Let $E_6$ be the root lattice, and $G$ be its automorphism group as a lattice (i.e. as $\mathbb Z$-module together with the inner product). Let $W(E_6)$ be the Weyl group. Apparently $W(E_6)\subset G$. Is it in fact an equality?
I know in general we cannot expect this for every root lattice, but is it true for this case? Thanks in advance.
No, $W(E_6) \subsetneq G$. The Dynkin diagram of the root system $E_6$ (likewise, $A_{n \ge 2}$ and $D_{n \ge 4}$) has a non-trivial diagram automorphism, which corresponds to an automorphism of the root system (hence the root lattice) which is not in the Weyl group. In this case, to construct such automorphism one can choose any set of simple roots, and then take the map which flips the simple roots $\alpha_1 \leftrightarrow \alpha_5, \alpha_2 \leftrightarrow \alpha_4$, and fixes $\alpha_3$ and $\alpha_6$, if $\alpha_6$ is the one that corresponds to the "lower" vertex of the Dynkin diagram, and $\alpha_1, ..., \alpha_5$ make up the "upper row".