CHAPTER 2: SMOOTH MAPS. 1. Introduction In this chapter we introduce smooth maps between manifolds, and some important

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1 CHAPTER 2: SMOOTH MAPS DAVID GLICKENSTEIN 1. Introduction In this chater we introduce smooth mas between manifolds, and some imortant concets. De nition 1. A function f : M! R k is a smooth function if for each 2 M; there exists a smooth coordinate neighborhood (U; ) of such that f 1 is smooth. More generally, we have the following. De nition 2. A ma F : M! N is a smooth ma if for each oint 2 M there exist smooth coordinate neighborhoods (U; ) of and (V; ) of F () such that F (U) V and F 1 : (U)! R n is smooth. The ma ^F = F 1 is called a coordinate reresentation of F: Basically, we have that F is smooth if and only if it is smooth in its coordinate reresentations. Note that we have to be a bit careful of the condition F (U) V: Consider some examles. Examle 1. We can use olar coordinates on R 2 : The function f (x; y) = x 2 + y 2 is equal to ^f (r; ) = r 2 in a olar coordinate chart (usually one takes a set such as (x; y) 2 R 2 : y > 0 for the coordinate domain, otherwise one could nd that the coordinate chart is not smooth). It follows that f is smooth. Lemma 3. Let F : M! N be a ma. If for each 2 M there exists an oen set U such that F j U is continuous, then F is continuous. If for each 2 M there exists an oen set U such that F j U is smooth, then F is smooth. Proof. Suose that for each 2 M there exists an oen set U such that F j U is continuous. So there is a cover fu g 2M with this roerty. Consider an oen set V N: We note that 1 F j U (V ) = F 1 (V ) \ U so F 1 (V ) = [ 1 F j U (V ) ; which is oen. Suose that for each 2 M there exists an oen set U such that F j U is smooth. Then if 2 U such that F j U is smooth, there is a coordinate neighborhood U 0 U Date: Setember 21,

2 2 DAVID GLICKENSTEIN of and V N of F () such that the coordinate reresentation is smooth. Since U 0 is oen in M; it follows that F is smooth (note that we used F (U 0 ) F (U) V ). Lemma 4. Let M; and N be smooth manifolds and let fu g be an oen cover of M: Suose that for each we are given a smooth ma F : U! N such that F j U\U = F j U\U Then there is a unique smooth ma F : M! N such that F j U = F : Proof. We can de ne F () = F () if 2 U : By the assumtion, this is wellde ned, i.e., if 2 U ; then F () = F () : Since F is smooth in a neighborhood of each oint, it is smooth. Continuous??? Lemma 5. Every smooth ma between manifolds is continuous. Proof. We know that smooth mas between subsets of R n and R m are continuous. Suose F : M! N is smooth. Thus for each oint 2 M there exist smooth coordinate neighborhoods (U; ) of and (V; ) of F () such that F 1 is smooth. If W N is oen, then W \ F (M) is covered by coordinate charts f(v ; )g 2A such at there exist nonemty oen coordinate charts (U ; ) such that F (U ) V and F 1 is smooth. Since W \V is oen, (W \ V ) is oen, and since F = F 1 is smooth, F 1 ( (W \ V )) is oen. However, F 1 ( (W \ V )) = F 1 (W \ V ) ; so F 1 (W \ V ) is oen. Since 1 (W ) is oen, and since W is arbitrary it follows that F is con- we have that F tinuous. F 1 (W ) = [ F 1 (W \ V ) ; 2A De nition 6. A di eomorhism M! N between smooth n-dimensional manifolds is a bijective, smooth ma with smooth inverse. Two manifolds are di eomorhic if there exists a di eomorhism between them. 2. Examles of smooth mas In this section we resent some examles of smooth mas. Examle 2. Consider the inclusion ma S n! R n+1 : We need to check the coordinate reresentations, which are of the form x 1 ; : : : ; x n! x 1 ; : : : ; x n ; q1 jxj 2 ; with domain the oen unit disk. Since the coordinate reresentations are smooth, the function is smooth. Examle 3. Consider the quotient ma R n+1 n f0g! RP n : The coordinate reresentations look like x 0 ; : : : ; x n x 1! x 0 ; : : : ; xn x 0 : If we restrict to where x 0 > 0; we see that this is a smooth ma.

