For a dominant integral weight $\lambda$, the <Weyl character formula> is
$$
\operatorname{ch}V(\lambda)=
\frac{\sum_{w\in W}(-1)^{\ell(w)}e^{w(\lambda+\rho)}}
{\sum_{w\in W}(-1)^{\ell(w)}e^{w\rho}},
$$
where $W$ is the <Weyl group>, $\ell$ is <Coxeter length>, and $\rho$ is the <half-sum of positive roots>. The <Weyl denominator formula> is
$$
\sum_{w\in W}(-1)^{\ell(w)}e^{w\rho}
=e^\rho\prod_{\alpha\in\Phi^+}(1-e^{-\alpha}).
$$

Set $\lambda=k\rho$. Apply the denominator identity after replacing every formal exponential $e^\mu$ by $e^{(k+1)\mu}$:
$$
\sum_w(-1)^{\ell(w)}e^{w((k+1)\rho)}
=e^{(k+1)\rho}
\prod_{\alpha\in\Phi^+}(1-e^{-(k+1)\alpha}).
$$
Dividing this by the ordinary denominator gives
$$
\begin{aligned}
\operatorname{ch}V(k\rho)
&=e^{k\rho}\prod_{\alpha\in\Phi^+}
\frac{1-e^{-(k+1)\alpha}}{1-e^{-\alpha}}\\
&=\boxed{e^{k\rho}\prod_{\alpha\in\Phi^+}
(1+e^{-\alpha}+\cdots+e^{-k\alpha})}.
\end{aligned}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 102 / 5 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-102/5/b/solution