Let $w_0$ be the longest element in the Weyl group of a semisimple Lie algebra $\mathfrak{g}$. How does $w_0$ act on the simple roots $\{ \alpha_1, \ldots, \alpha_n \}$? If $L_{\lambda}$ is an irreducile $\mathfrak{g}$-module with highest weight $\lambda$, is $w_0(\lambda)=-\lambda$? Are there some references about properties of $w_0$? Thank you very much.


It depends on the Lie algebra. We have that $-w_0$ induces some permutation of the simple roots since it sends the positive Weyl chamber to itself. Since $-w_0$ respects inner products, it must induce an automorphism of the Dynkin diagram of $\mathfrak g$. This immediately tell us that $-w_0$ is the identity for the simple Lie algebras $B_n$, $C_n$, $E_7$, $E_8$, $F_4$, and $G_2$.

  • For $A_n$, the root system can be taken to have simple roots $$ \boxed{e_1-e_2}-\boxed{e_2-e_3}-\cdots -\boxed{e_n-e_{n+1}} $$ where the $e_i$ are standard basis vectors, and the inner product is the standard one. The Weyl group is generated by reflections through $e_i-e_{i+1}$, which switch $e_i$ and $e_{i+1}$, so the Weyl group is $S_{n+1}$, acting by permuting the $e_i$. If I can find an element of the Weyl group which takes the set of simple roots to the set of negatives of the simple roots, it must be $w_0$. The permutation sending $e_1,\dots, e_{n+1}$ to $e_{n+1},\dots, e_1$ does the trick, so we see that $-w_0$ reverses the order of the simple roots. That is, it sends the above Dynkin diagram to $$ \boxed{-e_{n+1}+e_n}-\boxed{-e_n+e_{n-1}}-\cdots -\boxed{-e_2+e_1} $$

  • For $D_n$, we can take the simple roots to be $$ \boxed{e_1-e_2}-\cdots -\boxed{e_{n-2}-e_{n-1}}<{\boxed{e_{n-1}-e_n}\atop\boxed{e_{n-1}+e_n}} $$ Here, the Weyl group again acts by permuting the $e_i$, but we also have reflection through $e_n+e_{n+1}$, which sends $e_n$ and $e_{n+1}$ to $-e_{n+1}$ and $-e_n$, respectively. Thus, the Weyl group can arbitrarily permute the $e_i$, and it can negate an even number of them. If $n$ is even, we can take $w_0$ to negate all the $e_i$, in which case $-w_0$ is the identity. If $n$ is odd, we can take $w_0$ to negate all the $e_i$ except $e_n$, in which case, $-w_0$ switches the two "horns" of the diagram.

  • For $E_6$, we can take the root system to be $$ \begin{align*} \boxed{e_1-e_2}-\boxed{e_2-e_3}&-\boxed{e_3-e_4}-\boxed{e_4-e_5}-\boxed{e_5-e_6}\\ &\qquad\qquad|\\ &\boxed{e_4+e_5+e_6} \end{align*} $$ Again, we can permute $e_1,\dots, e_6$ arbitrarily, and this time we have the additional operation of reflecting through $e_4+e_5+e_6$. I don't see an easy way to negate the set of simple roots using these operations. If you do, please add a comment. I feel like we might be able to embed the root system into $E_7$ (by adding a $\boxed{e_6-e_7}$ to the end), then say that the longest word negates all the simple roots (since there are no diagram automorphisms), then argue that the longest word for $E_6$ should "do the same thing" (this part is not clear to me), so $-w_0$ should be the identity.

Incidentally, since the dual of $V_\lambda$ is $V_{-w_0(\lambda)}$, this answers the following question: Which simple $\mathfrak g$ have the property that every finite-dimensional representation is isomorphic to its dual?

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    $\begingroup$ Is it equivalent to say that for $A_n$, $D_{2n+1}$, and $E_6$, there is no Weyl group element inducing $x \mapsto -x$ on the Cartan subalgebra? How far is that from forcing the Weyl group to lack central elements? $\endgroup$ – jdc Jul 15 '14 at 10:29
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    $\begingroup$ $-w_0$ is not the identity for $E_6$, but corresponds to the non-trivial automorphism of the Dynkin diagram. According to mathoverflow.net/a/54977/27465, $w_0 = (s_1 s_4 s_6 s_3 s_2 s_5)^6$ where $s_i$ is the reflection corresponding the $i$-th simple root in Bourbaki's labelling of the diagram (that is, $\alpha_2$ is the one at the bottom, and the top row is $1-3-4-5-6$). $\endgroup$ – Torsten Schoeneberg Nov 17 '17 at 5:51
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    $\begingroup$ By the way, Bourbaki's table (end of ch. VI in Lie Groups and Algebras) also states for $E_6$ that $-w_0$ is not the identity, but the "diagram flip". Would you please edit this answer accordingly -- it is great otherwise, and I have linked to it in several other answers already. $\endgroup$ – Torsten Schoeneberg May 11 '18 at 19:31

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