# Exponential map is surjective for compact connected Lie group

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!

• 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.
– user99914
Jan 3, 2015 at 16:42
• 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.
– aes
Jan 3, 2015 at 20:47
• 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. Jan 5, 2015 at 14:17
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.