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 $\Delta$.