# The Unapologetic Mathematician

## Intertwinors from Semistandard Tableaux Span, part 1

Now that we’ve shown the intertwinors that come from semistandard tableaux are independent, we want to show that they span the space $\hom(S^\lambda,M^\mu)$. This is a bit fidgety, but should somewhat resemble the way we showed that standard polytabloids span Specht modules.

So, let $\theta\in\hom(S^\lambda,M^\mu)$ be any intertwinor, and write out the image

$\displaystyle\theta(e_t)=\sum\limits_Tc_TT$

Here we’re implicitly using the fact that $\mathbb{C}[T_{\lambda\mu}]\cong M^\mu$.

First of all, I say that if $\pi\in C_t$ and $T_1=\pi T_2$, then the coefficients of $T_1$ and $T_2$ differ by a factor of $\mathrm{sgn}(\pi)$. Indeed, we calculate

$\displaystyle\pi\left(\theta(e_t)\right)=\theta\left(\pi(\kappa_t\{t\})\right)=\theta(\mathrm{sgn}\kappa_t\{t\})=\mathrm{sgn}(\pi)\theta(e_t)$

This tells us that

$\displaystyle\pi\sum\limits_Tc_TT=\mathrm{sgn}(\pi)\sum\limits_Tc_TT$

Comparing coefficients on the left and right gives us our assertion.

As an immediate corollary to this lemma, we conclude that if $T$ has a repetition in some column, then $c_T=0$. Indeed, we can let $\pi$ be the permutation that swaps the places of these two identical entries. Then $T=\pi T$, while the previous result tells us that $c_T=\mathrm{sgn}c_T=-c_T$, and so $c_T=0$.

February 11, 2011

## Independence of Intertwinors from Semistandard Tableaux

Let’s start with the semistandard generalized tableaux $T\in T_{\lambda\mu}^0$ and use them to construct intertwinors $\bar{\theta}_T:\hom(S^\lambda,M^\mu)$. I say that this collection is linearly independent.

Indeed, let’s index the semistandard generalized tableaux as $T_1,\dots,T_m$. We will take our reference tableau $t$ and show that the vectors $\bar{\theta}_{T_i}(e_t)\in M^\mu$ are independent. This will show that the $\bar{\theta}_{T_i}$ are independent, since any linear dependence between the operators would immediately give a linear dependence between the $\bar{\theta}_{T_i}(v)$ for all $v\in S^\lambda$.

Anyway, we have

$\displaystyle\bar{\theta}_{T_i}(e_t)=\theta_{T_i}\left(\kappa_t\{t\}\right)=\kappa_t\theta_{T_i}(\{t\})$

Since we assumed $T_i$ to be semistandard, we know that $[T_i]\triangleright[S]$ for all summands $S\in\theta_{T_i}(\{t\})$. Now the permutations in $\kappa_t$ do not change column equivalence classes, so this still holds: $[T_i]\triangleright[S]$ for all summands $S\in\kappa_t\theta_{T_i}(\{t\})$. And further all the $[T_i]$ are distinct since no column equivalence class can contain more than one semistandard tableau.

But now we can go back to the lemma we used when showing that the standard polytabloids were independent! The $\kappa_t\theta_{T_i}(\{t\})=\bar{\theta}(e_t)$ are a collection of vectors in $M^\mu$. For each one, we can pick a basis vector $[T_i]$ which is maximum among all those having a nonzero coefficient in the vector, and these selected maximum basis elements are all distinct. We conclude that our collection of vectors in independent, and then it follows that the intertwinors $\bar{\theta}_{T_i}$ are independent.

February 9, 2011

## Dominance for Generalized Tabloids

Sorry I forgot to post this yesterday afternoon.

You could probably have predicted this: we’re going to have orders on generalized tabloids analogous to the dominance and column dominance orders for tabloids without repetitions. Each tabloid (or column tabloid) gives a sequence of compositions, and at the $i$th step we throw in all the entries with value $i$.

For example, the generalized column tabloid

$\displaystyle[S]=\begin{array}{|c|c|c|}2&1&1\\3&2&\multicolumn{1}{c}{}\end{array}$

gives the sequence of compositions

\displaystyle\begin{aligned}\lambda^1&=(0,1,1)\\\lambda^2&=(1,2,1)\\\lambda^3&=(2,2,1)\end{aligned}

while the semistandard generalized column tabloid

$\displaystyle[T]=\begin{array}{|c|c|c|}1&1&1\\2&3&\multicolumn{1}{c}{}\end{array}$

gives the sequence of compositions

\displaystyle\begin{aligned}\mu^1&=(1,1,1)\\\mu^2&=(2,1,1)\\\mu^3&=(2,2,1)\end{aligned}

and we find that $[S]\trianglelefteq[T]$ since $\lambda^i\trianglelefteq\mu^i$ for all $i$.

