GEOMETRY FINAL CLAY SHONKWILER

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1 GEOMETRY FINAL CLAY SHONKWILER 1 Let X be the space obtained by adding to a 2-dimensional sphere of radius one, a line on the z-axis going from north pole to south pole. Compute the fundamental group and cohomology ring of X. Answer: Let l denote the line on the z-axis from north pole to south pole, and let γ be a longitudinal arc from the north pole to the south pole on the surface of the sphere (e.g. γ = the intersection of the sphere and the xz-plane). Then we can give X a cellular structure where the north and south poles are the two 0-cells, l and γ are the two 1-cells, and the single 2-cell is given by attaching the boundary of a disc to γ. Now, since γ is contractible, we see that X and X/γ are homotopy equivalent (this is Hatcher s Proposition 0.17). Now, X/γ is just that is, X/γ (and, therefore, X) is homotopy equivalent to S 1 S 2. Therefore, π 1 (X) = π 1 (S 1 S 2 ) = π 1 (S 1 ) π 1 (S 2 ) = Z 0 = Z. Now, to compute the cohomology ring, we use the following simplicial complex for S 1 S 2 (I know this could probably be done directly from the 1

2 2 CLAY SHONKWILER CW complex, but I m more sure of this way, even if it s more tedious): Then we have the following boundary computations: (v i ) = 0 (a) = v 1 v 0 (b) = v 2 v 1 (c) = v 2 v 0 (d) = v 0 v 0 = 0 (σ 1 ) = a + b c (σ 2 ) = a + b c For each simplex, let denote its dual. Then, since δ (σ) = (σ), the above give rise to the following coboundary computations: δ(v 0) = c a δ(v 1) = a b δ(v 2) = b + c δ(a ) = σ 1 + σ 2 δ(b ) = σ 1 + σ 2 δ(c ) = σ 1 σ 2 δ(d ) = 0 δ(σ i ) = 0. Therefore, we see that v0 + v 1 + v 2 is a generator of H0 (X); further, since X is connected, we know that H 0 (X) = Z, so this is the only generator. Also, since d is not in the image of any linear combination of the vi and δ(d ) = 0, we see that d is a generator of H 1 (X). Since H 1 (X) is the free part of the abelianization H 1 (X) of the fundamental group π 1 (X) = Z, we see that H 1 (X) = Z, so this is the only generator. Finally, σ1 is a generator of H 2 (X); since [σ 1 ] = [σ 2 ] in H 2 (X), we see that this is the only generator. Thus, in the cohomology ring, if we make the renamings

3 GEOMETRY FINAL 3 α = d and β = σ 1, then we see that H (X) is a quotient of Z[α, β]. Now, since d does not lie in the boundary of σ 1 or σ 2, and [d ] [d ](σ 1 ) = 0 [d ] [d ](σ 2 ) = 0, so we see that α 2 = [d ] [d ] = 0. Also, since the product of α and β would have grade three and the square of β would have grade 4 and there is no cohomology of X in either of these dimensions, we see that αβ = 0 and β 2 = 0. Therefore, the cohomology ring H (X) = Z[α, β]/(α 2, β 2, αβ). 2 Define a natural embedding of GL(n, C) into GL + (2n, R), where + are the matrices with positive determinant. Show that the homogeneous space GL + (2, R)/GL(1, C) is contractible. Proof. For each complex number z = x + iy, we can define the associated 2 2 real matrix ( ) x y. y x This ( ) is a natural association because, if we identify C with R 2 by x + iy x, then y ( ) ( ) ( ) a b x ax by =, b a y bx + ay which we associate with (ax by) + i(bx + ay) = (a + ib)(x + iy). In fact, if we multiply purely as matrices, we see that ( ) ( ) ( a b x y ax by ay bx = b a y x ay + bx ax by which, again, we associate with (ax by) + i(bx + ay) = (a + ib)(x + iy). Hence, for each entry in M GL(n, C), we associate a 2 2 block in a matrix M GL(2n, R) (i.e. the top left entry associates with the top left block, the entry to the right associates with the block to the right, etc.). It s clear that this is an injective map. Furthermore, if we consider an element a 11 + ib 11 a 12 + ib 12 a 1n + ib 1n a 21 + ib 21 a 22 + ib 22 a 2n + ib 2n..... GL(n, C). a n1 + ib n1 a n2 + ib n2 a nn + ib nn ),

