3.1. Tangent space. Let M be a smooth manifold, and let Y ( M )be the ring of smooth functions on M .
Definition: A tangent vector of M at a point a map 5: Y ( M )+ R such that
t(f&= 5 ( f )g ( 4 +f(4 5(g),
E
M is a linear
f,g E Y(W*
T h e tangent vectors form a real vector space, T,(M), under the linear operations
( A t + P7)(f)
+ P7(f),
=
A, P E R 597 E Ta(M),
fEY(M)
T , ( M ) is called the tangent space of M at a. I n sec. 3.3 it will be shown that dim T J M ) = dim M. Lemma I: Let f E T,(M),f E Y ( M ) .Then the number f(f) depends only on the germ off at a. Proof: It has to be shown that ((f) = 0 iff is zero in some neighbourhood U of a. Choose g E Y ( M )so that g(u) = 0 and gf =f. Then
t ( f )= 5 W ) = t ( g ) f W + g ( 4 5 ( f ) = 0.
Q.E.D.
Corollary: I f f is constant in a neighbourhood of a, then [(f) = 0. Proof: Set f ( a ) = h and let h also denote the constant function
M 3 A. Then
5(f)
=
[(A)
= At(1) = 87
At(1 * 1)
=
2A5(1) = 0.
Q.E.D.
111. Tangent Bundle and Differential Forms
88
Example: Let t ct x ( t ) ( t o < t < t l ) be a smooth path on M. For ) setting each t E (to , t l ) we obtain 4 E T z ( t ) ( Mby
That f does in fact belong to T z ( t ) ( Mfollows ) from elementary calculus. I t is called the tangent vector to the path at x ( t ) and we will denote it by x ( t ) . 3.2. The derivative of a smooth map. Let rp: M -+ N be a smooth map. Recall that rp induces a homomorphism I*: Y ( M )+ Y ( N )
given by
f e Y ( N ) , x E Ma
(V*f)(X) = f ( v ( x ) ) ,
Lemma 11: Let 6 E T,(M). Then f 0 rp* E Tw(,)(N),and the correspondence f ct f o rp* defines a linear map from T,(M) to Tor(,)(N).
6
Proof:
0
rp* is a linear map from Y ( N )to R. Moreover,
( 5 v*)(fg) = &*f O
*
(f,g E Y ( N ) )and so 5 rp* 0
v*g) E
=
4(v*f).g(v(a)) +f(P(QN
Tw(,)(N).Clearly f
ct
f
o
*
4(v*g)
rp* is linear.
Q.E.D.
Definition: Let rp: M + N be a smooth map and let a E M. The linear map T , ( M ) + Tv(,)(N)defined by f I+ 5 0 rp* is called the derivative o f v at a. It will be denoted by (drp), , ((drp)&(g)
If
4: N -+
=
4(v*g),
g
E
YW), 4 E Ta(M).
Q is a second smooth map, then
( 4 4 v)), = O
(WW(U)
Moreover, for the identity map
-
L:
O
(dvh
9
Q
E M.
M + M, we have
(4, = LT,(M)
9
Q
EM
In particular, if rp: M + N is a diffeomorphism, then ( W o :
T,(M)
TdiV
are inverse linear isomorphisms.
and
(4J-1)w(a):
Tda)(W
+
Ta(M)
1. Tangent bundle
89
Examples: 1. Let t H x ( t ) (to < t < t l ) be a smooth path on a manifold M . Let f r E (R) be the tangent vector given by
&(f) =
yP(w
!€
9”
Then the tangent vector i ( t ) E T z ( i ) ( M(cf. ) example, sec. 3.1) is given by
“4
= (dx), 5 t
*
2, Let 0 be an open subset of a manifold M and let j : 0 --t M denote the inclusion map. Then
(di),:
TdO)
-
Taw)
is a linear isomorphism for each a E 0. We shall prove this by constructing an inverse map. Fix a E 0 and choose p E Y ( M )so that p = 1 in a neighbourhood of a, and carrp C 0. Define /I: Tu(M)-+T J O ) by
(18(5))(f)= 5(Pf),
!€
w4, 5 E T a w )
That /I is well defined, and inverse to (d& follows easily from Lemma I, sec. 3.1. Using this example, we obtain 3. Let q ~ :M + N be a smooth map which sends a neighbourhood U of a point a E M diffeomorphically onto a neighbourhood V of ~ ( ain) N . Then (d94a: T a w )
is a linear isomorphism.
