I wonder whether it is true that *the composition of two [GIT-quotients](https://en.wikipedia.org/wiki/GIT_quotient) 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.




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**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~~](https://en.wikipedia.org/wiki/Geometric_quotient) [categorical quotient](https://en.wikipedia.org/wiki/Categorical_quotient) $p: X\_{ss,\phi} \to X\_{ss,\phi}/G$. It is called (semistable) GIT-quotient.




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**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!



Supposed generalization of $X/(G \times H)\simeq (X/G)/H$ for GIT-quotients
*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?*](http://homepages.warwick.ac.uk/staff/G.Brown/reid.pdf)


