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According to the remark 6.14 in Shigeru Mukai's An introduction to invariants and moduli (unfortunately, the page is not available on Google Books, so I explain it below), the GIT-quotient of an affine variety $X$ by a reductive group $G$ with respect to a nontrivial character $\chi: G \to \mathbb G_m$ may be considered as

• first taking an affine quotient $\operatorname{Spec} (\mathbb k[X]^{G_\chi})$ by $G_\chi:=\operatorname{Ker}(\chi: G \to \mathbb G_m)$,
• then taking a quotient by $\mathbb G_m=G / G_\chi$ via replacing $\operatorname{Spec}$ to $\operatorname{Proj}$ of non-negative gradings.

As a result, I wonder why it is so much better then just taking an affine quotient $\operatorname{Spec}(\mathbb k[X]^G)$ if $G$ has much more characters like $\mathrm{GL}_{k_1} \times \ldots \times \mathrm{GL}_{k_r}$. It seems that replacing $\operatorname{Spec}$ to $\operatorname{Proj}$ of non-negative gradings is a correct way to take a quotient by $\mathbb G_m$, but the first step looks fairly stupid.

Update: Yes, as Will Sawin said, the question is approximately why one does not iterate it (why there is no need/why nothing better happens)?

Details of Mukai's description. For simplicity, work over an algebraically closed field $\mathbb k$ of characteristic $0$. Let $X$ be an affine space or any other affine irreducible algebraic variety with $\operatorname{Pic} X=0$, and let $G$ be a connected reductive group acting on $X$. Then a $G$-linearized line bundle $\mathcal L$ is equivalent to a character $\chi: G \to \mathbb G_m$, so Mumford's GIT quotient is $$X //_{\mathcal L} G:=\operatorname{Proj} \left( \bigoplus_{n \geqslant 0} \Gamma(X, \mathcal L^{\otimes n}) \right) = \operatorname{Proj} \left( \bigoplus_{n \geqslant 0} \mathbb k[X]^G_{\chi^n} \right).$$

Here $$\mathbb k[X]^G_{\chi^n}=\{ f \in \mathbb k[X]: f(gx)=\chi(g)^nf(x) \}$$ are $\chi^n$-semi-invariants. If $G_\chi:=\operatorname{Ker}(\chi: G \to \mathbb G_m)$, then $\mathbb k[X]^{G_\chi}$ has an action of $\mathbb G_m=G / G_\chi$, so is graded -- exactly by $(\mathbb k[X]^{G_\chi})_n=\mathbb k[X]^G_{\chi^n}$. From this follows Mukai's description of $X //_{\mathcal L} G$.

Generality of Mumford's GIT-quotient. By the way, I believe that I have seen somewhere that any geometric quotient arises as an open subset in Mumford's GIT-quotient for some line bundle or something like this. Could you give me a reference?

According to remark 6.14 in Shigeru Mukai's An introduction to invariants and moduli (unfortunately, the page is not available on Google Books, so I explain it below), the GIT-quotient of an affine variety $X$ by a reductive group $G$ with respect to a nontrivial character $\chi: G \to \mathbb G_m$ may be considered as

• first taking an affine quotient $\operatorname{Spec} (\mathbb k[X]^{G_\chi})$ by $G_\chi:=\operatorname{Ker}(\chi: G \to \mathbb G_m)$,
• then taking a quotient by $\mathbb G_m=G / G_\chi$ via replacing $\operatorname{Spec}$ to $\operatorname{Proj}$ of non-negative gradings.

As a result, I wonder why it is so much better then just taking an affine quotient $\operatorname{Spec}(\mathbb k[X]^G)$ if $G$ has much more characters like $\mathrm{GL}_{k_1} \times \ldots \times \mathrm{GL}_{k_r}$. It seems that replacing $\operatorname{Spec}$ to $\operatorname{Proj}$ of non-negative gradings is a correct way to take a quotient by $\mathbb G_m$, but the first step looks fairly stupid.

Update: Yes, as Will Sawin said, the question is approximately why one does not iterate it (why there is no need/why nothing better happens)?

