Middle East Technical University Spring 1996 Math 505 Differentiable Manifolds

MIDTERM EXAM

APRIL 20

SOLUTIONS

Problem 1. (2 pts) Prove that the quotient map S^{2} → ℝp^{2} is differentiable.

Solution: This map is the product of the inclusion S^{2} ↪ ℝ^{3} and the projection p: ℝ^{3} − {0} → ℝp^{2}, (x, y, z) ↦ [x : y : z]. The both maps are smooth: the first one because S^{2} is a subamifold in ℝ^{3}, the second because the coordinate functions in any of three charts covering ℝp^{2} are smooth. For example, in chart (U_{x}, φ_{x}) they are (\frac{y}{x}, \frac{z}{x}). The product of differentiable maps is differentiable.

Problem 2. (3 pts) Let D_{0} be the standard smooth structure on ℝ. Construct another smooth structure D on the same topological space so that the identity map id: (ℝ, D_{0}) → (ℝ, D) is not differentiable. Construct a diffeomorphism (ℝ, D_{0}) → (ℝ, D).

Solution: Take any homeomorphism f: ℝ → ℝ which is not differentiable at some point, for example, f(x) = x^{\frac{1}{3}} , which is not smooth at 0, and consider f as a chart, which defines a new differentiable structure, D, on ℝ. Then id: (ℝ, D_{0}) → (ℝ, D) is not differentiable, because it is given in the charts, id for D_{0} and f for D as f ∘ id ∘ id^{−1} = f. (Note that if we take f(x) = x^{3} then id is smooth, although not a diffeomorphism).

The diffeomorphism is f^{−1} : (ℝ, D_{0}) → (ℝ, D). To check it note that it is given in our charts as f ∘ f^{−1} ∘ id^{−1} = id.

Problem 3. (3 pts) Consider function f: U_{0} → ℂ defined in the coordinate system (U_{0}, φ_{0}) on the Riemannian sphere ℂp^{1}, U_{0} = {[z_{0} : z_{1}] ∈ ℂp^{1} | z_{0} ≠ 0}, z = φ_{0}([z_{0} :z_{1}]) = \frac{z_{1}}{z_{0}} , as f(z) = z^{3} + z − 2.

(1) Write down f in another chart on ℂp^{1}.

(2) Determine if f can be extended to a smooth function defined on the whole ℂp^{1}.

(3) Determine if f can be extended to a smooth map f̄: ℂp^{1} → ℂp^{1}.

Solution: (1) In the other coordinate system (U_{1}, φ_{1}), z′ = φ_{1}([z_{0} : z_{1}]) = \frac{z_{0}}{z_{1}} the function f is the product of the overlap map φ_{0} ∘ φ_{1}^{−1}(z′) = \frac{1}{z′} and f, that is z′ ↦ (z′)^{−3} +

(z′)^{−1} − 2.

(2) f is not defined on the whole ℂp^{1}, as in the previous formula it is not defined at z′ = 0.

(3) However, it can be extended to f̄: ℂp^{1} → ℂp^{1}, f̄([z_{0} : z_{1}]) = [z_{1}^{3} + z_{1}z_{0}^{2} − 2z_{0}^{3} : z_{0}^{3}], which send the “infinity” point [0 : 1] (z′ = 0) to itself. To check that f̄ is smooth we write it in the charts (U_{1}, φ_{1}) (for the domain and the image): f̄(z′) = ((z′)^{−3} + (z′)^{−1} − 2)^{−1} = \frac{(z′)^{3}}{1+(z′)^{2}−2(z′)^{3}} , and note that it is well defined and smooth at z′ = 0.

Problem 4. (3 pts) Consider the map f: ℝ^{2} → ℝp^{2}, (x,y) ↦ [x^{2} −y^{2} : x+y : x^{2} + 1]. Write down f using some chart in ℝp^{2} and the polar coordinate system in ℝ^{2}.

Solution: Since x = r cos ϕ, y = r sin ϕ in polar coordinates, we have f(r, ϕ) =

[r^{2}(cos(2ϕ)) : r(cos ϕ + sin ϕ) : r^{2} cos^{2} ϕ + 1], and in the chart (U_{z}, φ_{z}),

f(r, ϕ) = (\frac{r^{2}(cos(2ϕ))}{r^{2} cos^{2} ϕ + 1}, \frac{r(cos ϕ + sin ϕ)}{r^{2} cos^{2} ϕ + 1}) .

Problem 5. (4 pts) Consider point p ∈ ℝp^{2} with coordinates [1:3:1], covector α ∈ T_{p}^{∗}(ℝp^{2}) with coordinates (−1, 1) in the chart φ_{x}: U_{x} → ℝ^{2}, φ_{x}[x: y: z] = (\frac{y}{x}, \frac{z}{x}), where U_{x} = {[x:y:z] | x ≠ 0}, and vector v tangent to the curve γ: [−1, 1] → ℝp^{2}, γ(t) = [t+1 : 2t^{2}+3 : e^{t}] at p (t = 0).

(1) Determine the coordinates of p in the charts φ_{x}, φ_{y} and φ_{z}.

(2) Determine the coordinates of α in the chart φ_{y}.

(3) Determine the coordinates of v in the chart φ_{z} and evaluate α(v).

Solution: (1) (3, 1), (\frac{1}{3}, \frac{1}{3}) and (1, 3) respectively.

(2) If x_{1} = \frac{y}{x}, x_{2} = \frac{z}{x} coordinates in chart φ_{x} and y_{1} = \frac{x}{y}, y_{2} = \frac{z}{y} coordinates in chart φ_{y}, then x_{1} = \frac{1}{y_{1}} , x_{2} = \frac{y_{2}}{y_{1}} and

α = −dx_{1} + dx_{2} = −(\frac{∂x_{1}}{∂y_{1}} dy_{1} + \frac{∂x_{1}}{∂y_{2}} dy_{2}) + (\frac{∂x_{2}}{∂y_{1}} dy_{1} + \frac{∂x_{2}}{∂y_{2}} dy_{2}) = 6dy_{1} + 3dy_{2}

(3) γ(t) = (\frac{t+1}{e^{t}}, \frac{2t^{2}+3}{e^{t}}) in chart φ_{z}. Calculations gives v = γ′(0) = (0, −3). To evaluate α(v) we should know the both α and v in the same chart. Foe example, we can find v in the

chart φ_{x}. γ(t) = \frac{2t^{2}+3}{t+1}, \frac{e^{t}}{t+1}), which gives v = γ′(0) = (−3, 0) and α(v) = (−1)(−3) + (0)(1) = 3.

Problem 6. (3 pts) Given covectors α_{1}, …, α_{k} ∈ V^{∗}, where V is n-dimensional vector space, n ⩾ k, prove that a_{1} ∧ … ∧ α_{k} = 0 if and only if α_{1}, …, α_{k} are linear dependent.

Problem 7. (3 pts) Consider a tensor field T of type (p, q) on ℝ, which has (its unique) coordinate T (x) = T_{1,...,1}^{1,...,1}(x) = x^{2} in the identity chart id: ℝ → ℝ. Determine the coordinate of T in the chart φ: ℝ → ℝ, φ(x) = 2x.

Problem 8. (5 pts) Let F_{t}, G_{s}: M → M, t, s ∈ ℝ, be 1-parameter group of diffeomorphisms defined by the vector fields X and Y on a manifold M. Assuming that F_{t} ∘ G_{s} = G_{s} ∘ F_{t}, ∀t, s ∈ ℝ, prove that [X, Y] = 0 at any point.