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Ian Agol
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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 ofFrom 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)=\mathbb{Z}$$\pi_k(U)=\pi_0(U)=0$. So for $m$ large, $\pi_{2k+1}(BU(m)) = \mathbb{Z}$$\pi_{2k+1}(BU(m)) = 0$.

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 $k$, $\pi_k(U)=\pi_0(U)=\mathbb{Z}$. So for $m$ large, $\pi_{2k+1}(BU(m)) = \mathbb{Z}$.

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, $\pi_{2k+1}(BU(m)) = 0$.

Source Link
Ian Agol
  • 68.9k
  • 3
  • 194
  • 358

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 $k$, $\pi_k(U)=\pi_0(U)=\mathbb{Z}$. So for $m$ large, $\pi_{2k+1}(BU(m)) = \mathbb{Z}$.