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A follow-up question to Alternating subgroups of $\mathrm{SU}_n $.

$\DeclareMathOperator\PU{PU}\DeclareMathOperator\PSU{PSU}\DeclareMathOperator\PSL{PSL}$Let $ \PU_m $ be the projective unitary group, a compact simple adjoint Lie group corresponding to the root system $ A_{m-1} $.

Let $ \PSL_n(q) $ be the finite simple group of Lie type $ A_{n-1}(q) $ given by taking the special linear group with entries from the field with $ q $ elements and modding out by the center.

Let $ \PSU_n(q^2) $ be the finite simple group of Lie type $ ^2 A_{n-1}(q^2) $ given by taking the special unitary group with entries from the field with $ q^2 $ elements and modding out by the center.

$ \PU_2 $ contains a $ 60 $ element subgroup isomorphic to $ A_5 \cong \PSL_2(4) \cong \PSL_2(5) $. It is a maximal closed subgroup of $ \PU_2 $ (the only closed subgroup containing it is the whole group).

$ \PU_3 $ contains a subgroup of order $ 360 $ isomorphic to $ A_6 \cong \PSL_2(9) $, it is maximal. Also $ \PU_3 $ contains a group of order $ 168 $ isomorphic to $ \PSL_2(7)\cong \PSL_3(2) $, it is maximal.

There is also a 60 element $ A_5 \cong \PSL_2(4) \cong \PSL_2(5) $ subgroup of $ \PU_3 $ but that is already in $ SO_3(\mathbb{R}) $ so it is not maximal. For more details see

https://math.stackexchange.com/questions/497853/closed-lie-subgroups-of-su3

The reference Hanany and He - A Monograph on the Classification of the Discrete Subgroups of SU(4) from (The finite subgroups of SU(n)) shows that $ \PU_4 $ contains a subgroup of order $ 25{,}920 $ isomorphic to $ \PSU_4(2)\cong PSp_4(3) $, it is maximal.

Also $ \PU_4 $ contains a maximal $ A_7 $ which in turn contains a group of order $ 360 $ isomorphic to $ \PSL_2(9) \cong A_6 $, and a group of order $ 168 $ isomorphic to $ \PSL_2(7)\cong \PSL_3(2) $,as well as a group of order 60 isomorphic to $ A_5 \cong \PSL_2(4) \cong \PSL_2(5) $. For more details see

https://math.stackexchange.com/questions/4535647/maximal-closed-subgroups-of-su-4

That leads me to ask: Does $ \PU_m $ always have a maximal $ \PSL_n(q) $ or $ \PSU_n(q) $?

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    $\begingroup$ Not an answer, but I suspect there are too many small values of $m$ for this to be true. In particular the smallest non-trivial representation of $PSL_n(q)$ grows like $q^n$. So varying $n$ and $q$ it's roughly like counting all numbers vs counting prime powers. $\endgroup$
    – Nate
    Commented Jan 20, 2022 at 20:13
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    $\begingroup$ $\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\SL{SL}\DeclareMathOperator\Z{Z}$I'm not sure if it's reasonable to call $\PSL_n(q)$ "of Lie type $A_{n - 1}(q)$". I'm not up on the definitions, but I'd expect this to refer only to $\mathbb F_q$-rational points of reductive groups over $\mathbb F_q$, whereas your $\PSL_n(q)$ is not of that type. (If you take not $\SL_n(q)/\Z(\SL_n(q))$ but $(\SL_n/\Z(\SL_n))(q)$, then you get $\operatorname{PGL}_n(q)$.) Similarly for $\operatorname{PSU}$. Also, TeX: $25,920$ needs curly brackets: $25{,}920$ 25{,}920. I edited accordingly. $\endgroup$
    – LSpice
    Commented Jan 31, 2022 at 17:14

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One can consult the tables of Bray-Holt-Roney Dougal to work out the subgroups of $\mathrm{PSU}_m$.

