[proofplan] We define the product $D:=e^\rho\prod_{\alpha\in\Phi^+}(1-e^{-\alpha})$ and prove that it is anti-invariant under every simple reflection. Since the simple reflections generate $W$, this makes $D$ a $W$-anti-invariant element of the weight group algebra. We then use the standard Weyl-polytope uniqueness principle for anti-invariants: an anti-invariant supported in the denominator polytope $\operatorname{conv}(W\rho)$ is determined by its coefficient at the dominant regular vertex $\rho$. The product $D$ has coefficient $1$ at $e^\rho$, so it must equal the alternating orbit sum $A_\rho$. [/proofplan] [step:Define the denominator product and the Weyl group action on it] Let \begin{align*} D := e^\rho\prod_{\alpha\in\Phi^+}(1-e^{-\alpha}) \end{align*} be the denominator product in the integral group algebra of the weight lattice. The Weyl group acts on formal exponentials by \begin{align*} w(e^\lambda) := e^{w\lambda} \end{align*} for $w\in W$ and weights $\lambda$, and this action extends multiplicatively and linearly to the group algebra. For a simple root $\alpha_i$, let $s_i\in W$ denote the corresponding simple reflection. We will prove that \begin{align*} s_iD=-D \end{align*} for every simple reflection $s_i$. [/step] [step:Show that each simple reflection changes the denominator product by a sign] Fix a simple root $\alpha_i$. We use the standard root-system facts that \begin{align*} s_i(\rho)=\rho-\alpha_i \end{align*} and that $s_i$ sends $\alpha_i$ to $-\alpha_i$ while permuting $\Phi^+\setminus\{\alpha_i\}$. Therefore \begin{align*} s_iD=e^{\rho-\alpha_i}(1-e^{\alpha_i})\prod_{\alpha\in\Phi^+\setminus\{\alpha_i\}}(1-e^{-s_i\alpha}). \end{align*} Since $s_i$ permutes $\Phi^+\setminus\{\alpha_i\}$, the remaining product is unchanged after reindexing: \begin{align*} \prod_{\alpha\in\Phi^+\setminus\{\alpha_i\}}(1-e^{-s_i\alpha})=\prod_{\alpha\in\Phi^+\setminus\{\alpha_i\}}(1-e^{-\alpha}). \end{align*} Finally, \begin{align*} e^{\rho-\alpha_i}(1-e^{\alpha_i})=-e^\rho(1-e^{-\alpha_i}). \end{align*} Combining these identities gives \begin{align*} s_iD=-e^\rho\prod_{\alpha\in\Phi^+}(1-e^{-\alpha})=-D. \end{align*} [guided] Fix a simple root $\alpha_i$ and let $s_i$ be the corresponding reflection. The goal is to compute $s_iD$ explicitly and see exactly where the minus sign appears. By definition, \begin{align*} D=e^\rho\prod_{\alpha\in\Phi^+}(1-e^{-\alpha}). \end{align*} Applying $s_i$ to each formal exponential gives \begin{align*} s_iD=e^{s_i\rho}\prod_{\alpha\in\Phi^+}(1-e^{-s_i\alpha}). \end{align*} Now we use two standard facts about simple reflections. First, \begin{align*} s_i\rho=\rho-\alpha_i. \end{align*} Second, $s_i(\alpha_i)=-\alpha_i$, and $s_i$ permutes the positive roots other than $\alpha_i$. Thus the single factor corresponding to $\alpha_i$ becomes \begin{align*} 1-e^{-s_i\alpha_i}=1-e^{\alpha_i}, \end{align*} while the remaining factors are merely reindexed. Therefore \begin{align*} s_iD=e^{\rho-\alpha_i}(1-e^{\alpha_i})\prod_{\alpha\in\Phi^+\setminus\{\alpha_i\}}(1-e^{-\alpha}). \end{align*} The sign is contained in the elementary formal-exponential identity \begin{align*} e^{\rho-\alpha_i}(1-e^{\alpha_i})=e^{\rho-\alpha_i}-e^\rho=-e^\rho(1-e^{-\alpha_i}). \end{align*} Substituting this into the previous expression gives \begin{align*} s_iD=-e^\rho(1-e^{-\alpha_i})\prod_{\alpha\in\Phi^+\setminus\{\alpha_i\}}(1-e^{-\alpha}). \end{align*} This is exactly \begin{align*} s_iD=-D. \end{align*} So