Let $S$ be a scheme.

Definition. Let $X$ be an $S$-scheme and $G$ a smooth affine group $S$-scheme acting on $X.$ An $S$-scheme $Y$ is a geometric quotient of $X$ by $G$ if there exists a morphism $\pi_X\colon X\rightarrow Y$ such that

  1. $\pi_X$ is $G$-invariant,

  2. the geometric fibers of $\pi_X$ are orbits,

  3. $\pi_X$ is universally submersive, i.e., $U\subset Y$ is open iff $\pi_X^{-1}(U)\subset X$ is open, and this property is preserved by base change,

  4. $(\pi_X)_*(\mathcal{O}_X)^G=\mathcal{O}_Y.$

Let $X$ and $X'$ be $S$-schemes with a $G$-action, where $G$ is the same introduced before. Assume that there exist geometric quotients $\pi_X\colon X\rightarrow Y$ and $\pi_{X'}\colon X'\rightarrow Y'.$

Question. Let $g\colon Y\rightarrow Y'$ be a morphism. Is there a $G$-equivariant morphism $f\colon X\rightarrow X'$ such that $\pi_{X'}\circ f = g\circ \pi_{X}$?


It is not true : for instance, $X$ and $X'$ might be two non-isomorphic $G$-torsors on the same $S$-scheme $T$.

Here is a precise example. Consider $S=T=Spec(\mathbb{R})$ and $G=\mathbb{Z}/2\mathbb{Z}$, let $X=Spec(\mathbb{C})$ and $X'=Spec(\mathbb{R})\cup Spec(\mathbb{R})$ viewed as $S$-schemes, and let $G$ act on $X$ and $X'$ respectively by complex conjugation and by exchanging the connected components. Both actions admit $Spec(\mathbb{R})$ as a geometric quotient. But an isomorphism between these quotients obviously doesn't lift to an equivariant morphism between $X$ and $X'$.

Here is another example : take $S=Spec(\mathbb{C})$, $G=\mathbb{G}_m$, $T=\mathbb{P}^1_{\mathbb{C}}$. Let $X$ (resp. $X'$) be the total space of the line bundle $\mathcal{O}$ (resp. $\mathcal{O}(1)$) on $T$ minus the zero-section, with the natural $G$-action. Both actions admit $T$ as a geometric quotient, but there is no equivariant morphism between $X$ and $X'$ lifting identity.

  • $\begingroup$ Thanks for your answer. Is it not possible to impose some constraints on $S$ or $X, X'$ to obtain a positive answer? For example, what does it happen if $S=Spec(k)$ with $k$ an algebraically closed field of characteristic zero? $\endgroup$ – Francesco Sep 6 '12 at 15:16
  • $\begingroup$ My edit should answer your question (negatively). If $G$ acts with trivial stabilizers, this question really is a question about torsors, and you could obtain results in concrete situations using the classification of torsors by $H^1$. $\endgroup$ – Olivier Benoist Sep 6 '12 at 15:30

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