What is the Baker-Campbell-Hausdorff (BCH) formula?

For non-commuting matrices X and Y, it expresses log(e^X e^Y) as an infinite series of nested Lie brackets: Z = X + Y + (1/2)[X,Y] + (1/12)([X,[X,Y]] + [Y,[Y,X]]) + …. When X and Y commute ([X,Y]=0) it reduces to e^X e^Y = e^{X+Y}, and on nilpotent Lie algebras it truncates to finitely many terms. It is used to analyze composed transformations on SO(3)/SE(3) and in quantum mechanics.

Why is the differential of the adjoint representation important?

The adjoint action ad_X(Y) = [X,Y] is the building block of both the BCH formula and the differential of the exponential map (the dexp formula). The relation Ad(e^X) = e^{ad_X} reduces perturbation analysis to the exponential of the adjoint action, which appears in robotics attitude estimation and SE(3) sensitivity computations.

What are the Killing form and the Maurer-Cartan form used for?

The Killing form B(X,Y) = tr(ad_X ad_Y) is an invariant bilinear form on a Lie algebra, used to test semisimplicity (Cartan's criterion). The Maurer-Cartan form is the left-invariant 1-form θ = g^{-1}dg on a Lie group; the Maurer-Cartan equation dθ + θ∧θ = 0 expresses that θ is a flat connection (zero curvature). It is used to describe connections and curvature and in geometric control theory.

BCH Formula and Perturbation | Proofs of Matrix Calculus Formulas

11.12 Differential of the Lie Bracket

Formula: $\displaystyle\dfrac{\partial}{\partial X_{ij}} [X, Y] = [J^{ij}, Y]$, $\displaystyle\dfrac{\partial}{\partial Y_{ij}} [X, Y] = [X, J^{ij}]$

Conditions: $X, Y \in \mathbb{R}^{n \times n}$, $[X, Y] = XY - YX$ (Lie bracket / commutator), $J^{ij}$ is the single-entry matrix with a $1$ only in position $(i,j)$.

Proof

By the definition of the Lie bracket (commutator),

\begin{equation}[X, Y] = XY - YX \label{eq:11-12-1}\end{equation}

Differentiate with respect to $X_{ij}$, assuming $Y$ does not depend on $X$.

\begin{equation}\dfrac{\partial [X, Y]}{\partial X_{ij}} = \dfrac{\partial (XY)}{\partial X_{ij}} - \dfrac{\partial (YX)}{\partial X_{ij}} \label{eq:11-12-2}\end{equation}

From 4.3, $\displaystyle\dfrac{\partial X}{\partial X_{ij}} = J^{ij}$.

\begin{equation}\dfrac{\partial (XY)}{\partial X_{ij}} = J^{ij} Y \label{eq:11-12-3}\end{equation}

\begin{equation}\dfrac{\partial (YX)}{\partial X_{ij}} = Y J^{ij} \label{eq:11-12-4}\end{equation}

Substituting $\eqref{eq:11-12-3}$ and $\eqref{eq:11-12-4}$ into $\eqref{eq:11-12-2}$,

\begin{equation}\dfrac{\partial [X, Y]}{\partial X_{ij}} = J^{ij} Y - Y J^{ij} = [J^{ij}, Y] \label{eq:11-12-5}\end{equation}

Similarly, differentiating with respect to $Y_{ij}$,

\begin{equation}\dfrac{\partial [X, Y]}{\partial Y_{ij}} = X J^{ij} - J^{ij} X = [X, J^{ij}] \label{eq:11-12-6}\end{equation}

Note: The Lie bracket is bilinear and antisymmetric, $[X, Y] = -[Y, X]$, and satisfies the Jacobi identity $[X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y]] = 0$. This differential formula is important in optimization problems on a Lie algebra $\mathfrak{g}$ (e.g., geodesic computation).

11.13 Differential of the Adjoint Representation

Formula: $\displaystyle\dfrac{d}{dt} \text{Ad}_{e^{tX}}(Y) \Big|_{t=0} = \text{ad}_X(Y) = [X, Y]$

Conditions: $X, Y \in \mathfrak{g}$ (Lie algebra), $\text{Ad}_g(Y) = gYg^{-1}$ (adjoint representation), $\text{ad}_X(Y) = [X, Y]$ (adjoint action).

