For natural numbers $a_1,\dots,a_n$, Freeman Dyson conjectured (and it was eventually proven) that the Laurent polynomial $$ \prod_{i,j=1\atop i\neq j}^n\left(1-\frac{x_i}{x_j}\right)^{a_i} $$ has constant term the multinomial coefficient $\binom{a_1+\cdots+a_n}{a_1,\dots,a_n}$.

A.C. Dixon proved Dixon's Identity: $$ \sum_{k=-a}^a(-1)^k\binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k}=\frac{(a+b+c)!}{a!b!c!}. $$

This last quantity is just the multinomial coefficient $\binom{a+b+c}{a,b,c}$. So letting $a_1=a,a_2=b,a_3=c$, it should be the case by Dyson's conjecture that $$ \prod_{i,j=1\atop i\neq j}^3\left(1-\frac{x_i}{x_j}\right)^{a_i} $$ that is, $$ \left(1-\frac{x_1}{x_2}\right)^a\left(1-\frac{x_1}{x_3}\right)^a\left(1-\frac{x_2}{x_1}\right)^b\left(1-\frac{x_2}{x_3}\right)^b\left(1-\frac{x_3}{x_1}\right)^c\left(1-\frac{x_3}{x_2}\right)^c $$ has constant term $\binom{a+b+c}{a,b,c}$. Does anyone see a clever way to conclude that the constant term is also calculated as $ \sum_{k=-a}^a(-1)^k\binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k}$ to get another proof of the result? Many thanks.

My guess is that one will want to "choose" say the same number of factors say $-\frac{x_1}{x_2}$ from $(1-\frac{x_1}{x_2})^a$ as the number of factors $-\frac{x_2}{x_1}$ from $(1-\frac{x_2}{x_1})^b$, when expanding. So I assume the sum fro $-a$ to $a$ will count how many ways there are to choose a carefully from the six terms in the product to get a constant when it's all multiplied out.


1 Answer 1


Setting $a_1=a$, $a_2=b$, $a_3=c$, we can start with $ \prod_{1\le i,j\le 3, \ i\ne j}\left(1-\frac{x_i}{x_j}\right)^{a_i} $ and combine pairs of factors involving the same pair of variables to find that $$ \prod_{1\le i,j\le 3, \ i\ne j}\left(1-\frac{x_i}{x_j}\right)^{a_i} $$ $$ = (-1)^{a+b+c} x_1^{a-c} x_2^{b-a} x_3^{c-b} (1-\frac{x_2}{x_1})^{a+b} (1-\frac{x_3}{x_2})^{b+c} (1-\frac{x_1}{x_3})^{c+a}. \qquad (1) $$ Using the binomial theorem, we can expand (1) as $$ \sum_{i,j,\ell} (-1)^{a+b+c+i+j+\ell} {a+b\choose i} {b+c\choose j} {c+a\choose \ell} x_1^{\ell-i+a-c} x_2^{i-j+b-a} x_3^{j-\ell+c-b}.\qquad (2) $$ To find the constant term in (2), we need to look at all terms where $$ \ell-i+a-c=0,\ i-j+b-a=0,\ j-\ell+c-b=0.\qquad (3) $$ If we set $i:=a+k$ in (3), we immediately get $\ell=c+k$ and $j=b+k$ from the first and second equations. These choices of $i$, $j$ and $\ell$ also satisfy the third equation. Therefore, the constant term in (2) is $$ \sum_k (-1)^{2a+2b+2c+3k} {a+b\choose a+k} {b+c\choose b+k} {c+a\choose c+k}, $$ where the sum is extended over all $k$ where the binomial coefficients are nonzero. Since $(-1)^{2a+2b+2c+3k}=(-1)^k$, we can now use Dixon's identity and prove that the constant term of (1) is $\left({a+b+c\atop a\ \ \ b\ \ \ c}\right)$.

  • $\begingroup$ Wonderful, thank you! $\endgroup$
    – Buble
    Feb 6, 2012 at 19:04

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