4 4 CLAY SHONKWILER as a vector (a 11 +ib 11, a 12 +ib 12,..., a nn +ib nn ) C n2, then there s a natural identification with (a 11, b 11, a 12, b 12,..., a nn, b nn ) R 2n2. In turn, the map described above gives a map R 2n2 R 4n2 given by (a 11, b 11,..., a nn, b nn ) (a 11, b 11, a 12, b 12,..., b 11, a 11, b 12, a 12,..., a nn, b nn ), which is certainly smooth; hence, this map is an embedding of GL(n, C) into M(2n, 2n, R) = R 4n2. Further, for any M GL(n, C), M has an inverse M 1 ; the images of these two matrices in M(2n, 2n, R) are inverses, since we can just do matrix multiplication blockwise. Hence, we ve defined an embedding GL(n, C) GL(2n, R). To see that this really is the desired embedding, we need to see that the determinants of all matrices in the image are positive. However, this is clear, for the following reason: our embedding is certainly continuous, and the continuous image of a connected set is connected; GL(2n, R) has two connected components, so the image of GL(n, C) must lie in one of them; since the identity element in GL(n, C) is mapped to the identity in GL(2n, R), the image of GL(n, C) must be contained in GL + (2n, R). Now, we turn to the case where n = 1. If we take a generic element z C, then we can re-write z = re iθ = r cos θ + ir sin θ, so the image of z under the embedding is ( r cos θ r sin θ r sin θ r cos θ ) ( cos θ sin θ = r sin θ cos θ The rightmost term is just a scalar multiple of a generic element of SO(2); since r > 0 and θ can be completely arbitrary, we see that the image of GL(1, C) under this embedding is precisely {ra : r R +, A SO(2)}. Now, suppose B GL(2, R). Then the columns of B form a basis for R 2, {c 1, c 2 }. If we apply the Gram-Schmidt process to the columns of B, then we get a map from B to some element of SO(2) (since the elements of SO(2) are precisely those matrices in GL + (2, R) whose columns form an orthonormal basis for R 2 ). Now, the Gram-Schmidt process simply sends c 1, column 1 of B, to b 1 = c 1 c 1 and c 2 (column 2) to b 2 = c 2 (c 2 b 1 )b 1. In coordinates, if ( ) a b M =, then the Gram Schmidt process maps c d M ( a a 2 +c 2 c a 2 +c 2 b a2 b+acd a 2 +c 2 d abc+c2 d a 2 +c 2 Since a and c cannot simultaneously be zero, we see that this map is continuous. Furthermore, the Gram-Schmidt process fixes elements of SO(2), so we see that this defines a deformation retraction of GL(2, R) to SO(2). Since SO(2) is contained in the image of GL(1, C) under our embedding, we see that the homogeneous space GL(2, R)/CL(1, C) is contractible. ). ).

5 GEOMETRY FINAL 5 3 Let U be an open subset of R 3. Show that H 1 (U) is torsion free. What if U is open in R n with n > 3? Proof. Note that U is naturally a 3-manifold and inherits the standard orientation on R 3, so U is a non-compact, orientable 3-manifold. Therefore, by Poincaré Duality with compact support, Now, by definition, H 2 c (U) H 1 (U). Hc 2 (U) = lim H 2 (U, U K) K U where the K are compact; therefore, if we can show that H 2 (U, U K) is torsion-free, then this will suffice to show that H 2 c (U) is torsion-free and, therefore, that H 1 (U) is torsion-free. Now, U = R 3 (R 3 U) and U K = (R 3 K) (R 3 U), so, by excision of R 3 U, H 2 (R 3, R 3 K) H 2 (U, U K). Now, from the long exact sequence of the pair (R 3, R 3 K), H 1 (R 3 ) H 1 (R 3 K) H 2 (R 3, R 3 K) H 2 (R 3 ). Now, since R 3 is contractible, H 1 (R 3 ) = 0 = H 2 (R 3 ), so the exactness of the above sequence implies that H 2 (R 3, R 3 K) H 1 (R 3 K). Since there is never any torsion in H 1, we see that H 2 (R 3, R 3 K) is torsionfree. Therefore, we conclude that H 1 (U) is torsion-free. Clearly, the above proof depends on the fact that we re in dimension 3. For n > 3, this statement is not true in general. To see a counterexample, note that RP 2 is a compact 2-manifold which can be embedded into R 4 and, hence, into S 4. Therefore, by Alexander Duality, H 1 (S 4 RP 2, Z) H 2 (RP 2 ) = Z/2. Now, since S 4 RP 2 does not comprise the entire sphere and is open, it is homeomorphic (via stereographic projection) to an open subset U of R 4. Then H 1 (U, Z) = H 1 (S 4 RP 2, Z) = Z/2, which clearly has torsion. Let M 2n+1 be compact with boundary M. characteristics, we have χ( M) = 2χ(M). 4 Show that for the Euler Proof. From the long exact sequence for the pair (M, M), H k ( M) H k (M) H k (M, M) H k 1 ( M)

