GEOMETRY FINAL CLAY SHONKWILER

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1 GEOMETRY FINAL CLAY SHONKWILER 1 a: If M is non-orientale and p M, is M {p} orientale? Answer: No. Suppose M {p} is orientale, and let U α, x α e an atlas that gives an orientation on M {p}. Now, let V, y e a coordinate chart on M in some atlas containing the point p. If U, x is a coordinate chart such that U V, then, since x and y are oth smooth maps with smooth inverses, the composition x y 1 : yu V xu V is a smooth map. Hence, the collection {U α, x α, V, α} gives an atlas on M. Now, since M {p} is orientale, det x i α x j β > 0 for all α, β. However, since M is non-orientale, there must exist coordinate charts where the Jacoian is negative on the overlap; clearly, one of each such pair must e V, y. Hence, there exists β such that det 0. x j β Suppose this is true for all β such that U β V. Then, since yq = y 1 q,..., y n q, let ỹ e such that ỹp = y 2 p, y 1 p, y 3 p, y 4 p,..., y n p. Then ỹ : V ỹv is a diffeomorphism, so V, ỹ gives a coordinate chart containing p and ỹ i det = det 0 x j β x j β for all β such that U β V. Hence, unless equality holds in some case, {U α, x α, V, ỹ} defines an orientation on M, contradicting the fact that M is non-orientale. Additionally, it s clear that equality can never hold, since y : U β V yu β V is a diffeomorphism and, thus, has full rank. 1

2 2 CLAY SHONKWILER Therefore, the aove contradiction implies that there exist β, γ such that det > 0 det < 0. x j β x j γ Let c e a path in V from q 1 U β V to q 2 U γ V such that c does not pass through p. Then the U α cover c and c is compact, so a finite sucollection U 1,..., U k covers c, where U 1 = U β and U k = U γ. Since det x j 1 > 0 and det l {1,..., k} such that det > 0 det x j l x j l+1 x j k < 0, there must exist < 0. Now, W = U l U l+1 V, since U l and U l+1 overlap on some segment of c, which is contained in V. Note that x i l = x j l+1 x j l x j l+1 y the chain rule, so 0 > det = det det x j l+1 x j l x i l x j l+1 > 0. From this contradiction, then, we conclude that M {p} is not orientale. : Let f : M N e a local diffeomorphism. If one of M or N is orientale, is the other orientale? Answer: If N is orientale, then so is M. If M is orientale, then N need not e. Suppose f : M N is a local diffeomorphism and that N is orientale. Then, at each point p M, there exists a neighorhood U p of p such that f : U p fu p is a diffeomorphism. Let V p, y p e a coordinate chart containing the point fp and let V p = fu p V p. Then fp V p fu p, so U p = f 1 V p is a neighorhood of p such that f U p : U p V p is a diffeomorphism. Furthermore, y p V p f U p : U p y p V p V p R n is a diffeomorphism, since it is the composition of diffeomorphisms. Now, suppose p, q M such that U p U q. Then y p V p f U p y q V q f U q 1 = y p V p f U p f 1 U y q 1 q V q = y p V p y q 1 V q

3 GEOMETRY FINAL 3 is a differentiale map from y q V p V q to y p V p V q. Furthermore, since we can find such neighorhoods for each p M, U p = M. p M Therefore, U p, y p V p f U p defines a differentiale structure on M. Finally, note that if U p U q, then detdy p V p f U p y q V q f U q 1 = detdy p V p y q 1 V > 0 q since N is orientale. Therefore, M is orientale. On the other hand, if M is orientale and f : M N is a ijective local diffeomorphism, then f 1 : N M is a local diffeomorphism and we can apply the aove result to see that N is orientale. However, if f is not surjective, then N is not necessarily orientale. To see why, let N e any non-orientale manifold. Let U, x e a coordinate chart containing some point p N. Then x : U xu R n is a diffeomorphism; let M = xu. Since M = xu R n is open, M is an orientale manifold and the map x 1 : M N is a local diffeomorphism. However, despite M eing orientale, N is not orientale. Furthermore, if f : M N is not injective, then the orientaility of M does not guarantee the orientaility of N. For example, consider the standard projection π : S 2 RP 2. Let M = S 2 and N = RP 2. Then M is orientale and N is not. Let U α, x α e an oriented atlas on M, let p M and let U, x e a coordinate chart in this atlas containing p. Then, since N = S 2 /{±1}, we know that π α U α, x α π α 1 U α gives an atlas on N. Hence, oth U and πu are diffeomorphic to xu and so π U : U πu is a diffeomorphism. Since our choice of p M was aritrary, we see that such a neighorhood exists for all p M, so π is a local diffeomorphism. Therefore, we conclude that f : M N eing a local diffeomorphism and M eing orientale does not necessarily mean that N is orientale. 2 Let X e a vector field on a manifold M such that Xp 0 for all p M. Show that there exists a diffeomorphism f : M M without any fixed points.

