Differential Geometry - Solution 1 Yossi Arjevani k Proposition (1). Let f = (f1; : : : ; fk): M ! R be a submerssion. Then for a given point p 2 M, the differentials dfi are independent. Proof Let (U; V; ) be a chart for p, and define φ := f ◦ , and φi = πi ◦ φ, where π is the projection map on the i’th coordinate. Then we have dφ = df ◦ d , and since f is a submerssion and is a diffeomorphism, it follows that φ is n k also a submerssion. Hence, since φ : R ! R , it follows that rank(dφ) = k. Furthermore, dφi = dπi ◦ dφ = eidφ, k and since the k columns of dφ are independent, we have that dφi are independent. Now, assume there exists α 2 R Pk such that i=1 αidfi = 0, then since d is linear, it follows that k ! k k X X X αidfi ◦ d = αidfi ◦ d = αidφi = 0; i=1 i=1 i=1 a contradiction! Proposition. (2) Let M and N be smooth manifolds. Introduce a natural smooth structure for M × N Proof First, endow M × N with the standard topological product between space. Clearly, M × N is Hausdorff and second countable. To define a smooth structure, we use the following charts: Given two charts (Ui;Vi; φi); i = 1; 2 for M and N, we let (U1 × U2;V1 × V2; φ1 × φ2) to be a chart for M × N. Our goal now is to show that these charts form a smooth atlas. Clearly, this set of charts covers M × N and φ1 × φ2 : U1 × U2 ! V1 × V2 are homeomorphism. So we need to show that the transition maps are smooth. But this follows directly by the fact that for φ1 × φ2 and 1 × 2, we have −1 −1 −1 (φ1; φ2) ◦ ( 1; 2) = (φ1 ◦ 1; φ2 ◦ 2): Proposition (3). Define a natural smooth structure on RPn. Proof First, we endow RPn with the standard quotient topology induced by Rn+1nf0g which is known to be Haus- dorff and second-countable. That is, we define π : Rn+1nf0g ! RPn by π(x) = [x] where [·] denotes the equivalence class under x ∼ y () x = λy for some λ 6= 0. We define the following atlas: for any i 2 [n] we define n Ui := R ;Vi := f[(x1; : : : ; xi−1; 1; xi; : : : ; xn]g; φi(x1; : : : ; xn) := π((x1; : : : ; xi; 1; xi+1; : : : ; xn)) −1 n+1 −1 Note that Vi are indeed open as π (Vi) = fx 2 R j xi 6= 0g. Also, note that for φi : Vi ! Ui : −1 −1 ([(x1; : : : ; xi; 1; xi+1; : : : ; xn]) 7! (x1; : : : ; xn), we have φi ◦ φi = φi ◦ φi = Id and therefore φ is homeomor- phism (both maps are continuous as composition of continuous functions). It remains to show that φi are smoothly compatible. Let i > j then −1 −1 φj φi(x1; : : : ; xn) = φj [(x1; : : : ; xi; 1; xi+1; : : : ; xn)] −1 x1 xi 1 xi+1 xn = φj [( ;:::; ; ; ;:::; )] xj xj xj xj xj 1 −1 n which is a diffeomorphism over φi (Vi \ Vj) = fx 2 R j xj 6= 0g. Proposition (4). Show that π : Sn ! RPn defined as in the previous questions if 2-to-1 local diffeomorphism. Proof First note that π(x) = π(−x). Also, if π(x) = π(y) then x = λy for some λ 6= 0, and since x; y 2 Sn, it follows that λ = ±1, thus π is 2-to-1. To show that π is a local diffeomorphism, we use the smooth structure 2 2 ± for RP shown above, and the standard smooth structure for S defined as follows. For i = 1; 2, we define Ui := 2 2 ± 2 2 2 f(x; y) j x + y < 1g;Vi := f(x1; x2; x3) j x1 + x2 + x3 = 1; ±xi > 0g, and ± ± ± 2 p 2 2 2 i : Ui ! Vi :(x; y) 7! (x ; ± 1 − x − y ; y ): | {z } i0th coordinate For simplicity, assume that (x; y; z) 2 S2 satisfies z > 0. We have π(x; y; z) = [(x=z; y=z; 1)], whose local represen- + −1 tation π~ = 1 ◦ π ◦ φ1 is ! p x y x y (x; y) 7! φ−1([( 1 − x2 − y2; x; y)]) = φ−1([(1; ; )]) = ; 1 1 p1 − x2 − y2 p1 − x2 − y2 p1 − x2 − y2 p1 − x2 − y2 Thus, this local representation is smooth. We also ! −1 2 2 x y π~ : R ! D :(x; y) 7! ; p1 + x2 + y2 p1 + x2 + y2 which is also smooth. Proving this for the rest of the sphere follows similar lines. Proposition (5). Let f : N ! M be an embedding map. Then, the smooth structure induced on f(N) through f is the same one induced by the ambient manifold M. Proof Let q 2 f(N) and p 2 N such that f(p) = q (note that, there exists exactly such p). Let (U; V; φ) be a chart with p 2 V . Since f is an embedding it is also an homeomorphism, thus f(V ) is open in the topology of M. Now, let (X; W; ) be a chart on M such that q 2 W . Since Y = W \ f(V ) is open, it follows that jY is an −1 homeomorphism on Z = jY (Y ). All in all, jY ◦ f ◦ φ is homeomorphism (after restricting V to be f (Y ) and −1 −1 U = φ (f (Y ))). Therefore, Y is of dimension n, which implies that χ := jY ◦ f ◦ φ is not only an immersion, but rather a diffeomorphism (Since the dimension of then domain coincides with the dimension of the image). In particular, φY is a diffeomorphism with a subspace of dimension n, showing that f(N) is indeed a submanifold. Since χ and χ−1 are in fact the transition maps which cross the smooth structures induced by f and by M, it follows that both structures are compatible. Proposition (6). Let f : R2 ! R2 be a smooth map satisfying rank(Df(x)) = 2 for all x 2 R2, and assume that limjxj!1 jf(x)j = 1. Prove that f is a diffeomorphism. Proof First, we show that f is onto. Suffices it to show that there exists x 2 R2 such that f(x) = 0 (since for any y 2 R2, the function f(x) − y maintains our hypothesis). We define F (x) := 1=2jf(x)j2. Since f is proper, it follows that F is proper. Let c 2 R+ be some value for which F −1(c) is non-empty, then F −1([0; c]) is compact, and F attains its minimum inside this preimage set. Since this point would also be a global minima (as F is non-negative), it follows that DF must vanish at this point. Now, since 2 @F X @fj = f @x j @x i j=1 i 2 and since f is local diffeomorphism (i.e., ( @fJ ) is invertible), we have DF (x) = 0 () f(x) = 0 () F (x) = 0, @xi thus concluding the first part. We now turn to show that f is injective. As before, suffices it to show that there is only solution for f(x) = 0. Let S = f −1(0). If S is infinite, then since S is compact (as f as proper), it follows that S contains an accumulation point q. But, since f is local diffeomorphism, there must exist a neighborhood of this point in which f is 1-to-1, a contradiction. Now, let S = fp1; : : : ; png. Our strategy to show that jSj = 1 is as follows: First, we consider the differential equations @x = −∇F (x(t)): (1) @t 2 Next, we define Wi to be the set of all q 2 R such that x(t) converges to pi. We show that Wi is a disjoint open cover of R2, which combined with the fact R2 is connected, implies that n = 1. 2 2 Wi covers R Let q 2 R . and let x(t) be a solution for Equation (1) with initial condition x(0) = q. Since F is proper, and F forms a Lyapunov function for x(t), it follows that x(t) must converge to a point where the differential of F vanishes, namely, one of the point fpig. Wi are open Since F (x(t)) is strictly monotonic decreasing (unless initialized at a point where DF vanishes), we can find balls of radius > 0 around pi, such that any path x(t) that enters Bpi () remains in this ball. Now, by continuity of the solutions of Equation (1), if the solution initialized at q enters one of these ball, and q0 is sufficiently close to q, then the solution initialized at q0 also enters this ball and consequently converges to the same point. 3.