Understanding equation manipulation on manifolds I cannot understand a specific equation manipulation of the paper:
C. Forster, L. Carlone, F. Dellaert and D. Scaramuzza, "On-Manifold Preintegration for Real-Time Visual--Inertial Odometry," in IEEE Transactions on Robotics, vol. 33, no. 1, pp. 1-21, Feb. 2017, doi: 10.1109/TRO.2016.2597321
Specifically, I do not understand the equation manipulation from line 1 to line 2 of equation (35) in the paper. In this question i stick to the same equation numbering as in the paper and simply copy the equations. They define the following properties of the exponential maps of the $\operatorname{SO}(3)$ manifold:
\begin{align}
 R \operatorname{Exp} (\phi) R^\top &= \operatorname{exp} (R \mathbf{\phi}^\wedge R^\top ) = \operatorname{Exp} (R\phi) \tag{10}\label{eq10} \\
\Leftrightarrow \operatorname{Exp}(\mathbf{\phi}) R &= R \operatorname{Exp} (R^\top \phi) 
\tag{11}\label{eq11}
\end{align}
They say that they rearrange line 1 of equation \eqref{eq35} by using the relation \eqref{eq11}:
\begin{align} 
&\dotsc \Pi_{k=i}^{j-1} 
\big[ 
\operatorname{Exp} ( ( \mathbf{\tilde{\omega}}_k - \mathbf{b}_i^g ) \Delta t)  
\operatorname{Exp} \big( -J_r^k\mathbf{\eta}^{gd}_k  \Delta t \big)
\big] \\
&\stackrel{\eqref{eq11}}{=} \Delta \tilde{R}_{ij} \Pi_{k=i}^{j-1} 
\big[
\operatorname{Exp} \big( - \color{red}{\Delta \tilde{R}_{k+1 j}^\top } J_r^k \mathbf{\eta}^{gd}_k \Delta t \big)
\big] \tag{35}\label{eq35}  , 
\end{align}
where
$\Delta \tilde{R}_{ij} \doteq \Pi_{k=i}^{j-1} \operatorname{Exp} (( \mathbf{\tilde{\omega}} - \mathbf{b}_i^g ) \Delta t ) $.
But I do not understand where the red marked term
$ \color{red}{\Delta \tilde{R}_{k+1 j}^\top } $
comes from. As far as I am concerned the red term appears magically since they summarize all factors
$\Delta \tilde{R}_{ij} \doteq \Pi_{k=i}^{j-1} \operatorname{Exp} (( \mathbf{\tilde{\omega}} - \mathbf{b}_i^g ) \Delta t ) $
into one term and it simply remains
$ \Pi_{k=i}^{j-1} 
\big[ 
\operatorname{Exp} \big( -J_r^k\mathbf{\eta}^{gd}_k  \Delta t \big)
\big] $
without $ \color{red}{\Delta \tilde{R}_{k+1 j}^\top } $ . What am I missing here?
 A: I found the answer after working on it for a while. Using
\begin{align} 
\operatorname{Exp} ( \mathbf{\phi} ) R = R \operatorname{Exp} (R^\top \phi) \tag{1}\label{eq1} 
\end{align}
Weird the above formula is displayed correctly in editing mode but is wrongly displayed in the answer for me.
and defining
$(\mathbf{\tilde{\omega}}_k- \mathbf{b}_i^g)\Delta t = a_k$
and
$-J_r^k\mathbf{\eta}_k^{gd} \Delta t = b_k $ we have:
\begin{align}
&\dotsc \Pi_{k=i}^{j-1} 
\big[ 
\operatorname{Exp} ( ( \mathbf{\tilde{\omega}}_k - \mathbf{b}_i^g ) \Delta t)  
\operatorname{Exp} \big( -J_r^k\mathbf{\eta}^{gd}_k  \Delta t \big) \\
&= \Pi_{k=i}^{j-1} 
\big[ \operatorname{Exp} (a_k) \operatorname{Exp} (b_k) \big] \\
&=  \operatorname{Exp} (a_i) \operatorname{Exp} (b_i) \operatorname{Exp} (a_{i+1}) \operatorname{Exp} (b_{i+1}) .. \operatorname{Exp} (a_{j-2}) \operatorname{Exp} (b_{j-2}) \operatorname{Exp} (a_{j-1}) \operatorname{Exp} (b_{j-1}) \\
&= R_i \operatorname{Exp} (b_i) R_{i+1} \operatorname{Exp} (b_{i+1}) .. R_{j-2} \operatorname{Exp} (b_{j-2}) R_{j-1} \operatorname{Exp} (b_{j-1}) \\
&\stackrel{\eqref{eq1}}{=} R_i \operatorname{Exp} (b_i) R_{i+1} \operatorname{Exp} (b_{i+1}) .. R_{j-2} R_{j-1} \operatorname{Exp} (R_{j-1}^\top b_{j-2})  \operatorname{Exp} (b_{j-1}) \\
&\stackrel{\eqref{eq1}}{=} R_i \operatorname{Exp} (b_i) R_{i+1}..R_{j-1} \operatorname{Exp} (R_{i+2}^\top..R_{j-1}^\top b_{i+1}) .. \operatorname{Exp} (R_{j-1}^\top b_{j-2})  \operatorname{Exp} (b_{j-1}) \\ 
&\stackrel{\eqref{eq1}}{=} R_i..R_{j-1} \operatorname{Exp} (R_{i+1}^\top..R_{j-1}^\top b_i)\operatorname{Exp} (R_{i+2}^\top..R_{j-1}^\top b_{i+1}) .. \operatorname{Exp} (R_{j-1}^\top b_{j-2})  \operatorname{Exp} (b_{j-1}) \\
&= \Delta \tilde{R}_{ij} \Pi_{k=i}^{j-1} \operatorname{Exp} 
\big( \Delta \tilde{R}_{k+1 j}^\top b_k \big) \\
&= \Delta \tilde{R}_{ij} \Pi_{k=i}^{j-1} \operatorname{Exp} 
\big( -\Delta \tilde{R}_{k+1 j}^\top J_r^k\mathbf{\eta}_k^{gd} \Delta t \big)
\end{align}
The last simplification is done using $\Delta \tilde{R}_{ij} \doteq \Pi_{k=i}^{j-1} \operatorname{Exp} (( \mathbf{\tilde{\omega}} - \mathbf{b}_i^g ) \Delta t ) = \Pi_{k=i}^{j-1} \operatorname{Exp} (a_k)  $.
