Let $H$ be a 1-dimensional connected and closed subgroup of $SO(3)$. Denote $\mathfrak h$ the lie algebra of $H$. We know that the exponential map $\exp:\mathfrak h\to H$ must be a local diffeomorphism. My question is: $\exp$ is also a global diffeomorfism?
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1$\begingroup$ Not necessarily. If your subgroup is a circle $\exp$ will not be injective. However it will be surjective, since it is always surjective for connected compact groups. See terrytao.wordpress.com/2011/06/25/… $\endgroup$– Maik PicklJan 31, 2017 at 14:54
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No. $H$ is compact since it is closed in $SO(3)$ which is compact. Moreover, since $H$ is $1$-dimensional and connected, it must be a circle $H\cong S^1$. Thus, the exponential map is $$\mathbb{R}\to S^1,\quad x\mapsto e^{ix},$$ which is surjective but has kernel $2\pi\mathbb{Z}$.
Added later: It is also worth mentioning that if $G$ is a compact Lie group, then the exponential map is never injective. Simply take a maximal torus $T\subseteq G$. Then, $T\cong S^1\times\cdots\times S^1$ ($k$-times for some $k>0$) and the exponential map restricted to the Lie algebra $\mathfrak{t}\cong\mathbb{R}^k$ of $T$ is $(x_1,\ldots,x_k)\mapsto(e^{ix_1},\ldots,e^{ix_k})$, which has a non-trivial kernel as above.
However, if $G$ is compact and connected, one can show that $\exp:\mathfrak{g}\to G$ is always surjective (but this is a bit difficult to show, see for example Knapp's book Lie groups beyond an introduction).