4 4 CLAY SHONKWILER 3 Let f : SL2, R R where fa = tra. Find the regular values of f. What is the diffeomorphism type of f 1 q for q regular? What does f 1 q look like for q singular? a Answer: Let SL2, R. Then ad c = 1, so either a and d c d are oth non-zero or and c are oth non-zero. Let { } { a a U 1 = a, d 0, U c d 2 = c d Define the map φ 1 : U 1 R 3 y a c d a,, d. }, c 0. Since 0, and ad c = 1, c = ad 1, so we can define ψ 1 : R 3 {a, 0, d} SL2, R y a a,, d. d ad 1 Then detψ 1 a,, d = ad ad 1 = ad ad 1 = 1 and a φ 1 ψ 1 a,, d = φ 1 ad 1 = a,, d d and a ψ 1 φ 1 c a = ψ d 1 a,, d = d ad 1 since ad c = 1, we see that c = ad 1. Therefore, φ 1 and ψ 1 are inverses; since we can view SL2, R as a suset of R 4 and φ 1 and ψ 1 are oth clearly smooth when viewed as maps etween R 4 and R 3 {a, 0, d}, we conclude that, in fact, φ 1 is a homeomorphism of U 1 with R 3 {a, 0, d}. Similarly, define the map φ 2 : U 2 R 3 y a a,, c. c d Then, since a 0 and ad c = 1, d = 1+c a, so we can define ψ 2 : R 3 {0,, c} SL2, R y a a,, c c Clearly, oth φ 2 and ψ 2 are smooth. Furthermore, y a similar argument to that given aove, φ 2 ψ 2 = id and ψ 2 φ 2 = id, so φ 2 : U 2 R 3 {0,, c} is a homeomorphism. Now, φ 1 U 1 φ 2 U 2 = R 3 {a, 0, d} {0,, c}, so we need to check that φ 1 φ 1 2 is smooth on this overlap. To that end, let a,, c R 3 such 1+c a. ;

5 that a, 0. Then φ 1 φ 1 2 a,, c = φ 1 GEOMETRY FINAL 5 a c 1+c a = Since a 0, this is a smooth map. Similarly, φ 2 φ 1 a 1 a,, c = φ 2 = c ad 1 a,, 1 + c. a a,, ad 1. Since 0, this is also a smooth map. Therefore, we conclude that U i, φ i gives an atlas on SL2, R for i = 1, 2. Now, consider the trace map. If x 1 0, then tr φ 1 1 x 1, x 2, x 3 = tr x1 x 2 x 1 x 3 1 x 2 x 3 = x 1 + x 3, so tr φ 1 1 Dtr = tr φ 1 1, tr φ 1 1, = 1, 0, 1, x 1 x 2 x 3 which has rank 1, so none of these points is critical. On the other hand, if 0, then tr φ 1 2 x x1 x 2 1, x 2, x 3 = tr 1+x x 2 x 3 = x x 2x 3, 3 x 1 x 1 so tr φ 1 2 Dtr = tr φ 1 2, tr φ 1 2, = x 1 x 2 x x 2x 3 x 2, x 3, x 2 1 x 1 x 1 which almost always has rank 1. The only way this matrix can have rank 0 is if x 2 = x 3 = 0 and 1 + x 2 x 3 x 2 = 1 x 2 1 = 1 + x 2 x 3 = 1, 1 i.e. x 1 = ±1. Since 1+x 2x 3 ±1 = ±1, we see that the only critical points of the trace map are ±1 0 0 ±1 so the critical values of the trace map are ±2. ±1 = 1 4 True or false? a: Every one form on S 1 = {x R 2 x = 1} can e extended to a one form on R 2. Answer: True. Let ω Ω 1 S 1. Define the map g : R 2 {0} S 1 y gx = x x.,

