It is classical that every alternating polynomial is (uniquely) the product of a symmetric polynomial with the Vandermonde polynomial, in particular the alternating polynomials are a free rank-one module over the symmetric polynomials.

For some cohomology computations I am doing, I would like to know some analogous statement, not for the polynomial ring, but for the free graded-commutative algebra $$\mathfrak{F}(n)=\mathbb{F}_\ell[X_1,\dots,X_n]\langle A_1,\dots,A_n\rangle,$$ i.e., polynomial variables $X_i$ and exterior variables $A_j$. For peace of mind, $\ell$ is a prime different from $2$, but I would be willing to assume $\ell> n$. Now consider the "diagonal permutation" action of the symmetric group $\Sigma_n$, i.e., an element $\sigma$ acts simultaneously on $\{X_1,\dots,X_n\}$ and $\{A_1,\dots,A_n\}$ via the natural permutation of indices. I am interested in the submodule of $\mathfrak{F}(n)$ where $\Sigma_n$ acts via the sign representation (may I call the elements alternating superpolynomials?).

I would like to know that this submodule is free over the ring of symmetric polynomials in $\{X_1,\dots,X_n\}$, and I would like to have a general method of computing the rank and explicit generators.

So far, the case $n= 2$ is clear and I also worked out the case $n=3$. This is based on first working out the $\Sigma_3$ decomposition of the polynomial ring and the exterior algebra separately and then tensoring the representations (using the assumption $\ell\neq 2,3$). The result is a free rank 8 module. This could of course be done in further cases, but will probably fail to produce a general result because it involves tensoring representations of the symmetric group. So the parts of my question are:

  • Has this question already been considered somewhere in the literature? (I would guess, it seems natural enough).

  • Is there a slick general argument dealing with all $n$ at once (and possibly with $\ell\leq n$ as well), like in the classical case of alternating polynomials?

Any ideas or literature references would be much appreciated.

  • $\begingroup$ Do you happen to know if your submodule is finitely generated as module over the ring of sym. polynomials ? $\endgroup$ – Todd Leason Oct 20 '15 at 17:57
  • $\begingroup$ I think the module is finitely generated. First, the polynomial part is finitely generated over the symmetric polynomials (actually free of rank equal to the order of the corresponding symmetric group) and the alternating part is a finite-dimensional $\mathbb{F}_\ell$-vector space. This means $\mathfrak{F}(n)$ is finitely generated, but the ring of symmetric polynomials is noetherian. $\endgroup$ – Matthias Wendt Oct 21 '15 at 7:06

If you assume $\ell>n$, then you may as well work over the field $F=\mathbb{Q}$, since the representation theory is exactly the same (the group algebra is semisimple, all Young symmetrisers in the group algebra over $\mathbb{Q}$ have denominators dividing $n!$, etc.).

Under this (much) simplifying assumption, we have $F[x_1,\ldots,x_n]\cong FS_n\otimes F[x_1,\ldots,x_n]^{S_n}$, and the exterior algebra $\Lambda(a_1,\ldots,a_n)$ can be written as $\Lambda(V\oplus F)$, where $V$ is the standard $(n-1)$-dimensional representation of $S_n$, and $F$ is the trivial one, and that is the same as $\Lambda(V)\otimes\Lambda(F)$, where $\Lambda(F)\cong \Lambda(a_1,\ldots,a_n)^{S_n}$. Now, we see that your ring is, as a representation of $S_n$, $$ FS_n\otimes\Lambda(V)\otimes F[x_1,\ldots,x_n]^{S_n}\otimes \Lambda(a_1,\ldots,a_n)^{S_n} . $$ To extract the isotypic component of the sign representation, you should take that component in $FS_n\otimes\Lambda(V)$, and tensor with the invariants. This already gives freeness as a module over the symmetric polynomials, of course. But also, note that $FS_n$ is the sum of all irreducible modules with multiplicities equal to dimensions, and $\Lambda(V)$ is the sum of all irreps corresponding to hooks with all multiplicities equal to one. Also, the sign representation multiplicity in $U\otimes V$ is the trivial representation multiplicity in $U\otimes V\otimes\mathrm{sign}$ is the dimension of $\mathop{\mathrm{Hom}}(U,V\otimes\mathrm{sign})$ because of self-duality of representations of symmetric groups. Finally, tensoring with sign replaces each Young diagram by its transpose, which does not change either $FS_n$ or $\Lambda(V)$ as a representation. Thus, the corresponding multiplicity is equal to the sum of all dimensions of irreps corresponding to hooks; those dimensions are $\binom{n-1}{k}$ for various $k$, so the multiplicity is $2^{n-1}$. Multiplying that by $2=\dim\Lambda(a_1,\ldots,a_n)^{S_n}$, we finally see that the module is free of rank $2^n$.

I am not sure what to expect in the case $\ell\le n$: most of the statements I made above would fail for rather trivial reasons.

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