4
$\begingroup$

Let $X$ be an affine spherical variety for some reductive algebraic group $G$. Let $X^0$ be the open orbit in $X$ under a fixed Borel subgroup $B \subseteq G$. Does there exists a function $f$ on $X$ such that $X^0$ is the same as $\{ \, x \in X \ | \ f(x) \neq 0 \, \}$?

I'm interested in any proof, counter-example or reference.

$\endgroup$
8
$\begingroup$

The answer is yes, even if $X$ is not normal. It even works for any connected solvable group acting on an affine variety with an open orbit.

To see this let $Y_1,\ldots,Y_r$ be the irreducible components of $X\setminus X^0$. Since the connected solvable group $B$ acts rationally on the ideal $\mathcal I(Y_i)\subset\mathcal O(X)$ it contains a non-zero $B$-semiinvariant $f_i$. Put $f:=\prod_if_i$ which is also a non-zero semiinvariant. Then $f$ vanishes on $X\setminus X^0$ by construction. On the other hand, if $f(x)=0$ with $x\in X^0$ then $f(X^0)=f(Bx)=\chi_f(B)f(x)=0$ which is impossible since $f\ne0$ and $X^0$ is dense in $X$. Hence $X^0$ is precisely the non-vanishing set of $X$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.