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If I understand correctly, in geometric invariant theory, polystable points can be defined as those which have a closed orbit. Is it true that semistable points can be characterized as those whose orbit closure contains a unique polystable orbit?

Edit: It would seem that polystable points are actually those whose orbit are closed in the semistable locus. So my pseudo-definition of semistability is circular, unless an independent definition of polystability is given. Also, I suppose that since the unstable (non semistable) locus is closed, there is no way the closure of an unstable orbit could intersect a polystable (or semistable) orbit. I guess that answers my question. But if someone has more insights, please feel free to write them.

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  • $\begingroup$ I suspect will be clearly layed out for the reader in Richard Thomas's notes on GIT: arxiv.org/pdf/math/0512411v3.pdf. Apologies if this doesn't help $\endgroup$ Commented Mar 7, 2015 at 2:40
  • $\begingroup$ @DavidSteinberg No need to apologize. I came across that paper but had not read it carefully enough. Please see edited question if you're interested. $\endgroup$
    – seub
    Commented Mar 7, 2015 at 4:23

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I believe it's the case that if you have $X \subseteq \mathbb P^n$ with an action of $G$ on $X$ and $\mathbb P^n$ you can understand the difference between stable, semistable, and unstable using the affine cone on $X$ in $\mathbb A^{n+1}$.

Semistable ones are the points such that, for any nonzero point on the corresponding line in the affine clone, its orbit closure does not include $0$. Stable is, moreover, that the orbit is closed in $\mathbb A^{n+1}$, and the stabilizer is finite.

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If I may answer my own question:

  1. Polystable points can be characterized as points that are semistable and with closed orbit in the semistable locus.
  2. Yes, If the orbit closure of a point contains a polystable orbit (necessarily unique), then that point is semistable.

However 2. is pretty much saying nothing. In particular unstable orbits can be closed.

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