Let $X$ be an algebraic variety over a field $\mathbb{K}$ equipped with a right action of a smooth algebraic group $G$. One can form the quotient stack $[X/G]$. My question is probably quite elementary for algebraic geometers: what is the tangent complex of such a quotient stack over a given $\mathbb{K}$-point $Spec(\mathbb{K})\rightarrow [X/G]$ ? Is there a good reference where such an example is well explained ?
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3$\begingroup$ I think these notes answer your question. $\endgroup$– abxCommented Jun 4, 2014 at 16:40
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1$\begingroup$ @abx If you could apply your expertise to make your comment more precise and self-contained, then that would be a welcome answer (i.e., in an answer box). $\endgroup$– Todd TrimbleCommented Jun 15, 2014 at 3:19
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1 Answer
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First of all remember that differentiating the action of $G$ at the identity gives you a Lie algebra morphism $\mathfrak{g}\to\Gamma(T_X)$, and thus, for any point $x\in X$, a map $\mathfrak g\to T_xX$.
Now pick a lift $x:Spec(\mathbb{K})\to X$ of your point $[x]:Spec(\mathbb{K})\to [X/G]$.
Claim: $\mathbb{T}_{[x]}[X/G]$ is (quasi-isomorphic to) the two term complex $(\mathfrak{g}\to T_xX)$.
Note that when the action is formally locally free at $x$ then $\mathfrak g\to T_xX$ is injective and thus $\mathbb{T}_{[x]}[X/G]=T_xX/\mathfrak g$ is the naive tangent space to the "local orbit space" at $x$.
More generally, the tangent complex of $[X/G]$ is the follwing two term complex of $G$-equivariant $\mathcal O_X$-modules: $(\mathfrak g \otimes \mathcal O_X\to T_X)$.
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$\begingroup$ Is $\frak g$ the Lie algebra of $G$? So $G$ finite would mean that the tangent complex of $[X/G]$ is $\cal O_X\to T_X$, correct? $\endgroup$ Commented Jan 26, 2016 at 21:29
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$\begingroup$ @Brian: no, if $G$ is finite then $\mathfrak{g}$ vanishes. In any case there isn't a natural map from $\mathcal{O}_X$ to $\mathcal{T}_X$. $\endgroup$ Commented Feb 11, 2016 at 4:03
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$\begingroup$ @QiaochuYuan So $G$ finite implies that the tangent complex of $[X/G]$ is the same as the tangent sheaf of $X$? $\endgroup$ Commented Feb 11, 2016 at 4:05
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$\begingroup$ @Brian: it's the tangent sheaf of $X$ regarded as a $G$-equivariant $\mathcal{O}_X$-module, as DamienC says. $\endgroup$ Commented Feb 11, 2016 at 4:54
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$\begingroup$ @QiaochuYuan So this would imply that $X$ smooth affine implies that $[X/G]$ has no non-trivial deformations? $\endgroup$ Commented Mar 16, 2016 at 7:15