# Super-plethysm?

Let $U$ be a representation of $S_m$ and $V$ a representation of $S_n$. Then the representation $\operatorname{Ind}_{S_m\wr S_n}^{S_{mn}}(U^{\otimes{n}}\otimes V)$ has a nice interpretation in terms of symmetric functions. If $\operatorname{ch}(U)$ and $\operatorname{ch}(V)$ are the Frobenius characteristics of $U$ and $V$ (symmetric functions of degree $m$ and $n$), then $$\operatorname{ch}\operatorname{Ind}_{S_m\wr S_n}^{S_{mn}}(U^{\otimes{n}}\otimes V)= \operatorname{ch}(V)[\operatorname{ch}(U)],$$ where the square brackets denote plethysm. I am interested in a graded version of this result.

More specifically, let $X$ be a space with an action of $S_m$ and $Y$ a space with an action of $S_n$. I am interested in the representation $$\operatorname{Ind}_{S_m\wr S_n}^{S_{mn}}\left(H^*(X^n)\otimes H^*(Y)\right)$$ of $S_{mn}$. The naive statement would be that $$\operatorname{ch}\operatorname{Ind}_{S_m\wr S_n}^{S_{mn}}\left(H^*(X^n)\otimes H^*(Y)\right) = \operatorname{ch}(H^*(Y))[\operatorname{ch}(H^*(X))],$$ where $\operatorname{ch}(H^*(X))$ and $\operatorname{ch}(H^*(Y))$ are symmetric functions of degree $m$ and $n$ with coefficients in the polynomial ring $\mathbb{Z}[t]$.

I'm pretty sure that this statement is correct when the cohomology of $X$ is concentrated in even degree, but otherwise it is wrong. The issue is that, at the level of $S_n$-representations, the Kunneth isomorphism $H^*(X^n)\cong H^*(X)^{\otimes n}$ needs to be interpreted in terms of super-vector spaces: the action of the simple transposition $(i,i+1)$ picks up a sign when the two cohomology classes being swapped have odd degree. For example, if $X=S^1$ and $n=2$, then $H^2(S^1\times S^1)\cong H^1(S^1)\otimes H^1(S^1)$ is the sign representation of $S_2$ rather than the trivial representation.

So my question is: Is there a "super version" of plethysm for symmetric functions with coefficients in $\mathbb{Z}[t]$ for which the last displayed equation is correct?

I know that the answer is tautologically "yes"--one can just translate from symmetric functions to representations, do the induction, and translate that, and take this as the definition of super-plethysm. But I'm looking for something explicit enough to allow me to do calculations in SAGE.

• You just need to use an auxiliary variable $t$ for homological degree, and use the weight $(-t)^i$ for the $H^i$. The plethysm must act on this variable by $p_n(t)=t**n$. – F. C. Mar 16 '16 at 15:45
• Ah, that makes sense, and it indeed seems to work in a simple example. Thanks! (If you want to post that as an answer, I'll accept it.) – Nicholas Proudfoot Mar 16 '16 at 16:26
• Does this provide an answer to my MO question mathoverflow.net/questions/33681/… with t=q? – Abdelmalek Abdesselam Mar 16 '16 at 19:27

The correct way to handle these computations using plethysm is to introduce an auxiliary variable $t$ and to weight the cohomology $H^i$ with the weight $(-t)^i$. The plethysm must act on $t$ by $p_n(t)=t^n$.
• I was recently thinking about the same question. I've got a little bit different formula, but it is equivalent to the one above modulo the change of variables $q=-t$. If one introduces an auxiliary variable $q$ to weight the cohomology $H^i$ with the weight $q^i$, then the plethysm acts on $q$ by $p_n(q)=(-1)^{n-1}q^n$. – Victor May 19 '17 at 17:25