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Where do people essentially use the reductive groups in the theory of GIT? Or how does reductive groups simplify the constructions in GIT?

I found that the property of completely reducible of reductive groups (in character 0) is fascinating (see Mumford's GIT book page 26-27), but I don't know if it is this property that makes the reductive groups so pervasively in GIT, and how does it used in the theory.

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    $\begingroup$ My understanding is this: If $G$ is a reductive group acting on an affine (projective) variety, then the invariants in the coordinate ring (homogeneous coordinate ring) are finitely generated. This is not true for non reductive groups. I will let someone with more knowledge than me fill in the details $\endgroup$
    – Daniel Barter
    Commented Jan 20, 2014 at 16:59
  • $\begingroup$ I see, this is a good point. But is reductive group a suitable notion just for such property? $\endgroup$
    – Li Yutong
    Commented Jan 20, 2014 at 17:03
  • $\begingroup$ Reductive groups are important in their own right also. Their representation categories are completely reducible (in fact this is why the coordinate rings are finitely generated). Also, every affine group variety you probably care about is reductive, in paticular, all of the classical groups $\endgroup$
    – Daniel Barter
    Commented Jan 20, 2014 at 17:05
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    $\begingroup$ My guess is that nobody knows how to get (in general) a quotient variety with respect to the other algebraic groups: the lack of general methods. "Mathematicians prefer what they are able to do." $\endgroup$
    – Sasha Anan'in
    Commented Jan 20, 2014 at 17:13
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    $\begingroup$ As mentioned before reductivity implies the invariant coordinate ring is finitely generated. Frances Kirwan has worked on (and I believe is working on) non-reductive GIT along with her student(s). The focus is on quotients by unipotent groups, by structure theorems for general groups. See for example "Towards non-reductive geometric invariant theory". $\endgroup$
    – Ruadhaí Dervan
    Commented Jan 20, 2014 at 17:50

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Since the comments are already getting long, I'll add this in community-wiki format to clarify a few points. I should emphasize that I'm not at all a specialist in GIT but have dealt with neighborhing problems involving algebraic groups.

First, there are fundamental differences between characteristic 0 (where Mumford mostly worked) and prime characteristic. In the former case, a linear algebraic group splits as a semidirect product of its unipotent radical and a reductive (Levi) subgroup which is unique up to conjugacy. In the latter case, such a splitting may fail. More important here is the fact that "reductive" is equivalent in characteristic 0 to "linearly reductive": all finite dimensional representations as an algebraic group are completely reducible. This fails badly in prime characteristic.

On the other hand, unipotent groups have the nice property that their algebraic actions have only closed orbits. (And their only irreducible representation is the trivial one.)

In terms of classical invariant theory, all reductive groups (even in prime characteristic) turn out to have the good property that the associated rings of polynomial invariants are finitely generated. (This was settled in prime characteristic by Haboush's proof using algebfaic geometry of the Mumford Conjecture, that reductive implies "geometrically reductive". There is an algebraic proof in Jantzen's book Representations of Algebraic Groups.)

Anyway, Mumford's initial slim volume has grown in its third edition to a larger book, with contributions first by Fogarty and then by Kirwan. But the prefaces Mumford wrote show pretty clearly what he had in mind and why he wanted to work especially with reductive groups. While unipotent groups have their own interest, much classical work on moduli problems involves reductive groups. Their actions involve orbits which need not be closed, as already seen in the adjoint representation (where only the semisimple elements live in closed orbits). So Mumford's ideas about stable and semistable points are delicate, but important to sort out.

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  • $\begingroup$ Thank you! So your point is that using reductive group is because of the invariant ring is f.g? Then how far is it from a group with its invariant ring being f.g to a group being a reductive group? $\endgroup$
    – Li Yutong
    Commented Jan 20, 2014 at 21:14
  • $\begingroup$ Some non-reductive groups act on a vector space and thus on its ring of polynomial functions with f.g. ring of invariants. An old example is given by the natural 2-dimensional module for an additive group in char. 0, but there are many more examples. Still, only reductive groups are known to produce f.g. invariant rings for any representation. (Mumford's transition to geometry is then a further big step.) I wrote an exposition of Hilbert's 14th Problem: Amer. Math. Monthly 85 (1978), but there are many sources dealing with this kind of invariant theory. $\endgroup$
    – Jim Humphreys
    Commented Jan 20, 2014 at 22:55
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    $\begingroup$ If $G$ acts on $X$ (affine, say) and $G$ is reductive of char $0$, and $N$ is a maximal unipotent subgroup, then $X//N$ exists (i.e. its ring of invariants is f.g.). It's enough to prove this for $X=G$, then obtain $X//N = (X \times G//N)//G$. So, there's a family of nonreductive yet f.g. quotients. $\endgroup$
    – Allen Knutson
    Commented Jan 21, 2014 at 10:13

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