# How to define the equivalence of Maurer-Cartan elements in an $L_{\infty}$-algebra?

First let $L^{\bullet}$ be a pro-nilpotent differential graded Lie algebra (dgla). We have the set of Maurer-Cartan elements in $L^{\bullet}$ ($MC(L^{\bullet})$) which are $\alpha \in L^1$ such that it satisfies the Maurer-Cartan equation $$\partial \alpha+ \frac{1}{2}[\alpha,\alpha]=0.$$ We have a definition of gauge equivalance: $\alpha_0,\alpha_1\in MC(L^{\bullet})$ are called gauge equivalent if and only if there exists a $\xi \in L^0$ such that $$e^{\text{ad}\xi}\circ(\partial +\text{ad}\alpha_0)\circ e^{-\text{ad}\xi}=\partial +\text{ad}\alpha_1$$ or in other words $$e^{\text{ad}\xi}\alpha_0-\frac{e^{\text{ad}\xi}-1}{\text{ad}\xi}\partial\xi=\alpha_1.$$ From the first definition it is easy to see that gauge equivalence is really an equivalent relation. From the second definition we can define a path between $\alpha_0$ and $\alpha_1$. Let $$\alpha(t)=e^{t\text{ad}\xi}\alpha_0-\frac{e^{t\text{ad}\xi}-1}{\text{ad}\xi}\partial\xi.$$ Then $\alpha(t)$ is a power series of $t$ in $L^{\bullet}$, $\alpha(0)=\alpha_0$, $\alpha(1)=\alpha_1$ and we can prove $\partial\alpha(t)+ \frac{1}{2}[\alpha(t),\alpha(t)]=0.$

Now we come to $L_{\infty}$ algebra $L^{\bullet}$ with higher bracket $[\cdot,\ldots,\cdot]_n$ with $n$ auguments. We still have Maurer-Cartan elements in $L^{\bullet}$ ( $MC(L^{\bullet})$) which are $\alpha \in L^1$ such that it satisfies the Maurer-Cartan equation $$\partial \alpha+ \sum \frac{1}{k!}[\alpha,\ldots,\alpha]_k=0.$$

My question is how to define equivalence of Maurer-Cartan elements in $L^{\bullet}$?

Of course we can define $\alpha_0,\alpha_1\in MC(L^{\bullet})$ are "equivalent" if and only if there exists a power series $\alpha(t)\in L^{\bullet}$ such that $\alpha(0)=\alpha_0$, $\alpha(1)=\alpha_1$ and $\alpha(t)$ satisfies the $L_{\infty}$ Maurer-Cartan equation. However, it is difficult to show that this is an equivalent relation, for example, how to connect two paths?

It seems that a generalization of gauge equivalence is what we want. But $e^{\text{ad}\xi}$ is not enough since we have higher bracket.

To add a bit to what Damien says, addressing your question on how to generalise the gauge approach (which is equivalent to the approach outlined by Damien, as proved by several people):

You can view gauge symmetries in DGLAs via solving the differential equation $$\frac{d\alpha}{dt}=-\partial\xi-[\alpha,\xi],$$ where $\xi$ is the given element of degree $0$. This generalises to homotopy Lie algebras as follows: consider the differential equation $$\frac{d\alpha}{dt}=-\partial\xi-[\alpha,\xi]-\frac12[\alpha,\alpha,\xi]-\ldots-\frac{1}{p!}[\underbrace{\alpha,\ldots,\alpha,}_{p \text{ times}}\xi]_{p+1}-\ldots,$$ where the right hand side is simply the negative of $[\xi]_1^\alpha$, the first structure map of the twisted Lie-infinity structure $$[x_1,\ldots,x_k]_k^\alpha:=\sum_{p\ge0}\frac{1}{p!}[\underbrace{\alpha,\ldots,\alpha,}_{p \text{ times}}x_1,\ldots,x_k]_{k+p}.$$ From that it is almost obvious that moving along the integral curves of this equation preserves the property of being Maurer--Cartan, since the Maurer--Cartan condition for $\alpha+\beta$, where $\alpha$ is a Maurer--Cartan element, and $\beta$ is infinitesimal becomes $$\partial\beta+[\alpha,\beta]+\frac12[\alpha,\alpha,\beta]+\ldots+\frac{1}{p!}[\underbrace{\alpha,\ldots,\alpha,}_{p \text{ times}}\beta]_{p+1}+\ldots,$$ that is $[\beta]_1^\alpha=0$, and so $\beta=[\xi]_1^\alpha$ satisfies that, $[\cdot]_1^\alpha$ being a differential of the twisted structure. This circle of ideas is explained in many places, one important reference is Lie theory for nilpotent $L\_\infty$-algebras'' by Ezra Getzler (Ann. of Math. (2) 170 (2009), no. 1, 271--301.).

• Thank you very much! I will look at paper you suggested. By the way, since the equivalent class of Maurer-Cartan elements forms an $\infty$-groupoid, does it means that the "composition" of two equivalences is not unique? Aug 31, 2012 at 4:24
• Yes, that is a good point. In particular, if you look at the formulas above, you of course realise that unlike the case of DGLAs, $L_0$ is not a Lie subalgebra of $L$, so you cannot expect the words "gauge equivalence" to be interpreted via honest Lie group symmetries. However, it's not too bad, since you have the desired properties hold up to homotopy. Aug 31, 2012 at 7:19

This is explained in Section 4.5.2 of "deformation quantization of poisson manifolds" by Kontsevich (http://arxiv.org/abs/q-alg/9709040).

The way you wrote the homotopy between two Maurer-Cartan elements is not enough : as it is explained in the above reference you also need a 1-parameter family of infinitesimal gauge equivalences.

A quick reformulation of Kontsevich definition is the following. An equivalence between two Maurer-Cartan elements $a$ and $b$ in $\mathfrak g$ is a Maurer-Cartan element $c$ in $DR([0,1])\otimes\mathfrak g$ such that $a=c(0)$ and $b=c(0)$.

Note that $DR(...)$ stands for the de Rham algebra of "...".