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 ?



A corollary to Weyl's formula on roots of Lie algebras
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 $z$).


