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How do I show that for every compact connected group $G$, the exponential map $\exp \colon\mathfrak{g} \rightarrow G$ is surjective?

I tried to find the proof on the internet but most of them are either just a short note or "left as an exercise for reader" with some hints like: use invariant inner product and existence of geodesic but I don't really understand.

So if someone could point out where to find a complete proof of this or give me a more extensive hints on how to start the proof that would be great.

Thank you!

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    $\begingroup$ One can find some results here: cuhkmath.wordpress.com/2011/06/28/… basically it said that when given a bi-invariant metric on $G$, the two notions of exponential maps coinside. Then by Hopf Rinow theorem in Riemannian geometry, the exponential map is surjective. $\endgroup$ – user99914 Jan 3 '15 at 16:42
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    $\begingroup$ The outline in the comment above is a good one. Note that you can't have too simple a proof, as the exponential map is not surjective $\mathfrak{sl}_2 \mathbb{C} \rightarrow SL_2 \mathbb{C}$ (and the Lie group $SL_2 \mathbb{C}$ is connected, but not compact). Note the corresponding metric is not Riemannian. $\endgroup$ – aes Jan 3 '15 at 20:47
  • $\begingroup$ See also here for an answer. $\endgroup$ – Dietrich Burde Jan 5 '15 at 13:18
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    $\begingroup$ Note that the linked MSE question is not a duplicate of this one, as the link deals with debunking a false proof of surjectivity of $\exp$. Another sketch of a proof is in Terry Tao's blog: terrytao.wordpress.com/2011/06/25/…. In the blong you can also find an interesting alternative argument which uses symplectic geometry instead of Hopf-Rinow. $\endgroup$ – Moishe Kohan Jan 5 '15 at 14:17

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