Geometric invariant theory doesn't work so well for non-reductive groups, since invariant rings are not generally finitely generated. However, in many cases the action of a non-reductive group has a finitely generated invariant ring (e.g. when the additive group acts on a polynomial ring).

How do I think of quotient varieties in this case?

I have a specific example in mind:

Let $G=\mathbb G_a$ act on the polynomial ring $k[x_0,\ldots,x_r,y_0,\ldots,y_r]$ by $$tx_i=x_i+ty_i, \text{ and } ty_i=y_i.$$

How should one think of a quotient $\mathbb P^{2r+1}/G$?

Does it make sense to consider $\mathrm{Proj}$ of the invariant ring (i.e. the subring generated by $y_0,\ldots,y_r$ and the minors of the $2\times(r+1)$-matrix of variables)?

If so, what does the resulting projective variety look like?

  • $\begingroup$ Note that the theorem (Weitzenb\”{o}ck’a theorem) which says that the invariant ring of a linear representation of $ \mathbb{G}_{a} $ is only known to hold in characteristic zero. In positive characteristic it is an open problem. $\endgroup$ – schemer Oct 3 '19 at 6:09

In general, the only definition I know of GIT quotient is $Proj$ of the invariant ring. The obvious statements one can make about the rational map $Proj\ R\to Proj\ R^G$ are that it collapses $G$-orbits, and if one semistable orbit is in the closure of several others, they all collapse together.

Your example is of a very special type, where the action of $G$ can be extended to that of a reductive group $H \geq G$, so instead of worrying directly about the nonreductive GIT quotient $X//G$ we can instead look at $(X \times H//G)//H$ and punt the nonreductivity issue to the more-universal problem of computing $H//G$. In the case at hand, $H = SL_2$, and $H//G \cong \mathbb A^2$. (Note that the map $H \to H//G$ is not onto!)

Instead of taking $X = \mathbb P^{2r+1}$, let me keep scaling in abeyance to quotient by later, and take $X = \mathbb A^{2r+2}$. Then $(X \times H//G)//H$ is the affine cone over $Gr(2,r+1)$. What's a little tricky, then, is that this remaining $\mathbb G_m$ action is not acting on said cone by dilation, because it doesn't act on the factor of $\mathbb A^2$ we just attached. Instead we get one of the "weighted Grassmannians" of Corti and Reid. I don't have a more detailed answer than that, but at least this replaces your original $G$-quotient question by a $\mathbb G_m$-quotient question.

  • $\begingroup$ You mean $(X \times H//G)//H$, not $(X \times H//G)//G$, right? $\endgroup$ – David E Speyer Aug 10 '14 at 15:22
  • $\begingroup$ Fixed. $\ \ \ \ \ \ \ \ $ $\endgroup$ – Allen Knutson Aug 10 '14 at 23:06

The following response is not really an answer, as the second (and probably most pertinent) part of your question (the specific example) has already been addressed.

However, with respect to your general question:

How do I think of quotient varieties in this case?

I would recommend:

Towards non-reductive geometric invariant theory by Doran and Kirwan.

They work out many good examples, and rather precisely address the case when the invariant ring is finitely generated.


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