# Does Induction Work Well with Augmentation Ideals?

Let $\Sigma_n$ be the $n$th symmetric group. Then the complex representation ring $R\Sigma_n$ has the augmentation ideal $I_n$, with the $k$th power ideal $I_n^k$. Since $\Sigma_n\times\Sigma_m\leq \Sigma_{n+m}$ we have an induction map

$$\operatorname{Ind}: R[\Sigma_n\times\Sigma_m]\longrightarrow R\Sigma_{n+m}$$

given by taking $M\in R\Sigma_n$ and $N\in R\Sigma_m$ and inducing up $M\otimes N$. Suppose $M\in I_n^k$ and $N\in I_m^l$. Then the induced module is isomorphic to a direct sum of copies of $M\otimes N$ with an induced action. Therefore the induced module still has virtual dimension 0. My question is whether $\operatorname{Ind}(M\otimes N)\in I_{n+m}^{k+l}$.

-
Very interesting question, but please: it's "$k$-th power ideal", not "ideal of $k$-th powers". –  darij grinberg Apr 11 '11 at 22:02
What is virtual dimension? Also, when you write $M \in R\Sigma_n$, I assume you mean that $M$ is an $R\Sigma_n$-module, and when you write $M \in I_n^k$, you mean that $M$ is an $R\Sigma_n$-submodule of $I_n^k$. –  Christopher Drupieski Apr 11 '11 at 22:11
@Chris: The elements of the augmentation ideal are those that map to 0 under the dimension homomorphism. I say virtual dimension 0 since they are not representations. By $M\in R\Sigma_n$ I mean $M$ is in the Grothendieck group generated by the complex representations of $\Sigma_n$. By $M\in I_n^k$, I mean that $M$ is an element of the $k$th power ideal of $R\Sigma_n$. –  Joe Johnson Apr 11 '11 at 23:06
@Joe I see, I misinterpreted the Grothendieck ring for the group ring. –  Christopher Drupieski Apr 12 '11 at 1:07