Consider the action of $S^1$ on the product of 3-spheres $S^3\times S^3$ defined by:
$$e^{it}.(z_1, z_2)=(e^{2it}z_1, e^{3it}z_2)$$
where $z_1, z_2\in S^3$. Here we understant $e^{2it}z_1$ as the multiplication by $e^{2it}$ in each component of $z_1$ when we look at $S^3$ as a subspace of $\mathbb{C}^2$ (more precisely, $S^3=\{(a,b)\in \mathbb{C}^2: |a|^2+|b|^2=1\}$).
So my question is what is (topologically, for instance) the quotient of $S^3\times S^3$ by this action? I'm convinced that this should give $S^3\times S^2$ because this action is in some sense a "twisted" Hopf fibration, but I've not been able to show this (at least in an explicit way).
What about using quaternions ? Identify $S^3$ with the Lie group quaternions of norm 1, we see that the manifold $X$ is the homogeneous space $T $ \ $ G $ where $T$ is a circle, and $G=S'\times S"$; $S'$ and $S"$ being two samples of $S$. Now the group $S$ freely acts on the right on $G$ by the digaonal action, and this action pass to the quotient on a free right action of $S$, making $X$ a $S$ principal bundle over something $T $ \ ($ S'\times S" $) \ $S$. To identify this quotient with $S^2$ note that this manifold is homogeneous under the right action of the simply connected group $ S' \times S"$ with a connected isotropy group $T\times S$. Now a principal $G$ bundle over $S^2$ is trivial if $G$ is simply connected, therefore the manifold is the trivial $S$ bundle over $S^2$. It seems rather non trivial to get an explicit trivialization as $S\times S$ viewed as a $T\times S$ bundle over $S^2$ is not trivial.