Timeline for Why people usually consider reductive groups in GIT?

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Jan 21, 2014 at 10:13	comment	added	Allen Knutson		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.
Jan 20, 2014 at 23:28	vote	accept	Li Yutong
Jan 20, 2014 at 22:55	comment	added	Jim Humphreys		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.
Jan 20, 2014 at 21:14	comment	added	Li Yutong		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?
S Jan 20, 2014 at 18:40	history	answered	Jim Humphreys	CC BY-SA 3.0
S Jan 20, 2014 at 18:40	history	made wiki			Post Made Community Wiki by Jim Humphreys