-
T d a W
3.3. Open subsets of vector spaces. Let E be an n-dimensional real vector space. Let 0 be an open subset of E and let a E 0. We shall define a linear isomorphism A,: E
5 T,(O).
First recall that if 9:0 --t F is a smooth map of 0 into a second vector space F, then the classical derivative of q~ at a is the linear map ~’(aE ) :-+F given by
111. Tangent Bundle and Differential Forms
90
Moreover, in the special case F
=
R we have the product formula
This shows that the linear map
Ch: Y ( 0 )
-+
is a tangent vector of 0 at a . Hence we have a canonical linear map A:,
R given by
E -+ T,(O) given by
If q : 0 -+F is a smooth map into a second vector space, then the diagram
commutes, as follows easily from the ordinary chain rule. Proposition I: isomorphism.
T h e canonical linear map A,: E-+ T,(O) is an
Lemma 111: Let ei(i Then
=
1,
..., a ) be
a basis for E and let j~ Y ( E ) .
c n
f = f ( 4+ i = l hi&
?
where (1) the functions h, E Y ( E ) are given by n
x -a =
hi(x) e, ,
XE
E,
i=l
(2) the functions gi E Y ( E ) satisfy gi(u) = f ' ( u ; ei),
i = 1, ..., n.
91
1. Tangent bundle
Proof:
By the fundamental theorem of calculus we have
=
p ( a
=
+ t ( x - a ) ; x - a ) dt
hi(x) r . f ' ( a 0
i=l
+ t ( x - a ) ; ei) dt.
Thus the lemma follows, with gi(x) = / l f ' ( a 0
+ t(x
-
a ) ; ei) dt,
xE
E.
Q.E.D.
Proof of Proposition I: Consider first the case 0 = E. We show first that A, is surjective. Let f E T,(E) and let f E Y ( E ) . Write
where the hi ,gi satisfy conditions (1) and (2) of Lemma 111. By the corollary to Lemma I, sec. 3.1, f maps the constant function f ( a ) into zero. Thus,
n
\
Since the functions hi are independent off, we can write
Thus A, is surjective. T o show that A, is injective, let f be any linear function in E. Then for h E E U h ) ( f )= f ( 4 .
92
111. Tangent Bundle and Differential Forms
Now suppose A,(h) = 0. Then f E E*.
f ( h ) = 0,
Hence h = 0, and so A, is injective. Finally, let 0 be any open subset of E and let j : 0 + E be the inclusion map. Then j’(a):E --f E is the identity map and so formula (3.1) yields the commutative diagram E
According to Example 2, sec. 3.2, (dj), is a linear isomorphism. Hence so is the map A,: E + T,(O). Q.E.D. Corollary: Let M be a smooth manifold and let a E M . Then dim T,(M)
=
dim M .
Proof: Let ( U , u, 0 ) be a chart for M such that a E U. Using the result of Example 2, see. 3.2, and the Proposition we find dim T a ( M )= dim Ta(U)= dim U
=
dim M.
Q.E.D. Proposition 11: The derivative of a constant map is zero. Conversely, let cp: M -+ N be a smooth map such that (dq), = 0, a E M . Then, if M is connected, cp is a constant map. Proof: Assume that cp is the constant map M g E Y(N), cp*g is the constant function given by (P*g)(X) = g ( h
XE
Hence (Corollary to Lemma I, see. 3.1), for
N . Then, for
M.
4 E T,(M), x E M ,
(dP)rn(t)(d= t(P*E) = 0.
It follows that each (dcp)%= 0.
--+ b E
1. Tangent bundle
Conversely, assume that
q ~ :M
4
93
N is a smooth map satisfying
(dp), = 0, a E M ; and let M be connected. Then, given two points xo E M and x1 E M , there exists a smooth curve f:R --t M such that f (0) = xo and f (1) = x1 (cf. Proposition IX, sec. 1.1I). Consider the map g = q~ 0 f:R -+ N . We have
(4%= (ddm (df),= 0, O
t
E R.
Now using an atlas for N and applying formula (3.1) we see thatg must be constant. I n particular g(0) = g( 1) and so
+
3.4. Example. Let Sn be the unit sphere in an (n 1)-dimensional Euclidean space E. We shall establish a canonical isomorphism between T,(Sn) and the subspace &, the orthogonal complement of u in E. The injection i: Sn 4 E determines a linear map
(di),: T,(Sn)
-+
T,(E).
Combining this map with the linear isomorphism Xi1: T J E ) + E, we obtain a linear map
j,: T,(Sn) + E. We show first that I m j , C a l . Consider the function f E Y ( E ) given by x
f ( ~= ) (x, x),
E E.
i*f is the constant function, 1, on Sn. It follows that whence
5(i*f)
= 0,
f'(a;j,(E)) =
5 E Ta(Sn),
&*f)
= 0.
An elementary computation shows that f ' ( u ; h) = 2 ( ~h), ,