every simple reflection acts on $D$ by multiplication by $-1$. [/guided] [/step] [step:Extend anti-invariance from simple reflections to the whole Weyl group] The simple reflections generate $W$, and each simple reflection has determinant $-1$. If $w=s_{i_1}\cdots s_{i_m}$ is any expression of $w$ as a product of simple reflections, then repeated application of the previous step gives \begin{align*} wD=(-1)^mD. \end{align*} Since $\det(w)=(-1)^m$ for any such expression, we obtain \begin{align*} wD=\det(w)D \end{align*} for every $w\in W$. Thus $D$ is $W$-anti-invariant. [/step] [step:Record the top coefficient and the support of the product] Expanding the product gives \begin{align*} D=\sum_{S\subseteq\Phi^+}(-1)^{|S|}e^{\rho-\sum_{\alpha\in S}\alpha}. \end{align*} The subset $S=\varnothing$ contributes the term $e^\rho$ with coefficient $1$. Every exponent appearing in this expansion has the form \begin{align*} \rho-\beta \end{align*} where $\beta$ is a sum of distinct positive roots. Equivalently, the support of $D$ lies in the standard denominator polytope \begin{align*} \operatorname{conv}(W\rho). \end{align*} This last containment is the usual Weyl denominator polytope fact: the weights $\rho-\sum_{\alpha\in S}\alpha$, for $S\subseteq\Phi^+$, are precisely contained in the convex hull of the Weyl orbit of $\rho$. [/step] [step:Use anti-invariant uniqueness in the denominator polytope] We use the following standard anti-invariant uniqueness principle. If $F$ is a $W$-anti-invariant element of the weight group algebra, if the support of $F$ is contained in $\operatorname{conv}(W\rho)$, and if the coefficient of $e^\rho$ in $F$ is $c$, then \begin{align*} F=cA_\rho. \end{align*} Indeed, decompose the support of $F$ into $W$-orbits. A singular orbit contributes nothing to an anti-invariant element: if $s\in W$ is a reflection fixing a weight $\lambda$, then anti-invariance gives \begin{align*} a_\lambda=-a_\lambda, \end{align*} so $a_\lambda=0$. Thus only regular orbits can contribute. Let $\eta$ be the dominant representative of a regular orbit appearing in the support. Since $\eta\in\operatorname{conv}(W\rho)$, the Weyl-polytope criterion gives \begin{align*} \rho-\eta\in Q_+, \end{align*} where $Q_+$ is the nonnegative integral span of the positive simple roots. Since $\eta$ is dominant and regular integral, $\eta-\rho$ is dominant. If $\eta\ne\rho$, then $\eta-\rho$ is a nonzero dominant weight, while $\rho-\eta\in Q_+$. With the $W$-invariant [inner product](/page/Inner%20Product), this would imply \begin{align*} 0<|\eta-\rho|^2=-(\rho-\eta,\eta-\rho)\le 0, \end{align*} because every simple root has nonnegative inner product with a dominant weight. This contradiction shows that $\eta=\rho$. Hence the only regular orbit that can appear is $W\rho$. On that orbit anti-invariance forces the coefficient of $e^{w\rho}$ to be $\det(w)$ times the coefficient of $e^\rho$. Therefore $F=cA_\rho$. [/step] [step:Apply uniqueness to the denominator product] The element $D$ is $W$-anti-invariant by the previous steps, its support lies in $\operatorname{conv}(W\rho)$, and the coefficient of $e^\rho$ in $D$ is $1$. Applying the anti-invariant uniqueness principle with $F=D$ and $c=1$ gives \begin{align*} D=A_\rho. \end{align*} Substituting the definition of $D$ yields \begin{align*} e^\rho\prod_{\alpha\in\Phi^+}(1-e^{-\alpha})=\sum_{w\in W}\det(w)e^{w\rho}. \end{align*} This is precisely the Weyl denominator formula. [/step]