1
$\begingroup$

We work over the field of complex numbers. Let $X$ be a smooth projective variety such that $\operatorname{Pic}(X)\simeq \mathbb Z$. Suppose that $X$ admits a non trivial $\mathbb C^*$-action, that is a map $$ \beta: \mathbb C^* \times X \to X $$ or equivalently a map $$ \beta: \mathbb C^* \to \operatorname{Aut}(X)=:G. $$ What can we say about the properties of $G$?

Edit. If we suppose that $X$ is Fano, then $\operatorname{Aut}(X)$ is a linear algebraic group.

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ What kind of answer do you expect, apart from $G\supset \mathbb{C}^*$? $\endgroup$
    – abx
    Commented Jul 20, 2022 at 12:51
  • $\begingroup$ @abx: Is the proof of $\mathbb C^* \subset G$ straightforward? Because I cannot see one. Then, once I know that $\mathbb C^* \subset G$, can we say anything else about $G$? For example, under which hypothesis is $G$ reductive? $\endgroup$
    – Bobech
    Commented Jul 20, 2022 at 12:55
  • 1
    $\begingroup$ Question 1: I was assuming that your action is algebraic. If it is, the map $\mathbb{C}^*\rightarrow G^{\rm o}$ is a homomorphism of algebraic groups, its image is a (nontrivial) quotient of $\mathbb{C}^*$, hence isomorphic to $\mathbb{C}^*$ (if you only assume that the action is holomorphic, the image could be an elliptic curve). Question 2: I don't know any reasonable hypothesis which implies $G$ reductive. $\endgroup$
    – abx
    Commented Jul 20, 2022 at 13:54
  • $\begingroup$ Please say any further information you have in your situation. With the information given, the only result that I can think of is the following. If $h^0(X,T_X)$ equals $1$, then the image of $\beta$ has finite index in $\text{Aut}^0(X)$. Thus $\text{Aut}^0(X)$ is reductive. $\endgroup$
    – Jason Starr
    Commented Jul 20, 2022 at 22:30

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .