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
$$
K(\mathbb{Z},4) \to K(\mathbb{Z}/2,4) \xrightarrow{Sq^2} K(\mathbb{Z}/2,6)
$$

isthe map? $\endgroup$ – David Roberts Sep 20 '16 at 9:001more comment