,···,vk,vk+1,···,vk+l)
=(!^⌘)(f(p))(f⇤(v1),···,f⇤(vk),f⇤(vk+1),···,f⇤(vk+l)) by equation 7
=!(f(p))(f⇤(v1),···,f⇤(vk))^⌘(f(p))(f⇤(vk+1),···,f⇤(vk+l))
=f⇤!^f⇤⌘
Theorem-09: Iff:Rn!Rnis di↵erentiable, then
f⇤(hdx1^dx2^···^dxn)=( hf)(det f0)(dx1^dx2^···dxn).
Proof: By theorm 8( III), we can write,
f⇤(hdx1^dx2^···^dxn)=( hf)f⇤(dx1^dx2^···dxn).
then it su ces to show that
f⇤(dx1^dx2^···^dxn)=( d e t f0)dx1^dx2^···dxn.
Letp2Rnand let A=(aij)b et h em a t r i xo f f0(p). For convenience
we shall omit ” p”. Then
f⇤(dx1^dx2^···^dxn)(e1,e2,···,en)
=dx1^dx2^···^dxn(f⇤e1,f⇤e2,···,f⇤en)b y e q u a t i o n 7
=dx1^dx2^···^dxn(Df1(ei),Df 2(ei),···,Df n(ei)) by equation 5
=dx1^dx2^···^dxn nX
i=1ai1ei,nX
i=1ai2ei,···,nX
i=1ainei!
=d e t ( aij)·dx1^dx2^···^dxn(e1,e2,···,en)b y t h e o r e m 6
=d e t ( f0)·dx1^dx2^···^dxn(e1,e2,···,en)
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Example 1: Let!=xydx +2zdyydz2⌦k(R3)a n d ↵:R2!
R3is defined as ↵(u, v)=( uv, u2,3u+v). Calculate ↵⇤!.
Solution: Instead of thinking of ↵as a map, think of it as a substition
of varibles:
x=uv, y =u2,z=3u+v
Then,
dx=@x
@udu+@x
@vdv=vdu+udvand similarly,
dy=2uduanddz=3du+dv
Consider,
!=xydx +2zdyydz=(uv)(u2)(vdu+udv)+2 ( 3 u+v)2udu-
u2(3du+dv)
=(u3v2+9u2+4uv)du+(u4v-u2)dv
We conclude that,
↵⇤!=↵⇤(xydx +2zdyydz)=( u3v2+9u2+4uv)du+(u4v-u2)
dv.
Example 2: Consider a map F:R3!R2given as,
F(x, y, z )=( x2+yz,exyz)
and 2 form !=uv3du^dvonR2. Then calculate F⇤!.
Solution: F⇤!=(x2+yz)e3xyzd(x2+yz)^dexyz
=(x2+yz)e3xyz(2xdx+zdy+ydz)^(yzexyzdx+xz exyzdy+xyexyzdz)
=(x2+yz)e4xyz(2x2zdx^dy+2x2ydx^dz+z2ydy^dx+xyzdy ^dz+
y2zdz^dx+xyz dz ^dy)
=(x2+yz)e4xyz((2x2z-yz2)dx^dy+( 2x2y-zy2)dx^dz).
2.6 Chapter End Exercise
1. In R3, let !=xydx +2zdyydzand↵:R2!R3be given by
↵(u, v)=( uv, u2,3u+v). Calculate ↵⇤(!).
2. If f:Rn!Ris di↵erentiable then show that df=@f
@x1dx1+
@f
@x2x2+···+@f
@xndxn
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Chapter 3
Exterior Derivatives
Unit Structure :
3.1O b j e c t i v e
3.2E x t e r i o rD e r i v a t i v e
3.3C l o s e da n dE x a c tF o r m s
3.4C h a p t e rE n dE x e r c i s e
3.1 Objectives
After going through this chapter you will be able to:
1. Define and calculate Exterior Derivative.
2. Learn properties of Exterior Derivative.
3. Identify closed and exact forms.
4. Learn the concept of Star Shaped Set.
3.2 Exterior Derivatives
The operator dwhich changes 0 forms into 1 forms. If
!=X
i1be a given kform, we define a ( k+1)form d!which is the di ↵erential
of!,b y
d!=X
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d!=X
i1,i2,···iknX
↵=1D↵(!i1,i2,···ik)·dx↵^dxi1^dxi2^···^dxik(3.1)
Theroem-10
(1) d(!+⌘)=d!+d⌘.
(2) If !is akform and ⌘is alform, then
d(!^⌘)=d!^⌘+(1)k!^d⌘.
(3) Cocycle condition: d(d!)=0 . B r i e fl y , d2=0 .
(4) If !is akform on Rmandf:Rn!Rmis di↵erentiable,
then f⇤(d!)=d(f⇤!).
Proof: (1) Let !and⌘arekform. From equation (3), We have
!=X
i1and
⌘=X
i1From equation (9), We have
d!=X
i1↵=1D↵(!i1,i2,···ik)·dx↵^dxi1^dxi2^···^dxik
d⌘=X
i1↵=1D↵(⌘i1,i2,···ik)·dx↵^dxi1^dxi2^···^dxik
)
d(!+⌘)=X
i1↵=1D↵(!i1,i2,···ik+⌘i1d(!+⌘)=X
i1↵=1D↵(!i1,i2,···ik)·dx↵^dxi1^dxi2^···^dxik
+X
i1↵=1D↵(⌘i1,i2,···ik)·dx↵^dxi1^dxi2^···^dxik
d(!+⌘)=d(!)+d(⌘)
(2) Let !is akform and ⌘is alform.
Claim: d(!^⌘)=d!^⌘+(1)k!^d⌘.
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CHAPTER 3. EXTERIOR DERIVATIVES
Case I: Let!and⌘both are 0 form. Then !=fand⌘=gfor some
scalar field fandg. Consider
d(!^⌘)=d(f^g)=nX
i=1Di(f·g)dxi
=nX
i=1(Dif)·gdxi+nX
i=1f·(Dig)dxi
=(df)^g+f^(dg)
=(df)^g+(1)0f^(dg)
Case II: If!=dxi1^dxi2^···^dxikand
⌘=dxj1^dxj2^···^dxjlthen
since D(1) = 0 all terms vanish, formula is true.
Case III: Let!is a 0 form and ⌘is alform.
Since !is a 0 form, let !=f, for some scalar field f.
Since ⌘is alform, we have
⌘=X
j1d(!^⌘)=d(f^⌘)=d(f·⌘)
=X
j1=1D(f·⌘j1,j2,···jl)dx^dxj1^dxj2^···^dxjl
=X
j1=1[(Df)·⌘j1,j2,···jl+f·(D⌘j1,j2,···jl)]dx^dxj1^dxj2^···^dxjl
=X
j1=1[(Df)·⌘j1,j2,···jldx^dxj1^dxj2^···^dxjl
+f·(D⌘j1,j2,···jl)dx^dxj1^dxj2^···^dxjl]
=df^⌘+f^d⌘
=df^⌘+(1)0f^d⌘
Case IV: Let!is akform and ⌘is alform. Let !iskform, We
have
!=X
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Since ⌘is alform, we have
⌘=X
j1)
!^⌘=(X
i1^(X
j1)
!^⌘=X
i1j1d(!^⌘)=X
i1j1↵=1D↵(!i1,i2,···ik·⌘j1,j2,···jl)
dx↵^dxi1^dxi2^···^dxik^dxj1^dxj2^···^dxjl
=X
i1j1↵=1[D↵(!i1,i2,···ik)^(⌘j1,j2,···jl)+( !i1,i2,···ik)^D↵(⌘j1,j2,···jl)]
dx↵^dxi1^dxi2^···^dxik^dxj1^dxj2^···^dxjl
=X
i1j1↵=1[D↵(!i1,i2,···ik)^(⌘j1,j2,···jl)
dx↵^dxi1^dxi2^···^dxik^dxj1^dxj2^···^dxjl
+(!i1,i2,···ik)^D↵(⌘j1,j2,···jl)dx↵^dxi1^dxi2^···^dxik^dxj1^dxj2^···^dxjl]
=X
i1j1↵=1[D↵(!i1,i2,···ik)dx↵^dxi1^dxi2^···^dxik]
^[(⌘j1,j2,···jl)dxj1^dxj2^···^dxjl]
+(1)k[(!i1,i2,···ik)dxi1^dxi2^···^dxik]^[D↵(⌘j1,j2,···jl)dx↵^dxj1^dxj2^···^dxjl]
d(!^⌘)=d!^⌘+(1)k!^d⌘
The sign ( 1)kadded since dxi1^dxi2^···^dxikiskform and
D↵(⌘j1,j2,···jl)i s1 form.
(3) Let !iskform. From equation (3), We have
!=X
i1From equation (9), We have
d!=X
i1,i2,···iknX
↵=1D↵(!i1,i2,···ik)·dx↵^dxi1^dxi2^···^dxik
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Operating dagain on d!we have
d(d!)=X
i1↵=1nX
=1D↵,(!i1i2···ik)dx^dx↵^dxi1^dxi2^···^dxik.
In this sum the terms
D↵,(!i1i2···ik)dx^dx↵^dxi1^dxi2^···^dxikand
D,↵(!i1i2···ik)dx↵^dx^dxi1^dxi2^···^dxikcancel in pairs since
D↵,(!i1i2···ik)dx^dx↵^dxi1^dxi2^···^dxik
=D,↵(!i1i2···ik)dx↵^dx^dxi1^dxi2^···^dxik
and hence
d(d!)=0
(4)Claim: If!is akform on Rmandf:Rn!Rmis di↵eren-
tiable, then
f⇤(d!)=d(f⇤!).
