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For $R$ any ring and $H R$ its Eilenberg-MacLane spectrum -- a ring spectrum -- there is an equivalence between the $\infty$-categories of $H R$-module spectra and that of unbounded chain complexes of $R$-modules - a stable version of the Dold-Kan correspondence.

At least in good cases such as $R = \mathbb{Z}$ this refines also to an equivalence between $H \mathbb{Z}$-algebra spectra and unbounded dg-rings -- a stable version of the monoidal Dold-Kan correspondence. (A presentation of this by a Quillen equivalence has been given by Shipley, see http://ncatlab.org/nlab/show/algebra+spectrum).

Now for $X$ a sufficiently nice topological space, one would hope that under this equivalence the "spectrum of integral homology chains" $(\Sigma^\infty_+ \Omega X) \wedge H \mathbb{Z}$ on the loop space of $X$ is identified, up to equivalence, with the ordinary chain complex of singular homology chains $C_\bullet(\Omega X, \mathbb{Z})$, both as $\infty$-modules and as $\infty$-algebra objects.

This must be easy to see, but I am a little stuck. Can anyone help?

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  • $\begingroup$ Homology is not an algebra. Cohomology is... $\endgroup$ – André Henriques Jun 7 '11 at 18:02
  • $\begingroup$ Er, sorry, I am thinking of X being a loop space. This is motivated from looking for chain versions of string topology. If I may, I'll edit the question to fix this... $\endgroup$ – Urs Schreiber Jun 7 '11 at 18:10
  • $\begingroup$ You just have to check that, up to equivalence, they go to one another under the chain of weak equivalences. This might be tedious, but I bet it0s straightforward. I'd give it for granted. $\endgroup$ – Fernando Muro Jun 7 '11 at 18:39
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    $\begingroup$ Fernando, thanks, I know what I'd have to check. I did go the road you indicate: first I took it for granted, then I thought that it is easy to check, then I found it was a bit more of a pain maybe. By the way, on p. 15 here arxiv.org/PS_cache/arxiv/pdf/0906/0906.5198v1.pdf#page=15 it is stated in a way that makes it true by definition. I thought it would be nice to know if somebody had actually checked it. $\endgroup$ – Urs Schreiber Jun 7 '11 at 19:36
  • $\begingroup$ Do you have a specific model for $H\mathbb{Z}$-modules that you are trying to work with? $\endgroup$ – Tyler Lawson Jun 7 '11 at 20:30

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