GEOMETRY HW Consider the parametrized surface (Enneper s surface)

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1 GEOMETRY HW 4 CLAY SHONKWILER Consider the parametrized surface (Enneper s surface φ(u, v (x u3 3 + uv2, v v3 3 + vu2, u 2 v 2 show that (a The coefficients of the first fundamental form are E G ( + u 2 + v 2 2, F. Answer: We know that E Φ, Φ, F Φ, Φ 2 G Φ 2, Φ 2, so we need to calculate Φ Φ 2. Now, Φ ( u 2 + v 2, 2uv, 2u Φ 2 (2uv, v 2 + u 2, 2v. E Φ, Φ ( u 2 + v u 2 v 2 + 4u 2 + 2u 2 + 2v 2 + u 4 + v 4 2u 2 v 2 ( + u 2 + v 2 2 F Φ, Φ 2 2uv( u 2 + v 2 + 2uv( v 2 + u 2 4uv 2uv 2u 3 v + 2uv 3 + 2uv 2uv 3 + 2u 3 v 4uv G Φ 2, Φ 2 4u 2 v 2 + ( v 2 + u v 2 + 2u 2 + 2v 2 + u 4 + v 4 2u 2 v 2 ( + u 2 + v 2 2. (b The coefficients of the second fundamental form are Answer: Recall that e 2, g 2, f. N Φ Φ 2 Φ Φ 2 Φ Φ 2 EG F 2 ( 2uv2 2u 2u 3, 2u 2 v + 2v + 2v 3, 2u 2 v 2 u 4 v 4 ( + u 2 + v 2 4

2 2 CLAY SHONKWILER e N, 2 φ u 2 ( 2uv 2 2u 2u 3, 2u 2 v + 2v + 2v 3, 2u 2 v 2 u 4 v 4, (+u 2 +v 2 2 ( 2u, 2v, 2 2( + 2u 2 + 2v 2 + u 4 + v 4 + 2u 2 v 2 (+u 2 +v (+u2 +v 2 2 (+u 2 +v Now, f N, 2 φ u v ( 2uv 2 2u 2u 3, 2u 2 v + 2v + 2v 3, (+u 2 +v 2 2 2u 2 v 2 u 4 v 4, (2v, 2u, 2( 2uv 3 2uv + 2u 3 v 2u 3 v + 2uv + 2uv 3 (+u 2 +v 2 2 ( (+u 2 +v 2 2. g N, 2 φ v 2 ( 2uv 2 2u 2u 3, 2u 2 v + 2v + 2v 3, (+u 2 +v 2 2 2u 2 v 2 u 4 v 4, (2u, 2v, 2 ( 2uv 2 2u 2u 3, 2u 2 v + 2v + 2v 3, (+u 2 +v 2 2 2u 2 v 2 u 4 v 4, (2u, 2v, 2 e 2. That is to say, e 2, g 2, f. (c The principal curvatures are Answer: Now, Since f F, k k a k 2 ( + u 2 + v 2 2, k 2 ff + eg EG F 2, k eg EG e E 2 ( + u 2 + v 2 2 k 2 2 ( + u 2 + v 2 2. k 2 a 22 k 2 ge EG g G. 2 (d The lines of curvature are the coordinate curves. ff + ge EG F 2. ( + u 2 + v 2 2.