3 SMOOTH MAPS 3 Examle 4. Consider the restriction of the revious ma S n! RP n : The coordinate reresentation is x 1 ; : : : ; x n! x 1 1 jxj ; : : : ;! x n 1 jxj or x 0 ; : : : ; x n 1! x1 x 0 ; : : : ; xn 1 1 x 0 ; x 0! jxj : Examle 5. For a roduct M N; consider the rojection ma : M N! M: This ma is smooth. 3. Partitions of unity Partitions of unity are used to glue together two smooth mas in such a way that the new ma is smooth. Note that one can easily glue together continuous mas to be continuous, but if alied to smooth mas, the new ma is rarely smooth. Consider, for examle, the absolute value function. We need a function that smoothly transitions between the constant function 1 and the constant function 0: To do this, we need a nonzero function with all zero derivatives at a oint. Consider the following function e 1=x if x > 0 f (x) = 0 if x 0 Lemma 7. The function f : R! R described above is smooth. Proof. We need to show that the derivatives all exist at x = 0 and are continuous. Suose we could show that the right-sided derivatives were zero, i.e., lim f (k) (x) = 0 x!0 + for all k 0: Since the derivatives from the left are clearly zero, this is su cient to show that all derivatives are zero. We have lim f (0) (x) = lim e 1=x = 0: x!0 + x!0 + Now we will rove by induction that for x > 0; f (k) (x) = k (x) 1=x e x2k for some olynomial k : This is clearly true for k = 0: Now suosing the formula for k; we get f (k+1) (x) = 0 k (x) x 2k e 1=x 2k k (x) 1=x k (x) 1=x e e x2k+1 x2k+2 = x2 0 k (x) 2kx k (x) k (x) e 1=x ; x 2(k+1) comleting the induction. Now we recall that for any integer k; e 1=x lim x!0 + x k = 0 (this is roved using L Hosital and induction). It follows that lim x!0 + f (k) (x) = 0:

4 4 DAVID GLICKENSTEIN Using f we can construct smooth cuto functions. Lemma 8. There exists a smooth function h : R! R such that h is identically zero on ( 1; 1]; h is identically one on [2; 1); and 0 < h (x) < 1 for h 2 (1; 2) : Proof. One way to construct such a function is to consider h (x) = f (2 x) f (2 x) + f (x 1) : Notice that h (x) 1 and h 0 everywhere. if 2 x 0; i.e., x 2; then h (x) = 0; and if x 1 0; i.e., x 1; then h (x) = 1: Note that the denominator is never zero, since f (2 x) is ositive if x < 2 and f (x 1) is ositive if x > 1: De nition 9. A smooth function as in the lemma is called a cuto function. The other thing we often need is a bum function. De nition 10. Let f : X! R be a continuous function. The suort of f; denoted su f is de ned as the closure of the set of oints where f is nonzero, i.e., su f = fx 2 X : f (x) 6= 0g: The function f is said to be comactly suorted if the suort is a comact set. Lemma 11. There exists a smooth function H : R n! R such that su H B (0; 2) and Hj B(0;1) 1: Proof. We can take H (x) = h (jxj) : Clearly this is smooth away from 0 since it is a comosition of smooth functions. Notice that Hj B(0;1) 1; so it must be smooth at 0 as well. De nition 12. Let X be a toological sace and let S = fs g 2A be a collection of subsets of X. S is said to be locally nite if each oint x 2 X has a neighborhood that intersects at most nitely many sets in S. De nition 13. Let M be a toological sace and let X = fx g 2A be an oen cover of M: A artition of unity subordinate to X is a collection of continuous functions f : M! Rg 2A such that (1) 0 (x) 1 for all 2 A and all x 2 M: (2) su X : (3) The set of suorts fsu g 2A is locally nite. (4) P 2A (x) = 1 for all x 2 M: Note that it is imortant that the set is locally nite so that the sum makes sense (without any notion of series). De nition 14. If are all smooth, then we call the collection a smooth artition of unity. One key fact is that manifolds admit artitions of unity. We have already introduced a number of the concets, but the roblem is ensuring that the suorts are locally nite. This will follow from the fact that manifolds are aracomact. De nition 15. Let U be an oen cover of a toological sace X: A re nement is another oen cover V such that for every V 2 V there exists a U 2 U such that V U: The Hausdor sace X is aracomact if every oen cover has a locally nite re nement.

