Can an abelian subgroup of $SO(4)$ be diagonalised over $\mathbb{C}$?

Question

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

Answer 1

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.