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Differentiable Manifolds Prof. A. Cattaneo Institut f¨urMathematik FS 2018 Universit¨atZ¨urich Differentiable Manifolds Solutions to Exercise Sheet 1 p Exercise 1 (A non-differentiable manifold). Consider R with the atlas f(R; id); (R; x 7! sgn(x) x)g. Show R with this atlas is a topological manifold but not a differentiable manifold. p Solution: This follows from the fact that the transition function x 7! sgn(x) x is a homeomor- phism but not differentiable at 0. Exercise 2 (Stereographic projection). Let f : Sn − f(0; :::; 0; 1)g ! Rn be the stereographic projection from N = (0; :::; 0; 1). More precisely, f sends a point p on Sn different from N to the intersection f(p) of the line Np passing through N and p with the equatorial plane xn+1 = 0, as shown in figure 1. Figure 1: Stereographic projection of S2 (a) Find an explicit formula for the stereographic projection map f. (b) Find an explicit formula for the inverse stereographic projection map f −1 (c) If S = −N, U = Sn − N, V = Sn − S and g : Sn ! Rn is the stereographic projection from S, then show that (U; f) and (V; g) form a C1 atlas of Sn. Solution: We show each point separately. (a) Stereographic projection f : Sn − f(0; :::; 0; 1)g ! Rn is given by 1 f(x1; :::; xn+1) = (x1; :::; xn): 1 − xn+1 (b) The inverse stereographic projection f −1 is given by 1 f −1(y1; :::; yn) = (2y1; :::; 2yn; kyk2 − 1): kyk2 + 1 2 Pn i 2 Here kyk = i=1(y ) . One can then easily check by direct computation that f and the map above are left and right inverses of each other. (c) The transition map g ◦ f −1 : Rn − f0g ! Rn − f0g is given by 1 g ◦ f −1(y1; :::; yn) = (y1; :::; yn); kyk2 which is smooth, and moreover has a smooth inverse, given by the same formula. The geometric explanation for this is that g ◦ f −1 represents inversion with respect to the sphere Sn−1 ⊂ Rn − f0g, which is an involution (an involution τ of a set X is a map X to X such that τ 2 = id). Hence Sn has a smooth atlas consisting of just two patches (U; f) and (V; g). Exercise 3 (Real projective space). Let RPn = (Rn+1 − f0g)= ∼, where x ∼ tx for all t 2 R − f0g; x 2 Rn+1, be the n-dimensional real projective space. (a) Prove that RPn is a differentiable manifold (which is Hausdorff and second countable). Solution: We first show that RPn is Hausdorff and second countable. For this, let π : Rn+1 ! RPn be the canonical projection. Claim 1. π is an open map. Proof. Let U ⊆ Rn+1 − f0g be open. We have to show that π(U) is open in RPn. By definition of the quotient topology, we have to show that π−1(π(U)) is open in Rn+1 − f0g. But π−1(π(U)) = fx 2 n+1; 9y 2 U : x ∼ yg = S t · U is a union of open sets (since R t2R−{0g multiplication by non-zero t 2 R is a homeomorphism of Rn+1 − f0g: Hence π is open. Now we get a countable basis for RPn by taking the image of a countable basis for Rn+1 under π. To show RPn is Hausdorff, we use the following lemma: Lemma 1. Let ∼ be an open equivalence relation on a topological space X(i.e. the canonical projection X ! X= ∼ is open). Then X= ∼ is Hausdorff if and only if Graph(∼) = f(x; y) 2 X × X : x ∼ yg is closed. n+1 n+1 P 2 To apply the lemma, define f : R × R ! R by f(x; y) = i6=j(xiyj − xjyi) and convince yourself that f(x; y) = 0 , x = ty; t 6= 0 , x ∼ y and hence Graph(∼) = f −1(0) is closed. Therefore RPn is hausdorff. n Next we are going to present a system of charts for RP . Let Vi := f(x1; : : : ; xn+1) 2 n+1 n 1 R : xi 6= 0g, Ui = π(Vi) and 'i : Ui ! R ; [(x1; : : : ; xn+1)] 