1

It's well known that any affine algebraic group $G$ over $\mathbb{C}$ can be viewed as a closed subgroup of $\mathbb{G}\mathbb{L}(n,\mathbb{C})$.

But suppose $G$ is an affine algebraic group acting on $\mathbb{C}\mathbb{P}^2$. Does this imply that the action of $G$ on $\mathbb{C}\mathbb{P}^2$ is equivalent to the action of a closed subgroup of $\mathbb{P}\mathbb{G}\mathbb{L}(3,\mathbb{C})$ on $\mathbb{C}\mathbb{P}^2$?

MR_Q
  • 158
  • The action is in what category? – Mohan Dec 30 '17 at 23:31
  • $G$ acts algebraically, the action of $G$ on $\mathbb{C}\mathbb{P}^2$ can be expressed as a morphism

    $$ G\times \mathbb{C}\mathbb{P}^2\rightarrow \mathbb{C}\mathbb{P}^2 $$

     – MR_Q Dec 31 '17 at 00:30
  • 1
    I don't know the answer, but what we want is something like: $$$$ Let us work over $\Bbb C$ and let $G$ be a linear algebraic group. Then, $G$ is isomorphic as an algebraic group, to a closed subgroup of $\text{GL}_n(\Bbb C)$. Now, let $G$ act morphically on $\Bbb P^2$, so that this induces an action of $\text{GL}_n$ on $\Bbb P^2$, where you want this to be $\text{GL}_3$ I suppose. I think without the standard action of $\text{GL}_3$ on $\Bbb C^3$, we won't induce an action of $\text{PGL}_3$ on $\Bbb P^2$? – Horizon Dec 31 '17 at 01:31
  • That's one thing I'm not sure about. Does $G$ acting morphically on $\mathbb{C}\mathbb{P}^2$ induce a well-defined action of $G$ on $\mathbb{C}^3$? – MR_Q Dec 31 '17 at 04:43

1 Answers1

1

The answer is yes. The action gives a morphism of algebraic groups $G\to \mathrm{Aut}(\mathbb{P}^2)=\mathbb{PGL}(3,\mathbb{C})$. The image is closed and the action is via this closed subgroup.

Mohan
  • 17,980
  • Thanks! For anyone reading this question later, a reference for Aut($\mathbb{P}^2)=\mathbb{P}\mathbb{G}\mathbb{L}(3,\mathbb{C})$ is: https://books.google.com/books?id=3rtX9t-nnvwC&lpg=PA151&ots=XN892EOHsa&dq=Hartshorne%20automorphisms&pg=PA151#v=onepage&q=Hartshorne%20automorphisms&f=false – MR_Q Jan 01 '18 at 17:55