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Suppose I have a morphism $f:X\to Y$ which is a GIT quotient of $X$ with respect to some reductive, linear group. Does the semistable $X^{ss}$ and stable locus $X^s\subset X$ determine completely the linearization (maybe up to taking a power of the linearization itself)?

Or, in better words, can two different linearizations $L$ and $L'$ of different GIT quotients of $X$ give the same semi-stable and stable locus?

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  • $\begingroup$ I think I have found an example that shows that the answer is YES. Two differente linearizations may give the same (semi-)stable locuses. I would be glad if someone could prove me wrong, though! $\endgroup$
    – IMeasy
    Commented Sep 18, 2011 at 17:49
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    $\begingroup$ Let $X = ({\mathbb P}^1)^2$, $Y = pt$, $G = T^2$. Let ${\mathcal O}(a) \boxtimes {\mathcal O}(b)$ carry the natural action. Then for all $a,b>0$, the stable locus is the open $T^2$-orbit. $\endgroup$
    – Allen Knutson
    Commented Sep 24, 2011 at 1:35

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The paper you want to have a look at is Thaddeus, GIT and Flips (JSTOR), or Dolgachev–Hu, Variation of GIT quotients (Numdam).

There are finitely many polyhedral chambers within the space of possible linearizations. In the interior of each chamber, the sets $X^s$ and $X^{ss}$ are constant. Crossing a wall, these sets change; in nice situations, the two GIT quotient $X^{ss}/G$ on each side of the wall are related by a flip.

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