Let $X$ be a topological space, and $Homeo(X)$ the group of self-homeomorphisms of $X$.

(1) What is the exact meaning of: $H^*(X)$ is a an $A_\infty$-module over $Homeo(X)$?

(2) Does $H_*(X)$ also have an $A_\infty$-module structure? Is it the same as that of $H^*(X)$?

Added later: Jeff Giansiracusa gave a nice answer to (1). But his answer uses the ring structure in cohomology, leaving (2) open: Is there an $A_\infty$ structure on homology as well?

  • $\begingroup$ I guess $C_\ast(X)$ has a dg coalgebra structure, and thus $H_\ast(X)$ has a "co-$A_\infty$" structure ... ? $\endgroup$ – Kevin H. Lin Nov 25 '10 at 6:34
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    $\begingroup$ yup. In fact, you can get an $E_\infty$ coalgebra structure on $C_*X$ and hence on $H_*X$, and an $E_\infty$ algebra structure on $C^*X$ and hence on $H^*(X)$ by transfer of structure (probably you need field coefficients, but not necessarily char zero). $\endgroup$ – Jeffrey Giansiracusa Nov 25 '10 at 9:52

Your category $X^X$ is just the group of homeomorphisms of $X$. This group certainly acts on the homology and cohomology, making them strict modules. But since the group of homeomorphisms is actually acting on $X$, it gives automorphisms of the rational homotopy type. The rational homotopy type of $X$ can be encoded in an $A_\infty$ algebra structure on the rational cohomology ring (technically, it is a $C_\infty$ structure, which is a special kind of $A_\infty$ structure). Thus the group of homeomorphisms of $X$ gives homotopy self-equivalences of the $A_\infty$ algebra $H^*(X)$. That is, a homeomorphism $\phi: X \to X$ gives an $A_\infty$ map $H^*(X) \to H^*(X)$ that is an equivalence. Note that such a map contains potentially more information than simply an automorphism of $H^*(X)$ as an ordinary ring.

  • $\begingroup$ Cool, thanks. Can this reasoning be applied to $H_*(X)$ as well? $\endgroup$ – Romeo Sep 17 '10 at 16:38

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