5 SMOOTH MAPS 5 The key fact is that aracomact follows from second countable, so every manifold is aracomact. Lemma 16. Every toological manifold admits a countable, locally nite cover by recomact oen sets. Proof. Let M be a manifold. Recall that M admits a countable basis of recomact oen sets fb i g 1 i=1 : We will now construct a cover U = fu ig 1 i=1 such that (1) U i is comact. (2) U i 1 U i if i 2: (3) B i U i : Once we have such a cover, we see that the set V = 1 V i = U i+2 n U i i=1 [fu 1; U 2 g is the aroriate cover. Since V i is a closed subset of U i+2 ; which is comact, it follows that V i are recomact. Also, we have that V i \V j is emty unless ji jj 1; so V is locally nite. It is clear that [ V = [ U; V 2V so V is a cover. To construct U, we roceed inductively. Let U 1 = B 1 : Then U 1 satis es all three roerties. Inductively, let s suose we have U 1 ; : : : ; U k 1 (k 2 ) satisfying all three roerties. Note that since U k 1 is recomact, there is a nite number m k such that U k 1 B 1 [ B 2 [ [ B mk : Set U k to be This will satisfy all the axioms. U k = U2U max(m k ;k) [ i=1 B i : Now a slightly stronger version of aracomactness. De nition 17. An oen cover U is regular if (1) U is countable and locally nite. (2) Each U i 2 U is the domain of a smooth coordinate ma i : U i! R n whose image is B (0; 3) R n : (3) The (countable) collection fv i g still covers M; where V i = 1 i (B (0; 1)) : Lemma 18. Let M be a manifold. Then every oen cover admits a regular re nement. In articular, M is aracomact. Proof. Let X be an oen cover of M and let fv j g be a countable, locally nite cover of M by recomact sets guaranteed by Lemma 16. For each 2 M; there exists an oen neighborhood W 0 of that intersects only nitely many V j. Denote by V the set of V j containing : Now let 0 1 W = W 0 \ A : V 2V V Note that this is an oen set, since V is a nite set. It follows that if 2 V i ; then W V i : Since 2 X for some X 2 X, for each we can consider W 00 W \X

