The Unapologetic Mathematician

The Branching Rule, Part 2

We pick up our proof of the branching rule. We have a partition $\lambda$ with inner corners in rows $r_1,\dots,r_k$. The partitions we get by removing each of the inner corner $r_i$ is $\lambda^i$. If the tableau $t$ (or the tabloid $\{t\}$ has its $n$ in row $i$, then $t^i$ (or $\{t^i\}$) is the result of removing that $n$.

We’re looking for a chain of subspaces

$\displaystyle0=V^{(0)}\subseteq V^{(1)}\subseteq\dots\subseteq V^{(k)}=S^\lambda$

such that $V^{(i)}/V^{(i-1)}=S^{\lambda^i}$ as $S_{n-1}$-modules. I say that we can define $V^{(i)}$ to be the subspace of $S^\lambda$ spanned by the standard polytabloids $e_t$ where the $n$ shows up in row $r_i$ or above in $t$.

For each $i$, define the map $\theta_i:M^\lambda\to M^{\lambda^i}$ by removing an $n$ in row $r_i$. That is, if $\{t\}\in$latex M^\lambda\$ has its $n$ in row $r_i$, set $\theta_i(\{t\})=\{t^i\}$; otherwise set $\theta_i(\{t\})=0$. These are all homomorphisms of $S_{n-1}$-modules, since the action of $S_{n-1}$ always leaves the $n$ in the same row, and so it commutes with removing an $n$ from row $r_i$.

Similarly, I say that $\theta_i(e_t)=e_{t^i}$ if $n$ is in row $r_i$ of $t$, and we get $\theta_i(e_t)=0$ if it’s in row $r_j$ with $j. Indeed if $n$ shows up above row $r_i$, then since it’s the bottommost entry in its column that column can have no entries at all in row $r_i$. Thus as we use $C_t$ to shuffle the columns, all of the tabloids that show up in $e_t=C_t^-\{t\}$ will be sent to zero by $\theta_i$. Similar considerations show that if $n$ is in row $r_i$, then of all the tabloids that show up in $e_t$, only those leaving $n$ in that row are not sent to zero by $\theta_i$. The permutations in $C_t$ leaving $n$ fixed are, of course, exactly those in $C_{t^i}^-$, and our assertion holds.

Now, since each standard polytabloid $e_{t^i}\in S^{\lambda^i}$ comes from some polytabloid $e_t\in M^{\lambda}$, we see they’re all in the image of $\theta_i$. Further, these $e_t$ all have their $n$s in row $r_i$, so they’re all in $V^{(i)}$. That is, $\theta_i(V^{(i)})=S^{\lambda^i}$. On the other hand, if $t$ has its $n$ above row $r_i$, then $\theta(e_t)=0$, and so $V^{(i-1)}\subseteq\mathrm{Ker}(\theta_i)$.

So now we’ve got a longer chain of subspaces:

$\displaystyle0=V^{(0)}\subseteq V^{(1)}\cap\mathrm{Ker}(\theta_1)\subseteq V^{(1)}\subseteq\dots\subseteq V^{(k)}\cap\mathrm{Ker}(\theta_k)\subseteq V^{(k)}=S^\lambda$

But we also know that

$\displaystyle\dim\left(\frac{V^{(i)}}{V^{(i)}\cap\mathrm{Ker}(\theta_i)}\right)=\dim(\theta_i(V^{(i)}))=\dim(S^{\lambda^i})=f^{\lambda^i}$

So the steps from $V^{(i)}\cap\mathrm{Ker}(\theta_i)$ to $V^{(i)}$ give us all the $f^{\lambda^i}$ as we add up dimensions. Comparing to the formula we’re categorifying, we see that this accounts for all of $f^\lambda=\dim(S^\lambda)$. And so there are no dimensions left for the steps from $V^{(i-1)}$ to $V^{(i)}\cap\mathrm{Ker}(\theta_i)$, and these containments must actually be equalities!

$\displaystyle V^{(i-1)}=V^{(i)}\cap\mathrm{Ker}(\theta_i)$

And thus

$\displaystyle \frac{V^{(i)}}{V^{(i-1)}}=\frac{V^{(i)}}{V^{(i)}\cap\mathrm{Ker}(\theta_i)}\cong S^{\lambda^i}$

as asserted. The branching rule then follows.

January 28, 2011