The Chevalley-Shephard-Todd Theorem

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 10 November 2011

The Chevalley-Shephard-Todd theorem

Let $V$ be a finite dimensional vector space over a field $𝔽$. Let $W$ be a finite subgroup of $\mathrm{GL}\left(V\right)$. If ${S\left(V\right)}^W$ is a polynomial algebra then $W$ is generated by reflections.

Proof. Let

$I=⟨f\in {S\left(V\right)}^W\left|f\right(0)=0⟩$

be the ideal in $S\left(V\right)$ generated by polynomials without constant term. Let $e_1,\dots ,e_r$ be homogeneous generators of $I$ (which exist, by Hilbert).

Step 1. Every $f\in {S\left(V\right)}^{W}is\; a\; polynomial\; in$ e_1,\dots ,e_r$.$

Proof. The proof is by induction on the degree of $f$. Assume $f$ is homogeneous and $\mathrm{deg}\left(f\right)>0$. Since $f\in I$,

${\displaystyle f=\sum _{i=1}^r{p}_i{e}_i,}\text{with}{p}_i\in S\left(V\right),$

and so $$f=\frac{1}{\left|W\right|}\sum _{w\in W}wf=\sum _{i=1}^r\left(\frac{1}{\left|W\right|}\sum _{w\in W}w{p}_i\right)e_i,$$ and since the internal sum has lower degree it can be written as a polynomial in $e_1,\dots ,e_r$. $\square$

Step 2. $r=\dim \left(V\right)$.

Proof. Let $n=\dim \left(V\right)$, let $x_1,\dots ,x_n$ be a basis of $V$ and let $ℂ(x_1,\dots ,x_n)$ be the field of fractions of $S\left(V\right)=ℂ[x_1,\dots ,x_n]$. Since $x_i$ is a root of $$m_i\left(t\right)=\prod _{w\in W}(t-w{x}_i)\in {S\left(V\right)}^W\left[t\right],$$ the variable $x_i$ is algebraic over $ℂ(e_1,\dots ,e_r)$, the field of fractions of ${S\left(V\right)}^W$. Thus $$0=\mathrm{trdeg}\left(\frac{ℂ(x_1,\dots ,x_n)}{ℂ(e_1,\dots ,e_r)}\right)=\mathrm{trdeg}\left(\frac{ℂ(x_1,\dots ,x_n)}{ℂ}\right)=\mathrm{trdeg}\left(\frac{ℂ(e_1,\dots ,e_r)}{ℂ}\right)=n-r.\square$$

Step 3. The Jacobian of a map $$\begin{array}{cccc}\phi :& V& ⟶& V\\ & x& ⟼& \left(\phi \right(x),\dots ,\phi (x\left)\right)\end{array}\text{is}{J}_{\phi }\left(x\right)=\det \left(\frac{\partial {\phi }_i}{\partial x_j}\right).$$ If $\phi$ is linear then there are ${\phi }_{ij}\in ℂ$ such that $${\phi }_i\left(x\right)=\sum _{j=1}^n{\phi }_{ij}{x}_j\text{and}{J}_{\phi }=\det \left({\phi }_{ij}\right).$$ The chain rule is the identity $$J_{\theta \circ \phi }=J_{\theta }\left(\phi x\right)J_{\phi }\left(x\right).$$ Let $$\begin{array}{cccc}\theta :& V& ⟶& V\\ & x& ⟼& \left(e_1\right(x),\dots ,e_n(x\left)\right)\end{array}\text{and}\begin{array}{cccc}w:& V& ⟶& V\\ & x& ⟼& wx\end{array}$$ for $w\in W$. Then $\theta \circ w=\theta$ and so $$J_{\theta }\left(x\right)=J_{\theta \circ w}\left(x\right)=J_{\theta }\left(wx\right)J_w\left(x\right)=J_{\theta }\left(wx\right)\det \left(w\right)=\det \left(w\right)\left(w^{-1}{J}_{\theta }\right)\left(x\right).$$ Thus $J_{\theta }$ is $W$-alternating and so $J_{\theta }$ is divisible by $$\Delta =\prod _{\alpha \in R^{+}}{\alpha }^{s_{\alpha }-1}.\text{Since}\mathrm{deg}\left(J_{\theta }\right)=\sum _{i=1}^n(d_i-1)=\mathrm{Card}\left(R^{+}\right),$$ it follows that $J_{\theta }=\lambda \cdot \Delta$ for some $\lambda \in ℂ$. $\square$

Step 4. The polynomials $e_1,\dots ,e_n$ are algebraically independent if and only if $J_{\theta }\ne 0$.

