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Let $\Lambda$ be a simply laced root lattice and $w$ a Coxeter element of the Weyl group of $\Lambda$.

Question: Is it true that the action of $w$ on the $\mathbb{F}_2$-vector space $\Lambda/2\Lambda$ is semisimple?

Motivation: I want to show that, if $F$ denotes the subset of $w$-fixed points on $\Lambda/2\Lambda$, then $w$ has no nonzero fixed points on $\left(\Lambda/2\Lambda\right)/F$. This is more or less equivalent to showing that $w$ acts semisimply on its $1$-eigenspace. A proof of the latter claim would already be great!

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    $\begingroup$ An element of even order never acts semisimply on a vector space in characteristic $2$. A Coxeter element has even order except in type $A_{2m}$. The only caveat is that sometimes the Weyl group has a kernel for its action on the roots, so it's the order of the Coxeter element in this quotient that matters. $\endgroup$
    – Paul Levy
    Commented Oct 23, 2021 at 10:15

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The answer seems to be negative.

According to this answer, an operator $X\in\operatorname{M}(n,\mathbb{F}_q)$ is semisimple if and only if $X^{q^m}=X$, where $m=\operatorname{lcm}(2,\ldots,n)$.

Now consider the standard realization of the root system of type $\mathsf{A}_n$ inside $\mathbb{Z}^{n+1}$, where the simple roots are $\alpha_i = e_i-e_{i+1}$ for $i=1,\ldots,n$. Then $$\Lambda = \{ (x_1,\ldots,x_{n+1})\in\mathbb{Z}^{n+1} \mid x_1+\ldots+x_{n+1}=0 \},$$ and the action of $W(\mathsf{A}_n)\cong S_{n+1}$ is by permutations of the entries. One possible choice of the Coxeter element $w$ is the long cycle $$w = (1\ \ 2\ \ \ldots\ \ n+1).$$ For $n=3$ one gets $\dim(\Lambda/2\Lambda)=3$ and $m=\operatorname{lcm}(2,3)=6$. Since $\operatorname{ord}(w)=n+1=4$ and $2^6 = 64 \equiv 0 \pmod{4}$, one has $w^{2^m} = \operatorname{id}$, while $w$ act non-trivially on $\Lambda/2\Lambda$.

The same considerations work for $\mathsf{A}_n$, $n$ odd, and for $\mathsf{D}_n$ (where the Coxeter number equals $2n-2$ and $2^m$ always gives an even remainder modulo $2n-2$).

The conjecture holds for $\mathsf{A}_n$, $n$ even, because in this case $2^m\equiv 1\pmod{n+1}$.

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  • $\begingroup$ To be concrete, in the type $A$ example, $\Lambda/2\Lambda$ is $\{(x_1, x_2, \dots, x_{n+1}) \in \mathbb{F}_2^{n+1} : \sum x_i=0 \}$. For $n$ odd, the fixed $F$ space of $w$ is spanned by $(1,1,\ldots,1)$. But $(1,0,1,0,\ldots)$ is fixed in $\Lambda/(2 \Lambda+F)$. $\endgroup$
    – David E Speyer
    Commented Oct 22, 2021 at 15:56
  • $\begingroup$ The previous comment works for $n \equiv 3 \bmod 4$; I don't see a comparable issue when $n \equiv 1 \bmod 4. $\endgroup$
    – David E Speyer
    Commented Oct 22, 2021 at 16:19

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