# Jacobian of exponential mapping in SO3/SE3

Following this post Jacobian matrix of the Rodrigues' formula (exponential map)

What if I really need the Jacobian of the exponential mapping function in $\omega \neq 0$?

Basically, I want to optimize

$argmin_{\mathbf T} \sum_i \mathbf{e}_i(\mathbf{T}), i = 0..N$ with

• $\mathbf{T} \in SE(3)$
• $\mathbf{e}_i(\mathbf{T}) = exp(a_i log(\mathbf{T}))\cdot P_i - P^*_i$
• $a_i \in [0,1]$
• $P_i, P_i^* \in \mathbb{R}^3$

This can be tought as a variation of the classical ICP (http://en.wikipedia.org/wiki/Iterative_closest_point) problem, where each point $P_i$ is captured from a different pose linearly interpolated (with known factor $a_i$) between the origin and the pose $\mathbf{T}$ that is subject of the estimation.

I want to use a Gauss-Newton like approach, thus I can reformulate the problem in terms of the algebra $\mathfrak{se3}$, i.e.,

$argmin_{\mathbf \omega} \sum_i \mathbf{e}_i(\mathbf{\omega}), i = 0..N$ with

• $\mathbf{\omega} \in \mathfrak{se3}$ (i.e., $\in \mathbb{R}^6$)
• $\mathbf{e}_i(\mathbf{\omega}) = exp(a_i log(exp(\omega)\cdot\mathbf{T}))\cdot P_i - P^*_i$

and I need to evaluate $\left.\frac{\partial \mathbf{e}_i(\mathbf{\omega})}{\partial \mathbf{\omega}}\right|_{\mathbf{\omega} = 0}$

the problem w.r.t. the original question is that the most external $exp (\cdot)$ function is evaluated in a generic point and not in $0$.

I tried succesfully to use the Pade approximation theorem to compute the generic $\frac{\partial \mathbf{e}_i(\mathbf{\omega})}{\partial \mathbf{\omega}}$, but I want something in closed form, if possible!

This comment (Jacobian matrix of the Rodrigues' formula (exponential map)) in the original question states that $\frac{\partial \mathbf{e}_i(\mathbf{\omega})}{\partial \mathbf{\omega}_k} = exp(\omega) \cdot \mathbf{G}_k$. Emplyoing this formula the convergence is not always reached (while the approximated Jacobians works fine), thus I think there is something wrong with this Jacobian. Notice that to simplify the problem, everything may be formulated in $SO(3)$ at the moment.

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• Velcome to the site! – kjetil b halvorsen May 28 '14 at 13:51
• I think you are right, and the equation $\partial \frac{e_i(\omega)}{\partial \omega_k} = \exp(\omega)\cdot G_k$ is wrong. But why do you think: $e_i(\omega)=\exp(\omega)$ The comment you mentioned states $\partial \frac{\exp(\omega)}{\partial \omega_k} = \exp(\omega)\cdot G_k$, but not $\partial \frac{\exp(a_i\log(\exp(\omega)T))P_i-P_i^*}{\partial \omega_k} = \exp(\omega)\cdot G_k$. – B0rk4 May 29 '14 at 8:29
• I believe the author used the chain rule to recover the jacobian of $\mathbf{e}_i(\omega) = \mathbf{e}_i(\omega_{ai}(\omega))$ where $\omega_{ai} = a_i log(exp(\omega)\cdot\mathbf{T})$ and used the partial derivative formula in question only to calculate the term $\frac{\partial \omega_{ai}(\mathbf{\omega})}{\partial \mathbf{\omega}_k} = exp(\omega_{ai}) \cdot \mathbf{G}_k$ – Pierluigi May 30 '14 at 8:57

## 1 Answer

I suggest reading this :

http://vision.in.tum.de/_media/spezial/bib/kerl2012msc.pdf

It uses a variation of Lucas-Kanade to calculate the pose between two cameras.

There is also the formula for the Jacobian of the exponential map not on $0$ in the paper (I don't know how it's derived though).

• I want to know how to derive in the paper. – jakeoung Apr 26 '15 at 9:55