# Does the semi-stable set determine the linearization of a GIT quotient?

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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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! –  IMeasy Sep 18 '11 at 17:49
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. –  Allen Knutson Sep 24 '11 at 1:35

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.