Given a smooth manifold $M$ and a Lie algebroid $E\rightarrow M$ we can consider the $E$-polydifferential operators $D_E$ and the $E$-polyvectorfields $T_E$ as $$D_E:=\bigoplus_{k=-1}^\infty\mathcal{U}(E)^{\otimes k+1}\hspace{0.3cm}\mbox{and}\hspace{0.3cm}T_E=\bigoplus_{k=-1}^\infty\Gamma\left(M;\bigwedge^{k+1}E\right) $$ where $\mathcal{U}(E)$ denotes the universal enveloping Hopf algebroid of the Lie-Rinehart pair $(\Gamma(E),C^\infty(M))$ and the tensor products are over $C^\infty(M)$ on the left hand side. One now finds the usual DGLA structures given by the analogues of the Gerstenhaber bracket (expressed in terms of the coproduct) and the Hochschild differential (expressed as Gerstenhaber bracket with a certain degree 1 element) on the left and the Schouten-Nijenhuis beacket on the right. Calaque Showed in http://link.springer.com/article/10.1007/s00220-005-1350-5 that one can construct a Kontsevich type formality $L_\infty$-quasi-isomorphism $$T_E\longrightarrow D_E$$ by using the Fedosov type techniques developed by Dolgushev in his thesis.

Reccently I had been studying this situation and this made me wonder about the functorial properties of these maps. Namely, consider the functors $$T,D\colon LA(M)\longrightarrow L_\infty[W^{-1}]$$ from the category of Lie algebroids over $M$ to the category of $L_\infty$-algebras localized at the quasi-isomorphisms.

Q: Is there a natural equivalence $T\rightarrow D$?

The work by Calaque shows that certainly $T_E$ is isomorphic to $D_E$ for all $E\rightarrow M$, but, after searching and thinking for a while, I could not really obtain any results about this kind of functoriality of these isomorphisms.

One thing that seems necessary (as Stefan pointed out in the comments) is that we consider the category of pairs $(E\rightarrow M,\nabla_E)$ where $\nabla_E$ is a torsion-free $E$-connection, since such connections are essential in constructing Calaque's quasi-isomorphism. Morphisms in this category would then have to somehow preserve these connections though. This seems to warrant considering only inclusions (considering only isomorphisms doesn't serve any purpose it seems to me).

In case there is some "stupid" reason that this question either makes no sense or is not worth investigating I would also welcome an explanation of this stupid reason, of course.

  • 1
    $\begingroup$ Hi Niek: perhaps a small (stupid?) comment. Would funcotriality perhaps be better behaved if you also take the (lie algebroid) connection into account which you need for the Fedosov part of the construction (once having fixed the fiberwise formality)? $\endgroup$ Mar 31, 2017 at 13:02
  • $\begingroup$ Hi Stefan, this is of course an option (I will make an edit). In fact it seems very probable, since such a connection fixes the quasi-isomorphism (to a high degree). I still didn't quite manage to find the natural iso though. $\endgroup$ Apr 1, 2017 at 13:19

1 Answer 1


The construction is functorial with respect to algebraic morphisms of Lie algebroids (as opposed to geometric ones): see for instance my paper with Van de Bergh https://arxiv.org/pdf/0708.2725.pdf (it encompasses the globalization techniques that are used in the paper you cite).

As Stefan pointed out, the choice of a connection is rather important for the Dolgushev-Fedosov resolutions. In the paper with Van den Bergh we use much bigger resolutions that are independant of such a choice (they are somehow universal). A torsion free connection allows one to linearize the jet bundle. The resolution we construct in the paper with Van den Bergh involves an algebra that is universal among the ones that can linearize the jet bundle (and as such it somehow carries a universal connection). This in particular shows that the homotopy class of the $L_\infty$-morphism that you get from my paper does not depend on the choice of the connection involved in the construction.

Note that the algebraic morphisms from the paper with Van den Bergh are called comorphisms by several people (see e.g. https://arxiv.org/pdf/1210.4443.pdf).

Let me also observe that if you fix the base and only consider morphisms that are the identity on the base, then the categoy of Lie algebroids with comorphisms/algebraic morphisms is just the opposite category to the category of Lie algebroids with morphisms/geometric morphisms.

So, combining all these comments, I would say that the answer to your question is yes.

EDIT JULY 25 2017: 1. the functoriality only works for isomorphisms of Lie algebroids (see comments below), and more generally for étale morphisms (like open embeddings in the tangent Lie algebroid case).
2. there are probably other situations where it works. As suggested in the question, one could expect it to work for inclusions of Lie algebroids.

  • $\begingroup$ This sounds exactly like what I was looking for. It will take a little time to go through the article of course, but I am excited to read it! Thanks for the answer! $\endgroup$ Jul 3, 2017 at 10:41
  • $\begingroup$ I have (finally) managed to go through the article and I was left with one question. Namely in the first paper you mention the functionality in theorem 7.1 holds only for morphisms that induce an isomorphism $T\otimes_R L\simeq M$ , could you say something about what this condition means? It seems like it is quite restrictive. In the case that $T=R$ this forces $L\simeq M$! Does it mean we only have functoriality for pull-backs? $\endgroup$ Jul 19, 2017 at 10:45
  • $\begingroup$ wooops. Sorry. You are absolutely right. The meaning of the condition that is required in Theorem 7.1 is "étaleness" (in differential geometric terms, and if we were to talk about tangent bundles, this would mean local diffeomorphism). This means that our work with Michel only implies functoriality for isomorphisms of Lie algebroids :-( Sorry about that, I overlooked it when I answered. $\endgroup$
    – DamienC
    Jul 25, 2017 at 19:05
  • $\begingroup$ That's too bad, do you think that arguments from that paper may be used to extend the functoriality result to a larger class of ``well-behaved" algebraic morphisms? $\endgroup$ Jul 28, 2017 at 11:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.