Sorry for the misleading title, I actually mean the following:

The $n$-th hyperoctahedral group, also known as the Weyl group of $\mathrm{Sp}_{2n}$ and of $\mathrm{SO}_{2n+1}$, is isomorphic to the group of $n\times n$ signed permutation matrices.

On the other hand, viewing $\mathrm{O}_n$, the usual orthogonal group, as a (non-connected) reductive group scheme over $\mathbb{Z}$, we see that $\mathrm{O}_n(\mathbb{Z})$ is also the group of signed permutation matrices. So, surprisingly, the rational point subgroup of a reductive group becomes the Weyl group of another reductive group.

Is there a conceptual (reductive group theoretic or representation theoretic) explanation for this coincidence?

  • 5
    $\begingroup$ Not that I know of. Other examples: the Weyl groups of type E_n are closely related to orthogonal groups O(n,F_2). See Bourbaki, Lie Groups and Lie Algebras, Chapter 6, Exercises for Section 4. $\endgroup$ – Jeffrey Adams May 11 '18 at 20:47
  • $\begingroup$ The title is out of focus, since "reductive" usually applies to linear algebraic groups (typically connected ones) rather than to discrete groups. Aside from this, note that the Weyl group of type $C_n$ here is the same as the Weyl group of type $B_n$ (belonging for example to the group of odd orthogonal matrices of size $2n+1$); but this doesn't answer your question. $\endgroup$ – Jim Humphreys May 11 '18 at 23:10
  • $\begingroup$ @JimHumphreys Sorry about the title...I just editted a bit. $\endgroup$ – user148212 May 12 '18 at 2:17
  • $\begingroup$ One might also wonder, probably fatuously, if there is any connection between this realisation of a Weyl group as a group of $\mathbb Z$-points of a reductive group scheme over $\mathbb Z$, and between the classical idea that Weyl groups should be viewed as the $\mathbb F_1$-rational points of algebraic groups. $\endgroup$ – LSpice Feb 12 '19 at 0:29

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.