question of macdonald'book "symmetric functions and hall polynomials' on page 3

Question

For each partition λ,define n(λ)=$\sum_{i\ge1}(i-1)λ_{i}$.And it can induce that n(λ)=$\sum_{i\ge1}\binom{λ'_{i}}{2}$(λ'is the cnjugate of λ),I tried to prove it but failed.Any hint or relevent reference is welcome.Thanks for your help.

Answer 1

for partition 5,5,4,4,2,1 fill a grid as indicated:

0   0   0   0   0
1   1   1   1   1
2   2   2   2   
3   3   3   3   
4   4           
5   

the row sums are $\lambda_i (i-1)$ and the column sums are $\sum_{j=0,\lambda'_j -1} j$ or $\lambda'_j (\lambda'_{j} -1)/2$
or, 5*0+5*1+4*2+4*3+2*4+1*5 = 6*5/2+5*4/2+4*3/2+4*3/2+2*1/2

The aim is to teach you to look at partitions as 2-dim structures.

Answer 2

The argument by Wouter M. should immediately convince your right brain hemisphere of the truth of this equation. In case your left hemisphere wants something more formal, try the following.

The result is obviously true for the (empty) partition of$~0$. Otherwise assume the result by induction for smaller partitions (in the sense of $|\lambda|$). There is some nonzero part of $\lambda$, and some $k$ such that $\lambda_k$ is the last part of its size. Then by decreasing $\lambda_k$ by$~1$ one finds a smaller partition$~\mu$ (as usual a part that has become $0$ is interpreted as no part at all). By induction $n(\mu)=\sum_j\binom{\mu'_j}2$. Also $n(\lambda)=\sum_i(i-1)\lambda_i$ and since $\lambda_i=\mu_i+\delta_{i,k}$ this gives $n(\lambda)=n(\mu)+(k-1)$. Similarly $\sum_j\binom{\lambda'_j}2=\sum_{j\neq\lambda_k}\binom{\mu'_j}2+\binom k2$ since $\lambda'_j=\mu'_j+\delta_{j,\lambda_k}$ and $\lambda'_{\lambda_k}=k$ (as $\lambda_k$ was the last part of its size); also one has $\mu'_{\lambda_k}=k-1$ and $\binom k2=k-1+\binom{k-1}2$, so that one gets $\sum_j\binom{\lambda'_j}2=k-1+\sum_j\binom{\mu'_j}2$. Finally $n(\lambda)=\sum_j\binom{\lambda'_j}2$ as desired.