Invariant Theory

Introduction

Invariant theory is the study of what a group action fixes. Let a group $G$ act on a polynomial ring $K[x_1,\ldots,x_n]$ by ring automorphisms — for instance, a finite group of linear substitutions $x_i \mapsto \sum_j a_{ij}x_j$. The ring of invariants is the subring

$$ K[x_1,\ldots,x_n]^G = \{f : \sigma f = f \ \text{for every } \sigma \in G\}, $$

and the two questions of the subject are: is it finitely generated, and if so by what? Hilbert's answer, that for the classical groups the ring of invariants is finitely generated, was the origin of the modern theory of commutative rings — the ascending chain condition, the basis theorem and the Nullstellensatz were devised for it — and the finiteness theorem is now a consequence of the ring-theoretic machinery of this Part: integrality over a subring and the finiteness of integral closures for finite groups, as in Integral Extensions and Krull Dimension.

For a finite group the theory is complete and sharp. The extension $K[x_1,\ldots,x_n]^G \subseteq K[x_1,\ldots,x_n]$ is integral, so the two rings have the same Krull dimension $n$; the invariants are finitely generated, and in characteristic zero or in characteristic larger than $\lvert G\rvert$ they are generated in degrees at most $\lvert G\rvert$; the Molien series computes the dimension of each graded piece from the linear action alone; and the ring of invariants is a polynomial ring exactly for the groups generated by reflections, by the theorem of Chevalley, Shephard and Todd. Examples — the invariants of a cyclic group, of the sign group on two variables, of the symmetric group — are computed below, and the symmetric group's invariants, the symmetric polynomials, are not covered here.

The general linear group and the other classical groups are not finite, and their invariant theory requires the notion of a linearly reductive group together with the complete reducibility of its representations; that theory, and the geometric invariant theory of quotients by group actions, belongs to Part II, where the objects needed to state it exist. This article keeps to finite groups acting on a polynomial ring, to the finiteness theorems that follow from integrality, to the Molien series, and to the sharp classification of the polynomial cases.

Throughout, $K$ is a field, $V = K^n$ with basis $x_1,\ldots,x_n$, the polynomial ring is $R = K[x_1,\ldots,x_n]$ regarded as a graded ring with $\deg x_i = 1$, $G$ is a group acting on $R$ by graded $K$-algebra automorphisms induced by a linear action on $V$, and $R^G$ is the ring of invariants. The Hilbert series of a graded $K$-algebra $A = \bigoplus_{d \geq 0}A_d$ with $\dim_K A_d < \infty$ is the formal power series

$$ H(A, t) = \sum_{d \geq 0}(\dim_K A_d)\,t^d , $$

regarded as a formal object; no convergence is used anywhere below, and all manipulations of Hilbert series are manipulations of formal power series with rational coefficients.

The Ring of Invariants

Invariants are a Graded Subring

Definition. The action of $G$ on $R$ is induced by its action on $V$: for $\sigma \in G$ the automorphism of $R$ is the unique one extending $x_i \mapsto \sum_j \sigma_{ij}x_j$, and it is a graded ring automorphism in the sense of Ring and Field Automorphisms. An element $f \in R$ is invariant if $\sigma f = f$ for all $\sigma$, and $R^G$ is the set of invariant elements.

Proposition. Let $G$ act on $R$ by graded automorphisms.

(a) $R^G$ is a graded $K$-subalgebra: if $f = \sum f_d$ is invariant then each homogeneous component $f_d$ is invariant, and a product of invariants is invariant.

(b) $R^G$ is the fixed ring of the action, and the quotient map $R \to R$ restricted to $R^G$ is the identity; the map $f \mapsto \frac{1}{\lvert G\rvert}\sum_{\sigma \in G}\sigma f$, the Reynolds operator, is a $K$-linear projection of $R$ onto $R^G$ when $\lvert G\rvert$ is invertible in $K$, and the dimension of $R^G_d$ equals the dimension of the space of fixed vectors of the action of $G$ on $R_d$.

(c) If $H \leq G$ then $R^G \subseteq R^H$.

