Let $(V,\cdot)$ be an associative algebra and $W$ be a vector space endowed with a bimodule structure $\triangleright:V\otimes W\to W$ and $\triangleleft:W\otimes V\to W$ such that the following relations hold:

$(f\cdot g)\cdot h=f\cdot (g\cdot h)$

$(f\cdot g)\triangleright X=f\triangleright(g\triangleright X)$

$X\triangleleft(f\cdot g)=(X\triangleleft f)\triangleleft g$

$(f \triangleright X)\triangleleft g=f \triangleright(X \triangleleft g)$.

One defines the associated Hochshild differential $\delta:H^i(V,W)\to H^{i+1}(V,W)$ where $H^i(V,W):=Hom(V^{\otimes i+1},W)$ as: $ \delta\Phi(a_0,\ldots,a_{i+1}):=(-1)^i a_0\triangleright\Phi(a_1,\ldots, a_{i+1})$

$-(-1)^i\underset{k,l\geqslant 0}{\sum_{k+l=i}}(-1)^{k}\,\Phi(a_0,\ldots,a_{k-1}\cdot a_k,\ldots,a_{i+1})+\Phi(a_0,\ldots,a_{i})\triangleleft a_{i+1}$.

for all $\Phi\in H^i(V,W)$, $a_i\in V$.

Question: Whenever $W=V$ and $\triangleright=\triangleleft=\cdot$, the differential on $H(V,V)$ can be extended to a differential graded Lie algebra by defining the Gerstenhaber bracket, for which the Hochschild differential is a derivation.

Can one generalise this construction for an arbitrary bimodule $W$ and define a graded Lie bracket on $H(V,W)$ such that the Hochschild differential is a derivation thereof?

  • 2
    $\begingroup$ A good start would be to think what happens in degree 1. $\endgroup$ Dec 12, 2017 at 4:14

1 Answer 1


Not in general, no. However $H^*(V,W)$ is a representation of the Lie algebra $H^*(V,V)$, by letting, for $f \in \hom(V^{\otimes m}, W)$ and $g \in \hom(V^{\otimes n}, V)$, $$f \circ g := \sum_{i=1}^m \pm f \circ_i g,$$ just like in the definition of the Lie bracket on $H^*(V,V)$. This module structure already appeared in Gerstenhaber's 1963 paper The cohomology structure of an associative ring.

  • $\begingroup$ I think there's something wrong with this formula. Take for example $n = m = 1$. Then $HH^1(V, V)$ is just a Lie algebra of outer derivations of $V$ and $HH^1(V, W)$ is the space of outer derivations of $V$ with values in $W$. Then this formula for action just tells to compose one with another. This will be a map $V \rightarrow W$, but it won't be a derivation. $\endgroup$ Apr 15, 2021 at 19:39
  • $\begingroup$ I think there's no way to define a Lie action of $HH^*(V, V)$ on $HH^*(V, W)$. For example, if $V$ is an algebra of functions on an (affine algebraic) smooth manifold, and $W$ is a skyscraper bimodule corresponding to some point on this manifold, then $HH^1(V, V)$ will be the Lie algebra of vector fields and $HH^1(V, W)$ will be the tangent space to the point in question. There is not any non-trivial action of the global vector fields on the tangent space to a point. $\endgroup$ Apr 15, 2021 at 19:46
  • $\begingroup$ @GrishaPapayanov I'm somewhat confused. I presume that by $HH^1(V,W)$ you mean Hochschild cohomology. I'm talking about OP's $H^1(V,W) = \hom(V,W)$ and I didn't write anything about the differential. $\endgroup$ Apr 16, 2021 at 13:48
  • $\begingroup$ Indeed! I didn't notice that $H$ in the question is Hom and not Hochschild cohomology, sorry. I agree with you. However, Gerstenhaber himself in the paper you cited claims that this induces an action on the level of cohomology (Theorem 4, section 8, page 282), and this cannot be right. $\endgroup$ Apr 16, 2021 at 17:46

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