I wonder whether it is true that the composition of two GIT-quotients is another GIT-quotient. It should be an analogue of a set-theoretic formula $X/(G \times H)\simeq (X/G)/H$ but with GIT-quotients instead.

Background. Mumford's theorem (GIT, thm. 1.10). Let $G$ be a reductive group acting on an algebraic variety $X$ and (linearly) on a vector space $V$, and let $\phi: X \to \mathbb PV$ be a $G$-morphism. Then there is a non-empty open subset $$X_{ss,\phi}=\{ x \in X : \text{for any $v\in V$ representing $\phi(x)\in\mathbb PV$}$$ $$\text{the closure $\overline{Gv}$ of the $G$-orbit of $v$ does not contain $0\in V$ } \}$$ such that there exists a geometric quotient categorical quotient $p: X_{ss,\phi} \to X_{ss,\phi}/G$. It is called (semistable) GIT-quotient.

Problem. Let $G \times H$ be a reductive group acting on an algebraic variety $X$ and (linearly) on a vector space $V$, and let $\phi: X \to \mathbb PV$ be a $G$-morphism. By Mumford's theorem, there is the GIT-quotient $p_1: X_{ss, \phi} \to X_{ss, \phi} / G$ and it satisfies a universal of categorical quotient, $H$ also acts on $Y:=X_{ss, \phi} / G$.

Now let $H$ act (linearly) on a vector space $W$ and $\psi: Y \to \mathbb PW$ be an $H$-morphism, so that there is a GIT-quotient $p_2: Y_{ss,\psi} \to Y_{ss,\psi} / H$. So let us denote $Z:=Y_{ss,\psi} / H$.

Now suppose that not only $H$ but $G \times H$ acts (linearly) on $W$. Let $\chi: X \to \mathbb PW$ be a $G \times H$-morphism such that the following diagram is commutative: $$\begin{array}{ccc} X & \xrightarrow{\chi} & \mathbb PW \\ \uparrow i && \uparrow \psi \\ X_{ss,\phi} & \xrightarrow{p} &Y \end{array}$$

So there is a GIT-quotient $X_{ss, \chi} \to X_{ss, \chi} / (G \times H)$. I wonder whether $$X_{ss, \chi}=X_{ss, \phi} \cap p_1^{-1}(Y_{ss, \psi})$$ and $$X_{ss, \chi} / (G \times H) \simeq Z.$$

Any suggestions: references, counterexamples or corrections are welcome!


1 Answer 1


My question should have consisted of two. I will formulated both using linearization via line bundles, not embeddings into projective space, as it makes them more clear.

Question 1. Suppose that $\pi: X \to Y$ is a quotient by $G$, that $H$ acts on $X$ commuting with $G$ (so that $H$ also acts on $Y$) and that $\mathcal L$ is $H$-line bundle on $Y$. Is it true that $$X //_{\pi^* \mathcal L} (G \times H) = Y //_{\mathcal L} H?$$

Answer 1. First, $\pi^* \mathcal L$ is $G \times H$-line bundle. Second, if the action of $G$ is free, holds $$\bigoplus_{n \geqslant 0} H^0(X, \pi^*\mathcal L^{\otimes n})^{G \times H}=\bigoplus_{n \geqslant 0} H^0(Y, \mathcal L^{\otimes n})^H,$$ and taking $Proj$'s, one proves the fact. If the action is not free, there should be a counterexample, I believe.

Question 2. Suppose that $\mathcal L_1$ and $\mathcal L_2$ are $G$- and $G \times H$-line bundles on $M$. Is it true that $$(M^{ss}(\mathcal L_1)) //_{\mathcal L_2} (G \times H) = M //_{\mathcal L_2} (G \times H)?$$

(To get the situation of the question 1, let $X=M^{ss}(\mathcal L_1)$ and $\mathcal L_2=\pi^*\mathcal L$.)

Answer 2. Of course not, why should it be, even if $H=0$? For example, consider Atiyah flop: let $$M=\mathrm{Spec}\; \mathbb C[x_i, y_j] \quad (i=1\ldots k,\;j=1\ldots l)$$ with $G=\mathbb C^*$ acting via $t(x_i, y_j)=(tx_i, t^{-1}y_j)$. Choose $\mathcal L_1=\mathcal O(t)$ and $\mathcal L_2=\mathcal O(t^{-1})$. Then $M^{ss}(\mathcal L_1)=\{ \overline x \neq 0 \}$ and $M^{ss}(\mathcal L_2)=\{ \overline y \neq 0 \}$ and the quotients are geometric, therefore $$(M^{ss}(\mathcal L_1)) //_{\mathcal L_2} (G \times H) \subsetneq M //_{\mathcal L_2} (G \times H).$$ For the details on this beautiful construction, see Miles Reid's What is a flip?


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