Middle East Technical University Fall 1995 Math 471 Elementary Topology and Geometry

FINAL EXAM

JANUARY 21, 1996

Problem 1. (5 pts) Suppose that manifold X is a union of open subsets X = X_{1} ⋃ X_{2} and A_{i} is an atlas in X_{i}, i = 1, 2, which defines a differential structure, so that in Y = X_{1} ⋂ X_{2} structures agree. Define a differential structure in X, which agrees with the structures in X_{1} and in X_{2}.

Solution: A = A_{1} ⋃ A_{2} is the atlas on X. This is because the union of its domains covers X = X_{1} ⋃ X_{2} and its charts agree on the common part Y . A agree with any of its charts and in particular with the ones from A_{i}, i = 1, 2.

Problem 2. (5 pts) Give explanations why S^{n} is orientable.

Solution: There is a normal vector field to S^{n} which assigns to a point x ∈ S^{n} the vector x ∈ ℝ^{n+1}. It is continuous (as the identity map), hence, S^{n} is orientable. In fact, any closed hypersurface in ℝ^{n} is orientable, as it was mentioned in the lectures, e.g., if a surface is defined by an equation f(x_{1}, …, x_{n}) = 0, then ∇f provides a normal vector field.

A different solution: ℝ^{n} is orientable as a manifold (since T_{x}(ℝ^{n}) ≅ ℝ^{n}), the unit ball in ℝ^{n} is orientable since any submanifold of an orientable manifold is orientable, finally, S^{n−1} is the bounday of an orientable manifold, hence, is orientable as well.

Another solution is based on the fact (discussed in the class) that H_{n}(S^{n}) ≅ ℤ (just triangulate S^{n} as the boundary of a (n + 1)-simplex), which is also a criterion for orientability.

Problem 3. (5 pts) Suppose that n-dimensional differential manifold X has nform ω which does not vanish at x for any x ∈ X. Prove that X is orientable.

Solution: n-form ω is a polylinear function ω = ω(v_{1}, …, v_{n}), v_{i} ∈ T_{x}(X) (or a linear function on Λ^{n}(T_{x}(X))). Nonvanishing of ω at x ∈ X means that ω(v_{1}, …, v_{n}) ≠ 0 for any basis v_{1}, …, v_{n} ∈ T_{x}(X). We can assign to such a basis value +1 if ω(v_{1}, …, v_{n}) is positive and value −1 if negative. This gives an orientation of X.

Problem 4. (5 pts) Assume that f: X → Y is a differentiable map, ω_{1} ∈ Ω^{p}(X), ω_{2} ∈ Ω^{p}(Y), η = ω_{1} ∧ f^{∗}(ω_{2}). Assuming that ω_{1} is closed and ω_{2} is exact can we claim that

(1) η is closed ?

(2) η is exact ?

(3) η = 0 ?

If we can, give an explanation.

Solution: First, note that f^{∗}(ω_{2}) is exact, because if ω_{2} = dω_{3}, then f^{∗}(ω_{2}) = f^{∗}(dω_{3}) = df^{∗}(ω_{3}). We have proved in the class that the product of a closed and an exact form is exact, hence, η is exact. It is closed because any exact form is closed, but not neccessarily zero, as the following simple example shows: consider f: ℝ^{2} → ℝ, the projection to the axis X, and choose ω_{1} = dy, ω_{2} = dx. Both forms are exact (and closed), but η = dy ∧ dx ≠ 0.

Problem 5. (5 pts) Consider curve γ: [0, 1] → ℝ^{3}, t ↦ (t, t^{2}, cos(πt)), and 1form ω = p_{z}^{∗}(z dz), where p_{z}: ℝ^{3} → ℝ is the projection to the axis Z. Determine ∫_{γ} ω.