6 6 CLAY SHONKWILER Then g is certainly a smooth map, and so g ω is a 1-form on R 2 {0}. Now, let f : R 2 R e a smooth map such that f0, 0 = 0 and f = 1 outside B1/2, the all of radius 1/2. Then fg ω is certainly a form on R 2 {0}, since f : R 2 {0} R is smooth. Furthermore, since fg ω is zero in a neighorhood of the origin, can e extended to a form η on R 2 simply y defining η = fg ω away from the origin and η = 0 at the origin. Then, since f = 1 on S 1 and g S 1 is the identity, η = fg ω = ω on S 1, so we see that η is an extension of the form ω to all of R 2. : Every closed one form on S 1 can e extended to a closed one form on R 2. Answer: False. Let ω Ω 1 S 1 e non-zero and closed. Suppose there existed an extension η of ω to all of R 2 such that dη = 0. Let B1 e the all of radius 1 in R 2. Then, y Stokes Theorem, 0 = dη = η = ω 0, B1 B1 S 1 as we showed on Homework 12, prolem 3. From this contradiction, we conclude that there is no extension of the non-zero closed form ω Ω 1 S 1 to a closed form on R 2. 5 Let M n e compact and orientale. Show that if ω is an n 1-form, then dω is zero at some point. Proof. Since M has no oundary, we know, y Stokes Theorem, that dω = ω = ω = 0. M M Let U α, x α e an atlas on M that yields an orientation. We may assume that, for each α, x α U α = R n since x α U α R n is open and, hence, diffeomorphic to R n. Since M is compact, there exists a finite suset U 1,..., U k of the U α s such that k M = U i. Let {φ i } k i=1 e a partition of unity suordinate to the U i. Then k k k 0 = dω = φ i dω = x 1 i φ i dω = x 1 i φ i dω. M U i R n R n i=1 i=1 i=1 Now, suppose dω were everywhere non-zero. Suppose that dω > 0 at some p M and dω < 0 at some q M. Then let c : [0, 1] M e a path from p to q. Then dω must e zero at some point cs for s [0, 1]. Therefore, we 0 i=1

7 GEOMETRY FINAL 7 see that dω > 0 or dω < 0 on all of M. Assume, without loss of generality, that dω > 0 on all of M. Then dω x 1 i > 0 on all of R n. Therefore, x 1 i φ i dω 0 at all points and, furthermore, must e strictly positive at some point, since φ i 0 and is a partition of unity and, therefore, non-negative at every point. Thus, we see that k x 1 i φ i dω > 0. R n i=1 From this contradiction, then, we conclude that dω cannot e everywhere non-zero, and so dω is zero at some point. 6 Let f : M N e smooth, M compact, and M and N orientale. Suppose there exists an open set U N such that f : f 1 U U is a diffeomorphism. Show that N is compact. Proof. Suppose N is not compact. Note that, since f : f 1 U U is a diffeomorphism, M and N are manifolds of the same dimension; call this dimension n. Now, let ω e an n-form on M such that suppω W f 1 U where W, x is a coordinate chart contained in f 1 U and M ω 0 for example, take some volume form on W and multiply it y a smooth, non-negative function g : M R such that suppg W and g = 1 on some open set contained in W. Note that W, the closure of W, is compact and contained in f 1 U, so f W is compact and contained in U. Hence, W and f W are compact n-manifolds with oundary. Then, since f 1 W : f W W is smooth, η = f 1 W ω is an n-form on U. Furthermore, η takes the oundary of f W to the oundary of W. Now, extend η to a form η on all of N y letting η = 0 outside of f W. Since all n-forms are closed, [η ] H n N. Now, since N is not compact, H n N = 0, so [η ] = 0 in H n N. Now, since f is smooth, f induces a homomorphism f : H n N H n M. Now consider f η. On W, is zero on the oundary of f W, since f 1 W f W = f f 1 W ω = ω. Now, since f 1 f W = W ecause f : f 1 U U is a diffeomorphism and since η = 0 outside of f W, we see that f η = 0 fdx 1... dx n = 0 outside of W. However, this means that f η = ω oth inside and outside of W, so f η = ω. Since M is compact, H n M = R and [ω] ω gives the isomorphism etween H n M and R. Now, recall that M ω 0, so ω is a non-zero element of H n M. Thus, f : H n N H n M is a vector space homomorphism mapping zero i.e. [η ]to a non-zero element M

