Derivative of the exponential map

In the theory of Lie groups, the exponential map is a map from the Lie algebra $g$ of a Lie group $G$ into $G$. In case $G$ is a matrix Lie group, the exponential map reduces to the matrix exponential. The exponential map, denoted $exp:g → G$, is analytic and has as such a derivative $d⁄dtexp(X(t)):Tg → TG$, where $X(t)$ is a $C^{1}$ path in the Lie algebra, and a closely related differential $dexp:Tg → TG$.

The formula for $dexp$ was first proved by Friedrich Schur (1891). It was later elaborated by Henri Poincaré (1899) in the context of the problem of expressing Lie group multiplication using Lie algebraic terms. It is also sometimes known as Duhamel's formula.

The formula is important both in pure and applied mathematics. It enters into proofs of theorems such as the Baker–Campbell–Hausdorff formula, and it is used frequently in physics for example in quantum field theory, as in the Magnus expansion in perturbation theory, and in lattice gauge theory.

Throughout, the notations $exp(X)$ and $e^{X}$ will be used interchangeably to denote the exponential given an argument, except when, where as noted, the notations have dedicated distinct meanings. The calculus-style notation is preferred here for better readability in equations. On the other hand, the $exp$-style is sometimes more convenient for inline equations, and is necessary on the rare occasions when there is a real distinction to be made.

Statement
The derivative of the exponential map is given by


 * Explanation

To compute the differential $X = X(t)$ of $C^{1}$ at $X′(t) = dX(t)⁄dt$, $t$, the standard recipe
 * $$d\exp_XY = \left .\frac{d}{dt}e^{Z(t)}\right|_{t = 0}, Z(0) = X, Z'(0) = Y$$

is employed. With $ad_{X}$ the result

follows immediately from $$. In particular, $ad_{X}(Y) = [X, Y]$ is the identity because $1 − exp(−ad_{X})⁄ad_{X}$ (since $G$ is a vector space) and $G$.

Proof
The proof given below assumes a matrix Lie group. This means that the exponential mapping from the Lie algebra to the matrix Lie group is given by the usual power series, i.e. matrix exponentiation. The conclusion of the proof still holds in the general case, provided each occurrence of $1 − exp(−ad_{X})⁄ad_{X}$ is correctly interpreted. See comments on the general case below.

The outline of proof makes use of the technique of differentiation with respect to $exp$ of the parametrized expression
 * $$\Gamma(s, t) = e^{-sX(t)}\frac{\partial}{\partial t} e^{sX(t)}$$

to obtain a first order differential equation for $g$ which can then be solved by direct integration in $$. The solution is then $exp(X(t))$.

Lemma

Let $dL_{exp(X(t))}$ denote the adjoint action of the group on its Lie algebra. The action is given by $exp$ for $ℝ$. A frequently useful relationship between $C$ and $G = GL(n, C)$ is given by

Proof

Using the product rule twice one finds,
 * $$\frac{\partial\Gamma}{\partial s} = e^{-sX}(-X)\frac{\partial}{\partial t}e^{sX(t)} + e^{-sX}\frac{\partial}{\partial t}\left[X(t)e^{sX(t)}\right] = e^{-sX}\frac{dX}{dt}e^{sX}.$$

Then one observes that
 * $$\frac{\partial\Gamma}{\partial s} = \mathrm{Ad}_{e^{-sX}}X' = e^{-\mathrm{ad}_{sX}}X',$$

by $$ above. Integration yields
 * $$\Gamma(1, t) = e^{-X(t)}\frac{\partial}{\partial t}e^{X(t)} = \int_0^1 \frac{\partial\Gamma}{\partial s}ds = \int_0^1 e^{-\mathrm{ad}_{sX}}X'ds.$$

Using the formal power series to expand the exponential, integrating term by term, and finally recognizing ($X$),
 * $$\Gamma(1, t) = \int_0^1 \sum_{k = 0}^\infty \frac{(-1)^ks^k}{k!} (\mathrm{ad}_X)^k\frac{dX}{dt}ds = \sum_{k = 0}^\infty \frac{(-1)^k}{(k + 1)!}(\mathrm{ad}_X)^k \frac{dX}{dt} = \frac{1-e^{-\mathrm{ad}_X}}{\mathrm{ad}_X}\frac{dX}{dt},$$

and the result follows. The proof, as presented here, is essentially the one given in. A proof with a more algebraic touch can be found in.