6 6 CLAY SHONKWILER Now, from Lefschetz Duality, we have that H k (M, M) H 2n+1 k (M), so we can re-write the long exact sequence as H k ( M) H k (M) H 2n+1 k (M) H k 1 ( M) Since we re interested in the Euler characteristic, which only takes into account the free parts of each homology, we may as well take these homologies with coefficients in R; in this situation, since there is no torsion, by universal coefficients we have that H j (M, R) H j (M, R). Hence, we have the further reduction H k ( M, R) H k (M, R) H 2n+1 k (M, R) H k 1 ( M, R) Since each term in the sequence is free and the sequence is exact, the alternating sum of the ranks must be equal to zero, which is to say that 0 = 2n+1 i=0 2n+1 i=0 [ ( 1) i rk H i ( M, R) + ( 1) i+1 rk H i (M, R) + ( 1) i+2 rk H 2n+1 i (M) ] 2n+1 2n+1 = ( 1) i rk H i ( M, R) + ( 1) i+1 rk H i (M, R) + ( 1) i rk H 2n+1 i (M, R) i=0 2n+1 = χ( M) χ(m) + ( 1) i rk H 2n+1 i (M, R). i=0 Now, since 2n + 1 is odd, we see that all the terms in the last sum will have the opposite sign they would in the computation of χ(m), so the last sum is just χ(m); hence, from the above, we see that i=0 0 = χ( M) χ(m) χ(m) = χ( M) 2χ(M). Therefore, we see that χ( M) = 2χ(M). 5 Let M be compact with boundary M. If M is contractible, show that M has the homology of a sphere. Proof. From the long exact sequence of the pair (M, M), we have that H k+1 (M, Z) H k+1 (M, M, Z) H k ( M, Z) H k (M, Z) Since M is contractible, H k (M, Z) = 0 for all k > 0, so we see that H k+1 (M, M, Z) H k ( M, Z) for all k > 0. Now, M may or may not be Z-orientable, but it is definitely Z/2-orientable; since M is compact, we can use Lefschetz duality to see that (1) H k (M, M, Z/2) H n k (M, Z/2) for all k. Hence, for k < n, H k (M, M, Z/2) = 0. By universal coefficients, this implies that H k (M, M, Z) = 0 for k < n and, hence, that H k ( M, Z) = 0 for 0 < k < n 1. Assuming M is connected (I m guessing this must be

7 GEOMETRY FINAL 7 a necessary assumption), H 0 ( M, Z) = Z, so it remains only to determine H n 1 ( M, Z) (since H k ( M, Z) = 0 for k > n 1, since M is n 1 dimensional). To that end, note that (1) implies that H n (M, M, Z/2) = Z/2, meaning that H n 1 ( M, Z/2) = Z/2. By universal coefficients, then, either H n 1 ( M, Z) = Z or H n 1 ( M, Z) = Z/2. However, M is a compact (n 1)-dimensional topological manifold, so M is either orientable or nonorientable. If it is orientable, then H n 1 ( M, Z) = Z; if non-orientable, then H n 1 ( M, Z) = 0. Thus, we see that Z/2 is not a viable possibility for H n 1 ( M, Z), so we conclude that H n 1 ( M, Z) = Z. Thus, consolidating our results, we see that { Z k = 0 or k = n 1 H k ( M, Z) = 0 0 < k < n 1 so M does indeed have the homology of an (n 1)-sphere. 6 Describe all 3-fold covers of the figure eight. Answer: We know that the cover of a connected, path connected, locally semi-simply connected space like the figure eight S 1 S 1, is completely determined by how the loops based at a given point lift to the cover. That is, by the element of the permutation group of the fiber determined by the lifting of the loops. We let our basepoint x 0 be the point of intersection of the two circles and consider the generators, a and b, of the fundamental group π 1 (S 1 S 1 ) as pictured: Throughout the following, we make use of the fact that two covering spaces in which the lifts of loops correspond to the same (or even conjugate) permutations of the fiber over x 0 are isomorphic; hence, we only need exhibit one covering space that satisfies the condition and we know that all others satisfying the condition are isomorphic to it. Now, a 3-fold cover of the figure eight can either be connected or disconnected. If disconnected, then such a cover is either the trivial cover (that is, three disjoint copies of S 1 S 1 ) or the disjoint union of a copy of S 1 S 1 and a connected 2-fold cover of S 1 S 1. Now, in a connected 2-fold cover, there are only a couple of possibilities. We might have that the lifts of b don t permute the fiber over x 0, but the lifts of a do. Such a situation is the