To prove this result let’s apply induction on k.
Step I: Subclaim: Result is true when k=0 ,i . e . i f !is a 0
form.
Since !is a 0 form, !=ffor some scalar field f.
Consider f⇤(d!)=f⇤(df)=d(f⇤(f)) == d(f⇤!).
Step II: Suppose result is true when !is akform.
i.e. if !is akform on Rmthen f⇤(d!)=d(f⇤!).
Subclaim: Result is true when !is (k+1)form of the type !^dxi.
Consider
f⇤(d(!^dxi)) = f⇤(d!^dxi+(1)k!^d(dxi)) by theorm 10(II)
=f⇤(d!^dxi)b y t h e o r m 1 0 ( I I I )
=f⇤(d!)^f⇤(dxi) by theorm 8(IV)
=d(f⇤!)^f⇤(dxi)) result is true for k-form
=d(f⇤(!^dxi))
Example I: Calculate exterior derivatives of the 1 forms z2dx^
dy+(z2+2y)dx^dzinR3.
Solution: Consider !=z2dx^dy+(z2+2y)dx^dzbe given 2 forms.
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Consider
d!=2zdz^dx^dy+( 2zdz+2dy)^dx^dz
d!=2zdx^dz^dy+2zdz^dx^dz+2dy^dx^dz
d!=2zdx^dy^dz2zdz^dz^dx2dx^dy^dz
d!=2zdx^dy^dz02dx^dy^dz
d!=2 (z1)dx^dy^dz
Example II: Calculate exterior derivatives of fdgwhere fandgare
functions.
Solution: Letf=f(x, y, z )a n d g=g(x, y, z )
)dg=gxdx+gydy+gzdz
Thus we have fdg=f(x, y, z )·(gxdx+gydy+gzdz)
Consider
d(f·dg)=df^dg+f^d(dg)fis 0form
=df^dg+f^d(dg)s i n c e d(dg)=0
=(fxdx+fydy+fzdz)^(gxdx+gydy+gzdz)
=fxdx^(gxdx+gydy+gzdz)+fydy^(gxdx+gydy+gzdz)
+fzdz^(gxdx+gydy+gzdz)
=fx·gxdx^dx+fx·gydx^dy+fx·gzdx^dz+fy·gxdy^dx
+fy·gydy^dy+fy·gzdy^dz+fz·gxdz^dx+fz·gydz^dy+fz·gzdz^dz
=0+ fx·gydx^dy+fx·gzdx^dzfy·gxdx^dy+0
+fy·gzdy^dzfz·gxdx^dzfz·gydy^dz+0
=(fx·gyfy·gx)dx^dy+(fx·gzfz·gx)dx^dz+(fy·gzfz·gy)dy^dz
Example III: IfFis a vector field on R3, define the forms
!1
F=F1dx+F2dy+F3dz
!2
F=F1dy^dz+F2dz^dx+F3dx^dy
Prove that
(1) df=!1
grad f where fis a scalar field in R3
(2) d(!1
F)=!2
curl F
(3) d(!2
F)=( div F )dx^dy^dz
(4) curl grad f =0
(5) div curl F =0
Solution:
(1) Let f=f(x, y, z )b eas c a l a rfi e l di n R3.
)
df=@f
@xdx+@f
@ydy+@f
@zdz
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where (@f
@x,@f
@y,@f
@z)=grad f
by definition of !1
F, we can write dfasdf=!1
grad f.
(2) Let !1
F=F1dx+F2dy+F3dzbe a 1 form. Consider
d(!1
F)=F1
xdx^dx+F1
ydy^dx+F1
zdz^dx
+F2
xdx^dy+F2
ydy^dy+F2
zdz^dy
+F3
xdx^dz+F3
ydy^dz+F3
zdz^dz
=0F1
ydx^dy+F1
zdz^dx
+F2
xdx^dy+0F2
zdy^dz
F3
xdz^dx+F3
ydy^dz++ 0
=(F2
xF1
y)dx^dy+(F3
yF2
z)dy^dz+(F1
zF3
x)dz^dx
where (( F2
xF1
y),(F3
yF2
z),(F1
zF3
x)) = curl F
by definition of !2
F, we can write d(!1
F)a sd(!1
F)=!2
curl F.
(3) Let !2
F=F1dy^dz+F2dz^dx+F3dx^dybe given 2 form.
Consider
d(!2
F)=dF1^dx^dy^dz+dF2^dy^dz^dx+dF3^dz^dx^dy
=dF1^dx^dy^dz+dF2^dx^dy^dz+dF3^dx^dy^dz
=(dF1+dF2+dF3)^dx^dy^dz
=(div F )dx^dy^dz
(4) By (2), we have !2
curl F =d(!1
F)
Replace Fbygrad f ,w eo b t a i n
!2
curl grad f =d(!1
grad f)
By (1), we have !2
curl grad f =d(d(f)) = 0
)curl grad f =0 .
(5) By (3), we have ( div F )dx^dy^dz=d(!2
F)
Replace Fbycurl F , we obtain
(div curl F )dx^dy^dz=d(!2
curl F)
By (2), we have ( div curl F )dx^dy^dz=d(d(!1
F)) = 0
)div curl F =0 .
Example 1: Let↵=xdx+ydy+zdz,=zdx+xdy+ydzand=
xydz in the following problems.
1. Calculate
(a)↵^
(b)↵^
(c)^
(d) ( ↵+)^(↵+)
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2. Calculate
(a)d↵
(b)d
(c)d(↵+)
(d)d(x↵)
Example 2: Consider the forms,
!=xydx +3dyyzdz,
⌘=xdx-yz2dy+2xdzinR3.
Verify by direct computation that
d(d!)=0a n d d(!^⌘)=( d!)^⌘-!^d⌘.
Example 3: InR3, let !=xydx +2zdyydz
Let↵:R2!R3be given by the equation,
↵(u, v)=( uv, u2,3u+v)
Calculate d!,↵⇤!,↵⇤(d!)a n d d(↵⇤!)d i r e c t l y .
3.3 Closed and Exact Form
Closed Form: A form !is called closed if d!=0 .
Exact Form: A form !is called exact if !=d⌘, for some ⌘.
Note: Theorem 10( III)s h o w st h a te v e r ye x a c tf o r mi sc l o s e ds i n c e
d!=d(d⌘)=0 .
Note: Is every closed form is exact?
In general every closed form is not exact.
If!is the 1 form Pdx+QdyonR2and is closed, then
d!=(D1Pdx+D2Pdy)^dx+(D1Qdx+D2Qdy)^dy
d!=D1Pdx^dx+D2Pdy^dx+D1Qdx^dy+D2Qdy^dy
d!=0D2Pdx^dy+D1Qdx^dy+0
d!=(D1QD2P)dx^dy
Thus since !is closed d!=0 )0=( D1QD2P)dx^dythen
D1Q=D2PThus we have !=Pdx+Qdyis exact if D1Q=D2Pi.e.
@Q
@x=@P
@y.
Example II: LetA=R20a n d
!=y
x2+y2dx+x
x2+y2dy
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CHAPTER 3. EXTERIOR DERIVATIVES
inA.S h o wt h a t , !is closed but not exact.
Star Shaped Set: Suppose that !=nP
i=1!idxiis a 1 form on Rn. If
!is exact then !=df=nP
i=1Difdxiwith assumption f(0) = 0. We
have
f(x)=1Z
0d
dtf(tx)dt
=1Z
0nX
i=1Dif(tx)xidt
=1Z
0nX
i=1!i(tx)xidt
)To find f, for a given !such that !=df, we consider the function
I!, defined by
I!(x)=1Z
0nX
i=1!i(tx)·xidt,
Note that the I!is well defined if !is defined only on an open set
A⇢Rnwith the property that whenever x2A, the line segment from
0t o xis contained in A. Such an open set is called star shaped with
respect to 0.
Theorem-11 : Poincar´ e Lemma IfA⇢Rnis an open set star-
shaped with respect to 0, then every closed form on Ais exact.
Proof: Let!belform
!=X
i1Define a function ( l1)forms Ifrom lforms !(for each l), such
that I(0) = 0 and !=I(d!)+d(I!)f o ra n yf o r m !.
