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?

  • 1
    $\begingroup$ The reference you are looking for for the Generality of Mumford's GIT-quotient is perhaps Converse 1.13 in the GIT-book of Mumford, Fogarty and Kirwan (for $X$ smooth connected over $\mathbb{k}$ and $G$ reductive). $\endgroup$
    – JoS
    Nov 10, 2016 at 8:40
  • $\begingroup$ Jos, thank you! To be precise, it is Converse 1.13 of chapter 1.4. $\endgroup$
    – evgeny
    Nov 12, 2016 at 19:26

1 Answer 1


By my understanding, your question is not "Why is Mumford's construction better than the affine quotient". As you note, Proj is better than Spec of invariants for taking quotients by $\mathbb{G}_m$. Instead, your question is "Why can't we get an even better quotient by going further, involving more characters somehow?"

I would say an answer is that any construction involving more characters would not be much different from Mumford's construction, as long as $\chi$ lies in the interior of the cone of all one-dimensional characters of $G$ appearing in $k[X]$.

Take a character $\chi$ and consider another character $\chi'$. Suppose there is a function $f$ which transforms according to $\chi'$. Should we incorporate this function into our quotient? No, as it is not an invariant function. On the other hand if we have two functions $f_1,f_2$ that transform according to $\chi'$, we might hope that $f_1/f_2$ descends to a rational function on our quotient.

However, as long as there is a natural number $n$ and function $g$ such that $g$ transforms under the character $\chi^n/\chi'$, then the rational function $f_1/f_2$ actually lives on the GIT quotient with character $\chi$, because it can be represented as $(gf_1)/(gf_2)$.

So at least birationally Mumford's GIT gives you all the information you could hope to get from the extra characters. For more precise information, one normally studies the spaces you get from varying the character $\chi$ ("variation of GIT") rather than trying to construct a single space incorporating all the information. But this is not very surprising as, gesturing very vaguely, a family of spaces can carry more information than a single space.

  • $\begingroup$ Will, thank you, your ideas are quite clarifying! Could you give me a reference about "...as long as $\chi$ lies in the interior of the cone..."? $\endgroup$
    – evgeny
    Nov 2, 2016 at 9:37
  • $\begingroup$ @evgeny, re, the characters $X(G)$ of $G$ are the same as the characters of its Abelianisation. This latter is (commutative, connected, reductive and so) a torus if $G$ is connected, so $X(G)$ is a lattice (free $\mathbb Z$-module). We may therefore consider the real vector space $X(G) \otimes_{\mathbb Z} \mathbb R$ and, in particular, may speak of cones in it in the usual Euclidean sense. I suspect the answer meant to refer to the cone generated by the weights (or else used some integer-valued generalisation of ‘cone’, probably equivalent). $\endgroup$
    – LSpice
    Jan 27, 2023 at 1:03

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