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?
 A: 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 ${\frak{sl}}(3,\mathbb{R})$.
You'll run into the same problem with the conformal group in this example, because the space of conformal transformations in this example is the same as the space of isometries.
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.
