## 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.

*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 for . 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.

*The Hook-Length Formula*

We dive right into the proof:

**Theorem: ***Let ( is not fixed) be a -frame and, as usual, the number of standard Young tableaux with . Then,*

*(where we take the empty product to be ) where 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 is the only one frame and with our convention on the empty product we see that

So, suppose the result is true for all -frames and let be a -frame. We use our relations between and subordinate frames to write

For each end block for let denote the collection of boxes obtained as, and so the sum is really taken over all such that is still a frame. Thus, we may rewrite it as

but, by our induction hypothesis we may replace each of these by the supposed rational function of hook-lengths. Note though that if is not a Ferrer’s diagram for some then by definition (clearly) we must have that is not a bottom right corner and so clearly and so it’s easy to see by definition that if denotes the difference of the hook lengths for then and so the formula in the induction hypothesis is zero. Consequently, we may rewrite as

where evidently the number of rows in and the hook-length of the square . Now, using the fact that for all but in which case and the fact that (in conjunction with the fact that if and ) one can how (although it is admittedly slightly tedious) that this can be rewritten

but, by our contrived lemma we know that

and thus the induction is complete.

**References:**

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

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