Proof

The adjoint representation of a Lie group $G$ is, for $g \in G$ and $Y \in \mathfrak{g}$,

\begin{equation}\text{Ad}_g(Y) = gYg^{-1} \label{eq:11-13-1}\end{equation}

Set $g = e^{tX}$ and compute the derivative at $t = 0$.

\begin{equation}\dfrac{d}{dt} \text{Ad}_{e^{tX}}(Y) = \dfrac{d}{dt} \left( e^{tX} Y e^{-tX} \right) \label{eq:11-13-2}\end{equation}

Applying the product rule, with $\displaystyle\dfrac{d}{dt} e^{tX} = X e^{tX}$ and $\displaystyle\dfrac{d}{dt} e^{-tX} = -X e^{-tX}$,

\begin{equation}\dfrac{d}{dt} \left( e^{tX} Y e^{-tX} \right) = X e^{tX} Y e^{-tX} + e^{tX} Y (-X) e^{-tX} \label{eq:11-13-3}\end{equation}

\begin{equation}= X e^{tX} Y e^{-tX} - e^{tX} Y X e^{-tX} \label{eq:11-13-4}\end{equation}

Evaluate at $t = 0$. Since $e^{0 \cdot X} = I$,

\begin{equation}\dfrac{d}{dt} \text{Ad}_{e^{tX}}(Y) \Big|_{t=0} = XY - YX = [X, Y] \label{eq:11-13-5}\end{equation}

This is the definition of the adjoint action $\text{ad}_X$ of the Lie algebra.

\begin{equation}\text{ad}_X(Y) = [X, Y] \label{eq:11-13-6}\end{equation}

Note: The adjoint representation $\text{Ad}: G \to \text{GL}(\mathfrak{g})$ is a homomorphism from the Lie group to the automorphism group of the Lie algebra. Its differential is the adjoint action $\text{ad}: \mathfrak{g} \to \text{End}(\mathfrak{g})$. This means $\text{ad}_X$ is a linear transformation on $\mathfrak{g}$ and admits a matrix representation.

11.14 Exponential of the Adjoint Action

Formula: $\displaystyle e^{\text{ad}_X}(Y) = \text{Ad}_{e^X}(Y) = e^X Y e^{-X}$

Conditions: $X, Y \in \mathfrak{g}$ (Lie algebra), $e^{\text{ad}_X} = \displaystyle\sum_{k=0}^{\infty} \displaystyle\dfrac{(\text{ad}_X)^k}{k!}$

Proof

The $k$-th power of $\text{ad}_X$ is expressed as a nested Lie bracket.

\begin{equation}(\text{ad}_X)^k(Y) = \underbrace{[X, [X, \cdots [X}_{k \text{ times}}, Y] \cdots ]] \label{eq:11-14-1}\end{equation}

Let $f(t) = e^{tX} Y e^{-tX}$ and expand it in a Taylor series.

\begin{equation}f(t) = \displaystyle\sum_{k=0}^{\infty} \dfrac{t^k}{k!} f^{(k)}(0) \label{eq:11-14-2}\end{equation}

We have $f(0) = Y$. Since $f'(t) = X f(t) - f(t) X = [X, f(t)]$,

\begin{equation}f'(0) = [X, Y] = \text{ad}_X(Y) \label{eq:11-14-3}\end{equation}

By induction, $f^{(k)}(0) = (\text{ad}_X)^k(Y)$.

\begin{equation}f^{(k)}(t) = [X, f^{(k-1)}(t)] \label{eq:11-14-4}\end{equation}

\begin{equation}f^{(k)}(0) = [X, f^{(k-1)}(0)] = \text{ad}_X(f^{(k-1)}(0)) = (\text{ad}_X)^k(Y) \label{eq:11-14-5}\end{equation}

Evaluating at $t = 1$,

\begin{equation}e^X Y e^{-X} = f(1) = \displaystyle\sum_{k=0}^{\infty} \dfrac{(\text{ad}_X)^k(Y)}{k!} = e^{\text{ad}_X}(Y) \label{eq:11-14-6}\end{equation}

Note: This formula can be written as $\text{Ad}_{e^X} = e^{\text{ad}_X}$, a fundamental relation connecting the exponential map of the Lie group with the adjoint action of the Lie algebra. Expanded, $e^X Y e^{-X} = Y + [X, Y] + \displaystyle\dfrac{1}{2!}[X, [X, Y]] + \displaystyle\dfrac{1}{3!}[X, [X, [X, Y]]] + \cdots$.

