CHAPTER 3: TANGENT SPACE
DAVID GLICKENSTEIN
1. Tangent space We shall de…ne the tangent space in several ways. We …rst try gluing them together. We know vectors in a Euclidean space require a basepoint x U Rn n 2 n and a vector v R : A C1-manifold consists of a number of pieces of R glued together via coordinate2 charts, so we can de…ne all tangents as follows. Consider what happens during a change of parametrization : V U: It will take a vector v to d (v) : This motivates the following: !
glue n n De…nition 1. T M = (Ui R ) = where for (x; v) Ui R ; (y; w) Uj i 2 2 n 1 1 R we have (x; v) (y; w)Fif and only i¤ y = j (x) and w = d j (v) : i i x The nice thing about this de…nition is it puts things together and gives the glue vectors in a good way. We de…ne the tangent space at a point p M as Tp M = n glue 2 [p; v]: v R : It is easy to see that Tp M is an n-dimensional vector space. It f 2 g is also easy to see that there is a map : T glue M M de…ned by ([p; v]) = p (since the parts of M are really equivalence classes! modulo equivalence. It also glue makes it clear that T M is a C1 manifold. We can de…ne tangent spaces in two other ways.
path De…nition 2. Tp M = paths :( "; ") M such that (0) = p = where f ! g path if ( i )0 (0) = ( i )0 (0) for every i such that p Ui:T M = path 2 Tp M: p M 2 F This is a more geometric de…nition. Note that there is a map : T path M M de…ned by ( ) = (0) : ! We shall show that T path M and T glue M are equivalent. The maps are : T path M T glue M p ! p de…ned by
([ ]) = i (0) ; ( i )0 (0) : The inverse map is : T glue M T path M ! de…ned by 1 ([ i (p) ; v]) = t ( i (p) + tv) : ! i It is clear that if well de…ned, they are inverses of each other. We need to show that and are well-de…ned. Clearly is well de…ned because i (0) =
Date: September 29, 2010. 1 2 DAVIDGLICKENSTEIN
i (0) ; ( i )0 (0) = ( i )0 (0) for any [ ] : Also for any ( j (p) ; w) 1 2 2 [ i (p) ; v] must satisfy d i j v = w: Notice that j (p) 1 0 1 j i ( i (p) + tv) (0) = d j i v = w = ( j (p) + tw)0 (0) : i(p) The third way is in terms of germs of functions. A germ of a function is an equivalence class of functions.
De…nition 3. Germsp is the set of functions f C1 (Uf ) for p Uf M modulo 2 2 the equivalence that [f] = [g] i¤ f (x) = g (x) for all x Uf Ug: Note that Germsp are an algebra since [f] + [g] = [f + g] is well-de…ned,2 etc. \
De…nition 4. A derivation of germs is an R-linear map X :Germsp R which satis…es ! X (fg) = f (p) X (g) + X (f) g (p) :
der De…nition 5. We de…ne Tp M to be the set of derivations of germs at p:
der Proposition 6. Alternately, we may de…ne the Tp M to be the set of derivations of smooth functions at p:
Proof. Suppose X : C1 (M) R is a derivation at p: Then it determines a deriva- tion of germs in the obvious! way. Conversely, suppose [f] is a germ at p: Then there is a representative f : U R, and within that open set is a coordinate ball B ! centered at p: Taking a smaller ball, we have a compact (closed) coordinate ball B0 around p within the domain U of f: We can consider the function x b (x) f (x) ; ! where b is a smooth bump function supported in U that is one on the ball B0. These
This de…nition is nice because it shows how tangent vectors act on functions. We note derivations are a vector space since (X + Y )(fg) = X (f) g (p) + f (p) X (g) + Y (f) g (p) + f (p) Y (g) = (X + Y )(f) g (p) + f (p)(X + Y )(g) :
n @ A good example of a germ on U R is i since @x p @ @f @g (fg) = (p) g (p) + f (p) (p) : @xi @xi @xi p
@ j j These are linearly independent since @xi p x = Ii : We see that X (1) = 1 X (1) + X (1) 1 so X (1) = 0: Similarly,
X xi pi xj pj = 0: So by Taylor series: @f f (x) = f (p) + xi pi + O x p 2 : @xi j j p
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@ der We have formally that @xi p span Tp U: To make this argument rigorous, we know that
1 d f (x) = f (p) + f (tx + (1 t) p) dt dt Z0 1 @f = f (p) + xi pi dt: @xi Z0 tx+(1 t)p
Hence if we apply a derivation X we have 1 @f 1 @f X (f) = dt X xi pi + X dt pi pi @xi @xi Z0 p Z0 tx+(1 t)p ! @f = X xi pi : @xi p Hence for U Rn we have a correspondence der n Tp U R ! given by X X x1 p1 ;:::;X (xn pn) ! which is an invertible linear map with inverse
n der R Tp U ! @f s1; : : : ; sn X (f) = si : ! @xi p !
