# Are there geometric $\mathbb{G}_a$-quotients with trivial stabilizers, not being principal bundles?

Consider algebraic $$\mathbb{C}$$-schemes. The group scheme $$\mathbb{G}_a$$ is the scheme $$\mathbb{A}^1$$ with the addition. This is not a reductive group. Here I want to know some examples of $$\mathbb{G}_a$$-actions with nice quotient schemes.

I mimic notions from GIT to define geometric $$\mathbb{G}_a$$-quotients for $$\mathbb{G}_a\curvearrowright X$$

1. $$X\to Y$$ is affine surjective and $$\mathbb{G}_a$$-invariant;
2. $$\mathcal{O}_Y\cong (\mathcal{O}_X)^{\mathbb{G}_a}$$, where we assume the necessary finite generation;
3. $$\mathbb{G}_a\times X\to X\times X$$ induces a surjective morphism $$\mathbb{G}_a\times X\to X\times_YX$$;
4. the topology on $$Y$$ is the quotient topology given by $$X\to Y$$.

Principal $$\mathbb{G}_a$$-bundles are geometric $$\mathbb{G}_a$$-quotients. It is well know that isomorphism classes of principal $$\mathbb{G}_a$$-bundles over $$X$$ can be identified with elements of $$H^1(X,\mathcal{O}_X)$$, giving plenty of examples. Besides, identity morphisms are geometric quotients for trivial actions.

The question is about the converse. I wonder whether the following is true.

Conjecture. If $$\mathbb{G}_a\curvearrowright X$$ and $$X\to Y$$ is a geometric $$\mathbb{G}_a$$-quotient such that stabilizers of geometric points are trivial, then $$X\to Y$$ is a principal bundle.

Counterexamples are also welcome.

I think that it is possible to give a positive answer to this, in a slightly larger level of generality.

Let me start by giving the hypotheses that we shall work with (the arguments work in an even more general setting, but I think that the following is reasonable for the purposes here).

Context 1 Let $$k$$ be a field, and let $$G$$ be an affine group scheme over $$k$$. Let $$X$$ be a finite type $$k$$-scheme with a $$G$$-action, and suppose that there is a morphism $$\pi: X \to Y$$ satisfying the conditions 1-3 of the post. Furthermore, we assume that the $$G$$-stabilizer of any geometric point of $$X$$ is trivial.

Just for completeness, the conditions 1-3 that we are assuming are:

(1) $$\pi$$ is affine, surjective, and $$G$$-invariant.

(2) We have $$\mathcal{O}_Y = (\pi_*\mathcal{O}_X)^{G}$$.

(3) $$G \times X \to X \times X$$ induces a surjective morphism $$G \times X \to X \times_Y X$$.

The following condition will be useful:

Definition We say that the action of $$G$$ on $$X$$ is proper if the induced morphism $$G \times X \to X \times X$$ is proper.

With this in mind, the following is not too difficult.

Proposition 1 In Context 1, assume that the action is proper. Then, the morphism $$\pi: X \to Y$$ is a principal $$G$$-bundle.

Proof: The condition on the stabilizers means that the quotient stack $$X/G$$ is actually an algebraic space; equivalently $$G \times X \to X\times X$$ is a monomorphism. By definition, the quotient morphism $$X \to X/G$$ is a principal $$G$$-bundle. There is an induced morphism $$f: X/G \to Y$$; it suffices to show that $$f$$ is an isomorphism. This can be checked Zariski locally on $$Y$$, so we may assume that $$Y$$ and $$X$$ are affine for concreteness. The assumption of that the action is proper is equivalent to $$X/G$$ being separated, and so in particular the morphism $$f: X/G \to Y$$ is separated. On the other hand, (3) implies that $$f$$ induces an injection on geometric points, and so $$f$$ is separated and quasi-finite. By https://stacks.math.columbia.edu/tag/03XX, it follows that $$X/G$$ is a scheme. Notice that $$H^0(\mathcal{O}_{X/G}) = H^0(\mathcal{O}_X)^G = H^0(\mathcal{O}_Y)$$, where the last equality is by (2). Therefore $$f_*(\mathcal{O}_{X/G}) = \mathcal{O}_Y$$. By Zariski's main theorem https://stacks.math.columbia.edu/tag/02LR, it follows that $$f$$ is an open immersion. The surjectivity in (1) implies that the open immersion $$X/G \to Y$$ is surjective, and therefore $$X/G \to Y$$ is an isomorphism. QED

Now, I think(?) that the condition of properness of $$G$$ is automatic in Context 1. This seems quite strong, I am including an argument below, which we may have to double-check (as it seemed a bit surprising to me).

