The role of Reynolds operator in GIT has always been a little mystery to me. Actually I see that in some proofs it gets used in an efficient way, but what I cannot grasp is the general philosophy. I see of course it is a projector onte the invariant subspace of a G-representation. But, as I asked, when do I want to use the R.O. and what for? I guess that one of the most important features is the Reynolds identity...
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$\begingroup$ As far as I remember, you need both $(1)$ Hilbert's basis Theorem and $(2)$ Reynolds Identity in order to show that the invariant subspace $\mathcal{A}^G$ is a finitely generated $\mathbb{C}$-algebra. $\endgroup$– Francesco PolizziNov 8, 2013 at 11:00
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1$\begingroup$ @IMeasy: You would probably get some insight from earlier questions raised on MO. Search for 'Reynolds operator'. $\endgroup$– Jim HumphreysNov 8, 2013 at 13:45
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$\begingroup$ I searched but could not find very much $\endgroup$– IMeasyNov 8, 2013 at 14:58
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$\begingroup$ Did you find for instance: mathoverflow.net/questions/139153/… $\endgroup$– Abdelmalek AbdesselamNov 8, 2013 at 16:47
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$\begingroup$ Of course, I found it. But it does not seem to answer my question. $\endgroup$– IMeasyNov 8, 2013 at 17:49
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It is used to prove that a uniform categorical quotient by the action of a reductive group on an affine scheme exists and that the quotient inherits basic properties of the original scheme such as affine, algebraic, noetherian. In particular in that proof it is used to prove that the ring of invariants of a quotient by an invariant ideal is just the image of the original ring of invariants. See Thm 1.1 of Mumford's book.
It is also an important indicator of the singularities of the quotient, although this aspect is not strictly "in GIT". See this and this papers for more.
answered Nov 13, 2013 at 16:40