1
$\begingroup$

Suppose that $G$ is a reductive group that acts algebraically on an affine variety $X$ over an algebraically closed field $k$.

Suppose also that $G$ is equipped with a normal abelian subgroup $N$ such that the quotient group $G/N$ is finite abelian.

Example: $G = \text{O}(2,k)$ and $N= \text{SO}(2,k)$, and $k = \Bbb C$ is the complex numbers.

My question: Under what circumstances can I conclude that a categorical/good quotient $X/\!\!/G$ coincides with a categorical/good quotient for $G/N$ acting on $X/\!\!/N$?

That is, under what circumstances is there a birational equivalence between $X/\!\!/G$ and $$ (X/\!\!/N)/\!\!/(G/N) \,\, ? $$ If there is such a result, can someone point me to a place in the literature where it can be found?

$\endgroup$
3
  • 1
    $\begingroup$ I'm confused about your condition. You seem just to be assuming that $G$ is an extension of $T$ by a finite, commutative group, which is either automatically reductive if you don't assume that reductive groups are connected (and then do we know that $X//G$ exists?), or never reductive unless $G$ equals $T$. In either case, what's the point of invoking reductivity? \\ See also Supposed generalization of $X/(G \times H)\simeq (X/G)/H$ for GIT-quotients. $\endgroup$
    – LSpice
    Commented Oct 29, 2023 at 17:09
  • $\begingroup$ I have edited the question. Thanks. $\endgroup$
    – John Klein
    Commented Oct 29, 2023 at 18:20
  • $\begingroup$ Doesn't this follow simply from $(A^N)^{G/N}=A^G$ for invariants? $\endgroup$ Commented Oct 30, 2023 at 9:07

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.