If $G$ is a discrete or a Lie Group acting smoothly on a manifold $M$, we can define the algebra of $G$-invariant de Rham classes, $H(M)^G$, and we can also consider the cohomology of the sub-complex of $G$-invariant forms and its cohomology $H(\Omega(M)^G)$ which I note simply $H(M^G)$.

The injection $j:\Omega(M)^G \hookrightarrow \Omega(M)$ passes to cohomology as $j:H(M^G) \to H(M)$, and it is clear that $j(H(M^G)) \subset H(M)^G$, so there is the induced morphism $$\bar j \colon H(M^G) \to H(M)^G$$ But $\bar j$ not injective since for $[\alpha]_M^G \in H(M^G)$, $j([\alpha]_M) = 0$ implies only that $\alpha = d\beta$ and $\beta$ has no reason to be $G$-invariant, so no reason why $[\alpha]_M^G$ should be null.

I think that $\bar j$ neither surjective since for $[\alpha]_M \in H(M)^G$, $g \cdot [\alpha]_M = [g \cdot \alpha]_M = [\alpha]_M$ for all $g \in G$ implies only that $g \cdot \alpha = \alpha + d\beta$, so $\alpha$ is not necessarily $G$-invariant.

But is it possible that $[\alpha]_M$ contains another $G$-invariant representant ?

Is there a simple counter-example showing that in general $\bar j$ is neither injective nor surjective ? For injectivity the example in @David's answer is OK, but quid for an example where surjectivité fails ?


1 Answer 1


If $G$ is compact, the inclusion $H(M^G) \to H(M)^G$ is an isomorphism. The inverse map is defined as follows: Take a class $\omega$ in $H(M)^G$ and lift it to a closed form $\alpha \in \Omega(M)$. Put $\beta = \int_{g \in G} g^{\ast} \alpha$, where the integral is with respect to Haar measure normalized to have volume $1$. Clearly, $\beta \in \Omega(M)^G$. Our lift will map $\omega$ to $[\beta]$.

We must check that $\beta$ is closed, is a de Rham representative of $\omega$, and that its class in $H(M^G)$ is independent of the choice of $\alpha$. Since all $g^{\ast} \alpha$ are closed, so is $\beta$. Since all $g^{\ast} \alpha$ are de Rham representatives of $\omega$, so is $\beta$. Finally, let $\alpha' = \alpha + d \eta$ be another lift of $\omega$. Then $$\int_{g \in G} g^{\ast} \alpha' = \int_{g \in G} g^{\ast} \alpha + d \int_{g \in G} g^{\ast} \eta$$ and $\int_{g \in G} g^{\ast} \eta$ is in $\Omega(M)^G$.

When $G$ is not compact, both injectivity and surjectivity can fail.

Failure of injectivity: Consider $M = \mathbb{R}$ and $G = \mathbb{Z}$ acting by translations. The $1$-form $dx$ is closed and $G$-invariant on $\mathbb{R}$, but has no $G$-invariant integral. So it gives a nonzero class in $H^1(M^G)$, but of course $H^1(M)^G \subseteq H^1(M) = 0$.

Failure of surjectivity Let $M = S^1$. For $\theta \in \mathbb{R}$, define $$\tilde{\phi}(\theta) = \theta + \tfrac{1}{2} \sin \theta.$$ Then $\tilde{\phi}: \mathbb{R} \to \mathbb{R}$ descends to a diffeomorphism $\phi: S^1 \to S^1$, with repelling and attracting fixed points at $\theta = 0$ and $\theta = \pi$ respectively. We let $G = \mathbb{Z}$, acting by $\phi$.

The map $\phi$ acts trivially on $H^1(S^1)$. I claim that there is no nonzero smooth $\phi$-invariant $1$-form on $S^1$. Suppose for the sake of contradiction that $\omega$ is a $\phi$ invariant $1$-form.