We of course have a dominance lemma: if $k, $k$ occurs in a column to the left of $l$ in $T$, and $S$ is obtained from $T$ by swapping these two entries, then $[T]\triangleright[S]$. As an immediate corollary, we find that if $T$ is semistandard and $S\in\{T\}$ is different from $T$, then $[T]\triangleright[S]$. That is, $[T]$ is the "largest" (in the dominance order) equivalence class in $\theta_T{t}$. The proofs of these facts are almost exactly as they were before.

February 9, 2011

## Semistandard Generalized Tableaux

We want to take our intertwinors and restrict them to the Specht modules. If the generalized tableau $T$ has shape $\lambda$ and content $\mu$, we get an intertwinor $\bar{\theta}_T\in\hom(S^\lambda,M^\mu)$. This will eventually be useful, since the dimension of this hom-space is the multiplicity of $S^\lambda$ in $M^\mu$.

Anyway, if $t$ is our standard “reference” tableau, then we can calculate

\displaystyle\begin{aligned}\bar{\theta}_T(e_t)&=\bar{\theta}_T(\kappa_t\{t\})\\&=\kappa_t\theta_T(\{t\})\\&=\kappa_t\left(\sum\limits_{S\in\{T\}}S\right)\\&=\sum\limits_{S\in\{T\}}\kappa_t(S)\end{aligned}

We can see that it will be useful to know when $\kappa_t(S)=0$. It turns out this happens if and only if $S$ has two equal elements in some column.

Indeed, if $\kappa_t(S)=0$, then

$\displaystyle S+\sum\limits_{\substack{\pi\in C_t\\\pi\neq e}}\mathrm{sgn}(\pi)\pi S=0$

Thus for some $\sigma\in C_t$ with $\mathrm{sgn}(\sigma)=-1$ we must have $S=\sigma S$. But then we must have all the elements in each cycle of $\sigma$ the same, and these cycles are restricted to the columns. Since $\sigma$ is not the identity, we have at least one nontrivial cycle and at least two elements the same.

On the other hand, assume $S(i)=S(j)$ in the same column of $S$. Then $[e-(i\,j)](S)=0$. But then the sign lemma tells us that $(e-(i\,j))$ is a factor of $\kappa_t$, and thus $\kappa_t(S)=0$.

This means that we can eliminate some intertwinors $\bar{\theta}_T$ from consideration by only working with things like standard tableaux. We say that a generalized tableau is semistandard if its columns strictly increase (as for standard tableaux) and its rows weakly increase. That is, we allow repetitions along the rows, but only so long as we never have any row descents. The tableau

$\displaystyle\begin{array}{ccc}1&1&2\\2&3&\end{array}$

is semistandard, but

$\displaystyle\begin{array}{ccc}2&1&1\\3&2&\end{array}$

is not.

February 8, 2011

## Intertwinors from Generalized Tableaux

Given any generalized Young tableau $T$ with shape $\lambda$ and content $\mu$, we can construct an intertwinor $\theta_T:M^\lambda\to M^\mu$. Actually, we’ll actually go from $M^\lambda$ to $\mathbb{C}[T_{\lambda\mu}$, but since we’ve seen that this is isomorphic to $M^\mu$, it’s good enough. Anyway, first, we have to define the row-equivalence class $\{T\}$ and column-equivalence class $[T]$. These are the same as for regular tableaux.

So, let $t$ be our reference tableau and let $\{t\}$ be the associated tabloid. We define

$\displaystyle\theta_T\left(\{t\}\right)=\sum\limits_{S\in\{T\}}S$

Continuing our example, with

$\displaystyle T=\begin{array}{ccc}2&1&1\\3&2&\end{array}$

we we define

\displaystyle\begin{aligned}\theta_T\left(\begin{array}{ccc}\cline{1-3}1&2&3\\\cline{1-3}4&5&\\\cline{1-2}\end{array}\right)&=\begin{array}{ccc}2&1&1\\3&2&\end{array}+\begin{array}{ccc}1&2&1\\3&2&\end{array}+\begin{array}{ccc}1&1&2\\3&2&\end{array}\\&+\begin{array}{ccc}2&1&1\\2&3&\end{array}+\begin{array}{ccc}1&2&1\\2&3&\end{array}+\begin{array}{ccc}1&1&2\\2&3&\end{array}\end{aligned}

Now, we extend in the only way possible. The module $M^\lambda$ is cyclic, meaning that it can be generated by a single element and the action of $\mathbb{C}[S_n]$. In fact, any single tabloid will do as a generator, and in particular $\{t\}$ generates $M^\lambda$.