Solution: By definition ∫_{γ} ω = ∫_{0}^{1} γ^{∗}ω = ∫_{0}^{1} γ^{∗}(p_{z}^{∗}(z dz)), but γ^{∗}(p_{z}^{∗}(z)) = γ^{∗}(z) = z(t) = cos(πt), z dz = \frac{1}{2}d(z^{2}), so, γ^{∗}(p_{z}^{∗}(z dz)) = \frac{1}{2}d(cos^{2}(πt)). Hence,

∫_{γ} ω = ∫_{0}^{1} d(cos^{2}(πt)) = cos^{2}(πt)|_{0}^{1} = 1 − 1 = 0.

Problem 6. (8 pts) Assume that U, V are smooth vector fields on a differentiable manifold X.

(1) Prove that there exist a vector field W on X such that for any function f ∈ ℂ^{∞}(X) we have W (f) = V (U(f)) − U(V (f)). (Hint: Show that the right hand side of the equality is a differentiation at any point).

(2) Assuming that dim X = 2 find coordinates of the vector field W is some coordinate system in terms of coordinates of the fields U and V.

Solution: Consider the map L: C^{∞}(X) → C^{∞}(X), L(f) = V (U(f)) − U(V (f)). It is linear, L(f + g) = L(f) + L(g), L(λf) = λL(f) for λ ∈ ℝ, since U and V give

linear operators C^{∞}(X) → C^{∞}(X). Moreover, L satisfies the Leibnitz iquality L(fg) = fL(g) + gL(f), as the direct calculation shows:

U(V (fg)) = U(fV (g) + gV (f)) = U(f)V (g) + fU(V (g)) + U(g)V (f) + gU(V (f)) V (U(fg)) = V (fU(g) + gU(f)) = V (f)U(g) + fV (U(g)) + V (g)U(f) + gV (U(f)).

Hence, L is a differentiation and is defined by some vector field W, so that L(f) = W (f).

To write W in coordinates, assume that V = V_{1} \frac{d}{dx} + V_{2} \frac{d}{dy} and U = U_{1} \frac{d}{dx} + U_{2} \frac{d}{dy}. Then

U(f) = U_{1}f_{x} + U_{2}f_{y},

\begin{matrix}V (U(f)) & = V_{1} \frac{d}{dx}(U_{1}f_{x} + U_{2}f_{y}) + V_{2} \frac{d}{dy} (U_{1}f_{x} + U_{2}f_{y}) = \\ V_{1}(U_{1x}f_{x} + U_{1}f_{xx} + U_{2x}f_{y} + U_{2}f_{yx}) + V_{2}(U_{1y}f_{x} + U_{1}f_{xy} + U_{2y}f_{y} + U_{2}f_{yy}) \end{matrix}

and similarly for U(V (f)). After mutual cancellation of the terms containing f_{xx}, f_{xy} and f_{yy}

we get W (f) = V (U(f)) − U(V (f)) = W_{1}f_{x} + W_{2}f_{y}, where W_{1} = V_{1}U_{1x} + V_{2}U_{1y} − U_{1}V_{1x} − U_{2}V_{1y} and W_{2} = V_{1}U_{2x} + V_{2}U_{2y} − U_{1}V_{2x} − U_{2}V_{2y}.

Problem 7. (8 pts) Consider 1-form ω which is defined as ω = \frac{dv}{u^{2}} in the chart with coordinates u = \frac{x}{z}, v = \frac{y}{z} in ℝp^{2}. Consider the curve γ: [0, 1] → ℝp^{2}, γ(t) = [1 : t : 0]. Extend by continuity form ω outside the given chart and determine ∫_{γ} ω.

Solution: Consider a new chart, φ_{x}, which contains the curve γ, with the coordinates α = \frac{y}{x}, β = \frac{z}{x}. Then u = \frac{1}{β}, v = \frac{α}{β}, dv = \frac{dα β−α dβ}{β^{2}} and ω = \frac{dv}{u^{2}} = dα β − α dβ. By definition,

∫_{γ} ω = ∫_{0}^{1} γ^{∗}(dα β − α dβ) = ∫_{0}^{1}(dα(t) β(t) − α(t) dβ(t)), where (α(t), β(t)) = γ(t) =

(t, 0). Since β(t) = 0 and dβ(t) = 0, the integral is zero.