6
$\begingroup$

Let $V$ be a vector space over $k$ of dimension $m$. (I'm only interested in the case $k=\mathbb{Q}$.) Let $R:=\Lambda^*V$ be the exterior algebra. It carries the structure of a supercommutative ring: $R=R^+\oplus R^{-}$, where both the odd and the even part have $k$-dimension $2^{m-1}$. We define a multiplication on $R^{\otimes n}$ by $(\alpha_1\otimes\cdots\otimes\alpha_n)\cdot (\beta_1\otimes\cdots\otimes\beta_n) =\pm (\alpha_1\beta_1\otimes\cdots\alpha_n\beta_n)$, where the sign comes from reordering factors of odd degree. The symmetric group $\mathfrak{S}_n$ acts on $R^{\otimes n}$ by permuting the factors such that a transposition of two factors of odd degrees yields a negative sign.

The supersymmetric power $S^nR$ is defined as the quotient of $R^{\otimes n}$ by this action. Over $k=\mathbb{Q}$ this is the same as taking the subspace of invariants. The canonical inclusion of $V$ in $S^nR$ gives $S^nR$ the structure of a $R$-algebra.

Now I conjecture that $S^nR$ is free as a module over $R$ for $n\geq 1$.

Let me give evidence to this. As a $k$-vector space, $ S^nR$ can be decomposed as: $$ S^nR = \bigoplus_{p+q=n} \text{Sym}^p(R^+)\otimes \Lambda^q(R^{-}). $$ The generating function of the dimensions of $S^nR$ is therefore $ \sum_{n\geq 0} \dim_k(S^nR)\, t^n= \left(\frac{1+t}{1-t}\right)^{2^{m-1}}, $ which can be seen to have all coefficients divisible by $2^m=\dim_k R$ except for the first one.

For instance, if $V$ is one-dimensional, then $R$ is isomorphic to $k[x]/(x^2)$ and $S^nR = R$ for all $n\geq1$. If $V$ is two-dimensional with basis $\{\theta_1,\theta_2\}$, then $S^2R$ is generated over $R$ by $1\otimes 1$ and $\theta_1\otimes \theta_2$.

Any idea of proof or references will be gratefully welcomed.

$\endgroup$

1 Answer 1

1
$\begingroup$

I found a proof for the desired freeness result in the case $\text{char}(k)=0$, but it is a bit technical.

The idea is to show first, that $R^{\otimes n}$ is a free $R$-module in a way that the action of $\mathfrak S_n$ is $R$-linear. The second step is to find a basis of $S^n R$, when the basis of $R^{\otimes n}$ already is constructed.

For some $v\in V$ we denote $v^{(i)} := 1^{\otimes i-1}\otimes v\otimes 1^{\otimes n-i+1} \in R^{\otimes n}$. Then $R^{\otimes n}$ is generated as a $k$-algebra by the elements $\{v_j^{(i)}\} $ for a $k$-basis $\{v_j\}$ of $V$. Now consider the ring automorphism $$ \sigma : R^{\otimes n} \longrightarrow R^{\otimes n}, \\ v^{(1)} \longmapsto v^{(1)} +v^{(2)} + \ldots + v^{(n)}, \quad v^{(i)} \longmapsto v^{(i)} \text{ for } i>1. $$ Now look at the injection $$ \iota: R \rightarrow R^{\otimes n},\quad \theta \mapsto \theta\otimes 1\otimes\cdots\otimes 1. $$ It is clear that this turns $R^{\otimes n}$ into a $R$-algebra and in particular into a free $R$-module. Denote $\{b_i\}$ a $R$-linear basis of $R^{\otimes n}$. But $R^{\otimes n}$ can also be considered as a free $R$-module with the $R$-module structure given by $\sigma\iota : R \rightarrow R^{\otimes n}$.

The image of $\sigma\iota$ is fixed by the action of $\mathfrak S_n$. Thus the kernel and the image of any $\pi\in \mathfrak S_n$ become free $R$-modules as well. In particular, the space of invariants, given by the image of $\frac{1}{n!}\sum\limits_{\pi\in\mathfrak S_n}\pi$ which is isomorphic to $S^nR$, is a free $R$-module.

$\endgroup$

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.