The quaternion group of order 8 has an irreducible two dimensional representation over $\mathbb{C}$ but how does one show that this representation cannot be defined over $\mathbb{R}$?
Representation of quaternion group over $\mathbb{C}$ and $\mathbb{R}$
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1Please don't give orders to the group (you are using the imperative mode). If you have a question, *ask*, don't *tell*. – 2011-02-23
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0I was wondering, if you know that quaternions already admits an 2x2 matrix representation over C, and had there been another 2x2 matrix representation over R, these two must be isomorphic. But that must mean then there's a (linear) isomorphic map $f: R \rightarrow C$. But that's impossible. Thanks for all the comments. – 2011-02-23
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2No, that's not right. The dihedral group of a square has a 2-dimensional irreducible representation over $\mathbb{C}$ which can be conjugated to be over $\mathbb{R}$. – 2011-02-23
1 Answers
There are sophisticated ways you can do this, but I think its best to push it through in a straightforward way. Here are two approaches; I'm deliberately not giving the details because this is a good homework problem:
Algebraic: Let $C$ be the subgroup $\{ 1, i, -1, -i \}$ of the quaternions. Let $V$ be the real representation of $C$ where $i$ acts by $\left( \begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix} \right)$.
Suppose that the two dimensional representation of the quaternions could be defined over $\mathbb{R}$. Show that the restriction of this representation to $C$ is isomorphic to $V$. In other words, you can always choose a basis where $i$ acts by $\left( \begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix} \right)$. Now write down the equations $ij=ji^3$ and $j^2=i^2$ and try to solve them
Geometric: Any representation on $\mathbb{R}^2$ can be conjugated to preserve the standard inner product; meaning that the action is by rotations and reflections. A case by case analysis should show you pretty quickly that you can't find rotations and reflections of the plane that give an action of the quaternions.
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0Many thanks. I still had to use the geometric argument for the algebraic one: that there is a conjugation that takes $\rho(i)$ to $\left( \begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix} \right)$, where $\rho: Q \rightarrow M_{2}(\mathbb{R}^2)$ is a representation of Q over the real. – 2011-02-23