Details of Mukai's description. For simplicity, work over an algebraically closed field $\mathbb k$ of characteristic $0$. Let $X$ be an affine space or any other affine irreducible algebraic variety with $\operatorname{Pic} X=0$, and let $G$ be a connected reductive group acting on $X$. Then a $G$-linearized line bundle $\mathcal L$ is equivalent to a character $\chi: G \to \mathbb G_m$, so Mumford's GIT quotient is $$X //_{\mathcal L} G:=\operatorname{Proj} \left( \bigoplus_{n \geqslant 0} \Gamma(X, \mathcal L^{\otimes n}) \right) = \operatorname{Proj} \left( \bigoplus_{n \geqslant 0} \mathbb k[X]^G_{\chi^n} \right).$$

Here $$\mathbb k[X]^G_{\chi^n}=\{ f \in \mathbb k[X]: f(gx)=\chi(g)^nf(x) \}$$ are $\chi^n$-semi-invariants. If $G_\chi:=\operatorname{Ker}(\chi: G \to \mathbb G_m)$, then $\mathbb k[X]^{G_\chi}$ has an action of $\mathbb G_m=G / G_\chi$, so is graded -- exactly by $(\mathbb k[X]^{G_\chi})_n=\mathbb k[X]^G_{\chi^n}$. From this follows Mukai's description of $X //_{\mathcal L} G$.

Generality of Mumford's GIT-quotient. By the way, I believe that I have seen somewhere that any geometric quotient arises as an open subset in Mumford's GIT-quotient for some line bundle or something like this. Could you give me a reference?

Problem. According to the remark 6.14 in Shigeru Mukai's An introduction to invariants and moduli (unfortunately, the page is not available on Google Books, so I explain it below), the GIT-quotient of an affine variety $X$ by a reductive group $G$ with respect to a nontrivial character $\chi: G \to \mathbb G_m$ may be considered as

• first taking an affine quotient $\operatorname{Spec} (\mathbb k[X]^{G_\chi})$ by $G_\chi:=\operatorname{Ker}(\chi: G \to \mathbb G_m)$,
• then taking a quotient by $\mathbb G_m=G / G_\chi$ via replacing $\operatorname{Spec}$ to $\operatorname{Proj}$ of non-negative gradings.

As a result, I wonder why it is so much better then just taking an affine quotient $\operatorname{Spec}(\mathbb k[X]^G)$ if $G$ has much more characters like $\mathrm{GL}_{k_1} \times \ldots \times \mathrm{GL}_{k_r}$. It seems that replacing $\operatorname{Spec}$ to $\operatorname{Proj}$ of non-negative gradings is a correct way to take a quotient by $\mathbb G_m$, but the first step looks fairly stupid.

Details of Mukai's description. For simplicity, work over an algebraically closed field $\mathbb k$ of characteristic $0$. Let $X$ be an affine space or any other affine irreducible algebraic variety with $\operatorname{Pic} X=0$, and let $G$ be a connected reductive group acting on $X$. Then a $G$-linearized line bundle $\mathcal L$ is equivalent to a character $\chi: G \to \mathbb G_m$, so Mumford's GIT quotient is $$X //_{\mathcal L} G:=\operatorname{Proj} \left( \bigoplus_{n \geqslant 0} \Gamma(X, \mathcal L^{\otimes n}) \right) = \operatorname{Proj} \left( \bigoplus_{n \geqslant 0} \mathbb k[X]^G_{\chi^n} \right).$$

Here $$\mathbb k[X]^G_{\chi^n}=\{ f \in \mathbb k[X]: f(gx)=\chi(g)^nf(x) \}$$ are $\chi^n$-semi-invariants. If $G_\chi:=\operatorname{Ker}(\chi: G \to \mathbb G_m)$, then $\mathbb k[X]^{G_\chi}$ has an action of $\mathbb G_m=G / G_\chi$, so is graded -- exactly by $(\mathbb k[X]^{G_\chi})_n=\mathbb k[X]^G_{\chi^n}$. From this follows Mukai's description of $X //_{\mathcal L} G$.

Generality of Mumford's GIT-quotient. By the way, I believe that I have seen somewhere that any geometric quotient arises as an open subset in Mumford's GIT-quotient for some line bundle or something like this. Could you give me a reference?

