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Invariants of a localization at an invariant element

Statement

Assume AC inherited from the Reynolds-operator suppliers. Let $G$ be a complex reductive affine algebraic group acting rationally on a commutative $C$ -algebra $A$ by algebra automorphisms, with Reynolds operator $R_{A} : A \to A^{G}$ (Complete reducibility and the Reynolds operator for a complex reductive group). Suppose $A = ⨁_{n \in Z} A_{n}$ is graded and $G$ acts by graded algebra automorphisms, with $R_{A}$ preserving degrees. Then for every homogeneous $f \in A^{G}$ one has $(A_{f})^{G} = (A^{G})_{f}$ compatibly with the grading, and consequently $((A_{f})_{0})^{G} = (A^{G})_{(f)}$ , where $(A^{G})_{(f)} = ((A^{G})_{f})_{0}$ is the degree-zero part of the localization.

Facts & Assumptions

Given: A complex reductive affine algebraic group $G$ , a rational $G$ -algebra $A$ that is graded with $G$ acting by graded algebra automorphisms, the Reynolds operator $R_{A}$ , and a homogeneous invariant element $f \in A^{G}$ .

[F1]

Reynolds operator. $R_{A} : A \to A^{G}$ is $G$ -equivariant, restricts to the identity on $A^{G}$ , is $A^{G}$ -linear and idempotent, and its image is exactly $A^{G}$ ; moreover every rational $G$ -module is a direct sum of simple submodules, so every $G$ -stable submodule of a rational $G$ -module has a $G$ -stable complement. (Complete reducibility and the Reynolds operator for a complex reductive group, The Reynolds operator and the ideal theory of the invariant subring)

[F2]

Rational modules. A rational $G$ -module is one in which every vector lies in a finite-dimensional $G$ -stable subspace on which $G$ acts by a morphism; a $G$ -stable subspace of a rational $G$ -module is again rational, and a direct sum of rational modules is rational. (Classical complex affine algebraic actions and rational modules)

[F3]

Graded conventions. In a graded ring, multiplication by a homogeneous element shifts degrees, so the kernel of multiplication by $f^{k}$ on a graded module is a graded submodule; the localization $A_{f}$ of a graded ring at a homogeneous element carries the induced $Z$ -grading, and the action of $G$ by graded automorphisms on $A$ extends to $A_{f}$ because $f$ is invariant. (Nonnegatively graded rings and modules, homogeneous elements, and twists)

[F4]

AC. The Axiom of Choice is inherited from the Reynolds-operator and complete-reducibility suppliers and is used only through them. (The Axiom of Choice)

Proof

technique · direct

1.1F1F3algebra

The localized Reynolds operator. Define $\tilde{R} : A_{f} \to A_{f}$ by $\tilde{R} (a / f^{n}) = R_{A} (a) / f^{n}$ . This is well defined: if $a / f^{n} = b / f^{m}$ , then $f^{k} (f^{m} a - f^{n} b) = 0$ in $A$ for some $k \geq 0$ , and applying the $A^{G}$ -linear operator $R_{A}$ to this relation (with the invariant elements $f^{k + m}$ , $f^{k + n}$ pulled out) gives $f^{k + m} R_{A} (a) = f^{k + n} R_{A} (b)$ , so $R_{A} (a) / f^{n} = R_{A} (b) / f^{m}$ in $A_{f}$ . The map $\tilde{R}$ is $(A^{G})_{f}$ -linear: for $h \in A^{G}$ and $r \geq 0$ , one has $h / f^{r} \in (A^{G})_{f}$ and $\tilde{R} ((h / f^{r}) (a / f^{n})) = R_{A} (h a) / f^{r + n} = (h / f^{r}) \tilde{R} (a / f^{n})$ .

1.2F1F2F3F4

The converse inclusion. Let $K = {b \in A : f^{k} b = 0 for some k \geq 0}$ be the kernel of the localization map $A \to A_{f}$ ; it is a $G$ -stable submodule because $f$ is invariant and $G$ acts by automorphisms, and it is a rational $G$ -module as a submodule of the rational module $A$ by [F2]. By complete reducibility there is a $G$ -stable complement $C$ with $A = K \oplus C$ ; in particular $K \cap C = 0$ . The complement is supplied by the complete-reducibility theorem, so this step inherits the Axiom of Choice and makes no new selection [F4].

2.1F1step 1.1

Image and fixed points. $\tilde{R}$ is idempotent and has image exactly $(A^{G})_{f}$ : the image is contained in $(A^{G})_{f}$ because $R_{A} (a) \in A^{G}$ , and an element $h / f^{n}$ with $h \in A^{G}$ is fixed by $\tilde{R}$ . Consequently $(A^{G})_{f} \subseteq (A_{f})^{G}$ , since $(A^{G})_{f}$ consists of invariant fractions.

3.1F1step 1.2algebra

Let $x \in (A_{f})^{G}$ and write $x = a / f^{n}$ with $a = k + c$ , $k \in K$ , $c \in C$ ; then $x = c / f^{n}$ . For every $g \in G$ the element $g c - c$ lies in $C$ , while the equality $g c / f^{n} = c / f^{n}$ in $A_{f}$ says precisely that $f^{m} (g c - c) = 0$ for some $m$ , i.e. $g c - c \in K$ . Hence $g c - c \in K \cap C = 0$ , so $g c = c$ for all $g$ and $c \in A^{G}$ ; thus $x = c / f^{n} \in (A^{G})_{f}$ . With step 2.1 this gives $(A_{f})^{G} = (A^{G})_{f}$ .

4.1F3step 3.1algebra

Gradings. The action is by graded automorphisms, so the invariant subspace of a graded rational $G$ -module is graded; both sides of $(A_{f})^{G} = (A^{G})_{f}$ are graded submodules of the graded ring $A_{f}$ (the localization of the graded subalgebra $A^{G}$ at the homogeneous element $f$ is graded, and $(A_{f})^{G}$ is graded). Taking degree-zero parts of the equality gives $((A_{f})_{0})^{G} = ((A^{G})_{f})_{0} = (A^{G})_{(f)}$ . The hypothesis that $R_{A}$ preserves degrees is what makes the Reynolds projection compatible with the grading in the computation of step 1.1, and the identity above is compatible with the gradings.

5.1step 2.1step 3.1step 4.1∎

Steps 2.1 and 3.1 establish $(A_{f})^{G} = (A^{G})_{f}$ , and step 4.1 gives the graded consequence $((A_{f})_{0})^{G} = (A^{G})_{(f)}$ , as claimed.

Remarks

No domain hypothesis. The proof uses complete reducibility to split off the $f$ -torsion of $A$ ; this replaces the clearing-denominators step of the classical treatment and makes the identity valid for an arbitrary graded rational $G$ -algebra, without assuming that $A$ is a domain.
Degree preservation. The hypothesis that $R_{A}$ preserves degrees enters only through the compatibility of the invariant identifications with the $Z$ -grading; it holds for the natural graded actions used on this page.

Depends on

Used by

Dependency tree · two levels

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Sources

P. E. Newstead, Geometric Invariant Theory, lecture notes, CIMAT Guanajuato 2006 (CEL/HAL; archived copy) (standard reference, not scraped)
Victoria Hoskins, Moduli Problems and Geometric Invariant Theory, FU Berlin lecture notes (2015/16) (standard reference, not scraped)