This question is almost a duplicate of a question of Christian Stump, except that Christian seems to ask about an isomorphism to irreducible representations rather than the regular representation: What does the regular representation of the coinvariant ring of a unitary reflection group look like?

If $W\subseteq GL(V)$ is a finite reflection group, then the Chevalley-Shephard-Todd theorem says that the invariant algebra $\mathbb{C}[V]^W$ is isomorphic to a polynomial algebra $\mathbb{C}[V]^W\simeq \mathbb{C}[e_1,e_2,\ldots,e_n]$ and that the coinvariant algebra $\mathbb{C}[V]^{\mathrm{co}W} := \mathbb{C}[V]/\mathbb{C}[V]^W_+$ (where $\mathbb{C}[V]^W_+$ is the set of invariant polynomials of positive degree) is isomorphic as a $W$-module to the left regular representation.

Question: can we write down some explicit basis of $\mathbb{C}[V]^{\mathrm{co}W}$ on which $W$ acts as the regular representation?

The only proofs I know of the fact that $\mathbb{C}[V]^{\mathrm{co}W}$ carries the regular representation use character computations which have a "non-constructive" flavor.

Really I am most interested just in the case of the symmetric group $W=S_n$.

Let me give a quick example of what this looks like for $W=S_2$. Then $\mathbb{C}[V] = \mathbb{C}[x_1,x_2]$ and we get $\mathbb{C}[V]^W = \mathbb{C}[x_1+x_2,x_1x_2]$ (these are the ``elementary symmetric polynomials''). (Maybe strictly speaking because $S_2$ acts on $\mathbb{R}^2/(1,1)$ I should write $\mathbb{C}[V] = \mathbb{C}[x_1,x_2]/\langle x_1+x_2 \rangle$ and $\mathbb{C}[V]^W =\mathbb{C}[x_1x_2]$ but I don't think this technicality matters.) At any rate we have that the coinvariant ring is $\mathbb{C}[V]^{\mathrm{co}W}=\mathbb{C}[x_1,x_2]/\langle x_1+x_2,x_1x_2 \rangle$. There are standard bases of the coinvariant ring for the symmetric group, like the staircase monomials or the Schubert polynomials. In this case both of those bases would be $\{x_1,1\}$ (note that those bases are homogeneous). But the symmetric group $S_2$ does not act on that basis as in the regular representation. Instead I would want a basis like $\{x_1+1,-x_1+1\}$.

EDIT: Christian Gaetz asked me what taking $W=C_n=\langle c \rangle$ the cyclic group of order $n$ looks like. Here $c$ acts on $\mathbb{C}$ by $c\cdot x = \xi x$ with $\xi$ a primitive $n$th root of unity. Thus $\mathbb{C}[V]^{\mathrm{co}W} = \mathbb{C}[x]/\langle x^n \rangle$. A good choice of basis here is $\{c^m\cdot f\colon m=0,1,\ldots,n-1\}$ where $f=1+x+x^2+\cdots+x^{n-1}$. Indeed, in this case you can check that all the $c^m\cdot f$ are linearly independent by writing them in the standard basis $\{1,x,\ldots,x^{n-1}\}$ and then evaluating a Vandermonde determinant where you plug in $x_i=\xi^i$.

  • 1
    $\begingroup$ Even for $W = S_n$, this question is extremely interesting! The descent basis seems to be a step in the right direction, but still not what we're looking for. $\endgroup$ Commented Apr 30, 2018 at 18:10
  • 1
    $\begingroup$ @darijgrinberg: the descent basis still has the problem that it is homogenous, correct? $\endgroup$ Commented May 1, 2018 at 13:56
  • $\begingroup$ True... I guess this is why those bases are so hard to find. $\endgroup$ Commented May 1, 2018 at 15:03
  • $\begingroup$ I think that maybe deforming the harmonic polynomials / space of all iterated partial derivatives of the Vandermonde in the case of $S_n$ is promising. At least it is easy to see how to try and make things inhomogeneous, by doing things like $\partial_i+a$ instead of $\partial_i$. $\endgroup$ Commented May 1, 2018 at 16:35
  • 1
    $\begingroup$ Another remark: while indeed many/most proofs compute characters, for $S_n$ this can alternatively be proved using Galois theory. Namely, take $L=\mathbb{C}(x_1,\ldots,x_n)$. The field of $S_n$-invariants of this would be rational symmetric functions which are $K=\mathbb{C}(e_1,\ldots,e_n)$. We see that $[L:K]=n!$ and in fact is the regular representation by the normal basis theorem. Passing from that result to the statement about the rings before taking fields of fractions is an easy exercise :-) $\endgroup$ Commented May 1, 2018 at 16:42

2 Answers 2


Any basis $B$ on which $W$ acts as the regular representation has the form $\{w\cdot f\,\mid\, w\in W\}$ for some $f$ in the coinvariant ring. Hence it suffices to take $f$ generic. It would be interesting to find some "nice" nongeneric $f$.

  • 3
    $\begingroup$ I agree with this but was hoping there might be a more canonical choice. $\endgroup$ Commented Apr 30, 2018 at 18:55

Let $W$ be a finite reflection group. Set $R={\mathbb C}[V]$, $R^W\subset R$ the invariant subring, and $R_W={\mathbb C}[V]^{coW}$ the coinvariant ring. For a linear character $\chi\colon W\rightarrow {\mathbb C}^*$, define the relative invariant ring $R^W_\chi=\left\{g\in R\left|w\cdot g=\chi(w)\cdot g, \ \forall w\in W\right.\right\}$. For reflection groups, this is always a cyclic module over $R^W$ (e.g. see Professor Stanley's paper). If $\chi_1,\ldots,\chi_m$ are the distinct linear characters of $W$, let $f_1,\ldots,f_m$ be any generators of the respective relative invariant rings. For $f\in R$, let $\bar{f}\in R_W$ denote its equivalence class.

Claim: If $W$ is abelian, then the elements $\left\{\sum_{i=1}^m\chi_i(w)\cdot \bar{f}_i\left|w\in W\right.\right\}$ form a basis for $R_W$ on which $W$ acts by the regular representation.

Proof: Note that the equivalence classes $\left\{\bar{f}_1,\ldots,\bar{f}_m\right\}$ are linearly independent because each is a simultaneous eigenvector for $W$ corresponding to a distinct linear character. Since $W$ is abelian this must be a basis for $R_W$. To see that the elements $\left\{\left.\sum_{i=1}^m\chi_i(w)\bar{f}_i\right|w\in W\right\}$ are linearly independent in $R_W$, use linear independence of distinct linear characters.

For non-abelian groups the elements $\left\{\bar{f}_1,\ldots,\bar{f}_m\right\}$ won't be a basis (since $m<|W|$), which will force the set $\left\{\sum_{i=1}^m\chi_i(w)\bar{f}_i\left|w\in W\right.\right\}$ to be linearly dependent.

  • $\begingroup$ The definition of $R_\chi^W$ is not correct. $\endgroup$ Commented May 1, 2018 at 2:03
  • $\begingroup$ @RichardStanley thanks, I think the definition is correct now. But the conjecture is still not true in general, although maybe it's true for Abelian groups... $\endgroup$ Commented May 1, 2018 at 2:33
  • $\begingroup$ @SamHopkins I edited my answer. $\endgroup$ Commented May 2, 2018 at 3:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.