# Action of the Weyl group on the symmetric algebra $S\mathfrak{h}$

Let $\mathfrak{g}$ be a complex semi-simple Lie algebra. Let $\mathfrak{h}$ be a cartan subalgebra. Let $\Delta$ be the resulting root system. Denote by $V$ the real span of the roots. Let $\alpha_1,...,\alpha_r$ be simple roots and $\check{\alpha_i}$ the corresponding coroots. Let $W$ be the Weyl group. Recall that it generated by the reflections $s_i:=s_{\alpha_i}$. $W$ can also be seen on the Lie group level as $N_G(T)/T$. Denote by $S \mathfrak{h}$ the symmetric algebra on $\mathfrak{h}$. It can be a seen as a polynomial ring on $\mathfrak{h}^*$.

Let $P \in S \mathfrak{h}$, $w \in W$ and $H \in \mathfrak{h}^*$. My questions are the following.

1) Is it correct that $w$ act on $P$ as $(w.P)(H)=P(w^{-1}H)$ ?

2) How does $W$ act on $H \in \mathfrak{h}^*$ if $H$ isn't in $V$ ?

3) Is it true that $s_i \check{\alpha_j} = \check{s_i \alpha_j}$ ? Edit: after some computations, I think that this is only true iff the Cartan matrix is symmetric. In particular it is true for types A,D,E.

For question one, that's certainly a way for $W$ to act on $P$. Whether or not it is correct depends on the context.
2): The roots are a subset of $\mathfrak{h}^*$, and the Weyl group acts on those roots. So you extend that action by linearity. Whether you extend to real scalars or complex ones (or whatever) doesn't really make a difference. Do you maybe mean to ask how it acts on elements of $S\mathfrak{h}$ that aren't in $\mathfrak{h}$? If so, the answer is that if $h_1\cdots h_k$ is a monomial with each $h_i \in \mathfrak{h}$ and $w\in W$ then $w\cdot (h_1\cdots h_k) = (w\cdot h_1)(w\cdot h_2)\cdots(w\cdot h_k)$.
3) Yes: use that $\check{\alpha} = 2\alpha/(\alpha,\alpha)$ and the fact that $W$ preserves the inner product, i.e. $(\alpha,\alpha) = (s\alpha, s\alpha)$ for any $s\in W$.
• Thank you for your answer. For 3), I agree that $\frac{2s(\alpha)}{(s(\alpha),s(\alpha))}=s(\frac{2 \alpha}{(\alpha,\alpha)})$. However, is it clear that the Weyl group action is compatible with the identification of $\check{\alpha}$ with $\frac{2 \alpha}{(\alpha,\alpha)}$ ? – Lepanais Dec 17 '13 at 10:10
• I am afraid I don't understand: Denote by $B^\flat: \mathfrak{h}^* \longrightarrow \mathfrak{h}$ the isomorphism induced by the Killing form. On one hand $s(\check{a})=s(B^\flat(\frac{2\alpha}{(\alpha,\alpha)}))=s(\frac{2B^\flat(\alpha)}{(\alpha,\alpha)})=\frac{2s(B^\flat(\alpha))}{(\alpha,\alpha)})$. On the other hand $\check{s(\alpha)}=B^\flat(\frac{ (2s(\alpha)}{(s(\alpha),s(\alpha))}=\frac{2B^\flat(s(\alpha)}{(\alpha,\alpha)}$. But why is $s(B^\flat (\alpha))$ is equal to $B^\flat (s(\alpha)$ ? – Lepanais Dec 19 '13 at 11:38
• I don't get it. $\check \alpha$ is an element of $\mathfrak{h}^*$ whereas $B$ has its image in $\mathfrak{h}$, so what does your first equation mean? For me both roots and coroots are elements of $\mathfrak{h}^*$, so I don't understand why you need $B$ -- could you explain? – Matthew Towers Dec 19 '13 at 12:46
• I thought that $\check{\alpha}$ belonged to $\mathfrak{h}$. Quoting for instance definition 1 page 2 of math.columbia.edu/~woit/notes11.pdf "The co-root $H_\alpha$ associated to a root $\alpha$ is the unique element in $[ \mathfrak{g}_\alpha , \mathfrak{g}_{-\alpha} ]$ satisfying $\alpha(H_\alpha) = 2$.". So it seemed to me that the coroot $\check{a}$ (also written $H_\alpha)$ belonged to $\mathfrak{h}$ but could be identified, via $B^\sharp$ with $\frac{2\alpha}{(\alpha,\alpha)} \in \mathfrak{h}^*$. – Lepanais Dec 19 '13 at 13:12