Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 19 / 6 / 1 / 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 ${\sum }_{k\ge 0}{e}^{-k\alpha }$ 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 yieldsThis 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 $1-e^{-\alpha }$, provingEquivalently, 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 equalssince ${\prod }_{\alpha >0}(1-e^{\alpha })=(-1{)}^N{e}^{2\rho }{\prod }_{\alpha >0}(1-e^{-\alpha })$. Thus the correction is substantive, not merely a choice of Fourier convention.