3 Proof. As given on pg. curvature is GEOMETRY HW 4 3 6, the differential equation giving the lines of (α 2 α α 2 (α 2 2 E F G e f g where α(t φ(α (t, α 2 (t is a regular curve. Now, recalling that F f, this reduces to the equation α α 2(Eg eg α α 2( 2( + u 2 + v 2 2 2( + u 2 + v 2. Since 4(+u 2 +v 2 2, it follows that either α or α 2. However, this occurs precisely when α is a coordinate curve, so we see that the lines of curvature are the coordinate curves. (e The asymptotic curves are u + v const., u v const. Proof. From pg. 6, we see that a connected regular curve C is an asymptotic curve if only if, for any parametrization α(t φ(α (t, α 2 (t, e(α 2 + fα α 2 + g(α 2 2. Now, plugging in our values for e, f g, we see that this reduces to the condition that 2(α 2 2(α 2 2. Reducing, we see that (u 2 (v 2, which is to say that u ±v. Integrating both sides of the above, we see that, u ±v + C. In other words, u v C. curves of this form are precisely the asymptotic curves φ(u, v (γ(v cos u, γ(v sin u, ψ(v is given as a surface of revolution with constant Gaussian curvature K. To determine the functions γ ψ, choose the parameter v in such a way that (γ 2 + (ψ 2. Show that (a γ satisfies γ + Kγ ψ is given by ψ (γ 2 dv; thus, < u < 2π, the domain of v is such that the last integral makes sense. Proof. We start by computing Φ, Φ 2 : Φ ( γ(v sin u, γ(v cos u,, From this, then, we can compute Φ 2 (γ (v cos u, γ (v sin u, ψ (v. E Φ, Φ (γ(v 2 (sin 2 u + cos 2 u (γ(v 2, F Φ, Φ 2 γ(vγ (v sin u cos u + γ(vγ (v sin u cos u

4 4 CLAY SHONKWILER G Φ 2, Φ 2 (γ (v 2 (sin 2 u + cos 2 u + (ψ (v 2 (γ (v 2 + (ψ (v 2. Recall that e /M(Φ, Φ 2, 2 φ u 2 f /M(Φ, Φ 2, 2 φ u v g /M(Φ, Φ 2, 2 φ v 2 where M EG F 2 (a, b, c denotes the determinant of the matrix with columns given by a, b c. Now, EG F 2 (γ(v 2 φ(v. Furthermore, e γ(v γ sin u γ cos u γ cos u γ cos u γ sin u γ sin u ψ γ(v ψ (γ 2 sin 2 u + γ 2 cos 2 u γ2 ψ γ γψ. f γ(v γ sin u γ cos u γ sin u γ cos u γ sin u γ cos u ψ γ ψ ( γγ sin u cos u + γγ sin u cos u. g γ(v γ sin u γ cos u γ cos u γ cos u γ sin u γ sin u ψ ψ γ [ ψ ( γγ sin 2 u + γγ cos 2 u + ψ ( γγ sin 2 u γγ cos 2 u] γ (ψγγ ψ γγ ψγ ψ γ.

5 GEOMETRY HW 4 5 Now, if we differentiate both sides of the equation (γ 2 + (ψ 2, we see that γ γ ψ ψ. K eg f 2 EG F 2 γψ (ψ γ ψ γ γ 2 ψ (ψ γ ψ γ γ (ψ 2 γ +ψ ψ γ γ (ψ 2 γ +(γ 2 γ γ γ γ. Re-arranging, we see that γ satisfies γ + kγ. Now, on the other h, since (γ 2 + (ψ 2, we can re-arrange to see that ψ (γ 2. ψ (γ 2 dv, where v is such that this integral makes sense. (b All surfaces of revolution with constant curvature K which intersect perpindicularly the plane xoy are given by v γ(v C cos v, ψ(v C 2 sin 2 vdv, where C is a constant. Determine the domain of v draw a rough sketch of the profile o f the surface in the xz plane for the cases C, C >, C <. Observe that C gives the sphere. Proof. Recall that in part (a we showed that γ satisfies the equation γ + Kγ. Plugging in K, we see that γ + γ. This has characteristic equation λ 2 +, meaning the solutions are λ ±. γ(v C e v cos v + C 2 e v sin v C cos v + C 2 sin v. After translating the curve, we may assume that at v the curve (γ(v, ψ(v passes through (C, (γ (, ψ ( (, z. Then γ ( C sin( + C 2 cos( C 2. γ(v C cos v. Now, using our formula for ψ given in part (a plugging in the above for γ, we see that v v ψ(v (γ 2 dv C 2 sin 2 vdv.