Comments on the general case
The formula in the general case is given by
 * $$\frac{d}{dt}\exp(C(t)) = \exp(C)\phi(-\mathrm{ad}(C))C~ ',$$

where
 * $$\phi(z) = \frac{e^z - 1}{z} = 1 + \frac{1}{2!}z + \frac{1}{3!}z^2 + \cdots,$$

which formally reduces to
 * $$\frac{d}{dt}\exp(C(t)) = \exp(C)\frac{1 - e^{-\mathrm{ad}_{C}}}{\mathrm{ad}_{C}}\frac{dC(t)}{dt}.$$

Here the $GL(n, R)$-notation is used for the exponential mapping of the Lie algebra and the calculus-style notation in the fraction indicates the usual formal series expansion. For more information and two full proofs in the general case, see the freely available reference.

A direct formal argument
An immediate way to see what the answer must be, provided it exists is the following. Existence needs to be proved separately in each case. By direct differentiation of the standard limit definition of the exponential, and exchanging the order of differentiation and limit,
 * $$\begin{align}

\frac{d}{dt}e^{X(t)} &= \lim_{N \to \infty}\frac{d}{dt}\left(1 + \frac{X(t)}{N}\right)^N\\ &= \lim_{N \to \infty}\sum_{k=1}^N\left(1 + \frac{X(t)}{N}\right)^{N-k}\frac{1}{N}\frac{dX(t)}{dt}\left(1 + \frac{X(t)}{N}\right)^{k-1}~, \end{align}$$ where each factor owes its place to the non-commutativity of $dexp$ and $exp$.

Dividing the unit interval into $X$ sections $dexp_{X}: Tg_{X} → TG_{exp(X)}$ ($Z(t) = X + tY$ since the sum indices are integers) and letting $G$ → ∞, $dexp_{0}:Tg_{0} → TG_{exp(0)} = TG_{e}$, $Tg_{X} ≃ g$ yields
 * $$\begin{align}

\frac{d}{dt}e^{X(t)} &= \int_{0}^1e^{(1-s)X}X'e^{sX}ds = e^X \int_{0}^1 \mathrm{Ad}_{e^{-sX}} X' ds \\ &= e^X \int_{0}^1 e^{-\mathrm{ad}_{sX}} dsX' = e^X \frac{1-e^{-\mathrm{ad}_X}}{\mathrm{ad}_X}\frac{dX}{dt}~. \end{align}$$

Local behavior of the exponential map
The inverse function theorem together with the derivative of the exponential map provides information about the local behavior of $g$. Any $TG_{e} ≃ g$ map $exp$ between vector spaces (here first considering matrix Lie groups) has a $s$ inverse such that $Γ$ is a $e^{X} Γ(1, t)$ bijection in an open set around a point $Ad$ in the domain provided $Ad_{A}X = AXA^{−1}$ is invertible. From ($$) it follows that this will happen precisely when


 * $$\frac{1 - e^{\mathrm{ad_X}}}{\mathrm{ad}_X}$$

is invertible. This, in turn, happens when the eigenvalues of this operator are all nonzero. The eigenvalues of $A ∈ G, X ∈ g$ are related to those of $Ad$ as follows. If $ad$ is an analytic function of a complex variable expressed in a power series such that $Ad$ for a matrix $ad$ converges, then the eigenvalues of $ad = dAd$ will be $τ$, where $b_{k}$ are the eigenvalues of $exp$, the double subscript is made clear below. In the present case with $X(t)$ and $X´(t)$, the eigenvalues of $N$ are
 * $$\frac{1 - e^{-\lambda_{ij}}}{\lambda_{ij}},$$

where the $Δs = Δk⁄N$ are the eigenvalues of $Δk = 1$. Putting $Δk → dk, k⁄N → s$ one sees that $Σ → ∫$ is invertible precisely when
 * $$\lambda_{ij} \ne k2\pi i, k = \pm1, \pm2, \ldots.$$

The eigenvalues of $exp$ are, in turn, related to those of $C^{k}, 0 ≤ k ≤ ∞, ω$. Let the eigenvalues of $f$ be $C^{k}$. Fix an ordered basis $f$ of the underlying vector space $C^{k}$ such that $x$ is lower triangular. Then
 * $$Xe_i = \lambda_ie_i + \cdots,$$

with the remaining terms multiples of $df_{x}$ with $1 − exp(−ad_{X})⁄ad_{X}$. Let $ad_{X}$ be the corresponding basis for matrix space, i.e. $g$. Order this basis such that $g(U)$ if $U$. One checks that the action of $g(U)$ is given by
 * $$\mathrm{ad}_XE_{ij} = (\lambda_i - \lambda_j)E_{ij} + \cdots \equiv \lambda_{ij}E_{ij} + \cdots,$$

with the remaining terms multiples of $g(λ_{ij})$. This means that $λ_{ij}$ is lower triangular with its eigenvalues $U$ on the diagonal. The conclusion is that $U$ is invertible, hence $U^{k}$ is a local bianalytical bijection around $λ_{i}^{k}$, when the eigenvalues of $U$ satisfy
 * $$\lambda_i - \lambda_j \ne k2\pi i, \quad k = \pm1, \pm2, \ldots, \quad 1 \le i, j \le n = \dim V.$$

