this is my first post on math overflow so I hope it goes well. I believe I have a fairly simple bijective proof for Stanley's Hook-Content Formula in the case of hook shapes. I wanted to see if anyone more experienced could try poking a hole in it for me, or validating it.

Let $\lambda = (1+r,1,\dots,1)$ be a partition of $d = r+c+1$, and $n$ a positive integer (greater than $c$). We will begin by constructing a filling of the Young diagram of $\lambda$ by an unusual totally ordered set. Start by selecting a number from 1 to $n$ to place in the top left corner. Work your way down the column of $\lambda$ placing distinct numbers in each position. So far we have selected one of $(n)_{c+1}$ possible fillings of the column.

Now we fill the row. From left to right choose a number from 1 to $n$ to place in each position in the row (no longer distinct). Additionally, if the number chosen agrees with any of the other numbers appearing in the row, choose a comparison between them. For example, if the number 1 already appears twice in the row, and you chose 1, you must decide if this is the smallest 1, the middle 1, or the greatest 1. It is easy to verify that, once complete, we have chosen from $(n+r)_r$ possible fillings of the row. So in total we have chosen from $(n+r)_{d}$ possible fillings of $\lambda$.

Time for the action. Select the smallest entry from the filling (breaking ties along the row by the order defined above, and breaking row-column ties by choosing the entry in the column) and swap it with the entry in the top left corner. This is a $d$-to-1 mapping and can easily be reversed given a position to return to. Now sort the other $r$ entries in the row according to the total ordering, and sort the $d$ entries in the column which are distinct. You have a Semistandard Young Tableau!

Final Result: $$ \frac{(n+r)_d}{r!c!d}. $$

Let me know what you think. If a simpler bijection is out there, please point me to it. This would be very useful to me. I really think this kind of thinking could lead to a simple bijective proof for the general result.

  • $\begingroup$ Hook shapes are very special shapes; they are in general much easier to handle compared to a general tableau. $\endgroup$ – Per Alexandersson Jun 7 '14 at 9:28

For the hook shapes the following bijection becomes especially simple.

Christian Krattenthaler, Another involution principle-free bijective proof of Stanley's hook-content formula, see here.


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