GEOMETRY HW (t, 0, e 1/t2 ), t > 0 1/t2, 0), t < 0. (0, 0, 0), t = 0

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1 GEOMETRY HW CLAY SHONKWILER Consider the map.5.0 t, 0, e /t ), t > 0 αt) = t, e /t, 0), t < 0 0, 0, 0), t = 0 a) Prove that α is a differentiable curve. Proof. If we denote αt) = xt), yt), zt0), then it is clear that xt) is differentiable, with x t) = for all t. Now, yt) and zt) are certainly differentiable for t 0. To check that they are differentiable at zero as well, it suffices to show that lim t 0 d dt e /t ) = lim t 0 t 3 e /t = 0. To see this, we re-write and then apply L Hopital s Rule twice: lim t 0 e t 3 /t t = lim 3 t 0 e /t 6 t = lim t 0 t 3 e/t 3 t = lim t 0 e /t = lim t 0 3 t t 3 e/t = lim t 0 3t e /t = 0. Hence, we have agreement at t = 0, so α is differentiable and { α, 0, e t) = t 3 /t ) t 0, e t 3 /t, 0) t 0 b) Prove that α is regular for all t and that the curvature kt) 0, for t 0, t ± /3, and k0) = 0.

2 CLAY SHONKWILER Proof. As we can see from the explicit solution for α t) given above, α t) for all t, so α is a regular curve. Furthermore, since ) d dt t 3 e /t = t 6 6 ) t e /t, we see that Hence, so long as t 0 and 0, 0, 6 ) α t t) = 6 t e /t ) t 0 0, 6 ) ) t 6 t e /t, 0 t 0 t 6 6 t 0 i.e. t ± /3), then we see that kt) = α t) = t 6 6 ) ) t e /t = t 6 6 ) t e /t 0. Furthermore, a L Hopital argument similar to that given in a) above demonstrates that k0) = 0. c) Show that the limit of the osculating planes as t 0, t > 0, is the plane y = 0 but that the limit of the osculating planes as t 0, t < 0, is the plane z = 0. Proof. The osculating plane is the plane spanned by α t) and α t) kt), so it s normal vector is given by α t) α t) kt). Note that, as t 0, α t), 0, 0). When t > 0, so α t) kt) = 0, 0, kt)) = 0, 0, ), kt) α t) α t), 0, 0) 0, 0, ) = 0,, 0). kt) If x, y, z) is a vector in the osculating plane, then 0 = x, y, z), 0,, 0) = y, so we see that the limit of the osculating planes is given by y = 0. On the other hand, when t < 0, so α t) kt) = 0, kt), 0) = 0,, 0), kt) α t) α t) kt), 0, 0) 0, 0, ) = 0, 0, ).

3 If x, y, z) is a vector in this osculating plane, then GEOMETRY HW 3 0 = x, y, z), 0, 0, ) = z so we see that the limit of the osculating planes is given by z = 0. d) Show that τ can be defined so that τ 0, even though α is not a plane curve. Proof. Recall from the definition that nt) = α t) α t) and b t) = τn. Also, b is the normal vector to the osculating plane, so we see that, from part c) above, for 0 < t < /3 and t > /3, bt) = 0,, 0), meaning that b t) = 0, 0, 0) on this interval. On the other hand, again using our results from part c), we know that, for /3 < t < 0 and t < /3, bt) = 0, 0, ), meaning that b t) = 0, 0, 0) on these intervals. Hence, we see that, anywhere t 0, t ± /3, 0, 0, 0) = b t) = τ0,, 0) = τ0, 0, ), which is to say that τ 0 except possibly at these three points. However, since τ is continuous, we extend to these points by continuity to see that τ 0 for all t..6.3 Show that the curvature kt) 0 of a regular parametrized curve α : I R 3 is the curvature at t of the plane curve π α, where π is the normal projection of α over the osculating plane at t. Proof. We know that ks) = α s), so it suffices to show that ks) = α s) = π α) s) for all t I. Now, the osculating plane P s0 at s 0 is simply the plane given by the span of ts 0 ) = α s 0 ) α s 0 ) and ns 0 ) = α s 0 ) α s 0 ). Since t and n are both unit vectors and are necessarily orthogonal, we know that the projection of α onto this plane is given by Hence, π αs) = ts 0 ), αs) ts 0 ) + ns 0 ), αs) ns 0 ). π α) s) = t, α s) t + n, α s) n = n, α s) n = α s 0 ), α s) α n. s 0 )

4 CLAY SHONKWILER Hence, when s = s 0, Therefore, π α) s 0 ) = α s 0 ), α s) α n = α s 0 ) s 0 ) α s 0 ) n. π α) s 0 ) = α s 0 ) n = α s 0 ) = ks 0 ). Hence, we see that the curvature of α is the curvature of its projection onto the osculating plane Show that the mean curvature H at p S is given by H = π π 0 k n θ)dθ, where k n θ) is the normal curvature at p along a direction making an angle θ with a fixed direction. Proof. By definitions, H = k +k, where k and k are the maximum and minimum normal curvatures, respectively. Also, k n θ) = k cos θ+k sin θ. Therefore, π π π = π π = π π = π 0 k nθ)dθ = π 0 k cos θ + k sin w θdθ 0 k sin θ) + k sin θdθ [ 0 k + sin θk k )dθ [k θ] π 0 + k k ) π 0 sin θdθ ] = k + k k π cos θ π 0 dθ = k + k k [ π θ/ + sin θ = k + k k = k +k = H Describe the region of the unit sphere covered by the image of the Gauss map of the following surfaces: a) Paraboloid of revolution z = x + y. b) Hyperboloid of revolution x + y z =. Answer: Consider the map f : R 3 R given by fx, y, z) = x + y z. Then df p = f x, f y, f z ) = x, y, z), which is nonzero so long as x, y, z) 0. Since f0), we see that is a regular value of f. Hence, we see that the hyperboloid S is a regular surface. Now, note that f = x + y + z = x + y + z = + z ) + z = + z. ] π 0

