# Are two Lie algebra deformations with cohomologous tangents isomorphic?

Let $(\mathfrak{g},[-,-])$ be a finite-dimensional Lie algebra. I'm interested in real Lie algebras, but feel free to complexify if this helps.

Say I'm interested in classifying isomorphism classes of Lie algebra deformations of $\mathfrak{g}$, by which I (here, today) mean isomorphism classes of Lie algebra structures on the vector space $\mathfrak{g}[[t]]$ of formal power series in $t$ with coefficients in $\mathfrak{g}$, which agree with $(\mathfrak{g},[-,-])$ for $t=0$.

The way I would normally go about doing this is the following:

1. Calculate $H^2(\mathfrak{g};\mathfrak{g})$ and choose representative cocycles $\varphi_1,\varphi_2,\dots$ for each cohomology class.
2. Study the integrability of the infinitesimal deformation with cocycle $t_1 \varphi_1 + t_2 \varphi_2 + \cdots$.
3. Profit.

The hope is that in doing this I would arrive at all isomorphism classes of formal deformations of my original Lie algebra. But for this hope to be realised, one seems to require that the choice of cocycle made in the first step would not matter.

Recent explicit calculations have made me doubt this long-held belief (misconception?). Of course, it's very likely I could have made an error, but before checking yet again, I thought I would ask the MO community to make sure I'm not in fact making a conceptual error.

Question If two (integrable) deformations of $\mathfrak{g}$ are infinitesimally equivalent (so that the cocycles defining the infinitesimal deformations are cohomologous) are they necessarily isomorphic? (You may assume that the deformations are polynomial in $t$.)

Let me try to rephrase the problem in a different language. If we let $\phi_0 \in \Lambda^2\mathfrak{g}^*\otimes \mathfrak{g}$ denote the original Lie bracket, I consider a formal power series $\phi(t)$ with $\phi(0) = \phi_0$ and such that $[\phi(t),\phi(t)] = 0$ for all $t$, where the bracket here is the Nijenhuis-Richardson bracket on $\Lambda^{\bullet + 1}\mathfrak{g}^*\otimes \mathfrak{g}$.
If I write $\phi(t) = \phi_0 + \alpha(t)$, with $\alpha(0)= 0$, then the condition that $\phi(t)$ defines a Lie bracket for all $t$ is the Maurer-Cartan equation for $\alpha(t)$: $$[\phi_0,\alpha(t)] + \frac12 [\alpha(t),\alpha(t)] = 0$$
Suppose I have two such solutions $\alpha_1(t)$ and $\alpha_2(t)$ of the Maurer-Cartan equation with $\alpha_1(0) = \alpha_2(0)= 0$. Then my question is the following: if $\alpha_1'(0)$ and $\alpha_2'(0)$ are gauge equivalent, will $\alpha_1(t)$ and $\alpha_2(t)$ be gauge equivalent for all $t$?
• Can't you get a counterexample to the rephrased question by starting from an abelian Lie algebra, taking $\alpha_1=0$ and $\alpha_2$ to be any non-trivial bracket divisible by $t^2$? – Jon Pridham Aug 10 '17 at 9:03
• Hi @JonPridham. Indeed. My rephrasing needs work. In your example, if a nontrivial bracket is divisible by $t^2$ then the first nonzero term is still a cocycle, so then the question is whether if the $t^2$ term is a coboundary, whether the deformation is trivial. I need to think more about this, because I'm starting to think that the answer is negative. – José Figueroa-O'Farrill Aug 10 '17 at 11:48