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 $\Lambda^{op}$.



Homotopy colimits of cyclic spaces
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 $BS^1$.


