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I do not know much about Geometric Invariant Theory. My question is the following:

Let $X$ and $Y$ be two complex affine or projective varieties. Let $G$ be a reductive group which acts on both $X$ and $Y$. Let $\pi_X: X \to X//G$ and $\pi_Y: Y \to Y//G$ be the "projection" maps to the G.I.T quotients. If $Z\subset X$ and $W \subset Y$ are closed $G$-invariant subsets and $f: Z \to W$ is a bijection, under which conditions $f$ descends to a bijection $\bar{f}: \pi_X(Z) \to \pi_Y(W)$ ?

flag	asked Jan 29 at 23:23 christian 157●3

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In the projective setting you also need line bundles to define GIT quotients, so you also need to have the extra assumption that $f$ pulls back one to the other. – Misha Jan 30 at 13:17

IMeasy · Answer 1 · 2013-01-31 07:48:14Z UTC

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as long as $f$ passes to the quotient (i.e. sends orbits on orbits) it has the required property. Moreover, as Misha remarked, the semistable locus of the first quotient should be sent to the semistable locus of the second quotient. This basically mean that the pullback of the invariants on the codomain must give the invariants of the domain.

edited Jan 31 at 7:48

answered Jan 30 at 12:48

IMeasy
1,355●4●12

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On the set-theoretic level, GIT quotients are not defined as orbit spaces, but "extended" orbit spaces, where two semistable orbits are identified whenever their closures (in the set of semistable points) have nonempty intersection. Thus, in addition to orbit preservation, you also need the requirement that semistable points go to semistable points as well as, say, continuity of $f$. (In all natural examples, of course, $f$ will be a regular map.) – Misha Jan 30 at 16:41

correct. I had forgotten to add that point. – IMeasy Jan 30 at 17:42

When does a G-invariant one to one map between two closed algebraic G-set descend to a one to one map on the G.I.T quotient ?

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