8 8 CLAY SHONKWILER i.e. [ω]. This is clearly impossile, so, from this contradiction, we conclude that, in fact, N is compact. 7 Define what one means y the product of two Lie algeras. Show that the Lie algera of SO4 is the direct product of 2 Lie algeras each one of which is R 3, with its cross product. Answer: Let g and h e two Lie algeras. Define the operation [, ] : g h g h y [X 1, Y 2, X 2, Y 2 ] = [X 1, X 2 ], [Y 1, Y 2 ]. Then we see that, for X, Y g h, [X, Y, X, Y ] = [X, X], [Y, Y ] = 0, 0 since g and h are Lie algeras. Furthermore, if X 1, Y 1, X 2, Y 2, X 3, Y 3 g h, [[X 1, Y 1, X 2, Y 2 ], X 3, Y 3 ] + [[X 2, Y 2, X 3, Y 3 ], X 1, Y 1 ] + [[X 3, Y 3, X 1, Y 1 ], X 2, Y 2 ] = [[X 1, X 2 ], [Y 1, Y 2 ], X 3, Y 3 ] + [[X 2, X 3 ], [Y 2, Y 3 ], X 1, Y 1 ] + [[X 3, X 1 ], [Y 3, Y 1 ], X 2, Y 2 ] = [[X 1, X 2 ], X 3 ], [[Y 1, Y 2 ], Y 3 ] + [[X 2, X 3 ], X 1 ], [[Y 2, Y 3 ], Y 1 ] + [[X 3, X 1 ]X 2 ], [[Y 3, Y 1 ], Y 2 ] = [[X 1, X 2 ], X 3 ] + [[X 2, X 3 ], X 1 ] + [[X 3, X 1 ], X 2 ], [[Y 1, Y 2 ], Y 3 ] + [[Y 2, Y 3 ], Y 1 ] + [[Y 3, Y 1 ], Y 2 ] = 0, 0, since the Jacoi identity holds in g and h. Therefore, we see that g h with racket defined as aove is a Lie algera. Now, let G and H e Lie groups with corresponding Lie algeras g and h. Then G H is also a Lie group with identity e G, e H. Let X 1, Y 1, X 2, Y 2 T eg,e H G H. Then X 1, Y 1 = X 1, Ỹ1 and X 2, Y 2 = X 2, Ỹ2, so [X 1, Y 1, X 2, Y 2 ] = [ X1, Y 1, X2, Y 2 ]e G, e H = X 1, Y 1 X 2, Y 2 e G, e H X 2, Y 2 X 1, Y 1 e G, e H = X 1, Ỹ1 X 2, Ỹ2e G, e H X 1, Ỹ1 X 2, Ỹ2e G, e H = X 1 X2 e G, Ỹ1Ỹ2e H X 2 X1 e G, Ỹ2Ỹ1e H = X 1 X2 e G X 2 X1 e G, Ỹ1Ỹ2e H Ỹ2Ỹ1e H = [ X 1, X 2 ]e G, [Ỹ1Ỹ2]e H = [X 1, X 2 ], [Y 1, Y 2 ], so we see that g h as defined aove is the Lie algera for G H. Now, turning to the case of SO4, we showed in Homework 8 Prolem 5 that there exists a Lie group homomorphism F : S 3 S 3 SO4, which in turn induced a diffeomorphism G : S 3 S 3 /x, y x, y. Now, since {1, 1, 1, 1} is a discrete normal sugroup, S 3 S 3 /x, y x, y is also a Lie group. Furthermore, since the group operation on