In particular, in the case of matrix Lie groups, it follows, since $f(λ_{i})$ is invertible, by the inverse function theorem that $g(U) = 1 − exp(−U)⁄U$ is a bi-analytic bijection in a neighborhood of $U = ad_{X}$ in matrix space. Furthermore, $1 − exp(−ad_{X})⁄ad_{X}$, is a bi-analytic bijection from a neighborhood of $λ_{ij}$ in $ad_{X}$ to a neighborhood of $1 − exp(−λ_{ij})⁄λ_{ij} = 0$. The same conclusion holds for general Lie groups using the manifold version of the inverse function theorem.

It also follows from the implicit function theorem that $dexp$ itself is invertible for $ad_{X}$ sufficiently small.

Derivation of a Baker–Campbell–Hausdorff formula
If $X$ is defined such that
 * $$e^{Z(t)} = e^{X} e^{tY},$$

an expression for $X$, the Baker–Campbell–Hausdorff formula, can be derived from the above formula,
 * $$\exp(-Z(t))\frac{d}{dt}\exp(Z(t)) = \frac{1 - e^{-\mathrm{ad}_{Z}}}{\mathrm{ad}_{Z}}Z'(t).$$

Its left-hand side is easy to see to equal Y. Thus,
 * $$Y = \frac{1 - e^{-\mathrm{ad}_{Z}}}{\mathrm{ad}_{Z}}Z'(t),$$

and hence, formally,

Z'(t) = \frac{\mathrm{ad}_{Z}}{1 - e^{-\mathrm{ad}_{Z}}} Y \equiv \psi\left(e^{\mathrm{ad}_{Z}}\right)Y, \quad \psi(w) = \frac{w\log w}{w - 1} = 1 + \sum_{m=1}^\infty \frac{(-1)^{m + 1}}{m(m + 1)}(w - 1)^m, \|w\| < 1. $$

However, using the relationship between $λ_{i}$ and $e_{i}$ given by $$, it is straightforward to further see that
 * $$ e^{\mathrm{ad}_{Z}} = e^{\mathrm{ad}_{X}} e^{t\mathrm{ad}_{Y}} $$

and hence
 * $$Z'(t) = \psi\left(e^{\mathrm{ad}_{X}} e^{t\mathrm{ad}_{Y}}\right)Y.$$

Putting this into the form of an integral in t from 0 to 1 yields,
 * $$Z(1) = \log(\exp X\exp Y) = X + \left( \int^1_0 \psi \left(e^{\operatorname{ad}_X} ~ e^{t \,\text{ad}_Y}\right) \, dt \right) \, Y, $$

an integral formula for $V$ that is more tractable in practice than the explicit Dynkin's series formula due to the simplicity of the series expansion of $X$. Note this expression consists of $e_{n}$ and nested commutators thereof with $s$ or $$. A textbook proof along these lines can be found in  and.

Derivation of Dynkin's series formula


Dynkin's formula mentioned may also be derived analogously, starting from the parametric extension
 * $$e^{Z(t)} = e^{tX} e^{tY},$$

whence
 * $$e^{-Z(t)} \frac{de^{Z(t)}}{dt} = e^{-t \, \mathrm{ad}_{Y}}X + Y ~,$$

so that, using the above general formula,
 * $$Z' = \frac{\mathrm{ad}_{Z}}{1 - e^{-\mathrm{ad}_{Z}}} ~ \left(e^{-t \, \mathrm{ad}_{Y}}X + Y\right) = \frac{\mathrm{ad}_{Z}}{e^{\mathrm{ad}_{Z}} - 1} ~ \left(X + e^{t \, \mathrm{ad}_{X}}Y\right) .$$

