M8130 Algebraic topology, tutorial 05, 2024 24. 10. 2024 Exercise 1. Prove that Sn has a nonzero vector field if and only if n is odd. Solution. First note, that we have v. Sn —> Wa+1 such that v(x) _L x. Consider the case of S1 and (xo,xi) 1—>• (xi,—x0). Take (x0, xi, x2, x3,..., x2n+i) G S2n+1 C IR2n. Then we get (xi, —xo,x3, —x2, . . . ,x2n+i, —x2n) as image and there is nothing more obvious than that the product is zero, i.e. it's perpendicular. Note, for S1 C C it is z H> ez, where e is the complex unit, usually denoted as i. Now we want to prove that if Sn has a nonzero vector field, then n is odd. We use the fact that deg(id) = 1 and deg(-id) = (-l)n+1. Take v: Sn -> Sn. If we show id ~ —id, then 1 = (—l)n+1 =>• n is odd. The homotopy is h(x,t) we are looking for is h(x,t) = cos(t)x + sin(t)v(x), where t G [0,7r]. Also note || h(x,t) ||= cos2t + sin2t = 1. We are done. □ Exercise 2. Compute homology groups of oriented two dimensional surfaces using a suitable structure of CW-complex. Solution. Denote Mg surface of genus g (it is the same as sphere with g handles, i.e. Mi is torus and M2 is double torus (homework 4). The CW-model is e° U e\ U • • • U e\ U e2 and we have 23 0 -> z -> 0z Z -> 0. 1 Second differential is zero, e° — e° = 0. The first one is zero as well (glue the model, the arrows go with + and then —). Then we get H0 = Z, Hi = Z, H2 = Z. For nonorientable surfaces, Ng is modelled by one 2-dimensional disc which has boundary composed with g segments every of which repeats twice with the same orientation. So we have one cell in dimensions 2 and 0 and g cells in dimension one. We get (quite similarly) 0 -> z -> 0z -> Z -> 0. This equality holds: Sn map of degree k. (such map always exists). Let X -Dn+1 Uf Sn and compute homology of X and the projection p: X —> X/Sn in homology. Solution. Easy, X = e° U en U en+1 andO^Z^Z^O^----^O^Z^O, also Sn _^ x(n) _^ x(n) j^n) _ ^ = gr^ M8130 Algebraic topology, tutorial 05, 2024 24. 10. 2024 We get Hn+1(X) = 0,Hn= Zk, H0(X) = Z. Note, X/Sn ^ Sn+1. So, for the we have p,: Hn+1(X) = 0 A Hn+1(Sn+1) = Z and p„: Hn(X)=Zk\ Hn(Sn+1) = 0 and at H0 it is identity Z —>• Z. □ Let X be a topological space with finitely generated homological groups and let Hi(X) = 0 for each sufficiently large i. Every finitely generated abelian group can be written as Z © Z © • • • © Z © Tor; where Tor denote torsion part of the group. The number k is k—times called the rank of the group. Euler characteristic X of X is defined by: x(X) = ^(-l)irank^(X) i=0 {% % = 0 n Thus x(Sn) = 1 - (-l)n. 0, otherwise. Exercise 4. Let (C*,<9) be a chain complex with homology H*{C*). Prove that x{C*), where 00 x(C*) = ^(-irranka. i=0 Solution. We have two short exact sequences: 0 -> Zi ^ d 4 -> 0 0^Bt^Zt^ Zt/Bt = Ht^0, where Ci, cycles Z;t and boundaries B;t are free abelian groups, thus rankCj = rank Z;t rank5j_! and rankiJj = rankZj — rank5j. Thus we have = ^(-irrankZ, + ^(-l)*rankJB,_1 = ^(-lyrankZi - ^(-1)* rank^ = X(X). □ i=0 i=0 Let X be a topological space with finitely generated homological groups and let Hi(X) = 0 for every sufficiently large i. Let f: X —>• X be a continuous map. Map f induces ho-momorphism on the chain complex /*: C*(X) —> C*(X) and on the homologiy groups M8130 Algebraic topology, tutorial 05, 2024 24. 10. 2024 H*f: H*(X) —> H*(X), where H^f (Tor H*(X)) C ToriJ*(X). Thus it induces homomor-phism HJ: H*(X)/TorH*(X)^-H*(X)/TorH*(X). Since H*(X)/Tor H*(X) = Z 0 Z 0 ■ ■ ■ 0 Z; map H*f can be written as a matrix, thus we rankH»(X) can compute its trace. So we can define the Lefschetz number of a map f: HD^i-iytxHif. i=0 Similarly to the case of the Euler characteristic, it can be proved that1 X)(-l)ťtr^/ = X;(-l)ťtr/ť. Theorem. If L(f) ^ 0, then f has a fixed point. Exercise 5. Use the theorem above to show, that every cts map f on Dn and M.Pn where n is even has a fixed point. ' % ~ Because H0f: H0(Dn) = Z -> 0, otherwise. Z = H0(Dn) can be only the identity, we have L(f) = 1, thus / has a fixed point. {Z, i = 0, Z/2, i M which satisfies X(t,x) = v(X(t,x)) for every x E M and X(0,x) = x. There exists t0 such that X(t0,x) ^ x. Denote f(x) = X(t0,x), thus / has no fixed point, thus L(f) = 0. Because / is homotopic to id and triJjid = rankiJj(M), we get from homotopy invariance 0 = L(f) = L(id) = x(M). □ V: d(X)^Ci(X)