For the 3D rotation operation $R^{-1}(R(\omega_0)*R(\omega))$, how can we compute the derivative wrt $\omega$?

Question

How can we compute the Jacobian derivative of the function:

$$f(\omega) = R^{-1}(R(\omega_0) R(w))$$

with respect to $\omega$, where $\omega_0 \in \mathbb{R}^3$ is some fixed/constant vector, the function: $$R(\omega) \triangleq \exp \big( [\omega]_{\times} \big) \triangleq \exp \left( \begin{bmatrix} 0 & -w_z & w_y \\ w_z & 0 & -w_x \\ -w_y & w_x & 0 \\ \end{bmatrix} \right), $$ is the Rodrigues-vector-to-rotation mapping, and $R^{-1}(\cdot)$ is the corresponding inverse function?

For the sake of this question, we are primarily interested in taking the derivative about the point $\omega=0$, since other values of $\omega$ can be absorbed into the constant $\omega_0$ with a little extra effort. So everything should be expressible in terms of $\omega_0$.

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Edit: Using some dirty empirical methods, I was able to get the first few coefficients of the Taylor expansion as: $$ \frac{d f}{d \omega} = Z^0 + \frac{1}{2!} Z^1+ \frac{1}{2 (3!)}Z^2 - \frac{1}{6 (5!)}Z^4 + \frac{1}{6 (7!)} Z^6 - (...) $$ where $Z=[\omega_0]_{\times}$. (Evidently, there no higher order odd terms beyond the first.) So this appears to have a similar form to Rodrigues, as: $$\frac{d f}{d \omega} = I + \frac{1}{2} Z + \beta(\omega_0) Z^2$$ but getting a functional form for $\beta$ seems nontrivial. (I'd venture a guess than $\beta$ is purely a function of $\|\omega_0\|$.)

Answer 1

To begin with, let's get rid of the logarithm map, and hit both sides by an arbitrary test vector $\mathbf{v}$: $$R_{f(\omega)}\mathbf{v} = R_{\omega_0} R_{\omega}\mathbf{v}.$$

Now we can differentiate both sides, making use of the formulas (derived using geometric arguments) in Gallego and Yezzi, A Compact Formula for the Derivative of a 3-D Rotation in Exponential Coordinates, Journal of Mathematical Imaging and Vision, Volume 51 Issue 3, March 2015, 378--384:

$$\left[\left(d R_{\omega}\right) \delta \omega\right]\mathbf{v} = -R_{\omega} [\mathbf{v}]_\times T(\omega)\delta \omega,$$ for variation $\delta\omega$ and $$T(\omega) = \begin{cases}I, & \|\omega\| = 0\\ \frac{\omega\omega^T + (R_{-\omega}-I)[\omega]_\times}{\|\omega\|^2}, & \|\omega\| > 0.\end{cases}$$

We have $$\left[dR_{f(\omega)} df(\omega)\delta\omega\right] \mathbf{v} = -R_{\omega_0} R_{\omega} [\mathbf{v}]_{\times} T(\omega) \delta \omega$$ $$-R_{f(\omega)} \left[\mathbf{v}\right]_{\times} T[f(\omega)] df(\omega) \delta \omega =-R_{\omega_0} R_{\omega} [\mathbf{v}]_{\times} T(\omega) \delta \omega.$$ The rotations at the beginning cancel, and since the variation $\delta\omega$ and test vector $\mathbf{v}$ are arbitrary, we must have equality of matrices $$df(\omega) = T[f(\omega)]^{-1}T(\omega),$$ with special case $$df(0) = T(\omega_0)^{-1}.$$

Answer 2

$ \def\a{\alpha} \def\b{\beta} \def\g{\gamma} \def\m{\lambda} \def\e{\varepsilon} \def\t{\theta} \def\s{\sigma} \def\w{\omega} \def\o{{\tt1}} \def\c#1{\color{red}{#1}} \def\l{\left}\def \r{\right}\def\lr#1{\l(#1\r)} \def\LR#1{\bigg(#1\bigg)} \def\fracLR#1#2{\lr{\frac{#1}{#2}}} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\gradLR#1#2{\fracLR{\p #1}{\p #2}} \def\cay#1{\operatorname{cay}\lr{#1}} \def\skew#1{\operatorname{skew}\lr{#1}} \def\pp#1{\operatorname{psi}\lr{#1}} \def\ndot{\mathop{\bullet}\limits} $First, a sketch of the Rodrigues and Gallego & Yezzi formulas.

