I have in several places now seen it tossed off as common knowledge that there is a homeomorphism
$$V_2(\mathbb{C}^4)\cong S^5\times S^7$$
between the Stiefel manifold of complex $2$-frames in $\mathbb{C}^4$ and the product $S^5\times S^7$. Of course $V_2(\mathbb{C}^4)\cong SU_4/SU_2$.
I suspect that there is a nice way to get this by identifying $V_2(\mathbb{C}^4)$ with some geometric object, and perhaps it is really a diffeomorphism? Or maybe it stems somehow from the identification of $\mathbb{C}^4$ with the octonions $\mathbb{O}$?
Where does the homeomorphism come from?
$V_2(\mathbb{C}^4)$ is an $S^5$-bundle over $S^7$. But every $S^5$-bundle over $S^7$ is trivial:
$S^5$-bundles over $S^7$ are constructed by gluing two trivial $S^5$-bundle over $D^7$ along the boundary $\partial D^7=S^6$. So they are classified by $$ [S^6,\operatorname{Aut}S^5]=[S_6,SO(6)]=\pi_6(SO(6))=0. $$
We can also see this by identify $\mathbb{C}^4$ with $\mathbb{O}$ as with the proof of parallelisability of $S^7$. Then selecting $x\in S^7$ as the first vector, the second vector has to be in the span of $i_2x,i_3x,i_4x,i_5x,i_6x,i_7x$ since $\mathbb{C}x=\mathbb{R}x+\mathbb{R}i_1x$. But this is precisely what gives the trivialisation of the bundle: $S^5\subset\operatorname{span}_\mathbb{R}\{i_2,i_3,i_4,i_5,i_6.i_7\}$.