Lifting maps of classifying bundles In the paper of Okonek and Van de ven "cubic forms and complex 3-folds". In Proposition 8 there is the claim.
Let $M$ be a closed, orientable smooth 6-manifold, and $t_{x}: X \rightarrow BSO(6)$ be the map ascossiated with the tangent bundle (I guess this is refering to the fact that every smooth manifold has a Riemannian metric).
Then, the obstruction to the existance of an almost complex structure lies in $H^{i}(X, \pi_{i-1}(SO(6)/U(3)))$.
Could somebody provide an answer or reference explaining this fact at a reasonably elementary level? (i.e. someone comfortable with most of Hatcher but not a hardcore algebraic topologist)
 A: The space $BSO(2n)$ is the classifying space for principal $SO(2n)$-bundles. Similarly, the space $BU(n)$ is the classifying space for principal $U(n)$-bundles.
One can take a unitary matrix and write it as a special orthogonal matrix. The complex number $x + iy$ gets sent to the matrix \begin{pmatrix}x 
 & y \\ -y & x\end{pmatrix} and one does this similarly for higher-dimensional matrices. Thus we have a map $U(n) \to SO(2n)$, and by functoriality a map $BU(n) \to BSO(2n)$. By a little bit of work, the (homotopy) fiber of this map is $SO(2n)/U(n)$.
Suppose you have an oriented manifold $M$ of dimension $2n$. Then the oriented orthonormal frame bundle associated to its tangent bundle is a principal $SO(2n)$-bundle, and $M$ admits an almost complex structure if one can reduce the structure group of this bundle to $U(n)$. Put in the language of classifying spaces, this says that one can lift the classifying map $M \to BSO(2n)$ of the oriented orthonormal frame bundle against the map $BU(n) \to BSO(2n)$.
Obstruction theory says that when you want to solve this kind of lifting problem, the obstructions live in cohomology groups of $M$ with coefficients in the (homotopy) fiber of the map you're trying to lift against. The precise groups for this situation are $H^i(M; \pi_{i-1}(SO(2n)/U(n))$.
Maybe this is enough for your situation, but for this specific case it's worth remarking that $\pi_k(SO(2n)/U(n)) \cong \pi_k(SO/U)$ for $k < 2n-1$, where $SO$ and $U$ are the direct limits of $SO(n)$ and $U(n)$ under the usual inclusions $SO(n) \to SO(n+1)$ (and similarly for $U(n)$), respectively. So the coefficient groups for these obstructions live in the stable range except for the very last one that's relevant, $\pi_{2n-1}(SO(2n)/U(n))$. This is nice because the stable groups are either $\mathbb{Z}, \mathbb{Z}/2\mathbb{Z}$, or zero. Also, for these obstructions where the coefficient group works out to $\mathbb{Z}$ the obstruction is actually (up to multiplication by an integer) an integral Stiefel-Whitney class. The first unstable group is not as pretty, but it's either cyclic (sometimes finite, sometimes infinite) or the direct sum $\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$.
Edit: For the case $n=3$.
From the above comments, one can compute that $\pi_j(SO/U)$ is zero for $j=0,1,3,4$ and $\pi_2(SO/U) = \mathbb{Z}$ and the first unstable group is $\pi_5(SO(6)/U(3))$, which actually turns out to be zero! Thus an oriented $6$-manifold $M$ admits an almost complex structure if and only if the obstruction in $H^3(M;\mathbb{Z})$ vanishes. This obstruction is, up to multiplication by some integer, the third integral Stiefel-Whitney class, so $M$ admits an almost complex structure if and only if $W_3(M)$ vanishes.
