Algebraic Topology exam

Transcription

1 Instituto Superior Técnico Departamento de Matemática Algebraic Topology exam June 12th Let X be a square with the edges cyclically identified: X = [0, 1] 2 / with (a) Compute π 1 (X). (x, 0) (1, x) (1 x, 1) (0, 1 x) for 0 x 1 (b) Describe a -complex structure on X and compute the corresponding simplicial homology. (a) X has a cell structure with one 0-cell [(0, 0)], one 1-cell {[(x, 0)]: 0 < x < 1} and one two cell {[(x, y)]: 0 < x < 1, 0 < y < 1}. Hence the 1-skeleton of X is S 1. The Van Kampen theorem implies that π 1 (X) is the quotient of π 1 (S 1 ) = Z by the normal subgroup generated by the image of the attaching map f : S 1 S 1 of the two cell. But f wraps S 1 around itself four times, therefore π 1 (X) = x x 4 = Z/4 (b) There is a -complex structure on X formed by identifying the faces of four 2- simplices as indicated in the following picture. The chain complex C (X) has four generators A, B, C, D in degree 2, five generators a, b, c, d, e in degree 1 and two generators x, y in degree 0 (see the picture) and in terms of these bases, the boundary maps are as follows [ ] C2 (X) = Z 4 C1 (X) = Z 5 C0 (X) = Z 2 The image of 1 is {m(1, 1): m Z} and since (1, 1) is part of a Z-basis for Z 2, we have H0 (X) = Z2 /(Im 1 ) = Z. The matrix representing 2 has full rank therefore H2 (X) = ker 2 = 0. A basis for ker 1 is given by {a, b e, c e, d e}. In terms of this basis the map 2 : Z 4 Z 4 = ker 2 is given by the matrix

2 Applying row and column reduction we see that this matrix can be transformed in so we conclude that H 1 (X) = Z 4 /(Z 3 4Z) = Z/4. 2. Let f : X Y be a map and C(f) denote its mapping cone. Recall this means that C(f) = (Y X [0, 1])/ with the equivalence relation generated by (x, 0) (x, 0) and (x, 1) f(x) for all x X. Recall that the canonical map Y C(f) is an embedding. (a) Show that a map g : Y Z extends to C(f) if and only if g f is null. (b) Show there are at least four distinct homotopy classes of maps S 1 S 1 RP 2. (a) By definition of the quotient topology, to give a map out of C(f) is the same as giving maps g : Y Z and H : X [0, 1] Z which are constant on the equivalence classes of. This means H(x, 0) = H(x, 0) for all x, x X (i.e. x H(x, 0) is constant) and H(x, 1) = g(f(x)). Hence a map out of C(f) is the same as a map g out of Y and a null homotopy of g f. In particular, a map g extends to C(f) if and only if g f is null. (b) S 1 S 1 = C(f) for f : S 1 S 1 S 1 the attaching map of the 2-cell in S 1 S 1. Since π 1 (RP 2 ) = Z/2, there are exactly four pointed homotopy classes of maps in [S 1 S 1, RP 2 ] = π1 (RP 2 ) π 1 (RP 2 ). Since π 1 (RP 2 ) is abelian, the action of the fundamental group of RP 2 on the set of pointed homotopy classes is trivial and hence the orbit set of the action of π 1 (RP 2 ), which is [S 1 S 1, RP 2 ], also has four elements. Writing a, b for standard generators of π 1 (S 1 S 1 ), the attaching map f represents the commutator [a, b] and hence, for a pointed homotopy class g [S 1 S 1, RP 2 ] corresponding to (x, y) (π 1 (RP 2 )) 2 we have g f = [x, y] = 0 π 1 (RP 2 ). By (a), all 4 homotopy classes in [S 1 S 1, RP 2 ] extend to S 1 S 1. The extensions are still all homotopically distinct because they induce distinct homomorphisms on π 1 (and the conjugation relation is trivial). We conclude that there are at least four distinct homotopy classes of maps from S 1 S 1 to RP Let {U, V, W } be an open cover of X by contractible open sets such that U V S 3, U W =, V W S 1. Compute the homology of X. Let Y = U V. The Mayer-Vietoris sequence for the reduced homology of Y takes the form H k+1 (U) H k+1 (V ) H k+1 (Y ) H k (U V ) H k (U) H k (V ) As U, V are contractible it follows that is an isomorphism for every k and hence { Z if i = 0, 4 H i (Y ) =