So, any other module element in $M^\lambda$ is of the form $\pi\{t\}$ for some $\pi\in\mathbb{C}[S_n]$. And so if $\theta_T$ is to be an intertwinor we must define

$\displaystyle\theta_T\left(\pi\{t\}\right)=\pi\theta_T(\{t\})=\sum\limits_{S\in\{T\}}\pi S$

Remember here that $\pi$ acts on generalized tableaux by shuffling the entries by place, not by value. Thus in our example we find

\displaystyle\begin{aligned}\theta_T\left(\begin{array}{ccc}\cline{1-3}2&4&3\\\cline{1-3}1&5&\\\cline{1-2}\end{array}\right)&=\theta_T\left((1\,2\,4)\begin{array}{ccc}\cline{1-3}1&2&3\\\cline{1-3}4&5&\\\cline{1-2}\end{array}\right)\\&=(1\,2\,4)\theta_T\left(\begin{array}{ccc}\cline{1-3}1&2&3\\\cline{1-3}4&5&\\\cline{1-2}\end{array}\right)\\&=(1\,2\,4)\begin{array}{ccc}2&1&1\\3&2&\end{array}+(1\,2\,4)\begin{array}{ccc}1&2&1\\3&2&\end{array}+(1\,2\,4)\begin{array}{ccc}1&1&2\\3&2&\end{array}\\&+(1\,2\,4)\begin{array}{ccc}2&1&1\\2&3&\end{array}+(1\,2\,4)\begin{array}{ccc}1&2&1\\2&3&\end{array}+(1\,2\,4)\begin{array}{ccc}1&1&2\\2&3&\end{array}\\&=\begin{array}{ccc}3&2&1\\1&2&\end{array}+\begin{array}{ccc}3&1&1\\2&2&\end{array}+\begin{array}{ccc}3&1&2\\1&2&\end{array}\\&+\begin{array}{ccc}2&2&1\\1&3&\end{array}+\begin{array}{ccc}2&1&1\\2&3&\end{array}+\begin{array}{ccc}2&1&2\\1&3&\end{array}\end{aligned}

Now it shouldn’t be a surprise that since so much of our construction to this point has depended on an aribtrary choice of a reference tableau $t$, the linear combination of generalized tableaux on the right doesn’t quite seem like it comes from the tabloid on the left. But this is okay. Just relax and go with it.

February 5, 2011

## Modules of Generalized Young Tableaux

We can obviously create vector spaces out of generalized Young tableaux. Given the collection $T_{\lambda\mu}$ of tableaux of shape $\lambda$ and content $\mu$, we get the vector space $\mathbb{C}[T_{\lambda\mu}]$. We want to turn this into an $S_n$-module.

First, given any tabloid $\{s\}$ of shape $\mu$, we can product a (generalized) tableau $T\in T_{\lambda\mu}$ by defining $T(i)$ to be the number of the row in $s$ that contains the entry $i$. As an example, consider the tabloid

$\displaystyle\{s\}=\begin{array}{cc}\cline{1-2}2&3\\\cline{1-2}1&5\\\cline{1-2}4&\\\cline{1-1}\end{array}$

This gives us the function $T(2)=T(3)=1$, $T(1)=T(5)=2$, and $T(4)=3$. If $\lambda=(3,2)$ and we use the usual reference tableau $t$, this gives us the generalized tabloid

$\displaystyle T=\begin{array}{ccc}T(1)&T(2)&T(3)\\T(4)&T(5)&\end{array}=\begin{array}{ccc}2&1&1\\3&2&\end{array}$

The shape of $T$ is obviously $\lambda$, and it’s easy to see that the content is exactly $\mu$. Indeed, there are $\mu_i$ entries in $T$ with the value $i$, just as there are $\mu_i$ entries in the first row of $\{s\}$.

It should also be clear that this correspondence is a bijection. That is, given any generalized tableau $T$ of shape $\lambda$ and content $\mu$ we can get a tabloid of shape $\mu$ by turning $T$ into a function and then putting $k$ on row $i$ of $\{s\}$ if $T(k)=i$.

That means that the basis of generalized tableaux $T_{\lambda\mu}$ of the vector space $\mathbb{C}[T_{\lambda\mu}]$ is in bijection with the basis of $\mu$-tabloids of the vector space $M^\mu$. And this space carries an action of $S_n$ — the linear extension of the action on tabloids. We want to pull this action across the bijection we just set up to get an action on $T_{\lambda\mu}$.

