MATH 31CH SPRING 2017 MIDTERM 2 SOLUTIONS

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1 MATH 3CH SPRING 207 MIDTERM 2 SOLUTIONS (20 pts). Let C be a smooth curve in R 2, in other words, a -dimensional manifold. Suppose that for each x C we choose a vector n( x) R 2 such that (i) 0 n( x) for all x; (ii) n( x) T x (C) for all x, and (iii) n : C R 2 is a continuous function. (a) (0 pts). Prove directly from the definition of orientation that the formula Ω x ( v) = sgn det( n( x), v) defines an orientation of C. { (b) (5 pts). Let C = x }. y < x <, y = x2 Suppose we choose the constant function n( x) = 0. Show that part (a) applies and thus defines an orientation of C. (c) (5 pts). With the oriented curve C given in part (b), and the -form field ϕ = y dx + x dy, calculate ϕ. C Solution. (a). We need Ω( x, v) = Ω x ( v) to be a continuous function of x C, v T x (C). By definition n( x) is a continous function of x, so the entries of the matrix ( n( x), v) are continuous functions of x and v. The determinant is a continuous function of the entries of a matrix (it is even a polynomial in the entries of the matrix), so det( n( x), v) is a continuous function of x, v. Finally, by assumption n( x) is not in the -dimensional subspace T x (C) of R 2, so necessarily n( x) and v span R 2, and so they are also independent. Thus det( n( x), v) 0. Since sgn : R {0} R is continuous, we conclude that sgn det( n( x), v) is a continuous function of x, v as required. We also need that for fixed x, Ω x ( v) = sgn det( n( x), v) defines an orientation of the vector space T x (C). Since T x (C) is -dimensional, any nonzero vector in this space is a basis. If v and v are 2 nonzero vectors in T x (C), then v = λ v for some λ 0, and the change of basis matrix is P v v = [λ]. We have det( n( x), v ) = det( n( x), λ v) = λ det( n( x), v) Date: June 7, 207.

2 Thus Ω x ( v) and Ω x ( v ) are equal if and only if λ = det P v v > 0, which is the definition of an orientation on a vector space. (b). We need n to satisfy the conditions in (a). Since it is constant nonzero function, it is clearly continous and nonzero. We just need to check that n( x) T x (C). Since γ : [, ] R 2 given by γ(t) = t is a parameterization of C, Dγ(t) = spans t 2 2t the tangent space to C at γ(t). The vector 0 is never in the space spanned by, as 2t required. (c). Consider the parameterization γ of part (b). Then for each t, Ω γ(t) (Dγ(t)) = sgn det( n, Dγ(t)) = sgn det 0 =. This shows that the given parameterization 2t reverses the orientation. Rather than changing the parameterization, we simply calculate the integral then multiply by at the end. We have Thus [γ] ydx + xdy = C 2 (5 pts). Given two vectors v = (t 2 )() + (t)(2t)dt = 3t 2 dt = t 3 ydx + xdy = ydx + xdy = 2. [γ] a a 2 a 3 and v 2 = b b 2 b 3 we define = 2. a 4 b 4 ϕ( v, v 2 ) = (3a + a 3 )(2b 2 + b 4 ) (3b + b 3 )(2a 2 + a 4 ). (a) (5 pts). Prove directly from the definition that ϕ is a 2-form on R 4. (b) (5 pts). Write ϕ as an explicit linear combination of elementary 2-forms. (c). (5 pts) Show that ϕ dx 4 dx 3 = λ det for some scalar λ, and find λ. Solution. 2

