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!

  • 11
    $\begingroup$ 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
    Commented Jan 3, 2015 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
    Commented Jan 3, 2015 at 20:47
  • $\begingroup$ See also here for an answer. $\endgroup$ Commented Jan 5, 2015 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. $\endgroup$ Commented Jan 5, 2015 at 14:17

1 Answer 1


This is following from the maximal torus theorem. Let $T$ be a maximal torus in the compact, connected Lie group $G$. Then, the $\exp$ map is surjective on $T$. The theorem says that every element in $G$ is contained in some maximal torus and any two maximal tori are conjugate to each other. The surjectivity of $\exp$ on $G$ follows from here.

For reference, Lie groups, Lie algebras, and Representations: An Elementary Introduction 2ed by Brian C. Hall chapter 11 is very useful.


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