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Home , Real projective line, Real projective space

Prof. A. Cattaneo Institut f¨urMathematik FS 2018 Universit¨atZ¨urich

Differentiable Solutions to Exercise Sheet 1

√ Exercise 1 (A non-differentiable ). Consider R with the {(R, id), (R, x 7→ sgn(x) x)}. Show R with this atlas is a but not a differentiable manifold.

√ 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 ). 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 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 − {0} → Rn − {0} 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 Sn−1 ⊂ Rn − {0}, 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 ). Let RPn = (Rn+1 − {0})/ ∼, where x ∼ tx for all t ∈ R − {0}, x ∈ Rn+1, be the n-dimensional .

(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 , we have to show that π−1(π(U)) is open in Rn+1 − {0}. But π−1(π(U)) = {x ∈ n+1, ∃y ∈ U : x ∼ y} = S t · U is a union of open sets (since R t∈R−{0} multiplication by non-zero t ∈ R is a 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 on a 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.

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 := {(x1, . . . , xn+1) ∈ n+1 n 1 R : xi 6= 0}, Ui = π(Vi) and ϕi : Ui → R , [(x1, . . . , xn+1)] 7→ x (x1,..., xˆi, . . . , 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 , ϕi is well-defined (since ϕi([tx]) = ϕ(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 ϕi are n and {(Ui, ϕi)} is an atlas for RP . Now we check that the atlas is smooth. For i < j,

−1 ϕi ◦ ϕj (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 −1 and ψβ on N, the composition ψβ ◦ f ◦ ϕα is smooth in its domain. Also recall that a diffeomorphism is a smooth bijective map with a smooth inverse. Show that RP1, the real , 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)) 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 ◦ ψW : (0, 2π) − {π} → R θ 7→ φ1(cos(θ/2), sin(θ/2)) = tan(θ/2) −1 fW,U2 = φ2 ◦ f ◦ ψW : (0, 2π) → R θ 7→ φ2(cos(θ/2), sin(θ/2)) = cot(θ/2) −1 fE,U1 = φ1 ◦ f ◦ ψE :(−π, π) → R θ 7→ φ1(cos(θ/2), sin(θ/2)) = tan(θ/2) −1 fW,U2 = φ2 ◦ f ◦ ψE : (0, 2π) − {0} → R θ 7→ φ2(cos(θ/2), sin(θ/2)) = cot(θ/2)

−1 where we have to restrict the compositions to ψA(f (Ui) for A = E,W . For example, −1 f ◦ ψW (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) + 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 = ψW ◦ g ◦ φ1 : R − {0} → (0, 2π) ( 2 arctan(y) y > 0 y 7→ ψW (2 arctan(y/1)) = 2 arctan(y) + 2π y < 0 −1 gU2,W = ψW ◦ g ◦ φ2 : R → (0, 2π)  2 arctan(1/x) x > 0  x 7→ π x = 0 π − 2 arctan(1/x) x < 0 −1 gU1,E = ψE ◦ g ◦ φ1 : R → (−π, π) y 7→ 2 arctan(y) −1 gU2,E = ψE ◦ g ◦ φ2 : R − {0} → (−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.