# Is the class (resp. Picard) group of a $G$-variety generated by invariant divisors?

Let's work over the complex numbers. Let $S$ be a normal surface, $\mathrm{A}^1(S)$ the class group of divisors on $S$ and $\mathrm{Pic}(S)$ its Picard group. Let $G$ be a reductive group acting on $S$.

Q1. When is $\mathrm{A}^1(S)$ generated by $G$-invariant divisors?

Q2. Assume $S$ smooth. When is $\mathrm{Pic}(S)$ generated by (line bundles associated to) $G$-invariant divisors?

Q3. What if $G$ is the multiplicative group $\mathbb{C}^*$?

Remark: I'm looking for answers in which $S$ is not toric with torus $G$, because I consider that case already known.

• What do you consider to be a $G$-invariant divisor? If $G$ acts transitively on $S$, e.g., for the standard $\textbf{PGL}_3$ action on $\mathbb{P}^2$, then the only $G$-stabilized irreducible closed subset of $S$ is $S$ itself. – Jason Starr Jul 21 '17 at 16:34

The obstruction to $\mathrm{Pic}(S)$ being generated by invariant divisors is pretty much the Picard group of the generic $G$-orbit which, in turn, is controlled by the character group of the generic isotropy group. So for $S=\mathbf{P}^2$ acted on transitively by $G=GL(2)$ or $G=SL(2)$ the Picard group is certainly not generated by invariant divisors since there aren't any.

On the positive side let me tackle Q3 which might be a blueprint for for general cases. Let $G$ be a torus (not necessarily of dimension $1$) acting effectively on a normal variety $S$ (not necessarily of dimension $2$). Then I claim that $A^1(S)$ is generated by invariant divisors. To see this, observe that $S$ contains an invariant open subset $S_0$ of the form $Y\times T$ where $Y=S_0/T$ (use Rosenlicht for the existence of $Y$ and Hilbert Satz 90 for the triviality of the bundle $S_0\to Y$). Then $A^1(S_0)=A^1(Y)$ is generated by invariant divisors of the form $D\times T$. Thus $A^1(S)$ is genrated by their closures plus the irreducible components of $S\setminus S_0$ which are of codimension $1$ in $S$. The latter are $G$-invariant, as well.