Suppose $\lambda\vdash n$ is a partition and $S^\lambda$ is the associated irreducible representation of $S\_n$. As we know from the branching rule, we have an isomorphism of $S\_{n-1}$ modules
$$S^\lambda\cong\bigoplus\_{\mu}S^{\mu}$$
where $\mu$ are the partitions obtained by deleting an outer corner from $\lambda$.


In particular, I am concerned with the case when $S^\lambda$ is spanned by basis elements $\{B\_Q\mid Q\in\text{SYT}(\lambda)\}$ of the Kazhdan-Lusztig basis (with the relevant left-action of the generators). Can the isomorphism from the branching rule be realised in such a way that preserves the KL-basis in some form? That is, if $\phi$ is a realisation of the isomorphism, then
$$\phi\cdot B\_Q=\sum\_{\mu}\sum\_{P\in\text{SYT}(\mu)}z\_pB\_P$$
for some constants $z\_P$. What can we say about these constants $z\_P$?


For example, when $\lambda$ is a rectangular partition (and so $S^\lambda\cong S^\mu$), then the isomorphism can be realised as $B\_Q\mapsto B\_{d(Q)}$, where $d(Q)$ is the tableau obtained by deleting the $n$-box from $Q$. I am unsure as to the theory for arbitrary partitions.


Any help would be appreciated!



Branching rule for Specht modules over Kazhdan-Lusztig basis
It turns out that this is indeed true. The relevant proofs are given in Chapter 3 of [this extended abstract](https://arxiv.org/abs/2111.09510).


