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

  • $\begingroup$ One might also wonder how this relates to the fifth k-invariant for $BSU(2)$, since $BSU(2) \to BSO(3)$ is an isomorphism in homotopy above dimension 2 -- in particular, are the "same"? $\endgroup$
    – David Roberts
    Sep 20, 2016 at 8:38
  • $\begingroup$ Another thought: it's most likely the sixth Stiefel-Whitney class $w_6$, but what is the map? $\endgroup$
    – David Roberts
    Sep 20, 2016 at 9:00
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    $\begingroup$ In "On universal relations in gauge theory", Trans. Amer. Math. Soc. 348 (1996) 1329-1355, Selman Akbulut computes the 6th stage Postnikov tower of BU(2). While not exactly what you are asking, the techniques might be useful. $\endgroup$
    – Tom Leness
    Sep 20, 2016 at 13:35
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    $\begingroup$ What do you need this for, David? I'm a little puzzled by your comment that "it's most likely the sixth Stiefel-Whitney class $w_6$", since $w_6$ doesn't exist in $H^*(BSO(3);\mathbb{Z}/2)$. Anyway, the computation shouldn't be too difficult, but the result will probably involve indeterminacy. A similar calculation for $MO(2)$ is carried out in "On the realization of the Stiefel-Whitney characteristic classes by submanifolds" by H. Suzuki. $\endgroup$
    – Mark Grant
    Sep 20, 2016 at 17:08
  • $\begingroup$ @MarkGrant well, it was a guess. :-) I didn't know that about $w_6$! $\endgroup$
    – David Roberts
    Sep 20, 2016 at 20:35

1 Answer 1


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

  • $\begingroup$ So they come to the conclusion I come to, then actually prove it (I was working on instinct (!)) and then, in essence, show the relevant k-invariant is not $w_2^3$ or $(Sq^1 w_2)^2$, but the third basis element for $H^6(BSO(3)[4],\mathbb{Z}/2)$, which has dimension 3. Hmm. $\endgroup$
    – David Roberts
    Sep 20, 2016 at 23:18

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