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For a partition $\mu$ of $n$, let $S^{\mu}$ be the associated Specht module, defined over $\mathbb{Z}$. For any field $k$, we can tensor $S^{\mu}$ with $k$ to get a representation $S^{\mu}_k$ of the symmetric group $S_n$ over $k$. Finally, let $A_n \cong \mathbb{Z}$ be the sign representation of $S_n$ over $\mathbb{Z}$ and let $A_{n,k} \cong k$ be the sign representation over $k$.

In James's book on the representation theory of the symmetric group, he proves that $S^{\mu}_k \otimes A_{n,k} \cong (S^{\mu'}_k)^{\ast}$, where $\mu'$ is the conjugate partition and the $\ast$ indicates that we are taking the dual. My question is whether this is true over $\mathbb{Z}$ in the sense that there is a nondegenerate bilinear pairing $$\omega : S^{\mu} \times S^{\mu'} \rightarrow \mathbb{Z},$$ and assuming this is true, how can we calculate $\omega(e_t,e_s)$, where $t$ (resp. $s$) is a tableau of shape $\mu$ (resp. $\mu'$) and $e_t$ (resp. $e_s$) is the associated polytabloid.

If the above is false, I'm still interested in explicit pairings $$S^{\mu}_k \times S^{\mu'}_k \rightarrow k.$$ I've had trouble extracting them from James's book.

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I do not understand your $A_n \cong \mathbb{Z}$ and $A_{n,k} \cong k$. – Duchamp Gérard H. E. Jul 28 '12 at 7:00
This simply means that if one forgets the action one gets a free module of rank one. – Wilberd van der Kallen Jul 28 '12 at 7:52
up vote 11 down vote accepted

Yes, this works over $\mathbb Z$, and the pairing can be explicitly realised with polytabloids. See Section 4 of my paper "On the structure of Specht modules", J. London Math. Soc. 67 (2003) 85–102. (In retrospect, I wish I'd given this paper a more helpful title.)

Briefly: start by defining a map

$[,]: M^\mu\times M^{\mu'}\longrightarrow\mathbb Z$

as follows. Fix a $\mu$-tableau $r$. Now given a $\mu$-tabloid $S$ and a $\mu'$-tabloid $T$, we want to define $[S,T]$. If there are two numbers in the same row of $S$ and in the same row of $T$, then set $[S,T]=0$. If not, then there is a unique $\mu$-tableau $v$ such that $v$ is in $S$ and $v'$ (the transpose of $v$) is in $T$. Set $[S,T]$ to equal the sign of the permutation that takes $r$ to $v$.

Now for the map

$S^\mu\times S^{\mu'}\longrightarrow\mathbb Z$

you just take the above map $[,]$, restrict to $S^\mu\times S^{\mu'}$, and divide by the product of the hook lengths in $\mu$.

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Thank you very much, Matt! – Ben Jul 28 '12 at 14:46

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