11.15 Baker-Campbell-Hausdorff Formula

Formula: $\displaystyle e^X e^Y = e^{Z}$, where $Z = X + Y + \displaystyle\dfrac{1}{2}[X, Y] + \displaystyle\dfrac{1}{12}[X, [X, Y]] - \displaystyle\dfrac{1}{12}[Y, [X, Y]] + \cdots$

Conditions: $X, Y \in \mathfrak{g}$ (Lie algebra), with $\|X\|, \|Y\|$ sufficiently small.

Proof

[Strategy] We derive the result two ways. First, (A) expand the series directly to watch the low-order terms reorganize from plain matrix products into commutators (most concrete). Then, (B) use a differential equation to systematize this to all orders.

[(A) Low-order terms by direct expansion]

Expand $e^X$ and $e^Y$ as power series and multiply.

\begin{equation}e^X e^Y = \Bigl(I + X + \tfrac{1}{2}X^2 + \cdots\Bigr)\Bigl(I + Y + \tfrac{1}{2}Y^2 + \cdots\Bigr) \label{eq:11-15-a1}\end{equation}

Collecting terms up to second order,

\begin{equation}e^X e^Y = I + (X + Y) + \Bigl(\tfrac{1}{2}X^2 + XY + \tfrac{1}{2}Y^2\Bigr) + O(3) \label{eq:11-15-a2}\end{equation}

To obtain $Z = \log(e^X e^Y)$, set $W = e^X e^Y - I$ and use $\log(I + W) = W - \tfrac{1}{2}W^2 + \tfrac{1}{3}W^3 - \cdots$. The lowest-order part of $W$ is $X + Y$, so the contribution of $W^2$ up to second order is

\begin{equation}W^2 = (X + Y)^2 + O(3) = X^2 + XY + YX + Y^2 + O(3) \label{eq:11-15-a3}\end{equation}

Substituting $\eqref{eq:11-15-a2}$ and $\eqref{eq:11-15-a3}$ into $\log(I+W) = W - \tfrac{1}{2}W^2 + \cdots$,

\begin{equation}Z = (X + Y) + \Bigl(\tfrac{1}{2}X^2 + XY + \tfrac{1}{2}Y^2\Bigr) - \tfrac{1}{2}\bigl(X^2 + XY + YX + Y^2\bigr) + O(3) \label{eq:11-15-a4}\end{equation}

The $X^2$ and $Y^2$ terms cancel exactly, leaving only $XY$ and $YX$.

\begin{equation}Z = X + Y + \Bigl(XY - \tfrac{1}{2}XY - \tfrac{1}{2}YX\Bigr) + O(3) = X + Y + \tfrac{1}{2}[X, Y] + O(3) \label{eq:11-15-a5}\end{equation}

The second-order term collapses into the commutator $[X, Y]$—this is the heart of the BCH formula. Continuing the same computation to third order gives $Z_3 = \tfrac{1}{12}\bigl(X^2Y + XY^2 - 2XYX + Y^2X + YX^2 - 2YXY\bigr)$ (as listed e.g. on Wikipedia), which likewise reorganizes into nested commutators.

\begin{equation}Z_3 = \tfrac{1}{12}\bigl([X, [X, Y]] + [Y, [Y, X]]\bigr) \label{eq:11-15-a6}\end{equation}

That every order is expressible using commutators alone holds in general (existence theorem). The reason the coefficient $\tfrac{1}{12}$ appears is explained naturally in (B) below, via the Bernoulli number $B_2 = \tfrac{1}{6}$.

[(B) Differential equation giving all orders]

Tracking low-order terms by hand as in (A) becomes rapidly tedious from fourth order on. We therefore introduce $t$, derive the differential equation satisfied by $Z(t) = \log(e^{tX} e^Y)$, and handle all orders at once. Differentiate both sides of $e^{Z(t)} = e^{tX} e^Y$ with respect to $t$.

\begin{equation}\dfrac{d}{dt} e^{Z(t)} = X e^{tX} e^Y = X e^{Z(t)} \label{eq:11-15-1}\end{equation}