  1. For $m=5$, we have copies of $PSL_2(11)$ and $PSU_4(4)$.
  2. For $m=6$ we have $PSL_2(11)$, $PSL_3(4)$ and $PSU_4(9)$.
  3. For $m=7$ we have $PSU_3(9)$.
  4. For $m=8$ we have $PSL_3(4)$.
  5. For $m=9$ we have $PSL_2(19)$.
  6. For $m=10$ we have $PSL_2(19)$ and $PSL_3(4)$.
  7. For $m=11$ we have $PSL_2(23)$ and $PSU_5(4)$.
  8. For $m=12$ we have $PSL_2(23)$ and $PSL_2(9)$.

So far, so good. Bray-Holt-Roney Dougal stops at dimension 12. For dimensions 13 to 15 one consults tables in Anna Schroeder's PhD thesis (St Andrews, 2015) and there are examples in those dimensions. For dimensions 16 and 17 one consults the thesis of Daniel Rogers (Warwick, 2017). For 16 there is an example, but there are no interesting subgroups (i.e., finite subgroups not contained in a closed positive-dimensional subgroup) of $\mathrm{PSU}_{17}$ at all, never mind linear or unitary ones.

So it seems $m=17$ is the first case where things go wrong.

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  • $\begingroup$ I looked at the references and they seem to all be about maximal subgroups of other finite groups? (maybe I'm missing something?) How did you extract information about maximal subgroups of the compact real Lie group $ PSU_m $ based on these sources? Something about the fact that every complex rep of a finite group is unitary? By $ PSU_5 $ you mean the 24 real dimensional compact Lie group of $ 5 \times 5 $ unitary matrices with complex entries mod its center, right? $\endgroup$
    – Ian Gershon Teixeira
    Commented Feb 1, 2022 at 13:50
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    $\begingroup$ Yes, there were a few details hidden under the carpet here. Irreducible representations in characteristic 0 are the same as irreducible representations in characteristic $p$ for any $p$ not dividing the order of the group. So from these tables you can deduce what happens in char 0. I only used them as they are a handy reference. One can also use the tables in Hiss--Malle, which go up to dimension 250, to find degrees where there are no irreducible representations of linear/unitary groups of a given dimension either. Their tables explicitly list char 0. $\endgroup$
    – David A. Craven
    Commented Feb 1, 2022 at 14:00
  • $\begingroup$ Ok so an irrep in char $ p $ with $ p $ not dividing order of the group is same as char $ 0 $. And a rep of a simple group always has trivial kernel so it is an embedding. And again simple means the embedded subgroup is disjoint from the center of $ U_m $ so also embeds in $ PSU_m $. But a maximal closed subgroup is always image of an irrep so if you can find a dimension $ m$ such that no $ PSL_n(q) $ or $ PSU_n(q^2) $ has a (non modular) irrep of dimension $ m $ then it must be that $ PSU_m $ contains no $ PSL_n(q) $ or $ PSU_n(q^2) $subgroups. same for any simple $\endgroup$
    – Ian Gershon Teixeira
    Commented Feb 1, 2022 at 14:22
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    $\begingroup$ Yes. You always have to look at central extensions. They give the centre in the tables. If the centre of the quasisimple group is not the centre of $SU$ then it is contained in the centralizer of a semisimple element and this is ignored. So these quasisimple groups are always simple mod $Z(SU)$, and are all such quasisimple groups. $\endgroup$
    – David A. Craven
    Commented Feb 1, 2022 at 20:57
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    $\begingroup$ Hiss--Malle exclude $PSL_2(q)$ in their tables as there are too many of them. Their degrees are given elsewhere in the paper: $(q\pm1)/2$, $q$, $q\pm 1$ for $q$ odd, (with the obvious excluded two for $q$ even). For all finite subgroups you have to be a bit more careful, as there can be normalizers of $r$-subgroups that are not contained in the normalizer of a torus. You can already see this in dimension 3: $3^{1+2}:Q_8$ appears in Bray et al, in Table 8.5, in class $C_6$. These should be the only possible exceptions. (These are called extraspecial-type subgroups.) $\endgroup$
    – David A. Craven
    Commented Feb 2, 2022 at 19:58

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