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let $X$ be an algebraic variety over a field $k$. Let $G$ be an algebraic group over $k$ and $P$, $Q$ $G$-torsors over $X$. Assume I have a morphism $\phi:P\to Q$ of $G$ torsors. Let $A\to k$ be a ring extension with square zero kernel $I$, $X_A$ a variety over $Spec(A)$ such that $X_A\otimes k\cong X$ and assume that $P$, $Q$ lifts to $X_A$ as $G_A$ torsors. What are the obstructions to lift $\phi $ to a morphism of $G_A$ torsors $\phi_A: P_A\to Q_A$ ?


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Are $G_A$ and $X_A$ flat over $A$? – Jason Starr Jul 5 '12 at 17:17
@Starr yes. we can also assume everything smooth – okko Jul 6 '12 at 7:46

1 Answer 1

up vote 6 down vote accepted

Assuming that $G_A$ is smooth over $A$ and that $X_A$ is flat over $A$, the "naive" obstruction to lifting $\phi$ is an element in $H^1(X,\mathfrak{g}_P)$, where $\mathfrak{g}_P$ is the locally free $\mathcal{O}_X$-module obtained from the trivial bundle $\mathfrak{g}\otimes_k \mathcal{O}_X$ by "twisting" via the adjoint action of $G$ on $\mathfrak{g}$ with respect to $P$ (or $Q$, since they are isomorphic). I will say more precisely what this means in a moment.

The simplest way to think of this is to consider the graph $\Gamma_\phi$ as a closed subscheme of $P\times_X Q$. Consider the diagonal action of $G$ on $P\times_X Q$, $g\cdot(p,q) = (gp,gq)$. Then $\Gamma_\phi$ is $G$-equivariant. Equivalently, $\Gamma_\phi$ is the pullback of a closed subscheme of the quotient $P\times^G Q := (P\times_X Q)/\Delta(G)$, which is itself a smooth scheme over $X$, i.e., $\rho:P\times^G Q \to X$ is smooth.

In fact, $\Gamma_\phi$ is the pullback of the image of a section $\widetilde{\phi}:X\to P\times^G Q$. This establishes a one-to-one correspondence between $G$-equivariant morphisms $\phi:P\to Q$ and sections $\widetilde{\phi}$ of $P\times^G Q$. So now your original problem, deforming $\phi$ to $\phi_A$, is equivalent to the problem of deforming the section $\widetilde{\phi}$ to a section $\widetilde{\phi}_A$ of $P_A \times^{G_A} Q_A$. Since $P\times^G Q$ is smooth over $X$, the obstruction to such a deformation is an element in $H^1(X,\widetilde{\phi}^* T_\rho)$. And $\widetilde{\phi}^*T_\rho$ equals the locally free sheaf $\mathfrak{g}_P$ from above. The fastest way to see this is to check that they give isomorphic descent data with respect to the fppf cover $P\to X$.

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Don't you mean $H^1(X,\mathfrak{g}_P\otimes I)$ instead of $H^1(X,\mathfrak{g}_P)$ ? – Niels May 27 at 17:42
@Niels: Yes, I meant $H^1(X,\mathfrak{g}_P\otimes I)$. – Jason Starr May 27 at 20:22

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