3 Now consider the open cover Y, W of X. We have Y W = V W = S 1. Therefore the Mayer-Vietoris for the reduced homology of X associated to this cover gives H k+1 (Y ) H k+1 (W ) H k+1 (X) H k (Y W ) H k (Y ) H k (W ) If k 3, 4 the two ends of the sequence above are 0 and hence is an isomorphism. For k = 3, is surjective and for k = 4 it is injective. When k = 4, the group in whcih H 5 (X) injects is 0. When k = 3, since H k+1 (Y W ) = 0, the first map in the sequence above is an isomorphism. We conclude that { Z if k = 0, 2, 4 H i (X) = 4. Let p i : E i S 1 S 1 for i = 1, 2 be 2-sheeted connected covering spaces. Show there are homeomorphisms ψ : E 1 E 2 and φ: S 1 S 1 S 1 S 1 such that E 1 p 1 S 1 S 1 ψ φ E 2 p 2 S 1 S 1 commutes. By the classification of covering spaces, p 1 and p 2 correspond to index 2 subgroups of π 1 (S 1 S 1 ) = Z 2. There are exactly three of these (all isomorphic to Z 2 ). Indeed, such a group must contain 2Z 2Z (as twice any element in the quotient group is 0) and this sets up a bijective correspondence between index 2 subgroups of Z 2 and index 2 subgroups of Z 2 /(2Z) 2 = Z/2 Z/2. Thus the index 2 subgroups of Z 2 are the groups generated by the columns of the matrices [ ] [ ] [ ] ,, and These matrices are all equivalent under row and column operations (including permutations), which means that the index 2 subgroups are all equivalent under the action of the automorphism group GL(2; Z) of Z 2. Now, an element of A GL(2; Z) determines a homeomorphism of R 2 which passes to the quotient by Z 2 translations and induces a homeomorphism φ A of S 1 S 1. Picking A GL(2; Z) such that A(Im p 1 ) = Im p 2 we have that φ A p 1 and p 2 are covering maps with connected domains and correspond to the same subgroup of π 1 (S 1 S 1 ). Setting φ = φ A, it follows from the classification of covering spaces that there exists ψ : E 1 E 2 such that the diagram in the statement commutes. 5. Let X be the space obtained from CP 3 by attaching a 3-cell to CP 1 by a map of degree 10. (a) Compute H (X) and H (X; Z/p) for each prime p. (b) Compute the cohomology rings H (X) and H (X; Z/2). (c) Compute H 5 (X RP 2 ).

4 (a) X is a cell complex with one cell in dimensions 0, 2, 3, 4 and 6. The degree of the attaching map of the 3-cell is 10 by assumption. The degree of the attaching map of the 4-cell is 0 since the cell is actually attached to CP 2. Thus the cellular chain complex of X (in dimensions 6 is and hence Z 0 Z 0 Z 10 Z 0 Z Z if i = 0, 4, 6, H i (X) = Z/10 if i = 2, To do the computation with Z/p coefficients we can tensor the cellular chain complex with Z/p. If p = 2 or 5 the differentials are all 0. Otherwise 3 is an isomorphism. Hence, for all prime p we have Z/p if i = 0, 4, 6 H i (X; Z/p) = Z/p if i = 2, 3 and p {2, 5} Alternatively, we could have reached this answer by applying the universal coefficient theorem. (b) By the universal coefficient theorem, we have therefore H k (X) = Hom(H k (X); Z) Ext(H k 1, Z) Z if k = 0, 4, 6 H k (X) = Z/10 if k = 3 Other than products with the identity class in degree 0, all products are 0. This is true for degree reasons except for the classes in degree 3. But the latter are torsion classes so the same is true of their cup products with any other class and H 6 (X) is torsion free. We conclude that the cup product of any two classes is 0 unless one of them is a (non-zero) multiple of the identity. Also by the universal coefficient theorem we have H k (X; Z/2) = Hom(H k (X); Z/2) Ext(H k 1 (X), Z/2) and hence H k (X; Z/2) = { Z/2 if k = 0, 2, 3, 4, 6 Using cellular homology we see that the inclusion CP 3 X determines an isomorphism on H i (X; Z/2) for i 3. This gives the product structure (disregarding the products with identity) for the even dimensional classes: If x H 2 (X; Z/2) is the generator, then x 2 and x 3 generate H 4 and H 6 respectively.