Proof. (a) The homogeneous components of $f$ are the images of $f$ under the projection onto $R_d$, which commutes with the action since the action is graded; hence $\sigma f_d = f_d$. Products and $K$-multiples of invariants are invariant, so $R^G$ is a subalgebra, and it is graded because it is the sum of its graded pieces by (a). (b) The Reynolds operator is well defined because $\lvert G\rvert$ is invertible, and it fixes exactly the invariants; the fixed space of $G$ acting on $R_d$ is $R^G_d$ by definition. (c) Clear.

Finiteness and Integrality

Theorem (integrality). Let $G$ be a finite group acting on $R = K[x_1,\ldots,x_n]$ by graded automorphisms. Then $R$ is integral over $R^G$, and consequently:

(a) the Krull dimensions agree, $\dim R^G = \dim R = n$, and $R^G$ is a domain;

(b) $R^G$ is a finitely generated $K$-algebra;

(c) $R$ is a finitely generated $R^G$-module, and the field of fractions of $R^G$ is $K(V)^G$, the fixed field of the induced action on the rational function field.

Proof. Let $f \in R$. The polynomial

$$ P(T) = \prod_{\sigma \in G}(T - \sigma f) \in R[T] $$

is monic and its coefficients are symmetric functions of the conjugates $\sigma f$, hence fixed by every $\tau \in G$ (which permutes the factors), so $P(T) \in R^G[T]$ and $P(f) = 0$: every element of $R$ is integral over $R^G$. Part (a) then follows from the theorem that an integral extension has the same Krull dimension, as in Integral Extensions and Krull Dimension, and $R$ is a domain, so its subring $R^G$ is one. Part (b) is the theorem of Artin and Tate: if $R$ is a finitely generated $K$-algebra and $G$ is finite, then $R^G$ is a finitely generated $K$-algebra; its proof uses the integrality just established together with the finiteness of the integral closure of a finitely generated algebra in a finite extension of its fraction field. Part (c) is the same integrality: the elements $x_i$ satisfy monic equations over $R^G$, so $R$ is a finite $R^G$-module; and the fraction field of $R^G$ is contained in $K(V)^G$ and contains it, since an invariant rational function can be averaged.

Corollary (Hilbert's finiteness theorem, finite case). For a finite group acting on a polynomial ring the ring of invariants is a finitely generated graded $K$-algebra; likewise for a group whose action is obtained by restriction from a group that is linearly reductive, by Weyl's theorem, whose statement requires the theory of group representations and is deferred.

Example. For $G = \{\pm1\}$ acting on $K[x,y]$ with $-1$ acting as $x \mapsto -x$, $y \mapsto -y$, the invariants are spanned by the monomials of even total degree, so $R^G = K[x^2, xy, y^2]$. The three generators are not independent: $(x^2)(y^2) = (xy)^2$, and the ring is the quotient $K[a,b,c]/(ac-b^2)$ with $a,b,c$ of degree $2$. The invariant ring is finitely generated by three elements, but it is not a polynomial ring, and the relation is forced by the dimension count: a polynomial ring in three generators of degree $2$ would have $\binom{n+2}{2}$ monomials of degree $2n$, whereas the invariants have exactly $2n+1$ of them, computed below.

Theorem (Noether's bound). Let $G$ be a finite group acting on $R$ and suppose that $\lvert G\rvert$ is invertible in $K$. Then $R^G$ is generated as a $K$-algebra by its elements of degree at most $\lvert G\rvert$. If $\lvert G\rvert$ is not invertible in $K$, the invariant ring is still finitely generated by the previous theorem, but no bound depending only on $\lvert G\rvert$ holds; the modular theory, in which $p$-groups give the extreme behaviour, is outside this Part.

Proof sketch. The Reynolds operator expresses any invariant as the average of its translates, and averaging a product of $t > \lvert G\rvert$ invariant elements expresses it in terms of invariants of lower degree; iterating gives generation in degrees at most $\lvert G\rvert$.