By the differential formula for the matrix exponential (11.7, right trivialization $d(e^{Z})\,e^{-Z} = \dfrac{e^{\text{ad}_Z}-1}{\text{ad}_Z}(dZ)$),

\begin{equation}\dfrac{d}{dt} e^{Z(t)} = \dfrac{e^{\text{ad}_{Z(t)}} - 1}{\text{ad}_{Z(t)}} \left( \dfrac{dZ}{dt} \right) e^{Z(t)} \label{eq:11-15-2}\end{equation}

Comparing $\eqref{eq:11-15-1}$ and $\eqref{eq:11-15-2}$,

\begin{equation}\dfrac{e^{\text{ad}_{Z(t)}} - 1}{\text{ad}_{Z(t)}} \left( \dfrac{dZ}{dt} \right) = X \label{eq:11-15-3}\end{equation}

Apply the inverse operator $\displaystyle\dfrac{\text{ad}_{Z(t)}}{e^{\text{ad}_{Z(t)}} - 1}$ to both sides. This function expands as $\displaystyle\dfrac{z}{e^{z} - 1} = \displaystyle\sum_{k=0}^{\infty} \displaystyle\dfrac{B_k}{k!} z^k$ ($B_k$ the Bernoulli numbers, $B_0=1,\ B_1=-\tfrac{1}{2},\ B_2=\tfrac{1}{6},\ B_4=-\tfrac{1}{30},\dots$).

\begin{equation}\dfrac{dZ}{dt} = \dfrac{\text{ad}_{Z(t)}}{e^{\text{ad}_{Z(t)}} - 1}(X) \label{eq:11-15-4}\end{equation}

The initial condition is $Z(0) = Y$. Substituting the expansion $\displaystyle\dfrac{z}{e^{z}-1} = 1 - \dfrac{z}{2} + \dfrac{z^2}{12} + O(z^4)$ into the right-hand side of $\eqref{eq:11-15-4}$ and evaluating at $t = 0$ (where $Z = Y$), the initial rate of change is

\begin{equation}\left.\dfrac{dZ}{dt}\right|_{t=0} = \dfrac{\text{ad}_Y}{e^{\text{ad}_Y} - 1}(X) = X - \dfrac{1}{2}[Y, X] + \dfrac{1}{12}[Y, [Y, X]] + O(4) \label{eq:11-15-5}\end{equation}

\begin{equation}= X + \dfrac{1}{2}[X, Y] + \dfrac{1}{12}[Y, [Y, X]] + O(4) \label{eq:11-15-5b}\end{equation}

Here one sees the coefficient $\tfrac{1}{2}[X,Y]$ agreeing with $\eqref{eq:11-15-a5}$ of (A), and the $\tfrac{1}{12}$ arising from the Bernoulli number $B_2 = \tfrac{1}{6}$.

[Result at $t=1$ (up to fifth order)]

Integrating from $t = 0$ to $t = 1$ determines $Z = Z(1)$. Carrying out the order-by-order expansion and integration (mechanical, but with many terms), the complete expansion up to fifth order is the following known result (Dynkin series). Orders 1–4 are $\eqref{eq:11-15-7}$–$\eqref{eq:11-15-8}$ and the fifth order is $\eqref{eq:11-15-9}$–$\eqref{eq:11-15-11}$; the latter is the asymmetric nested-bracket form of $C_5$ below.

\begin{equation}Z = X + Y + \dfrac{1}{2}[X, Y] + \dfrac{1}{12}\bigl([X, [X, Y]] + [Y, [Y, X]]\bigr) \label{eq:11-15-7}\end{equation}

\begin{equation}- \dfrac{1}{24}[Y, [X, [X, Y]]] \label{eq:11-15-8}\end{equation}

\begin{equation}- \dfrac{1}{720}\bigl([Y, [Y, [Y, [Y, X]]]] + [X, [X, [X, [X, Y]]]]\bigr) \label{eq:11-15-9}\end{equation}

\begin{equation}+ \dfrac{1}{360}\bigl([X, [Y, [Y, [Y, X]]]] + [Y, [X, [X, [X, Y]]]]\bigr) \label{eq:11-15-10}\end{equation}

\begin{equation}+ \dfrac{1}{120}\bigl([Y, [X, [Y, [X, Y]]]] + [X, [Y, [X, [Y, X]]]]\bigr) + O(6) \label{eq:11-15-11}\end{equation}

[Dynkin form (symmetric form)]

