Hello to all,

If $(A,\mu)$ is an algebra, it is very well known that set of deformations mod equivalence is isomorphic to the of Maurer-Cartan set of the DG Lie algebra of the hochschild cocomplex mod gauge equivalence. One usually finds this result proven by brute force in the case of an infinitesimal deformation, but it seems to me that the problem becomes a little more meaty in the general case: clearly a Maurer-Cartan element must be a map $\mu'$ such that $\mu+\mu'$ is an associative multiplication as expected, but the gauge action requires a little more work: two such Maurer-Cartan elements are equivalent if $\mu''= e^{[f, -]}(\mu')+ \sum \frac{1}{n!}[f,df]^{n-1}$. Can one prove using an elementary computation that the above equation is the same as $\mu+\mu'$ being an equivalent multiplication to $\mu+\mu''$?


1 Answer 1


A good way of rewriting the gauge action is $e^{f}(\mu+\mu')e^{-f}=\mu+e^{f}*\mu'$, which makes the equivalence manifest. There are several references for the computation above. The first coming to my mind is Marco Manetti's Lectures on deformations of complex manifolds where, if I'm not wrong, it's spelled out in detail.

  • $\begingroup$ You're right, it was implicitely written in Manetti. Thank you! $\endgroup$
    – Louis
    Aug 2, 2011 at 14:12

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