Proposition 2 (please double-check) In Context 1, the action is automatically proper. In particular, in Context 1 the morphism $$\pi: X \to Y$$ is always a principal $$G$$-bundle.

Proof: By the triviality of stabilizers we have that $$G \times X \to X \times X$$ is a monomorphism, and it factors through the closed subscheme $$X \times_Y X \subset X \times X$$, thus inducing a monomorphism $$i: G \times X \to X \times_Y X$$. It suffices to show that $$i$$ is a closed immersion, which by item (6) in https://stacks.math.columbia.edu/tag/04XV is equivalent to showing that $$i$$ is universally closed. The morphism $$i: G\times X \to X \times_Y X$$ is $$G$$-equivariant, where $$G$$ acts on the first coordinate of the source by multiplication, and it acts on the second coordinate of the target by the action on $$X$$. These actions are free, and taking quotients we get a monomorphism of algebraic spaces $$\widetilde{i}: X \to X \times_Y (X/G)$$. Note that $$G \times X \to X$$ and $$X \times_Y X \to X \times_Y (X/G)$$ are principal $$G$$-bundles, and by working flat locally on $$X \times_Y (X/G)$$ one can see that the following square is Cartesian: $$\require{AMScd}$$ $$\begin{CD} G \times X @>i>> X \times_Y X\\ @V V V @VV V\\ X @>>\widetilde{i}> X \times_Y (X/G) \end{CD}$$ Therefore it suffices to show that $$\widetilde{i}$$ is universally closed. Notice that by (3) the morphism $$i$$ is a surjective monomorphism, and it follows that $$\widetilde{i}$$ is also a surjective monomorphism, in particular it is universally bijective on points. Furthermore, the morphism $$\widetilde{i}$$ has a section $$p: X \times_Y (X/G) \to X$$ given by the first projection. To check that $$\widetilde{i}$$ is universally closed, choose a morphism $$T \to X \times_Y (X/G)$$ from a scheme $$T$$ and form the base-change $$\widetilde{i}_T: X_T \to T$$; we need to show that $$\widetilde{i}_T$$ is closed. But it is still the case that $$\widetilde{i}_T$$ is a surjective monomorphism (so it is bijective on topological points) and has a section $$p_T: T \to X_T$$. Using this, we see that, for given closed subset $$Z \subset |X_T|$$of the topological space, the image $$\widetilde{i}_T(Z)$$ coincides with the preimage $$(p_T)^{-1}(Z)$$, and so it is closed, as desired. QED

• It seems plausible. As a non-expert, I wonder if you are suspecting the generalization of 04XV to algebraic spaces? Commented Apr 1 at 6:43
• Can you say more about the implication from the triviality of stabilisers of geometric points to monomorphism of $G\times X\to X\times X$? The conclusion is exactly triviality of stabilisers of $T$-points for any $T$. Does it suffice to check only geometric points? Commented Apr 1 at 6:51
• It is true that an analogue of stacks.math.columbia.edu/tag/04XV holds for algebraic spaces, but this is not what is used in the argument above.
– afh
Commented Apr 1 at 11:39
• For the stabilizers, you want to show that inertia (aka the relative stabilizer group scheme over X) is trivial. I think that this can be checked on geometric points by Lrmma 5.2 in arxiv.org/abs/2211.06754
– afh
Commented Apr 1 at 11:41
• @DisplayName sorry, I forgot to ping in the previous comments.
– afh
Commented Apr 1 at 11:52