Let $G$ be a reflection group acting on $\mathbb{K}^n$ (where $\mathbb{K}$ is a field of characteristic zero). Then $G$ induces an action on the ring of polynomials $\mathbb{K}[x_1, \ldots, x_n]$ by acting linearly on the $x_i$. Chevalley's theorem says that there are $n$ polynomials $f_1, \ldots, f_n$ which are invariant under the action of $G$ such that the space of all $G$ invariant polynomials can be written as $$\mathbb{K}[x_1, \ldots, x_n]^G = \mathbb{K}[f_1, \ldots, f_n].$$ A further version of this theorem (see Humphreys "Reflection Groups and Coxeter Groups", Proposition 3.13) says that the alternating polynomials under the $G$-action can be written as $$Alt_G[x_1, \ldots, x_n] = \mathbb{K}[f_1, \ldots, f_n] \cdot J$$ where $J$ is the Jacobian determinant $J = det ( \frac{\partial f_i}{\partial x_j} )_{1 \leq i,j \leq n}$. That is, a polynomial $f$ is alternating iff it can be written as a product of an invariant polynomial and $J$.

I'm interested in the case $G = S_n$. Now simply plugging $S_n$ in for $G$ one can calculate $J$ to be the Vandermonde determinant $\prod_{1 \leq i < j \leq n}{x_j - x_i}$ and the $f_i$ can be chosen to be either the elementary symmetric polynomials or for example $f_i = x_1^{i+1} + \cdots + x_n^{i+1}$ (which makes the calculation of $J$ easier).

The symmetric group $S_n$ however, also acts on $\mathbb{K}^{n-1}$ (since the action of $S_n$ on the subspace generated by $(1, \ldots, 1)$ is trivial). The space of invariant polynomials under this action can then be expressed as $$\mathbb{K}[f_2, \ldots, f_n]$$ where $f_2, \ldots, f_n$ are the $2^{\text{nd}}$ to $n^{\text{th}}$ elementary symmetric polynomials in $n$ variables. But how can I calculate the alternating polynomials under this action?

I mean, one can't just plug in $f_2, \ldots, f_n$ in the formula for $J$ cause these are $n-1$ polynomials in $n$ variables ($\rightarrow$ non-square matrix).

Suggestions: There are two ways to proceed that I can think of but I'm not sure which one leads to the right answer.

  1. Just take $J$ to be the Vandermonde determinant as above and write the alternating polynomials under the $S_n$ action on $\mathbb{K}^{n-1}$ as $$\mathbb{K}[f_2, \ldots, f_n] \cdot J.$$ I have no idea if this is true or not and why...

  2. Proceed as Humphreys does in his Example 3.12: The action of $S_n$ on $\mathbb{K}^{n-1}$ induces an action on $\mathbb{K}[x_1, \ldots, x_n]$ by permuting $x_1, \ldots, x_n$ subject to the relation $x_n = -(x_1 + \cdots + x_n)$.He then uses the $n-1$ polynomials $$g_i := x_1^{i+1} + \cdots + x_n^{i+1} \hspace{2mm} (1 \leq i \leq n-1)$$ and calculates the corresponding determinant $$J = det ( \frac{\partial g_i}{\partial x_j} )_{1 \leq i,j \leq n-1} = (n+1)! \prod_{1 \leq i,j \leq n-1}(x_j - x_i) \prod_{i=1}^{n-1}(x_i+z)$$ where $z := x_1 + \cdots + x_{n-1}$. So far, so good, but what do I get in the end? Are the alternating polynomials something like $$\mathbb{K}[g_1(x_1, \ldots, x_{n-1}, -z), \ldots, g_{n-1}(x_1, \ldots, x_{n-1}, -z)] \cdot J \hspace{1mm}?$$


1 Answer 1


What is going on is that you restrict the reflection representation of $S_n$ on $K^n$ to the hyperplane given by $x_1+\cdots+x_n=0$. This hyperplane is stable under all reflections of the representation, and not itself a reflection hyperplane. It is then sufficient to apply the projection of $K[x_1,\ldots,x_n]$ to $K[x_1,\ldots,x_n]/I$ where $I$ is the ideal generated by $x_1+\cdots+x_n$. That generator was the first elementary symmetric function$~f_1$, so it simple disappears in the quotient; all other things are replaced by their image in the quotient (which quotient is clearly isomorphic to $K[x_1,\ldots,x_{n-1}]$).

  • $\begingroup$ Thanks for your answer! So if I understand you correctly you would just mod out $I$ in my first approach to get $(K[f_2, \ldots, f_n] \cdot J)/I$ where $J$ is the usual vandermonde determinant? But then I would actually get something with a relation which seems to contradict the theorems I stated above? Or can the result be reformulated to something not involving a relation (i.e. a free algebra times $J$)? $\endgroup$ Commented Dec 13, 2016 at 19:53
  • $\begingroup$ I would prefer to write it $K[\overline{f_2},\ldots,\overline{f_n}]\cdot\overline J$ where the overlines indicate the image in the quotient. Since the images of $f_2,\ldots,f_n$ are algebraically independent, that is a free algebra time $\overline J$. So I don't see any contradiction, though I imagine that the Jacobi determinant might end up being a scalar multiple of $\overline J$. $\endgroup$ Commented Dec 14, 2016 at 6:12
  • $\begingroup$ So, clearly you can identify $K[\bar{f_2}, \ldots, \bar{f_n}]$ with $K[f_2, \ldots, f_n]$ (since you essentially just mod out $f_1$ and $f_1, f_2, \ldots, f_n$ are algebraically independent). But how do you know that $\bar{J}$ is just a scalar multiple of $J$? If this were indeed the case then we could identify $K[\bar{f_2}, \ldots, \bar{f_n}] \cdot \bar{J}$ with $K[f_2, \ldots, f_n] \cdot J$ and we'd be done, right? $\endgroup$ Commented Dec 14, 2016 at 8:44
  • $\begingroup$ I think I got it after all: It looks like my comment above is correct, I found an argument why $\bar{J}$ is a scalar multiple of $J$. $\endgroup$ Commented Dec 19, 2016 at 19:41
  • $\begingroup$ Just recording what I believe to be unjustly silenced: The main fact being used in this answer is that if $G$ is a finite group and $M$ and $N$ are two representations of $G$ over $\mathbb{K}$ and $f : M \to N$ is a surjective $G$-equivariant homomorphism, then $f$ is surjective on each isotypic component (i.e., for each irreducible representation $\rho$ of $G$, the restriction of $f$ to the $\rho$-isotypic component of $M$ is a surjective homomorphism onto the $\rho$-isotypic component of $N$). This follows fairly ... $\endgroup$ Commented Nov 25, 2018 at 22:58

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