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Let $G$ be a reductive group over an algebraically closed field $k$ of characteristic $p>0$. If $V$ is a rational $G$-module then we can define the Hilbert nullcone $\mathcal{N}(V)$ to be the zero locus of all positive degree homogeneous invariant polynomials. By Haboush's theorem you can regard this as the trivial fibre of the quotient map $V \to V// G$. Let $S$ be the set of all points $v\in V$ such that the $G$-orbit of $v$ is $k^\times$-stable, where the scalar action is coming from the vector space structure.

Since the $k^\times$-action permutes the non-trivial fibres of $V \to V// G$ one can show that $S \subseteq \mathcal{N}(V)$. If $G$ acts with finitely many orbits on $\mathcal{N}(V)$ then you can show that $S = \mathcal{N}(V)$, by repeating the argument of Lemma 2.10 from Jantzen's book ``Nilpotent Orbits in Representation Theory''. In general however this argument doesn't work.

I have two questions:

1) does $S = \mathcal N(V)$ in general?

2) if not then is $S$ always a closed subvariety of $V$?

Thanks in advance for answers!

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Edit: I'm especially interested in the representation $\operatorname{Hom}(V \otimes V, V) \cong (V \otimes V)^* \otimes V$.

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    $\begingroup$ Consider the case where $G$ is the multiplicative group acting on a $2$-dimensional vector space by $t\cdot (x,y) = (t^ax,t^by)$ for $a,b$ distinct positive integers. Then $\mathcal{N}$ is the entire vector space, yet $S$ is the union of the two coordinate axes. $\endgroup$ Nov 19, 2020 at 16:33
  • $\begingroup$ Thanks Jason, that's a negative answer to q1! I should have considered torus actions. The most important part for my intended application is q2 $\endgroup$ Nov 20, 2020 at 7:44

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