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Let $G$ be a connected reductive algebraic group and $X$ be a $G$-variety. Let $Y$ be a $G$-invariant irreducible sub variety of $X$ which has non-trivial intersection with the semi stable locus $X^{ss}$ (for a fixed line bundle $L$ on $X$). Let $Z$ be an irreducible codimension one sub variety of $Y$ which is not $G$-invariant and such that $Z$ lies strictly in $X\setminus X^{ss}$.

Question: What can be said about the $G$-sweep $G\cdot Z$, of $Z$. Under what conditions is it equal to $Y$?

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  • $\begingroup$ Any examples? It seems to me that your scenario is impossible. $\endgroup$ Commented May 14, 2017 at 20:12

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First, if $Z$ is contained in the unstable locus $X^{us}=X\setminus X^{ss}$ then the ``$G$-sweep'' $G\cdot Z$ is still in $X^{us}$ since the latter is $G$-invariant. In particular, it can't be all of $Y$.

Secondly, if $Z$ is of codimension one in $Y$ then it is an irreducible component of $Y\cap X^{us}$. The latter being $G$-stable forces $Z$ to be $G$ stable, as well. So $G\cdot Z=Z$.

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