Math 426H (Differential Geometry) Final Exam April 24, 2006.

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1 Math 426H Differential Geometry Final Exam April 24, Let M be a surface and let : [0, 1] M be a smooth loop. Let φ be a 1-form on M. a Suppose φ is exact i.e. φ = df for some f : M R. Show that φ = 0. Solution: φ = df f [1] f [0] f [f0] f [f0] = 0. Here, is the Fundamental Theorem of Calculus for 1-forms Theorem 4.6.2, p.170, while is because is a loop, so 0 = 1. b Let M := R 2 \ {0, 0} be the punctured plane. Compute dφ, where φ := fx, y dy gx, y dx, with fx, y := Solution: First, observe that x fx, y = x2 + y 2 2x 2 x 2 + y 2 2 = By switching x with y, we also get y fx, y = y2 x 2 x 2 + y 2 2. Thus, x x 2 + y 2 and gx, y := y x 2 + y 2. x 2 y 2 x 2 + y 2 2. dφ = x fx, y dxdy y gx, y dydx = x fx, y + y gx, y dxdy = x2 y 2 + y 2 x 2 x 2 + y 2 2 dxdy = 0. c Let : [0, 2π] M be the unit circle: t := cost, sint. Compute Solution: If t [0, 2π], then t = sint, cost and φt = cost dy sint dx. Thus, φ[ t] = [cost dy sint dx] sint, cost = sint 2 + cost 2 = 1. 2π 2π Thus, φ = φ[ t] dt = 1 dt = 2π. 0 0 d Is φ closed? Is φ exact? Is nullhomotopic in M? Justify your answers. Solution: φ is closed because part b says dφ = 0. φ is not exact because if it was, then a would imply that γ φ = 0, which contradicts c. Likewise, γ is not nullhomotopic, because if it was, then Lemma p.182 would imply that γ φ = 0, which contradicts c. The existence of a nonexact closed 1-form shows that the de Rham cohomology group H 1 M is nontrivial. The existence of a loop which is not nullhomotopic shows that the fundamental group π 1 M is nontrivial. Each of these indicates the presence of the hole in M. e Let L := {x, 0 ; x 0} be the positive x axis, and let N := R 2 \L be the plane with this line removed, as in Figure 1A below [thus, N M]. Is φ exact when restricted to N? Why or why not? You may use any theorem from the text. Solution: Yes, φ is exact on N. The surface N is simply connected. Thus, any closed 1-form on N is exact, by Poincaré s Lemma Lemma 4.7.9, p.183. Thus, φ is exact on N, because φ is closed by part b. φ.

2 replacements R 2 \ L RP 2 V α 0,0 A B Figure 1: A Question #1e B Question #4c Let M be a surface and let p M. Suppose v p, w p T p M are two orthonormal tangent vectors, and that S v p S w p = 0. What can you conclude about any special geometric properties of v p and w p, or about the principal/gaussian/mean curvatures of M at p? Solution: We conclude that either H = 0, or v p and w p are principal vectors. Case 1: If p is an umbilic point, then every vector is a principal vector including v p and w p. In this case, S is just multiplication by some scalar k, so S v p = k v p and S w p = k w p, so that S v p S w p = k v p k w p = k 2 v p w = k 2 0 = 0, Case 2: Suppose that p is not an umbilic point. Let e 1 and e 2 be the unique principal vectors at p; then e [ 1 and e 2 ] are orthonormal eigenvectors of S, so the matrix of S relative to the basis k1 0 { e 1, e 2 } is for some k 0 k 1 > k 2 note that k 1 k 2 because we assume p not umbilic. 2 Let v p = v 1 e 1 + v 2 e 2 and w p = w 1 e 1 + w 2 e 2 for some v 1, v 2, w 1, w 2 R. Then S v p = Sv 1 e 1 + v 2 e 2 = v 1 S e 1 + v 2 S e 2 = k 1 v 1 e 1 + k 2 v 2 e 2. Likewise, S w p = k 1 w 1 e 1 + k 2 w 2 e 2. Thus, S v p S w p = k 1 v 1 e 1 + k 2 v 2 e 2 k 1 w 1 e 1 + k 2 w 2 e 2 = k 2 1v 1 w 1 e 1 e 1 + k 1 k 2 v 1 w 2 e 1 e 2 + k 1 k 2 v 2 w 1 e 2 e 1 + k 2 2v 2 w 2 e 2 e 2 k 2 1v 1 w 1 + k 2 2v 2 w 2 = k 2 1v 1 w 1 + k 2 1v 2 w 2 + k 2 1 k 2 2v 2 w 2 = k 2 1v 1 w 1 + v 2 w 2 + k 2 1 k 2 2v 2 w 2 = k 2 1 v p w p + k 2 1 k 2 2v 2 w 2 k 2 1 k 2 2v 2 w 2. where is because e i e j = δ ij, and is because v p w p = 0 because v p is orthogonal to w p. Thus, v 2 w 2 = S v p S w p 0 k1 2 = k2 2 k1 2 = 0. k2 2 At this point there are two possibilities: Case 2a: Suppose k 2 1 k2 2 = 0. Then k 1 + k 2 k 1 k 2 = 0. Thus, H = k 1 + k 2 = 0, because k 1 k 2 cannot be zero, because k 1 k 2 because we assume p is not umbilic.

