What is the maximal compact subgroup of $SO(2,2)$?

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Consider the orthogonal group corresponding to the quadratic form $Q(\vec{x}) = x_1^2 + x_2^2 - x_3^2 - x_4^2 $, then $SO(Q, \mathbb{R}) \simeq SO(2,2)$. We could think of this as the group of $4 \times 4$ matrices leaving the diagonal matrix invariant under conjugation:

$$ \left[ \begin{array}{cc|cc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \hline 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] $$

The main symmetries I can think of are the rotations that change the first two variables, and the second two variables.

$$ \left[ \begin{array}{rc|cc} \cos \theta & \sin \theta & 0 & 0 \\ -\sin \theta & \cos \theta & 0 & 0 \\ \hline 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] , \left[ \begin{array}{cc|rc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \hline 0 & 0 & \cos \theta & \sin \theta \\ 0 & 0 & -\sin \theta & \cos \theta \end{array}\right] $$

This is a copy of the rotation group $SO(2) \times SO(2) \hookrightarrow SO(2,2) $ and there are two distict copies.