Since Ais star-shaped we can define
I!(x)=X
i1↵=1(1)↵10
@1Z
0tl1!i1i2···il(tx)dt1
Axi↵dxi1···^ddxi↵^···^dxil
(3.2)
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Note that the symbol ddxi↵indicates that it is omitted. Now let’s con-
sider d(I!(x)), note that
d[(!i1i2···il(tx))xi↵dxi1···^ddxi↵^···^dxil]
=(!i1i2···il(tx))d[xi↵]^dxi1···^ddxi↵^···^dxil
+d(!i1i2···il(tx))xi↵dxi1···^ddxi↵^···^dxil
=(1)↵1·l·(!i1i2···il(tx))dxi1···^dxi↵^···^dxil
+nX
j=1t·Dj(!i1i2···il(tx))xi↵dxi1^···^ddxi↵^···^dxil
since ↵running from 1 to land
(1)↵1added because of ( ↵1) permutations of dxi↵
hence d(I!(x)) becomes
d(I!(x)) = l·X
i1@1Z
0tl1!i1i2···il(tx)dt1
Adxi1···^dxi↵^···^dxil
+X
i1↵=1nX
j=1(1)↵10
@1Z
0tlDj!i1i2···il(tx)dt1
Axi↵dxi1···^ddxi↵^···^dxil
(11)
Using equation (9), consider d!as
d!=X
i1j=1Dj(!i1i2···il)dxj^dxi1^dxi2^···^dxil
Applying Ito the ( l+1 )form d!, as per definition of Iwe obtain
lform as
I(d!)=X
i1j=10
@1Z
0tlxjDj(!i1i2···il)(tx)dt1
Adxi1^···^dxi↵^···^dxil
X
i1<···ilnX
j=1lX
↵=1(1)↵10
@1Z
0tlDj(!i1i2···il)(tx)dt1
Axi↵dxj^dxi1^···^ddxi↵^···^dxil
(12)
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Adding equations (11) and (12), the triple sums cancel, and we obtain
d(I!)+d(d!)=X
i1@1Z
0tl1(!i1i2···il)(tx)dt1
Adxi1^dxi2^···^dxil
+X
i1j=10
@1Z
0tlxjDj(!i1i2···il)(tx)dt1
Adxi1^dxi2^···^dxil
=X
i1@1Z
0d
dt[tl(!i1i2···il)(tx)]dt1
Adxi1^dxi2^···^dxil
=X
i1=!.
Thus we have !=d(I!)+d(d!)s i n c e !is closed d!=0 .
Thus !=d(I!)h e n c e !is exact.
3.4 Chapter End Exercise
1. Is the 1 form !=(x2+y2)dx+2xydy closed and exact? Justify
your answer.
2. Let !be a any 3 form. Prove or disprove: d(d!)=0 .
3. Let A=R20a n d !=(ydx+xdy)
(x2+y2)inA. Prove or disprove:
!is closed and exact in A.
4. In R3, let !=xydx +2zdyydzand↵:R2!R3be given by
↵(u, v)=( uv, u2,3u+v). Calculate ↵⇤(d!).
5. State the necessary condition for every closed form on A⇢Rnto
be exact. Is the 1 form !=( 1+ ex)dy+ex(yx)dyclosed and
exact? Justify your answer.
6. If !is a 0 form and ⌘is alform, then show that d(!^⌘)=
d!^⌘+(1)k!^d⌘.
7. If Fis a vector field on R3. Let !1
F=F1dx+F2dy+F3dzand
!2
F=F1dy^dz+F2dz^dx+F3dx^dythen show that d(!1
F)=
!2
curl F.
8. Show that every exact form is closed. Is the converse true? Justify
your answer.
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Chapter 4
Basics of Submanifolds of Rn
Unit Structure :
4.1O b j e c t i v e
4.2B a s i cP r e l i m i n a r i e s
4.3M a n i f o l d si n Rn
4.4M a n i f o l d si n Rnwithout boundary
4.5M a n i f o l d si n Rnwith boundary
4.6 Fields and Forms on Manifolds
4.7O r i e n t a t i o no fM a n i f o l d s
4.8C h a p t e rE n dE x e r c i s e
4.1 Objectives
After going through this chapter you will be able to:
1. Define a manifolds with and without boundary.
2. Learn the concepts of Coordinate system and M conditions.
3. Learn the properties of tangent space of manifolds and vector field
on manifolds.
4. Identify orientation of Manifolds.
4.2 Basic Preliminaries
Smooth map: Am a p p i n g fof an open set U⇢Rninto Rmis
called smooth if it has continuous partial derivatives of all orders.
Note: For partial derivatives domain of fis essentially required to be
open.
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Di↵eomorphism: A smooth map f:X!Yof subsets of two eu-
clidean spaces is a di ↵eomorphism if it is bijective and if the inverse
f1:Y!Xis also smooth. Xand Yare di ↵eomorphic if such a
map exists.
OR
IfUand Vare open sets in Rn,ad i ↵erentiable function h:U!V
with a di ↵erntiable inverse h1:V!U, will be called a di ↵eomor-
phism.
(“Di↵erntiable”hencefoth, means “ C1”.)
Exercise: Give an example of di ↵eromorphism.
4.3 Manifolds in Rn
A subset MofRnis called a kdimensional manifold in Rnif for
every point x2M, the following condition is satisfied
Condition M: If there is an open set Ucontaining x, an open set
V⇢Rn,a n dad i ↵eomorphism h:U!Vsuch that
h(U\M)=V\(Rk⇥{0})={y2V:yk+1=yk+2=···=yn=0}.
i.e. ( y1,···,yk,yk+1,···,yn)!(y1,···,yk,0,···,0)
OR
A subset Mof a euclidean space Rnis known as a kdimensional man-
ifold if it is locally di ↵eomorphic to Rk.
Note that, local referring to behaviour only in some neighborhood of a
point.
Submanifolds: IfM1andM2are both manifolds in RnandM1⇢
M2then M1is known as submanifold of M2.
Note:
(1)Mis itself submanifold of Rn.
(2) Any open set of Mis submanifold of M.
(3) A point in Rnis a 0 dimensional manifolds.
(4) An open subset in Rnis an ndimensional manifolds.
Theorem-01: LetA⇢Rnbe open and let g:A!Rpbe a di ↵eren-
tiable function such that g0(x)h a sr a n k pwhenever g(x)=0 .
Then g1(0) is an ( np)dimensional manifold in Rn.
Proof: Step I: Consider following theorem from Real Analysis
Subclaim: Theorem: Letf:Rn!Rpbe a continuously di ↵eren-
tiable function in an open set containing awhere pn. Iff(a)=0
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and the p⇥nmatrix Djfi(a)h a sr a n k pthen there is an open set
A⇢Rncontaining aand a di ↵erentiable function h:A!Rnwith
di↵erentiable inverse such that
foh(x1,x2,···,xn)=( xnp+1,xnp+2,···,xn).
Add proof of above theorem.
Step II: By applying above theorem and by definition of manifold we
can conclude that g1(0) is an ( np)dimensional manifold in Rn.
Example: Show that the nSphere Sn, defined as {x2Rn+1:|x|=
1}isndimensional manifold.
Solution: Apply above theorem (1) by considering Sn=g1(0),where
g:Rn+1!Ris defined by g(x)=|x|21.
Note that nis replaced by n+1 ,
p=1 ,
g(0) = 0.
By theorem (1), Sphere Snis (np)=( n+11) = ndimensional
manifold.
Theorem-02: A subset MofRnis akdimensional manifold if and
only if for each point x2Mthe following “coordinate condition”is
satisfied:
Coordinate condition C: There is an open set Ucontaining x,a n
open set W⇢Rk,a n da1 1d i↵erentiable function f:W!Rnsuch
that
(1) f(W)=M\U,
(2) f0(y)h a sr a n k kfor each y2W,
(3) f1:f(W)!Wis continuous.
note that, such a function fis called a coordinate system around x.
Proof: Step I: Assume that Mis akdimensional manifold in Rn.
Claim: Each point x2Msatisfies the coordinate condition.
Since Miskdimensional manifold in Rnby definition each point
x2Msatisfies the following condition
If there is an open set Ucontaining x, an open set V⇢Rn,a n da
di↵eomorphism h:U!Vsuch that
h(U\M)=V\(Rk⇥{0})={y2V:yk+1=yk+2=···=yn=0}.
LetW={a2Rk:(a,0)2h(M)}.
Define f:W!Rnbyf(a)=h1(a,0).
Clearly
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(1) Since h:U!V)h1(V)=Uand
(a,0)2h(M))h1(a,0) = M
hence f(W)=M\U,
(2) Since his di↵omorphism, flis continuous and
(3) If H:U!Rkis defined by H(z)=( h1(z),···,hk(z)),
then H(f(y)) = yfor all y2W(*Since f=h1)
Therefore on di ↵erentiating by using chain rule we obtain
H0(f(y))·f0(y)=Iandf0(y) must have rank k.
Thus each point x2Msatisfies the coordinate conditions.
Step II: Suppose that f:W!Rnsatisfies coordinate conditions.
Claim: Mis akdimensional manifold in Rn.
Letf(y)=x.
Assume that the matrix ( Djfi(y)), 1 i, jkhas a non-zero deter-
minant.
Define g:W⇥Rnk!Rnbyg(a, b)=f(a)+f(0,b).
Then det g0(a, b) = det ( Djfi(a)),
so det g0(y,0)6=0 .
Now lets use Inverse Function Theorem as
Inverse Function Theorem: Suppose that f:Rn!Rnis continu-
ously di ↵erentiable in an open set containing aand det f0(a)6= 0. Then
there is an open set Vcontaining aand open set Wcontaining f(a)
such that f:V!Whas a continuous inverse f1:W!Vwhich
is di↵erentiable and for all y2Wsatisfies ( f1)0(y)=[ f0(f1(y))]1.
By Inverse Function Theorem
There is an open set V0
1containing ( y,0) and an open set V0
2contain-
ingg(y,0) = xsuch that g:V0
1!V0
2has a di ↵erentiable inverse
h:V0
2!V0
1.
By third coordinate condition, f1is continuous,
{f(a):(a,0)2V0
1}=U\f(W)f o rs o m eo p e ns e t U(By first coordi-
nate condition).
LetV2=V0
2\UandV1=g1(V2).
