How does one show that $SO(6)/U(3)\cong \Bbb C\rm P^3$? The identification of $\Bbb R^6$ with $\Bbb C^3$ induces an inclusion $U(3)\hookrightarrow SO(6)$. The homogeneous space $SO(6)/U(3)$ can then be identified as the space of almost complex structures on $\Bbb R^6$ compatible with the standard inner product and a given orientation. I have seen several papers that claim $SO(6)/U(3)\cong \Bbb C \rm P^3$, i.e. that this manifold is diffeomorphic to three-dimensional complex projective space. In the paper "Harmonic and holomorphic maps" (chapter three in this book), for instance, Salamon writes
$$SO(6)/U(3)\cong SU(4)/S(U(3)\times U(1))\cong \Bbb C\rm P^3 $$
The second diffeomorphism is clear to me (this is the standard description of $\Bbb C \rm P^3$ as a homogeneous space, induced by the transitive action of $SU(4)$ on orthonormal bases of $\Bbb C^4$). 
However, I'm not sure how to understand that first step. I do know that there is an isomorphism of Lie groups $U(3)\cong S(U(3)\times U(1))$, defined by sending $A\mapsto (A,\det A^{-1})$. Since $SU(4)$ is the double (universal) cover of $SO(6)$, it seems like the "block-diagonal subgroup" $S(U(3)\times U(1))\subset SU(4)$ must also be a double cover of the subgroup defined by the natural inclusion of $U(3)$ into $SO(6)$. However, I seem to be stuck at that point. Therefore, my main question is:

How do I prove the claim $SO(6)/U(3)\cong \Bbb C\rm P^3$ (or $SO(6)/U(3)\cong SU(4)/S(U(3)\times U(1))$)?

 A: I'll answer this rather indirectly.  Let $\pi:SU(4)\rightarrow SO(6)$ denote the double covering map.  I'll use the notation $U$ to denote the subgroup $\pi^{-1}(U(3))$ where $U(3)\subseteq SO(6)$ is the one you identified previously.
Proposition 1:  The homogeneous space $SO(6)/U(3)$ is naturally diffeomorphic to the homogeneous space $SU(4)/U$.
Proof:  Consider the map $\psi:SU(4)/U\rightarrow SO(6)/U(3)$ given by $\psi(g \, U) = \pi(g) \, U(3)$.
The map $\psi$ is well defined:  for $h\in U$, we have $\psi(gh\, U) = \pi(gh) \, U(3) = \pi(g)\pi(h) U(3) = \pi(g) U(3) = \psi(g\, U)$ since $\pi(h)\in U(3)$.
The map $\psi$ is surjective:  for $g\, U(3) \in SO(6)/U(3)$, pick $h\in SU(4)$ with $\pi(h) = g$.  Then $\psi(h\, U) = g\, U(3)$.
The map $\psi$ is injective:  assume $\psi(g \, U) =\psi(h\, U)$ for some $g,h\in SU(4)$.  Then $\pi(g) \, U(3) = \pi(h)\, U(3)$, so $\pi(g)(\pi(h))^{-1} = \pi(gh^{-1})\in U(3)$.  It follows that $gh^{-1}\in \pi^{-1}(U(3)) = U$, so $g\, U = h\, U$.
The map $\psi$ is smooth: this follows from the fact that $SU(4)\rightarrow SU(4)/U$ is a submersion and the fact that $\pi$ is smooth.
The map $\psi^{-1}$ is smooth:  To see this, we show that $\psi$ has full rank everywhere, and then appeal to the inverse function theorem.  The map $SU(4)\rightarrow SO(6)$ is a submersion because its a covering.  Since $SO(6)\rightarrow SO(6)/U(3)$ is a submersion, the composition $SU(4)\rightarrow SO(6)\rightarrow SO(6)/U(3)$ is a submersion.
On the other hand, this composition is the same as the composition $SU(4)\rightarrow SU(4)/U\xrightarrow{\psi} SO(6)/U(3)$, so this composition is a submersion.  It follows from the chain rule that $\psi$ must be surjective on tangent spacces.
This completes the proof that $\psi$ is a diffeomorphism.  $\square$
(Of course, Proposition 1 is much more general - it works for any covering $G\rightarrow G'$ and subgroup $K\subseteq G'$.)
So we have reduced our problem to identifying $U = \pi^{-1}(U(3))$.  
Proposition 2:  $U$ is connected.
Proof:  We first note that $\pi^{-1}(I) = \{\pm I\}$, so it is enough to find a path in $U$ which connects $I$ with $-I$.
Consider the path $\gamma:[0,2\pi]\rightarrow U(3)$ given by $\gamma(t) = \operatorname{diag}(e^{it},1,1)$, which generates $\pi_1(U(3))$.  Under the inclusion $U(3)\rightarrow SO(6)$, this maps to a matrix having a block form $\begin{bmatrix} \cos\theta & \sin \theta \\ -\sin \theta & \cos\theta \end{bmatrix}$ with the other blocks being $0$ or the identity.  A matrix of this form generates $\pi_1(SO(6))$, so the loop $\gamma$ is homotopically non-trivial.
Now, lift $\gamma$ to a path $\hat{\gamma}$ in $SU(4)$ where $\hat{\gamma}(0) = I$.  If $\hat{\gamma}(2\pi) = I$, then the closed loop $\hat{\gamma}$, which is null homotopic, projects to a null homotopic loop in $SO(6)$.  But it projects to the image of $\gamma$, which we've already argued is an essential loop.  Thus, $\hat{\gamma}(2\pi) = -I$.
Finally, just note that $\pi(\hat{\gamma}) = \gamma\subseteq U(3)$, so $\hat{\gamma}\subseteq U$.  $\square$
So, $U$ is a connected subgroup of $SU(4)$.  Further, $\pi$, when restricted to $U$, is a covering of $U(3)$.  It follows that $U$ is full rank subgroup of $SU(4)$ and that is has dimension $9$.
One can now turn, for example, to the Borel - de Siebenthal classification of maximal connected max rank subgroups of simple Lie groups or use some simpler representation theory to see the following:
Proposition 3:  The only maximal connected max rank subgroups of $SU(4)$ are, up to conjugacy, $U(3)$ and $S(U(2)\times U(2))$.
Since $U$ is connected, $U$ must be a subgroup of one of these.  Since $\dim U = 9$, it must be a subgroup of $U(3)$.  Since $U(3)$ has dimension $9$, $U = U(3)$.
