# Derivative of exponential mapping of $\mathbb{SE}(3)$ and $\mathbb{SIM(3)}$ in vector/matrix of $\mathfrak{se}(3)$ and $\mathfrak{sim}(3)$.

This question was raised a while ago(here), but since it has not been solved yet and solving the problem is incredibly useful for robotics and 3-D vision etc., here I raised it again.

1. Suppose we have $\mathbf{T} = \begin{bmatrix} \mathbf{R} & \mathbf{t} \\ \mathbf{0} & 1 \end{bmatrix} \in \mathbb{SE}(3)$, where $\mathbf{R} \in \mathbb{SO}(3)$ and $\mathbf{t}\in \mathbb{R}^3$, and $\mathbf{x} = \begin{bmatrix}\mathbf{u}^\top & \mathbf{\omega}^\top \end{bmatrix}^\top \in \mathfrak{se}(3)$, where $\mathbf{u}\in\mathbb{R}^3$ and $\mathbf{\omega} \in \mathfrak{so}(3)$ and $\mathbf{T}=\exp(\mathbf{x})$. The problem is how to derive the gradient or Jacobian matrix of the $\mathbf{T}$ with respect to $\mathbf{x}$:

$$\frac{\partial \mathbf{T}}{\partial \mathbf{x}} = ?$$

1. we can expand the question further to include the derivatives of $\mathbb{SIM}(3)$ with respect to $\mathbb{sim}(3)$, where $\mathbf{S} = \begin{bmatrix} \exp(\lambda)\mathbf{R} & \mathbf{t} \\ \mathbf{0} & 1 \end{bmatrix} \in \mathbb{SE}(3)$, and its corresponding tangent vector is $\mathbf{\delta} = \begin{bmatrix}\mathbf{u}^\top& \mathbf{\omega}^ \top &\lambda\end{bmatrix}^\top$ and $\lambda\in\mathbb{R}$ is a scalar. More details are available here. The problem of calculating the derivatives of $\mathbb{SIM}(3)$ w.r.t. its $\mathfrak{sim}(3)$ is shown as: $$\frac{\partial \mathbf{S}}{\partial \mathbf{\delta}} = ?$$
2. The problem is also closely related to the problem of calculating the derivatives of matrix in $\mathbb{SO}(3)$ w.r.t. its exponential coordinates in $\mathfrak{so}(3)$ i.e. $\frac{\partial \mathbf{R}}{\partial \omega}$, where $\mathbf{R}\in\mathbb{SO}(3)$, $\omega\in\mathfrak{so}(3)$, and $\mathbf{R}=\exp([\omega]_\times)$. Luckily, unlike previous two problems where computational cost was not considered when deriving the derivatives, Gallego and Yezzi provide a neat and compact solutions. Briefly, when $||\omega||\approx0$: $$\frac{\partial \mathbf{R}}{\partial \mathbf{\omega}} = \begin{bmatrix}[\mathbf{e}_1]_\times \\ [\mathbf{e}_2]_\times \\ [\mathbf{e}_3]_\times \end{bmatrix},$$
where $e_i$ is the $i$-th column of an $\mathbf{I}$ identity matrix. otherwise, $$\frac{\partial \mathbf{R}}{\partial \omega_i} = \frac{\omega_i[\mathbf{\omega}]_\times+[\omega\times(\mathbf{I}-\mathbf{R})\mathbf{e}_i]_\times}{||\omega||^2}\mathbf{R}$$
• In the current form it is difficult to answer the question since you are trying to use the notation of classical analysis (in particular $\frac{\partial\mathbf{T}}{\partial\mathbf{x}}$ in a situation where it does not make sense. I think that essentially you want to know the deriviative of the exponential mapping in two cases. But the target in both cases is a manifold rather than an open subset in Euclidean space, so it is not as easy to write out "derivatives" as a matrix. So you have to clarify what you really want (how to view tangent spaces of the groups, etc.) – Andreas Cap Jan 1 '17 at 11:47
• Thank you very much for your comments. I understand the derivative does not belong to Euclidean space in both cases, can you point me out where I should look into for representing such a manifold? Can you also explain why $\frac{\partial T}{\partial x}$ does not make sense? – Shuda Li Jan 1 '17 at 12:02
• "does not make sense" probably was exaggerating a bit, it is just a very ambigous notation and using it in the way one knows from usual analysis carries a lot of danger of making mistakes. The last part you have edited into the question confirms the impression that what you are looking for is the left logarithmic derivative of the exponential map. This is known for general Lie groups for a long time (and can be found in textbooks) so the question is more about translating these results into a language that is understandable and useful for you. – Andreas Cap Jan 1 '17 at 13:21
• Talking about "existing naive solutions" is pretty strange. There are universal formulae for the left and right logarithmic derivatives which are valied for any abstract Lie group, which are much simpler than what you are trying to derive below and they are expressed in a language which is appropriate and efficient for Lie theory. What you are trying is to recast these formulae in very simple cases in a language you are used to. (I completely agree that people working in Lie theory usually do not think about computational efficacy, but still the languge they use has been polished well.) – Andreas Cap Jan 2 '17 at 7:35
• You can find a computation of the left logarithmic derivative of the exponential map in the book by Kolar, Michor and Slovak which is available via mat.univie.ac.at/~michor/listpubl.html#books . You will have to recast it in the language or matrix groups, and there are many introductory books on matrix groups. – Andreas Cap Jan 2 '17 at 7:38

