Suppose $G$ is a group acting freely on a topological space $X$. Take an element $g$ of $G$. My question is: If the induced action of $g$ on cohomology with $\Bbb Z$-coefficient is trivial then when does it follow that induce map on cohomology with $\Bbb Z_2$-coefficient is also trivial?
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$\begingroup$ A trivial sufficient condition: when $H^*(X,Z)$ is torsion-free. $\endgroup$– user43326Commented Nov 24, 2018 at 14:53
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$\begingroup$ In that case $H^*(X,Z/2)$ is naturally isomorphic to $H^*(X,Z)\otimes Z/2$, so $g_{Z/2}^*=g_Z^*\otimes Z/2$. $\endgroup$– user43326Commented Nov 27, 2018 at 17:19
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1 Answer
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Consider the group $A=\mathbb{Z}\times\mathbb{Z}/2$ and the automorphism $f(n,m)=(n,n+m)$. This gives an automorphism $g=Bf$ of the classifying space $X=BA$. Standard methods show that $H^*(X;\mathbb{Z}/2)=E[x]\otimes(\mathbb{Z}/2)[y]$ with $|x|=|y|=1$ and $g^*(x)=x$ and $g^*(y)=x+y$, so in particular $g^*$ is not the identity. On the other hand $H^*(X;\mathbb{Z})=E[x]\otimes\mathbb{Z}[z]/(2z)$ with $|z|=2$. Here $g^*(z)$ must be $z$, because there is no other nonzero element in $H^2(X)$ that it could be, and $g^*(x)=x$ because $x$ comes from $H^1(B\mathbb{Z})$. So $g^*$ is just the identity on integral cohomology.
If we want to make sure that the group $\langle g\rangle$ acts freely, we can replace $X$ by $\mathbb{R}\times X$ and $g$ by $(t,x)\mapsto (t+1,g(x))$.
answered Nov 22, 2018 at 10:00
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3$\begingroup$ $E[x]$ is the exterior algebra on the generator $x$, which is just $\mathbb{Z}\oplus\mathbb{Z}x$ and is the cohomology of $B\mathbb{Z}=S^1$. $\endgroup$ Commented Nov 22, 2018 at 11:48
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$\begingroup$ I guess if we set $f(n,m)=(n+m,m)$ instead, we get an action of $Z/2$ on $BA$, so you get an example with $G$ finite. $\endgroup$ Commented Dec 1, 2018 at 18:04