The Molien Series

Molien's Theorem

Theorem (Molien). Let $G$ be a finite group acting on $V = K^n$ and hence on $R = K[x_1,\ldots,x_n]$, and suppose $\lvert G\rvert$ is invertible in $K$. Then the Hilbert series of the ring of invariants is

$$ H(R^G, t) = \frac{1}{\lvert G\rvert}\sum_{\sigma \in G}\frac{1}{\det(1 - t\sigma)}, $$

where $\sigma$ acts on $V$ and $1-t\sigma$ is the endomorphism of $V \otimes_K K(t)$ that it defines; the right-hand side is computed as a formal power series in $t$.

Proof sketch. By the projection formula, $\dim_K R^G_d = \frac{1}{\lvert G\rvert}\sum_{\sigma}\operatorname{tr}(\sigma|_{R_d})$; the trace of $\sigma$ on the $d$-th symmetric power of $V$ is the complete homogeneous symmetric function $h_d$ of the eigenvalues of $\sigma$, and $\sum_{d \geq 0}h_d(\lambda_1,\ldots,\lambda_n)t^d = \prod_{i}(1-\lambda_it)^{-1} = \det(1-t\sigma)^{-1}$. Summing over $G$ and dividing by $\lvert G\rvert$ gives the identity as an identity of formal power series.

Example (a cyclic group). Let $G = \mathbb{Z}/n\mathbb{Z}$ act on $K[x]$ by $x \mapsto \zeta x$, where $\zeta$ is a primitive $n$-th root of unity in $K$; the invariants are $K[x^n]$, so $H(R^G,t) = 1/(1-t^n)$. Molien's formula reads

$$ \frac{1}{n}\sum_{j=0}^{n-1}\frac{1}{1-\zeta^jt} = \frac{1}{1-t^n}, $$

the partial-fraction decomposition of $1/(1-t^n)$, which one verifies by the roots-of-unity filter: expanding each term as a geometric series, the coefficient of $t^m$ is $\frac{1}{n}\sum_j\zeta^{jm}$, which is $1$ if $n \mid m$ and $0$ otherwise.

Example (the sign group on two variables). Let $G = \{\pm1\}$ act on $K[x,y]$ by negating both coordinates. The invariants are the even-degree polynomials, so $R^G$ has basis the monomials $x^{2i}(xy)^jy^{2k}$, of degree $2(i+j+k)$, and the ring is $K[a,b,c]/(ac-b^2)$ with $a,b,c$ of degree $2$. Its Hilbert series is

$$ H(R^G,t) = \frac{1-t^4}{(1-t^2)^3} = \frac{1+t^2}{(1-t^2)^2} = 1 + 3t^2 + 5t^4 + 7t^6 + \cdots , $$

the coefficient of $t^{2m}$ being $2m+1$: the number of monomials $x^{2i}(xy)^jy^{2k}$ of degree $2m$ is the number of triples with $i+j+k = m$, namely $\binom{m+2}{2}$, minus the number of those divisible by the relation $ac = b^2$, namely $\binom{m}{2}$, giving $\binom{m+2}{2}-\binom{m}{2} = 2m+1$. Molien's formula gives

$$ \frac12\left(\frac{1}{(1-t)^2} + \frac{1}{(1+t)^2}\right) = \frac{1+t^2}{(1-t^2)^2}, $$

in agreement. The verification of the coefficients was carried out symbolically: the expansion of the Molien sum to degree $9$ is $1, 0, 3, 0, 5, 0, 7, 0, 9$, matching $2m+1$ at even degrees and matching the quotient computation above.

Example (two variables, swap). Let $G = \mathbb{Z}/2$ act on $K[x,y]$ by $x \leftrightarrow y$. The invariants are the symmetric polynomials $K[e_1,e_2]$ with $e_1 = x+y$ of degree $1$ and $e_2 = xy$ of degree $2$, so $H(R^G,t) = 1/((1-t)(1-t^2))$. Molien's formula gives

$$ \frac12\left(\frac{1}{(1-t)^2} + \frac{1}{(1-t)(1+t)}\right) = \frac{1}{(1-t)^2(1+t)} = \frac{1}{(1-t)(1-t^2)}, $$

and the coefficient of $t^m$ in either series is $\lfloor m/2\rfloor+1$, the number of partitions of $m$ into parts $1$ and $2$.

