Let $G_0$ be the identity component of a (compact) Lie group $G$ and $G_1$ one of its connected components. Is there any relation between $G_0$ and $G_1$? For example, is there a $g\in{G_1}$ such that $L_g\left(G_0\right)=G_1$? Here $L_g$ is the left-translation-by-$g$ map. Also, let's consider the map $\varphi\equiv{L_g}\circ\exp\colon\mathfrak{g}\to{G_1}$. Is this map injective or surjective?
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All the connected components are diffeomorphic to $G_0$. Choose some $g \in G_1$. Then $L_g : G \to G$ is continuous, therefore $L_g(G_0)$ is connected; since it contains $g = L_g(e) \in G_1$, you get $L_g(G_0) \subset G_1$. Similarly $L_{g^{-1}}(G_1) \subset G_0$. The two maps $L_g : G_0 \to G_1$ and $L_{g^{-1}} : G_1 \to G_0$ are in fact inverse to each other, and they're both differentiable, so you get a diffeomorphism $G_0 \cong G_1$.
answered Jan 17 '17 at 9:45
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For the second question: if $G$ is compact, then so is $G_0$ and hence also $G_1$ (using that $G_0\cong G_1$ as Najib Idrissi showed). The exponential map of a compact connected Lie group is surjective (but that is difficult to show, see for example Knapp's book "Lie groups beyond an introduction"). So $\exp:\mathfrak{g}\to G_0$ is surjective and hence $L_g\circ\exp:\mathfrak{g}\to G_1$ is surjective. But it is not injective in general. (Take for example $G=S^1$).
