3
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ I guess $C_\ast(X)$ has a dg coalgebra structure, and thus $H_\ast(X)$ has a "co-$A_\infty$" structure ... ? $\endgroup$ Nov 25, 2010 at 6:34
  • 1
    $\begingroup$ 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). $\endgroup$ Nov 25, 2010 at 9:52

1 Answer 1

6
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Cool, thanks. Can this reasoning be applied to $H_*(X)$ as well? $\endgroup$
    – Romeo
    Sep 17, 2010 at 16:38

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.