8 8 CLAY SHONKWILER following: Alternatively, the lifts of b could permute the fiber while the lifts of a do not; in this case, we just switch a and b in the above picture. The only other possibility is that the lifts of both a and b define a non-trivial permutation of σ 1 (x 0 ); such a case is pictured below: Since the permutations induced on the fiber over x 0 determines the covering space up to isomorphism of covering spaces, we see that the above are all the possible connected 2-fold covers of S 1 S 1. Now, turning to the case of connected 3-fold covers, we again consider the permutations induced on the fiber over x 0 by the lifts of a and b. If the lifts of both a and b simply fix the fiber, then we re in the trivial case. If the lifts of either a or b simply induce the trivial permutation of the fiber while the lifts of the other simply define a transposition, then we re in the situation of picture (1) (and, hence, not connected). Now, if the lifts of a define a 3-cycle on the fiber while the lifts of b define the trivial permutation, then we have something like the following: Again, if the lifts of b define a 3-cycle and the lifts of a the trivial permutation, then we just switch the labels a and b in the above picture. Now, if the lifts of both a and b induce a transposition on the fiber, then we re either in the case of picture (2) (and, hence, not connected) or we have something

9 GEOMETRY FINAL 9 like the following: We might have that the lifts of a determine a 3-cycle and the lifts of b a transposition (or vice versa); then we re in a situation like this (respectively, the same picture with the labels flipped): Finally, we might have that the lifts of both a and b induce 3-cycles on the fiber; this yields the following covering space: Therefore, the following is a complete list of the 3-fold covers of the figure eight:

10 10 CLAY SHONKWILER 7 Let M 7 be a topological manifold with H 7 (M, Z) = Z, H 6 (M, Z) = Z, H 5 (M, Z) = Z/2, H 4 (M, Z) = Z Z/3. Determine as much of the cohomology ring of M as you can. Answer: If M is non-compact or non-orientable, then H n (M, Z) = 0, so we see immediately that M is a compact, orientable manifold. Therefore, we can apply Poincaré Duality to see that: H 0 (M, Z) H 7 (M, Z) = Z H 1 (M, Z) H 6 (M, Z) = Z H 2 (M, Z) H 5 (M, Z) = Z/2 H 3 (M, Z) H 4 (M, Z) = Z Z/3. Now, by universal coefficients, if T k denotes the torsion part of H k (M, Z), and F k denotes the free part of H k (M, Z), then H 7 (M, Z) F 7 T 6 = Z 0 = Z H 6 (M, Z) F 6 T 5 = Z Z/2 H 5 (M, Z) F 5 T 4 = 0 Z/3 = Z/3 H 4 (M, Z) F 4 T 3 = Z T 3. Taking universal coefficients the other way, we see that H 3 (M, Z) = F 3 T 3 = Fr(H 3 (M, Z)) T 3 = Z T 3 H 2 (M, Z) = F 2 T 2 = Fr(H 2 (M, Z)) To(H 3 (M, Z)) = 0 Z/3 = Z/3 H 1 (M, Z) = F 1 T 1 = Fr(H 1 (M, Z)) To(H 2 (M, Z)) = Z Z/2 H 0 (M, Z) = F 0 T 0 = Fr(H 0 (M, Z)) To(H 1 (M, Z)) = Z, where Fr(H k (M, Z)) denotes the free part of H k (M, Z) and To(H k (M, Z)) denotes the torsion part of H k (M, Z). Thus, except for T 3, we see that the homology and cohomology groups of M are completely determined. Now, consider the following list of generators: H 0 H 1 H 2 H 3 H 4 H 5 H 6 H 7 Z Z Z/2 Z Z/3 Z T 3 Z/3 Z Z/2 Z 1 α 1 α 2 α 3,1, α 3,2 α 4,1,? α 5 α 6,1, α 6,2 α 7 Then, since H (M, Z) is anticommutative in odd dimensions, we see that α 2 i = 0 for i odd. Also, α i α j = 0 for i + j > 7. Furthermore, note that if α is m-torsion, then α β is also m-torsion, since m(α β) = (mα) β = 0; similarly for β α, since β α = ±α β. Using these rules, we construct the following partial multiplication table for H (M, Z):

11 GEOMETRY FINAL 11 1 α 1 α 2 α 3,1 α 3,2 α 4,1? α 5 α 6,1 α 6,2 α α 1 α 2 α 3,1 α 3,2 α 4,1? α 5 α 6,1 α 6,2 α 7 α 1 α β 1,3 α 4,1? 0? β 1,6 α α 2 α 2 0? 0 0 β 2,4 α 6,1? α 3,1 α 3,1 β 1,3 α 4, β 3,4 α α 3,2 α 3,2? α 4,1 α 4,1 0 β 2,4 α 6,2 β 3,4 α ???? α 5 α α 6,1 α 6,1 β 1,6 α α 6,2 α 6, α 7 α The? s persist because we have no idea what T 3 looks like (in fact, it might have more than one generator). Also, the β i,j can be any integer, including 1 or 0. DRL 3E3A, University of Pennsylvania address: shonkwil@math.upenn.edu