5 GEOMETRY HW 5 Since the hyperboloid is a regular surface, we know that the normal vector field is given by x 0, y 0, z 0 ) = N = fp) ) x, y, z) = fp) x = + z, y, z. + z + z + z Hence, we see that when z is constant, the z-coordinate of N is also constant and the x- and y-coordinates of N are proportional to x and y, respectively. In other words, if a vector x 0, y 0, z 0 ) is in the image of the Gauss map, then the entire latitudinal cirle given by {x, y, z) S : z = z 0 } will also be in the image. Hence, to describe the image, we need only find the maximum and minimum possible values of z 0, which will, in turn, give the image of the Gauss map as a band around the equator of the sphere. To do so, we can assume that x = 0. Then, since they are the coordinates of a point lying on the hyperbola, y and z satisfy the following relation: y z =. Note that this implies that y. In other words, z = ± y, so we see that z 0 = y ± y As y, we see that z 0 ± / and that, furthermore, if we consider a single branch of the above expression, the possible values for z 0 start at zero and increase or decrease monotonically. Hence, z 0 [0, /). Based on our above argument that concluded that this solution holds for all x and y, we see that, in fact, the image of the hyperboloid under the Gauss map is given by {x, y, z) S : z < /}. c) Catenoid x + y = cosh z. Answer: Consider the function f : R 3 R given by fx, y, z) = x + y cosh z. Then, recalling that cosh z = ez + e z ), we see that f = f x, f y, f z ) = x, y, e z e z )). Then we see that the only critical value of f occurs at the point 0, 0, 0). Now, f0, 0, 0) = /, so 0 is a regular value of f. Hence, the catenoid S that we are considering is f 0), where 0 is a regular value. Therefore, S is a regular surface.

6 6 CLAY SHONKWILER Thus, the normal vector field is given by x 0, y 0, z 0 ) = N p = fp) fp) = x,y, ez e z )) q x +y + ez +e z ) = x,y, ez e z )) q cosh z+ ez +e z ) ) = x,y, ez e z )) q. cosh z+ cosh z Hence, as in part b) above, when z is constant then z 0 is constant and x 0 and y 0 are proportional to x and y, respectively. In other words, as before, if x 0, y 0, z 0 ) is in the image of the Gauss map, then the entire latitudinal circle given by {x, y, z) S : z = z 0 } will also be in the image. Therefore, we need only determine the possible values of z 0 in order to completely describe the image of the Gauss map. Now, let x, y, z) S and define α := q. Then Then In other words, cosh z+ cosh z x x 0, y 0, z 0 ) = t, y z, ) ez e z ). t = x 0 + y 0 + z 0 = x t + y t z 0 = ± ) t cosh z. Now, we see that, it must be the case that z 0 < lim z t cosh z ) cosh z = 0) =, + z 0 = cosh z t + z 0. ) cosh z+ cosh z ) ez +e z +) e z +e z ++ ez e z ) )) e z +e z + e z +e z ez z ) +e z +e z +e z + 3 e z + ez e 6z ) ) + ez )) where we arrived at the fourth equality by multiplying numerator and denominator by e z. Now, since z can be any element of R, we see that z 0

7 GEOMETRY HW 7 achieves every value on the sphere except z 0 = and z 0 =. Therefore, the image of the Gauss map is all of S except the north and south poles. Suppose F : R n R k is a C -function and r R k is a regular value of F ; that is, for each p F r) w have that df p : T p R n T F p) R k is onto. Show that for each p S = F r) there exists V S open neighborhood of p in S and φ : U R n k V such that a) φ : U V R n is differentiable. b) φ : U V is a homeomorphism. c) dφ q : T q R n k T φq) R n is injective for any q U. Proof. Denote r = r,..., r k ) and let p F r). Since r is a regular value, the matrix F F F x x x n F F F df p = x x x n..... F k F k F x x k x n has a minor with nonzero determinant. Assume, without loss of generality, that the rightmost k columns are such a minor. Now, define H : R n R n by Hx, x,..., x n ) = x, x,..., x n k, F x,..., x n ),..., F k x,..., x n )). We will denote by u,..., u n ) the elements in R 3 where H takes its values. Now, dh p = F F F p p p n.. F k F k F p p k p n Then F F p n k+... p n det dh p = F k p n k+... By the Inverse Function Theorem, then, there exists a neighborhood V of p and W of F p) such that H : V W is a diffeomorphism. Hence, the coordinate functions x = u,..., x n k = u n k, x n k+ = g u,..., u n ),..., x n = g k u,..., u n ) F k p n

8 8 CLAY SHONKWILER of H are differentiable. In particular, for i =,..., k, x n k+i = g i u,..., u n k, r,..., r k ) = h i u,..., u n k ) are differentiable functions defined on π V ), where π is the projection of F r) onto the set U = W {u,..., u n k, r,..., r k )}. Note that HF r) V ) = U, so, in fact, H U : U F r) V = V is a diffeomorphism. Now, if we let π : R n R n be the projection onto the set A = {x,..., x n k, 0, 0,..., 0}, then it is clear that π U is a diffeomorphism of U with π U U ). Define U := π U U ). Then φ : U V given by φu,..., u n k ) = u,..., u n k, h u,..., u n k ),..., h k u,..., u n k )) is simply the composition φ = H U π U. Since both of the functions in the composition are diffeomorphisms, so is φ. Hence, φ : U V precisely fulfills the desired requirements. DRL 3E3A, University of Pennsylvania address: shonkwil@math.upenn.edu