Example (the reflection representation of $S_3$). Let $S_3$ act on $K[x,y]$ through its two-dimensional irreducible representation: the transpositions act as $x \leftrightarrow y$, with eigenvalues $1,-1$, and the $3$-cycles act with eigenvalues $\zeta,\zeta^2$ for $\zeta$ a primitive cube root of unity. Molien's formula gives

$$ \frac16\left(\frac{1}{(1-t)^2} + \frac{3}{(1-t)(1+t)} + \frac{2}{1+t+t^2}\right) = \frac{1}{(1-t^2)(1-t^3)} , $$

whose coefficients $1,0,1,1,1,1,2,1,2,2,\ldots$ were verified against the Molien sum by expansion to degree $11$. The ring of invariants is therefore a polynomial ring in generators of degrees $2$ and $3$: the degrees multiply to $6 = \lvert S_3\rvert$, as the theorem of Chevalley, Shephard and Todd requires.

Groups Generated by Reflections

Definition. An element $\sigma$ of a finite group $G$ acting on $V$ is a reflection (or pseudo-reflection) if it fixes a hyperplane of $V$ pointwise, equivalently if its fixed space has dimension $n-1$; $G$ is a reflection group if it is generated by reflections. For $n = 2$ a reflection is a linear map fixing a line pointwise, and for $n = 2$ the finite reflection groups are the dihedral groups and the cyclic groups.

Theorem (Chevalley–Shephard–Todd). Let $G$ be a finite group acting on $V = K^n$ with $\lvert G\rvert$ invertible in $K$. Then $R^G$ is a polynomial ring if and only if $G$ is generated by reflections. In that case $R^G = K[f_1,\ldots,f_n]$ with the $f_i$ homogeneous and algebraically independent, of degrees $d_1,\ldots,d_n$ satisfying

$$ \lvert G\rvert = d_1d_2\cdots d_n, \qquad H(R^G,t) = \prod_{i=1}^{n}\frac{1}{1-t^{d_i}} , $$

and the number of reflections of $G$ is $\sum_i(d_i-1)$.

Proof sketch. If $G$ is generated by reflections, one shows that the Jacobian determinant $\det(\partial f_i/\partial x_j)$ is nonzero and divisible exactly by the product of the linear equations of the reflecting hyperplanes, from which the algebraic independence of the $f_i$ and the degree formula follow; conversely, if $R^G$ is polynomial with generators $f_1,\ldots,f_n$, then averaging the $f_i$ shows that they may be taken homogeneous, the Jacobian is a nonzero invariant divisible by the hyperplane product, and each factor must occur, forcing $G$ to be generated by the corresponding reflections. The invariant theory of the Jacobian is the same computation as the Jacobian criterion of Integral Extensions and Krull Dimension.

Example. $G = \{\pm1\}$ on $K[x,y]$: the nontrivial element is $-1$, which fixes only the origin and no hyperplane, so $G$ is not generated by reflections; consistently the invariant ring $K[x^2,xy,y^2]$ is not polynomial, and indeed $\lvert G\rvert = 2$ cannot be written as a product of $n = 2$ integers greater than $1$, whereas the degree formula would require it. $G = S_2$ acting by the swap: the swap is a reflection with fixed line $x = y$, and $R^G = K[e_1,e_2]$ is polynomial with degrees $1,2$ and $\lvert S_2\rvert = 2 = 1\cdot2$. $G = S_3$ in its reflection representation: degrees $2,3$ with $6 = 2\cdot3$ and $3$ reflections, as computed above. More generally the symmetric group $S_n$ acting on $K[x_1,\ldots,x_n]$ by permuting the variables is a reflection group, generated by the transpositions, and the invariant ring is the ring of symmetric polynomials $K[e_1,\ldots,e_n]$ with $\deg e_i = i$, so $\lvert S_n\rvert = n! = 1\cdot2\cdots n$; the elementary symmetric polynomials, and the algebra of symmetric functions they generate, arebeing.

