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.


2 Answers 2


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.).

  • $\begingroup$ 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? $\endgroup$ Aug 31, 2012 at 4:24
  • $\begingroup$ 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. $\endgroup$ 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 "...".


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