# Is the complement of the open $B$-orbit in a spherical variety cut out by one equation?

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.

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 $X$.