Suppose we have a connected compact Lie group $G$ with root system $\Sigma$ with respect to a maximal torus $T$. As usual, we identify $\mathfrak t=Lie(T)$ with its dual $\mathfrak t^*$ using the invariant bilinear form on $Lie(G)$ to obtain $\Sigma\subseteq\mathfrak a = i\mathfrak t$.
We define coroots $\alpha^\vee = \frac{2\alpha}{\langle \alpha, \alpha\rangle}$ and define the coroot lattice $Q^\vee = \mathbb{Z}.\Sigma^\vee$. Furthermore, we define the integral lattice $I = i\text{ker}(\exp\colon \mathfrak t\to T)$.
The Weyl group $W$ is generated by reflections
$$
s_\alpha(x) = x – \langle \alpha^\vee, x\rangle \alpha.
$$
Direct calculation shows $s_\alpha(\beta^\vee) \in Q^\vee$, so the Weyl group acts on the coroot lattice.
Question: Does the Weyl group also act on the integral lattice? Or is it possible that $s_\alpha(x)\notin I$ for some $x\in I$?
We know $\Gamma = 2\pi Q^\vee \subseteq I$ and $\pi_1(G) \cong I/\Gamma$. If $G$ is semisimple then the fundamental group $\pi_1(G)$ is finite.
In the case of $G$ being simply connected, we know $I=\Gamma$, so the answer is yes. But what about the other cases?
Using the classification of connected compact Lie groups we know $G=(H\times K)/C$ where $H$ is a torus, $K$ is simply connected compact and $C$ is a finite subgroup of the center of $H\times K$ with $C\cap H = \{e\}$. The Lie algebra $\mathfrak g$ of $G$ conincides with the direct sum of Lie algebra $\mathfrak h \oplus \mathfrak k$ of $H$ and $K$. Now $G$, $H\times K$ and $K$ all have the same roots, which act trivially on $\mathfrak h$, so i guess one could reduce the general situation to the semisimple situation [but it is not exactly clear to me how to do this; how to relate the maximal torus in $G$ to a maximal torus in $K$ etc.]. It then remains to see what happens in the semisimple situation.
Are there general results that guarantee that $W$ acts on $I$? If not, are there some handsome criteria when this works?
For example $SU(n)$ is simply connected, so there is nothing to show here, but we can also hands-on calculate for $T=\text{diag}(i\mathbb{R}^n_0)$ that $I=2\pi\mathbb Z^n_0$. Here, $\mathbb R^n_0$ denotes the subspace of $\mathbb R^n$ of vectors whose entries add to $0$ and $\mathbb Z^n_0 = \mathbb R^n_0 \cap \mathbb Z^n$.
In the situation of $U(n)$ one can simply calculate $I=2\pi\mathbb Z^n$ if we take $T=\text{diag}(i\mathbb{R}^n)$ and so on. We also see that $p(I)=\mathbb Z^n_0=I\cap \mathbb R^n_0$, the integral lattice of $U(n)$ simply by adding the free direction, i.e. $\mathbb Z^n = \mathbb Z^n_0 \oplus \mathbb Z \cdot (1,\ldots, 1)$. Here, $p$ denotes the orthogonal projection $p\colon\mathfrak g_\mathbb{C}\to\mathfrak k_\mathbb{C}$, i.e. $p\colon \mathbb C^n\to\mathbb C^n_0$.
In the case of $SO(4)$ the group is not simply connected and has a double cover, so $\# I/\Gamma = 2$, but the integral lattice still seems nice enough for the $s_\alpha$ to be lattice automorphisms…
Best Answer
If we use the analytic definition of the Weyl group, i.e. $W\cong N_K(T)/T$, we see for $x\in\text{ker}(\exp\colon\mathfrak t\to T)$ and $w=\text{Ad}(k)\in W$ that $$ \exp(\text{Ad}(k)x) = k\exp(x)k^{-1} = e, $$ so that indeed $w\colon I\to I$ is well-defined. As $w^{-1}\in W$ also maps $I$ into $I$, we see that $w$ is indeed a lattice automorphism.