I am reading https://limhyungtae.github.io/2022-04-01-IMU-Preintegration-(Easy)-4.-Derivation-of-Preintegrated-IMU-Measurements/
And trying to work my way through it, I already asked some questions wrt right-jacobians but I am also stuck on how you derive $Exp(\phi) R = R Exp(R^T \phi)$ We are working on SO(3).
$\phi$ is the vector representing a rotation where $||\phi||_2$ is the angle. I believe it is implicitly put in skew-symmetric form.
Naively expanding the series defintion hasn't lead me anywhere
The notation used in your question is confusing and is actually hiding important details. What I think is going on here is that vector argument of exponential map is transformed into skew-symmetric matrix under the hood. Let me explain:
Let $v$ be a 3d vector, $[v]_×$ be a $3 \times 3$ skew-symmetric matrix (a.k.a. the "cross product" matrix) of veector $v$ and $R$ be a $3 \times 3$ rotation matrix.
The skew-symmetric matrix $[v]_×$ belongs to the Lie algebra of the rotation group SO(3) so the expomential map $\exp([v]_×)$ gives a rotation matrix via Rodrigues formula.
The notation you're using is "wrong" in the sense thar you are applying the exponential map to a vector $v$ like $\exp(v)$. That doesn't make actual sense. But it seem the author you are following has stablished the following abuse of notation: $\exp(v) = \exp([v]_×)$.
The skew-symmetric matrices forms a vector space and rotations can be applied to them where the following identity holds:
$[R \ v]_× = R \ [v]_× \ R^T$
So the following relation holds:
$\exp([R \ v]_×) = \exp(R \ [v]_× \ R^T)$
But $\exp(R \ [v]_× \ R^T) = R \ \exp([v]_×) \ R^T$ as can be checked by expanding $\exp(R \ [v]_× \ R^T)$ using the Rodrigues formula, so:
$\exp([R \ v]_×) = R \ \exp([v]_×) \ R^T$
Then going back to your question, it is basically:
$\exp([v]_×) \ R = R \ \exp([R^T \ v]_×)$
Which can be rewrited as:
$R^T \ \exp([v]_×) \ R = \exp([R^T \ v]_×)$
Which clearly holds.