Root order on weights

Assume the Axiom of Choice. Let $\mathfrak{g}$ be a finite-dimensional complex semisimple Lie algebra with Cartan subalgebra $\mathfrak{h}$ and root system $\mathrm{\Phi }$, and let ${\mathrm{\Phi }}^{+}$ be a chosen positive system with base $\mathrm{\Delta }=\{{\alpha }_1,\dots ,{\alpha }_r\}$ of simple roots (Positive systems and simple roots, The roots form a reduced crystallographic Euclidean root system). By Simple roots form a signed integral basis $\mathrm{\Delta }$ is a basis of $E={\mathrm{span}}_{\mathbb{R}}\mathrm{\Phi }$, so every element of $E$ has unique real coefficients in this basis; the root lattice $Q$ consists of the integral combinations of $\mathrm{\Delta }$ (Root, coroot, weight, and coweight lattices). Put $Q_{+}:=\{{\sum }_{i=1}^r{n}_i{\alpha }_i:n_i\in {\mathbb{Z}}_{\ge 0}\}\subseteq Q\mathrm{.}$

For $\lambda ,\mu \in {\mathfrak{h}}^{\ast }$ write $\mu \le \lambda$ when $\lambda -\mu \in Q_{+}$; in words, when $\lambda -\mu$ is a nonnegative integral combination of the chosen simple roots. This is the root order on ${\mathfrak{h}}^{\ast }$. In particular $\mu \le \lambda$ forces $\lambda -\mu \in E$, so comparable functionals lie in the same affine coset of $E$ in ${\mathfrak{h}}^{\ast }$; neither functional need itself lie in $E$.

The root order is a partial order. Reflexivity holds with $n_i=0$. Antisymmetry: if $\mu \le \lambda$ and $\lambda \le \mu$, then ${\sum }_i(m_i+n_i){\alpha }_i=0$ with $m_i,n_i\ge 0$, and linear independence of the simple roots forces $m_i+n_i=0$, so $m_i=n_i=0$ and $\lambda =\mu$. Transitivity: if $\nu \le \mu$ and $\mu \le \lambda$, then $\lambda -\nu =(\lambda -\mu )+(\mu -\nu )$ is a sum of two elements of $Q_{+}$, hence lies in $Q_{+}$ and $\nu \le \lambda$. Thus $\le$ is a partial order on ${\mathfrak{h}}^{\ast }$; restricted to any set of weights it is a partial order on that set, and every comparison chain of weights is finite whenever the weight set is finite, because a strict increase adds a nonzero element of $Q_{+}$.