Have there been any computations of the higher homotopy groups of $MO(2)$, the Thom space of the universal $O(2)$-bundle? Thom himself noted in his landmark 1954 paper that $$ \pi_1(MO(2))=0,\quad \pi_2(MO(2))=\mathbb{Z}/2,\quad\pi_3(MO(2))=0,\quad \pi_4(MO(2))=\mathbb{Z}. $$ By the Pontrjagin-Thom construction $\pi_n(MO(2))$ is the group of cobordism classes of embeddings of closed $(n-2)$-manifolds in $\mathbb{R}^n$, where a cobordism is an embedding in $\mathbb{R}^n\times [0,1]$.

  • 2
    $\begingroup$ I haven't sat down to do the computation, but $\pi_5(MO(2))$ should be reasonably computable. All $3$-manifolds embed in $\mathbb R^5$ so it boils down to the question of which of their null-cobordisms embed in $\mathbb R^6$, which shouldn't be too hard... but I haven't finished my morning cup of tea yet. I suspect somebody has already done this computation. $\endgroup$ – Ryan Budney Nov 6 '14 at 16:07
  • 3
    $\begingroup$ A preliminary computation with the unstable Adams spectral sequence seems to support Ryan: it looks to me like $\pi_5(MO(2)) = 0$. $\endgroup$ – Tyler Lawson Nov 6 '14 at 16:56
  • $\begingroup$ Thank you, both. Which version of the unstable Adams spectral sequence are you using, Tyler? I would very much like to check what $\pi_7$ is, if possible. $\endgroup$ – Mark Grant Nov 7 '14 at 14:59
  • $\begingroup$ @MarkGrant The one based on Ext in the (nonabelian) category of unstable algebras over the Steenrod algebra; it looks like the Postnikov tower might be a more pedestrian approach. I think I'm getting that $\pi_7$ is finite with cyclic factors of size $2$, $4$, and $3$. Is that useful? $\endgroup$ – Tyler Lawson Nov 7 '14 at 22:21

There are a couple of references which show that $\pi_5(MO(2))=\mathbb{Z}/2$.

Suzuki, H., On the realization of the Stiefel-Whitney characteristic classes by submanifolds, Tohoku Math. J., II. Ser. 10, 91-115 (1958). ZBL0107.17001.

The proof uses the Postnikov tower. The conclusion is on page 108.

Zvagel’skij, M. Yu., Cobordisms of embeddings with codimension two, J. Math. Sci., New York 104, No. 4, 1276-1282 (2001); translation from Zap. Nauchn. Semin. POMI 252, 40-51 (2001). ZBL1052.57042.

The proof here is geometric, and (for me) harder to follow. This paper also contains a proof that $\pi_6(MO(2))=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.