The BCH formula has a closed expression due to Dynkin. Let $C_n$ be the term consisting of $n$-th order Lie brackets; then

\begin{equation}Z = \displaystyle\sum_{n=1}^{\infty} C_n \label{eq:11-15-12}\end{equation}

where

\begin{equation}C_1 = X + Y \label{eq:11-15-13}\end{equation}

\begin{equation}C_2 = \dfrac{1}{2}[X, Y] \label{eq:11-15-14}\end{equation}

\begin{equation}C_3 = \dfrac{1}{12}\bigl([X, [X, Y]] - [Y, [X, Y]]\bigr) \label{eq:11-15-15}\end{equation}

\begin{equation}C_4 = -\dfrac{1}{24}[X, [Y, [X, Y]]] \label{eq:11-15-16}\end{equation}

(Note: we used $[Y, [X, [X, Y]]] = -[X, [Y, [X, Y]]]$ via the Jacobi identity.)

[Fifth- and sixth-order terms]

\begin{equation}C_5 = -\dfrac{1}{720}\bigl([X,[X,[X,[X,Y]]]] + [Y,[Y,[Y,[Y,X]]]]\bigr) \label{eq:11-15-17}\end{equation}

\begin{equation}+ \dfrac{1}{360}\bigl([X,[Y,[Y,[Y,X]]]] + [Y,[X,[X,[X,Y]]]]\bigr) \label{eq:11-15-18}\end{equation}

\begin{equation}+ \dfrac{1}{120}\bigl([X,[Y,[X,[Y,X]]]] + [Y,[X,[Y,[X,Y]]]]\bigr) \label{eq:11-15-19}\end{equation}

\begin{equation}C_6 = \dfrac{1}{720}\bigl([X,[Y,[X,[X,[X,Y]]]]] - [Y,[X,[Y,[Y,[Y,X]]]]]\bigr) \label{eq:11-15-20}\end{equation}

\begin{equation}\quad + \dfrac{1}{240}[X,[Y,[Y,[X,[X,Y]]]]] \label{eq:11-15-20b}\end{equation}

\begin{equation}\quad + \dfrac{1}{1440}\bigl([X,[X,[Y,[X,[Y,X]]]]] - [Y,[Y,[X,[Y,[X,Y]]]]]\bigr) \label{eq:11-15-21}\end{equation}

\begin{equation}\quad - \dfrac{1}{720}[X,[X,[Y,[Y,[X,Y]]]]] \label{eq:11-15-21b}\end{equation}

Note: The BCH formula is an infinite series, and every term is expressed as nested Lie brackets. In the commutative case ($[X, Y] = 0$) it reduces to $e^X e^Y = e^{X+Y}$. On nilpotent Lie algebras (e.g., upper-triangular matrices) it truncates to finitely many terms. The coefficients are computed from Bernoulli numbers and combinatorial factors. Goldberg (1956) showed that the number of $n$-th order terms is bounded above by $\displaystyle\dfrac{2^n - 2}{n}$.
On convergence: $Z$ is always defined as a formal power series (its existence in the free Lie algebra is guaranteed independently of the coefficient computation). When summed analytically for matrices, it converges absolutely under a condition such as $\|X\| + \|Y\| < \ln 2$ in operator norm.
Non-uniqueness of higher-order terms: The free Lie algebra carries relations coming from the Jacobi identity, so the same element can be written with different nested commutators. The $C_n$ on this page (especially fifth and sixth order) are one standard choice; a different basis (e.g., a Hall basis) yields a different-looking expression.
References: H.F. Baker (1905) "Alternants and continuous groups"; J.E. Campbell (1897) "On a law of combination of operators"; F. Hausdorff (1906) "Die symbolische Exponentialformel in der Gruppentheorie". The Dynkin form is due to E.B. Dynkin (1947). For explicit higher-order terms see M.W. Reinsch (2000), "A simple expression for the terms in the Baker-Campbell-Hausdorff series", J. Math. Phys. 41, 2434. For a modern treatment see B.C. Hall, "Lie Groups, Lie Algebras, and Representations", Ch.5. The fifth- and sixth-order terms on this page have been verified by computer algebra.

