Let $\alpha: \mathfrak g_A \to T_{A/k}$ be a Lie algebroid over a $k$-algebra $A$. Numerous facts about and its universal enveloping algebra comes from the theory of ring differential operators on $A$. A generalization of theory of D-modules has been used to characterize modules over $\mathfrak g_A$ (eg. Sophi Chemla paper on Inverse image functor for Lie algebroids). But I couldn't find is a good reference for the notion itself. What is a $\mathfrak g_A$-module?
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2$\begingroup$ I don't know this stuff, so I will leave only a comment. I think that there is some disagreement over the correct notion of module. In particular, my friend Alfonso Garcia-Saz math.toronto.edu/alfonso has some papers where they look towards a more general notion of module. The test is whether the Lie algebroid is a module over itself; in some older definitions, it is not. $\endgroup$– Theo Johnson-FreydCommented May 14, 2010 at 20:09
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$\begingroup$ Yes, the adjoint action will not give a a module structure over the Lie algebroid itself because it is not $A$-linear. Thank you for the paper-- $\endgroup$– leminCommented May 14, 2010 at 23:31
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2 Answers
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A $\mathfrak{g}_A$-module $M$ is a $k$-module endowed with structures of $A$-module and $\mathfrak{g}_A$-module satisfying the compatibility equations $(ax)m = a(xm)$ and $x(am) = x(a)m + a(xm)$ for any $a\in A$, $x\in\mathfrak{g}_A$, and $m\in M$. Here $x(a)$ denotes the action of $\mathfrak{g}_A$ in $A$, while the three other actions are denoted by $ax$, $am$, and $xm$.
A $\mathfrak{g}_A$-module is the same that a module over the enveloping algebra $U_A(\mathfrak{g}_A)$.
answered May 14, 2010 at 12:24
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Since Leonid already gave the definition, let me give a reference: Beilinson and Bernstein's A Proof of the Jantzen Conjectures.
answered May 14, 2010 at 13:57