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Using the notation of my previous question, let $N(T)$ denote the normalizer of the maximal torus $T$ and hence the Weyl group $W(G,T) = N(T)/T$.

Here think of roots $\alpha$ as maps $T \rightarrow GL(g_\alpha)$ where $g_\alpha$ is the corresponding root-space. Then I can see that the Weyl group has a permuting action on the roots buy conjugating their arguments but there seems to be some different ways of extending this action with non-trivial implications which I can't understand very well.

  • But is something stronger true about the above action? - as in for any root $\alpha$ there seems to exist a $s_\alpha \in W(G,T)$ such that it fixes $ker(\alpha)$ - what is such an action?

  • What is the action of $W(G,T)$ on $t$ ? (..which fixes the Stiefel diagram of the group consisting of the sets $exp^{-1}(ker(\alpha))$..are these hyperplanes in $t$?..why?..)

  • Let $R(T)$ be the representation ring of $T$. What is the action of $W(G,T)$ on $R(T)$? The point being that if I think of the obvious conjugation action of $W(G,T)$ on the arguments of $R(T)$ ( conjugating the arguments..) then all elements in $R(T)$ seem to be $W(G,T)$-stable.

But it seems from literature that if one restricts the elements of the representation ring of $G$ to that of $T$ then the image lies in the $W(G,T)$ invariant subspace of $R(T)$ - which doesn't seem anything special since by the conjugation action everything in $R(T)$ is $W(G,T)$ stable! What am I missing?

  • One defines the "weight lattice" as those weights in $t^*$ which evaluate to integers when evaluated on the co-roots of the simple roots. Now it seems that there is a natural action for the Weyl group on this weight lattice such that the lattice is invariant. What is this action?
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In your last question, do you mean to say "when evaluated on the simple coroots" rather than "when evaluated on the root-spaces of the simple roots"? – Brad Jan 9 '12 at 20:28

The action of $W(G,T)$ is non-trivial on $R(T)$. For example, consider $G=U(n)$ with the standard torus $T$ of diagonal matrices. Then the representation ring $R(T)$ is isomorphic to the polynomial ring $\mathbb{Z}[x_1,\ldots,x_n]$ where the one-dimensional representation of $T$ with weight $(d_1,\ldots,d_n)$ gives the monomial $x_1^{d_1}\cdots x_n^{d_n}$. In this situation the Weyl group $W(G,T)$ is the symmetric group $S_n$, which acts on the weight lattice by permuting it in the usual way, and the induced action on the representation ring is also by permutation. The image of the restriction map $R(G)\to R(T)$ is isomorphic to the space $\mathbb{Z}[x_1,\ldots,x_n]^{S_n}$ of symmetric polynomials, containing for example $\sum_{i=1}^nx_i$ but not $x_1+x_3$.

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I am a bit confused by what you say. If $M \in T$ (of $U(n)$) as you say then on a polynomial $f$ of $n$-variables $x =(x_1,x_2,...,x_n)$ its action I guess is $f(x) \mapsto f(M^{-1}x)$. right? I am thinking of "weights" as irreducible representations of $T$ and those you say are spanned by monomials like $\prod _{i=1} ^n x_i ^{d_i}$ and this is what you are calling an irreducible representation of weight $(d_1,d_2,..,d_n)$..right? – Anirbit Jan 10 '12 at 21:24
So if $M = diag(e^{i\theta_1},e^{i\theta_2},...,e^{i\theta _n})$ then by this action the monomial above will pick up an eigenvalue of $\prod_{i=1}^n e^{-id_i \theta_i}$..right? Now what is the action of $W(G,T)$ on weights that you have in mind? Can you state that more explicitly? If I have a root $\alpha : T \mapsto GL(g_\alpha)$ then I will think that an coset $nT \in N(T)/T = W(G,T)$ acts on $\alpha$ to map it to say $n(\alpha)$ such that $n(\alpha)[t\in T] = \alpha[ntn^{-1}]$. Are you extending this action to the weights and to the weight-lattice? – Anirbit Jan 10 '12 at 21:29
Similarly on an element $\chi_\pi \in R(T)$ (for some representation $\pi$ of $T$) I would think that $[nT]$ (as defined above) acts such that it is mapped to $\chi'$ such that $\chi'(t) = \chi_\pi (ntn^{-1}) = Tr(\pi(ntn^{-1}))$. But $\chi'(t) = \chi_\pi(t)$ and hence I was thinking that everything in $R(T)$ is $W(G,T)$ stable. What am I missing? – Anirbit Jan 10 '12 at 21:35

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