According to the remark 6.14 in Shigeru Mukai's An introduction to invariants and moduli (unfortunately, the page is not available on Google Books, so I explain it below), the GIT-quotient of an affine variety $X$ by a reductive group $G$ with respect to a nontrivial character $\chi: G \to \mathbb G_m$ may be considered as

• first taking an affine quotient $\operatorname{Spec} (\mathbb k[X]^{G_\chi})$ by $G_\chi:=\operatorname{Ker}(\chi: G \to \mathbb G_m)$,
• then taking a quotient by $\mathbb G_m=G / G_\chi$ via replacing $\operatorname{Spec}$ to $\operatorname{Proj}$ of non-negative gradings.

As a result, I wonder why it is so much better then just taking an affine quotient $\operatorname{Spec}(\mathbb k[X]^G)$ if $G$ has much more characters like $\mathrm{GL}_{k_1} \times \ldots \times \mathrm{GL}_{k_r}$. It seems that replacing $\operatorname{Spec}$ to $\operatorname{Proj}$ of non-negative gradings is a correct way to take a quotient by $\mathbb G_m$, but the first step looks fairly stupid.

Update: Yes, as Will Sawin said, the question is approximately why one does not iterate it (why there is no need/why nothing better happens)?

Details of Mukai's description. For simplicity, work over an algebraically closed field $\mathbb k$ of characteristic $0$. Let $X$ be an affine space or any other affine irreducible algebraic variety with $\operatorname{Pic} X=0$, and let $G$ be a connected reductive group acting on $X$. Then a $G$-linearized line bundle $\mathcal L$ is equivalent to a character $\chi: G \to \mathbb G_m$, so Mumford's GIT quotient is $$X //_{\mathcal L} G:=\operatorname{Proj} \left( \bigoplus_{n \geqslant 0} \Gamma(X, \mathcal L^{\otimes n}) \right) = \operatorname{Proj} \left( \bigoplus_{n \geqslant 0} \mathbb k[X]^G_{\chi^n} \right).$$

Here $$\mathbb k[X]^G_{\chi^n}=\{ f \in \mathbb k[X]: f(gx)=\chi(g)^nf(x) \}$$ are $\chi^n$-semi-invariants. If $G_\chi:=\operatorname{Ker}(\chi: G \to \mathbb G_m)$, then $\mathbb k[X]^{G_\chi}$ has an action of $\mathbb G_m=G / G_\chi$, so is graded -- exactly by $(\mathbb k[X]^{G_\chi})_n=\mathbb k[X]^G_{\chi^n}$. From this follows Mukai's description of $X //_{\mathcal L} G$.

Generality of Mumford's GIT-quotient. By the way, I believe that I have seen somewhere that any geometric quotient arises as an open subset in Mumford's GIT-quotient for some line bundle or something like this. Could you give me a reference?

Problem. According to the remark 6.14 in Shigeru Mukai's An introduction to invariants and moduli (unfortunately, the page is not available on Google Books, so I explain it below), the GIT-quotient of an affine variety $X$ by a reductive group $G$ with respect to a nontrivial character $\chi: G \to \mathbb G_m$ may be considered as

• first taking an affine quotient $\operatorname{Spec} (\mathbb k[X]^{G_\chi})$ by $G_\chi:=\operatorname{Ker}(\chi: G \to \mathbb G_m)$,
• then taking a quotient by $\mathbb G_m=G / G_\chi$ via replacing $\operatorname{Spec}$ to $\operatorname{Proj}$ of non-negative gradings.

As a result, I wonder why it is so much better then just taking an affine quotient $\operatorname{Spec}(\mathbb k[X]^G)$ if $G$ has much more characters like $\mathrm{GL}_{k_1} \times \ldots \times \mathrm{GL}_{k_r}$. It seems that replacing $\operatorname{Spec}$ to $\operatorname{Proj}$ of non-negative gradings is a correct way to take a quotient by $\mathbb G_m$, but the first step looks fairly stupid.

