Let $\mathfrak{g}$ be a complex simple Lie algebra, and $W$ its Weyl group. We make a choice of simple roots, and note $\Phi_+$ the corresponding set of positive roots, while $\Phi$ is the set of all roots. Finally, I denote $z^{\lambda}$ the character associated to a weight $\lambda$.

I claim that $$\sum\limits_{w \in W} \prod\limits_{\alpha \in \Phi_+} (1-z^{w(\alpha)}) = \prod\limits_{\alpha \in \Phi}(1-z^{\alpha}) \, . $$

My questions are:

  • Where can I find this formula (I would particularly appreciate a textbook) ? Does it have a specific name ?
  • If the first question can't be answered, how would you prove the formula ?

I don't have a reference but one can deduce it from the denominator identity $$ \prod_{\alpha>0}(1-z^\alpha)=\sum_w(-1)^{\ell(w)}z^{\rho-w\rho}. $$ With $\beta=w\alpha$ write $$ \prod_{\alpha>0}(1-z^{w\alpha})=\prod_{\beta>0,w^{-1}\beta>0}(1-z^\beta) \prod_{\beta<0,w^{-1}\beta>0}(1-z^\beta) $$ The second factor equals (with $\beta=-\gamma$) $$ \prod_{\gamma>0,w^{-1}\gamma<0}(1-z^{-\gamma})=\prod_{\gamma>0,w^{-1}\gamma<0}(-z^{-\gamma})\prod_{\gamma>0,w^{-1}\gamma<0}(1-z^\gamma) $$ The first factor is easily identified as $(-1)^{\ell(w)}z^{w\rho-\rho}$ where $\rho=\frac12\sum_{\alpha>0}\alpha$. Combining everything, we get $$ \prod_{\alpha>0}(1-z^{w\alpha})=(-1)^{\ell(w)}z^{w\rho-\rho}\prod_{\gamma>0}(1-z^\gamma) $$ Now sum over $W$. Then the formula follows from the denominator formula (with $z^{-1}$ instead of $z$).

  • $\begingroup$ It is very elegant how you make appear the $(-1)^{\ell (w)}$ using the factorization in your third equation. This is the step I was missing, thank you for your answer! $\endgroup$
    – Antoine
    May 29 '17 at 8:44

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