3 Case 2b: If k 2 1 k2 2 0, then either v 2 = 0 or w 2 = 0. Suppose that v 2 = 0; then v 1 = 1 because v v2 2 = v p 2 = 1, so that v p = e 1. But then w p = ± e 2 because w p is normal to v p = e 1 and e 2 is also normal to e 1. We conclude that v p = e 1 and w p = e 2 are the two principal vectors Let M, M R 3 be two surfaces, and let f : M M be a local isometry. Let α : [0, 1] M be a unit-speed curve, and let α := f α : [0, 1] M. Show that α is a geodesic on M α is a geodesic on M. You may use any result from previous quizzes. Solution: Let E 1, E 2 be a tangent frame field on α so that E 1 = α and E 2 is orthogonal to E 1. Let E 1, E 2 be the corresponding frame field on α i.e. for all t [0, 1], E 1 = F E 1 = α t, and E 2 = F E 2 ; hence E 1, E 2 is a frame field along α, because F preserves inner products. Let ω 12 be the connection form for E 1, E 2 and let ω 12 be the connection form for E 1, E 2. Then Lemma 6.5.3b says that ω 12 = F ω 12. Thus, α is a geodesic on M ω 12 α 0 F ω 12 α 0 ω 12 F α 0 ω 12 α 0 α is a geodesic on M. Here, is by exercise #6.1.1 which appeared on Quiz #9. 1 Some people tried to map the second derivative through the tangent map. In other words, they wrote expressions like α = F α. There are two problems with this: α is not necessarily an element of T p M indeed, if α is a geodesic, then α is normal to T p M. Thus, F α is not well-defined, because F is only defined for tangent vectors. The correct equation α = F α is just a special case of the Chain Rule. It can be true because α and F are both first derivatives. However, α is a second derivative. Hence, if α was a tangent vector, then the proper equation would be something like α = F α, where F is the second derivative of F that is, the derivative of F that is, the tangent map of the tangent map i.e. the double tangent map. In fact, F is a well-defined object, but it is more complicated than you think. Let s adopt the old notation F p := T p F. Recall that the tangent bundle TM is a four-dimensional smooth manifold. If we glue together the tangent maps T p F : T p M T F p M for all p M, then we get a function TF : TM TM, and TF is a smooth mapping between these two four-dimensional manifolds. Thus, we can then compute the tangent map of TF itself, to get a function TTF : TTM TTM. This is a smooth function between the double-tangent bundles which are smooth, eight-dimensional manifolds and it is the correct generalization of the second derivative for smooth mappings on manifolds. 2 Many people argued as follows: α is a geodesic iff α is orthogonal to M. If F is an isometry, then F preserves inner products; hence F α will be orthogonal to M. The problem here is that F only preserves inner products between tangent vectors. Indeed, F is only defined on tangent vectors. If α is orthogonal to M, then F α isn t even well-defined. In princial, we could well-defined F α by smoothly extending F from M to a smooth map F : R 3 R 3. In this case, F α would be well-defined because α is a tangent vector to R 3.