On the one hand, this is as easy as saying it: if $T$ corresponds to $\{s\}$, we define $\pi T$ to be the generalized tableau corresponding to $\pi\{s\}$ and we’re done. To be a bit more explicit, we define $\pi T$ by considering it as a function and setting

$\displaystyle\left[\pi T\right](i)=T(\pi^{-1}i)$

So, for example, if

$\displaystyle T=\begin{array}{ccc}T(1)&T(2)&T(3)\\T(4)&T(5)&\end{array}$

then we can calculate

$\displaystyle (1\,2\,4)T=\begin{array}{ccc}T(4)&T(1)&T(3)\\T(2)&T(5)&\end{array}$

Even more explicitly, if

$\displaystyle T=\begin{array}{ccc}2&1&1\\3&2&\end{array}$

then we calculate

$\displaystyle (1\,2\,4)T=\begin{array}{ccc}3&2&1\\1&2&\end{array}$

We should be clear about a major distinction here: the permutation $\pi\in S_n$ acts on the entries in $\{s\}$ — replacing $i$ by $\pi i$ — but it acts on the places in $T$ — moving $T(i)$ to the position of $T(\pi i)$.

If we write the correspondence as $\theta(\{s\})=T$, then for $\theta$ to be an intertwinor we need $\theta(\pi\{s\})=\pi T$. This forces

\displaystyle\begin{aligned}\left[\pi T\right](i)&=\text{row number of }i\text{ in }\pi\{s\}\\&=\text{row number of }\pi^{-1}i\text{ in }\{s\}\\&=T(\pi^{-1}i)\end{aligned}

and so this explicit action is forced on us.

The really interesting thing is that when we use this action on the generalized tableaux in $T_{\lambda\mu}$, we always get a module $\mathbb{C}[T_{\lambda\mu}]\cong M^\mu$, no matter what shape $\lambda$ we start with.

February 3, 2011

## Generalized Young Tableaux

And now we have another generalization of Young tableaux. These are the same, except now we allow repetitions of the entries.

Explicitly, a generalized Young tableau $T$ — we write them with capital letters — of shape $\lambda$ is an array obtained by replacing the points of the Ferrers diagram of $\lambda$ with positive integers. Any skipped or repeated numbers are fine. We say that the “content” of $T$ is the composition $\mu=(\mu_1,\dots,\mu_m)$ where $\mu_i$ is the number of $i$ entries in $T$.

As an example, we have the generalized Young tableau

$\displaystyle\begin{array}{ccc}4&1&4\\1&3&\end{array}$

of shape $(3,2)$ and content $(2,0,1,2)$.

Notice that if $\lambda\vdash n$, then $\mu\vdash n$ as well, since both count up the total number of places in the tableau. Given a partition $\lambda$ and a composition $\mu$, both decomposing the same number $n$, we define $T_{\lambda\mu}$ to be the collection of generalized Young tableaux of shape $\lambda$ and content $\mu$. All the tableaux we’ve considered up until now have content $(1,\dots,1)=(1^n)$.

Now, pick some fixed (ungeneralized) tableau $t$. We can use the same one we usually do, numbering the rows from $1$ to $n$ across each row and from top to bottom, but it doesn’t really matter which we use. For our examples we’ll pick

$\displaystyle t=\begin{array}{ccc}1&2&3\\4&5&\end{array}$

Using this “reference” tableau, we can rewrite any generalized tableau as a function; define $T(i)$ to be the entry of $T$ in the same place as $i$ is in $t$. That is, any generalized tableau looks like

$\displaystyle\begin{array}{ccc}T(1)&T(2)&T(3)\\T(4)&T(5)&\end{array}$

and in our particular example above we have $T(1)=T(3)=4$, $T(2)=T(4)=1$, and $T(5)=3$. Conversely, any such function assigning a positive integer to each number from $1$ to $n$ can be interpreted as a generalized Young tableau. Of course the particular correspondence depends on exactly which reference tableau we use, but there will always be some such correspondence between functions and generalized tableaux.

February 2, 2011

## The Branching Rule, Part 4

What? More!?

Well, we got the idea to look for the branching rule by trying to categorify a certain combinatorial relation. So let’s take the flip side of the branching rule and decategorify it to see what it says!

