Let $G$ be a Lie group (I only care about finite dimensional connected simply connected nilpotent groups, if that makes the answer easier). Let $\mathfrak g$ be its Lie algebra and let $\exp:{\mathfrak g}\to G$ be the exponential map. The Baker-Campbell-Hausdorff formula expresses $\log(\exp X\cdot\exp Y)$ in terms of (iterated) Lie brackets of $X$ and $Y$ (and when $G$ is nilpotent, this formula is finite).

I'm interested in a similar formula for $\log\big(\exp X\cdot\exp Y\cdot\exp(-X)\big)$. One could in principle use BCH twice and simplify, but I was unsuccessful when attempting that. I don't even care about the coefficients, just about which brackets appear.


1 Answer 1


Suitably interpreted, conjugation commutes with exponentials and logarithms, so this is just $\exp(X) Y \exp(-X)$, or more precisely $\text{ad}_{\exp(X)}(Y)$, and this has a Taylor series expansion

$$\text{ad}_{\exp(tX)}(Y) = \sum_{n \ge 0} \text{ad}_X^n(Y) \frac{t^n}{n!}$$

where $\text{ad}_X(Y) = [X, Y]$, so $\text{ad}_X^2 (Y) = [X, [X, Y]]$ and so forth. This is much easier to see than BCH.

  • $\begingroup$ Took me a while to 'suitably interpret' your answer but I think it all makes sense now, thanks $\endgroup$ Nov 10, 2015 at 4:20
  • $\begingroup$ Ah, this solved my problem too! :D $\endgroup$
    – Ziofil
    Oct 8, 2016 at 2:51

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