4 However, the fact that F is an isometry only guarantees that F preserves inner products and hence, orthogonality between vectors tangent to the surface M, and α is not tangent to M, so there is no guarantee that F α will be orthogonal to anything. 3 Some people wrote down basically the correct proof, but they forgot to define the frame fields E 1, E 2 and E 1, E 2. In other words, they just said, Let ω 12 be the connection 1-form of M... and went from there. The problem is that the connection 1-form is only well-defined with respect to a specific frame field. There is not such thing as the connection 1-form of M you can only talk about the connection 1-form of the frame field E 1, E 2 defined on M Let M be an abstract surface. a Define what it means for M to be orientable. Solution: M is orientable if there is a nonvanishing smooth 2-form η on M. Here, nonvanishing means that, for all p M, and any pair of linearly independent tangent vectors v p, w p T p M, η[ v p, w p ] 0. 4 Some people said, M is orientable if there is a nonvanishing normal vector field on M. This definition is correct, but only for surfaces embeddeded in R 3. The problem is that normal vector fields are only well-defined on surfaces embedded in R 3, and here, M is an abstract surface. This is an important distinction, because some abstract surfaces [such as the real projective plane RP 2 in part c] simply cannot be embedded in R 3. Thus, it makes no sense to talk about putting a normal vector field on RP 2 nonvanishing or otherwise. In other words, you just can t do part c unless you use the abstract surface definition of orientability. b Suppose there is a smooth closed curve α : [0, 1] M [i.e. α0 = α1 and α 0 = α 1] and a smooth tangent vector field V : [0, 1] TM defined along γ, such that: i. V t and α t are linearly independent for all t [0, 1]. ii. V 1 = V 0. Show that M is not orientable. Solution: Suppose M was orientable, and let η be a nonvanishing two-form. Define the function f : [0, 1] R by ft := η[α t, V t]. Then f is a smooth function because α, V, and η are smooth and ft 0 for all t by hypothesis i, because η is nonvanishing. However, hypothesis ii implies that f1 = η[α 1, V 1] = η[α 0, V 0] = η[α 0, V 0] = f0. Thus, if f0 > 0, then f1 < 0, whereas if f0 < 0, then f1 > 0. Either way, the Intermediate Value Theorem says that ft = 0 for some t 0, 1. Contradiction. c Figure 1B is a picture of the real projective plane RP 2 as a disk with opposite boundary points identified. Draw a smooth closed path on RP 2 and a smooth tangent vector field along this path, satisfying the hypothesis of part b. Conclude that RP 2 is not orientable. Solution: See Figure 1B. 5 Some people drew some pretty weird pictures, where it really wasn t clear which way the vector field was pointing, etc. Indeed, several people drew a path along the perimeter of

5 the disk in the picture. This is a bad idea, because every point in RP 2 corresponds to two points on the perimeter Let V and W be two smooth tangent vector fields on M. Let U be a unit normal vector field, and let S : TM TM be the shape operator. a Show that U V W = S V W. Solution: First, note that U W 0 because W is a tangent field and U is a normal field. Thus, 0 = V [0] = V [ U W ] L V U W + U V W. Here L is the Liebniz rule for directional derivatives Theorem on p.79. Thus, U V W = V U W = S V W. b The Lie Bracket of V and W is the vector field defined by: [ V, W ] := V W W V. Show that [ V, W ] is also a tangent vector field i.e. show that [ V, W ]p T p M for all p M. Hint: Use the symmetry of S. Solution: [ V, W ] is a tangent vector field iff U is orthogonal to [ V, W ]. But U [ V, W ] = U V W W V = U V W U W V S V W S W V S V W S V W = is by part a and is because S is a symmetric operator, so S W V = S V W. Thus, U [ V, W ] = 0, so [ V, W ] is a tangent vector field, as desired Let M be a torus, and let α be the top circle of M, as shown in Figure 2A above ignore the scissors for now. Is α an asymptotic curve on M? Why or why not? Solution: Yes, α is an asymptotic curve, because α is a circle in a plane P tangent to M, so the acceleration vector α is always in P, and hence, always tangent to M. Thus, α is asymptotic by the remarks on page Let M R 3 be a surface and let γ : [0, 1] M be a unit-speed curve. Let T, N, B be the Frenet frame field along γ; that is, T := γ is the tangent vector field, N is the normal vector field, and B is the binormal vector field. Let τ be the torsion of the curve γ as defined by the Frenet equations. a Show: γ is an asymptotic curve B is orthogonal to M at all points on γ. Solution: γ is an asymptotic curve γ is tangent to M N is tangent to M B is orthogonal to M. Here, is by the criterion on p is because T B N; hence T and N are tangent to M iff B is normal to M.

