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For an algebra $A$ we can define its Hochschild cohomology (see this Wikipedia page) $HH^{\cdot}(A,A)$. It is well-known that the cup product makes $HH^{\cdot}(A,A)$ a (graded-commutative) algebra.

Now let $A$ be an $A_{\infty}$-algebra (see Mescher - A primer on $A_\infty$-algebras and their Hochschild homology Section 1 for definition) and we can still define its Hochschild cohomology (see the same preprint Section 3 for definition).

My question is: is $HH^{\cdot}(A,A)$ still an algebra for an $A_{\infty}$-algebra $A$? Or it is an $A_{\infty}$-algebra?

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    $\begingroup$ It is an $E_2$-algebra, which is a bit stronger than saying that it is an $A_{\infty}$-algebra. This was the Deligne conjecture, which has had several proofs over the years. There is a nice brief overview on this nLab page: nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/… $\endgroup$
    – user164898
    Commented Jan 26, 2022 at 4:25
  • $\begingroup$ @A.S. The Deligne conjecture says that the Hochschild cochain has the structure of an $E_2$-algebra, but my question is whether the Hochschild cohomology has the structure of an algebra. $\endgroup$
    – Zhaoting Wei
    Commented Jan 26, 2022 at 14:23
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    $\begingroup$ I am sorry if I misunderstand your question and my answer is not helpful. If I do not misunderstand, then it appears to me that your question gets an answer from the solutions to the Deligne conjecture given at the link I provided. The reason is this: once you know that the Hochschild cochain complex of an $A_\infty$-algebra $A$ has the structure of an $E_2$-algebra, then you know that the cohomology of that Hochschild cochain complex (i.e., the Hochschild cohomology of $A$) has the structure of a Batalin-Vilkovisky algebra, which is an algebra plus a bit more structure. $\endgroup$
    – user164898
    Commented Jan 26, 2022 at 15:28
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    $\begingroup$ I think maybe the answer to the question is that it's both (and more, as the previous comments say). You can always take the cohomology of an $A_\infty$ algebra to get an ordinary algebra. You can also give the cohomology the structure of an $A_\infty$ algebra that is quasiisomorphic to the original $A_\infty$ algebra. The higher order products are basically the Massey products. See the Nlab. $\endgroup$
    – Aaron Bergman
    Commented Jan 26, 2022 at 18:03

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