# Homotopy groups of $MO(2)$

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 $\mathbb{R}^n\times [0,1]$.

• 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. – Ryan Budney Nov 6 '14 at 16:07
• A preliminary computation with the unstable Adams spectral sequence seems to support Ryan: it looks to me like $\pi_5(MO(2)) = 0$. – Tyler Lawson Nov 6 '14 at 16:56
• 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. – Mark Grant Nov 7 '14 at 14:59
• @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? – Tyler Lawson Nov 7 '14 at 22:21

There are a couple of references which show that $\pi_5(MO(2))=\mathbb{Z}/2$.
The proof here is geometric, and (for me) harder to follow. This paper also contains a proof that $\pi_6(MO(2))=0$.