5 If y H 3 (X; Z/2) is the generator, it remains to compute y 2 (as all other products of non-identity classes with y are 0 by degree reasons). Using cellular homology we see that this class is the image of a generator of H 3 (X) under the ring map H (X) H (X; Z/2) determined by the ring map Z Z/2. Since the generator of H 3 (X) squares to 0, the same is true of y. (c) By the Künneth formula H 5 (X RP 2 ) = 2 i=0h 5 i (X) H i (RP 2 ) 2 i=0 Tor(H 4 i (X), H i (RP 2 )) = Z/2 6. Let 0 G H K 0 be a short exact sequence of abelian groups. (a) Show that there is a natural long exact sequence on homology H n (X; G) H n (X; H) H n (X; K) H n 1 (X; G) (b) Show that in the long exact sequence for X = RP 2 determined by the short exact sequence 0 Z/2 Z/4 Z/2 0, the boundary map is an isomorphism. H 2 (RP 2 ; Z/2) H 1 (RP 2 ; Z/2) (a) Since C (X) is free, tensoring with the short exact sequence in the statement gives a short exact sequence of chain complexes 0 C (X) G C (X) H C (X) K 0 The snake lemma applied to this exact sequence gives rise to the desired long exact sequence on homology. (b) The short exact sequence of chain complexes in (a) for the cellular chain complex of RP 2 looks as follows in dimensions 1 and 2: 0 Z/2 i Z/4 j Z/ Z/2 i Z/4 2 = 2 j 0 Z/2 0 where i: Z/2 Z/4 sends [1] to [2] and j : Z/4 Z/2 sends [1] to [1]. By definition of the boundary map in the long exact sequence, if x C2 CW (RP 2 ; Z/2) is a generator then ([x]) = [i 1 2 (j 1 (x))] is the homology class of the generator of C 1 (RP 2 ; Z/2) = Z/2, which completes the proof. 7. Let M and N be oriented closed connected manifolds of dimension n. The degree of a map f : M N is the integer deg(f) such that f : H n (M) H n (N) satisfies f ([M]) = deg(f)[n]

6 (a) Show that if f is not surjective then deg(f) = 0. (b) For M = N = S 1 S 1 S 1, show that deg(f) = det A where A is a 3 3 matrix representing f : H 1 (M) H 1 (M) (with respect to any basis). (a) If f is not surjective then there exists p N such that f factors as a composite M N \ p N Since N \ p is a connected non-compact manifold of dimension n we have that H n (N \ p) = 0 hence f induces the 0 homomorphism on H n, i.e. deg(f) = 0. (b) Let x H 1 (S 1 ) be a generator. The Künneth formula for cohomology implies that H 1 (M) = Z 3 is generated by the cross products x 1 = x 1 1, x 2 = 1 x 1 and x 3 = 1 1 x, and that H 2 (M) = Z 3 is generated by x x 1, x 1 x and 1 x x. Finally x x x generates H 3 (M). Since (a b) (c d) = ( 1) b c (a c) (b d) we have that and x x 1 = x 1 x 2, x 1 x = x 1 x 3, 1 x x = x 2 x 3 x x x = x 1 x 2 x 3 Since the x i have odd degree, x i x j = x j x i. Hence H i (M) can be identified via the cup product with the exterior product Λ i (H 1 (X)). If f : H 1 (M) H 1 (M) is represented by a matrix A, then, with respect to the dual basis, f : H 1 (M) H 1 (M) is represented by the transpose matrix A t. Then Λ 3 (f ) will be multiplication by det(a t ) = det(a). This computation can also be done explicitly: If ([ f 1 ]) 2 = x 3 [ a b c d e f g h i ] [ x1 ] 2 x 3 then (since x i x i = 0, x i x j = x j x i ) we have f (x 1 x 2 x 3 ) = (ax 1 + bx 2 + cx 3 ) (dx 1 + ex 2 + fx 3 ) (gx 1 + hx 2 + ix 3 ) = (ax 1 + bx 2 + cx 3 ) ((dh eg)x 1 x 2 + (di gf)x 1 x 3 +(ei fh)x 2 x 3 ) = (a(ei fh) b(di gf) + c(dh eg))x 1 x 2 x 3 = det(a)x 1 x 2 x 3 8. Let M be a closed connected manifold with dimension a multiple of 4. Show that χ(m) 0. Solution : This is wrong as pointed out by Miguel Moreira. A counterexample is S 2 Σ 2 as χ(s 2 Σ 2 ) = χ(s 2 )χ(σ 2 ) = 2 ( 2) = 4. In a moment of insanity I must have believed that the contributions of dimensions i and 4n i cancelled leaving only dimension 2n but this makes no sense... I promise to be more careful in the remaining exams.