Example (the symmetric group in three variables). For $S_3$ acting on $K[x_1,x_2,x_3]$ by permuting the variables, Molien's formula with the eigenvalue data of the six permutation matrices — the identity with eigenvalues $1,1,1$, three transpositions with $1,1,-1$, and two $3$-cycles with $1,\zeta,\zeta^2$ — gives

$$ \frac16\left(\frac{1}{(1-t)^3} + \frac{3}{(1-t)^2(1+t)} + \frac{2}{(1-t)(1+t+t^2)}\right) = \frac{1}{(1-t)(1-t^2)(1-t^3)} , $$

whose coefficients $1,1,2,3,4,5,7,8,10,\ldots$ were verified against the Molien sum by expansion to degree $11$; the coefficient of $t^m$ is the number of partitions of $m$ into parts at most $3$, which is the number of monomials $e_1^ae_2^be_3^c$ of degree $m$.

Example (Dickson's theorem). Let $G = GL_n(\mathbb{F}_q)$ act on $\mathbb{F}_q[x_1,\ldots,x_n]$ by linear substitution. The ring of invariants is a polynomial ring, generated by the Dickson invariants $c_{n,k}$ of degree $q^n-q^k$ for $0 \leq k \leq n-1$, so that the degrees satisfy

$$ \prod_{k=0}^{n-1}(q^n-q^k) = q^{\binom{n}{2}}\prod_{k=0}^{n-1}(q^{n-k}-1) = \lvert GL_n(\mathbb{F}_q)\rvert , $$

in agreement with the product formula for a polynomial invariant ring. The group is generated by transvections, each fixing a hyperplane pointwise and having order $p = \operatorname{char}\mathbb{F}_q$; the group is therefore a reflection group in the extended sense of the modular theory, but $\lvert G\rvert$ is divisible by $p$, so the classification theorem above, whose hypothesis requires $\lvert G\rvert$ invertible in $K$, does not apply to it, and the degrees are computed directly.

Invariants, Fields and Rationality

Theorem (field of invariants). Let $G$ be a finite group acting on $R$ and let $L = K(x_1,\ldots,x_n)$ be the fraction field. Then $L/L^G$ is a Galois extension with group $G$, and $L^G$ is the fraction field of $R^G$; the invariants of the action and the fixed field of the Galois correspondence of Galois Theory coincide.

Proof. Each $\sigma \in G$ acts on $L$ by automorphisms fixing $L^G$; the fixed field of the action is $L^G$ by definition; the extension is finite of degree $\lvert G\rvert$ because $L$ is integral over $L^G$ (the same monic polynomial argument applies to the fraction field) with at most $\lvert G\rvert$ conjugates for each element, and Artin's theorem on the degree of a fixed field gives $[L:L^G] = \lvert G\rvert$.

Theorem (the Noether problem, statement). Let $G$ be a finite group and let $G \hookrightarrow GL_n(K)$ be a faithful representation over a field $K$. Then $G$ acts on $K(x_1,\ldots,x_n)$ and the question of the Noether problem is whether the fixed field $K(x_1,\ldots,x_n)^G$ is a purely transcendental extension of $K$, that is, a field of rational functions. For $K$ algebraically closed and $G$ abelian the answer is yes; for general finite groups the answer is no, the first counterexamples being the groups constructed by Swan and Saltman. The problem is the rationality question behind the construction of Galois extensions with prescribed group, and it is treated with the inverse Galois problem.

Example. For $G = S_3$ in the reflection representation on $K(x,y)$: the invariant ring is a polynomial ring in degrees $2$ and $3$, so $K(x,y)^{S_3} = K(f_2,f_3)$ is purely transcendental, with transcendence degree $2$, in agreement with the abelian-and-reflection cases of the Noether problem. For the sign group $\{\pm1\}$ acting on $K(x,y)$ by negating both variables, the invariant field is $K(x^2,xy,y^2) = K(x^2,xy)$, a rational function field in the two algebraically independent elements $x^2$ and $xy$, although the invariant ring was not polynomial; the Noether problem concerns the field, not the ring, and the difference is exactly this.

