# Taking quotient of a variety by the additive group

1. Let $$X$$ be a smooth irreducible $$\Bbb C$$-variety, on which the algebraic $$\Bbb C$$-group $$G={\bf G}_{a,{\Bbb C}}$$ (the additive group) acts freely on the right: $$X\times _{\Bbb C} G\to X,\quad (x,g)\mapsto x\cdot g.$$ Assume that there exists a surjective morphism onto a smooth $$\Bbb C$$-variety $$Y$$ $$\begin{equation}\label{e:*} \varphi\colon X\to Y \tag{*} \end{equation}$$ whose fibres are the orbits of $$G$$ in $$X$$. Then the morphism $$\varphi$$ is smooth, from which one can deduce that $$\varphi$$ induces a locally trivial fibre bundle (in the usual topology) of $$C^\infty$$-manifolds $$\varphi\colon X(\Bbb C)\to Y(\Bbb C).$$

Question 1. Does it follow that $$(*)$$ is locally trivial in the flat topology, that is, a $$Y$$-torsor under $$G$$? In other words, is the morphism $$X\times_Y G\to X\times_Y X,\quad (x,g)\mapsto (x,x\cdot g)$$ an isomorphism of $$\Bbb C$$-varieties?

2. Assume that $$(*)$$ is a torsor. Since $$H^1({\Bbb C}(Y),G)=\{1\}$$, we know that $$(*)$$ admits a rational section.

Question 2. Does $$(*)$$ admit a regular section? In other words, does there exist a regular map (morphism) $$s\colon Y\to X$$ such that $$\varphi\circ s={\rm id}_Y\,$$?

For 2, no. Any variety $$Y$$ with $$H^1 (Y, \mathcal O_Y) \neq 0$$ is a counterexample, as every class gives a torsor $$X$$.
• @MikhailBorovoi of $X$. For example, take $y\in Y$, $x \in X$ with $\varphi(x)=y$, and choose any function on (a neighborhood of $x$ in) $X$ vanishing at $x$ and whose first derivative is nonzero on the relative tangent space of $X$ over $Y$. The vanishing locus of this function is still smooth over $Y$ at the point $x$, thus smooth over $Y$ in a neighborhood of $x$. Apr 29, 2021 at 17:28
• Does it follow that a morphism $\varphi$ as in my question induces a locally trivial fibre bundle of complex analytic manifolds (with the usual topology)? May 1, 2021 at 18:05