jacobian involving SO(3) exponential map: $\log(R \exp(m))$ I would like to compute the 3 × 3 Jacobian of
$$
\log(R \exp(m))
$$
with respect to the 3-vector $m$, evaluated at $m=0$. In the above, $\exp$ is the exponential map from so(3) to SO(3), $\log$ is the inverse of the exponential map, and $R$ is a constant 3 × 3 rotation matrix.
I understand that the inverse of the exponential is not well-defined everywhere. Is there an expression for this Jacobian that holds almost everywhere?
 A: This Jacobian is well understood through the concept of 'Right Jacobian of SO(3)', noted $J_r(\theta)$, where $\theta \in R^3$ is a rotation vector. The right Jacobian is defined by:
$$
J_r(\theta) = \frac{\partial \exp([\theta]_\times)}{\partial \theta}
$$ 
and can be computed as
$$
J_r(\theta) = I - \frac{1-\cos||\theta||}{||\theta||^2}[\theta]_\times + \frac{||\theta||-\sin||\theta||}{||\theta||^3}[\theta]_\times^2
$$
and its inverse as
$$
J_r^{-1}(\theta) = I + \frac12[\theta]_\times + \left(\frac1{||\theta||^2} - \frac{1+\cos||\theta||}{2||\theta||\sin||\theta||}\right)[\theta]_\times^2
$$
The matrix $J_r$ is $3\times3$ and has among others the following property:
$$
\log(\exp([\theta]_\times)\exp([m]_\times))=\theta+J_r^{-1}(\theta)\cdot m
$$
where $R=\exp([\theta]_\times)$ is computed e.g. with the Rodrigues formula.
Therefore your Jacobian is
$$
\frac{\partial\log(R\exp(m))}{\partial m}=J_r^{-1}(\theta)
$$
where
$$
\theta = (\log(R))^\vee
$$
is the rotation vector such that $R=\exp([\theta]_\times)$.
Note 1: the operator $\cdot^\vee$ is the inverse of the skew operator $[\cdot]_\times$.
Note 2: the strict expression for exponential $R=\exp(\theta)$ used in the question, is $R=\exp([\theta]_\times)$ which I used in the answer. They are therefore the same exponential. The same happens with the logs.
You can find references to the right Jacobian with its formulae in "Chirikjian, G. S. (2012). Stochastic Models, Information Theory, and Lie Groups, volume 2: Analytic Methods and Modern Applications of Applied and Numerical Harmonic Analysis. Birkhauser, Basel.", at page 40.
A: Just to clarify to @rsp1984 a little better.
When we speak of Jacobians in Lie groups, we usually mean this:
$$
df(X)/dX = lim_{d\to0}\frac{Log(f(X)^{-1}f(X*Exp(d)))}{d}
$$
This means that the numerator is always a vector (due to Log -- see my comment above for its definition). The denominator is always a vector, which gets 'added' to the group element X by exponentiation and composition. We can in effect define $\oplus,\ominus$ operators as:
$$
Y=X\oplus d = X * Exp(d) 
$$
and
$$
d=Y \ominus X = Log(X^{-1} * Y)
$$
which respectively 'add' a vector to a group element, or subtract group elements to express their difference in vector form.
Then the derivative above reads
$$
df(X)/dX = \lim_{d\to0}\frac{f(X\oplus d) \ominus f(X)}{d}
$$
which now ressembles the typical derivative in vector spaces, that is, the Jacobian matrix, 
$$
dg(x)/dx = \lim_{dx\to0}\frac{g(x + dx) - g(x)}{dx} .
$$
with the +,- signs replaced by the new operators.
Notice however how in the Lie group definition of the Jacobian, the notation has been forced to mean actually a different definition. This is probably the cause of the confusion.
Regarding the question "how is $J_r(\theta)$ a 3x3 matrix?", just apply the definition to the identified function $f:R^3\to SO(3); \theta\mapsto f(\theta)=Exp(\theta)$:
$$
J_r(\theta)\triangleq\frac{dExp(\theta)}{d\theta} = \lim_{d\to0} \frac{Log(Exp(\theta)^{-1} Exp(\theta+d))}{d} 
$$
where we use $\theta+d$, indeed, because the space of departure is $R^3$ and in $R^3$ you have $X\oplus d = X+d$. 
The rest of the answer follows from above: the $Log$ produces a 3-vector on the numerator, and the denominator $d$ is also a 3-vector, thus the 3x3 Jacobian matrix.
It has to be said that finding the closed-forms of $J_r(\theta)$ is not trivial. See Chirikjian's book for reference.
