String topology, as well as Hochschild (co)homology, give a rich perspective on the homology and cohomology of a free loop space $LM$ of a manifold $M$.

Let $k$ be a field and let $M$ be $n$-dimensional. There is a graded ring $H^*(LM, k)$ with multiplication the usual multiplication on cohomology of a space. If $M$ is oriented, there is a graded ring $\mathbb{H}_*(LM) = H_{*+n}(LM, k)$ with multiplication given by the Chas-Sullivan product.

Is there any relation between these graded rings? One should feel free to assume that $M$ is simply connected and also that $TM$ is stably trivial if that improves the situation. Of course, in that case one has that $H_*(LM) = HH_*(C^*(M), C^*(M))$ while $\mathbb{H}_*(LM) = HH^*(C^*(M), C^*(M))$ (where $C^*(M)$ denotes the algebra of cochains on $M$) but from the fact perspective the cup-product structure on $H^*(LM)$ is wholly not self-evident (for $k = \mathbb{Q}$, where $C^*(M)$ can be made into a cdga, it arises from the product on $C^*(M)$ and the shuffle product).


1 Answer 1


I'll address your question for a reason why the Hochschild homology $HH_*(C^*(M),C^*(M))$ should have a cup product, without resorting to the fact that it is the cohomology of a space. We assume everything implicitly rationalized and $M$ simply connected.

The Hochschild homology of $C^*(M)$ can be computed as the factorization homology $\int_{S^1} C^*(M)$. The relation between Hochschild homology and the free loop space is an artifact of the more homotopical claim that $\int_{S^1} \Sigma^\infty M^\vee \simeq \Sigma^\infty_+ \operatorname{Map}(S^1,M)$. So to address your question, we should supply $\int_{S^1} \Sigma^\infty M^\vee$ with a diagonal without actually computing the factorization homology.

Factorization homology over $N$ can be computed as a bar construction of the little disks modules $E_N$ with an $E_n$ algebra $A$. Let $(S^1)^\times$ denote the right $\mathrm{com}$ module which in degree $i$ is $(S^1)^{\times i}$ with right module structure induced by the diagonal. This is equivalent to $\mathrm{ind}^{\mathrm{com}}_{E_1} E_{S^1}$. We need 3 facts: the zLie algebra model of $M$ is $K(\Sigma^\infty M^\vee)$, the enveloping algebra of this is $\Sigma^\infty \Omega M$, and $K(\Sigma^\infty_+(S^1)^\times) \simeq \Sigma^\infty_+ E_{S^1}$ as $\mathrm{lie}$ modules. The first two are classical, and the third is a consequence of the Koszul self duality of the maps of operads $\mathrm{lie} \rightarrow \mathrm{Ass} \rightarrow \mathrm{com}$. Combined with basic facts about induction and restriction, we may compute:

\begin{align}\int_{S^1} \Sigma^\infty_+ M ^\vee &\simeq B\big(\Sigma^\infty_+E_{S^1},\Sigma^\infty_+E_1, \Sigma^\infty M^\vee\big)\\ &\simeq B\big(\mathrm{ind}_{E_1}^{\mathrm{com}} \Sigma^\infty_+E_{S^1},\mathrm{com}, \Sigma^\infty M^\vee\big)\\ &\simeq B\big(\Sigma^\infty_+(S^1)^\times,\mathrm{com},\Sigma^\infty M^\vee \big)\\ &\simeq B\big(K(\Sigma^\infty_+(S^1)^\times),K(\mathrm{com}),K(\Sigma^\infty M^\vee\big)\\ &\simeq B\big(\Sigma^\infty_+ E_{S_1},\Sigma^\infty_+E_1,\mathrm{ind}_{\mathrm{lie}}^{E_1} K(\Sigma^\infty M^\vee)\big)\\ &\simeq B\big(\Sigma^\infty_+ E_{S_1},\Sigma^\infty_+ E_1,\Sigma^\infty \Omega M\big)\simeq \int_{S^1} \Sigma^\infty \Omega M \end{align}

This last algebra has a diagonal since it is the suspension spectrum of a space, hence the factorization homology has one as well. Note: I have played fast and loose with suspensions, so some of these equivalences might only hold plus or minus some desuspensions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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