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I posted this on math.se a week ago, currently it has 23 views and no other feedback.

Here on MO there are several questions about orbit closures but I could not find anything about what I need.

To be more specific, let $G\subseteq GL(V)$ be a reductive linear algebraic group over an algebraically closed field of characteristic zero. For $v\in V$, can the (Zariski) closure of $Gv$ in $V$ contain infinitely many $G$-orbits? If no, why? If yes, what would be a "simplest" example?

In case this matters, still more specifically I am interested in the case of nilpotent orbits, i. e. when that closure contains the zero vector.

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    $\begingroup$ Yes, it can contain infinitely many orbits. Let $G$ be $\textbf{GL}_3$ with its action on $\mathbb{P}^2$ and $\mathcal{O}_{\mathbb{P}^2}(1)$. There is an induced action on $V=H^0(\mathbb{P}^2,\mathcal{O}_{\mathbb{P}^2}(4))$, the vector space of homogeneous polynomials of degree $4$. Consider the orbit of a product of $4$ linear forms, no three of which are linearly dependent (the closure contains the origin). The closure of this orbit contains the locus parameterizing products of $4$ linear forms in a pencil of linear forms. Consider the cross-ratio of these $4$ members of the pencil. $\endgroup$
    – Jason Starr
    Commented Apr 19, 2022 at 13:26
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    $\begingroup$ For an even easier example, consider the postcomposition action of $\textbf{GL}_2$ on the vector space $W=\text{Mat}_{2\times 2}$ of $2\times 2$ matrices. There is an open orbit, namely $\textbf{GL}_2$. However, for the induced action on the locus of rank $1$ matrices, the kernel of the rank $1$ matrix gives an invariant of the action that is an element in the projective line. Using this example, you can induce examples for all noncommutative reductive groups (obviously this cannot happen for multiplicative group schemes). $\endgroup$
    – Jason Starr
    Commented Apr 19, 2022 at 14:20
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    $\begingroup$ To add some context: an action of a reductive group is called visible if each orbit closure has only finitely many orbits. For a linear representation this is equivalent to there being finitely many nilpotent orbits. Visible representations are relatively rare in some sense. There are some tables in sciencedirect.com/science/article/pii/0021869380901416 . $\endgroup$
    – Sam Gunningham
    Commented Apr 19, 2022 at 14:37
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    $\begingroup$ @SamGunningham wait what I wrote is certainly not true - take the standard multiplicative group action on a vector space. It has infinitely many orbits (lines through the origin with the origin removed, and the one point orbit at the origin) but each orbit closure contains only itself and the origin. Every orbit is nilpotent for sure. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Apr 21, 2022 at 6:16
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    $\begingroup$ @მამუკა ჯიბლაძე: sorry, what I wrote was probably not quite correct. An affine G-variety is visible if the fibers of the affine GIT quotient contain finitely many orbits. In other words, there are finitely many orbits in each S-equivalence class. For a representation, it is enough to check the fiber over the 0 orbit, i.e. the nilcone. As you note, visibility is not implied by the statement that the closure of any orbit contains only finitely many orbits. Sorry for any confusion! $\endgroup$
    – Sam Gunningham
    Commented Apr 21, 2022 at 21:02

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