Then V2\Mis exactly {f(a):(a,0)2V1}={g(a,0) : ( a,0)2V1},
where M⇢RnSo
h(V2\M)=g1(V2\M)s i n c e h=g1
=g1({g(a,0) : ( a,0)2V1})=( {(a,0) : ( a,0)2V1})
=V1\(Rk⇥{0}).
hence by definition Mis akdimensional manifold in Rn.
Note: Iff1:W1⇢Rk!Rnandf2:W2⇢Rk!Rnare two
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CHAPTER 4. BASICS OF SUBMANIFOLDS OF RN
coordinate systems, then
f1
2f1:f1
1(f2(W2))!Rk
is di↵erentiable with non-singular Jacobian. If fact, f1
2(y)c o n s i s t so f
the first kcomponents of h(y).
4.4 Manifolds of Rnwithout boundary
Manifolds in Rnwithout boundary: Letk>0. Suppose that
Mis a subspace of Rnhaving the following property:
For each p2M, there is an open set Vcontaining pthat is open in M,
as e t Uthat is open in Rk, and a continuous map f:U!Vcarrying
Uonto Vin a 1 1f a s h i o ns u c ht h a t
(1) fis of class Cr
(2) Df(x)h a sr a n k kfor each x2U,
(3) f1:V!Uis continuous.
Then Mis called a kmanifold without boundary Rnof class Cr. The
map fis called a coordinate patch on Mabout p.
Example 1: Let↵:R!R2be given by ↵(t)=( t3,t2). Let M
be image set of ↵. IsM1manifold without boundary in R2? Justify
your answer.
Solution: Let↵:R!R2be given by ↵(t)=( t3,t2)i sa1 1m a p .
Clearly
(1) ↵is of class C1
(2) ↵1:V!Uis continuous where Uis open in RandVis open in R2,
(3) D↵(t)=( 3 t2,2t)h a sn o tr a n k1a t t=0 .
hence Mnot 1 manifold without boundary in R2.
Example 2: Let:R2!R3be given by (x, y)=( x(x2+
y2),y(x2+y2),(x2+y2),). Let Mbe image set of . IsM2manifold
without boundary in R3? Justify your answer.
Solution: Let:R2!R3be given by (x, y)=( x(x2+y2),y(x2+
y2),(x2+y2),)i sa1 1m a p .C l e a r l y
(1) is of class C1
(2) 1:V!Uis continuous where Uis open in RandVis open in R2,
(3) D(t)=(x2+y2)+2 x22xy 2x
2xy (x2+y2)+2 y22y
D(t)h a sn o tr a n k2a t0 .
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hence Mnot 2 manifold without boundary in R3.
Example 3: Let:R!R2be given by (t)=( s i n 2 t)(|cost|
,sint)f o r0 without boundary in R3? Justify your answer.
Solution: Let:R!R2be given by ↵(t)=( s i n 2 t)(|cost|,sint)
is a 1 1m a pf o r0 (1) is of class C1
(2) D(t)=( s i n2 t)(|sint|,cost)+( 2c o s2 t)(|cost|,sint)h a sr a n k1f o ra l l t.
(3) Since image of smaller interval Uwhich contains⇡
2is not open in Mhence
1:V!Uis not continuous where Vis open in R2,
hence Mnot 1 manifold without boundary in R3.
M Uo T/2 0 T
4.5 Manifolds of Rnwith boundary
Half Space: The half-space Hk⇢Rkis defined as {x2Rk:xk
0}.
Manifold with Boundary: A subset MofRnis a kdimensional
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manifold-with boundary if for every point x2Meither condition ( M)
or the following condition is satisfied:
Condition M’: There is an open set Ucontaining x, an open set
V⇢Rn,a n dad i ↵eomorphism h:U!Vsuch that
h(U\M)=V\(Hk⇥{0})={y2V:yk0,andyk+1=yk+2=···=yn=0}
andh(x)h a s kthcomponent = 0.
The set of all points x2Mfor which condition M0is satisfied is called
the boundary of Mand denoted @M.
Note: Conditions ( M)a n d( M0)c a n n o tb o t hh o l df o rt h es a m e x.
Examples: (1) Let ↵:R!R2be the map ↵(t)=( t, t2). Let
Mbe image set of ↵.S h o w t h a t M1manifold in R2covered by the
single coordinate patch ↵.
(2) Let :H1!R2be the map (t)=( t, t2). Let Nbe image set of
.S h o wt h a t Nis 1manifold in R2.
(3) Show that unit circle S1is a 1 manifold in R2.
(4) Show that the function ↵:[ 0,1]!S1given by ↵(t)=( c o s2 ⇡t,sin 2⇡t)
is not a coordinate patch on S1.
4.6 Fields and Forms on Manifolds
Tangent Space of M:LetMbe a kdimensional manifold in Rn
and let
f:W!Rnbe a coordinate system around x=f(a).
Since f0(a)h a sr a n k k, the linear transformation f⇤:Rk
a!Rn
x,i s1 1,
andf⇤(Rk
a)i sa kdimensional subspace of Rn
x.
Ifg:V!Rnis another coordinate system, with x=g(b), then
g⇤(Rk
b)=f⇤(f1g)⇤(Rk
b)=f⇤(Rk
a)
Thus the kdimensional subspace f⇤(Rk
a)d o e sn o td e p e n do nt h ec o -
ordinate system f. This subspace is denoted Mx, and is called the
tangent space of Matx.
Note: There is a natural inner product Tx,o nMx, induced by that on
Rn
x,
ifv, w2Mx, define Tx(v, w)=hv, wix.
Vector field on M:Suppose that Ais an open set containing M,
and Fis a di ↵erentiable vector field on Asuch that F(x)2Mx, for
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each x2M. Iff:W!Rnis a coordinate system, there is a unique
di↵erentiable vector field GonWsuch that f⇤(G(a)) = F(f(a)) for
each a2W. such a function Fis called a vector field on M.
Note: (1) we define Fto be di ↵erentiable if Gis di↵erentiable.
(2) Note that our definition does not depend on the coordinate system
chosen.
ifg:V!Rnandg⇤(H(b)) = F(g(b)) for all b2V, then the component
functions of H(b) must equal the component functions of G(f1(g(b))),
soHis di↵erentiable if Gis di↵erentiable.
pform on M:A function !which assigns !(x)2⇤p(Mx)f o re a c h
x2Mis called a pform on M.
Iff:W!Rnis a coordinate system, then f⇤!is apform on W.
Note: (1) We define !to be di ↵erentiable if f⇤!is di↵erentiable.
(2) A pform !onMcan be written as
!=X
i1here the functions !i1i2···ipare defined only on M.
Theorem-03: There is a unique ( p+1 )form d!onMsuch that for
every coordinate system f:W!Rnwe have f⇤(d!)=d(f⇤!).
Proof: Iff:W!Rnis a coordinate system with x=f(a)a n d
v1,v2,···,vp+12Mx, there are unique !1,!2,···,!p+1inRk
asuch that
f⇤(!i)=vi.
Define d!(x)(v1,v2,···,vp+1)=df⇤(!)(a)(!1,!2,···,!p+1).
One can check that this definition of d!(x)d o e sn o td e p e n do nt h e
coordinate system f, so that d!is well-defined.
Moreover, it is clear that d!has to be defined this way, so d!is unique.
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4.7 Orientable Manifolds
Consistent: For each tangent space Mxof a manifold M,i ti s
necessary to choose an orientation µx. Such choices are called consistent
provided that for every coordinate systems f:W!Rnanda, b2W
the relation
[f⇤((e1)a),f⇤((e2)a),···,f⇤((ek)a)=µf(a)
holds if and only if
[f⇤((e1)b),f⇤((e2)b),···,f⇤((ek)b)=µf(b).
Orientation Preserving: Suppose orientations µxhave been cho-
sen consistently. If f:W!Rnis a coordinate system such that
[f⇤((e1)a),f⇤((e2)a),···,f⇤((ek)a)=µf(a)
for one, and hence for every a2W, then fis called orientation-
preserving.
Note: (1) If fis not orientation-preserving and T:Rk!Rkis a linear
transformation with det T=1, then fTis orientation-preserving.
(2) Therefore there is an orientation-preserving coordinate system around
each point.
(3) If fandgare orientation-preserving and x=f(a)=g(b), then the
relation
[f⇤((e1)a),f⇤((e2)a),···,f⇤((ek)a)] = µx=[g⇤((e1)b),g⇤((e2)b),···,g⇤((ek)b)]
implies that
[(g1f)⇤((e1)a),(g1f)⇤((e2)a),···,(g1f)⇤((ek)a)] = [( e1)b,(e2)b,···,(ek)b],
so that det ( g1f)0>0.
Orientable Manifold: A manifold for which orientations µxcan
be chosen consistently is called orientable, and a particular choice of
theµxis called an orientation µofM. A manifold together with an
orientation µis called an oriented manifold.
Outward Unit Normal: IfMis akdimensional manifold-with-
boundary and x2@M, then ( @M)x,i sa( k1)dimensional subspace
of the kdimensional vector space Mx. Thus there are exactly two
unit vectors in M, which are perpendicular to ( @M)x. They can be
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distinguished as follows.
Iff:W!Rnis a coordinate system with W⇢Hkand f(0) = x,
then only one of these unit vectors is f⇤(v0)f o rs o m e v0with vk<0.
This unit vector is called the outward unit normal n(x).
Note: Outward unit normal does not depend on the coordinate system
f.
