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.
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 128130. MR0107238 (21 #5963); Swan, Richard G. A new method in fixed point theory. Comment. Math. Helv. 34 1960 116. 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 nontrivial 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.
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.)

$\begingroup$ You probably want to square the $d$ appearing in $\phi$, no? $\endgroup$ – Mariano SuárezÁlvarez Oct 6 '11 at 21:02

$\begingroup$ 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 Morseliketheory 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. $\endgroup$ – some guy on the street Oct 7 '11 at 2:43

$\begingroup$ @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. $\endgroup$ – user2035 Oct 7 '11 at 7:41

$\begingroup$ Ah. I had misrad the question as asking for a diffeo. In any case, cute :) $\endgroup$ – Mariano SuárezÁlvarez Oct 8 '11 at 1:42