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What is the induced map on fundamental group of the inclusion of unitary group $U(n)$ in the orthogonal group $SO(2n)$?(Note that the unitary group $U(n)$ can only embedded in the group $SO(2n)$, not $SO(n)$(its dimension is greater than $SO(n)$)!)

Lao-tzu
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1 Answers1

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Given that $\pi_1(U(n))\cong \mathbb{Z}$, and $\pi_1(SO(n))\cong\mathbb{Z}/2\mathbb{Z}$, either $i_*\colon\pi_1(U(n))\to \pi_1(SO(n))$ is trivial or surjective.

Note that $U(1)\subset U(n)$ for all $n$ and in fact this inclusion induces an isomorphism on fundamental groups, and so we can instead consider the inclusion $U(1)\cong S^1$ into $SO(n)$ which factors through $U(n)$. As a subgroup, $U(1)$ can be seen as the set of linear maps representing rotations about some fixed axis, and in fact this subset represents the image of a loop which generates $\pi_1(SO(n))$ (as described in this answer). So we have a surjective group homomorphism (factoring through the fundamental group of $U(n)$) $$\pi_1(U(1))\stackrel{\cong}{\to}\pi_1(U(n))\stackrel{i_*}{\to}\pi_1(SO(n))$$ and then as the first map is an isomorphism $\pi_1(U(1))\to\pi_1(U(n))$, we see that $i_*$ must be a surjective homomorphism onto $\pi_1(SO(n))\cong\mathbb{Z}/2\mathbb{Z}$.

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Dan Rust
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  • Just a small correction: the above is correct of course except in the case $n=1$, we have $\pi_1(U(1)) \cong \pi_1(SO(2)) \cong \mathbf{Z}$. The map is still surjective as indicated in the argument above since in this case the Lie groups are isomorphic. – user44441 Nov 25 '13 at 00:59