Induced Orientation: Suppose that µis an orientation of a k
dimensional manifold with-boundary M. Ifx2@M, choose v1,v2,··
·,vk12(@M)x, so that [( n(x),!1,!1,···,!k1]= µx. If it is also
true that [( n(x),!1,!1,···,!k1]=µx, then both [ v1,v2,···,vk1]a n d
[(!1,!1,···,!k1] are the same orientation for ( @M)x. This orientation
is denoted ( @µ)x. The orientations ( @µ)x, for x2@M, are consistent
on@M. Thus if Mis orientable, @Mis also orientable, and an orienta-
tion µforMdetermines an orientation @µfor@M, called the induced
orientation.
Note: If we apply these definitions to Hkwith the usual orientation,
we find that the induced orientation on Rk1={(x2Hk:xk=0}is
(1)ktimes the usual orientation.
Example: Show that the M¨ obius strip is a non-orientable manifold.
4.8 Chapter End Exercise
1. Define di ↵eomorphism and give an example of di ↵eomorphism.
Justify your answer.
2. Show that unit circle S1is a 1 manifold in R2.
3. Let :R!R2be given by (t)=( s i n 2 t)(|cost|,sint)f o r
0boundary in R3? Justify your answer.
4. Let f:R1!R1is given by
f(x)=8
<
:e1
x2,x>0,
0,x0
Prove or disprove: fis di↵eomorphism.
5. Let :H1!R2be the map (t)=( t, t2). Let Nbe image set
of.S h o wt h a t Nis 1manifold in R2.
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6. Prove or disprove: the M¨ obius strip is a orientable manifold.
7. Is the nSphere Sn, defined by {x2Rn+1:|x|=1}andimensional
manifold? Justify your answer.
8. Let :R!R2be given by (t)=( s i n 2 t)(|cost|,sint)f o r
0boundary in R3? Justify your answer.
9. Show that there is a unique ( p+1 )form d!onMsuch that for
every coordinate system f:W!Rnwe have f⇤(d!)=d(f⇤!).
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Chapter 5
Stokes’s Theorem
Unit Structure :
5.1O b j e c t i v e
5.2B a s i cP r e l i m i n a r i e s
5.3T h eI n t e g r a lo f kforms
5.4S t o k e s ’ sT h e o r e mf o rI n t e g r a lo f kforms
5.5S t o k e s ’ sT h e o r e mo nM a n i f o l d s
5.6T h eV o l u m eE l e m e n t
5.7C h a p t e rE n dE x e r c i s e
5.1 Objectives
After going through this chapter you will be able to:
1. Define a integral of kforms.
2. Learn the concepts of line integral, surface integral and volume
integral.
3. Learn the properties of the volume element.
5.2 Basic Preliminaries
nfold product: [0,1]ndenotes the nfold product and is given
by
[0,1]n=[ 0,1]⇥[0,1]⇥···⇥[0,1]
Singular ncube: A singular ncube in A⇢Rnis a continuous
function C:[ 0,1]n!A.
Note: LetR0and [0 ,1]0both denote {0}.
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Standard ncube: The standard ncube In:[ 0,1]n!Rndefined
byIn(x)=xforx2[0,1]n.
Definitions and Properties:
1. The vector field ~Fis known as solenoidal if Div ~F=0 .
2. The vector field ~Fis known as irrotational if Curl ~F=0 .
3. If the vector field ~Fis solenoidal then by Divergence theorem
Z
MdivFdv=Z
@MhF,nidA=0.
4. If the vector field ~Fis irrotational then by Stokes theorem
Z
Mh(O⇥F),nidA=Z
@MhF,Tids=0.
5. If the line integral of a vector field is independent of path then such
av e c t o rfi e l d sa r ec a l l e dc o n s e r v a t i v e .
6. A conservative vector fields are irrotational and an irrotational vec-
tor fields are also conservative if domain is simply connected.
5.3 The Integral of kform
The Integral of kform on the cube [0,1]k:If!is akform
on [0 ,1]k, then !=fdx1^dx2^···^dxkfor a unique function f. We
defineZ
[0,1]k!=Z
[0,1]kf.
We could also write this as
Z
[0,1]kfdx1^dx2^···dxk=Z
[0,1]kf(x1,x2,···,xk)dx1dx2···dxk.
The Integral of kform on the singular kcube c:If!is a
kform on Aandcis a singular kcube in A, we define
Z
c!=Z
[0,1]kc⇤!.
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Note, in particular, that
Z
Ikfdx1^dx2^···^dxk=Z
[0,1]k(Ik)⇤f(dx1^dx2^···^dxk)
=Z
[0,1]kf(x1,x2,···,xk)dx1dx2···dxk.(1)
Note: (1) A 0 form !is a function; if c:{0}!Ais a singular
0cube in A. We define
Z
c!=!(c(0))
(2) The integral of !over a kchain c=Paiciis defined by
Z
c!=X
aiZ
ci!
(3) The integral of a 1 form over a 1 chain is often called a line
integral.
IfPdx+Qdyis a 1 form on R2andc:[ 0,1]!R2is a singular 1 cube
(a curve), then one can prove that
Z
cPdx+Qdy=l i mnX
i=1[c1(ti)c1(ti1)]·P(c(ti))+[c2(ti)c2(ti1)]·Q(c(ti))
where t0,t1,···,tnis a partition of [0 ,1], the choice of tiin [ti1,ti]i s
arbitrary, and the limit is taken over all partition as the maximum of
[ti1,ti] goes to 0.
5.4 Stokes’s Theorem for Integral of
kforms
Theorem-15: Stokes Theorem If!is a ( k1)form on an open
setA⇢Rnandcis akchain in A, then
Z
cd!=Z
@c!.
Proof: Suppose first that c=Ikand!is a ( k1)form on [0 ,1]k.
Then !is the sum of ( k1)forms of the type
!=fdx1^dx2^···cdxi^···dxk
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Note that
Z
[0,1]k1Ik
(j,↵)⇤(fdx1^dx2^···^cdxi^···^dxk)
=8
<
:0i f i6=j,R
[0,1]kf(x1,x2,···,↵,···,xk)dx1dx2···dxkifj=i.
Therefore
Z
@Ikfdx1^dx2^···cdxi^···^dxk
=kX
j=1X
↵=0,1(1)j+↵Z
[0,1]k1Ik
(j,↵)⇤(fdx1^dx2^···cdxi^···^dxk)
on expanding summation and using equation (1)
=(1)i+1Z
[0,1]kf(x1,x2,···,1,···,xk)dx1dx2···dxk
+(1)iZ
[0,1]kf(x1,x2,···,0,···,xk)dx1dx2···dxk. (2)
On the other hand,
Z
Ikd(fdx1^dx2^···cdxi^···^dxk)=Z
[0,1]kDifdxi^dx1^dx2^···cdxi^···^dxk
=(1)i1Z
[0,1]kDif.
By Fubini theorem and the fundamental theorem of calculus in one
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dimension
Z
Ikd(fdx1^dx2^···cdxi^···dxk)
=(1)i1Z
[0,1]Z
[0,1]···(Z
[0,1]Dif(x1,x2,···,↵,···,xk)dxi)dx1dx2···cdxi···dxk
=(1)i11Z
01Z
0···1Z
0[f(x1,x2,···,1,···,xk)f(x1,x2,···,0,···,xk)]dx1dx2···dxk.
=(1)i1Z
[0,1]kf(x1,x2,···,1,···,xk)dx1dx2···dxk
+(1)iZ
[0,1]kf(x1,x2,···,0,···,xk)dx1dx2···dxk.
Thus by equation (2) we have
Z
Ikd!=Z
@Ik!.
Note: Ifcis an arbitrary singular kcube, working through the
definitions will show that
Z
@c!=Z
@Ikc⇤!.
Therefore
Z
cd!=Z
Ikc⇤(d!)=Z
Ikd(c⇤!)=Z
@Ikc⇤!=Z
@c!.
Finally, if cis akchainPaici,w eh a v e
Z
cd!=X
aiZ
cid!=X
aiZ
@ci!=Z
@c!.
5.5 Stokes’s Theorem on Manifolds
If!is apform on a kdimensional manifold with boundary M
andcis a singular pcube in M, we define
Z
c!=Z
[0,1]pc⇤! (3)
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Note: (1) In the case p=kit may happen that there is an open set
W[0,1]kand a coordinate system f:W!Rnsuch that c(x)=f(x)
forx2[0,1]k.
(2) If Mis oriented, the singular kcube cis called orientation-preserving
iffis orientation-preserving.
Theorem (16): Ifc1,c2:[ 0,1]k!Mare two orientation preserving
singular kcubes in the oriented kdimensional manifold Mand!is
akform on Msuch that !=0o u t s i d eo f c1([0,1]k)\c2([0,1]k)), then
Z
c1!=Z
c2!
Proof: We have
Z
c1!=Z
[0,1]kc⇤
1(!)b y e q u a t i o n ( 3 )
Z
c1!=Z
[0,1]k(c1
2c1)⇤c⇤
2(!)
Note that c1
2c1is defined only on a subset of [0 ,1]kand the sec-
ond equality depends on the fact that !=0o u t s i d eo f c1([0,1]k)\
c2([0,1]k)).)
It therefore su ces to show that
Z
[0,1]k(c1
2c1)⇤c⇤
2(!)=Z
[0,1]kc⇤
2(!)=Z
c2!.