Solution to problem 1

Inspired by Gallego's paper and Eade's notes about Lie Groups, I tried to derive a solution to my own problem 1 (correct me please if you see any error, thanks in advance):

According to Eqn (64) in Eade's notes, we have $\mathbf{T}=\begin{bmatrix}\exp([\omega]_\times) & \mathbf{Wu}\\ \mathbf{0} & 1 \end{bmatrix}=\begin{bmatrix}\mathbf{R} & \mathbf{Wu}\\ \mathbf{0} & 1 \end{bmatrix}$ where $\mathbf{W}=\mathbf{Id}+\frac{1-\cos\theta}{\theta^2}[\omega]_\times+\frac{\theta-\sin\theta}{\theta^3}[\omega]^2_\times$ and $\theta = ||\omega||$. Let's take a look at the partial derivatives w.r.t. to $\mathbf{t}$ first, $$\frac{\partial \mathbf{T}}{\partial t_i} = \begin{bmatrix}\frac{\partial \mathbf{R}}{\partial t_i} & \frac{\partial \mathbf{Wu}}{\partial t_i}\\ \mathbf{0} & 1 \end{bmatrix},$$ where $u_i$ is the $i$-th component of $\mathbf{u}$. It can be seen that $\frac{\partial \mathbf{R}}{\partial u_i}=\mathbf{0}$, since the rotation matrix has nothing to do with translation. $$\frac{\partial \mathbf{Wu}}{\partial u_i}=\mathbf{W}\frac{\partial \mathbf{u}}{\partial u_i}=\mathbf{W},$$ Then we can take a look at the partial derivatives w.r.t. to $\mathbf{\omega}$,

$$\frac{\partial \mathbf{T}}{\partial \omega_i} = \begin{bmatrix}\frac{\partial \mathbf{R}}{\partial \omega_i} & \frac{\partial \mathbf{Wu}}{\partial \omega_i}\\ \mathbf{0} & 1 \end{bmatrix},$$ where $\omega_i$ is the $i$-th component from $\omega$.

