# Homology of free loop space and Hochschild cohomology

I am looking for honest proof of the following isomorphism. For a simply connected space $X$, let $LX$ be its free loop space. Then $$H_{\ast}(LX) \simeq HH^\ast(C^\ast(X),C_\ast(X))$$

I have looked Jones' approach on ''Cyclic homology and equivariant homology'' that given a map $$f_k : \Delta^k \times LX \to X^{k+1}$$ defined by $$(x_1,\cdots,x_k)\times \gamma \mapsto (\gamma(0),\gamma(x_1),\cdots,\gamma(x_k))$$

its cochain map $$f_k^\ast : C^\ast(X)^{\otimes k+1} \to C^{\ast -k}(LX)$$ fit together to define a chain map from the Hochschild complex to the free loop space $$f^\ast : CH_\ast(C^\ast(X)) \to C^\ast(LX)$$ which is a chain homotopy equivalence, so that we get an isomorphism $$HH_\ast(C^\ast(X),C^\ast(X)) \simeq H^\ast(LX)$$

Many people simply say that ''dualizing'' the statement we get the relation between homology of loop space and Hochschild cohomology, but what is the honest procedure to dualize?

On Cohen and Voronov's Notes on String topology, they just take dual of $f^\ast_k$ and they claim that $$C_\ast(LX) \simeq Hom(C^{\ast}(LX),\mathbb{Z})$$ and $$Hom(C^\ast(X)^{\otimes k+1},\mathbb{Z}) \simeq Hom(C^\ast(X)^{\otimes k},C_\ast(X))$$ which also means $$C_\ast(X) \simeq Hom(C^{\ast}(X),\mathbb{Z})$$

However, there is no reason to be true that those groups are either isomorphic or quasi-isomorphic.

• I think this is proved as Theorem 1.5.2 in Cohen and Voronov's "Notes on String Toplogy", arxiv.org/pdf/math/0503625.pdf . Commented May 16, 2018 at 9:04
• @Tyrone and that is exactly what I am looking for. They dualizes cochain complex to get chain complex but those are not isomorphic. I think those are quasi-isomorphic if I use field coefficients and the homology is finitely generated. Commented May 16, 2018 at 9:26
• I was also confused about such question and just wrote my thinking as an answer (too long to be a comment). I also want to mention that people mostly construct a chain map $CH_*(C^*(X),C^*(X))\to C^*(LX)$, whether they use singular cochains or differential forms. Such a map is basically the same as a map $C_*(LX)\to CH^*(C^*(X),C^*(X)^\vee)$. It seems much harder to directly define a chain map from $C_*(LX)$ to $CH^*(C^*(X),C^*(X))$. There is a construction by K. Irie over $\mathbb{R}$: arxiv.org/pdf/1404.0153v5.pdf Commented Mar 5, 2021 at 5:07

• On a dual complex, $$((C_*)^\vee)^i=Hom(C_*,\mathbb{Z})^i=Hom(C_i,\mathbb{Z}).$$
• On the Hochschild chain complex of $$C^*(X)$$, $$CH_i(C^*(X),C^*(X))=\bigoplus_{i_1+\cdots+ i_k=i+k} C^{i_1}(X)\otimes\cdots\otimes C^{i_k}(X).$$ Therefore $$CH_*$$ is honestly $$\mathbb{Z}$$-graded, and the Hochschild differential increases degree: $$\partial^*+\delta: CH_*\to CH_{*+1}$$.

With the above grading, there is an isomorphism $$H^i(LX)\cong HH_i(C^*(X),C^*(X))\ \ \ \forall i$$ in case $$X$$ is simply-connected.

The validity of dualizing the above isomorphism could come from the Universal Coefficient Theorem. For simplicity let’s work over $$\mathbb{Z}$$, and assume that homology groups are finitely generated in each degree.

For a degree-decreasing complex $$\{\cdots \to C_*\to C_{*-1}\to \cdots\}$$, form its dual complex $$C^*=(C_*)^\vee$$, then $$F(H^i)\cong F(H_i),\ \ \ T(H^i)\cong T(H_{i-1}),$$ where $$F,T$$ denotes free part and torsion part, respectively. Similarly, if $$\{\cdots \to C^*\to C^{*+1}\to \cdots\}$$ is a degree-increasing complex, its dual complex $$\bar{C}_*=(C^*)^\vee$$ has $$F(\bar{H}_i)\cong F(H^i),\ \ \ T(\bar{H}_i)\cong T(H^{i+1}).$$ As a consequence, take $$\bar{C}_*= C^{\vee\vee}_*$$ to be the cocochain complex obtained by doubly dualizing $$C_*$$, we have $$F(\bar{H}_i)\cong F(H_i),\ \ \ T(\bar{H}_i)\cong T(H^{i+1})\cong T(H_i).$$ Therefore $$C_*\to C_*^{\vee\vee}$$ is a quasi-isomorphism.

By this MO question I think the finite generation assumption is satisfied in our situation.

Since $$CH^*(C^*(X),C^*(X)^\vee)$$ is exactly the dual complex of $$CH_*(C^*(X),C^*(X))$$, where the differential on the latter is degree-increasing, we have \begin{align*} T(H_i(LX))&\cong T(H^{i+1}(LX))\\ &\cong T(HH_{i+1}(C^*(X),C^*(X)))\\ &\cong T(HH^{i}(C^*(X),C^*(X)^\vee)). \end{align*}

The free parts are of course identified in the same degree, so $$H_i(LX)\cong HH^{i}(C^*(X),C^*(X)^\vee).$$

Finally, the $$C^*(X)$$-bimodule quasi-isomorphism $$C_*(X)\to C_*(X)^{\vee\vee}=C^*(X)^{\vee}$$ induces an isomorphism $$HH^{i}(C^*(X),C_*(X)))\cong HH^{i}(C^*(X),C^*(X)^{\vee}),$$ and we are done.