= Solution
The printed identity has a sign error. With the defined <Kostant partition function> and the exponent $e^{-\nu}$, the cancelling factors must be $1-e^{-\alpha}$. This is already visible for $SU(2)$ with <positive root> $\alpha$: the formal series $\sum_{k\geq0}e^{-k\alpha}$ multiplied by $1-e^\alpha$ is $-e^\alpha$, not $1$.
For the corrected formula, a partition of $\nu$ is a tuple of <nonnegative integers> $(k_\alpha)_{\alpha>0}$ satisfying $\nu=\sum_{\alpha>0}k_\alpha\alpha$. The order of summands is not counted. Multiplying the formal <geometric series> yields
$$
\prod_{\alpha>0}\left(\sum_{k\geq0}e^{-k\alpha}\right)
=\sum_\nu p(\nu)e^{-\nu}.
$$
This product is coefficientwise meaningful. Choose a <linear functional> strictly positive on every <positive root>. For a fixed $\nu$ it bounds each $k_\alpha$, so only finitely many tuples contribute. We are working in the completion supported on the negative positive-root cone, not claiming convergence of an ordinary <Fourier series> on the <torus>.
Each <geometric series> cancels its factor $1-e^{-\alpha}$, proving
$$
\boxed{\left(\sum_\nu p(\nu)e^{-\nu}\right)\prod_{\alpha>0}(1-e^{-\alpha})=1.}
$$
Equivalently, reversing all the exponential signs gives $(\sum_\nu p(\nu)e^{\nu})\prod_{\alpha>0}(1-e^\alpha)=1$ in the opposite completion. More generally, if $N$ is the number of <positive roots>, the expression literally printed in the paper equals
$$
(-1)^N e^{2\rho},
$$
since $\prod_{\alpha>0}(1-e^\alpha)=(-1)^N e^{2\rho}\prod_{\alpha>0}(1-e^{-\alpha})$. Thus the correction is substantive, not merely a choice of Fourier convention.
Back to article page
