Let $U$ be the infinite unitary group $\lim\_{n\to\infty}U(n)$. It is well known that, over the rationals, $BU$ is homotopy equivalent to $\prod\_{n=1}^\infty K(\mathbb{Z}, 2n)$.


**Question:** Is it true that, for any $k\geq 0$, the homotopy group $\pi\_{2k+1}(BU(m))$ is 0 for $m$ sufficiently large?



Odd homotopy group of $BU(m)$
From the [long exact sequence of homotopy groups](https://en.wikipedia.org/wiki/Homotopy_group#Long_exact_sequence_of_a_fibration) associated to the fibration $U\to EU\to BU$, one has that $\pi\_{2k+1}(BU(m))=\pi\_{2k}(U(m))$ (since $EU$ is contractible, and for $k>0$). [From the proof of](https://ncatlab.org/nlab/show/unitary+group#homotopy_groups) the [Bott periodicity theorem](https://en.wikipedia.org/wiki/Bott_periodicity_theorem), $\pi\_k(U(m))$ stabilizes for $m$ large (and equals $\pi\_k(U)$). Moreover, $\pi\_k(U)=\pi\_{k+2}(U)$. So for even $k$, $\pi\_k(U)=\pi\_0(U)=0$. So for $m$ large, $\pi\_{2k+1}(BU(m)) = 0$. 