6 A Torus B Tire Surface α α E Cylinder γ γ δ C D Barrel Surface Sphere Figure 2: Questions #6, #7, and #8. δ 5 b Suppose γ is asymptotic. Deduce from a that S T = τ N at all points on γ. Solution: Part a says that B is a unit normal vector field along γ. Thus, S T T B = B τ N. Here, is the definition of the shape operator Defn , p.196. is because B = τn by the third Frenet formula Theorem 2.3.2, p.58. c Suppose γ is asymptotic. Show that K = τ 2 at all points on γ where K is the Gaussian curvature of M. Solution: Fix t [0, 1]. Let p := γt, and let e 1 and e 2 be the two principal vectors at p. Thus, S e 1 = k 1 e 1 and S e 2 = k 2 e 2, where k 1, k 2 are the principal curvatures. Suppose T = c e 1 + s e 2, for some c, s R; then c 2 + s 2 = 1, because T is a unit vector. Also N = s e 1 ± c e 2, because N is a unit vector orthogonal to T, and the only unit vectors orthogonal to c, s are s, c and s, c. Assume WOLOG that N = s e 1 + c e 2 ; then S T = Sc e 1 + s e 2 = cs e 1 + ss e 2 = ck 1 e 1 + sk 2 e 2, and also, S T τ N = τs e 1 + τc e 2. Here is by part b. This gives two different orthonormal expansions for S T in terms of the basis e 1, e 2. But any vector has a unique orthonormal expansion, so we conclude that ck 1 = τs and sk 2 = τc. Hence Hence K = k 1 k 2 = τs c k 1 = τs c τc s and k 2 = τc s. = τ 2, as desired.

7 d Let M be a torus, and let α be the top circle of M as in Figure 2A above ignore the scissors. Compute the Gaussian curvature of M at points on α. Solution: K 0 on α. To see this, recall that α is an asymptotic curve by question #6. Thus, part c says that K τ 2 along α. However, the torsion of α is everywhere equal to zero, because α is a circle, so it lies entirely in a single plane; hence τ 0 by Corollary p Let α and be the top and bottom circles of a torus, as shown in Figure 2A. If we cut the torus along α and, and remove the inner half, then we are left with the tire surface of Figure 2B. Similarly, let γ and δ be two parallels of a unit sphere, as shown in Figure 2C; if we cut the sphere along γ and δ, and remove the top and bottom domes, then we are left with the barrel surface of Figure 2D. Finally, consider the cylinder in Figure 2E. Are there any isometries between any two surfaces in the collection {Tire, Barrel, Cylinder}? If so, then describe these isometries. If not, then prove that no such isometries can exist. Solution: There are no isometries from any member of {Tire, Barrel, Cylinder} to any other member. To see this, we use Theorema Egregium. First, recall that the cylinder has constant Gaussian curvature K 0. Also, the unit sphere has constant Gaussian curvature K 1, so the Barrel surface a subset of the sphere also has constant Gaussian curvature K 1. Thus, Theorema Egregium says the Barrel cannot be isometric to the Cylinder. In #6d we showed that the Gaussian curvature of the torus along the curve α is equal to zero. Hence the Gaussian curvature of the Tire is arbitrarily close to zero near the boundaries α and, so the Tire cannot be isometric to the Barrel which has constant curvature K 1. Note, however, that the Gaussian curvature of the Tire is never equal to zero. The curves α and themselves are not part of the Tire. Indeed, the torus has positive Gaussian curvature on its outer side; hence the Tire has positive curvature everywhere; hence Theorema Egregium says the Tire is not isometric to the Cylinder.

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