• $G$ be a real semisimple Lie group;
  • $\rho$ be an irreducible representation of $G$ on a finite-dimensional real vector space;
  • $A$ be a "Cartan subspace" of $G$ (a Lie subalgebra which is abelian, composed of hyperbolic elements, and maximal);
  • $L$ be the centralizer of $A$ in $G$ (this group is often called "$MA$");
  • $W = N_G(A)/Z_G(A)$ be the restricted Weyl group of $G$, and $w_0$ its longest element.

I would like to classify the representations that satisfy the following two conditions:

  1. the space $V^L$ of points of $V$ fixed by every element of $L$ is nonzero;
  2. the action of $w_0$ on $V^L$ is nontrivial.

An important particular case (which it would already be nice to have) is when $G$ is split. In that case, $V^L$ is simply the zero weight space, which is nonzero whenever the highest weight is an integer combination of roots. So we just need to find the action of $w_0$ on the zero weight space.

This looks like it should have already been done somewhere, but I have been unable to find any references so far. Any pointers would be appreciated!

  • 1
    $\begingroup$ In your special case, which is equally about algebraic groups and their finite dimensional rational representations, there has been some scattered work done which might be relevant. I wrote up notes on this a couple of years ago at people.math.umass.edu/~jeh/pub/zero.pdf $\endgroup$ Jul 21, 2016 at 14:08
  • $\begingroup$ Yes, I have seen this paper at the question mathoverflow.net/questions/185797/… . It does not seem to contain the answer to my question but it is certainly useful to have in mind. $\endgroup$ Jul 21, 2016 at 14:35
  • 2
    $\begingroup$ Most of the time (not $A_{n>1}, D_{odd}, E_6$), $w_0$ acts as $-1$ on the weight lattice. In that case (and when $V$ is finite-dimensional) the trace of $w_0$ on $V^L$ is the trace of $w_0$ on $V$, i.e. computable from WCF. $\endgroup$ Jul 26, 2016 at 13:04
  • 2
    $\begingroup$ Yes, I have thought about this. I agree with your comment up to the last comma. Unfortunately however, the Weyl Character Formula yields $0/0$ (at least for $B_2$). $\endgroup$ Jul 27, 2016 at 12:48
  • $\begingroup$ Together with Bruno Le Floch, we solved this question for split $G$: arxiv.org/abs/1806.00347 . We are now working on the general case. $\endgroup$ Apr 15, 2019 at 10:15

1 Answer 1


The action of $w_0$ can be computed by Allen Knutson's suggestion (whether or not $-1\in W$). As Allen says, suppose $g\in N(T)$ is a representative of $w_0$. Then

$\text{trace}(\pi(w_0)) \text{ on } {V_0}=\text{trace}(\pi(g))\text{ on } V$

(since the only element of the root lattice fixed by $w_0$ is $0$).

It is known that $g$ is conjugate to $\exp(\pi i\rho^\vee)$. Note that $g^2=\exp(2\pi i\rho^\vee)\in Z(G)$, and $\lambda(g^2)=e^{2\pi i\langle\lambda,\rho^\vee\rangle}=1$. Write $\Delta(V)$ for the set of weights of $V$. These are in the root lattice; write $\text{ht}(\mu)$ for the height (writing $\mu$ as $\pm$ a sum of simple roots). Then

$$\text{trace}(\pi(w_0))=|\{\mu\in\Delta(V)\mid \text{ht}(\mu)\text{ even}\}|-|\{\mu\in \Delta(V)\mid ht(\mu)\text{ odd}\}|$$

Since $\pi(w_0)$ on $V_0$ is diagonalizable with $\pm1$ on the diagonal this computes the number of $1/-1$ eigenvalues.

  • $\begingroup$ One remark: the statement that "the only element of the root lattice fixed by $w_0$ is $0$" is not true in general! For example with $G = SL_3(\mathbb{R})$, the vector $e_1 - 2e_2 + e_3 = (e_1 - e_2) + (e_3 - e_2)$ is a nonzero $w_0$-invariant element of the root lattice. $\endgroup$ May 15, 2017 at 12:41
  • $\begingroup$ When we can ignore this issue (for example if we exclude from consideration $A_{n > 1}$, $D_{2n+1}$ and $E_6$), I agree that this gives an algorithm to compute the trace of $w_0$ for one given representation. (I already knew this algorithm, and a few others.) However, what I am looking for is a description of the set of all representations for which $w_0$ has maximal trace. If course it may be possible to use the former in order to get the latter, but I do not know how to do it, and I am afraid it is not so easy to do. $\endgroup$ May 15, 2017 at 12:41

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