6 6 CLAY SHONKWILER where v [ sin (/C, sin (/C]. (c All surfaces of revolution with constant curvature K may be given by one of the following types:. γ(v C cosh v, ψ(v v C 2 sinh 2 vdv. 2. γ(v C sinh v, ψ(v v C 2 cosh 2 vdv. 3. γ(v e v, ψ(v v e 2v dv. Determine the domain of v draw a rough sketch of the profile of the surface in the xz plane. Proof. Plugging the value K into the expression derived in (a, we see that γ γ. The characteristic equation, then, is λ 2, so the solution of the differential equation is γ(v C e v + C 2 e v since λ ± is the solution of the characteristic equation. Since there are no initial conditions given, we cannot necessarily reduce γ to one of the above three forms. If we had initial conditions that dictated that C ±C 2 or C or C 2, then we could clearly reduce to one of the above three cases. (d The surface of type 3 in part (c is the pseudosphere of Exercise 6. Proof. Now, the coordinates of a point p S are given by v (e v cos u, e v sin u, e 2v dv. Now, the pseudosphere is given by taking a curve in the xz-plane such that the length of th tractrix is equal to at every point on the curve, then rotating this curve about the z-axis. So we first demonstrate that the γ ψ in 3 agree with such a curve in the xz-plane. To that end, let y. Since y e v sin u, this implies that u. a point a in the plane that lies on this surface is given by v (e v,, e 2v dv. We let x e v, making z v x 2 dv v x 2. Since v ( so < x. Now, differentiating, we see that the tangent to the curve is given by (e v,, e 2v (x,, x 2. This tangent, then, clearly has length x 2 + ( x 2 2 x 2 + x 2. Therefore, the surface agrees with the pseudosphere on the xz-plane.

7 GEOMETRY HW 4 7 Now, if we fix v let u vary, then we see that the coordinates of points on the surface are given by v (e v cos u, e v sin u, e 2v. this describes a circle of radius e v, so we see that our surface is just the surface of revolution of the xz-plane curve we described above. However, this is precisely the pseudosphere. (e The only surface of revolution with K are the right circular cylinder, the right circular cone, the plane. Proof. If K, then, using the differential equation established in (a, then γ. The only possibilities for γ, then, are that γ is a line, a constant not equal to zero, or zero. If γ is a line, then we see that the coordinates give a right circular cone, with its peak at the point where γ intersects the axis. On the other h, if γ is a nonzero constant, then the coordinates give a right circular cylinder. Finally, if γ, then the surface is completely determined by z, which is itself constant, so the surface is simply a plane Let h : S R be a differentiable function on a surface S, let p S be a critical point of h (i.e., dh p. Let w T p S let α : ( ɛ, ɛ S be a parametrized curve with α( p, α ( w. Set H p h(w d2 (h α dt 2 t. (a Let φ : U S be a parametrization of S at p, show that H p h(u Φ + v Φ 2 h uu (p(u 2 + 2h uv (pu v + h vv (p(v 2. Conclude that H p h : T p S R is a well-defined quadratic form on T p S. H p h is called the Hessian of h at p. Proof. First, we note that so φ α(t d(φ α t ( u(t v(t ( u (t v (t.

8 8 CLAY SHONKWILER Note, furthermore, that dh p dφ q ( h x, h y, h x x z u v y y u v z z u v x x u + h y y u + h z z u, h x x v + h y y v + h z z v ( h (h u, h v. Since p is a critical point, recall that dh p so h u h v when evaluated at the values associated with p. Putting all of this together, recalling that w u Φ + v Φ 2, we see that H p h(u Φ + v Φ 2 d2 (h α d2dt2 t dt 2 (h α d2 ((h φ (φ α dt 2 d dt (dh [ pdφ q d(φ ( ] α t d u dt (h u, h v v d dt (h uu + h v v dhu dt u + h u u + dhv dt v + h v v ( h u du u dt + hu dv v dt u + h u u + ( h v du u dt + hv dv v dt + hv v (h uu u + h uv v u + h u u + (h vu u + h vv v v + h v v h uu (u 2 + 2h uv u v + h vv (v 2 + h u u + h v v h uu (u 2 + 2h uv u v + h vv (v 2 + u + v h uu (u 2 + 2h uv u v + h vv (v 2. Since H p h : T p S R does not depend on the choice of α, we see that H p h is indeed a well-defined quadratic form. (b Let h : S R be the height function of S relative to T p S; that is, h(q q p, N(p, q S. Verify that p is a critical point of h thus that the Hessian H p h is well defined. Show that if w T p S, w, then H p h(w normal curvature at p in the direction of w. Conclude that the Hessian at p of the height function relative to T p S is the second fundamental form of S at p. Proof. Denote q : φ(u, v. Then h(u, v φ(u, v p, N(p φ(u, v, N(p p, N(p. h u Φ, N(p + Φ, N(p h v Φ 2, N(p + Φ 2, N(p. However, since Φ, Φ 2 T p S, Φ, N(p Φ 2, N(p.

