# Odd homotopy group of $BU(m)$

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?

From the long exact sequence of homotopy groups 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 the 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, $\pi_{2k+1}(BU(m)) = 0$.