Strictly speaking, decategorification means passing from a category to its set of isomorphism classes. That is, in the case of our categories of $S_n$-modules we should go from the Specht module $S^\lambda$ to its character $\chi^\lambda$. And that is an interesting question, but since the original relation turned into the dimensions of the modules in the branching rule, let’s do the same thing in reverse.

So the flip side of the branching rule tells us how a Specht module decomposes after being induced up to the next larger symmetric group. That is:

$\displaystyle S^\lambda\!\!\uparrow_{S_n}^{S_{n+1}}\cong\bigoplus\limits_{\lambda^+}S^{\lambda^+}$

To “decategorify”, we take dimensions

\displaystyle\begin{aligned}\dim\left(S^\lambda\!\!\uparrow_{S_n}^{S_{n+1}}\right)&=\dim\left(\bigoplus\limits_{\lambda^+}S^{\lambda^+}\right)\\&=\sum\limits_{\lambda^+}\dim\left(S^{\lambda^+}\right)\\&=\sum\limits_{\lambda^+}f^{\lambda^+}\end{aligned}

As we see, the right hand side is obvious, but he left takes a little more work. We consult the definition of induction to find

$\displaystyle S^\lambda\!\!\uparrow_{S_n}^{S_{n+1}}=\mathbb{C}\left[S_{n+1}\right]\otimes_{S_n}S^\lambda$

Calculating its dimension is a little easier if we recall what induction looks like for matrix representations: the direct sum of a bunch of copies of $S^\lambda$, one for each element in a transversal of the subgroup. That is, there are

$\displaystyle\lvert S_{n+1}/S_n\rvert=\lvert S_{n+1}\rvert/\lvert S_n\rvert=\frac{(n+1)!}{n!}=n+1$

copies of $S^\lambda$ in the induced representation. We find our new relation:

$\displaystyle(n+1)f^\lambda=\sum\limits_{\lambda^+}f^{\lambda^+}$

That is, the sum of the numbers of standard tableaux of all the shapes we get by adding an outer corner to $\lambda$ is $n+1$ times the number of standard tableaux of shape $\lambda$.

This is actually a sort of surprising result, and there should be some sort of combinatorial proof of it. I’ll admit, though, that I don’t know of one offhand. If anyone can point me to a good one, I’d be glad to post it.

February 1, 2011

## The Branching Rule, Part 3

“Part 3”? Didn’t we just finish proving the branching rule? Well, yes, but there’s another part we haven’t mentioned yet. Not only does the branching rule tell us how representations of $S_n$ decompose when they’re restricted to $S_{n-1}$, it also tells us how representations of $S_{n-1}$ decompose when they’re induced to $S_n$.

Now that we have the first statement of the branching rule down, proving the other one is fairly straightforward: it’s a consequence of Frobenius reciprocity. Indeed, the branching rule tells us that

$\displaystyle\dim\left(\hom_{S_{n-1}}(S^\mu,S^\lambda\!\!\downarrow)\right)=\bigg\{\begin{array}{cl}1&\mu=\lambda^-\\{0}&\mathrm{otherwise}\end{array}$

That is, there is one copy of $S^\mu$ inside $S^\lambda$ (considered as an $S_{n-1}$-module) if $\mu$ comes from $\lambda$ by removing an inner corner, and there are no copies otherwise.

So let’s try to calculate the multiplicity of $S^\lambda$ in the induced module $S^\mu\!\!\uparrow$:

\displaystyle\begin{aligned}\hom_{S_n}(S^\lambda,S^\mu\!\!\uparrow)^*&\cong\hom_{S_n}(S^\mu\!\!\uparrow,S^\lambda)\\&\cong\hom_{S_{n-1}}(S^\mu,S^\lambda\!\!\downarrow)\end{aligned}

Taking dimensions, we find

\displaystyle\begin{aligned}\dim\left(\hom_{S_n}(S^\lambda,S^\mu\!\!\uparrow)\right)&=\dim\left(\hom_{S_{n-1}}(S^\mu,S^\lambda\!\!\downarrow)\right)\\&=\bigg\{\begin{array}{cl}1&\mu=\lambda^-\\{0}&\mathrm{otherwise}\end{array}\\&=\bigg\{\begin{array}{cl}1&\lambda=\mu^+\\{0}&\mathrm{otherwise}\end{array}\end{aligned}

since if $\mu$ comes from $\lambda$ by removing an inner corner, then $\lambda$ comes from $\mu$ by adding an outer corner.

We conclude that

$\displaystyle S^\mu\!\!\uparrow_{S_{n-1}}^{S_n}\cong\bigoplus\limits_{\mu^+}S^{\mu^+}$

which is the other half of the branching rule.

January 31, 2011

## 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