CHAPTER 3: TANGENT SPACE

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1 CHAPTER 3: TANGENT SPACE DAVID GLICKENSTEIN 1. Tangent sace We shall de ne the tangent sace in several ways. We rst try gluing them together. We know vectors in a Euclidean sace require a baseoint x 2 U R n and a vector v 2 R n : A C 1 -manifold consists of a number of ieces of R n glued together via coordinate charts, so we can de ne all tangents as follows. Consider what haens during a change of arametrization : V U: It will take a vector v to d (v) : This motivates the following: De nition 1. T glue M = (U i R n ) = where for (x; v) 2 U i R n ; (y; w) 2 U j i R n we have (x; v) (y; w) if and only i y = j 1 i (x) and w = d j 1 i (v) : x The nice thing about this de nition is it uts things together and gives the vectors in a good way. We de ne the tangent sace at a oint 2 M as T glue M = f[; v] : v 2 R n g : It is easy to see that T glue M is an n-dimensional vector sace. It is also easy to see that there is a ma : T glue M M de ned by ([; v]) = (since the arts of M are really equivalence classes modulo equivalence. It also makes it clear that T glue M is a C 1 manifold. We can de ne tangent saces in two other ways. De nition 2. T ath M = faths : ( "; ") M such that () = g = where if ( i ) () = ( i ) () for every i such that 2 U i : T ath M = T ath M: 2M This is a more geometric de nition. Note that there is a ma : T ath M M de ned by () = () : We shall show that T ath M and T glue M are equivalent. The mas are de ned by The inverse ma is : T ath M T glue M ([]) = i () ; ( i ) () : : T glue M T ath M de ned by ([ i () ; v]) = t 1 i ( i () + tv) : 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 () = Date: Setember 29, 21. 1

2 2 DAVID GLICKENSTEIN i () ; ( i ) () = ( i ) () for any 2 [] : Also for any ( j () ; w) 2 [ i () ; v] must satisfy d i 1 j v = w: Notice that j() j 1 i ( i () + tv) () = d j 1 i v = w = ( i() j () + tw) () : The third way is in terms of germs of functions. equivalence class of functions. A germ of a function is an De nition 3. Germs is the set of functions f 2 C 1 (U f ) for 2 U f M modulo the equivalence that [f] = [g] i f (x) = g (x) for all x 2 U f \ U g : Note that Germs are an algebra since [f] + [g] = [f + g] is well-de ned, etc. De nition 4. A derivation of germs is an R-linear ma X :Germs R which satis es X (fg) = f () X (g) + X (f) g () : De nition 5. We de ne T der M to be the set of derivations of germs at : Proosition 6. Alternately, we may de ne the T der M to be the set of derivations of smooth functions at : Proof. Suose X : C 1 (M) R is a derivation at : Then it determines a derivation of germs in the obvious way. Conversely, suose [f] is a germ at : Then there is a reresentative f : U R, and within that oen set is a coordinate ball B centered at : Taking a smaller ball, we have a comact (closed) coordinate ball B around within the domain U of f: We can consider the function x b (x) f (x) ; where b is a smooth bum function suorted in U that is one on the ball B. These This de nition is nice because it shows how tangent vectors act on functions. We note derivations are a vector sace since (X + Y ) (fg) = X (f) g () + f () X (g) + Y (f) g () + f () Y (g) = (X + Y ) (f) g () + f () (X + Y ) (g) : A good examle of a germ on U R n is x since i x i (fg) = f g () g () + f () xi x i () : These are linearly indeendent since so X (1) = : Similarly, x x j = I j i i : We see that X (1) = 1 X (1) + X (1) 1 X x i i x j j = : So by Taylor series: f (x) = f () + f x i x i i + O jx j 2 :

