Let $G$ be a reductive algebraic group over an algebraically closed field $K$ of characteristic zero. Assume that $G$ acts on an affine variety $X$. Assume that $X$ contains an open orbit $U$ (so $\bar{U}=X$) which is also affine.

Question 1: Is it necessarily true that there exists an $f\in K[X]$ such that $U=X_f$?

More precisely, I would like to get an answer to the following more specific question:

Question 2: Is it necessarily true that there exists $f\in K[X]^G$ such that $U=X_f$? Or at least that $G\cdot f$ is one dimensional over $K$?

Of course, $X\backslash U$ is closed, and since $G$ is reductive, it is the zero set of some $G$-stable ideal. But this is still not enough to answer that question. In general, the affine open subsets in an affine scheme may not be just the principal open subsets (see for example Ring-theoretic characterization of open affines?). Since the situation here is much more particular I hope that there might also be a more particular answer.

  • $\begingroup$ If $G=SL(2)$ (or $GL(2)$) acting on $\mathbb{A}^2$ in the natural way, you can take $U=\mathbb{A}^2-\{(0,0)\}$ and then there is no such $f$. $\endgroup$
    – Mohan
    Jul 23, 2016 at 12:31
  • 3
    $\begingroup$ @Mohan: this $U$ is not affine. $\endgroup$ Jul 23, 2016 at 12:52

1 Answer 1


As you suspected, the answer is negative: Let $G=SL(2,\mathbb C)$ and $U$ the orbit of the $3$-form $c:=x_1^2x_2$ in $S^3\mathbb C^2$ and let $X$ be its closure. Clearly $X$ is affine. One checks that the stabilizer of $c$ is trivial, so $U\cong G$ is affine, as well. One also checks that $0\in X$, so $U\subsetneq X$. But $U\ne X_f$ for any $f$ since $U$ ($\cong G$) does not carry non-constant invertible functions.


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