9 GEOMETRY HW 4 9 dh p, so we see that p is indeed a critical point of h. Now, h uu u φ u, N(p φ 2, N(p e, u2 h uv v φ u, N(p 2 φ, N(p f, u v h vv v φ v, N(p φ 2, N(p g. v2 by the work we did in (a above, H p h(w e(u 2 + 2fu v + g(v 2 II p (w. Now, since the normal curvature at p in the direction of a unit vector v is simply II p (v, we see that, in fact H p h(w normal curvature at p in the direction of w A critical point p S of a differentiable function h : S R is nondegenerate if the self-adjoint linear map A p h associated to the quadratic form H p h is nonsingular. Otherwise, p is a degenerate critical point. A differentiable function on S is a Morse function if all its critical points are nondegenerate. Let h r : S R 3 R be the distance function from S to r; i.e., h r (q q r, q r, q S, r R 3, r / S. (a Show that p S is a critical point of h r if only if the straight line pr is normal to S at p. Proof. Let α(t be a curve on S with α( p, α ( w. Then p is a critical point if only if dh r (w d dt ( α( r, α( r /2 t 2 α( r,α( r /2 α (, α( r α( r, α ( 2( p r,p r /2 w, p r p r, w 2 w,p r 2( p r,p r /2 w,p r p r,p r we see that it must be the case that w, p r. Since this implies that the vector on the line joining p r is orthogonal to w. Since w T p (S our choice of α was arbitrary, we see that the line joining p r is normal to S at p if only if p is a critical point of h r.

10 CLAY SHONKWILER (b Let p be a critical point of h r : S R. Let w T p S, w, let α : ( ɛ, ɛ S be a curve parametrized by arc length with α( p, α ( w. Prove that H p h r (w h r (p k n, where k n is the normal curvature at p along the direction of w. Conclude that the orthonormal basis {e, e 2 }, where e e 2 are alon gthe principal directions of T p S, diagonalizes the self-adjoint linear map A p h r. Conclude further that p is a degenerate critical point of h r if only if either h r (p /k or h r (p /k 2, where k k 2 are the principal curvatures at p. Proof. Let φ be a coordinate chart at the point p. Then h r (q q r, q r φ(u, v r, φ(u, v r φ, φ 2 φ, r + r, r If we let then since p is a critical point, l(u, v φ, φ 2 φ, r + r, r, h r l, l u 2 φ u, φ 2 φ u, r φ u, φ r l v 2 φ v, φ 2 φ v, r φ u, φ r. l uu 2 φ uu, φ r + 2 φ u, φ u, l vv 2 φ vv, φ r + 2 φ u, φ u, l uv 2 φ uv, φ + 2 φ u, φ v. h ru 2 l /2 l u h rv 2 l /2 l u. Therefore, h ruu 2 l /2 l uu 4 l 3/2 (l u 2 2 l /2 l uu l u 4l h r u 2 l /2 l uu, h rvv 2 l /2 l vv 4 l 3/2 (l v 2 2 l /2 l vv l v 4l h r v 2 l /2 l vv h ruv 2 l /2 l uv 4 l 3/2 l u l v 2 l /2 l uv l v 4l h r u 2 l /2 l uv. Now, since p is a critical point, p r is normal to the surface at p, so is equal to h r (pn p, since h r (p is definde to be the length of p r. Also,

11 GEOMETRY HW 4 notice that I p (w since w. by what we showed in 22(a above, H p h r (w h ruu (u 2 + 2h ruv u v + h rvv (v 2 2 l /2 l uu (u 2 + l /2 l uv u v + 2 l /2 l vv (v 2 h ( r(p 2 l uu(u 2 + l uv u v + 2 l vv(v 2 h (( φ r(p uu, p r + φ u, φ u (u 2 +2( φ uv, p r + φ u, φ v u v + ( φ vv, p r + φ u, φ v (v 2 h ((ke + r(p E(u 2 + (kf + F u v + (kg + G(v 2 h (I r(p p(w h r (pii p (w h II r(p p(w. Now, we need only recall that, since w is a unit vector, II p (w k n. Plugging this into the above equation, we see that H p h r (w h r (p k n. Since h r (p is fixed in this context, we see that the value of the quadratic form H p h r depends solely on the value of k n. Now, k k 2 are the maximum minimum values, respectively, of k n II p (w. With these values are associated the unit vectors e e 2, respectively, which are mutually orthogonal. Now, it is clear that e minimizes H p h r e 2 maximizes H p h r. e e 2 are the eigenvectors of the associated self-adjoint linear map A p h r. Since {e, e 2 } is an orthonormal set, these vectors diagonalize the map A p h r. Finally, p is a degenerate critical point of h r if only if A p h r is singular; A p h r is singular if only if H p h r (e or H p h r (e 2. Now, solving for h r (p, we see that H p h r (e h r (p k h r (p k H p h r (e 2 h r (p k 2 h r (p. k 2 Therefore, p is a degenerate critical point if only if h r (p k or h r (p k 2. (c Show that the set B {r R 3 : h r is a Morse function} is an open dense set in R 3 ; here dense in R 3 means that in each neighborhood of a given point of R 3 there exists a point of B. Proof. A point a R 3 is associated with a non-morse function h a if only if h a (p k or h a (p k 2 for some p S such that the line from p to a is parallel to N p. if we attach line segments of length k k 2 parallel to N p N p to each p S, then the endpoints of these line

12 2 CLAY SHONKWILER segments will comprise the entirety of the set R 3 \B. We want to show that, for any point q R, any ɛ-ball B ɛ (q centered at q, there is a point r B ɛ (q such that r B; that is, r is not an endpoint of one of these line segments. Now, suppose B is not dense in R 3. Then there exists a q R 3 an ɛ > such that the open ɛ-ball B ɛ (q centered at q is contained in the complement of B. That is to say, every point in B ɛ (q is the endpoint of one of the line-segments constructed above. Specifically, q is the endpoint of such a line-segment l; suppose, without loss of generality, that the length of l is k. In this situation, there are two possibilities: ( If we translate B ɛ (q by a distance k such that the image of q is in S, then the image of the ball is contained in S. However, this is clearly impossible, as S is locally diffeomorphic to R 2, as such, certainly cannot contain such a ball. (2 If, when we perform the above, there are points in the ball whose image is not contained in S. Denote by B ɛ(q the ball with all points whose image under this k -translation is contained in S deleted. Then, if we translate B ɛ(q by a distance of k 2, then its image will be contained in S. However, again this is impossible, since B ɛ(q is not locally diffeomorphic to R 2, either. From this contradiction, then, we conclude that, in fact, B is dense in R Prove that the differentiability of a vector field does not depend on the choice of a coordinate system. Proof. Suppose w is a vector field on a neighborhood U S with a coordinate chart φ(u, v such that a(u, v b(u, v are differentiable, where w(p a(u, vφ + b(u, vφ 2. Now, suppose there is another coordinate chart ψ(u, v on U such that w(p c(u, v Ψ + d(u, v Ψ 2. Then it suffices to show that c d are differentiable. Recall that Φ u u Ψ + v u Ψ 2 Φ 2 u v Ψ + v v Ψ 2.

13 GEOMETRY HW 4 3 w(p a(u, vφ( + b(u, vφ 2 a(u, v u u Ψ + v u Ψ 2 ( a(u, v u u u + b(u, v v c(u, v Ψ + d(u, v Ψ 2. ( + b(u, v u Ψ + v Ψ + v v Ψ 2 ( a(u, v v u + b(u, v v v Ψ 2 since we see that c d are merely sums of products of differentiable functions, c d are differentiable. Therefore, w is a differentiable vector field with respect to ψ as well. Since our choice of ψ was arbitrary, we see that differentiability of the vector field w is independent of the choice of coordinate system. DRL 3E3A, University of Pennsylvania address: shonkwil@math.upenn.edu