This is inevitably an imprecise question, but there are already several questions like this on the site so I thought i'd try anyway.

If I understand correctly, for any reductive algebraic group $G$ the points of $G$ over the field with one element should be $G(\mathbb{F}_1) = W$ where $W$ is the Weyl group of $G$. My question, stated briefly (and generally), is the following:

For a reductive group there are several notions which make sense over any base field, what are the analogs of these for $W$:

  1. Parabolic subgroups
  2. Levi subgroups
  3. Borel subgroups
  4. Bruhat decomposition

I will be satisfied with an answer for the case $G=GL_n$ so let's consider this case in a little more detail:

In this case: $GL_n(\mathbb{F}_1)=W=\Sigma_n$. It seems to me like the only reasonable notion of a Levi subgroup of $\Sigma_n$ would be a Young subgroup $\Sigma_{\lambda} := \Sigma_{\lambda_1} \times \dots \times \Sigma_{\lambda_l}$ for some partition $\lambda = (\lambda_1,...,\lambda_l)$. Is this correct? If so what are the corresponding parabolic subgroups?

Following this line of thought we get that the maximal torus should be trivial (which is somehow reasonable). Should the Borel be the trivial subgroup then? If so this is somehow disappointing...

  • 7
    $\begingroup$ There is an existing notion of a parabolic subgroup of a Coxeter group $(W, S)$, namely, a subgroup of the form $\langle I\rangle$ for some $I \subseteq S$. This corresponds to what you are calling a Levi subgroup. As you observe, the smallest among these is the trivial subgroup, so that's probably the one that one has to call the Borel. On the plus side, it is indisputable that the trivial group is soluble, and that the quotient of the finite group $W$ by its trivial subgroup is projective. :-) $\endgroup$ – LSpice Dec 25 '17 at 14:20
  • 1
    $\begingroup$ (Also, taking the Borel to be the trivial subgroup does give us the Bruhat decomposition $W = \bigsqcup_{w \in W} 1w1$ ….) $\endgroup$ – LSpice Dec 25 '17 at 20:33
  • 2
    $\begingroup$ The Borel should be trivial. The analog of the flag variety over $\mathbb{F}_1$ is the set of flags on $[n]$, meaning increasing sequences of subsets of length $n$, and the action of $S_n$ on these has trivial stabilizer. $\endgroup$ – Qiaochu Yuan Dec 25 '17 at 21:23
  • $\begingroup$ One more thought: a linear algebraic group over a field of positive characteristic $p$ is unipotent precisely if it is $p^\infty$-torsion; so this lends credence to the idea that a unipotent group over $\mathbb F_1$ should be trivial. (Of course, such reasoning can lead to silliness, but it's mostly analogising when working "over $\mathbb F_1$" anyway.) $\endgroup$ – LSpice Dec 25 '17 at 23:46
  • 1
    $\begingroup$ The parabolic decomposition $G=\sqcup_{w\in W^P} BwP$ suggests that the parabolic subgroups of $G$ over $\mathbb F_1$ are also the parabolic subgroup $W_P$. $\endgroup$ – Hugh Thomas Jan 6 '18 at 0:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.