4
$\begingroup$

The following theorem is known from a paper "Duality in Gerstenhaber Algebras" by Felix, Menichi, Thomas. Given a simply connected space X of finite type.

There is an equivalence of Gerstenhaber algebras

$HH^*(C_*(\Omega X,\mathbb{Q}), C_*(\Omega X,\mathbb{Q}) \cong HH^*(C^*(X,\mathbb{Q}),C^*(X,\mathbb{Q})$

On the left hand side we have Pontryagin product on the based loop space and on the right hand side rational cochains. $HH^*$ denotes Hochschild cohomology.

I have never seen anyone speak to the following enhanced statement, which makes me wonder if there is a counterexample or if I am simply missing some literature.

$HCH^*(C_*(\Omega X), C_*(\Omega X) \cong HCH^*(C^*(X),C^*(X))$

The question is: Is this statement true, false or unknown?

Here we are looking at Hochschild cochains in the homotopy category of $B(\infty)$ algebras. For the background police, a $B(\infty)$ algebra is a type of dg-Gerstenhaber structure, that naturally gives rise to a Gerstenhaber structure by passing to homology. For more info, see the paper of Keller mentioned below.

It is possible to prove this theorem when $C^*(X)$ is equivalent to a graded simply connected Koszul algebra( i.e. X is both formal and coformal). I believe this is due to Keller in a paper called the "Derived Invariance of Higher Structures of the Hochschild complex".

$\endgroup$

1 Answer 1

2
$\begingroup$

Looking closer at Keller's paper, the result seems to be in there. Namely, in his main theorem in section 3.3, he proves that fully faithful dg-functors $per(A) \to D(B)$ induced by an $A\otimes B^{op}$ module X induce $B(\infty)$ morphisms $\phi_X: HCH^*(B,B) \to HCH^*(A,A)$. Additionally, in the same theorem, he proves that if the map $per(B^{op}) \to D(A^{op})$ induced by X is also fully faithful, then $\phi_X$ is invertible.

These criterion all apply to M a simply connected space, $X= \mathbb{Q}$ the trivial local system, $A=C_*(\Omega(M))$, and $B= C^*(M)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .