2 trivial typo fix + retag

This question concerns the characteristic $0$ representation theory of the symmetric group $S_n$. I'm a topologist, not a representation theorist, so I apologize if I state it in an odd way.

First, a bit of background. The finite-dimensional irreducible representations of $S_n$ are given by the Specht modules $S^{\mu}$. Here $\mu$ is a partition of $n$, which is best visualized as a Young diagram. There are classical rules for restricting $S^{\mu}$ to $S_{n-1}$ and inducing $S^{\mu}$ to $S_{n+1}$ (these are known as branching rules). Namely, we have the following.

1. The restriction of $S^{\mu}$ to $S_{n-1}$ is isomorphic to the direct sum of the $S_{n-1}$-representations $S^{\mu'}$ as $\mu'$ goes over all ways of removing a box from the Young diagram for $\mu$ (while staying in the world of Young diagrams).

2. The induction of $S^{\mu}$ to $S_{n+1}$ is isomorphic to the direct sum of the $S_{n+1}$-representations $S^{\mu'}$ as $\mu'$ goes over all ways of adding a box to the Young diagram for $\mu$ (again while staying in the world of Young diagrams).

These two rules are equivalent by Frobenius reciprocity.

There is a nice concrete proof of the restriction rule (I believe that it is due to Peel, though I first learned about it from James's book "The representation theory of the symmetric groups"). Assume that the rows of the Young diagram for $\mu$ from which we can remove a box are $r_1 < \ldots < r_k$, and denote by $\mu_i$ the partition of $n-1$ obtained by removing a box from the $i^{\text{th}}$ r_i^{\text{th}}$row of$\mu$. There is then a sequence $$0=V_0 \subset V_1 \subset \cdots \subset V_k = S^{\mu}$$ of$S_{n-1}$-modules such that$V_i/V_{i-1} \cong S^{\mu_i}$. In fact, recalling that$S^{\mu}$is generated by elements corresponding to Young tableaux of shape$\mu$(known as polytabloids), the module$V_i$is the subspace spanned by the polytabloids in which$n$appears in a row between$1$and$i$. Question : Is there a similarly concrete proof of the induction rule (in particular, a proof which does not appeal to Frobenius reciprocity and the restriction rule)? 1 # Making the branching rule for the symmetric group concrete This question concerns the characteristic$0$representation theory of the symmetric group$S_n$. I'm a topologist, not a representation theorist, so I apologize if I state it in an odd way. First, a bit of background. The finite-dimensional irreducible representations of$S_n$are given by the Specht modules$S^{\mu}$. Here$\mu$is a partition of$n$, which is best visualized as a Young diagram. There are classical rules for restricting$S^{\mu}$to$S_{n-1}$and inducing$S^{\mu}$to$S_{n+1}$(these are known as branching rules). Namely, we have the following. 1. The restriction of$S^{\mu}$to$S_{n-1}$is isomorphic to the direct sum of the$S_{n-1}$-representations$S^{\mu'}$as$\mu'$goes over all ways of removing a box from the Young diagram for$\mu$(while staying in the world of Young diagrams). 2. The induction of$S^{\mu}$to$S_{n+1}$is isomorphic to the direct sum of the$S_{n+1}$-representations$S^{\mu'}$as$\mu'$goes over all ways of adding a box to the Young diagram for$\mu$(again while staying in the world of Young diagrams). These two rules are equivalent by Frobenius reciprocity. There is a nice concrete proof of the restriction rule (I believe that it is due to Peel, though I first learned about it from James's book "The representation theory of the symmetric groups"). Assume that the rows of the Young diagram for$\mu$from which we can remove a box are$r_1 < \ldots < r_k$, and denote by$\mu_i$the partition of$n-1$obtained by removing a box from the$i^{\text{th}}$row of$\mu$. There is then a sequence $$0=V_0 \subset V_1 \subset \cdots \subset V_k = S^{\mu}$$ of$S_{n-1}$-modules such that$V_i/V_{i-1} \cong S^{\mu_i}$. In fact, recalling that$S^{\mu}$is generated by elements corresponding to Young tableaux of shape$\mu$(known as polytabloids), the module$V_i$is the subspace spanned by the polytabloids in which$n$appears in a row between$1$and$i\$.

Question : Is there a similarly concrete proof of the induction rule (in particular, a proof which does not appeal to Frobenius reciprocity and the restriction rule)?