3 (a). (Note that since you are asked to prove (a) from the definition, you cannot prove (b) first and then claim you are done by the theorem that linear combinations of elementary 2-forms are 2-forms). We need the function ϕ : (R 4 ) 2 R to be multilinear and antisymmetric. If w = c c 2 c 3 λa + µc λa 2 + µc 2 and λ, µ R, then λ v + µ w = λa 3 + µc 3. Thus c 4 λa 4 + µc 4 ϕ(λ v + µ w, v 2 ) = (3(λa + µc ) + λa 3 + µc 3 )(2b 2 + b 4 ) (3b + b 3 )(2(λa 2 + µc 2 ) + λa 4 + µc 4 ) = λ[(3a + a 3 )(2b 2 + b 4 ) (3b + b 3 )(2a 2 + a 4 )] + µ[(3c + c 3 )(2b 2 + b 4 ) (3b + b 3 )(2c 2 + c 4 )] = λϕ( v, v 2 ) + µϕ( w, v 2 ) which shows linearity in the first coordinate. The proof of linearity in the second coordinate is analogous. For antisymmetry, ϕ( v 2, v ) = (3b + b 3 )(2a 2 + a 4 ) (3a + a 3 )(2b 2 + b 4 ) = [(3a + a 3 )(2b 2 + b 4 ) (3b + b 3 )(2a 2 + a 4 )] = ϕ( v, v 2 ) where the middle equality comes from an elementary comparison of the terms on each side after distributing. (b). If we notice that ϕ( v, v 2 ) = 6(a b 2 b a 2 ) 2(a 2 b 3 a 3 b 2 ) + 3(a b 4 a 4 b ) + (a 3 b 4 b 3 a 4 ) Then it is clear that ϕ = 6dx dx 2 2dx 2 dx 3 + 3dx dx 4 + dx 3 dx 4 by definition. (c). Since any wedge product of forms with dx i occurring twice is 0, the only term that does not disappear in the wedge product is 6dx dx 2 dx 4 dx 3. This is the same as 6dx dx 2 dx 3 dx 4. By definition dx dx 2 dx 3 dx 4 is the same as the determinant. Thus λ = 6. 3

4 3 (20 pts). Let S be the surface given by z = (x 2 + y 2 ) for 0 < z <. (a) (0 pts). Find a parameterization of this surface and calculate its surface area. (b) (0 pts). Let f(x, y, z) = z (x 2 + y 2 ), so that S is the set of points where f is zero. Orient S using the gradient vector f, in words using the formula Ω x ( v, v 2 ) = sgn det( f, v, v 2 ). Find ϕ for the 2-form field ϕ = z dx dy. S Solution. (a). This is a cone for which cylindrical coordinates are helpful for parametrizing. We ( find the parameterization γ : U S given by γ θ ) r cos θ = r sin θ r, where r { U = θ } 0 θ 2π, 0 r. r We calculate the derivative: r sin θ cos θ Dγ(θ, r) = r cos θ sin θ. 0 The surface area is by definition vol 2 S = det([dγ( u)]t [Dγ( u)]) d 2 u. U We have r sin θ cos θ r sin θ r cos θ 0 (Dγ) T Dγ = r cos θ sin θ cos θ sin θ = r and so det([dγ( u)] T [Dγ( u)]) = 2r 2 = 2r. Then vol 2 (S) = 2π rdrdθ = 2π.

5 (b). We already have a parametrization from part (a). We check to see if it preserves the specified orientation on S. Note that 2x(x 2 + y 2 ) /2 f = 2y(x 2 + y 2 ) /2. We calculate 2 cos θ r sin θ cos θ Ωγ(θ,r) (D γ(θ, r), D 2 γ(θ, r)) = det 2 sin θ r cos θ sin θ = 3r. 0 This is negative for all points on S, so the parametrization reverses orientation. Rather than changing parametrization, we just adjust the sign to account for this. Now 2π r sin θ z dx dy = z dx dy = r det cos θ drdθ S [γ(u)] 0 0 r cos θ sin θ 2π 2π = r( r) = r 2 = 2π/

6 . Some definitions and formulas Definition.. Suppose that M R n is a k-dimensional manifold. Let γ : U R n be a relaxed parametrization of M, where X is the set of bad points. Then vol k M = det([dγ( u)]t [Dγ( u)]) d k u. U X Definition.2. let U R k be a bounded open set with vol k U = 0. Let V be an open subset of R n and let γ : U R n be a C mapping with γ(u) V. Let ϕ be a k-form field defined on V. Then the integral of ϕ over [γ(u)] is ϕ = ϕ(p γ( u) ( D γ( u),..., D k γ( u))) d k u. [γ(u)] U Definition.3. Let M R n be a k-dimensional oriented manifold, ϕ a k-form field on a neighborhood of M, and γ : U R n an orientation-preserving parametrization of M. Then ϕ = ϕ as defined in the previous definition. M [γ(u)] 6