The <dominance order> is
$$
\mu\preceq\lambda
\quad\Longleftrightarrow\quad
\lambda-\mu=\sum_i n_i\alpha_i
\quad(n_i\in\mathbb Z_{\geq0}).
$$
Suppose $\lambda\ne\mu$. Put $\delta=\lambda-\mu$. Since $\mu$ is dominant,
$$
(\lambda,\delta)=(\mu,\delta)+(\delta,\delta)>0.
$$
Writing $\delta=\sum_i n_i\alpha_i$, some index with $n_i>0$ therefore satisfies $(\lambda,\alpha_i)>0$, equivalently $\langle\lambda,\alpha_i^\vee\rangle>0$. The $\mathfrak{sl}_2$ lowering operator
$$
f_i:V_\lambda\longrightarrow V_{\lambda-\alpha_i}
$$
is injective by the <Injectivity of sl2 lowering above weight zero>. Hence
$$
\operatorname{mult}(\lambda)
\leq\operatorname{mult}(\lambda-\alpha_i).
$$
The new weight still dominates $\mu$ in the partial order. Iterating until reaching $\mu$ gives
$$
\boxed{\operatorname{mult}(\mu)\geq\operatorname{mult}(\lambda)}.
$$
This is the <weight multiplicity decreases away from a dominant weight> property.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 102 / 5 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-102/5/c/solution