Showing $\frac{\partial}{\partial \Phi}\left(\Phi\circ\exp(\varphi)\circ\Phi^{-1}\right) = I - C(\Phi)C(\varphi)C(\Phi)^T$ for 3D rotations

Question

I am trying to reproduce a result from "A Primer on the Differential Calculus of 3D Orientations" - Bloesch, 2016

Consider equations (72) and (73):

(72): $\frac{\partial}{\partial \Phi}\left[\exp(\Phi(\varphi)) = \Phi\circ\exp(\varphi)\circ\Phi^{-1}\right]$

(73): $-\Gamma(\Phi(\varphi))\Phi(\varphi)^\times = I - C(\Phi)C(\varphi)C(\Phi)^T$

From that, I am having trouble figuring out, why this holds:

$\frac{\partial}{\partial \Phi}\left(\Phi\circ\exp(\varphi)\circ\Phi^{-1}\right) = I - C(\Phi)C(\varphi)C(\Phi)^T$

The article mentions using the chain rule, the product rule, and identities 29, 30, 28:

(28): $\frac{\partial}{\partial \Phi}(\Phi^{-1}) = -C(\Phi)^T$

(29): $\frac{\partial}{\partial \Phi_1} (\Phi_1 \circ \Phi_2) = I$

(30): $\frac{\partial}{\partial \Phi_2} (\Phi_1 \circ \Phi_2) = C(\Phi_1)$

Therefore, I have tried the following:

Set:

$f(\Phi) = \Phi \circ exp(\varphi)$

and

$g(\Phi) = \Phi^{-1}$

Then

$\frac{\partial}{\partial \Phi}\left(\Phi\circ\exp(\varphi)\circ\Phi^{-1}\right) = \frac{\partial}{\partial \Phi}\left(f(\Phi) \circ g(\Phi)\right)$

And from the identities:

$f'(\Phi) = I$

$g'(\Phi) = -C(\Phi)^T$

Then I apply the product rule (equation 38 in The Matrix Cookbook):

$\frac{\partial}{\partial \Phi}\left(f(\Phi) \circ g(\Phi)\right) = f'(\Phi) \circ g(\Phi) + f(\Phi) \circ g'(\Phi)$

$= I \circ \Phi^{-1} + \Phi \circ exp(\varphi) \circ (-C(\Phi)^T)$

I then simplify this. For that, I use, that I am working with 3D orientations, and for rotation matrices $C^{-1} = C^T$:

$= C(\Phi)^{T} - C(\Phi) \circ C(\varphi) \circ C(\Phi)^T$

This brings me to my problem, because I am left with $C(\Phi)^{T}$ as the first term in this equation, whereas I should end up with the identity matrix. Where did I go wrong here?

I have tried verifying this numerically, and from that it seems that I should in fact end up with the identity matrix term instead of the one I end up with. I also find it suspicious, that the author mentions the use of identity 30, which I do not seem to use.

Answer 1

Consider the following sequence of mappings:

$$ \Phi \mathop{\longmapsto}\limits^{f_1\strut} (g(\Phi), h(\Phi)) \mathop{\longmapsto}\limits^{f_2\strut} g(\Phi) \circ h(\Phi) $$

Where $\circ$ is the concatenation operation defined in the paper. Define $i = f_2 \bullet f_1$ where $\bullet$ denotes the function composition. The product rule is the expanded form of the expression $\frac{\partial}{\partial \Phi} i(\Phi)$: $$ \frac{\partial}{\partial \Phi} i(\Phi) = \frac{\partial}{\partial \Phi} (f_2 \bullet f_1)(\Phi) = \Big( \frac{\partial f_2}{\partial f_1(\Phi)} \Big) \Big( \frac{\partial f_1}{\partial \Phi} \Big) = \begin{bmatrix}\frac{\partial}{\partial g}(g \circ h)&\frac{\partial}{\partial h}(g \circ h)\end{bmatrix} \begin{bmatrix}\frac{\partial}{\partial \Phi} g(\Phi)\\ \\ \frac{\partial}{\partial \Phi} h(\Phi)\end{bmatrix} $$ According to the identities (29) and (30), this could be further simplified as $$ \begin{bmatrix}\boldsymbol{I}&\boldsymbol{C}(g(\Phi))\end{bmatrix} \begin{bmatrix}\frac{\partial}{\partial \Phi} g(\Phi)\\ \\ \frac{\partial}{\partial \Phi} h(\Phi)\end{bmatrix} $$ Performing the multiplication, the product rule is obtained: $$ \frac{\partial}{\partial \Phi} (g(\Phi) \circ h(\Phi)) = \frac{\partial}{\partial \Phi} g(\Phi) + \boldsymbol{C}(g(\Phi)) \frac{\partial}{\partial \Phi} h(\Phi) $$ Returning back to the problem of computing $\frac{\partial}{\partial \Phi}\left(\Phi \circ \exp (\varphi) \circ \Phi^{-1}\right)$, take $g=\Phi$ and $h= \exp (\varphi) \circ \Phi^{-1}$: \begin{align*} \frac{\partial}{\partial \Phi}\left(\Phi \circ \exp (\varphi) \circ \Phi^{-1}\right) &= \frac{\partial}{\partial \Phi} \Phi + \boldsymbol{C}(\Phi) \frac{\partial}{\partial \Phi} ( \exp (\varphi) \circ \Phi^{-1} ) \\ &= \boldsymbol{I} + \boldsymbol{C}(\Phi) ( \boldsymbol{C}(\varphi) \frac{\partial}{\partial \Phi} \Phi^{-1}) \\ \\ &= \boldsymbol{I} - \boldsymbol{C}(\Phi) \boldsymbol{C}(\varphi) \boldsymbol{C}(\Phi)^{T} \tag*{(identity (28)) } \end{align*} Which is the result presented in the paper.