Ifc⇤
2(!)=fdx1^fdx2^···^fdxkandc1
2c1, is denoted by g, then
by Theorem (9) we have
(c1
2c1)⇤c⇤
2(!)=g⇤(fdx1^fdx2^···^fdxk)
=(fg)·detg0.dx1^dx2^···^dxk
=(fg)·|detg0|.dx1^dx2^···^dxk,
where det g0=d e t ( c1
2c1)0>0.
On integrating both sides over [0 ,1]k,w eo b t a i n
Z
[0,1]k(c1
2c1)⇤c⇤
2(!)=Z
[0,1]k(fg)·|detg0|.dx1^dx2^···^dxk(4)
Now lets apply following theorem to equation (4)
LetA⇢Rnbe an open set and g:A!Rnis 11c o n t i n u o u s l y
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di↵erentiable function such that det g0(x)6=0f o ra l l x2A. If f:
g(A)!Ris integrable then
Z
g(A)f=Z
A(fog)|detg0|
Above theorem and equation (4) shows that
Z
[0,1]k(c1
2c1)⇤c⇤
2(!)=Z
[0,1]kfdx1^dx2^···^dxk
Z
[0,1]k(c1
2c1)⇤c⇤
2(!)=Z
[0,1]kc⇤
2(!)=Z
c2!
Note: (1) Let !be a kform on an oriented kdimensional manifold
M. If there is an orientation-preserving singular kcube cinMsuch
that !=0o u t s i d eo f c([0,1]k), we define
Z
M!=Z
c!.
Theorem (15) showsR
M!does not depend on the choice of c.
(2) Suppose that !is an arbitrary kform on M. There is an open cover
OofMsuch that for each U2Othere is an orientation-preserving
singular kcube cwith U⇢c([0,1]k). Let be a partition of unity for
Msubordinate to this cover. We define
Z
M!=X
'2Z
'·!
Theorem-16: Stokes Theorem on Manifolds: IfMis a com-
pact oriented kdimensional manifold with boundary and !is a ( k
1)form on M, thenZ
Md!=Z
@M!.
(Here Mis given the induced orientation.)
Proof: Case I: Suppose that there is an orientation-preserving singu-
larkcube in M@Msuch that !=0o u t s i d eo f c((0,1)k).
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By Theorem (15) and the definition of d!we have
Z
cd!=Z
[0,1]kc⇤(d!)b y e q u a t i o n ( 3 )
=Z
[0,1]kd(c⇤!)b y t h e o r e m ( 1 4 )
=Z
@Ik(c⇤!)b y t h e o r e m ( 1 5 )
=Z
@c!by equation (3)
ThenZ
Md!=Z
cd!=Z
@c!=0.
since !=0o n @c.
On the other hand,R
@M!=Osince !=0o n @M.
Suppose that there is an orientation-preserving singular kcube in M
such that c(k,0) is the only face in @M,a n d !=0o u t s i d eo f c([0,1]k)
ThenZ
Md!=Z
c(d!)=Z
@c!=Z
@M!.
Case II: The general case: There is an open cover OofMand a
partition of unity forMsubordinate to Osuch that for each '2
the form '·!is of one of the two sorts already considered. We have
0=d(1) = d X
'2'!
=X
'2d(')
so thatX
'2d(')^=0.
Since Mis compact, this is a finite sum and we have
Z
MX
'2d(')^=0.
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Therefore
Z
Md!=X
'2Z
M'·d!
=X
'2Z
Md'^!+'·d!since d'=0
=X
'2Z
Md('·!)
=X
'2Z
@M'·!
=Z
@M!.
5.6 The Volume Element
The Volume Element LetMbe a kdimensional manifold (or
manifold with boundary) in Rn, with an orientation µ. Ifx2M, then
µxand the inner product Txwe defined previously determine a volume
element !(x)2⇤k(Mx). We therefore obtain a nowhere-zero kform
!onM, which is called the volume element on M(determined by µ)
and denoted dV, even though it is not generally the di ↵erential of a
(k1)form.
The volume of Mis defined asR
MdV, provided this integral exists, which
is certainly the case if Mis compact.
Note: (1) Volume is usually called length or surface area for one
and two-dimensional manifolds, and dVis denoted ds(the ”element
of length”) or dA[ords] (the ”element of (surface) area”). (2) Con-
sider the volume element of an oriented surface (two-dimensional man-
ifold) MinR3. Let n(x)b et h eu n i to u t w a r dn o r m a la t x2M. If
!2⇤2(Mx)i sd e fi n e db y
!(v,w)=d e t2
4v
w
n(x)3
5,
then !(v,w)=1i f vand ware an orthonormal basis of Mxwith
[v,w]=µx. Thus dA=!.
On the other hand, !(v,w)=hv⇥w,n(x)iby definition of v⇥w. Thus
we have dA(v,w)=hv⇥w,n(x)i. Since v⇥wis a multiple of n(x)
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forv,w2M, we conclude that dA(v,w)=|v⇥w|if [v,w]=µx.( 3 )
If we wish to compute the area of M, we must evaluateR
[0,1]2c⇤(dA)f o r
orientation-preserving singular 2 cubes c. Define
E(a)=[ D1c1(a)]2+[D1c2(a)]2+[D1c3(a)]2.
F(a)=[ D1c1(a)·D2c1(a)] + [ D1c2(a)·D2c2(a)] + [ D1c3(a)·D2c3(a)]
.G(a)=[ D2c1(a)]2+[D2c2(a)]2+[D2c3(a)]2.
Then
c⇤(dA)((e1)a,(e2)a,)=dA(c⇤(e1)a,c⇤(e2)a,)
=|(D1c1(a),D1c2(a),D1c3(a))·(D2c1(a),D2c2(a),D2c3(a))|
=p
E(a)G(a)F(a)2
Thus Z
[0,1]2c⇤(dA)=Z
[0,1]2p
E(a)G(a)F(a)2.
Theorem-18: LetMbe an oriented two-dimensional manifold (or
manifold with boundary) in R3and let nbe the unit outward normal.
Then
(1) dA=n1dy^dz+n2dz^dx+n3dx^dy.
Moreover, on Mwe have
(2) n1dA=dy^dz.
(3) n2dA=dz^dx.
(4) n3dA=dx^dy.
Proof: Equation (1) is equivalent to the equation
dA(v,w)=d e t2
4v
w
n(x)3
5,
This is seen by expanding the determinant by minors along the bottom
row.
To prove the other equations, let z2R3
x. Since v⇥w=↵n(x)f o rs o m e
↵2R, we have
hz,n(x)i·hv⇥w,n(x)i=hz,n(x)i↵=hz,↵n(x)i=hz,v⇥wi.
Choosing z=e1,e2,a n d e3we obtain (2) ,(3) and (4).
A word of caution; if !2⇤2(R3
a)i sd e fi n e db y
!=n1(a)·dy(a)^dz(a)+n2(a)·dz(a)^dx(a)+n3(a)·dx(a)^dy(a),
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it is not true, for example, that n1(a).w=dy(a)^dz(a). The two sides
give the same result only when applied to v,w2Ma.
5.7 Chapter End Exercise
1. State and prove the Stokes theorem for any 3 forms !.
2. Consider vector field ~F=(y+z)i+(z+x)j+(x+y)k. Is vector
field ~Fsolenoidal and irrotational? Justify your answer.
3. Let Mbe a two-dimensional manifold in R3. Compute the area
ofMover orientation preserving singular 2 cubes c.
4. Consider an orientation-preserving singular kcube in M@M
such that !=0o u t s i d eo f c((0,1)k)w h e r e Mis a compact ori-
ented kdimensional manifold with boundary and !is a ( k
1)form on Mthen show thatR
Md!=R
@M!.
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Chapter 6
Classical Theorems
Unit Structure :
6.1O b j e c t i v e
6.2C l a s s i c a lT h e o r e m s
6.3 Applications of classical theorem
6.4C h a p t e rE n dE x e r c i s e
6.1 Objectives
After going through this chapter you will be able to:
1. Evaluation of a line integral using Green’s Theorem.
2. Evaluation of a volume integral using Divergence Theorem.
3. Evaluation of a surface integral using Stoke’s Theorem.
4. Learn a concept of conservative fields.
6.2 Classical Theorems
Theorem-19: Green’s Theorem: LetM⇢R2be a compact
two-dimensional manifold with boundary. Suppose that ↵,:M!R
are di ↵erentiable. Then
Z
@M↵dx+dy=Z
M(D1D2↵)dx^dy=ZZ
M✓@
@x@↵
@y◆
dxdy
(Here Mis given the usual orientation, and @Mthe induced orientation,
also known as the counter clockwise orientation.)
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Proof: We have the Stoke’s theorem on Manifolds as
IfMis a compact oriented kdimensional manifold with boundary and
!is a ( k1)form on M, then
Z
Md!=Z
@M!.
Let!=↵dx+dy
)d!=D1↵dx^dx+D2↵dy^dx+D1dx^dy+D2dy^dy
)d!=D2↵dx^dy+D1dx^dy
)d!=(D1D2↵)dx^dy
Substitute in above toke’s theorem on Manifolds we obtain
Z
@M↵dx+dy=Z
M(D1D2↵)dx^dy=ZZ
M✓@
@x@↵
@y◆
dxdy
Theorem-20: Divergence Theorem: LetM⇢R3be a compact
three-dimensional manifold with boundary and nthe unit outward nor-
mal on @M. Let Fbe a di ↵erentiable vector field on M. Then
Z
MdivFdv=Z
@MhF,nidA.
