Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 19 / 6

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.

Use the corrected generating series of the Kostant partition function and the permitted Weyl character formula:For a fixed $w$, the coefficient of $e^{\mu }$ arises exactly whenThere are only finitely many Weyl-group terms, and the partition function vanishes outside the cone of nonnegative root combinations. Coefficient extraction is therefore legitimate in the formal completion used in the preceding part. Summing the signed contributions provesThis proves the Kostant multiplicity formula for every compact connected Lie group, not just the unitary groups. In the presence of a central torus, roots have zero central component, so the partition function automatically gives zero for any incompatible central weight. Also $\rho$ need not itself be an integral weight for the given global group: $w\rho -\rho$ always belongs to the root lattice, so every argument in the formula is nevertheless integral whenever $\lambda$ and $\mu$ are integral group weights.

As a normalization check, for $SU(2)$ write a highest weight as the nonnegative integer $\ell$, with positive root $2$ and $\rho =1$. The two Weyl-group terms giveSince $p(a)=1$ precisely when $a$ is even and nonnegative, this is one for $\mu =\ell ,\ell -2,\dots ,-\ell$ and zero otherwise, including the cancellation below the lowest weight. This recovers the full familiar weight string.