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.

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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

• f(x¹, ..., xⁿ⁺¹) =
• (x¹, ..., xⁿ).

1 − xⁿ⁺¹

(b) The inverse stereographic projection f⁻¹ is given by
1

2
2

• f
• ⁻¹(y¹, ..., yⁿ) =
• (2y¹, ..., 2yⁿ, kyk − 1).

kyk + 1

P

n

2

Here 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

• g ◦ f⁻¹(y¹, ..., yⁿ) =
• ₂ (y¹, ..., yⁿ),

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}.

S

But 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.

P

To apply the lemma, define f : Rⁿ⁺¹ × Rⁿ⁺¹ → R by f(x, y) =

_i=j(x_iy_j − x_jy_i)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 V_i := {(x₁, . . . , x_n+1) ∈

1

R

ⁿ⁺¹ : x_i = 0}, U_i = π(V_i) and ϕ_i : U_i → Rⁿ, [(x₁, . . . , x_n+1)] → (x₁, . . . , xˆ , . . . , x_n+1).

i

x_i

Then one checks that: U_i are open in RPⁿ (since π is an open map) and their union is RPⁿ, ϕ_i is well-defined (since ϕ_i([tx]) = ϕ(x)), continuous, and bijective, for all i, with a continuous inverse given by (x₁, . . . , x_n) → [(x₁, . . . , x_i−1, 1, x_i, . . . , x_n+1)] (the continuity follows general topological arguments on quotients). Hence, the ϕ_i are homeomorphisms and {(U_i, ϕ_i)} is an atlas for RPⁿ. Now we check that the atlas is smooth. For i < j,

ϕ_i ◦ ϕ⁻_j ¹(x₁, . . . , x_n) = ϕ_i([(x₁, . . . , x_j−1, 1, . . . , x_n)])

1
=

(x₁, . . . , 1, . . . , x_n)

x_i

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 ϕ_α on M and ψ_β on N, the composition ψ_β ◦ 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 ψ_W , ψ_E 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:

f_W,U = φ₁ ◦ f ◦ ψ_W⁻¹ : (0, 2π) − {π} → R

1

θ → φ₁(cos(θ/2), sin(θ/2)) = tan(θ/2) f_W,U = φ₂ ◦ f ◦ ψ_W⁻¹ : (0, 2π) → R

2

θ → φ₂(cos(θ/2), sin(θ/2)) = cot(θ/2)

f_E,U = φ₁ ◦ f ◦ ψ_E⁻¹ : (−π, π) → R

1

θ → φ₁(cos(θ/2), sin(θ/2)) = tan(θ/2)

f_W,U = φ₂ ◦ f ◦ ψ_E⁻¹ : (0, 2π) − {0} → R

2

θ → φ₂(cos(θ/2), sin(θ/2)) = cot(θ/2) where we have to restrict the compositions to ψ_A(f⁻¹(U_i) for A = E, W. For example, f ◦ ψ_W⁻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 lim_t→±∞ 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 = id_S , 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,

2

we have to show that g is smooth. One computes

g_{U ,W} = ψ_W ◦ g ◦ φ⁻₁ ¹ : R − {0} → (0, 2π)

1

(

2 arctan(y) y > 0 y → ψ_W (2 arctan(y/1)) =
2 arctan(y) + 2π y < 0

g_{U ,W} = ψ_W ◦ g ◦ φ⁻₂ ¹ : R → (0, 2π)

2



2 arctan(1/x) x > 0 x = 0 π − 2 arctan(1/x) x < 0



• x →
• π



g_{U ,E} = ψ_E ◦ g ◦ φ⁻₁ ¹ : R → (−π, π)

1

y → 2 arctan(y)

g_{U ,E} = ψ_E ◦ g ◦ φ⁻₂ ¹ : R − {0} → (−π, π)

2

x → 2 arctan(1/x)
The only difficult case is g_{U ,W} . Here smoothness follows from the fact that all derivatives

2

dⁿ

dxⁿ

of arctan vanish at infinity, i.e. lim_x→±∞ diffeomorphism. arctan(x) = 0. Hence g is smooth, and f is a
Hence RPⁿ is diffeomorphic to S¹.