TITLE: jacobian involving SO(3) exponential map: $\log(R \exp(m))$
QUESTION [5 upvotes]: 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?

REPLY [6 votes]: 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.