5
$\begingroup$

Let $G=\mathrm U_n$ or $\mathrm{GL}_n(\mathbf C)$ and $H$ the subgroup of block diagonal matrices respecting a partition $n=n_1+\dots+n_k$. Is the normalizer $N=N_G(H)$ computed anywhere in the literature?

I guess, but haven’t proved, that it is generated by $H$ and the permutations (“transpositions”) exchanging the partition’s same-length segments ($\smash{n_i=n_j}$, if any). I also suspect this may be discussed in these papers, to which I don’t have access:

  • Koĭbaev, V. A., On subgroups of the full linear groups containing a group of elementary block diagonal matrices. ZBL0521.20027. (1982, translation 1983; other translation?)

  • Borevich, Z. I.; Vavilov, N. A., Ordering of subgroups, containing a group of block-diagonal matrices, in the general linear group over a ring. ZBL0512.20031. (1982; translation?)

  • Vavilov, N. A., Subgroups of the general linear group over a semi-local ring containing the group of block-diagonal matrices. ZBL0509.20035. (1983; ever translated?)

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ Your guess is correct. Because there are clearly elements in the normaliser arbitrarily permuting same-sized blocks, if you've got an element of the normaliser, you may assume that it fixes each block. But then it must induce an inner isomorphism on each block, and so you can further assume it fixes each block pointwise. But this forces it to be block scalar. $\endgroup$
    – LSpice
    Commented Jun 1, 2020 at 4:13
  • $\begingroup$ @LSpice Fantastic, thank you. I am still mildly interested in whether this is printed anywhere, but meanwhile will gladly accept your comment as an answer. $\endgroup$
    – Francois Ziegler
    Commented Jun 1, 2020 at 10:46

1 Answer 1

Reset to default
3
$\begingroup$

By request, from my comment: Your guess is correct. Because there are clearly elements in the normaliser arbitrarily permuting same-sized blocks, if you've got an element of the normaliser, you may assume that it fixes each block. But then it must induce an inner isomorphism on each block, and so you can further assume it fixes each block pointwise. But this forces it to be block scalar.

I'm not sure where this would be written down, but someone must have worked it out as part of an example in some notes. If you're willing to combine some theorems, then it may help to think of your block diagonal matrices are examples of Levi subgroups. Let $G$ be an appropriate general linear or unitary group, $M$ a block diagonal subgroup of the sort you describe, and $A$ the subgroup of diagonal matrices. If $g$ belongs to $\operatorname N_G(M)$, then $g A g^{-1}$ is (the group of rational points of) a split maximal torus in $M$, hence is conjugate in $M$ to $A$ (that's one of the theorems; p. 387 of Murnaghan); so, modulo $M$, it suffices to ask which elements of $\operatorname N_G(A)$ preserve $M$. This is an easy computation, since $\operatorname N_G(A)/A$ is a symmetric group (that's another theorem; Example 8.1 of Murnaghan). If this is an argument that it would be satisfactory to cite, I could easily dig up some references; here I've indicated where you can look in the notes of Murnaghan on linear algebraic groups in CMP 4 (MSN), which just state the results but were the first place to look that occurred to me.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Thanks again! Ironically the argument for $\operatorname N_G(M)\subset\operatorname N_G(A)M$ is one I forgot essentially making here. AFAICT the “easy computation” to then exclude Weyl group elements that don’t permute same-size segments still requires unpleasant though manageable bookkeeping. $\endgroup$
    – Francois Ziegler
    Commented Jun 1, 2020 at 23:23
  • 1
    $\begingroup$ I now see that Corollary 12.11 (= Exercise 20.4) of Malle-Testerman (2011) also has that argument. $\endgroup$
    – Francois Ziegler
    Commented Jun 2, 2020 at 0:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .