If $X$ is aspherical, we know that $H^*(X,k) = \text{Ext}_R(k,k)$, with $R = k\pi_1$. For non-aspherical spaces, do we ever have $H^*(X,k) = \text{Ext}_R(k,k)$ for some ring $R$? Obviously we need this algebra to be graded-commutative, so perhaps we want $R$ to be a Hopf algebra.

Or, instead of a ring $R$, maybe we want a tensor category $\mathcal{C}$? Whatever the right level of generality is.


EDIT: I should probably have guessed that "whatever the right level of generality" was too bold of a claim; I'm not looking for things as fancy as derived categories of sheaves or ring spectra. So for now I'll stick to the question of when it really is $\text{Ext}_R(k,k)$ for a ring $R$. I'm open to $k$-linear categories that are "not too much more complicated than group rings", but have no way of making this precise.

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    $\begingroup$ The sheaf cohomology $H^{\bullet}(X, k)$ is $\text{Ext}^{\bullet}(k, k)$ where the $\text{Ext}$ comes from taking derived hom in something like derived locally constant sheaves of $k$-modules on $X$ (note that this is not the derived category of locally constant sheaves). This agrees with the singular cohomology for reasonable $X$. $\endgroup$ – Qiaochu Yuan May 1 '15 at 4:58
  • $\begingroup$ The relevant "Hopf algebra," such as it is, is the "group algebra" $k[\Omega X]$, which is a ring spectrum over $k$ (assuming $X$ is path connected). If $k$ is a field maybe you can replace this with the dg algebra $C_{\bullet}(\Omega X, k)$, at least if $X$ is simply connected...? $\endgroup$ – Qiaochu Yuan May 1 '15 at 5:12
  • $\begingroup$ If you don't want to go to the differential graded seeing, at least, then you have to stick to aspherical spaces. $\endgroup$ – Fernando Muro May 1 '15 at 6:11
  • $\begingroup$ @Kevin: well, then it depends on what you want. Do you want $R$ to just be arbitrary and for there to just be some arbitrary isomorphism (which seems like it would both be hard to find and unhelpful once you found it), or do you want $R$ and the isomorphism to be functorial in $X$? If the latter then I bet this is impossible and I have no idea how to prove it. $\endgroup$ – Qiaochu Yuan May 1 '15 at 6:12
  • $\begingroup$ @Fernando I think I'm willing to go to the DGA setting - did you have in mind what Qiaochu had said? Could one of you provide a reference or write it down in a bit more detail? $\endgroup$ – Kevin Casto May 1 '15 at 7:10

Yes, for any connected X we can find a $\pi$ such that $H^*(X;k) \cong Ext_{{k}\pi}(k,k)$

This is the Kan-Thurston theorem:

Kan, D. M.; Thurston, W. P. Every connected space has the homology of a K(π,1). Topology 15 (1976), no. 3, 253–258.

combined with the universal coefficient theorem and the isomorphism $H^*(K(\pi,1);k) \cong Ext^*_{k\pi}(k,k)$

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