Middle East Technical University Fall 1998 Math 541 Differential Topology

MIDTERM EXAM

NOVEMBER 23

TIME: 70 MIN + ONE “TAKE TO HOME” PROBLEM

Problem 1. (10 pts) Consider a conic A = {x^{2} + y^{2} − 4xy + 6y + 2 = 0} in ℝ^{2}.

(1) Determine the homogeneous equation of its closure, \overline{A}, in ℝP^{2}.

(2) Determine if \overline{A} is a smooth manifold (if not, then determine the singular points of \overline{A}).

(3) Answer the same question as in (2) for the complexification of \overline{A}.

Solution: (1) Putting x = \frac{X}{Z}, y = \frac{Y}{Z} into the equation and multiplying it by Z^{2}, we obtain the homogeneous equation, X^{2} + Y^{2} − 4XY + 6Y Z + 2Z^{2}.

(2) Solving the equations

\{ \begin{matrix}f_{x} & = 2x − 4y = 0 \\ f_{y} & = 2y − 4x + 6 = 0 \end{matrix}

we find the only critical point, (2, 1), of f(x, y) = x^{2} + y^{2} − 4xy + 6y + 2, which does not belong to A, since f(2, 1) ≠ 0. So, A has no singularities, but its projective closure, Ā, may still have them on the line {Z = 0}. To check it, we analyze the equations, g(y, z) = 0 and h(x, z) = 0, of Ā in the other two standard charts, where g(y, z) = 1 + y^{2} − 4y + 6yz + 2z^{2} and h(x, z) = x^{2} − 4x + 6z + 2z^{2}, are obtained from the homogeneous function by putting X = 1 and Y = 1 respectively. The systems

\{ \begin{matrix}g_{y} & = 2y − 4 + 6z = 0 \\ g_{z} & = 6y + 4z = 0 \end{matrix}  \{ \begin{matrix}h_{x} & = 2x − 4 = 0 \\ h_{z} & = 6 + 4z \end{matrix}

have obviously no solutions with z = 0, therefore, Ā have no singularities on the line Z = 0 as well. (One can also easily solve the above two linear systems and find the critical points of g and h, then to notice that g and respectively h do not vanish at these points). The conclusion: Ā is non-singular.

(3) All the above equations are linear with real coefficients, and thus may have only real solutions. Thus, the complexification of a conic Ā cannot have singularities, if its real part is non-singular.

Problem 2. (10 pts) A vector v ∈ T_{p}X in X = ℝ^{3} is located at the point p = (1, 2, 3) and has coordinates (1, 1, −1) in the chart ϕ: ℝ^{3} → ℝ^{3}, ϕ(x, y, z) = (y, z, x).

(1) Determine the coordinate of v in another chart, ψ: ℝ^{3} → ℝ^{3}, ψ(x, y, z) = (z^{2}, x + y, x − y).

(2) Evaluate the derivative, D_{v}f, if f(x, y, z) = x + y^{2} + z^{3} (here x, y, z are the usual coordinates in X = ℝ^{3}).

Solution: (1) (a, b, c) = ϕ(x, y, z) = (y, z, x), so a = y, b = z, c = x and (x, y, z) = ϕ^{−1}(a, b, c) = (c, a, b). The overlap mapping, (s, r, t) = ψ∘ϕ^{−1}(a, b, c) = (b^{2}, c+a, c−a),

have Jacobi matrix

J = \frac{∂(s, r, t)}{∂(a, b, c)} = ( \begin{matrix}0 & 2b & 0 \\ 1 & 0 & 1 \\ −1 & 0 & 1 \end{matrix} )

Since b = z = 3 at the point p, the vector v in the chart ψ is Jv = ( \begin{matrix}6 \\ 0 \\ −2 \end{matrix} ).

(2) D_{v}f = 1f_{a} + 1f_{b} − 1f_{c}. In the chart ϕ, f(a, b, c) = c + a^{2} + b^{3}, and p = (a, b, c) = (y, z, x) = (2, 3, 1), so, we obtain D_{v}f|_{p} = 2a + 3b^{2} − 1|_{(2,3,1)} = 4 + 27 − 1 = 30.

Problem 3. (10 pts) Assume that equation f(x, y, z) = 0 defines in ℝ^{3} a smooth surface F homeomorphic to S^{2}. Let A ⊂ ℝ^{2} denote the set of points, (x_{0}, y_{0}), such that the line {(x_{0}, y_{0}, z)|z ∈ ℝ} intersects F in infinitely many points. Prove that

(1) Int(A) = ∅,

(2) Int Cl(A) = ∅.

Solution: Consider the projection, p: F → ℝ^{2}, of F to the xy-plane. For a regular value, a ∈ ℝ^{2}, F_{a} = p^{−1}(a) is a compact 0-dimensional manifold (compactness is due to that F_{a} is a closed subset of F). Thus, F_{a} is a finite set and A ⊂ C, where C is the set of the critical values of p.

By Sard’s theorem, C has measure 0 and thus Int(C) = ∅, which implies (1), since Int(A) ⊂

Int(C).

Furthermore, the set of critical points, C_{F} ⊂ F is closed and thus compact. Therefore, C = p(C_{F}) is also compact and thus is closed. This yields (2)

Int Cl(A) ⊂ Int Cl(C) = Int(C) = ∅

Problem 4. (10 pts) Consider a mapping f: ℂ → ℂ, defined by a polynomial f of degree n and denote by f̄: ℂP^{1} → ℂP^{1} the extension of f defined at the infinity point as f([1 : 0]) = [1, 0]. Prove that the degree of f as of a smooth mapping is also equal to n. (You do not need to prove that f̄ is smooth as it was the task in a homework.)

Solution: By the fundamental theorem of algebra (which we proved), a polynomial, f, of degree n has n roots (counted with multiplicities). If b ∈ ℂ is not a critical value (that is f′(a) ≠ 0 for any a ∈ f^{−1}(b)), then f − b has precisely n roots. So, to prove the satatement we should show that all the roots are counted with the sign “+” in the definition of degree of f. This is because the Jacobian of a complex map is always positive (in our case, the complex Jacobian is f′(a) ≠ 0, and the corresponding real Jacobian is |f′(a)|^{2} > 0).

Problem 5. (15 pts) Given a smooth manifold X with the boundary M = ∂X, we define the double, Y, of X as a disjoint union of two copies, X_{1}, X_{2}, of X, identified along their boundary, M = ∂X_{1} = ∂X_{2}.

(1) Show that Y is a manifold without boundary.

(2) Describe a smooth structure on Y.

(3) Prove that Y is orientable if and only if X is orientable.