TITLE: The fifth k-invariant of BSO(3)
QUESTION [12 upvotes]: From work of Pontryagin and Whitney, as I understand it, the homotopy 4-type of $BSO(3)$ is $K(\mathbb{Z}/2,2) \times_{K(\mathbb{Z}/4,4)} K(\mathbb{Z},4)$, where the pullback is along the maps $\mathfrak{P}_2\colon K(\mathbb{Z}/2,2) \to K(\mathbb{Z}/4,4)$ (Pontryagin square) and the obvious $K(\mathbb{Z},4) \to  K(\mathbb{Z}/4,4)$. In particular, the $k$-invariant $k_4$ is given by the composite 
$$
K(\mathbb{Z}/2,2)\to K(\mathbb{Z}/4,4)\to K(\mathbb{Z},5)
$$
where the latter map arises from the short exact sequence $\mathbb{Z} \stackrel{\times 4}{\to}\mathbb{Z} \to \mathbb{Z}/4$

Has the next $k$-invariant been identified in the literature? 

This would be a map
$$
K(\mathbb{Z}/2,2) \times_{K(\mathbb{Z}/4,4)} K(\mathbb{Z},4) \to K(\mathbb{Z}/2,6)
$$
as $\pi_5(BSO(3)) \simeq \mathbb{Z}/2$. An obvious option would be (reduction mod 2 then) cup product. But there may be some other cohomology operation floating around.

Edit: For $BSpin(3) = BSU(2)$ the analogous $k$-invariant is given by the map
$$
K(\mathbb{Z},4) \to K(\mathbb{Z}/2,4) \xrightarrow{Sq^2} K(\mathbb{Z}/2,6)
$$

REPLY [10 votes]: Recall that $SU(2)\cong Spin(3)$ and $PU(2)\cong SU(2)/\mathbb{Z}_2\cong SO(3)$. Originally Woodward calculated the lower stages of a Postnikov decomposition of $PU(n)$ in  “The Classification of Principal PU(n)-bundles Over a 4-complex.” If I recall, his cacluations did not apply fully to the case $n=2$ and they were amended by Antieau and Williams in On the classification of oriented 3-plane bundles over a 6-complex [1]. I think the calculations they perform are as complete as exist in the literature.
[1] Topology and its Applications, 173 (2014) pp 91-93, doi:10.1016/j.topol.2014.05.012, arXiv:1209.2219