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I consider a finite irreducible matrix group $\Gamma\subseteq\mathrm{GL}(\Bbb R^d)$. I am interested in the maximal size of $\Gamma$ depending on $d$. But this question makes only sense if there is an upper limit.

In even dimension there is no such limit. This is easiest seen in dimension $d=2$, where we have the cyclic groups or dihedral groups of arbitrarily large size. More generally, in dimension $d=2n$ we can consier the $n$-th cartesian power of a regular $k$-gon $P_k$:

$$\overbrace{P_k\times \cdots\times P_k}^{\text{n times}}.$$

Its symmetry group is irreducible and gets arbitrarily large with $k\to\infty$.

Question: What about odd dimensions? Can there be arbitrarily large finite irreducible matrix groups in dimension $d=2n+1$?

For example, in dimension $d=3$ we have the arbitrarily large symmetry groups of prisms and antiprisms, which are reducible. The largest irreducible group is probably the symmetry group of the icosahedron.

I have the feeling that in sufficiently large odd dimensions, the largest such group is the reflection group $B_d$.

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2 Answers 2

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Indeed, in odd dimension it's bounded.

Indeed, let $\Gamma$ be such a matrix group. By Jordan's theorem, it has a normal abelian subgroup $\Lambda$ of index $\le c_d$. (An explicit bound for $d\ge 71$ is $c_d=(d+1)!$, by work of Collins and Weisfeiler, see Breuillard - An exposition of Jordan's original proof of his theorem on finite subgroups of $\operatorname{GL}_n(\mathbb C)$.)

If $\Lambda$ acts diagonalizably, then it has cardinal $\le 2^d$ and hence $\Gamma$ has cardinal $\le 2^dc_d$.

Otherwise, $\Lambda$ has blocks of size 2. By irreducibility, the sum of blocks of size two being invariant, it equals $\mathbf{R}^d$. Hence $d$ is even.

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    $\begingroup$ Many thanks! Is it this "Jordan's theorem" you are referring to? What does "acts diagonalizably" mean? $\endgroup$
    – M. Winter
    Jan 14, 2021 at 15:27
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    $\begingroup$ Yes. I don't know why it's called Jordan-Schur on Wikipedia since it's 100% due to Jordan. $\endgroup$
    – YCor
    Jan 14, 2021 at 15:28
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    $\begingroup$ The bound for $d \geq 71$ is due to a combination of work of Weisfeiler and Collins ( essentially the same work I mentioned, since we were writing at the same time) and does require CFSG.. Frobenius, Schur and Burnside all generalized Jordan's Theorem in various explicit ways, but I agree that there is no doubt that the existence of an Abelian normal subgroup of bounded index was first proved by C. Jordan. $\endgroup$ Jan 14, 2021 at 15:39
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    $\begingroup$ Thanks, I added the reference for the explicit bound, which was just mentioned as a bonus. $\endgroup$
    – YCor
    Jan 14, 2021 at 15:53
  • $\begingroup$ The reflection group $B_d$ has order $2^dd!$, and you give an upper bound $2^d(d+1)!$. Is there a group of order $2^d (d+1)!$ for infinitely many (odd) $d$, or do we have the hope to improve the upper bound to $|B_d|$? $\endgroup$
    – M. Winter
    May 1, 2021 at 14:33
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View your group as a subgroup $\Gamma$ of ${\rm GL}(d,\mathbb{C})$ by extending scalars. If it remains irreducible as such , then it has an Abelian normal subgroup $A$ with the index $[\Gamma :A]$ bounded in terms of $d$ (by a Theorem of Jordan). The rank of $A$ is clearly at most $d$, so we only need to consider the exponent of $A$.

Let $\chi$ be the character afforded by $\Gamma$. By Clifford's Theorem $\chi$ decomposes on restriction to $a$ as a sum $e(\lambda_{1} + \lambda_{2} + \ldots + \lambda_{t})$, where $et = d$ and each $\lambda_{i}$ is a linear character of $A$, with $\Gamma$ transitively permuting the $\lambda_{i}$. Note that $e$ and $t$ are both odd. However, since $\chi$ is real-valued on restriction to $A$, we see that $\lambda_{i}$ and $\overline{\lambda_{i}}$ occur with equal multiplicity in ${\rm Res}^{\Gamma}_{A}(\chi)$ for each $i$. Since $t$ is odd, at least one $\lambda_{i}$ is real-valued.

But $[A:{\rm ker}\lambda_{i}]$ is independent of $i$ by the transitive action of $\Gamma$, and $A/{\rm ker} \lambda_{i}$ is cyclic for each $i$. Hence $[A: {\rm ker} \lambda_{i}] = 2$ for each $i$. Thus $A$ is an elementary Abelian $2$-group in the case that $\Gamma$ reamins irreducible as a complex linear group, and $|\Gamma|$ is bounded in terms of $d$ in that case. In this case, it is true for large enough $d$ that the maximum possible order is attained by the group $(\mathbb{Z}/2\mathbb{Z}) \wr S_{d}$, though this requires theorems of B. Weisfeiler and M.J. Collins, which require the Classification of finite simple groups.

More generally, if the character $\chi$ afforded by $\Gamma$ is a sum of real-valued irreducible complex characters, we get a bound in terms of $d$ for $|\Gamma|$ by the same argument (applied to each irreducible summand).

Otherwise, (since we are given that $\Gamma$ is irreducible as a real linear group) we may write $\chi = \mu + \overline{\mu}$ for some non-real irreducible complex character $\mu$, contrary to the fact that $d$ is odd.

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  • $\begingroup$ Do you mean "the maximum possible order is attained by the group $(\Bbb Z/2\Bbb Z)^d \wr S_d$? In (sufficiently large) even dimension, if we exclude this arbitrarily large groups from my question, we still get $(\Bbb Z/2\Bbb Z)^d \wr S_d$ as an upper bound, right? $\endgroup$
    – M. Winter
    Jan 14, 2021 at 15:48
  • $\begingroup$ I meant still in the case $d$ odd. $\endgroup$ Jan 14, 2021 at 16:08
  • $\begingroup$ I think that for $d$ even, you can wreath one of the arbitrairy large $2$-dimensional example with $S_{\frac{d}{2}}$, for example. $\endgroup$ Jan 14, 2021 at 16:13
  • $\begingroup$ Just to be clear, for $d$ odd, I was using the non-trivial $1$-dimensional representation of $\mathbb{Z}/2\mathbb{Z}$, so the group I meant was the group of "signed permutation $d \times d$-matrices", whch is the reflection group $B_{d}$. $\endgroup$ Jan 14, 2021 at 16:51
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    $\begingroup$ I think YCor is giving an upper bound, but that is not always the optimal upper bound. For large enough $d$ there is an Abelian normal subgroup of index at most $(d+1)!$ for a finite subgroup of ${\rm GL}(d,\mathbb{C}).$ $\endgroup$ Jan 15, 2021 at 9:15

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