7! x (x1;:::; x^i; : : : ; xn+1). n i Then one checks that: Ui are open in RP (since π is an open map) and their union is n RP , 'i is well-defined (since 'i([tx]) = '(x)), continuous, and bijective, for all i, with a continuous inverse given by (x1; : : : ; xn) 7! [(x1; : : : ; xi−1; 1; xi; : : : ; xn+1)] (the continuity follows general topological arguments on quotients). Hence, the 'i are homeomorphisms n and f(Ui;'i)g is an atlas for RP . Now we check that the atlas is smooth. For i < j, −1 'i ◦ 'j (x1; : : : ; xn) = 'i([(x1; : : : ; xj−1; 1; : : : ; xn)]) 1 = (x1;:::; 1; : : : ; xn) xi is clearly a smooth map (rational with nonzero denominator). Hence RPn is a smooth manifold. (b) Recall that a map between two manifolds f : M ! N is smooth if for all charts 'α on M −1 and β on N, the composition β ◦ f ◦ 'α is smooth in its domain. Also recall that a diffeomorphism is a smooth bijective map with a smooth inverse. Show that RP1, the real projective line, is diffeomorphic to the circle S1. Solution: Recall that we had a description of S1 in terms of R=2πZ (example 2.9 in the script). We define a map f : S1 ! RP1 by f(θ + 2πZ) = [(cos(θ=2); sin(θ=2)]. Claim 2. f is a diffeomorphism Proof. First we have to check that f is well-defined. For this, suppose that θ+2πZ = '+2πZ, then θ = ' + 2πk for some k 2 Z and f(' + 2πZ) = [(cos('=2); sin('=2)] = [(cos((θ + 2πk)=2); sin((θ + 2πk)=2)] = [(cos(θ=2 + πk); sin(θ=2 + πk)] = [(−1)k(cos(θ=2); sin(θ=2)] = f(θ + 2πZ). Next, we should check that f is smooth. Recall that on S1, we can define two charts W ; E which are the identity maps on different intervals (say (0; 2π) and (−π; π)) 2 and that we have two charts φ1; φ2 on RP given by φ1(x; y) = y=x; φ2(x; y) = x=y. So, we have to check the 4 representations of f in these charts are smooth: −1 fW;U1 = φ1 ◦ f ◦ W : (0; 2π) − fπg ! R θ 7! φ1(cos(θ=2); sin(θ=2)) = tan(θ=2) −1 fW;U2 = φ2 ◦ f ◦ W : (0; 2π) ! R θ 7! φ2(cos(θ=2); sin(θ=2)) = cot(θ=2) −1 fE;U1 = φ1 ◦ f ◦ E :(−π; π) ! R θ 7! φ1(cos(θ=2); sin(θ=2)) = tan(θ=2) −1 fW;U2 = φ2 ◦ f ◦ E : (0; 2π) − f0g ! R θ 7! φ2(cos(θ=2); sin(θ=2)) = cot(θ=2) −1 where we have to restrict the compositions to A(f (Ui) for A = E; W . For example, −1 f ◦ W (π) = (0; 1) 2= U1. Since all the representations are trigonometric functions, they are smooth. Now we have to find a smooth inverse for f. The correct map is ( 2 arctan(y=x) + 2π x 6= 0 g([x; y]) = Z π + 2πZ x = 0 Again, we first have to check the well-definedness of g. This follows from the fact that arctan(ty=tx) = arctan(y=x) and limt→±∞ 2 arctan(t) = ±π. g is the correct inverse because g◦f(θ+2πZ) = 2 arctan(tan(θ=2)) = θ for θ 6= π and g◦f(π+ 2πZ) = g(0; 1) = π+2πZ, so g◦f = idS1 , and f◦g([x; y]) = [(cos(arctan(y=x)); sin(arctan(y=x))] = 2 2 − 1 [±(x + y ) 2 (x; y)] = [(x; y)] for x 6= 0 and f ◦ g([0; y]) = [(0; 1)] = [(0; y)]. As a last step, we have to show that g is smooth. One computes −1 gU1;W = W ◦ g ◦ φ1 : R − f0g ! (0; 2π) ( 2 arctan(y) y > 0 y 7! W (2 arctan(y=1)) = 2 arctan(y) + 2π y < 0 −1 gU2;W = W ◦ g ◦ φ2 : R ! (0; 2π) 8 2 arctan(1=x) x > 0 <> x 7! π x = 0 :>π − 2 arctan(1=x) x < 0 −1 gU1;E = E ◦ g ◦ φ1 : R ! (−π; π) y 7! 2 arctan(y) −1 gU2;E = E ◦ g ◦ φ2 : R − f0g ! (−π; π) x 7! 2 arctan(1=x) The only difficult case is gU2;W . Here smoothness follows from the fact that all derivatives dn of arctan vanish at infinity, i.e. limx→±∞ dxn arctan(x) = 0. Hence g is smooth, and f is a diffeomorphism. Hence RPn is diffeomorphic to S1..
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