# Homology and cohomology of free loop spaces

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

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.