Let $\omega = g(\theta) d \theta$ by $\phi$ invariant, and suppose for the sake of contradiction that $g(\theta_0) \neq 0$ for some $\theta_0 \not \in \pi \mathbb{Z}$. Then $$g(\phi^n(\theta_0)) = \prod_{k=0}^{n-1} (\phi')(\phi^k(\theta_0))^{-1} g(\theta_0).$$ As $k \to \infty$, $\phi^k(\theta_0) \to \pi$ and $\phi'(\phi^k(\theta_0))^{-1} \to 2$. So $g(\phi^n(\theta_0)) \to \infty$ as $n \to \infty$, contradicting that $g(\theta) d \theta$ is supposed to be a continuous $1$-form.

A perhaps nicer failure of surjectivity Take a look at this great post by Robert Bryant. Let $T$ be the torus $S^1 \times S^1$. Let $C_0$ and $C_{\pi}$ be the circles $\{ 0 \} \times S^1$ and $\{ \pi \} \times S^1$. Bryant constructs a flow (in other words, action of $\mathbb{R}$) which takes $C_0$ and $C_1$ to themselves, and where every other orbit approaches $C_0$ as $t \to (-\infty)$, and $C_1$ as $t \to \infty$, with opposite orientations. A picture is worth a thousand words: Here is what the orbits of this flow looks like on one of the two cylinders of $T \setminus (C_0 \cup C_1)$. I claim that there is no flow invariant $1$-form with $\int_{C_0} \omega \neq 0$.

Let $f(t)$ be one of these nonclosed flows. For $T$ very positive, $f$ takes the interval $(T, T + 2 \pi)$ very close to $C_1$ and, for $T$ very negative, it is very close to $C_0$ with opposite orientation. So, if $\omega$ is a flow invariant continuous $1$-form, then $$\int_{C_0} \omega = \lim_{T \to - \infty} \int_T^{T+2 \pi} f^{\ast} \omega = \lim_{T \to \infty} \int_T^{T+2 \pi} f^{\ast} \omega = - \int_{C_1} \omega.$$

On the other hand, $C_0$ and $C_{1}$ are homologous so, if $\omega$ is closed, we have $$\int_{C_0} \omega = \int_{C_1} \omega.$$

Combining these, any closed flow-invariant $1$-form has $\int_{C_0} \omega=\int_{C_1} \omega=0$.

One more and I'll shut up Let $C$ be the cylinder $\mathbb{R}^2/\mathbb{Z}$, where we identify $(x,y)$ and $(x+k,y)$ for $k \in \mathbb{Z}$. Let the action of $\mathbb{Z}$ on $C$ be generated by $\phi(x,y) = (x,x+y)$. Let $\gamma$ be the image in $C$ of the line segment $(0,0)$ to $(0,1)$ in $\mathbb{R}^2$. Then $\phi(\gamma)$ is the image of the line segment from $(0,0)$ to $(1,1)$. So, if $\omega$ is a $\phi$-invariant $1$-form, then $\int_{\gamma} \omega = \int_{\phi(\gamma)} \omega$. But $\gamma - \phi(\gamma)$ is a $1$-cycle representing the nontrical class in $H_1(C)$, so we deduce that any $\phi$ invariant $1$-form integrates to $0$ against this class.

  • $\begingroup$ thx for that quick answer, but what I really need is an example where surjectivité fails, I have edited my question. $\endgroup$
    – ychemama
    Commented Mar 11, 2018 at 14:24
  • $\begingroup$ See my edit. I have an example of failure of surjectivity, but it uses an action which isn't free (and isn't even proper!) so I don't know if it is good enough for you. $\endgroup$ Commented Mar 11, 2018 at 14:31
  • $\begingroup$ yes it's good enough, thx ! Just to be sure, your Z action on S^1 is p.x = \phi \circ ... \circ \phi(x) = \phi^p (x) ? $\endgroup$
    – ychemama
    Commented Mar 11, 2018 at 14:56
  • $\begingroup$ Yes, that is right. $\endgroup$ Commented Mar 11, 2018 at 14:57
  • $\begingroup$ And I don't really understand where your formula for g(\phi^n(\theta_0)) come from... $\endgroup$
    – ychemama
    Commented Mar 11, 2018 at 15:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.