MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A$ be a commutative $\mathbb{C}$-algebra, and consider $C_{\bullet}(A,A)$ the simplicial Hochschild homology module of $A$ with respect to itself (i.e. $C_{n}(A,A)=A^{\otimes (n+1)}$). This is a simplicial commutative $\mathbb{C}$-algebra and we can take its $Spec$ levelwise, to get a cosimplicial $\mathbb{C}$-scheme $X_A :=Spec(C_{\bullet}(A,A))$. If $A$ is of finite type over $\mathbb{C}$, then we may take levelwise the associated topological space for the analytic topology, and get a cosimplicial topological space, denoted $X_{A}^{top}$. My question is whether somebody knows what the geometric realization of $X_{A}^{top}$ looks like (e.g. any relation to the topological free loop space of $(Spec(A))^{top}$?).

share|cite|improve this question
what happens if you take the geometric realization of the simplicial set underlying $C_\bullet(A,A)$? – Yosemite Sam Feb 28 '12 at 11:43
I thought it was a cyclic thing, not a simplicial one. Maybe I am mistaken though. – Sean Tilson Feb 29 '12 at 3:53
A cyclic is simplicial by forgetful. – Sereza Feb 29 '12 at 10:17
up vote 4 down vote accepted

Nice question. I did this computation a while ago, but I guess what you get, by taking the levelwise analytification of $X_A$, is the canonical cosimplicial model for the (topological) free loop space of $(SpecA)^{top}$ (i.e. its totalization is homeomorphic to the free loop space $L((SpecA)^{top})$).

share|cite|improve this answer
Thanks! where can I find this canonical model written down? – Sereza Feb 29 '12 at 10:17

Let $Y$ denote the associated topological space for $Spec(A)$. It looks like the cosimplicial space $X_A ^{top}$ is just

$ Y \to Y \times Y \to Y\times Y \times Y \to \ldots$,

where there should be two arrows at the first step which are both the diagonal, then the other arrows are given by various diagonal maps. This is the cosimplicial diagram which computes the homotopy fibre product

$ Y \times _{Y\times Y} Y$

i.e. the free loop space of $Y$.

I am writing this in a bit of a hurry, so I hope this makes sense!

share|cite|improve this answer
The standard cosimplicial space modeling the homotopy fiber product $X\times_Z Y$ in degree $n$ is $X\times Z^{n}\times Y$; you appear to be missing some terms. – Justin Noel Feb 29 '12 at 9:56

Sam Gunningham had the right idea, but I would like to fix it up for the construction of the homotopy pullback that I am aware of.

Alternatively one can compute $HH(A;A)$ as $\mathrm{Tor}^{A^e}(A,A)$ where $A^e$ is the enveloping algebra of $A$, $A\otimes A^{op}$, since your $A$ is commutative $A^{e}\cong A\otimes A$. These Tor groups can be calculated via the simplicial complex which in degree $n$ is $A\otimes (A^{e})^{\otimes n}\otimes A\cong A^{\otimes {2n+2}}$. As a side note the geometric realization of this complex gives a topological commutative ring (Because $A$ is commutative) which models $THH^{HZ}(HA;HA)$ via the Eilenberg-Maclane functor. Applying your contravariant functor to spaces to the simplicial algebra we obtain a cosimplicial space modeling the homotopy pullback $Spec(A)^{top}\times_{Spec(A)^{top}\times Spec(A)^{top}}\times Spec(A)^{top}$ (see 3.3 of ) which as Sam pointed out, models the free loop space.

share|cite|improve this answer
Ah, thanks! I still believe that the cosimplicial space that I wrote down calculates the free loopspace though (just as both complexes calculate the Hochshild homology). I think the idea is that my diagram sits inside yours as a retract. – Sam Gunningham Mar 1 '12 at 0:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.