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I am reading Macdonlad's book on "symmetric functions and Hall polynomials" and I have difficulty figuring out an identity which involves hook-lengths. I would like to ask for a hint.

Let $\lambda=(\lambda_1,\dots, \lambda_k)$ be a partition and define $\mu_i=\lambda_i+k-i$, for $1\leq i\leq k$. The hook-length of a box $x\in \lambda$ in the Young diagram is denoted by $h(x)$. One can show the following identity: $$ \sum_{x\in \lambda}t^{h(x)}+\sum_{1\leq i<j\leq k}t^{\mu_i-\mu_j}=\sum_{i=1}^k\sum_{j=1}^{\mu_i}t^j. $$ From this identity, Macdonald concludes the following identity which I am not able to verify: $$ \prod_{x\in \lambda}\left(1-t^{h(x)}\right)\cdot \prod_{1\leq i<j\leq k}\left(1-t^{\mu_i-\mu_j}\right)=\prod_{i=1}^k\prod_{j=1}^{\mu_i}\left(1-t^j\right) $$ How can I conclude the second identity from the first one?

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The first identity says that $$\{h(x) : x \in \lambda\} \cup \{\mu_i - \mu_j : 1 \leq i < j \leq k\} = \bigcup_{i=1}^k \{j : 1 \leq j \leq \mu_i\}$$ as multisets. The second identity then follows because the multisets of exponents are the same on the LHS and RHS.

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  • $\begingroup$ The first identity is fine. The question is how to obtain the second identity from the first one. $\endgroup$
    – MO B
    Commented Jul 26, 2017 at 16:32
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    $\begingroup$ @MO B if two multisets coincide, then tĥe products over them (of any function, in particular of $1-t^k$) coincide too. $\endgroup$
    – Fedor Petrov
    Commented Jul 26, 2017 at 16:47
  • $\begingroup$ Answer edited to be more explicit. $\endgroup$
    – John Machacek
    Commented Jul 26, 2017 at 16:55
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    $\begingroup$ Alternatively, you can get the second identity from the first one by applying the monoid homomorphism $\left(\mathbb{N}\left[t\right], +\right) \to \left(\mathbb{Z}\left[t\right], \cdot\right)$ sending each $t^i$ to $1-t^i$. (This is well-defined, since $\left(\mathbb{N}\left[t\right], +\right)$ is a free abelian monoid with basis $\left(t^0,t^1,t^2,\ldots\right)$.) $\endgroup$
    – darij grinberg
    Commented Jul 26, 2017 at 18:42
  • $\begingroup$ Sorry, it is absolutely obvious. Thanks. $\endgroup$
    – MO B
    Commented Jul 27, 2017 at 14:56

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