Remark. The general theory of invariants of a linearly reductive group — Hilbert's finiteness theorem for $SL_2$, the First and Second Fundamental Theorems for the classical groups, and the quotient construction of geometric invariant theory — requires the representation theory of such groups and the geometry of quotients. Both belong to Part II, where the topological and geometric language needed to state them is available.

Summary

For a group $G$ acting on the polynomial ring $R = K[x_1,\ldots,x_n]$ by graded $K$-algebra automorphisms, the ring of invariants $R^G$ is a graded subalgebra, equal to the image of the Reynolds operator $f \mapsto \frac{1}{\lvert G\rvert}\sum_\sigma \sigma f$ when $\lvert G\rvert$ is invertible in $K$. For a finite group, $R$ is integral over $R^G$ — the monic polynomial $\prod_\sigma(T-\sigma f)$ has invariant coefficients — so that $R^G$ is a domain of Krull dimension $n$, is finitely generated by the theorem of Artin and Tate, and has fraction field $K(x_1,\ldots,x_n)^G$, which is a Galois extension of the invariant field with group $G$ in the sense of Galois Theory. When $\lvert G\rvert$ is invertible in $K$, Noether's bound gives generation in degrees at most $\lvert G\rvert$; in the modular case the invariant ring is still finitely generated but admits no such bound.

Molien's theorem computes the Hilbert series of the invariant ring from the linear action,

$$ H(R^G,t) = \frac{1}{\lvert G\rvert}\sum_{\sigma \in G}\det(1-t\sigma)^{-1}, $$

and the computed cases are $K[x^n]$ for the cyclic group of order $n$ with Molien series $1/(1-t^n)$; $K[x^2,xy,y^2] = K[a,b,c]/(ac-b^2)$ for the sign group on two variables, with Hilbert series $(1-t^4)/(1-t^2)^3 = (1+t^2)/(1-t^2)^2$ and coefficient $2m+1$ in degree $2m$; the polynomial ring $K[e_1,e_2]$ for the swap, with series $1/((1-t)(1-t^2))$; the polynomial ring of degrees $2,3$ for the reflection representation of $S_3$, with series $1/((1-t^2)(1-t^3))$; and $K[e_1,e_2,e_3]$ for $S_3$ on three variables, with series $1/((1-t)(1-t^2)(1-t^3))$ and coefficients the partitions into parts at most $3$. The theorem of Chevalley, Shephard and Todd states that $R^G$ is a polynomial ring exactly when $G$ is generated by reflections, the degrees then satisfying $\lvert G\rvert = \prod d_i$ and the number of reflections being $\sum(d_i-1)$; the symmetric groups are the fundamental examples, their invariant rings being the rings of symmetric polynomials treated. Dickson's theorem gives the polynomial invariant ring of the finite general linear group over a finite field, and the Noether problem asks whether the invariant field is rational, its counterexamples lying beyond the finite abelian case. The invariant theory of the classical groups, and geometric invariant theory, belong to Part II.

Summary of Notation

Symbol	Meaning
$K$	Field of coefficients
$V = K^n$	The degree-one part of the polynomial ring
$R = K[x_1,\ldots,x_n]$	Polynomial ring with $\deg x_i = 1$
$G$	Group acting by graded automorphisms
$R^G$	Ring of invariants
$\sigma$	Element of $G$; its matrix on $V$
$H(A,t)$	Hilbert series of a graded algebra $A$
$e_i$	Elementary symmetric polynomial of degree $i$
$\zeta$	A primitive root of unity
$f_i$, $d_i$	Polynomial generators of a polynomial invariant ring and their degrees
$c_{n,k}$	Dickson invariant of degree $q^n-q^k$
$L = K(x_1,\ldots,x_n)$	Fraction field of $R$
$L^G$	Fixed field of the action