AI 中文总结
本文研究半单 Lie 代数支配权偏序集上的原子数非负性，证明当 Dynkin 图为路径时原子数非负且等于显式权空间维数，并刻画了所有不可约类型中非负的充要条件。
AI 中文摘要
设 $\mathfrak{g}$ 为复半单 Lie 代数，对于支配整权 $\lambda$，令 $\Theta_\lambda$ 为不可约模 $V(\lambda)$ 的权（按重数不计）之和。$\Theta_\mu$ 构成 $W$-不变量的 $\mathbb{Z}$-基，因此存在唯一整数 $a(\mu,\lambda)$（称为原子数）使得 $\mathrm{ch}\\,V(\lambda) = \sum_\mu a(\mu,\lambda)\Theta_\mu$（对支配 $\mu$ 求和）。我们研究这些原子数何时非负。它们是支配权偏序集上权重数函数的 Moebius 变换，通过 crosscut 定理可化为至多 $2^{\mathrm{rank}\\,\Phi}$ 个重数（由 $[\mu,\lambda]$ 中 $\mu$ 的覆盖子集索引）的交替和。我们证明：当 Dynkin 图的每个连通分支均为路径时，$a(\mu,\lambda)\ge 0$ 恒成立，并将该系数等同于一个显式权空间的维数：该空间由 $V(\lambda)$ 沿路径依次交替取升算子的核与降算子的余核（每个覆盖对应一个）而得到。类型 $D_4$ 表明该假设是必要的。令 $\beta = \alpha_1 + 2\alpha_2 + 2\alpha_3 + 2\alpha_4$；对于每个满足 $\langle\mu,\alpha_2^\vee\rangle = 1$ 且 $\mu+\beta$ 支配的支配 $\mu$，$a(\mu,\mu+\beta)$ 在 $\langle\mu,\alpha_1^\vee\rangle = 0$ 时为 $-2$，否则为 $-1$。限制到 $\lambda-\mu$ 的支撑集可将此族带入每个具有三叉节点的不可约类型。因此，一个不可约有限根系的所有原子数非负当且仅当其 Dynkin 图为路径，即类型 $A_n$、$B_n$、$C_n$、$F_4$、$G_2$。在支配腔深处的一个显式范围内，$a(\mu,\lambda)$ 等于 $\lambda-\mu$ 关于非单正根的 Kostant 配分数；因此负原子数仅限于边界薄层。
英文摘要
Let $\mathfrak{g}$ be a complex semisimple Lie algebra and, for a dominant integral weight $λ$, let $Θ_λ$ be the multiplicity-free sum of the weights of the irreducible module $V(λ)$. The $Θ_μ$ form a $\mathbb{Z}$-basis of the $W$-invariants, so there are unique integers $a(μ,λ)$, the atomic numbers, with $\mathrm{ch}\,V(λ) = \sum_μa(μ,λ)Θ_μ$ over dominant $μ$. We ask when they are nonnegative. They are the Moebius transform of the weight-multiplicity function on the dominant-weight poset, which the crosscut theorem turns into an alternating sum of at most $2^{\mathrm{rank}\,Φ}$ multiplicities indexed by subsets of the covers of $μ$ in $[μ,λ]$. We prove that $a(μ,λ)\ge 0$ whenever every connected component of the Dynkin diagram is a path, and identify the coefficient with the dimension of an explicit weight space: it is cut out of $V(λ)$ by alternately taking kernels of raising operators and cokernels of lowering operators, one per cover, in order along the path. Type $D_4$ shows the hypothesis is necessary. Put $β= α_1 + 2α_2 + 2α_3 + 2α_4$; for every dominant $μ$ with $\langleμ,α_2^\vee\rangle = 1$ and $μ+β$ dominant, $a(μ,μ+β)$ is $-2$ if $\langleμ,α_1^\vee\rangle = 0$ and $-1$ otherwise. Restriction to the support of $λ-μ$ carries this family into every irreducible type with a trivalent node. Thus an irreducible finite root system has all atomic numbers nonnegative exactly when its Dynkin diagram is a path, namely in types $A_n$, $B_n$, $C_n$, $F_4$, $G_2$. Deep in the dominant chamber, in an explicit range, $a(μ,λ)$ is the Kostant partition number of $λ-μ$ for the nonsimple positive roots; negative atomic numbers are therefore confined to boundary slabs.
Comments47 pages, 2 figures
