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To satisfy a referee, I need a reference for the following well-known fact (which is not hard to prove, but it seems silly to prove it in a paper). I can't find it in either Serre or Fulton-Harris's books on representation theory. Can anyone provide me a reference?

Let $Q$ be the $8$-element quaternion group, so we have a central extension $$1 \longrightarrow \mathbb{Z}/2 \longrightarrow Q \longrightarrow (\mathbb{Z}/2)^2 \longrightarrow 1.$$ Then the following constitute a complete list of irreducible representations of $Q$ over $\mathbb{R}$.

  1. One-dimensional representations that factor through $(\mathbb{Z}/2)^2$.

  2. A four-dimensional representation $W$ obtained from the left action of $Q$ on the real quaternions (viewed as a $4$-dimensional real vector space).

I remark that the representation $W$ is irreducible but not absolutely irreducible -- when we extend the field of scalars to $\mathbb{C}$, it breaks up into two copies of the (unique) two-dimensional irreducible complex representation.

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I much prefer to write (and read) papers that are as self-contained as possible... So it doesn't seem at all silly to reprove a well-known fact if it is useful and relevant. Given that the fact you mention can be proved in 2 lines - as per the answers below - that seems a much better solution to me than providing a reference. –  Nick Gill Oct 16 '12 at 9:40
    
Gill! I didn't know you were on MO. Well everybody is, I suppose. Hope you're well. –  Pierre Oct 16 '12 at 16:30

4 Answers 4

up vote 2 down vote accepted

Since $Q_8$ is an extraspecial group, if you want you can say that the result follows from Quillen's classification of their real representations in

Quillen, The mod 2 cohomology rings of extra-special 2-groups and the spinor groups

Mathematische Annalen, 1971

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Where exactly is this done in Quillen's paper? –  darij grinberg Oct 17 '12 at 4:36
    
As a reader, I would prefer not to have this result referred to that paper which is probably a few hundred orders of magnitude more complicated that the result itself! –  Mariano Suárez-Alvarez Oct 17 '12 at 4:53
    
@darij grinberg: this is Proposition 5.8 in Quillen's paper. @Mariano: it is my understanding that the only point is to satisfy a referee :) And yes, even understanding the notation takes longer than proving the result for oneself, but hey, somebody asked for a reference. –  Pierre Oct 17 '12 at 11:58
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There were a number of great answers, but I really like the passive-aggressive quality of this one :). –  Tina Oct 27 '12 at 3:29

This example is elementary (close to an exercise), which does make it tempting just to write a couple of lines of explanation in a paper. But including such a "proof" without further comment tends to leave the impression that the ideas were just discovered. It's easy to refer to an older textbook such as I.M. Isaacs Character Theory of Finite Groups (now an AMS reprint): Exercise (2.4) combined with page 145 on the real case. The easy fact, usually just stated without comment, is that the complex linear characters of a finite group are those of the quotient by the derived group (here a Klein 4-group). But the real case needs an explicit reminder.

There's no magic way to deal with "well-known" facts (if they are indeed well-known), but here it's easy to give in to a referee request without inflating the paper. By the way, neither Serre nor Fulton-Harris pretends to be as comprehensive a reference for finite group characters as the books by Curtis-Reiner and Isaacs.

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It seems easier simply to supply a proof of two lines. Th unique two dimensional complex representation $V$ is not a real representation, so the direct sum of $V$ and $V\simeq {\overline V}$ is defined over ${\mathbb R}$. By dimension count, you have the four characters (four one dimensional real representations), and this irreducible four dimensional one; so this gives the dimension of the group algebra over ${\mathbb R}$ (namely $8$).

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Here is a stupid proof which, I think, would not elicit any further questions:

Let $\left[1\right]$, $\left[-1\right]$, $\left[i\right]$, $\left[-i\right]$, $\left[j\right]$, $\left[-j\right]$, $\left[k\right]$, $\left[-k\right]$ be the elements of $Q$ which correspond to $1$, $-1$, $i$, $-i$, $j$, $-j$, $k$, $-k$ under the canonical embedding $Q\subseteq \mathbb H^{\times}$. Then, the $\mathbb R$-linear map $\mathbb H\oplus \mathbb R\oplus \mathbb R\oplus \mathbb R \oplus \mathbb R$ which sends every $\left(a+bi+cj+dk,e,f,g,h\right)$ to

$\dfrac{\left[1\right]-\left[-1\right]}{2}\left(a+b\left[i\right]+c\left[j\right]+d\left[k\right]\right) $ $+ \dfrac{\left[1\right]+\left[-1\right]}{2}\left(\left(e+f+g+h\right)\left[1\right]+\left(e-f-g+h\right)\left[i\right]+\left(e+f-g-h\right)\left[j\right]+\left(e-f+g-h\right)\left[k\right]\right)$

(for all reals $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$) is easily seen to be an $\mathbb R$-algebra isomorphism. Now, the representation theory of a direct product of algebras is well-known, and so is the representation theory of fields.

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The algebra $\mathbb RQ_8$ is semisimple. Since there are $|Q_8/Q_8'|=4$ modules of degree $1$ and the tautological representation on $\mathbb H$ is irreducible and has $\mathbb H$ as its endomorphism ring, as one can see at once, the wedderburn factorization must be $\mathbb R\times\mathbb R\times\mathbb R\times\mathbb R\times\times\mathbb H$, for there is no room for anything else. –  Mariano Suárez-Alvarez Oct 17 '12 at 5:05

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