# Why is the dual space of Cartan subalgebra an irreducible representation of Weyl group

it is proposition 14.31 in Fulton-Harris book. The proof goes like this. Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$, and assume $\mathfrak{z}\subseteq\mathfrak{h}^*$ were preserved by the action of the Weyl group. Let $\mathfrak{g}'$ be the subspace spanned by the subalgebras $\{\mathfrak{s}_{\alpha}\}_{\alpha\in\mathfrak{z}}$ ($s_{\alpha}$ is the subalgebra isomorphic to $sl_2(\mathbb{C})$ corresponding to the root $\alpha$). Then it shows that $\mathfrak{g}'$ is an ideal. Then it says

Thus either all the roots lie in $\mathfrak{h}$ or all the roots are perpendicular to $\mathfrak{h}$

I don't understand this sentence. Why is it true?

Thanks.

First notice that for this proposition, $\mathfrak{g}$ is assumed to be simple. (This is in the text two lines before the proposition.) Then you are a little quick in your description, and you make a typo (it must be $\mathfrak{z}$ instead of $\mathfrak{h}$). The proof in the edition I have goes like this:

If there is a subspace $\mathfrak{z} \subseteq \mathfrak{h}^*$ invariant under the Weyl group, then the following holds:

(*) Each root (the roots are elements of $\mathfrak{h}^*$) is either contained in $\mathfrak{z}$, or perpendicular to $\mathfrak{z}$.

(Note that this is not yet the assertion about all roots satisfying (simultaneously) either one or the other; this comes in a minute.)

They conclude from this that the $\mathfrak{g}'$ you describe is an ideal. Now because $\mathfrak{g}$ is simple, $\mathfrak{g}'$ is either $\mathfrak{g}$ or $0$.

In the first case, since every root has a root space in $\mathfrak{g}$, every root must be contained in $\mathfrak{z}$. But the roots span $\mathfrak{h}^*$, so $\mathfrak{z} = \mathfrak{h}^*$.

In the second case it is immediate that no root can be in $\mathfrak{z}$. So by (*), all roots are perpendicular to $\mathfrak{z}$, and so is their span $\mathfrak{h}^*$. But the Killing form is non-degenerate on $\mathfrak{h}$ and per transport of structure (see 14.27--14.30) on $\mathfrak{h}^*$, so $\mathfrak{z} = 0$.