I'm now interested in the modular representation of symmetric groups.

It is well-known that for a fixed prime $p$, there is a bijection between the irreducible representations of $S_{n}$ over a field of characteristic $p$ and $p$-regular partitions of $n$.

And it seems that the sign representation of $S_{n}$ exists for all odd primes $p$.

However, the partition $(1^{n})$ of $n$ which corresponds to the ordinary (i.e. characteristic 0) sign representation of $S_{n}$ is no longer $p$-regular if $p \leq n$

So I hope to know, which $p$-regular partition corresponds to the irreducible 1-dimensional sign representation?

Is there an easy answer?

Thank you very much!


The answer is the $p$-regularisation of $(1^n)$: this is $(1^n)$ if $n < p$ and otherwise the partition $$ (r^a, (r-1)^{p-1-a}) $$ with $p-1$ parts, where $r$ and $a \in \{0,1,\ldots, p-1\}$ are defined uniquely by $$(r-1)(p-1) + a = n, $$ taking $a = 0$ if $p-1$ divides $n$.

For more on this see G. D. James, On the decomposition matrices of the symmetric groups II, J. Algebra 43 (1976), 45--54. The Mullineux Conjecture (now proved) determines the partition labelling a general $D^\lambda \otimes \mathrm{sgn}$ for $\lambda$ a $p$-regular partition. The shortest proof I know is in Example 24.5(iii) in James' Springer lecture notes on the symmetric groups.

Edit. I'd like to record a slightly unusual proof here. By some general theory, the permutation module $M$ for a finite group $G$ acting on the cosets of $H \le G$ defined over a field $F$ of prime characteristic $p$ has a distinguished Scott module summand. In Green's theory of vertices and sources, this summand has vertex $Q$ where $Q \in \mathrm{Syl}_p(H)$ and trivial source. It is the unique summand of $M$ having the trivial module $F$ in its top and socle. In particular, $P_F$, the projective cover the trivial module, is the Scott module for trivial vertex.

Now let $\lambda$ be a partition of $n$ and let $M^\lambda$ be the Young permutation module for $FS_n$ induced from the Young subgroup $S_{\lambda_1} \times \cdots \times S_{\lambda_{\ell(\lambda)}}$. It is known that $M^\lambda$ has a distinguished summand $Y^\lambda$: this is the unique summand (in any given direct sum decomposition of $M^\lambda$) containing the Specht module $S^\lambda$. Moreover, $M^\lambda \cong Y^\lambda \oplus \bigoplus_{\mu > \lambda} c_\mu Y^\mu$ for some coefficients $c_\mu$.

Observe that $\lambda$ has a part of size $p$ or more if and only if $S_{\lambda_1} \times \cdots \times S_{\lambda_{\ell(\lambda)}}$ has a non-trivial Sylow $p$-subgroup and so if and only if the Scott module summand of $M^\lambda$ is non-projective. Therefore the lexicographically greatest partition $\mu$ such that $M^\mu$ has $P_F$ as a summand is the greatest partition with all parts $< p$, namely $((p-1)^s, t)$ where $0 \le t < p-1$ and $n = (p-1)s + t$. Moreover, since $P_F$ is not a summand of any $M^\lambda$ with $\lambda > \mu$ (they all have a Scott module summand, which cannot be project, so is not $P_F$), we have $Y^{((p-1)^s, t)} \cong P_F$.

This shows that the trivial module appears in the socle of the Specht module $S^{((p-1)^s,t)}$, and so using the basic result that $(S^\lambda)^\star \cong S^{\lambda'} \otimes \mathrm{sgn}$, the sign module appears at the top of $S^{((p-1)^s,t))'}$. Since $((p-1)^s,t)'$ is $p$-regular, this Specht module has a unique top composition factor and we get $D^{((p-1)^s, t)'} \cong \mathrm{sgn}$, as required. (And in agreement with the result from $p$-regularisation.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.