Details of Mukai's description. For simplicity, work over an algebraically closed field $\mathbb k$ of characteristic $0$. Let $X$ be an affine space or any other affine irreducible algebraic variety with $\operatorname{Pic} X=0$, and let $G$ be a connected reductive group acting on $X$. Then a $G$-linearized line bundle $\mathcal L$ is equivalent to a character $\chi: G \to \mathbb G_m$, so Mumford's GIT quotient is $$V //_{\mathcal L} G:=\operatorname{Proj} \left( \bigoplus_{n \geqslant 0} \Gamma(X, \mathcal L^{\otimes n}) \right) = \operatorname{Proj} \left( \bigoplus_{n \geqslant 0} \mathbb k[X]^G_{\chi^n} \right).$$

Here $$\mathbb k[X]^G_{\chi^n}=\{ f \in \mathbb k[X]: f(gx)=\chi(g)^nf(x) \}$$ are $\chi^n$-semi-invariants. If $G_\chi:=\operatorname{Ker}(\chi: G \to \mathbb G_m)$, then $\mathbb k[X]^{G_\chi}$ has an action of $\mathbb G_m=G / G_\chi$, so is graded -- exactly by $(\mathbb k[X]^{G_\chi})_n=\mathbb k[X]^G_{\chi^n}$. From this follows Mukai's description of $X //_{\mathcal L} G$.

Generality of Mumford's GIT-quotient. By the way, I believe that I have seen somewhere that any geometric quotient arises as an open subset in Mumford's GIT-quotient for some line bundle or something like this. Could you give me a reference?

Problem. According to the remark 6.14 in Shigeru Mukai's An introduction to invariants and moduli (unfortunately, the page is not available on Google Books, so I explain it below), the GIT-quotient of an affine variety $X$ by a reductive group $G$ with respect to a nontrivial character $\chi: G \to \mathbb G_m$ may be considered as

• first taking an affine quotient $\operatorname{Spec} (\mathbb k[X]^{G_\chi})$ by $G_\chi:=\operatorname{Ker}(\chi: G \to \mathbb G_m)$,
• then taking a quotient by $\mathbb G_m=G / G_\chi$ via replacing $\operatorname{Spec}$ to $\operatorname{Proj}$ of non-negative gradings.

As a result, I wonder why it is so much better then just taking an affine quotient $\operatorname{Spec}(\mathbb k[X]^G)$ if $G$ has much more characters like $\mathrm{GL}_{k_1} \times \ldots \times \mathrm{GL}_{k_r}$. It seems that replacing $\operatorname{Spec}$ to $\operatorname{Proj}$ of non-negative gradings is a correct way to take a quotient by $\mathbb G_m$, but the first step looks fairly stupid.

Details of Mukai's description. For simplicity, work over an algebraically closed field $\mathbb k$ of characteristic $0$. Let $X$ be an affine space or any other affine irreducible algebraic variety with $\operatorname{Pic} X=0$, and let $G$ be a connected reductive group acting on $X$. Then a $G$-linearized line bundle $\mathcal L$ is equivalent to a character $\chi: G \to \mathbb G_m$, so Mumford's GIT quotient is $$X //_{\mathcal L} G:=\operatorname{Proj} \left( \bigoplus_{n \geqslant 0} \Gamma(X, \mathcal L^{\otimes n}) \right) = \operatorname{Proj} \left( \bigoplus_{n \geqslant 0} \mathbb k[X]^G_{\chi^n} \right).$$

Here $$\mathbb k[X]^G_{\chi^n}=\{ f \in \mathbb k[X]: f(gx)=\chi(g)^nf(x) \}$$ are $\chi^n$-semi-invariants. If $G_\chi:=\operatorname{Ker}(\chi: G \to \mathbb G_m)$, then $\mathbb k[X]^{G_\chi}$ has an action of $\mathbb G_m=G / G_\chi$, so is graded -- exactly by $(\mathbb k[X]^{G_\chi})_n=\mathbb k[X]^G_{\chi^n}$. From this follows Mukai's description of $X //_{\mathcal L} G$.

Generality of Mumford's GIT-quotient. By the way, I believe that I have seen somewhere that any geometric quotient arises as an open subset in Mumford's GIT-quotient for some line bundle or something like this. Could you give me a reference?