11.16 Differential of the BCH Formula

Formula: $\displaystyle\dfrac{\partial Z}{\partial X} = \dfrac{\text{ad}_Z}{e^{\text{ad}_Z} - 1} \cdot \dfrac{e^{\text{ad}_X} - 1}{\text{ad}_X}$, $\displaystyle\dfrac{\partial Z}{\partial Y} = \dfrac{\text{ad}_Z}{e^{\text{ad}_Z} - 1}\, e^{\text{ad}_X}\, \dfrac{e^{\text{ad}_Y} - 1}{\text{ad}_Y}$

Conditions: $Z = \log(e^X e^Y)$ (BCH formula), where $\text{ad}_Z$ is the adjoint action by $Z$.

Proof

Differentiate $e^Z = e^X e^Y$ with respect to $X$ (using the right trivialization $d(e^A)=\dfrac{e^{\text{ad}_A}-1}{\text{ad}_A}(dA)\,e^A$). The right-hand side is

\begin{equation}\dfrac{\partial}{\partial X}(e^X e^Y) = \dfrac{e^{\text{ad}_X} - 1}{\text{ad}_X}(dX) \cdot e^X e^Y \label{eq:11-16-1}\end{equation}

The left-hand side is

\begin{equation}\dfrac{\partial}{\partial X} e^Z = \dfrac{e^{\text{ad}_Z} - 1}{\text{ad}_Z}(dZ) \cdot e^Z \label{eq:11-16-2}\end{equation}

Since $e^X e^Y = e^Z$, equate $\eqref{eq:11-16-1}$ and $\eqref{eq:11-16-2}$.

\begin{equation}\dfrac{e^{\text{ad}_Z} - 1}{\text{ad}_Z}(dZ) = \dfrac{e^{\text{ad}_X} - 1}{\text{ad}_X}(dX) \label{eq:11-16-3}\end{equation}

Apply $\displaystyle\dfrac{\text{ad}_Z}{e^{\text{ad}_Z} - 1}$ to both sides. Writing $\dfrac{e^{\text{ad}_A}-1}{\text{ad}_A}=\text{dexp}_A$ and $\dfrac{\text{ad}_A}{e^{\text{ad}_A}-1}=\text{dexp}_A^{-1}$,

\begin{equation}\dfrac{\partial Z}{\partial X} = \dfrac{\text{ad}_Z}{e^{\text{ad}_Z} - 1} \cdot \dfrac{e^{\text{ad}_X} - 1}{\text{ad}_X} = \text{dexp}_Z^{-1} \circ \text{dexp}_X \label{eq:11-16-5}\end{equation}

Next, differentiate with respect to $Y$. Since $\displaystyle\dfrac{\partial}{\partial Y}(e^X e^Y) = e^X \dfrac{e^{\text{ad}_Y} - 1}{\text{ad}_Y}(dY)\, e^Y$, equate with $\eqref{eq:11-16-2}$ and cancel $e^Y$ on the right.

\begin{equation}\dfrac{e^{\text{ad}_Z} - 1}{\text{ad}_Z}(dZ)\, e^X = e^X\, \dfrac{e^{\text{ad}_Y} - 1}{\text{ad}_Y}(dY) \label{eq:11-16-6}\end{equation}

Using $e^X (\,\cdot\,) e^{-X} = \text{Ad}_{e^X} = e^{\text{ad}_X}$ and applying $\dfrac{\text{ad}_Z}{e^{\text{ad}_Z}-1}$ from the left,

\begin{equation}\dfrac{\partial Z}{\partial Y} = \dfrac{\text{ad}_Z}{e^{\text{ad}_Z} - 1}\, e^{\text{ad}_X}\, \dfrac{e^{\text{ad}_Y} - 1}{\text{ad}_Y} = \text{dexp}_Z^{-1} \circ \text{Ad}_{e^X} \circ \text{dexp}_Y \label{eq:11-16-7}\end{equation}

Note: The right-trivialization convention ($d(e^A)e^{-A}$) is used throughout. Here $\dfrac{\partial Z}{\partial X}$ is not a giant Jacobian matrix but a linear operator (Fréchet derivative) acting on an increment $\Delta X \in \mathfrak{g}$; it is read precisely as $D_X Z[\Delta X] = \text{dexp}_Z^{-1}\!\bigl(\text{dexp}_X(\Delta X)\bigr)$ ($\circ$ denotes operator composition). This differential formula is important in optimization on Lie groups (geodesic gradient methods) and in robotics (attitude control). Using the expansion $\displaystyle\dfrac{z}{e^{z} - 1} = 1 - \displaystyle\dfrac{z}{2} + \displaystyle\dfrac{z^2}{12} - \displaystyle\dfrac{z^4}{720} + \cdots$ (Bernoulli numbers) makes numerical computation feasible.

