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.

Convergence comparisons between the Pade' approximation (blue) and the closed form formula

Notice that to simplify the problem, everything may be formulated in $SO(3)$ at the moment.


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  • $\begingroup$ Velcome to the site! $\endgroup$ – kjetil b halvorsen May 28 '14 at 13:51
  • $\begingroup$ 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 $\partial \frac{\exp(a_i\log(\exp(\omega)T))P_i-P_i^*}{\partial \omega_k} = \exp(\omega)\cdot G_k$. $\endgroup$ – B0rk4 May 29 '14 at 8:29
  • $\begingroup$ 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 $\frac{\partial \omega_{ai}(\mathbf{\omega})}{\partial \mathbf{\omega}_k} = exp(\omega_{ai}) \cdot \mathbf{G}_k$ $\endgroup$ – Pierluigi May 30 '14 at 8:57

I suggest reading this :


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).

  • 1
    $\begingroup$ I want to know how to derive in the paper. $\endgroup$ – jakeoung Apr 26 '15 at 9:55

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