This equation is also written in terms of three di ↵erentiable functions
↵,,:M!R:
ZZZ
M✓@↵
@x+@
@y+@
@z◆
dV=ZZ
@M(n1↵+n2+n3)dS.
Proof: Define !onMby!=Fldy^dz+F2dz^dx+F3dx^dy
Then d!=d i v FdV. See example III(3) of Unit 2
According to Theorem-18, on @Mwe have
n1dA=dy^dz,
n2dA=dz^dx,
n3dA=dx^dy.
Therefore on @Mwe have
hF,nidA=F1n1dA+F2n2dA+F3n3dA,
Since F=(F1,F2,F3)a n d n=(n1,n2,n3)
hF,nidA=F1dy^dz+F2dz^dx+F3dx^dy,
hF,nidA=!.
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We have the Stoke’s theorem on Manifolds as
IfMis a compact oriented kdimensional manifold with boundary and
!is a ( k1)form on M, then
Z
Md!=Z
@M!.
Thus using values of !andd!in the above theorem, we obtain
Z
MdivFd V =Z
@MhF,nidA.
Theorem-21: Stokes’ Theorem: LetM⇢R3be a compact ori-
ented two-dimensional manifold with boundary and nthe unit outward
normal on Mdetermined by the orientation of M. Let @Mhave the
induced orientation. Let Tbe the vector field on @Mwith ds(T)=1
and let fbe a di ↵erentiable vector field in an open set containing M.
Then Z
Mh(O⇥F),nidA=Z
@MhF,Tids.
This equation also written as
Z
@M↵dx+dy+dz=ZZ
M
n1✓@
@y@
@z◆
+n2✓@↵
@z@
@x◆
+n3✓@
@x@↵
@y◆
dS
Proof: Define !onMby!=Fldx+F2dy+F3dz.
Since O⇥F=(D2F3D3F2,D3F1D1F3,D1F2D2F1)
it follows that on Mwe have
h(O⇥F),nidA=(D2F3D3F2)n1dA+(D3F1D1F3)n2dA+(D1F2D2F1)n3dA
According to Theorem-18, on @Mwe have
n1dA=dy^dz,
n2dA=dz^dx,
n3dA=dx^dy.
Therefore on Mwe have
h(O⇥F),nidA
=(D2F3D3F2)dy^dz+(D3F1D1F3)dz^dx+(D1F2D2F1)dx^dy
=d!.See example III(2) of Unit 2
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On the other hand, since ds(T)=1 ,o n @Mwe have
T1ds=dx,
T2ds=dy,
T3ds=dz.
Therefore on @Mwe have
hF,Tids=FlT1ds+F2T2ds+F3T3ds=Fldx+F2dy+F3dz=!
We have the Stoke’s theorem on Manifolds as
IfMis a compact oriented kdimensional manifold with boundary and
!is a ( k1)form on M, then
Z
Md!=Z
@M!.
Thus using values of !andd!in the above theorem, we obtain
Z
Mh(O⇥F),nidA=Z
@MhF,Tids.
6.3 Applications of classical theorem
Example 1: State and verify Green’s Theorem in the plane forH
(3x2-8y2)dx+( 4 y6xy)dywhere Cis boundary of the region
bounded by x0,y0a n d2 x3y=6.
Solution: Here closed curve C consists of straight lines OB, BA and
AO, where coordinates of A and B are (3, 0) and (0, -2) respectively.
Let R be the region bounded by C.
Then by Green’s Theorem in plane, we have,
H
(3x2-8y2)dx+(4y6xy)dy=RR
R[@
@x(4y6xy)-@
@y(3x2-8y2)]dxdy........(1)
=RR
R(6y+1 6 y)dxdy
=RR
R(10y)dxdy
=1 0R3
0dxR0
1
3(2x6)ydy
=1 0R3
0dx=- 2 0
Now we evaluate L.H.S. of (1) along OB, BA and AO.
Along OB, x=0 , dx=0 a n d yvaries from 0 to -2.
Along BA, x=1
2(6 + 3 y),dx=3
2dy and yvaries -2 to 0.
and along AO, y=0 , dy=0a n d xvaries from 3 to 0
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L.H.S of (1) =H
(3x2-8y2)dx+( 4y6xy)dy
=R
OB(3x2-8y2)dx+(4y6xy)dy+R
BA(3x2-8y2)dx+(4y6xy)dy+R
AO(3x2
-8y2)dx+( 4y6xy)dy
=R2
04ydy+R0
2[9
8(6 + 3 y)2-1 2y2+4y-18y-9y2]dy+R0
33x2dx
=[ 2y2]2
0+R0
2[9
8(6 + 3 y)2-1 2y2+4y-18y-9y2]dy+[x3]0
3
=[ 2 ( 4 ) ]+R0
2[9
8(6 + 3 y)2-2 1y2-14y]dy+[ 0 - 2 7 ]
=- 1 9 + 2 7 - 5 6 + 2 8
=- 2 0
with help of (2) and (3), we find that (1) is true and so Green’s Theorem
is verified.
Example 2: Verify Stoke’s theorem for the vector field ~F=( 2xy)ˆi
-yz2ˆj-y2zˆkover the upper half of the surface x2+y2+z2=1 bounded
by its projection on xy-plane.
Solution: Let S be the upper half of the surface x2+y2+z2=1. The
boundary CorSis a circle in the xy plane of radius unity and centre O.
The equation of C are x2+y2=1 , z=0
whose parametric form is x=cos(t),y=sin(t),z=0 ,0 C~F·~dr=R
C[(2xy)ˆi-yz2ˆj-y2zˆk]·[dxˆi+dyˆj+dzˆk]
=R
C[(2xy)dx-yz2dy-y2zdz]
=R
C[(2xy)dxsince on C,z=0a n d2 z=0
=R2⇡
0[2cos(t)sin(t)]dx
dtdt
=R2⇡
0[2cos(t)sin(t)](sin(t))dt
=R2⇡
0[sin(2t)-sin2(t)]dt
=R2⇡
0[sin(2t)+1cos(2t)
2]dt
=[cos(2t)
2+t
2-sin(2t)
4]2⇡
0
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=1
2+⇡-1
2=⇡...........(1)
Consider,
Curl ~F=ˆi ˆj ˆk
@
@x@
@y@
@z
2xyyz2zy2=( - 2 yz+2yz)ˆi+(0-0) ˆj+(0+1) ˆk=ˆk
Curl ~F·ˆn=ˆk·ˆn=ˆn·ˆkRR
SCurl ~F·ˆnds =RR
Sˆn·ˆkds=RR
Rˆn·ˆkdx
ˆndy
ˆkwhere Ris the projection of Sonxy-plane.
=R1
1Rp
1x2
p
1x2dxdy
=R1
12p
1x2dx
=4R1
0p
1x2dx
=4 [x
2p
1x2+1
2sin1(x)]1
0
=4 [1
2][⇡
2]
=⇡
From (1) and (2), we have,R
C~F·~dr=C u r l ~F·ˆndswhich is the stoke’s theorem.
Example 3: Verify the divergence theorem for the function ~F=2x2yˆi-
y2ˆj+4xz2ˆktaken over the region in the first octant bounded by y2+z2
=9a n d x=2 .
Solution:RRR
Vr·~FdV =RRR
(@
@xˆi+@
@yˆj+@
@zˆk)·(2x2yˆi-y2ˆj+4xz2ˆk)dV
=RRR
(4xy2y+8xz)dxdydz
=R2
0dxR3
0dyRp
9y2
0(4xy2y+8xz)dz
=R2
0dxR3
0dy[(4xyz2yz+4xz2)]p
9y2
0
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=R2
0dxR3
0[(4xyp
9y22yp
9y2+4x(9y2)]dy
=R2
0dx[4x
22
3(9y2)3
2+2
3(9y2)3
2+3 6 xy4xy3
3]
=R2
0(0 + 0 + 108 x36x+3 6 x18)dx
=R2
0(108x18)dx
=2 1 6 - 3 6
=1 8 0
Here,RR
S~F·ˆnd s =RR
OABC~F·ˆnd s +RR
OCE~F·ˆnd s +RR
OADE~F·
ˆnd s +RR
ABD~F·ˆnd s +RR
BDEC~F·ˆnd s
Consider,RR
BDEC~F·ˆnd s=RR
BDEC(2x2yˆi-y2ˆj+4xz2ˆk)·ˆnd s.........................(1)
Normal vector
=r=(@
@xˆi+@
@yˆj+@
@zˆk)(y2+z29) = 2y ˆj+2 z ˆk
Unit normal vector = ˆ n=2yˆj+2zˆkp
4y2+4z2=yˆj+zˆkp
y2+z2=yˆj+zˆkp
9=
yˆj+zˆk
3
From (1),
RR
BDEC(2x2yˆi-y2ˆj+4xz2ˆk)·yˆj+zˆk
3ds=1
3RR
BDEC(-y3+4xz3)ds
=1
3RR
BDEC(-y3+4xz3)dxdy
z
3
=R2
0dxR3
0(-y3
z+4xz2)dy
by substitution, y = 3sin( ✓), z = 3cos( ✓))
=R2
0dxR⇡
2
0[27sin3(✓)
3cos(✓)+4x(9cos2✓)]
=R2
0dx((-27)(2
3)+108 x(2
3))
=R2
0(-18+72 x)dx
=1 0 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 2 )
Consider,RR
OABC(2x2yˆi-y2ˆj+4xz2ˆk)·(-ˆk)ds
=RR
OABC4xz2= 0.................(3) because in OABC xy-plane, z=0
Consider,RR
OADE(2x2yˆi-y2ˆj+4xz2ˆk)·(-ˆj)ds
=RR
OADEy2ds= 0.................(4) because in OADE xz-plane, y=0
Consider,RR
OCE(2x2yˆi-y2ˆj+4 x z2ˆk)·(-ˆi)ds
=RR
OCE-2x2yds=0 . . . . . . . . . . . . . . . . . ( 5 )b e c a u s ei nO C E yz-plane, x=0
Consider,RR
ABD(2x2yˆi-y2ˆj+4xz2ˆk)·(ˆi)ds
=RR
ABD2x2yds
=RR
ABD2x2ydydz
=R3
0dzRp
9z2
02(2)2ydybecause in ABD plane, x=2
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=8R3
0dz[y2
2]p
9z2
0
=4R3
0dz(9 -z2)
=4 [ 9 z-z3
3]3
0
=4 [ 2 7 - 9 ]
=7 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 6 )
on adding (2), (3), (4), (5) and (6), we getRR
S~F·ˆnds = 108 + 0 + 0 + 0 + 72 = 180......(7)
from (1) to (7), we have,RRR
Vr·~FdV =RR
S~F·ˆnd s
Hence the theorem is verified.
