Differential Geometry - Solution 1

Yossi Arjevani

k Proposition (1). Let f = (f1, . . . , fk): M → R be a submerssion. Then for a given point p ∈ M, the differentials dfi are independent.

Proof Let (U, V, ψ) be a chart for p, and deﬁne φ := f ◦ ψ, and φi = πi ◦ φ, where π is the projection map on the i’th coordinate. Then we have dφ = df ◦ dψ, and since f is a submerssion and ψ 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 ∈ R Pk such that i=1 αidfi = 0, then since dψ is linear, it follows that k ! k k X X X αidfi ◦ dψ = αidfi ◦ dψ = 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 ψ1 × ψ2, we have −1 −1 −1 (φ1, φ2) ◦ (ψ1, ψ2) = (φ1 ◦ ψ1, φ2 ◦ ψ2).

Proposition (3). Define a natural smooth structure on RPn. Proof First, we endow RPn with the standard quotient topology induced by Rn+1\{0} which is known to be Haus- dorff and second-countable. That is, we define π : Rn+1\{0} → 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 ∈ [n] we define n Ui := R ,Vi := {[(x1, . . . , xi−1, 1, xi, . . . , xn]}, φi(x1, . . . , xn) := π((x1, . . . , xi, 1, xi+1, . . . , xn)) −1 n+1 −1 Note that Vi are indeed open as π (Vi) = {x ∈ R | xi 6= 0}. 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 homeomorphism (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) = {x ∈ R | xj 6= 0}.

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 ∈ 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 {(x, y) | x + y < 1},Vi := {(x1, x2, x3) | x1 + x2 + x3 = 1, ±xi > 0}, and

± ± ± 2 p 2 2 2 ψi : Ui → Vi :(x, y) 7→ (x , ± 1 − x − y , y ). | {z } i0th coordinate For simplicity, assume that (x, y, z) ∈ S2 satisﬁes z > 0. We have π(x, y, z) = [(x/z, y/z, 1)], whose local represen- + −1 tation π˜ = ψ1 ◦ 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 ∈ f(N) and p ∈ N such that f(p) = q (note that, there exists exactly such p). Let (U, V, φ) be a chart with p ∈ 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, ψ) be a chart on M such that q ∈ W . Since Y = W ∩ f(V ) is open, it follows that ψ|Y is an −1 homeomorphism on Z = ψ|Y (Y ). All in all, ψ|Y ◦ 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 := ψ|Y ◦ 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 ∈ R2, and assume that lim|x|→∞ |f(x)| = ∞. Prove that f is a diffeomorphism. Proof First, we show that f is onto. Sufﬁces it to show that there exists x ∈ R2 such that f(x) = 0 (since for any y ∈ R2, the function f(x) − y maintains our hypothesis). We deﬁne F (x) := 1/2|f(x)|2. Since f is proper, it follows that F is proper. Let c ∈ 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 = {p1, . . . , pn}. Our strategy to show that |S| = 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 ∈ 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 ∈ 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 {pi}.

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.