# The surjectivity of the exponential map for the isometry group

Little is known on general conditions guaranteeing that the exponential map between a Lie algebra and an associated Lie group is surjective.

Let $M$ be a noncompact connected Riemann manifold, and $G$ be its (Lie) group of isometries. Does anybody know whether the exponential map between the Lie algebra of $G$ and the connected component of $G$ is surjective?

(If $M$ is compact then $G$ also is, which in this case answers my question in the affirmative.)

What about replacing the isometry group with the group of conformal transformations?

Edit: The question has been answered in the negative by @RobertBryant below. Does anybody know what conditions (other than compactness) should be added to get an affirmative answer?

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No. Consider, for example, $M=\mathrm{SL}(3,\mathbb{R})/\mathrm{SO(3)}$ endowed with its $\mathrm{SL}(3,\mathbb{R})$-invariant Riemannian metric $g$ (which is unique up to a positive constant multiple). This is an irreducible symmetric space of noncompact type. The identity component of the isometry group of $(M,g)$ is $\mathrm{SL}(3,\mathbb{R})$ itself (since $\mathrm{SL}(3,\mathbb{R})$ has trivial center), and the exponential map of $\mathrm{SL}(3,\mathbb{R})$ is not surjective.
To see this, note that, when all of the generalized eigenvalues of $a\in{\frak{sl}}(3,\mathbb{R})$ are real, then $A=\exp(a)$ has positive (generalized) eigenvalues, and, when $a$ has one real eigenvalue and two distinct complex conjugate eigenvalues, say, $z\not=\bar z$ and $-(z{+}\bar z)$, the eigenvalues of $A = \exp(a)$ are $\lambda = e^z$, $\bar\lambda = e^{\bar z}$, and $1/(\lambda\bar\lambda) = e^{-z-\bar z}>0$. Thus, $A = \exp(a)$, cannot have eigenvalues $(-\tfrac12, -2, 1)$ since $-\tfrac12$ and $-2$ are not complex conjugates. In particular $A = \mathrm{diag}(-\tfrac12, -2, 1)\in\mathrm{SL}(3,\mathbb{R})$ is not the exponential of anything in ${\frak{sl}}(3,\mathbb{R})$.
Added comment in response to OP's edit: Unfortunately, I think that, without more restrictions on the class of Riemannian metrics of interest, this is probably not a very sensible question. For example, for any connected semi-simple group $G$, the generic left-invariant metric on $G$ will have $G$ as the identity component of its isometry group, so there are many examples of such metrics for each $G$ for which the exponential map is not surjective (which happens for very many if not 'most' non-compact semi-simple Lie groups). Thus, for example, the generic left-invariant metric on $G=\mathrm{SL}(2,\mathbb{R})$ gives a $2$-parameter family of $3$-dimensional examples that are distinct up to homothety.