Being an integral weight can be checked on a base.

Question

Let $\Phi$ be a root system and $\Delta$ a base for a finite dimensional semi-simple complex Lie algebra. For $\lambda \in H^*$ we say that $\lambda $ is integral if $\frac{2(\lambda,\alpha)}{(\alpha,\alpha)}\in \mathbb{Z}$. Now my question is why we can check this condition on $\Delta$. Here is my approach so far: By definition of a base we can write $\alpha=\sum_{i=0}^nk_i \alpha_i$ wit $k_i\in \mathbb{Z}$. This transforms our condition into $$\frac{\sum_{i=0}^nk_i 2(\lambda,\alpha_i)}{\sum_{i,j=0}^nk_ik_j(\alpha_i,\alpha_j)}\in \mathbb{Z}.$$ Now we know that $\frac{2(\alpha_i,\alpha_j)}{(\alpha_i,\alpha_i)}\in \mathbb{Z}$. I feel like i should get somewhere with this but im a bit stuck here. Thanks for any hints or answers in advance!

Answer 1

Use the fact that every root $\alpha$ is conjugated to a simple root $\alpha_i \in \Delta$ by a product of simple reflections $\sigma = \prod_i \sigma_{\alpha_i} \in W$ where $W$ is the Weyl group of $\Phi$. That means $\sigma(\alpha) = \alpha_i \in \Delta$.

Now define $\langle \lambda, \alpha \rangle:= \frac{2 (\lambda, \alpha)}{(\alpha, \alpha)}$. This function is invariant under elements of the Weyl group. That means with the chosen $\sigma$ from above $\langle \lambda, \alpha\rangle = \langle \sigma(\lambda), \sigma(\alpha)\rangle$. Notice that $\sigma(\alpha)$ is now a simple root. That means $\langle \lambda, \sigma(\alpha) \rangle \in \mathbb{Z}$, since you have checked the condition on the simple roots. The last open question is: How can we go from $\langle \sigma(\lambda), \sigma(\alpha) \rangle$ to $\langle \lambda, \sigma(\alpha) \rangle$?