In this blog post, Terence Tao mentions that 'If $G$ is a compact connected Lie group, then the exponential map is surjective'. It seems to me from this statement that every element of a compact connected Lie group has some corresponding element in the Lie algebra $\mathfrak{g}$, though the exp map may be many to one. However I have also seen the following statements

1) 'A simply connected Lie group is determined by its Lie algebra...'

2) 'If $G$ is a compact, simply connected and connected, then arbitrary $g\in G$ can be written as $g=e^X$, for some $X\in \mathfrak{g}$.'

These seem to contradict Terence Tao's statement, since now it seems we need the extra condition of being simply connected for elements of $G$ to be determined by $\mathfrak{g}$. What is the reason for this contradiction? (Perhaps when $G$ is simply connected, the map is locally injective?)

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    $\begingroup$ Tao has compact and connected as hypothesis, your second statement has simply connected and connected. Where do you see the contradiction? $\endgroup$ Commented Sep 18, 2015 at 4:45
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    $\begingroup$ There is no contradiction. The subjectivity of the exponential map doesn't mean that the lie group determines the group in the same sense, since there is no clear way to define a map if you don't have a target. On the other hand, given a Lie algebra, I can, no questions asked, give you a simply connected compact lie group in unique way $\endgroup$
    – Pax
    Commented Sep 18, 2015 at 4:47
  • $\begingroup$ @Mariano I've made a correction. $\endgroup$ Commented Sep 18, 2015 at 4:53
  • $\begingroup$ @Chanler Do you mean '..lie algebra determines the group..' in the first sentence? Why do we not have a target if the map is only surjective? $\endgroup$ Commented Sep 18, 2015 at 4:57
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    $\begingroup$ @Meer A Lie algebra doesn't determine the group, there are many groups corresponding to a given Lie algebra, so that the first remark doesn't yield a contridiction. Though, note there is always a surjection from a simply connected group to any group which factors through the identity map (namely the universal cover) so that the two imply each other. $\endgroup$
    – Pax
    Commented Sep 18, 2015 at 7:17

1 Answer 1


You need simply connectedness for uniqueness. For example, both $SU(2)$ and $PSU(2)$ are compact connected group with lie algebra $\mathfrak{su}(2)$. One is covering of the other.

On the other hand, your statements 1) and 2) are called Lie's third fundamental theorem: For a lie algebra $\mathfrak{g}$ of finite dimension over $\mathbb{R}$ or $\mathbb{C}$, there is a connected simply connected lie group $G$ such that $\mathfrak{g}=lie(G)$, unique up to isomorphism.

The proof is by embedding the finite dimensional lie algebra into some $\mathfrak{gl}(n,\mathbb{R})$ and $exp(\mathfrak{g})$ can be seen as a subgroup of $GL(n,\mathbb{R})$. The uniqueness part involves Lie's secord fundamental theorem which uses simply connectedness.

  • $\begingroup$ Doesn’t simply connected imply connected? Why the repetition? $\endgroup$
    – Bcpicao
    Commented Jan 6 at 3:07

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