Let a finite group $G$ acts on an orientable manifold $X$ freely. Denote $\pi:X\rightarrow Y=X/G$ be the quotient map. This covering map defines two maps between cohomology groups $\pi^*=H^\ast(\pi):H^*(Y,\mathbb{Z})\rightarrow H^*(X,\mathbb{Z})$ and $\pi_!:H^*(X,\mathbb{Z})\rightarrow H^*(Y,\mathbb{Z})$. The latter map is given by $\pi_!=(PD_Y)^{-1}\circ \pi_*\circ PD_X$, where $PD_X:H^\ast(X,\mathbb{Z})\to H_{n-\ast}(X,\mathbb{Z})$ denotes Poincare-duality and $\pi_\ast=H_\ast(\pi)$.

Is it true that $\pi_!\circ \pi^*=|G|\cdot id_{H^*(Y,\mathbb{Z})}$ and $\pi^*\circ \pi_!=|G|\cdot id_{H^*(X,\mathbb{Z})}$?

  • $\begingroup$ Could you say what $PD_X,PD_Y$ are? $\endgroup$ – Spice the Bird Mar 7 '13 at 5:05
  • $\begingroup$ I would guess $PD$ stands for Poincaré duality $\endgroup$ – Sándor Kovács Mar 7 '13 at 5:09
  • $\begingroup$ Yes, PD stands for the Poincare duality map. $\endgroup$ – Juan Mar 7 '13 at 7:44
  • $\begingroup$ Have you tried just checking definitions at the chain-level?. One thing to quickly note is that we already have transfer maps that satisfy your conclusion. I'm not sure your construction produces the transfer, and so I wouldn't immediately expect your conclusion to hold. (sorry I'm being lazy right now) $\endgroup$ – Chris Gerig Mar 7 '13 at 7:53
  • $\begingroup$ @Chris π∗ should be the transfer map. If my memory serves, one of the equalities in the last line holds. It seems that Russel's example below shows us the latter does not hold in general... $\endgroup$ – Juan Mar 7 '13 at 8:29

The first identity $\pi_! \circ \pi^* = \vert G \vert \cdot \mathrm{Id}$ holds, and follows from knowing that $\pi_!$ is a $H^*(Y)$-module map via $\pi^*$, so $$\pi_!( \pi^*(x)) = \pi_!(1)\cdot x$$ and $\pi_!(1) = \vert G \vert$ as may be seen from the $G$-cover over a point.

The second proposed identity $\pi^* \circ \pi_! = \vert G \vert \cdot \mathrm{Id}$ is false, as may be seen in the example $X= G \times Y$ and $G$ acting by translation on the first factor. A cohomology class $x$ supported on $\{e\} \times Y$ is sent by $\pi^* \circ \pi_!$ to its `$G$-invariantisation" $\sum_{g \in G} g \cdot x$, which will never be a multiple of the original class, as it has support on each $\{g\} \times Y$.

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  • $\begingroup$ Thank you for the answer. I don't quite understand the last line. WHat about cohomology class corresponding to $e\times Y$ itself? Then isn't the "G-invariantization" |G|-times the original one? $\endgroup$ – Juan Mar 7 '13 at 10:36
  • $\begingroup$ No. Even simpler: consider the action of $G$ on $G$ by translation, with quotient space the point $*$. Let $[g] \in H_0(G)$ be the class of the point $g$, so the $[g]$ give a basis for $H_0(G)$, and let $[g]^\vee$ be the dual basis element. Then $\pi^*(\pi_!([e]^\vee)) = \sum_{g \in G} [g]^\vee \neq \vert G\vert \cdot [e]^\vee$. $\endgroup$ – Oscar Randal-Williams Mar 7 '13 at 13:45
  • $\begingroup$ You are right; I totally forgot that $G$ is finite. $\endgroup$ – Juan Mar 7 '13 at 21:24

Yes: Let $\alpha$ be a cohomology class in $Y$. For simplicity, let's assume that $PD_{Y}\alpha$ is represented by an embedded cycle in $Y$. Then $PD_{X}\pi^{*}\alpha=\pi^{-1}PD_{Y}(\alpha)$. Therefore, $\pi\circ PD_{X}\pi^{*}\alpha = \pi (\pi^{-1}PD_{Y}\alpha)=|G|PD_{Y}\alpha$ as the map $\pi: \pi^{-1} PD_{Y}\alpha\rightarrow PD_{Y}\alpha$ is a $|G|$-to-1 covering. Now apply $PD_{Y}^{-1}$ to the previous equation to obtain $\pi_{!}\circ \pi^{*}\alpha=|G|\alpha$.

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  • $\begingroup$ Be careful, we're invoking Poincare-duality and so your spaces have to be orientable. I assume the question wants $X$ oriented and $\pi$ to be orientation-preserving. $\endgroup$ – Chris Gerig Mar 7 '13 at 7:37
  • $\begingroup$ Ahh, good point! I'll modify the example so that everything is orientable. $\endgroup$ – Russell Mar 7 '13 at 7:42
  • $\begingroup$ Thank you for the answer, Russell. I just added orientability in my question. I would appreciate your counter-example. $\endgroup$ – Juan Mar 7 '13 at 7:46
  • $\begingroup$ @Juan that is not enough, because his original example had $S^{2k}$ orientable while $\mathbb{R}P^{2k}$ is nonorientable! You need a condition on the action (luckily for $S^{2k+1}$ the antipodal action has orientable quotient). $\endgroup$ – Chris Gerig Mar 7 '13 at 8:31
  • $\begingroup$ @Russell, this still doesn't work because $|\mathbb{Z}_2|$ will kill the 2-torsion, and so the conclusion would be satisfied trivially. $\endgroup$ – Chris Gerig Mar 7 '13 at 8:33

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