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Suppose $M$ is a smooth manifold and $\phi : M \to M$ is a homeomorphism whose fixed point set is a smooth submanifold $M_{\phi}$. Is there any relation between the cohomology ring of $M_{\phi}$ and the cohomology ring of $M$ augmented by its natural $\mathbf{Z}[\phi]$-algebra structure? Is it too much to try to compute the former in terms of the latter? I am sorry for the vagueness of this; I am not sure precisely what question I want to be asking, but I am posting this to find out if I am completely wasting my time by looking for such interactions.

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2 Answers 2

Suppose $G$ is a finite group on a space $X$. Then [Swan, Richard G. A new method in fixed point theory. Bull. Amer. Math. Soc. 65 1959 128--130. MR0107238 (21 #5963); Swan, Richard G. A new method in fixed point theory. Comment. Math. Helv. 34 1960 1--16. MR0115176 (22 #5978)] constructs a spectral sequence which has $$E_2^{i,j}=\hat H^i(G,H^j(X))$$ converging to a certain filtered graded group $J^\bullet(X)$. The spectral sequence degenerates at $E_2$ if the action of the group acts trivially. Here cohomology of spaces is Cech cohomology, with any coefficients you like, and $\hat H$ is Tate cohomology of groups.

Now, suppose $G$ is cyclic, that $X=M$ is a manifold, and that each point in $M$ with non-trivial stabilizer is actually fixed by the whole of $G$. Then Swan shows that $J^\bullet(X)$ is isomorphic to $J^\bullet(X^G)$.

Over a field, an under the above conditions, this gives information about $H^\bullet(X^G)$ from information on $H^\bullet(X)$ and its $G$-module structure.

Swan constructs a multiplicative structure on the spectral sequence, and this gives information on the cohomology ring of the invariants.

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Example: Let $N\subset\mathbb R^n$ be a compact submanifold. Choose a diffeomorphism $h\colon\mathbb R\to\mathbb R$ satisfying $h(t)\geq t$ for all $t$, and $h(t)=t$ iff $t=0$. Let $M=\mathbb R^n\times\mathbb R$ and $\phi\colon M\to M$, $(x,t)\mapsto(x,h(t)+d(x,N))$. The fixed point set of $\phi$ is $N\times0$, but the homotopy class of $\phi$ does not depend on $N$. (You can easily modify this example so that $M$ is compact.)

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You probably want to square the $d$ appearing in $\phi$, no? –  Mariano Suárez-Alvarez Oct 6 '11 at 21:02
True, but your diffeomorphism is rather delicate! Give it a generic perturbation, and the fixed point set isn't a manifold anymore. It also fits neatly into a contractible family of such homeomorphisms; it's like you've engineered a homological $0/0$. Which is Nifty. But it rather skirts around the lots of Morse-like-theory which should be accessible. I'd tell the Questioner to heed this unstable warning well, but consider Mariano's response as closer to the asker's intent. –  some guy on the street Oct 7 '11 at 2:43
@Mariano: $\phi$ is only required to be an homeomorphism, and squaring $d$ does not make it differentiable in general. But you can modify $d$ so that $\phi$ is a diffeomorphism, agreed. –  user2035 Oct 7 '11 at 7:41
Ah. I had misrad the question as asking for a diffeo. In any case, cute :) –  Mariano Suárez-Alvarez Oct 8 '11 at 1:42

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