9 GEOMETRY FINAL 9 this group is induced y the group operation on S 3 S 3 and F respects this group operation, we see that G is, in fact, a Lie group isomorphism. Since the coordinate chart on S 3 S 3 /x, y x, y is simply given y πu, x π 1 U where π is the standard projection and U, x is an atlas on S 3 S 3, we see that these two Lie groups have the same tangent plane at the identity and, thus, the same Lie algeras. If s 3 is the Lie algera of S 3, then our work aove indicates that this Lie algera is, in fact, s 3 s 3, which, given the aove Lie group isomorphism, is also the Lie algera of SO4, which we might call so4. Now, as we showed on Homework 8 Prolem 2, s 3 is isomorphic to R 3,, 3-space with its cross product. Therefore, we conclude that so4 R 3, R 3,. Show that every 2 form on R 6 has one of the following forms: 8 e 1 e 2, e 1 e 2 + e 3 e 4, e 1 e 2 + e 3 e 4 + e 5 e 6 with respect to some asis e 1,..., e 6 not necessarily the same asis for each form. Furthermore show that these 3 cases are mutually exclusive. 9 Compute the DeRham cohomology groups of the Klein ottle. Answer: Let K denote the Klein ottle. Then, since K is connected, H 0 K = R. Also, since K is non-orientale, H 2 R = 0. Now, consider the two open susets of K, U and V, indicated in the elow picture: Then oth U and V are oth cylinders and, as such, each is homotopic to S 1. Furthermore, U V consists of two extremely short cylinders and so is homotopy equivalent to the disjoint union of two copies of S 1. Now, form

10 10 CLAY SHONKWILER the Mayer-Vietoris sequence: 0 H 0 K H 0 U H 0 V H 0 U V H 2 K H 1 U V H 1 U H 1 V H 1 K Note that H 0 K = R, H 0 U = H 0 V = H 0 S 1 = R and H 1 U = H 1 V = H 1 S 1 = R. Furthermore, since U V has two components, an element of H 0 U V is a constant function on each component ut can e different constants on each component and since no zero form is exact, we see that H 0 U V = R R. Finally, the closed 1-forms on the disjoint union of two copies of S 1 consists of any two closed 1-forms on S 1 so, since H 1 S 1 = R, we see that H 1 U V = H 1 U V = R R. Let us indicate these calculations on the sequence: 0 H 0 K H 0 U H 0 V H 0 U V R R R R R H 2 K H 1 U V H 1 U H 1 V H 1 K 0 R R R R Therefore, we see that H 0 K H 0 U H 0 V is injective, so its image is 1-dimensional and, hence, the kernel of H 0 U H 0 V H 0 U V is 1-dimensional, meaning its image is also 1-dimensional. In turn, this implies that the kernel of H 0 U V H 1 K is 1-dimensional; since H 0 U V is 2-dimensional, the image of this map is also 1-dimensional, and so the kernel of H 1 K H 1 U H 1 V is 1-dimensional. On the other hand, since H 1 U H 1 V H 1 U V must e surjective and oth spaces are 2-dimensional, the kernel of this map must e trivial, and so the image of H 1 K H 1 U H 1 V is trivial. Therefore, since H 1 K H 1 U H 1 V has trivial image and 1-dimensional kernel, we see that H 1 K is 1-dimensional and, thus, must e isomorphic to R as a vector space. Therefore, we see that the complete cohomology of K is { H i R i = 0, 1 K = 0 i = 2 DRL 3E3A, University of Pennsylvania address: shonkwil@math.upenn.edu