11.20 Jacobi Identity and Differentiation

Formula: $\text{ad}_{[X,Y]} = [\text{ad}_X, \text{ad}_Y]$, i.e., $\text{ad}_{[X,Y]}(Z) = \text{ad}_X(\text{ad}_Y(Z)) - \text{ad}_Y(\text{ad}_X(Z))$

Conditions: $X, Y, Z \in \mathfrak{g}$ (Lie algebra).

Proof

We prove the Jacobi identity. By the definition of the Lie bracket,

\begin{equation}[X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y]] = 0 \label{eq:11-20-1}\end{equation}

Expand $\eqref{eq:11-20-1}$. Since $[A, B] = AB - BA$,

\begin{equation}[X, [Y, Z]] = X(YZ - ZY) - (YZ - ZY)X = XYZ - XZY - YZX + ZYX \label{eq:11-20-2}\end{equation}

Similarly,

\begin{equation}[Y, [Z, X]] = YZX - YXZ - ZXY + XZY \label{eq:11-20-3}\end{equation}

\begin{equation}[Z, [X, Y]] = ZXY - ZYX - XYZ + YXZ \label{eq:11-20-4}\end{equation}

Adding $\eqref{eq:11-20-2}$, $\eqref{eq:11-20-3}$, and $\eqref{eq:11-20-4}$, all terms cancel and the sum is $0$.

[Relation to the adjoint representation]

Rewrite the Jacobi identity using $\text{ad}$. Since $\text{ad}_X(Y) = [X, Y]$,

\begin{equation}\text{ad}_X(\text{ad}_Y(Z)) - \text{ad}_Y(\text{ad}_X(Z)) = [[X, Y], Z] = \text{ad}_{[X,Y]}(Z) \label{eq:11-20-5}\end{equation}

Thus $\eqref{eq:11-20-5}$ can be written as $\text{ad}_{[X,Y]} = [\text{ad}_X, \text{ad}_Y]$. This shows that $\text{ad}: \mathfrak{g} \to \text{End}(\mathfrak{g})$ is a Lie algebra homomorphism.

Note: The Jacobi identity is the axiom that characterizes a Lie algebra in place of associativity. That $\text{ad}$ is a Lie algebra homomorphism means that, using the structure constants $c_{ij}^k$ ($[e_i, e_j] = \displaystyle\sum_k c_{ij}^k e_k$), it admits the matrix representation $(\text{ad}_{e_i})_j^k = c_{ij}^k$.
References: Due to C.G.J. Jacobi (posthumous manuscripts, ca. 1862). The modern formulation is due to S. Lie (1888-1893), "Theorie der Transformationsgruppen".

11.21 Differential of the Killing Form

Formula: $\displaystyle\dfrac{\partial \kappa(X, Y)}{\partial X_{ij}} = \text{tr}(\text{ad}_{J^{ij}} \text{ad}_Y)$, $\displaystyle\dfrac{\partial \kappa(X, Y)}{\partial Y_{ij}} = \text{tr}(\text{ad}_X \text{ad}_{J^{ij}})$

Conditions: $\kappa(X, Y) = \text{tr}(\text{ad}_X \text{ad}_Y)$ (Killing form), $X, Y \in \mathfrak{g}$.

Proof

The Killing form is defined by

\begin{equation}\kappa(X, Y) = \text{tr}(\text{ad}_X \text{ad}_Y) \label{eq:11-21-1}\end{equation}

Since $\text{ad}_X$ is a linear function of $X$, we have $\displaystyle\dfrac{\partial}{\partial X_{ij}} \text{ad}_X = \text{ad}_{J^{ij}}$.

