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Background

Suppose we are given $L_\infty$-algebras $(g,Q)$ and $(g',Q')$ and an $L_\infty$-morphism $F$ from $(g,Q)$ to $(g',Q')$. Furthermore, we have a Maurer-Cartan element $\pi$ of $(g,Q)$. One can twist $(g,Q)$ with the Maurer-Cartan element $\pi$ and obtains a new $L_\infty$-algebra that we call $(g,Q_\pi)$. Furthermore, we can construct a Maurer-Cartan element $\pi'$ of $(g',Q')$ by the formula

$\pi' = \sum_{n=1}^\infty \frac{1}{n!} F_n(\pi, \ldots , \pi)$,

where $F_n$ is the $\bigwedge^n g \rightarrow g'$-part of $F$. I don't know whether there is a (better) term, so I call $\pi$ and $\pi'$ associated Maurer-Cartan elements

One can twist the morphism $F$ with the Maurer-Cartan elements $\pi$ and $\pi'$ and obtain an $L_\infty$-morphism $F_\pi$ from $(g,Q_\pi)$ to $(g',Q'_{\pi'})$. The references I found are Dolgushev: A Proof of Tsygan's Formality Conjecture for an Arbitrary Smooth Manifold (section 2.4) and Yekutieli: Continuous and Twisted L_infinity Morphisms (section 3).

Question

Given $L_\infty$-algebras $(g,Q)$ and $(g',Q')$, an $L_\infty$-morphism $F$ from $(g,Q)$ to $(g',Q')$, a Maurer-Cartan elements $\pi$ of $(g,Q)$ and a Maurer-Cartan element $\omega$, $\omega\neq \pi'$, of $(g',Q')$. Can one construct an $L_\infty$-morphisms between $(g,Q)$ twisted with the Maurer-Cartan element $\pi$ and $(g',Q')$ twisted with the Maurer-Cartan element $\omega$, where the Maurer-Cartan elements are not "associated"? I.e. can one construct an $L_\infty$-morphism between $(g,Q_\pi)$ to $(g',Q'_\omega)$?

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  • $\begingroup$ If I understand your question correctly, my guess is that the answer is yes. Maurer-Cartan elements are to be thought of as deformations of the "underlying formal manifold". $Q$ is a "vector field" on this formal manifold. "Twisting" on either side should just correspond to deforming the vector field... But I'm not sure. Again this is my guess. $\endgroup$ Apr 26, 2010 at 14:33
  • $\begingroup$ Unfortunately, I know too little about the formal manifold language of L-infinity algebras, so I did not get any intuition from there. Do you or anyone know a good introductory reference for this point of view? I only know the Kontsevich letter. (Maybe I should make this a separate question?) $\endgroup$
    – C. Jost
    Apr 26, 2010 at 16:35

1 Answer 1

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Here's an idea for simplifying the problem a bit. I don't know the technical details or whether it will work, but it wouldn't fit in the comment box, so I'll put it as an answer, even though it's incomplete.

First, answer this simplified question: If $\alpha$ is an MC-element of $(h,P)$, is there a morphism $(h,P_\alpha) \to (h,P)$ and a morphism $(h,P)\to (h,P_\alpha)$? (This corresponds to $\pi$ or $\omega$ being zero and $g=g'$, $Q=Q'$ in the original question.) Then compose the morphisms

$(g,Q_\pi) \to (g,Q) \to (g',Q') \to (g',Q'_\omega) $,

where the first arrow comes from $h=g$, $P=Q_\pi$, and $\alpha = -\pi$, the second is the given morphism, and the third comes from $h=g'$, $P = Q'$, and $\alpha = \omega$.

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  • $\begingroup$ Hm, it has not helped me yet. But thanks anyway, Peter! $\endgroup$
    – C. Jost
    Apr 27, 2010 at 8:43

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