Example 4: EvaluateRR
S~A·ˆnd swhere ~A=1 8 zˆi-1 2ˆj+3yˆkandS
is the part of the plane 2 x+3y+6z=1 2i n c l u d e di nt h efi r s to c t a n t .
Solution: Here ~A=1 8 zˆi-1 2ˆj+3yˆk
Given surface f(x, y, z )=2 x+3y+6z-1 2
Normal vector = rf=(@
@xˆi+@
@yˆj+@
@zˆk)(2x+3y+6z-1 2 )=2 ˆi+3ˆj+6ˆk
ˆn=u n i tn o r m a lv e c t o ra ta n yp o i n t( x, y, z )o f2 x+3y+6z=1 2
=2ˆi+3ˆj+6ˆkp4+9+1 6=1
7(2ˆi+3ˆj+6ˆk)
anddS=dxdy
ˆn·ˆk=dxdy
1
7(2ˆi+3ˆj+6ˆk)·ˆk=dxdy
6
7=7
6dxdy
Consider,RR
S~A·ˆnd s =RR
(18zˆi-1 2ˆj+3yˆk)1
7(2ˆi+3ˆj+6ˆk)7
6dxdy
=RR
(36z36 + 18 y)dxdy
6
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=RR
(6z6+3 y)dxdy
putting the value of 6 z=1 2 2x3y, we get,
=R6
0R1
3(122x)
0(12-2 x-3y-6+3 y)dxdy
=R6
0R1
3(122x)
0(62x)dxdy
=R6
0(62x)dxR1
3(122x)
0dy
=R6
0(62x)dx(y)1
3(122x)
0
=R6
0(62x)1
3(122x)dx
=1
3R6
0(4x236x+7 2 ) dx
=1
3[4x3
318x2+7 2 x]6
0
=72
3[4-9+6]
=2 4
Example 5: Show thatRR
S~F·ˆnd s =3
2, where ~F=4xzˆi-y2ˆj+yzˆk
andSis the surface of the cube bounded by the planes x=0 , x=1 ,
y=0 , y=1 , z=0a n d z=1 .
Solution:RR
S~F·ˆnd s
=RR
OABC~F·ˆnd s +RR
DEFG~F·ˆnd s+RR
OAGF~F·ˆnd s+RR
BCED~F·ˆn
ds+RR
ABDG~F·ˆnd s+RR
OCEF~F·ˆnd s.........(1)
Consider,RR
OABC~F·ˆnd s
=RR
OABC(4xzˆi-y2ˆj+yzˆk)(-ˆk)dxdy
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=R1
0R1
0(yz)dxdy
=0( a s z=0 )
Consider,RR
DEFG~F·ˆnd s
=RR
DEFG(4xzˆi-y2ˆj+y z ˆk)·(ˆk)dxdy
=RR
DEFGyzdxdy
=R1
0R1
0y(1)dxdy
=R1
0dx[y2
2]1
0
=1
2
Consider,RR
OAGF~F·ˆnd s
=RR
OAGF(4xzˆi-y2ˆj+yzˆk)·(-ˆj)dxdz
=0
Consider,RR
BCED~F·ˆnd s =RR
BCED(4xzˆi-y2ˆj+y z ˆk)·(ˆj)dxdz
=RR
BCED(-y2)dxdz
=R1
0R1
0(-1)dxdz......(as y=1 )
=- 1
Consider,RR
ABDG~F·ˆnd s
=RR
ABDG(4xzˆi-y2ˆj+y z ˆk)·(ˆi)dydz
=RR
4xzdydz =R1
0R1
04(1) zdydz ......(as x=1 )
=2
Consider,RR
OCEF~F·ˆnds =RR
OCEF(4xzˆi-y2ˆj+y z ˆk)·(-ˆi)dydz
=R1
0R1
0-4xzdydz ......(as x=0 )
=0
putting all values in equation (1),RR
S~F·ˆnd s =3
2.
Example 6: Using Green’s theorem, evaluateR
C(x2yd x +x2dy)
where Cis the boundary described counter clockwise of the triangle
with vertices (0,0), (1,0) and (1,1).
Solution: By Green’s theorem, we have,R
C(x2yd x +x2dy)=RR
R(2x-x2)d x d y
=R1
0(2x-x2)dxRx
0dy
=R1
0(2x-x2)dx[y]x
0
=R1
0(2x-x2)(x)dx
=5
12
Example 7: EvaluateH
C-y
x2+y2dx+x
x2+y2dywhere C=C1[
C2with C1:x2+y2=1a n d C2: x = 2, -2 and y=2 ,- 2 .
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Solution: ConsiderH
C-y
x2+y2dx+x
x2+y2dy
=RR@
@xx
(x2+y2)+@
@yy
(x2+y2)dxdy
=RR(x2+y2)12x(x)
(x2+y2)2+(x2+y2)12y(y)
(x2+y2)2dxdy
=RRx2+y22x2
(x2+y2)2+x2+y22y2
(x2+y2)2dxdy
=RRy2x2
(x2+y2)2+x2y2
(x2+y2)2dxdy
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=RR0
(x2+y2)2dxdy
=0
Example 8: Directly or by Stoke’s theorem, evaluateRR
Scurl ~v·ˆn
dS,~v=yˆi+zˆj+xˆk,Sis the surface of the paraboloid z=1- x2-y2,
z30a n dˆ nis the unit vector normal to S.
Solution:
r⇥~v=-ˆi-ˆj-ˆk
Obviously, ˆ n=ˆk
(r⇥~v)·ˆn=( - ˆi-ˆj-ˆk)·ˆk=- 1RR
S(r⇥~v)·ˆnd s =RR
S(-1)dxdy =-RR
Sdxdy =-⇡(1)2=-⇡.
6.4 Chapter End Exercise
1. If ~F=2yˆi-3ˆj+x2ˆkandSis the surface of parabolic cylinder
y2=8xin the first octant bounded by the planes y=4a n d z=6
then evaluateRR
S~F·ˆnd S.[ Ans. 132 ]
2. If ~F=( 2x2-3z)ˆi-2 x y ˆj-4 xˆkthen evaluateRRR
Vr⇥~Fd V where
Vis the closed region bounded by planes x=0 , y=0 , z=0a n d
2x+2y+z= 4.[ Ans.8
3(ˆj-ˆk)]
3. EvaluateRRR
V(2x+y)dVwhere Vis the closed region bounded
by the cylinder z=4x2and the planes x=0 , y=0 , y=2
andz=0 .[ Ans.80
3]
4. Either directly or by Green’s theorem, evaluate the line integralR
Cex(cos(y)dxsin(y)dy)w h e r e Cis the rectangle with vertices
(0, 0), ( ⇡,0 ) ,( ⇡,⇡
2)a n d( 0 ,⇡
2).[ Ans.2(1- e⇡)]
5. Use the Green’s theorem in a plane to the evaluate the integralR
C[(2x2-y2)dx+(x2+y2)dy] where Cis the boundary in the xy-
plane of the area enclosed by the x-axis and the semi-circle x2+
y2=1i nt h eu p p e rh a l f xy-plane. [ Ans.4
3]
6. If ~F=3yˆi-xyˆj+yz2ˆkandSis the surface of the parboloid 2 z
=x2+y2bounded by z=2 ,s h o wb yu s i n gS t o k e ’ st h e o r e mt h a tRR
Scurl⇥~F·dS=2 0 ⇡
7. If ~F=(xz)ˆi+(x3+yz)ˆj+3 x y2ˆkand Sis the surface
of the cone z=a-p
x2+y2above the xy-plane, show thatRR
S
curl ~F·dS=3⇡a4
4.
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8. Let M⇢R3be a compact three-dimensional manifold with bound-
ary and nthe unit outward normal on @M. Let Fbe a di ↵eren-
tiable vector field on M. Then show that
ZZZ
M✓@f1
@x+@f2
@y+@f3
@z◆
dV=ZZ
@M(n1f1+n2f2+n3f3)dS.
9. Let M⇢R3be a compact three-dimensional manifold with bound-
ary and nthe unit outward normal on @M. Let Fbe a di ↵eren-
tiable vector field on M. Then show that
Z
MdivFdv=Z
@MhF,nidA.
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