# Homotopy group of residual symmetry of SO(2n+1)

Dear all,

Is there a way to compute $\pi_2(\frac{SO(2n+1)}{SO(2n-1)\times SO(2)})$?

For the case $n=2$ the answer is $\mathbb{Z}$

How about $\pi_2(\frac{SO(2n)}{SO(2n-2)\times SO(2)})$?

Thanks.

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The answer is $\mathbb{Z}$: one could use the long exact homotopy sequence and the fact that $\pi_2(SO(k))$ is trivial for all $k$ and $\pi_1(SO(k))$ is $\mathbb{Z}/2$ when $k>2$ and $\mathbb{Z}$ for $k=2$, with the map of $\pi_1$'s induced by $SO(k)\subset SO(k+1)$ being non-trivial for all $k\geq 2$. –  algori Nov 3 '12 at 0:07
This helps a lot.Thanks! –  lokman Nov 3 '12 at 8:28