On a manifold, we de…ne @ @ f = (f i) @xi @xi p i(p)
1 n for coordinates x ; : : : ; x = i (p) : Notice that under a change of coordinates from y1; : : : ; yn = (p) we have that j @ @ = (f i) @xk @xk i(p)
@ 1 = k f j i i @x 1 (p) j i i @y` @ = (f j) @xk @y` i(p) j (p)
Also, we have the projection : T derM M: !
n @ Proposition 7. Let M = R : The derivations @xi p form a basis for the deriva- tions at p:
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Proof. We …rst see that X (c) = 0 if c is a constant function. By linearity of the derivation, we need only show that X (1) = 0: We compute: X (1) = X (1 1) = 1 X (1) + X (1) 1 = 2X (1) : We conclude that X (1) = 0: Now, let X be a derivation and f a smooth function. We can write f as 1 d f (x) = f (p) + f (tx + (1 t) p) dt dt Z0 1 @f = f (p) + xi pi dt: @xi Z0 tx+(1 t)p
By linearity and the derivation property, we have 1 @f X (f) = X (f (p)) + X xi pi dt @xi Z0 tx+(1 t)p ! 1 @f 1 @f = 0 + dt X xi pi + X dt pi pi @xi @xi Z0 tp+(1 t)p ! Z0 tx+(1 t)p ! @f = X xi pi : @xi p So, X xi p i are just some numbers, and so we see that X is a linear combination of @ ; meaning that these span the space of derivations! Since it is clear that @xi p @ and @ are linearly independent for each i = j (consider the functions @xi p @xj p i i 6 x p ), the result follows. De…nition 8. Given any smooth map F : M N; there is a push forward F : ! TpM T M given as follows: ! F (p) F path [ ] = [F ] F der X f = X (f F ) : De…nition 9. In any coordinate neighborhood (U; ) of p, we de…ne the derivation @ @xk p by @ @ = 1 @xk @xk p (p)
der path We may now see that Tp M is isomorphic to Tp M: The map is d [ ] f f ( (t)) : ! ! dt t=0
We note that
1 j d @ f i d ( i ) f ( (t)) = j dt t=0 @x dt i (0) 0
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1 n and hence it is well-de…ned up to equivalence of paths. Note that i (p + tek) k=1 form a basis for [ ] and map to @ so this is a linear isometry. @xk p We will now use whichever de…nition we wish. Also note the following:
Proposition 10. If p U M is an open set, then 2 TpM = TpU: Therefore, we will not make a distinction.
2. Computation in coordinates m @ Let’s compute the push-forward in coordinates. Recall that k is a @x p k=1 n @ n o basis for TpM: Now, suppose that @ya is a basis for TF (p) N: Given a F (p) a=1 smooth map F : M N; we should be able to compute the push forward in ! coordinates. If X TpM; we can write it in terms of the basis, 2 @ X = Xk @xk p
k for some numbers X R. To compute the push forward, which is a linear map, we have that 2 @ F X = XkF : @xk p m n @ First, let’s suppose M = R and N = R : To compute F k ; for f C1 (N) @x p 2 we need to compute
@ @ F f = (f F ) @xk @xk p! p
@f @ya = @ya @xk F (p) p
(note the summation) where, in the second expression, we really mean @ya @ya (F (x)) @F a = = @xk @xk @xk p p p
1 n if F = F ;:::;F is written in y-coordinates. Notice that once we have speci…ed the coordinates, we have an expression for F in terms of the di¤erential. Now suppose we are on a manifold, then
@ 1 1 @ F = F : @xk @xk p! ! m The middle map is known to us, as it is the di¤erential of a map between R and Rn; that is ^a 1 @F F = k ( (p)) @x !k=1;:::;m a=1;:::;n 6 DAVIDGLICKENSTEIN
1 where F^ = F : In particular, we get @ @F^a @ F = ( (p)) @xk @xk @ya p! F (p)
One can also consider change of coordinates. If (U; ) and (V; ) are coordinate charts with coordinates xi and x~i ; then any tangent vector can be written as @ @ X = X i = X~ i : @xi @x~i p p
How are Xi and X~ i related? We can compute:
@ @ X~ i = X~ i 1 @x~i @x~i p (p)
i 1 1 @ = X~ @x~i (p)
~ i 1 1 @ = X i @x~ (p) 1 k @ @ = X~ i 1 ( (p)) @x~i @xk " (p)#
1 k @ @ = X~ i ( (p)) @x~i @xk p and so 1 k @ Xk = X~ i ( (p)) : @x~i 1 Example 1. Calculate the di¤erential of the map F : C2 (0; 0) CP . n f g !