It can be seen that we can apply problem 3. for $\frac{\partial \mathbf{R}}{\partial \omega_i}$, and $\frac{\partial \mathbf{Wu}}{\partial \omega_i}=\frac{\partial \mathbf{W}}{\partial \omega_i}\mathbf{u}$. Now problem becomes how to solve $\frac{\partial \mathbf{W}}{\partial \omega_i}$.

\begin{align*} \frac{\partial \mathbf{W}}{\partial \omega_i}&=\frac{\partial (\mathbf{Id}+\frac{1-\cos\theta}{\theta^2}[\omega]_\times+\frac{\theta-\sin\theta}{\theta^3}[\omega]^2_\times)}{\partial \omega_i}\\ &=\frac{\partial(\frac{1-\cos\theta}{\theta^2})}{\partial \omega_i}[\omega]_\times+ \frac{1-\cos\theta}{\theta^2}\frac{\partial [\omega]_\times}{\partial \omega_i} + \frac{\partial(\frac{\theta-\sin\theta}{\theta^3})}{\partial \omega_i}[\omega]^2_\times+ \frac{\theta-\sin\theta}{\theta^3}\frac{\partial [\omega]^2_\times}{\partial \omega_i}\\ &=(\frac{-2(1-\cos\theta)}{\theta^3}+\frac{\sin\theta}{\theta^2})\frac{\partial \theta}{\partial \omega_i}[\omega]_\times+(\frac{1-\cos\theta}{\theta^2})[e_i]_\times+(\frac{-3(\theta-\sin\theta)}{\theta^4}+\frac{1-\cos\theta}{\theta^3})\frac{\partial \theta}{\partial \omega_i}[\omega]^2_\times+\frac{\theta-\sin\theta}{\theta^3}(\mathbf{e}_i\mathbf{v}^\top+\mathbf{v}^\top\mathbf{e}_i-2\omega_i\mathbf{I})\\ &=(\frac{-2(1-\cos\theta)}{\theta^3}+\frac{\sin\theta}{\theta^2})\frac{\omega_i}{\theta}[\omega]_\times+(\frac{1-\cos\theta}{\theta^2})[e_i]_\times+(\frac{-3(\theta-\sin\theta)}{\theta^4}+\frac{1-\cos\theta}{\theta^3})\frac{\omega_i}{\theta}[\omega]^2_\times+\frac{\theta-\sin\theta}{\theta^3}(\mathbf{e}_i\mathbf{v}^\top+\mathbf{v}^\top\mathbf{e}_i-2\omega_i\mathbf{I}) \end{align*}

Similar to the solution to $\mathbb{SO}(3)$, when $||\omega||\approx0$ the $\frac{\partial \mathbf{W}}{\partial\omega_i}$ becomes: $$\frac{\partial \mathbf{W}}{\partial\omega_i}=[\mathbf{e}_i]_\times$$

This derivation should be correct, but I am really not happy about its computational costs and it looks ugly as well in contrast to the beautiful soltion to SO(3) derived by Gallego et al.. Another cue might be useful to professionals is that according to Hausdorff formula the derivative of an exponential map should fulfill:

$$\frac{\partial \exp(\mathbf{X})}{\partial \mathbf{t}}=\mathbf{X}_d\exp(\mathbf{X}),$$ where $\mathbf{X}_d=\frac{\partial \mathbf{x}}{\partial t}+\frac{1}{2!}[\mathbf{X},\frac{\mathbf{x}}{\partial t}]+\frac{1}{3!}[\mathbf{X}, [\mathbf{X},\frac{\mathbf{x}}{\partial t}]] +...$ where $[.]$ is the Lie bracket operation (Page 73 from (1)). For example, in problem 3, we can see that $$\frac{\partial \mathbf{R}}{\partial \omega}=\mathbf{R}_d\mathbf{R},$$ where $\mathbf{R}_d=\frac{\omega_i[\mathbf{\omega}]_\times+[\omega\times(\mathbf{Id}-\mathbf{R})\mathbf{e}_i]_\times}{||\omega||^2}$ similarly, the derivative of exponential map of $\mathbb{SO}(3)$ and $\mathbb{SIM}(3)$ should have the same format: $$\frac{\partial \mathbf{T}}{\partial \omega}=\mathbf{T}_d\mathbf{T}=\begin{bmatrix}\mathbf{R}_d\mathbf{R} & \frac{\partial \mathbf{W}}{\partial \omega_i}\mathbf{W}^{-1}\mathbf{Wu} \\ \mathbf{0} & 1 \end{bmatrix} = \begin{bmatrix}\mathbf{R}_d & \frac{\partial \mathbf{W}}{\partial \omega_i}\mathbf{W}^{-1} \\ \mathbf{0} & 1 \end{bmatrix}\begin{bmatrix}\mathbf{R} & \mathbf{t} \\ \mathbf{0} & 1 \end{bmatrix}= \begin{bmatrix}\mathbf{R}_d & \frac{\partial \mathbf{W}}{\partial \omega_i}\mathbf{W}^{-1} \\ \mathbf{0} & 1 \end{bmatrix}\mathbf{T}$$

