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Let $\Lambda$ denote Connes's cyclic category. It is an extension of the simplex category $\Delta$ (of nonempty finite linearly ordered sets) obtained by adding an automorphism of order $n+1$ to the object $\textbf{n}$.

Question: Suppose $X: \Lambda^{op} \to Top$ is a cyclic space. What is a description of the homotopy colimit of this functor?

Just to put this in a bit of context, if $Y: \Delta^{op} \to Top$ is a simplicial space then it has a geometric realisation $|Y|$. One can also take the homotopy colimit of $Y$, and under some reasonable hypotheses there will be an equivalence $\mathrm{hocolim}_{\Delta^{op}} Y \simeq |Y|$.

There is an inclusion $\Delta \to \Lambda$, so a cyclic space $X$ can be considered as a simplicial space and one can thus make a geometric realisation $|X|$. This space is supposed to have a circle action. I suppose my question should be:

How is the hocolim of $X$ over $\Delta^{op}$ related to the hocolim of $X$ over $\Lambda^{op}$.

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The homotopy theory of cyclic spaces is equivalent to that of spaces over $BS^1$ (Dwyer-Hopkins-Kan). The colimit over the simplicial category is as you say a space $X$ with $S^1$ action, and the colimit over the cyclic category is the quotient (Borel construction) $X/S^1$ as a space over $BS^1$.

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    $\begingroup$ More precisely, a localization of cyclic spaces is equivalent to spaces over $BS^1$. Without the localization, it's more like spaces with an $S^1$ action, with actual fixed point spaces for subgroups (but not for the whole group). also see MO's suggestions. Spalinski: ams.org/mathscinet-getitem?mr=1325168 $\endgroup$
    – Ben Wieland
    May 13, 2010 at 16:45
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    $\begingroup$ Hi Ben! I'm not sure I understand your comment -- with the model structure that DHK give (as your link mentions) you do get the $\infty$-category of spaces over $BS^1$ on the nose. So sure, you can give them a more refined model structure (which Spalinski does) keeping track of extra fixed point data, and this is what is discussed in Reid's question, but in what sense does this is a "more precise" answer to the question as opposed to a different homotopy theory? $\endgroup$
    – David Ben-Zvi
    May 13, 2010 at 22:17

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