Suppose $G$ is an affine algebraic group acting linearly on a vector space $V$. A point $v \in V$ is *stable* if the orbit $Gv$ is closed and $v$ is *regular* (the dimension of the stabilizer of $v$ is locally constant, or equivalently, [locally minimum](https://mathoverflow.net/questions/193/when-is-fiber-dimension-upper-semi-continuous/202#202)). I would really like to say this is equivalent to the orbit $Gv$ being closed and *not* being in the closure of another orbit.


Since the orbit of a regular point has locally maximum dimension, it can't be in the closure of another orbit. But is the converse true? If a point is not in the closure of an orbit larger than its own, is it regular?


The answer is no ... we have to throw in some hypotheses. If you consider the action of $\mathbf G\_a$ on $\mathbf A^2$ given by $t(x,y)=(x,tx+y)$, then all the orbits are closed, but points of the form $(0,y)$ are irregular. So let's throw in the hypothesis that $G$ *is linearly reductive*. I feel like we might also want to insist that $v\neq 0$, but I'm not sure about that.


Linearly reductive seems like a strange hypothesis, so feel free to modify it. I was thinking that you could somehow show that if v is not in the closure of a larger-dimensional orbit, then $\operatorname{span}(Gv)$ would be an invariant subspace with no complement, but I haven't been able to get this argument to work.



Are irregular points of an action necessarily in the closure of a larger orbit?
Linearly reductive is a very good hypothesis; it tells you that the closed orbits of your group are the points of an affine variety (they are Spec of invariant polynomials). Maybe you can use some kind of semi-continuity of fiber dimension?