Proof. $⟹$: Assume $e_1,\dots ,e_r$ are algebraically independent. Since $$\mathrm{trdeg}\left(\frac{ℂ(x_1,\dots ,x_n)}{ℂ(e_1,\dots ,e_r)}\right)=\mathrm{trdeg}\left(\frac{ℂ(x_1,\dots ,x_n)}{ℂ}\right)=\mathrm{trdeg}\left(\frac{ℂ(e_1,\dots ,e_r)}{ℂ}\right)\ge n-r,$$ $x_1,\dots ,x_n$ are algebraic over $ℂ(e_1,\dots ,e_r)$ if and only if $0\ge n-r$, that is, if and only if $n=r$.

Let $m_i\left(t\right)\in {S\left(V\right)}^W\left[t\right]$ be the minimal polynomial of $x_i$ over $ℂ(e_1,\dots ,e_r)$, the field of fractions of ${S\left(V\right)}^W$. Then $$\frac{\partial m_i}{\partial x_k}=\sum _{j=1}^r\frac{\partial m_i}{\partial e_j}\frac{\partial e_j}{\partial x_k}+\frac{\partial m_i}{t}\frac{t}{\partial x_k}$$ and $$0=\frac{\partial m_i\left(x_i\right)}{\partial x_k}==\sum _{j=1}^r\frac{\partial m_i}{\partial e_j}\left(x_i\right)\frac{\partial e_j}{\partial x_k}+m_i^{\prime }\left(x_i\right){\delta }_{ik}.$$ Thus $$\det \left(\frac{\partial m_i}{\partial e_j}\left(x_i\right)\right)\cdot J_{\theta }=\det (-\mathrm{diag}\left({m\prime }_1\right(x_1),\dots {m\prime }_n(x_n\left)\right))={(-1)}^n\prod _{i=1}^r{m\prime }_i\left(x_i\right).$$ Since $m_i\left(t\right)$ is the minimal polynomial of $x_i$, each factor ${m\prime }_i\left(x_i\right)\ne 0$ and, thus, $J_{\theta }\ne 0$.

$⟸$: Assume $e_1,\dots ,e_n$ are algebraically independent. Let $f(y_1,\dots ,y_n)$ be of minimal degree such that $f(e_1,\dots ,e_n)=0$. Then $$\frac{\partial f}{\partial y_i}\ne 0\text{for some}{y}_i,\text{and so}{g}_i=\frac{\partial f}{\partial y_i}(e_1,\dots ,e_n)\ne 0\text{for some}i.$$ But $$0=\frac{\partial f(e_1,\dots ,e_n)}{\partial x_j}=\sum _{i=1}^n\frac{\partial f}{\partial y_i}(e_1,\dots ,e_n)\frac{\partial e_i}{\partial x_j},\text{and so}\sum _{i=1}^n{g}_i\frac{\partial e_i}{\partial x_j}=0.$$ So $g_i$ is a solution to the equation $(g_1,\dots ,g_n)\left(\partial e_i/\partial x_j\right)=0$ and so $J_{\theta }=0$. $\square$

Notes and References

These notes are a retyping of section 3 of http://researchers.ms.unimelb.edu.au/~aram@unimelb/Notespre2005/winvts1.26.04.pdf

References

[St] R. Steinberg, Lectures on Chevalley groups, Notes prepared by John Faulkner and Robert Wilson, Yale University, New Haven, Conn., 1968. iii+277 pp. MR0466335.

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