Abstract Nonsense

Crushing one theorem at a time

The Hook-length Formula


Point of Post: In this post we derive the hook-length which well tell us, given a frame, the number of standard Young tableaux that have that frame.

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Motivation

This is the big theorem that we discussed in our last post that will give us, using the hook-lengths of a frame, the number of standard Young tableaux with that frame. Consequently, as was previously mentioned this will also give us the degree of the irrep \rho^{(\mathcal{F})} for S_n. The idea of the proof is simple, we induct on the size of the frames (how many blocks it contains) and then use the relation between the number of standard Young tableaux on a frame and the number of standard Young tableaux on the subordinate frames to use our induction hypothesis in which we will use our so-called contrived lemma.

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The Hook-Length Formula

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We dive right into the proof:

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Theorem: Let \mathcal{F}=(m_1,\cdots,m_r) (r is not fixed)  be a n-frame and, as usual, f_{\text{st}}\left(\mathcal{F}\right) the number of standard Young tableaux \mathcal{T} with \text{Frame}\left(\mathcal{T}\right)=\mathcal{F}. Then,

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\displaystyle f_{\text{st}}\left(\mathcal{F}\right)=n! \frac{\displaystyle \prod_{1\leqslant i<j\leqslant r}\left(h_{i,1}-h_{j,1}\right)}{\displaystyle \prod_{i=1}^{r}(h_{i,1})!}=\frac{n!}{\displaystyle \prod_{(i,j)\in\mathcal{F}}h_{i,j}}

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(where we take the empty product to be 1) where h_{i,j} is the hook-length and the second equivalence follows from previous work.

Proof: We proceed by induction on the size of the frame. Evidently since \begin{array}{|c|}\cline{1-1}\hphantom{\times}\\ \cline{1-1}\end{array} is the only one frame and with our convention on the empty product we see that

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\displaystyle f_{\text{st}}\left(\begin{array}{|c|}\cline{1-1}\hphantom{\times}\\ \cline{1-1}\end{array}\right)=1=\frac{1!}{1}

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So, suppose the result is true for all n-frames and let \mathcal{F}=(m_1,\cdots,m_r) be a n+1-frame. We use our relations between f_{\text{st}} and subordinate frames to write

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\displaystyle f_{\text{st}}\left(\mathcal{F}\right)=\sum_{\mathcal{G}\triangleleft\mathcal{F}}f_{\text{st}}\left(\mathcal{G}\right)\quad\mathbf{(1)}

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For each end block (k,m_k) for k=1,\cdots,r let  \mathcal{F}-(k,m_k) denote the collection of boxes obtained as, and so the sum is really taken over all j such that \mathcal{F}-(k,m_k) is still a frame. Thus, we may rewrite it as

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\displaystyle f_{\text{st}}\left(\mathcal{F}\right)=\sum_{k\in[r]\text{ s.t. }\mathcal{F}-(k,m_k)\in \text{Frame}_n}f_{\text{st}}\left(\mathcal{F}-(k,m_k)\right)

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but, by our induction hypothesis we may replace each of these by the supposed rational function of hook-lengths. Note though that if \mathcal{F}-(k,m_k) is not a Ferrer’s diagram for some k then by definition (clearly) we must have that (k,m_k) is not a bottom right corner and so clearly m_k=m_{k+1} and so it’s easy to see by definition that if \widetilde{h_{k,1}}-\widetilde{h_{k+1,1}} denotes the difference of the hook lengths for \mathcal{F}-(k,m_k) then \widetilde{h_{k,1}}-\widetilde{h_{k+1,1}}=0 and so the formula in the induction hypothesis is zero. Consequently, we may rewrite \mathbf{(1)} as

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\displaystyle \sum_{k=1}^{r}n!\frac{\displaystyle \prod_{i=1}^{r^{(k)}}(h^{(k)}_{i,1})!}{\displaystyle \prod_{1\leqslant i<j\leqslant r^{(k)}}\left(h^{(k)}_{i,1}-h^{(k)}_{j,1}\right)}

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where evidently r^{(k)} the number of rows in \mathcal{F}-(k,m_k) and h^{(k)}_{t,1} the hook-length of the square (t,1)\in\mathcal{F}-(k,m_k). Now, using the fact that r^{(k)}=r for all but k=r in which case r^{(r)}=r-1 and the fact that h^{(k)}_{i,1}=m^{(k)}_i+r^{(k)}-i (in conjunction with the fact that m^{(k)}_i=m_i if k\ne i and m^{(i)}_i=m_i-1) one can how (although it is admittedly slightly tedious) that this can be rewritten

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\displaystyle (n-1)!\frac{\displaystyle \prod_{1\leqslant i<j\leqslant r}(h_{i,1}-h_{j,1})}{\displaystyle \prod_{i=1}^{r}(h_{i,1})!}\sum_{k=1}^{r}h_{i,k}\prod_{i\ne k}\left(1+\frac{1}{h_{k,1}-h_{i,1}}\right)

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but, by our contrived lemma we know that

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\displaystyle \sum_{k=1}^{r}h_{k,1}\prod_{i\ne k}\left(1+\frac{1}{h_{k,1}-h_{i,1}}\right)=\sum_{k=1}^{r}h_{k,1}-\frac{r(r-1)}{2}=n

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and thus the induction is complete. \blacksquare

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References:

1. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.

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May 14, 2011 - Posted by | Algebra, Algebraic Combinatorics, Representation Theory | , , , , , , , ,

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