6 6 DAVID GLICKENSTEIN such that where W 00 ; is a coordinate ball centered at ; choosing such that : W 00! B (0; 3) : Now let U = 1 (B (0; 1)) : The sets U cover M, and the sets W 00 form a re nement of X with the correct coordinate ma roerty. We need to restrict to a countable subcover and show that it is locally nite. De ne Notice that the set U : 2 V i forms a cover of V i ; and since V i is comact, there is a nite subcover n U i;1 ; : : : ; U i;k(i) o. This subcover corresonds to a collection of W 00 that we denote Wi;1 00 ; : : : ; W i;k(i) 00 : We can now take the union of these over all i 2 N, and call it U. This is countable (since it is the countable union of nite sets), a re nement of X, and has the aroriate coordinate ma conditions. We still need to show that U is locally nite. Notice that if Wi;k 00 \ W i 00 0 ;k0; then since there are j; j 0 such that Wi;k 00 V j and Wi 00 0 ;k V 0 j 0; we must have V j \ V j 0; V i \ V j ; and V i 0 \ V j 0 all nonemty. If we x i; there are only nitely many i 0 that satisfy this since fv j g is locally nite. It follows that U is locally nite. (Note: this argument is from the errata on J. Lee s website.) Theorem 19. If M is a smooth manifold and X = fx g 2A is any oen cover of M; there exists a smooth artition of unity subordinate to X. Proof. Let fw i g be a regular re nement of X, and let i : W i! B (0; 3) be the corresonding coordinate mas. By Lemma 11, there is a smooth function H that is 1 on B (0; 1) and zero outside B (0; 2) : For every i 2 N, let f i : M! R be the function H (i ()) if 2 W f i () = i 0 else These are clearly smooth and equal to one on U i = 1 i (B (0; 1)) : Now de ne g i () = f i () Pi f i () : The sum in the bottom makes sense because fw i g is locally nite, so the denominator is a nite sum. Furthermore, since every is in some U i ; the denominator is never zero. Note that P g i () = 1 for all 2 M: Finally, we need to re-index so that we are indexed by the set A: Every W i is contained in some X ; so there is a function a : N!A such that a (i) is the aroriate : It may be that more than one i corresonds to the same ; so we need to sum: () = X g i () i2a (if a 1 () =?; then the sum is zero). Again, there can be only nitely many nonzero g i (q) in a neighborhood of ; so the sum is nite and the function is smooth. It also follows that the suorts are locally nite for the same reason, thus f g is the artition of unity. It will follow that bum functions exist. 1 ()

7 SMOOTH MAPS 7 De nition 20. If A M is a closed subset and U M is an oen set containing A; a continuous function : M! R is called a bum function for A suorted in U if 0 (x) 1 on M; 1 on A and su U: Proosition 21. Let M be a smooth manifold. For any closed set A M and any oen set U containing A; there exists a smooth bum function for A suorted in U: Proof. Consider the cover fu; M n Ag of M: There is a smooth artition of unity f 0 ; 1 g subordinate to this cover. It is clear that 1 0 on A; so 0 1 on A since : Recall the de nition of a smooth function on a closed set A: De nition 22. Let A M be a closed set. A function F : A! N is smooth if for every 2 A there is a neighborhood W of and a smooth function ~ F : W! N such that ~ F W \A = F j W \A : Lemma 23. Let M be a smooth manifold, and let A M be a closed subset, and let f : A! R k be a smooth function. For any oen set U containing A; there exists a smooth function ~ f : M! R k such that ~ f A = f and su ~ f U: One might be temed to use the smooth bum function b and look at bf: The roblem is that we do not know there is a smooth extenion of f to U; so we cannot get a smooth function this way. The roof is slightly more involved. Proof. We know that for each ; there exists a neighborhood W and function ~f : W! R k with the aroriate roerties. We can relace W with W \ U to ensure W U: The set fw g 2A [ fm n Ag forms an oen cover, and we can nd a artition of unity f g [ f 0 g subordinate to it. We then de ne ~f (x) = X 2M (x) ~ f (x) : Since nsu f ~ o is locally nite, the sum is nite. Since su f ~ W U; we have that su f ~ U and f ~ is smooth. Finally, for x 2 A; we have ~f (x) = 0 (x) f (x) + X (x) ~ h f (x) = 0 (x) + X i (x) f (x) = f (x) : 2M De nition 24. Let M be a toological sace. An exhaustion function for M is a continuous function f : M! R such that M c = fx 2 M : f (x) cg is comact for each c 2 R. Some oular exhaustion functions are f (x) = jxj 2 for R n and f (x) = 1 1 jxj 2 for the oen ball of radius 1: Proosition 25. Every smooth manifold admits a smooth ositive exhaustion function.

8 8 DAVID GLICKENSTEIN Proof. Let fv j g j2n be a countable oen cover of a manifold M by recomact sets and let f j g j2n be a smooth artition of unity subordinate to fv j g : De ne f : M! R by 1X f (x) = j j (x) : j=1 This sum is nite (for each x) since the suorts are locally nite, and the function is smooth. It is ositive 1X f (x) j (x) = 1: j=1