Differentiate $\eqref{eq:11-21-1}$ with respect to $X_{ij}$. By the product rule and the linearity of the trace,

\begin{equation}\dfrac{\partial \kappa(X, Y)}{\partial X_{ij}} = \text{tr}\left( \dfrac{\partial \text{ad}_X}{\partial X_{ij}} \text{ad}_Y \right) = \text{tr}(\text{ad}_{J^{ij}} \text{ad}_Y) \label{eq:11-21-2}\end{equation}

Using the symmetry $\kappa(X, Y) = \kappa(Y, X)$, the derivative with respect to $Y$ follows similarly.

\begin{equation}\dfrac{\partial \kappa(X, Y)}{\partial Y_{ij}} = \text{tr}(\text{ad}_X \text{ad}_{J^{ij}}) \label{eq:11-21-3}\end{equation}

When it depends on both variables,

\begin{equation}d\kappa(X, Y) = \text{tr}(\text{ad}_{dX} \text{ad}_Y) + \text{tr}(\text{ad}_X \text{ad}_{dY}) \label{eq:11-21-4}\end{equation}

Note: The Killing form is $\text{Ad}$-invariant ($\kappa(\text{Ad}_g X, \text{Ad}_g Y) = \kappa(X, Y)$) and is non-degenerate on semisimple Lie algebras. For $\mathfrak{su}(n)$, $\kappa(X, Y) = 2n \cdot \text{tr}(XY)$, and for $\mathfrak{so}(n)$, $\kappa(X, Y) = (n-2) \cdot \text{tr}(XY)$. This invariant metric defines Riemannian geometry on the Lie group.
References: W. Killing (1888), "Die Zusammensetzung der stetigen endlichen Transformationsgruppen". Cartan's criterion is due to É. Cartan (1894).

11.22 Maurer-Cartan Form

Formula: $\omega = g^{-1} dg$, $d\omega + \omega \wedge \omega = 0$ (Maurer-Cartan equation)

Conditions: $g: M \to G$ (map into a Lie group), $\omega \in \Omega^1(M, \mathfrak{g})$ ($\mathfrak{g}$-valued 1-form).

Proof

Define the Maurer-Cartan form. For $g \in G$,

\begin{equation}\omega = g^{-1} dg \label{eq:11-22-1}\end{equation}

We check that $\omega$ takes values in the Lie algebra $\mathfrak{g}$: $g^{-1} dg$ is an element of $T_e G \cong \mathfrak{g}$.

[Derivation of the Maurer-Cartan equation]

Use $d(g^{-1}) = -g^{-1} (dg) g^{-1}$ (from differentiating $g \cdot g^{-1} = I$).

\begin{equation}d(g \cdot g^{-1}) = dg \cdot g^{-1} + g \cdot d(g^{-1}) = 0 \label{eq:11-22-2}\end{equation}

\begin{equation}d(g^{-1}) = -g^{-1} dg \cdot g^{-1} \label{eq:11-22-3}\end{equation}

Compute the exterior derivative of $\omega = g^{-1} dg$.

\begin{equation}d\omega = d(g^{-1}) \wedge dg = -g^{-1} dg \cdot g^{-1} \wedge dg \label{eq:11-22-4}\end{equation}

Compute $\omega \wedge \omega$. In the matrix-valued case, $\omega \wedge \omega$ combines matrix multiplication with the wedge product.

\begin{equation}\omega \wedge \omega = (g^{-1} dg) \wedge (g^{-1} dg) \label{eq:11-22-5}\end{equation}

Since $g^{-1}$ is a 0-form,

\begin{equation}\omega \wedge \omega = g^{-1} dg \cdot g^{-1} \wedge dg = g^{-1} dg \wedge g^{-1} dg \label{eq:11-22-6}\end{equation}

Comparing $\eqref{eq:11-22-4}$ and $\eqref{eq:11-22-6}$,

\begin{equation}d\omega = -\omega \wedge \omega \label{eq:11-22-7}\end{equation}

that is,

\begin{equation}d\omega + \omega \wedge \omega = 0 \label{eq:11-22-8}\end{equation}

Note: The Maurer-Cartan equation expresses that the curvature of the connection is zero (flat). The right-invariant form $\tilde{\omega} = dg \cdot g^{-1}$ satisfies the sign-flipped equation $d\tilde{\omega} - \tilde{\omega} \wedge \tilde{\omega} = 0$ (since $d(g^{-1}) = -g^{-1}\,dg\,g^{-1}$ gives $d\tilde\omega = +\tilde\omega\wedge\tilde\omega$). In components, $d\omega^k + \displaystyle\dfrac{1}{2} c_{ij}^k \omega^i \wedge \omega^j = 0$ (where $c_{ij}^k$ are the structure constants).
References: Due to L. Maurer (1888) and É. Cartan (1904), "Sur la structure des groupes infinis de transformations". For a modern treatment see S. Kobayashi & K. Nomizu, "Foundations of Differential Geometry", Vol.1, Ch.1.