The Unapologetic Mathematician

Mathematics for the interested outsider

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<i. 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 ns 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


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 - Posted by | Algebra, Representation Theory, Representations of Symmetric Groups


  1. […] 3″? Didn’t we just finish proving the branching rule? Well, yes, but there’s another part we haven’t mentioned yet. Not […]

    Pingback by The Branching Rule, Part 3 « The Unapologetic Mathematician | January 31, 2011 | Reply

  2. Hi, I really like your proof where everything is explained so well. But I dont get why the V^(i) as you defined them here are S^(n-1) modules. I tried showing it and the actual point missing is that the map from S_{n-1} x V^(i) gives results in V^(i). I think i.e. linear combinations of the polytabloids you used for V^(i) do still have n in one of the required rows after a permutation of S_{n-1} was applied. But how can this be true for an arbitrary permutation?

    Comment by pani puri | December 4, 2013 | Reply

  3. Sorry, it’s been a long time since I looked at this (almost three years now), so I might be a little rusty. I think the problem might be that I’m asserting that the quotient space V^{(i)}/V^{(i-1)} is an $S_{n-1}$-module, not that V^{(i)} itself is. Does that help?

    Comment by John Armstrong | December 4, 2013 | Reply

  4. Hi, thanks a lot for your answer! But I think there I have the same problem since now I need that n stays in the same row…

    Comment by pani puri | December 9, 2013 | Reply

  5. What I wrote was wrong. Now I need that all n stay in their line or are in a higher one by considering only polytabloids where the n in t is in one special line.

    Comment by pani puri | December 10, 2013 | Reply

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