Let $\mathbf r =(w,x,y,z)$ be a unit vector on $\mathbb R^4$. If I am not mistaken, any point on the unit $3$-sphere can be parametrized as
$\left(\begin{matrix}w & x\\ y & z \end{matrix}\right)=\left(\begin{matrix}\cos \theta_1 & -\sin \theta_1\\ \sin \theta_1 & \cos\theta_1 \end{matrix}\right).\left(\begin{matrix}\cos\theta_2 & 0\\ 0 & \sin\theta_2 \end{matrix}\right).\left(\begin{matrix}\cos \theta_3 & \sin \theta_3\\ -\sin \theta_3 & \cos\theta_3 \end{matrix}\right)$,
with $\,\,0\leq\theta_1,\theta_3\leq\frac{\pi}{4}$, $\,\,0\leq\theta_2\leq 2\pi$.
If we keep $\theta_2$ fixed, what is the resulting 2-dimensional manifold which is embedded in this $3$-sphere? I know that keeping $\theta_2$ fixed, breaks the $O(4)$ symmetry of the 3-sphere to $O(2)\times O(2)$. I am wondering, if the topology of this 2-manifold is something well-known like a $2$-sphere, $2$-torus or Klein bottle?
Edit
After reading Kyle Miller's comment, I think, a better way to parametrize the $3$-sphere is by taking the ranges $\,\,0\leq\theta_1,\theta_3\leq2\pi$, $\,\,-\frac{\pi}{4}\leq\theta_2\leq \frac{\pi}{4}$.