3 TANGENT SPACE 3 We have formally that x san T der i U: To make this argument rigorous, we know that d f (x) = f () + f (tx + (1 t) ) dt dt f = f () + x i x i i dt: tx+(1 t) Hence if we aly a derivation X we have f X (f) = x i dt X x i i + X = f x i X x i i : f x i tx+(1 t) dt i i Hence for U R n we have a corresondence given by T der U R n X X x 1 1 ; : : : ; X (x n n ) which is an invertible linear ma with inverse R n T der U s 1 ; : : : ; s n X (f) = f x i s i : On a manifold, we de ne x i f = x i i() (f i ) for coordinates x 1 ; : : : ; x n = i () : Notice that under a change of coordinates from y 1 ; : : : ; y n = j () we have that x k = x k (f i ) i() = x k f j 1 i i j 1 i i() y` = x k (f j ) y` i() j() Also, we have the rojection : T der M M: x i form a basis for the deriva- Proosition 7. Let M = R n : The derivations tions at :

4 4 DAVID GLICKENSTEIN Proof. We rst see that X (c) = if c is a constant function. By linearity of the derivation, we need only show that X (1) = : We comute: X (1) = X (1 1) = 1 X (1) + X (1) 1 = 2X (1) : We conclude that X (1) = : Now, let X be a derivation and f a smooth function. We can write f as f (x) = f () + = f () + d f (tx + (1 dt f t) ) dt x i x i i dt: tx+(1 t) By linearity and the derivation roerty, we have f X (f) = X (f ()) + X x i x i i dt tx+(1 t) f = + x i dt X x i i + X t+(1 t) = f x i X x i i : f x i tx+(1 t) dt i i So, X x i i are just some numbers, and so we see that X is a linear combination of x ; meaning that these san the sace of derivations Since it is clear that i x and i x are linearly indeendent for each i 6= j (consider the functions j x i i ), the result follows. De nition 8. Given any smooth ma : M N; there is a ush forward : T M T () M given as follows: ath [] = [ ] der X f = X (f ) : De nition 9. In any coordinate neighborhood (U; ) of, we de ne the derivation by x k x k = 1 x k () We may now see that T der M is isomorhic to T ath M: The ma is [] f d dt f ( (t)) : t= We note that d dt f ( (t)) = f 1 i t= x j i() d ( i ) j dt

5 TANGENT SPACE 5 and hence it is well-de ned u to equivalence of aths. Note that 1 i ( + te k ) n k=1 form a basis for [] and ma to so this is a linear isometry. x k We will now use whichever de nition we wish. Also note the following: Proosition 1. If 2 U M is an oen set, then T M = T U: Therefore, we will not make a distinction. 2. Comutation in coordinates n Let s comute the ush-forward in coordinates. Recall that basis for T M: Now, suose that x k o m k=1 n y is a basis for T a () N: Given a () a=1 smooth ma : M N; we should be able to comute the ush forward in coordinates. If X 2 T M; we can write it in terms of the basis, X = X k x k for some numbers X k 2 R. To comute the ush forward, which is a linear ma, we have that X = X k x k : irst, let s suose M = R m and N = R n : To comute we need to comute x k f = x k (f ) = f y a y a () x k (note the summation) where, in the second exression, we really mean y a x k = ya ( (x)) x k = a x k is a x k ; for f 2 C 1 (N) if = 1 ; : : : ; n is written in y-coordinates. Notice that once we have seci ed the coordinates, we have an exression for in terms of the di erential. Now suose we are on a manifold, then x k = 1 1 x k : The middle ma is known to us, as it is the di erential of a ma between R m and R n ; that is 1 = ^ a ( ()) xk k=1;:::;m a=1;:::;n

6 6 DAVID GLICKENSTEIN where ^ = 1 : In articular, we get x k = ^ a ( ()) xk y a () One can also consider change of coordinates. If (U; ) and (V; ) are coordinate charts with coordinates x i and ~x i ; then any tangent vector can be written as X = X i x i = X ~ i ~x i : How are X i and X ~ i related? We can comute: ~X i ~x i = X ~ i 1 ~x i () = X ~ i 1 1 ~x i () = X ~ i 1 1 ~x i () " = X ~ i 1 1 k ~x i ( ()) x k and so = ~ X i 1 k ~x i X k = ~ X i 1 k ( ()) x k ~x i ( ()) : () Examle 1. Calculate the di erential of the ma : C 2 n f(; )g CP 1. #