Let $\e$ denote the Levi-Civita tensor, then given a vector $(a),\,$ its associated magnitude $(\a)$, skew matrix $(A)$, and rotation matrix $(R)$ can be calculated as $$\eqalign{ \a &= \|a\|,\;\qquad A = -\e\cdot a \\ R &= e^A = I + \rho_1 A + \rho_2 A^2 \\ \rho_1 &= \frac{\sin(\a)}{\a} \quad\quad \rho_2=\frac{\o-\cos(\a)}{\a^2} \\ }$$ The skew matrix has an interesting periodic property for any $n\ge\o$ $$A^{n+2} = -\a^2A^n$$ allowing any analytic function of $A$ to be reduced to a quadratic polynomial.

Gallego & Yezzi utilize one such function $$\eqalign{ G &= \pp{A} = I + \g_1 A + \g_2 A^2 \\ \g_1 &= \frac{\cos(\a)-\o}{\a^2} = -\rho_2 \\ \g_2 &= \frac{\a-\sin(\a)}{\a^3} = \frac{\o-\rho_1}{\a^2} \\ }$$ where $\phi(x) = \fracLR{e^x-1}{x}\;$ is a function commonly seen in exponential integrators
$\;$ and $\;\pp{x} = \phi(x)\,e^{-x}$

They actually employ a different (but equivalent) expression for $G$ $$G = \fracLR{aa^T+(R^T-I)A}{\a^2}$$ and derive an amazing formula for the gradient of $Ru$ (where $u$ is an arbitrary vector)
$$\eqalign{ \grad{(Ru)}{a} &= R\cdot\lr{\e\cdot u}\cdot G \\ }$$ Multiplying each side by $e_k\,$ yields their so-called compact formula for the gradient of $R$ $$\eqalign{ &\grad{(Ru)}{a_k} &= +R\cdot\lr{\e\cdot u}\cdot(Ge_k) \\ &\grad{R}{a_k}\cdot u &= -R\cdot\lr{\e\cdot g_k}\cdot u \\ &\grad{R}{a_k} &= -R\cdot\e\cdot g_k \\ &\grad{R}{a} &= -R\cdot\e\cdot G \\ \\ }$$

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In the current problem, instead of a single vector $\{a\}$, there are three vectors $\{f,w_0,w\}$ whose respective rotation matrices satisfy the relationship $$R_f = R_0\,R_w $$ The differential of this relationship leads directly to the desired gradient $$\eqalign{ dR_f &= R_0\cdot dR_w \\ \gradLR{R_f}{f}\cdot df &= R_0\cdot\gradLR{R_w}{w}\cdot dw \\ \c{R_f\cdot\e}\cdot G_f\cdot df &= \c{R_0\cdot R_w\cdot\e}\cdot G_w\cdot dw \\ G_f\cdot df &= G_w\cdot dw \\ \grad{f}{w} &= G_f^{-1}G_w \\\\ }$$

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Update

There was a question in one of the comments on how to invert $G$.

Since the inverse function will also be a quadratic polynomial one can immediately write $$\eqalign{ B &= G^{-1} = I + \b_1 A + \b_2 A^2 \\ }$$ The equation $BG=I$ generates a $2\times 2$ system for the $\beta$-coefficients in terms of the $\g$-coefficients, which can be solved using Cramer's Rule. Then the $\{\g_k\}$ are substituted by the $\{\rho_k\}$ to obtain the result in terms of the Rodrigues coefficients. $$\eqalign{ \b_1 &= \frac{1}{2}, \quad \b_2 = \frac{2\rho_2-\rho_1}{2\a^2\rho_2} \\ }$$

The various coefficients are trigonometric functions of $\a$ which have finite limits as $\a\to 0$ $$\eqalign{ \rho_1 &= 1, \quad &\rho_2 &= \frac{1}{2}, \quad \g_1 &= \frac{-1}{2}, \quad &\g_2 &= \frac{1}{6}, \quad \b_1 &= \frac{1}{2}, \quad &\b_2 &= \frac{1}{12} \\ }$$