Solving problem 2.

According to Eqn(17) to (20) in Spyridon's ICRA15 paper, we have $$\mathbf{S}=\begin{bmatrix}\exp(\lambda+[\omega]_\times) & \mathbf{Vu}\\ \mathbf{0}& 1\end{bmatrix}=\begin{bmatrix}\exp(\lambda)\mathbf{R} & \mathbf{Vu}\\\mathbf{0}&1\end{bmatrix},$$ where

1) $\mathbf{V}=\mathbf{i}$ if $\theta=0$ and $\lambda=0$;

2) $\mathbf{V}= \frac{\exp(\lambda)-1}{\lambda}\mathbf{I}$ if $\theta=0$ but $\lambda\neq0$;

3) $\mathbf{V}=\mathbf{W}$ if $\theta\neq0$ and $\lambda=0$.

4) $\mathbf{V}=\frac{\exp(\lambda)-1}{\lambda}\mathbf{I}+\frac{\theta(1-\exp(\lambda)\cos\theta)+\exp(\lambda)\lambda\sin\theta}{\theta(\lambda^2+\theta^2)}[\omega]_\times+\left ( \frac{\exp(\lambda)-1}{\lambda\theta^2}-\frac{\exp(\lambda)\sin\theta}{\theta(\lambda^2+\theta^2)} -\frac{\lambda(\exp(\lambda)\cos\theta-1)}{\theta^2(\lambda^2+\theta^2)} \right )[\omega]_\times^2$, if $\theta\neq0$ but $\lambda\neq0$;

Now, let's take a look at the derivatives in $\mathbf{u}$ first:

(to be continued)

• References

(1) Bruno Siciliano, O. khatib(Eds. . (2008). Geometric Fundamentals of Robotics.

(2) Drummond, T. (2014). Lie groups, Lie algebras, projective geometry and optimization for 3D Geometry, Engineering and Computer Vision.

(3) Eade, E. (2013). Lie Groups for 2D and 3D Transformations, 1–24.

(4) Gallego, G., & Yezzi, A. (2015). A Compact Formula for the Derivative of a 3-D Rotation in Exponential Coordinates. Journal of Mathematical Imaging and Vision, 51(3), 378–384.

(5) Leonardos, S., Blanchette, C. A., & Gallier, J. (2015). The exponential map for the group of similarity transformations and applications to motion interpolation. In ICRA (Vol. 1).

• Acknowledgement

Cave

greg

Andrea

hans

• It appears that the scalar function you're trying to evaluate in Problem #2 is \eqalign{f(z)&=\phi(z+\lambda)\cr\phi(z)&=\frac{e^z-1}{z}\cr}Again, you can use the 3 distinct eigenvalues of $W$ to solve for the coefficients $$V=f(W)=\phi(W+\lambda I)=a_2W^2+a_1W+a_0I$$ – hans Jan 7 '17 at 21:20
• what is the name of the scalar function? characteristic polynomial? – Shuda Li Jan 7 '17 at 21:44
• Hi Thanks for this hint. In this way, I can verify $V$ and $V^{-1}$ I am wondering if it is possible to use it to calculate the $\frac{\partial V}{\partial \omega_i}$ by calculating $\frac{\partial f(z)}{\partial \omega_i}=\frac{\partial \phi(z+\lambda)}{\partial z}\frac{\partial z}{\partial \omega_i}$ as shown in this question? – Shuda Li Jan 8 '17 at 15:17