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Suppose an algebraic group $G$ transitively acts on a variety $X$. Then it is well known that $G$-equivariant vector bundles on $X$ are in correspondence with representations of the stabilizer of a point. Namely, let $x \in X$, and $H = G_x$. Then for a $G$-equivariant vector bundle $V$ on $X$, $V_x$ is naturally a $H$-module. In the opposite direction, given $V_x$, one gets a $G$-equivariant vector bundle $G \times^{H} V_x$ on $G/H = X$.

I wonder if there is a similar description for the case of transitive Lie algebra action (instead of Lie group).

Suppose a Lie algebra $\mathfrak{g}$ transitively acts on a smooth variety $X$. This means that the map $a: \mathfrak{g} \rightarrow Vect (X)$ is given, such that the map $a_x: \mathfrak{g} \rightarrow T_x X$ is surjective for any $x \in X$. Can one describe, what are the $\mathfrak g$-equivariant vector bundles on $X$ (equivalently, representations of the Lie algebroid $\mathfrak g \times X \rightarrow X$)? The role of the stabilizer of a point is played by the Lie algebra $\mathfrak h = \ker a_x$, and given a $\mathfrak g$-equivariant vector bundle $V$, $V_x$ is indeed a representation of $\mathfrak h$. However, I am not sure that this gives a bijection in this case.

Any partial results with restrictions on $X$ or $\mathfrak g$ would be appreciated. One may assume that everything is defined over an algebraically closed field of characteristic zero.

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  • $\begingroup$ I guess "subjective" means "surjective". $\endgroup$
    – Sasha
    Commented May 2, 2023 at 18:19
  • $\begingroup$ @Sasha thank you, fixed it. $\endgroup$ Commented May 2, 2023 at 18:21

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