Can an abelian subgroup of $SO(4)$ be diagonalised over $\mathbb{C}$? I encountered this while studying chapter 4 of Thruston's geometry and topology. I think the answer is yes but really don't know why!
Any help is appreciated!
Thanks
 A: I'm gonna expand on my comment above.  Your question seems to ask if there exists some abelian subgroup of $SO(4)$ which can be diagonalized over $\mathbb{C}$, and indeed we can see that $SO(2)$ consisting of matrices of the form
$$
\left [ \begin{array}{cc}
\cos\theta & -\sin \theta \\
\sin \theta & \cos\theta \\
\end{array} \right ]
$$
is both abelian and diagonalizable.  You can easily demonstrate that the eigenvalues are the solutions of the equation
$$
(\lambda - \cos\theta)^2 + \sin^2\theta \;\; =\;\; 0
$$
and you can show these will be $e^{i\theta}$ and $e^{-i\theta}$.  In particular the corresponding eigenvectors can be taken as
$$
\left (e^{i\theta}, \; \left [ \begin{array}{c}
i/\sqrt{2} \\ 1/\sqrt{2} \\
\end{array} \right ] \right ) \hspace{2pc} \left (e^{-i\theta}, \; \left [ \begin{array}{c}
-i/\sqrt{2} \\ 1/\sqrt{2} \\
\end{array} \right ] \right ).
$$
Thinking of elements $R(\theta) \in SO(2)$ we can consider the subgroup $SO(2)\oplus SO(2)\subset SO(4)$ of the form
$$
\left [ \begin{array}{cc}
R(\theta) & \textbf{0} \\
\textbf{0} & R(\phi) \\
\end{array} \right ].
$$
These matrices can be diagonalized simply by extending the above computations to the 4-dimensional case.
