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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 $\mathbb{R}$?

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    No, 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

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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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    Many 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