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Given is a partition $\lambda$, and the $\lambda$ also denote it's Young Diagram, and $\lambda'$ is the conjugate/transpose. Then the hook-length of $\lambda$ at $x = (i,j)$ is defined to be $$h(x) = h(i,j) = \lambda_i + \lambda'_j - i - j + 1$$We also have $$n(\lambda)=\sum_{i\geq 1} (i-1)\lambda_i=\sum_{i\geq 1}\binom{\lambda'_i}{2},\ \text{and}$$ $$|\lambda|=\sum_{i\geq 1}\lambda_i$$

I have to show that the sum of the hook-lengths of $\lambda$ is $$\sum_{x \in \lambda}h(x)=n(\lambda) + n(\lambda')+|\lambda|$$

My attempt: $$\sum_{x \in \lambda}h(x)=\sum_{i \geq 1}\sum_{j = 1}^{\lambda_i}h(i,j)=\sum_{i \geq 1}\sum_{j = 1}^{\lambda_i}(\lambda_i + \lambda'_j - i - j + 1)\\=\Big(\sum_{i \geq 1}\sum_{j = 1}^{\lambda_i}\lambda_j'\Big)+\sum_{i \geq 1}\Big(\lambda_i^{2}-i\lambda_i-\frac{\lambda_i(\lambda_i + 1)}{2}+\lambda_i\Big) \\ =\Big(\sum_{i \geq 1}\sum_{j = 1}^{\lambda_i}\lambda_j'\Big)+\sum_{i \geq 1}\bigg(\frac{\lambda_i(\lambda_i - 1)}{2}-(i-1)\lambda_i\bigg)\\=\Big(\sum_{i \geq 1}\sum_{j = 1}^{\lambda_i}\lambda_j'\Big)+n(\lambda') - n(\lambda)$$

I'm not really sure how to evaluate the first term of the final expression, and I'm getting a minus instead of a plus in the latter 2 terms; and hence I'm stuck. Can I get some kind of assistance?

Naweed G. Seldon
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  • Would you mind sharing how you found the first assumption? Why are the two formulas for n($\lambda$) equal? – simples123 Mar 26 '18 at 19:36
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    We see that $n(\lambda)$, from the first formula, is the sum of the numbers obtained by attaching a 0 to each node in the top row of the diagram of $\lambda$, a 1 to each node in the second row, and so on. Now, if we add the numbers in each column, we see that each column sum is $0 + 1 + \cdots + \lambda'_i = \binom{\lambda'_i}{2}$, summing all the columns gives us the second formula. – Naweed G. Seldon Mar 26 '18 at 19:51

1 Answers1

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$\lambda_j'$ is the number of parts in $\lambda$ that are greater than $j$. You can see that $$\sum_i\sum_{j=1}^{\lambda_i}\lambda'_j=\sum_{j \geq 1}(\lambda'_j)^2=\sum_{j \geq 1}\left(2\frac{\lambda'_j(\lambda'_j-1)}{2}+\lambda'_j\right)=2n(\lambda)+|\lambda|.$$

In the first step we have reversed the order of the sums by asking how many times each $\lambda_j'$ will appear. The answer is once for each $i$ such that $\lambda_i\ge j$ and this is precisely $\lambda_j'$ itself.

Naweed G. Seldon
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thedude
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  • Aren't we counting more than $|\lambda|$ though? Because $i=1$, just gives me $|\lambda|$, and then we get more for $i=2,3,\cdots,l(\lambda)$ – Naweed G. Seldon Feb 05 '18 at 14:06
  • For example, $\lambda = (3 \ 1 \ 1) = \lambda'$. I have my sum as $\sum_{i=1,2,3}\sum_{j=1}^{\lambda_i}\lambda'_j = (\lambda_1 + \lambda_2 + \lambda_3) + \lambda'_1 + \lambda'_1 = |\lambda| + 2 \cdot \lambda'_1$ – Naweed G. Seldon Feb 05 '18 at 14:12
  • Sorry, I have corrected the answer – thedude Feb 05 '18 at 14:31
  • Ah! Thank you! Could you explain me the first step a bit? Combinatorics isn't really my field. – Naweed G. Seldon Feb 05 '18 at 14:32
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    We are trying to reverse the order of the sums. How many $\lambda'_1$ do we have? As many as the parts of $\lambda$ greater than 1. And how many $\lambda'_2$? As many as the parts of $\lambda$ greater than 2. Each $\lambda'_j$ appears $\lambda'_j$ times. – thedude Feb 05 '18 at 14:38
  • @junkquill you are welcome to accept the answer if you are satisfied – thedude Feb 05 '18 at 14:53