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Suppose we have a Lie group $G$ with Lie algebra $\mathfrak{g}$. If $G$ is connected, is it true that $\exp\mathfrak{g} = G$? If $G$ is not connected, does $\exp\mathfrak{g} = G_0$ where $G_0$ is the identity component of $G$?

If we take $ A = \begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix} \in \mathsf{SL}(2; \mathbb{C}) $, I think there does not exist any $X\in M_2(\mathbb{C})$ such that $\mathrm{e}^X = A$ which hints that the Lie algebra does not "generate" the entire (connected) Lie group. That said, is this true, and are there are any other statements we can make about how much of a Lie group is generated by it's Lie algebra?

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  • $\begingroup$ I may have just found my answer in math.stackexchange.com/questions/475385/…. Still wrapping my head around it, so will leave this open for now. $\endgroup$
    – Nate Stemen
    Feb 16, 2021 at 2:35
  • $\begingroup$ "I think there does not exist any X" - why? $\endgroup$
    – lisyarus
    Feb 16, 2021 at 2:40
  • $\begingroup$ One of the eigenvalues of $X$ must exponentiate to -1, and $\operatorname{tr} X = 0$, so $X$ must have distinct eigenvalues. Hence $X$ is diagonalizable, and so should $\mathrm{e}^X$. But $A = \mathrm{e}^X$ is not diagonalizable, so there cannot exist an $X$. Does that make sense? $\endgroup$
    – Nate Stemen
    Feb 16, 2021 at 2:43
  • $\begingroup$ Yes, thank you! $\endgroup$
    – lisyarus
    Feb 16, 2021 at 8:48

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The image of the exponential mapping only "generates" the identity component of the group in the sense of group theory, i.e. the smallest subgroup of $G$ that contains all elements of the form $\exp(X)$ with $X\in\mathfrak g$ is the connected component $G_0$ of the identity. Otherwise put, any element of $G_0$ can be written as a products of finitely many exponentials.

In general, this does not imply that $\exp:\mathfrak g\to G_0$ is surjective and you have a nice counterexample in your question. There is a general result that $\exp:\mathfrak g\to G_0$ is surjective if $G$ is compact, but this is much more difficult.

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  • $\begingroup$ Do you have a reference for your last statement that $\exp$ is injective when $G$ is compact? $\endgroup$
    – Nate Stemen
    Feb 16, 2021 at 16:01
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    $\begingroup$ Sorry, I don't have an explicit reference at hand. Surjectivity of exp for compact groups is a consequence of the theory of maximal tori in compact Lie groups. There one shows that any element of a compact group is conjugate to an element of a fixed maximal torus. (This is rather hard, the proofs I know go via algebraic topolgy and mapping degree.) This in turn implies that any element lies in some maximal torus, and for tori surjectivity of exp is obvious. $\endgroup$
    – Andreas Cap
    Feb 17, 2021 at 10:19
  • $\begingroup$ I think I'm missing something, but: by Baker–Campbell–Hausdorff a product of exponentials is an exponential, thus the image of $\exp$ should already be a subgroup? $\endgroup$
    – lisyarus
    Feb 18, 2021 at 6:03
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    $\begingroup$ Baker-Campbell-Hausdorff only holds on some neighborhood of the identity (for which one already knows that it is in the image of exp). It says nothing about general products of exponentials. $\endgroup$
    – Andreas Cap
    Feb 18, 2021 at 7:10
  • $\begingroup$ @AndreasCap Indeed, thank you! $\endgroup$
    – lisyarus
    Feb 18, 2021 at 10:06

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