# Definition of the weight lattice for a nonreduced root system

Let $$(V,\Phi)$$ be a root system with dual root system $$(V^{\ast},\Phi^{\vee})$$. Let $$\Delta = \{\alpha_1, ... , \alpha_n\}$$ be a set of simple roots for $$V$$, and let $$\Delta^{\vee} = \{\alpha_1^{\vee}, ... , \alpha_n^{\vee}\}$$ be the coroots corresponding to $$\Delta$$.

We have the fundamental weights $$\hat{\Delta} = \{ \omega_1, ... , \omega_n\}$$, which is by definition the dual basis of $$\Delta^{\vee}$$, and we have the fundamental coweights $$\hat{\Delta}^{\vee} = \{ \omega_1^{\vee}, ... , \omega_n^{\vee}\}$$, which is by definition the dual basis of $$\Delta$$.

When $$\Phi$$ is reduced, the lattices $$\{v \in V : \langle v, \alpha^{\vee} \rangle \in \mathbb Z \textrm{ for all } \alpha^{\vee} \in \Phi^{\vee} \}$$ and $$\mathbb Z \omega_1 \oplus \cdots \mathbb Z \omega_n$$ are equal, and this is called the weight lattice. But for $$\Phi$$ not necessarily reduced, the second lattice is in general a proper subset of the first. This is because although $$\Delta$$ is a base of $$\Phi$$, $$\Delta^{\vee}$$ is not in general a base of $$\Phi^{\vee}$$ (since if $$\alpha \in \Delta$$ is such that $$2\alpha$$ is a root, then $$\alpha^{\vee}$$ is divisible by $$(2\alpha)^{\vee}$$).

How should the weight lattice be defined in the nonreduced case? Should we even bother calling the $$\omega_i$$ "fundamental weights?" I know that $$\hat{\Delta}$$ and $$\hat{\Delta}^{\vee}$$ are still dealt with in the nonreduced case, for example in the proof of Langlands classification here (section 2.2).

A possible alternative around this is to define $$\Delta^{\vee}$$ not as the set of coroots $$\alpha^{\vee}$$ for $$\alpha \in \Delta$$, but rather as the set of simple roots of $$\Phi^{\vee}$$ corresponding to same chamber as $$\Delta$$.