Given an exponential parameterization of a 3D rigid rotation $R\in SO(3)$ by the vector $v = (v_x, v_y, v_z)^T$ I would like to find its second derivatives at the point $v=(0,0,0)$.
Using the exponential parameterization rotations can be expressed using the matrix exponential $\exp$ and a function $J$ that maps vectors $v$ to skew-symmetric matrices corresponding to infinitesimal rotations such that $J(v) = [v]_{\times}$ are cross product matrices.
\begin{equation}
R = \exp\{J(v)\}
\end{equation}
In a paper [1] formulas for the first and second derivatives are provided. Results for the first derivatives are in alignment with well known results related to generators (see this question):
\begin{equation}
\left.\frac{\partial}{\partial v_x}R\,\right|_{v=0} = J(\hat x)
\end{equation}
where $\hat{x}$ is the first unit vector.
Additionally the authors give a formula for the second derivatives without much explanation:
\begin{equation}
\left.\frac{\partial}{\partial v_x \partial v_y}R\,\right|_{v=0} = \frac{1}{2}(J(\hat x)J(\hat y) + J(\hat y)J(\hat x))
\end{equation}
In [2] a formula for the second directional derivative is provided, when applied to directions corresponding to the unit vectors this gives the same result ($R_{xx}|_{v=0}=J(\hat{x})^2$).
My question is how the formula for the mixed derivatives could be rigorously derived, so that its correctness can be proven?
[1] C. J. Taylor and D. J. Kriegman, “Minimization on the Lie group SO (3) and related manifolds,” Yale University, 1994.
[2] M. Sarkis and K. Diepold, “Camera-pose estimation via projective Newton optimization on the manifold,” IEEE Trans. Image Process., vol. 21, no. 4, pp. 1729–1741, Apr. 2012.
Best Answer
Use the power series on skew-symmetric matrices $\omega\in\mathfrak{so(3)}$:
$$\exp(\omega) = \sum_{i\geq 0} \frac{\omega^i}{i!}$$
Term-wise differentiation gives: $$\newcommand{\dd}{\mathrm{d}} \dd\exp(\omega).\dd\omega = \sum_{i\geq 0} \frac{\dd\left(\omega^i\right).\dd\omega}{i!} = \dd \omega + \frac{\dd \omega.\omega + \omega.\dd \omega}{2} + O(\omega^2)$$
where the $O(\omega^2)$ denotes an expression that is (at least) quadratic in $\omega$. Now, differentiating once more gives:
$$\dd^2 \exp(\omega).\dd \omega_1.\dd \omega_2 = \frac{\dd \omega_1.\dd \omega_2 + \dd \omega_2. \dd \omega_1}{2} + O(\omega)$$
Setting $\omega = 0$ gives the expected answer. Note that the second derivative is a symmetric matrix and can no longer be represented as a 3-vector via the Lie algebra coordinates on $\mathfrak{so(3)}$.