4
$\begingroup$

I'm writing a software package to decompose group representations, and am struggling to find good examples of quaternionic-type representations of dimension > 4.

Reading MathOverflow, I found that the McLaughlin group has quaternionic representations of dimension 3520 and 4752. This is too big for my numerical experiments. Note that the aforementioned MathOverflow questions has links to two papers that I can't easily grasp.

The other answer I found describes a group where quaternionic/symplectic representations have dimension 4.

Are there finite groups with quaternionic type irreducible real representations of dimension > 4 and < 1000 (say)?

$\endgroup$
  • 2
    $\begingroup$ I guess this is more suitable for MathSE. Let $H$ be the algebra of real quaternions. If $Q\subset H^*$ is the quaternionic group of order 8, then the wreath product $Q^n\rtimes\mathfrak{S}_n$, of order $8^nn!$, naturally acts on $H^n$ and this is probably irreducible as $4n$-dimensional representation, hence yields an irreducible, $2n$-dimensional irreducible complex representation of quaternionic type. I haven't checked details. $\endgroup$ – YCor Feb 27 at 19:49
  • $\begingroup$ Thanks YCor. Indeed, it's on the fence for MathSE, and I should have thought of a wreath product representation. For the context, I'm expanding the support of real-type representations in the following library: replab.github.io/replab $\endgroup$ – Denis Rosset Mar 1 at 2:54
3
$\begingroup$

Let $\mathbf{H}$ be the skew field of real quaternions. Let $Q\subset \mathbf{H}^*$ be the quaternion subgroup of order 8, namely $Q=\{\pm 1,\pm i,\pm j,\pm k\}$.

Let the wreath product $G_n=\mathfrak{S}_n\ltimes Q^n$ (of order $n!8^n$) act on $\mathbf{H}^n$ (viewed as right $\mathbf{H}$-module) as monomial matrices (with nonzero entries in $Q$), namely $Q^n$ acts diagonally: $(q_1,\dots,q_n)\cdot (v_1,\dots v_n)=(q_1v_1,\dots,q_nv_n)$, while $\mathfrak{S}_n$ permutes the coordinates.

Claim: this representation on the real $4n$-dimensional vector space $\mathbf{H}^n$ is irreducible (for $n\ge 1$). Indeed, starting from any nonzero vector $v$, perform a permutation to ensure $v_1=1$, then act by $(-1,0,\dots,0)$ and subtract, to get $v_1\neq 0$ and $v_i=0$ for $i\ge 2$, and finally multiply by elements of $Q$ and use permutations to generate all other vectors.

Then, this can be viewed as an irreducible complex $2n$-dimensional representation, of quaternionic type (and conversely any irreducible complex representation of quaternionic type has positive even dimension).

Actually, in the above argument, $\mathfrak{S}_n$ can be replaced with any transitive subgroup (for the irreducibility argument to hold), and hence we can replace $G_n$ with a subgroup of order $n8^n$.

$\endgroup$
7
$\begingroup$

One source of examples in dimension $2^{n}$ for any positive integer $n$ is given by extraspecial $2$-groups of order $2^{2n+1}.$

I won't give all details, but one can argue as follows: if $S$ is an extraspecial $2$-group of order $2^{2n+1},$ then the number of elements not of order $4$ in $S$ is given by $2^{n}(2^{n}+ \nu(\chi)),$ where $\nu(\chi)$ is the Frobenius-Schur (henceforth F-S)indicator the unique non-linear complex irreducible character $\chi$ of $S$. This uses the well-known formula that the number of solutions of $x^{2} = 1$ in $S$ is given by $\sum_{\mu \in {\rm Irr}(S)} \nu(\mu) \mu(1).$

For any positive $n,$ there are two non-isomorphic extraspecial groups of order $2^{2n+1},$ and they are distinguishable because they have different numbers of elements of order $4$. Each of those groups have $2^{2n}$ linear characters (all realizable over $\mathbb{R}),$ and one real-valued irreducible character of degree $2^{n}.$

Hence one of the groups has the real-valued irreducible character of degree $2^{n}$ with F-S indicator $1$ and one has such a character with F-S indicator $-1$. The latter comes from a quaternionic representation.

If you want to know which extraspecial group has the quaternionic representation, you can argue as follows.

One of the extraspecial groups is (up to isomorphism) a central product of $n$ copies of $D_{8}$, and the other is (up to isomorphism) a central product of $n-1$ copies of $D_{8}$ with one copy of $Q_{8}.$ The first of these groups clearly has a representation of degree $2^{n}$ which is realizable over $\mathbb{R}$ as $D_{8}$ has a $2$-dimensional absolutely irreducible representation over $\mathbb{R}.$

Hence the $2^{n}$-dimensional faithful representation of a central product of $n-1$ copies of $D_{8}$ with one copy of $Q_{8}$ is a quaternionic representation.

Later edit: Note that (as is well-known) the central product of $n$ copies of $D_{8}$ therefore has $2^{2n}-2^{n}$ elements of order $4$ and that the central product of $n-1$ copies of $D_{8}$ with one copy of $Q_{8}$ has $2^{2n} +2^{n}$ elements of order $4$.

Later edit in a different (more general) direction: If you have enough information about the irreducible characters of a finite group $G,$ then it is easy to check whether $G$ has quaternionic representations or not. If the sum of the degrees of the non-trivial real-valued complex irreducible characters of $G$ is equal to the number of involutions of $G,$ then $G$ has no quaternionic irreducible representations, and in any other case, $G$ does have (necessarily non-trivial) irreducible quaternionic representations.

In the latter case, when $G$ is also simple, any quaternionic representation is likely to have relatively large degree, since for any integer $m >1$, there are only finitely many simple groups with an $m$-dimensional complex irreducible representation.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.