Differentiable Manifolds

Prof. A. Cattaneo Institut fu¨r Mathematik Universita¨t Zu¨rich FS 2018 Differentiable Manifolds Solutions to Exercise Sheet 1 √Exercise 1 (A non-differentiable manifold). Consider R with the atlas {(R, id), (R, x → sgn(x) x)}. Show R with this atlas is a topological manifold but not a differentiable manifold. √Solution: This follows from the fact that the transition function x → sgn(x) x is a homeomorphism but not differentiable at 0. Exercise 2 (Stereographic projection). Let f : Sn − {(0, ..., 0, 1)} → 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 C∞ atlas of Sn. Solution: We show each point separately. (a) Stereographic projection f : Sn − {(0, ..., 0, 1)} → Rn is given by 1<ul style="display: flex;"><li style="flex:1">f(x1, ..., xn+1) = </li><li style="flex:1">(x1, ..., xn). </li></ul>1 − xn+1 (b) The inverse stereographic projection f−1 is given by 12 2<ul style="display: flex;"><li style="flex:1">f</li><li style="flex:1">−1(y1, ..., yn) = </li><li style="flex:1">(2y1, ..., 2yn, kyk − 1). </li></ul>kyk + 1 Pn2Here kyk = i=1(yi)2. 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 − {0} → Rn − {0} is given by 1<ul style="display: flex;"><li style="flex:1">g ◦ f−1(y1, ..., yn) = </li><li style="flex:1">2 (y1, ..., yn), </li></ul>kyk 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 − {0}, which is an involution (an involution τ of a set X is a map X to X such that τ2 = 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 − {0})/ ∼, where x ∼ tx for all t ∈ R − {0}, x ∈ 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 − {0} 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 − {0}. SBut π−1(π(U)) = {x ∈ Rn+1, ∃y ∈ U : x ∼ y} = t∈R−{0} t · U is a union of open sets (since multiplication by non-zero t ∈ R is a homeomorphism of Rn+1 − {0}. 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(∼) = {(x, y) ∈ X × X : x ∼ y} is closed. PTo apply the lemma, define f : Rn+1 × Rn+1 → R by f(x, y) = i=j(xiyj − xjyi)2 and convince yourself that f(x, y) = 0 ⇔ x = ty, t = 0 ⇔ x ∼ y and hence Graph(∼) = f−1(0) is closed. Therefore RPn is hausdorff. Next we are going to present a system of charts for RPn. Let Vi := {(x1, . . . , xn+1) ∈ 1Rn+1 : xi = 0}, Ui = π(Vi) and ϕi : Ui → Rn, [(x1, . . . , xn+1)] → (x1, . . . , xˆ , . . . , xn+1). ixi Then one checks that: Ui are open in RPn (since π is an open map) and their union is RPn, ϕi is well-defined (since ϕi([tx]) = ϕ(x)), continuous, and bijective, for all i, with a continuous inverse given by (x1, . . . , xn) → [(x1, . . . , xi−1, 1, xi, . . . , xn+1)] (the continuity follows general topological arguments on quotients). Hence, the ϕi are homeomorphisms and {(Ui, ϕi)} is an atlas for RPn. Now we check that the atlas is smooth. For i < j, ϕi ◦ ϕ−j 1(x1, . . . , xn) = ϕi([(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 ϕα on M and ψβ on N, the composition ψβ ◦ f ◦ ϕ−α 1 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 = ϕ+2πZ, then θ = ϕ + 2πk for some k ∈ Z and f(ϕ + 2πZ) = [(cos(ϕ/2), sin(ϕ/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 ψW , ψE which are the identity maps on different intervals (say (0, 2π) and (−π, π)) and that we have two charts φ1, φ2 on RP2 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: fW,U = φ1 ◦ f ◦ ψW−1 : (0, 2π) − {π} → R 1θ → φ1(cos(θ/2), sin(θ/2)) = tan(θ/2) fW,U = φ2 ◦ f ◦ ψW−1 : (0, 2π) → R 2θ → φ2(cos(θ/2), sin(θ/2)) = cot(θ/2) fE,U = φ1 ◦ f ◦ ψE−1 : (−π, π) → R 1θ → φ1(cos(θ/2), sin(θ/2)) = tan(θ/2) fW,U = φ2 ◦ f ◦ ψE−1 : (0, 2π) − {0} → R 2θ → φ2(cos(θ/2), sin(θ/2)) = cot(θ/2) where we have to restrict the compositions to ψA(f−1(Ui) for A = E, W. For example, f ◦ ψW−1(π) = (0, 1) ∈/ 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) + 2πZ x = 0 g([x, y]) = 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 = π and g◦f(π+ 2πZ) = g(0, 1) = π+2πZ, so g◦f = idS , and f◦g([x, y]) = [(cos(arctan(y/x)), sin(arctan(y/x))] = 1 1[±(x2 + y2)− (x, y)] = [(x, y)] for x = 0 and f ◦ g([0, y]) = [(0, 1)] = [(0, y)]. As a last step, 2we have to show that g is smooth. One computes gU ,W = ψW ◦ g ◦ φ−1 1 : R − {0} → (0, 2π) 1(2 arctan(y) y > 0 y → ψW (2 arctan(y/1)) = 2 arctan(y) + 2π y < 0 gU ,W = ψW ◦ g ◦ φ−2 1 : R → (0, 2π) 22 arctan(1/x) x > 0 x = 0 π − 2 arctan(1/x) x < 0 <ul style="display: flex;"><li style="flex:1">x → </li><li style="flex:1">π</li></ul>gU ,E = ψE ◦ g ◦ φ−1 1 : R → (−π, π) 1y → 2 arctan(y) gU ,E = ψE ◦ g ◦ φ−2 1 : R − {0} → (−π, π) 2x → 2 arctan(1/x) The only difficult case is gU ,W . Here smoothness follows from the fact that all derivatives 2dn dxn of arctan vanish at infinity, i.e. limx→±∞ diffeomorphism. arctan(x) = 0. Hence g is smooth, and f is a Hence RPn is diffeomorphic to S1.

Differentiable Manifolds

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