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 $s\in W$.

  • $\begingroup$ 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)} $ ? $\endgroup$ – Lepanais Dec 17 '13 at 10:10
  • $\begingroup$ the Weyl group action is linear $\endgroup$ – Matthew Towers Dec 17 '13 at 10:20
  • $\begingroup$ 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 $B^\flat (s(\alpha)$ ? $\endgroup$ – Lepanais Dec 19 '13 at 11:38
  • $\begingroup$ 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? $\endgroup$ – Matthew Towers Dec 19 '13 at 12:46
  • $\begingroup$ 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}^*$. $\endgroup$ – Lepanais Dec 19 '13 at 13:12

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