M8130 Algebraic topology, tutorial 05, 2017
23. 3. 2017
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) i—> (xi,—x0). Take (x0, xi, x2, x3,..., x2n+i) G S2n+1 C ]R2n. 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. Prove deg(/) = Yi=i deg(f/xi).
Solution. We have /: Sn —>• Sn and y G Sn has a neighborhood (nbhd) V. Denote f~x{y) = {rri,... ,xk} and U nbhd's of xt, UdUj■ = 0. Let degree of f/xt be n,i. We have/: (Ui,Ui — Xi) -+ (V,V - y) and (f/Ui)*: Hn(Ut,Ut - x{) = Z -> #n(V, V - y) = Z, so z ^ n^. Consider the following diagram, its evolution (i.e. drawing order) was revealed in the tutorial class:
Hn(Sn,Sn
Af/Ui),
r\y))
■Hn(V,V-y)
h
■Hn(Sn,Sn —Hn(S'n
y)
We have jy* iso and h is iso by excision theorem. So, we have Hn(Sn, Sn- f(y)) £ © Hn(Ut,Ut-x,t
Hn(\_\i(Ui,Ui — Xi)). Iso between
Hn(Sn, Sn — f(y)) and iJn(|J1(f/j, Ui — Xi)) by excision. So k is also iso.
We can take jy* = h o Y(f/Ui)* ° k^1 o     and for generator 1 G Hn(Sn) we have
(&W.)(i)
deg(/) • 1. We are done.
Y,de9(f/xi
deg( f/xi
□
M8130 Algebraic topology, tutorial 05, 2017
23. 3. 2017
Exercise 3. Compute homology groups o/IRPn using CW-structure.
Solution. Let X be the CW-complex, we have C%w = 0Z[e™], Hn(C^w,d) = Hn(X). Also, d[e™] =      ^a[e/3_1]) where oPa is the degree of the map
Sn-1 Xn-1 _^ Xn-ll(Xn-1 - enfV) = S^1.
We know that IRPn = e° U e1 U ■ • • U e", cell in every dimension. Attaching map is
/: (KP)(i) = RV = S'l/(x ~ -x).
Then have
0 ^ Z ^ Z ^ —^z^z^z^o,
where Z appears n + 1 times. Now, Si 4 KF KPVKF-1 = S^, hence we can apply the attaching map. Class [y] has two preimages y, —y, there are nbhd's, deg(f/y) = ±1. Note that f / — y = f/y o (—id). Now we can compute the degree: deg(/) = deg(//y) + deg(// — y) = ±(1 + (—It is ±2 if i is odd and 0 otherwise. So, we can add to the sequence above:
o^z^z^ —^zAzAz^o,
and compute for X = RFn, n even: H0(X) = Z, H^X) = Z2, H2(X) = 0, H^X) = 0 for i even, Hi(X) = Z2 for « odd. The homology is zero for i > 0. For n even the only difference is Hn(X) = Z.
(Note: this is because for even they are not orientable, for odd it is oriented space.) □
Exercise 4. Compute homology groups of oriented two dimensional surfaces. What about nonorientable?
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
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)
9
0 -> z -> 0z -> Z -> 0.
M8130 Algebraic topology, tutorial 05, 2017
23. 3. 2017
This equality holds: d[e2] = 2[e\] + 2[el] + ..., so H2 = 0, H0 = Z and the only interesting case is
Z[ei,...,eji
Z[2e\ + ■■■ + 2e\] Y
□
Exercise 5. Have f: Sn —)• S*n map o/ 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^----> 0 ^ Z ^ 0, also
Sn _^ x(n) _^ x(n) j^n) _ gn) = ^
We get Hn+1(X) = 0,Hn= Zk, H0(X) = Z. Note, X/Sn ^ Sn+1. So, for the p* we have
p,: Hn+1(X) = 0 A Hn+1(Sn+1) = Z
and
p*:Hn(X)=Zk^Hn(Sn+1) = 0 and at iJ0 it is identity Z —>• Z.
□