Invariant cohomology vs cohomology of quotient

Question

Let $M$ be an oriented smooth manifold and $\varphi : M \to M$ an orientation preserving diffeomorphism. Suppose that action of $\mathbb{Z}$ on $M$ induced by $\varphi$ is free and proper so that the quotient $M/\mathbb{Z}$ is a smooth manifold. The map $\varphi$ also acts on de Rham cohomology $H^k(M, \mathbb{R})$; let $H^k(M, \mathbb{R})^{\mathbb{Z}}$ be the set of fixed points. We have a map $$H^k(M/\mathbb{Z}, \mathbb{R}) \to H^k(M, \mathbb{R})^{\mathbb{Z}}$$ given by pullback by the quotient map. Question: Is this map surjective?

Note that this is indeed true if the $\mathbb{Z}$-action descends to a $\mathbb{Z}/n\mathbb{Z}$-action, i.e. $\varphi^n = \mathrm{Id}$ for some $n$. More generally, this is true for an action of a compact Lie group; the argument uses averaging by the Haar measure.

Answer 1

Yes, I think so. There is a Hochschild-Serre spectral sequence $$E^{pq}_2=H^p(\mathbb{Z}, H^q(M,\mathbb{R}))\implies H^{p+q}(M/\mathbb{Z},\mathbb{R})\,.$$ Since the group $\mathbb{Z}$ has cohomological dimension 1, this spectral sequence degenerates at $E_2$. Therefore $E^{0,k}_{\infty}=E^{0,k}_2$, so your map is the surjective edge-homomorphism $H^k(M/\mathbb{Z},\mathbb{R})\rightarrow E^{0,k}_{\infty}=H^k(M,\mathbb{R})^{\mathbb{Z}}$.

Answer 2

Here is a version of the proof that avoids spectral sequences by spelling out all the definitions:

Define the mapping torus $T(M, \phi) := (M\times [0,1])/(x, 0)\sim (\phi(x),1)$. The projection to the first factor defines a map $p: T(M, f)\to M/\mathbb Z$, which induces an isomorphism on cohomology since the action is free and proper (cover $M$ with open sets $U_i$ which are equivariantly homeomorphic to $(U_i/\mathbb Z)\times Z$; the condition is easy to check for each $U_i$ and follows by induction for finite unions of $U_i$'s from the five-lemma applied to the Mayer-Vietoris sequence; for infinite unions, apply the five-lemma to the Milnor exact sequence). The pair $(T(M, \phi),[M\times\{0\} ])$ then gives a long exact sequence in cohomology. Since $[M\times\{0\}]\subset T(M,\phi)$ and $M\times\{0, 1\}\subset M\times [0,1]$ have the homotopy extension property, the quotient map induces an isomorphism $H^n(M\times [0, 1],M\times \{0,1\})\cong H^n(T(M, \phi), [M\times\{0\}])$. Since the inclusion $M\times \{0,1\}\subset M\times [0,1]$ induces the diagonal on cohomology, the long exact sequence of cohomology groups provides an isomorphism $\big(H^n(M)\oplus H^n(M)\big)/\Delta\cong H^{n+1}(M\times [0, 1],M\times \{0,1\})$. The source is of course isomorphic to $H^n(M)$ via $(x, y) \mapsto x - y$. Since the composition $M\times \{1\}\to [M\times \{1\}] = [M\times \{0\}] \cong M\times \{0\}$ is given by $f$, the boundary operator of the pair $(T(M, \phi), [M\times \{0\}])$ can be identified with the map $\operatorname{id} - f^*: H^n(M)\to H^n(M)$, and consequently there are short exact sequences

$$ 0\to \operatorname{coker}(H^{n-1}(M)\xrightarrow{\operatorname{id} - f^*} H^{n-1}(M)\big)\to H^n(T(M,\phi))\to \operatorname{ker}(H^{n}(M)\xrightarrow{\operatorname{id} - f^*} H^{n}(M)\big)\to 0 $$

The target are precisely the invariants $H^n(M)^\mathbb Z$.

Answer 3

There is something even more precise in the appendix to

Serre, Jean-Pierre Homologie singulière des espaces fibrés. Applications. (French) Ann. of Math. (2) 54 (1951), 425–505.

He states it for homology rather than cohomology, but here it is. Homology is taken with integer coefficients, and superscripts indicate invariants and subscripts indicate coinvariants.

Theorem: Let $X$ be a simplicial complex equipped with a free action of $G = \mathbb{Z}$. Then we have a short exact sequence $$0 \rightarrow H_i(X)_G \rightarrow H_i(X/G) \rightarrow H_{i-1}(X)^G \rightarrow 0.$$ In particular, if $G$ acts trivially on the homology of $X$ (which often holds, for instance if the action extends to an action of $\mathbb{R}$) then we have a short exact sequence $$0 \rightarrow H_i(X) \rightarrow H_i(X/G) \rightarrow H_{i-1}(X) \rightarrow 0.$$

Proof: Let $C_{\bullet}(X)$ be the singular chain complex for $X$. Since $G$ acts freely and $G$ is infinite, we have $$C_{\bullet}(X)_G = C_{\bullet}(X/G) \quad \text{and} \quad C_{\bullet}(X)^G = 0.$$ Let $t \in G$ be a generator. The above implies that we have a short exact sequence $$0 \longrightarrow C_{\bullet}(X) \stackrel{1-t}{\longrightarrow} C_{\bullet}(X) \longrightarrow C_{\bullet}(X/G) \longrightarrow 0.$$ The associated long exact sequence in homology contains the segment $$H_i(X) \stackrel{1-t}{\longrightarrow} H_i(X) \longrightarrow H_i(X/G) \longrightarrow H_{i-1}(X) \stackrel{1-t}{\longrightarrow} H_{i-1}(X).$$ Examining the outer terms, this reduces to a short exact sequence $$0 \rightarrow H_i(X)_G \rightarrow H_i(X/G) \rightarrow H_{i-1}(X)^G \rightarrow 0,$$ as desired.