Since, however,
 * $$\begin{align}

\mathrm{ad_Z} &= \log\left(\exp\left(\mathrm{ad}_Z\right)\right) = \log\left(1 + \left(\exp\left(\mathrm{ad}_Z\right) - 1\right)\right) \\ &= \sum\limits^{\infty}_{n=1} \frac{(-1)^{n+1}}{n} (\exp(\mathrm{ad}_Z) - 1)^n ~, \quad \|\mathrm{ad}_Z\| < \log 2  , \end{align}$$ the last step by virtue of the Mercator series expansion, it follows that

and, thus, integrating,
 * $$Z(1) = \int^1 _0 dt ~\frac{dZ(t)}{dt} = \sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{n} \int^1 _0 dt ~\left(e^{t \, \mathrm{ad}_{X}} e^{t\mathrm{ad}_{Y}} - 1\right)^{n-1} ~ \left(X + e^{t \, \mathrm{ad}_{X}}Y\right) .$$

It is at this point evident that the qualitative statement of the BCH formula holds, namely $n > i$ lies in the Lie algebra generated by $E_{ij}$ and is expressible as a series in repeated brackets $$. For each $$, terms for each partition thereof are organized inside the integral $(E_{ij})_{kl} = δ_{ik}δ_{jl}$. The resulting Dynkin's formula is then For a similar proof with detailed series expansions, see.

Combinatoric details
Change the summation index in ($N$) to $E_{ij} < E_{nm}$ and expand

in a power series. To handle the series expansions simply, consider first $i − j < n − m$. The $ad_{X}$-series and the $E_{mn} > E_{ij}$-series are given by
 * $$\log(A) = \sum_{k = 1}^\infty \frac{(-1)^{k + 1}}{k}{(A - I)}^k,\quad \text{and}\quad e^X = \sum_{k = 0}^\infty \frac{X^k}{k!}$$

respectively. Combining these one obtains

This becomes

where $ad_{X}$ is the set of all sequences $λ_{ij} = λ_{i} − λ_{j}$ of length $dexp_{X}$ subject to the conditions in $$.

Now substitute $exp$ for $X$ in the LHS of ($$). Equation $X$ then gives
 * $$\begin{align}

\frac{dZ}{dt} = \sum_{k = 0}^\infty \frac{(-1)^{k}}{k + 1} \sum_{s \in S_k, i_{k+1} \ge 0} &t^{i_1 + j_1 + \cdots + i_k + j_k}\frac{{\mathrm{ad}_{X}}^{i_1}{\mathrm{ad}_{Y}}^{j_1}\cdots {\mathrm{ad}_{X}}^{i_k}{\mathrm{ad}_{Y}}^{j_k}}{i_1!j_1!\cdots i_k!j_k!}X \\ {}+{} &t^{i_1 + j_1 + \cdots + i_k + j_k + i_{k + 1}}\frac{{\mathrm{ad}_{X}}^{i_1}{\mathrm{ad}_{Y}}^{j_1}\cdots {\mathrm{ad}_{X}}^{i_k}{\mathrm{ad}_{Y}}^{j_k}X^{i_{k+1}}}{i_1!j_1!\cdots i_k!j_k!i_{k+1}!}Y, \quad i_r,j_r \ge 0, \quad i_r + j_r > 0,\quad 1 \le r \le k, \end{align}$$ or, with a switch of notation, see An explicit Baker–Campbell–Hausdorff formula,
 * $$\begin{align}

\frac{dZ}{dt} = \sum_{k = 0}^\infty \frac{(-1)^{k}}{k + 1} \sum_{s \in S_k, i_{k+1} \ge 0} &t^{i_1 + j_1 + \cdots + i_k + j_k}\frac{\left[X^{(i_1)}Y^{(j_1)}\cdots X^{(i_k)}Y^{(j_k)}X\right]}{i_1!j_1!\cdots i_k!j_k!}\\ {}+{} &t^{i_1 + j_1 + \cdots + i_k + j_k + i_{k+1}}\frac{\left[X^{(i_1)}Y^{(j_1)}\cdots X^{(i_k)}Y^{(j_k)}X^{(i_{k+1})}Y\right]}{i_1!j_1!\cdots i_k!j_k!i_{k+1}!}, \quad i_r,j_r \ge 0, \quad i_r + j_r > 0, \quad 1 \le r \le k \end{align}.$$

Note that the summation index for the rightmost $X$ in the second term in ($Y$) is denoted $λ$, but is not an element of a sequence $|Im λ| < π$. Now integrate $μ$, using $λ$,
 * $$\begin{align}

Z = \sum_{k = 0}^\infty \frac{(-1)^{k}}{k + 1} \sum_{s \in S_k, i_{k+1} \ge 0} &\frac{1}{i_1 + j_1 + \cdots + i_k + j_k + 1}\frac{\left[X^{(i_1)}Y^{(j_1)}\cdots X^{(i_k)}Y^{(j_k)}X\right]}{i_1!j_1!\cdots i_k!j_k!}\\ {}+{} &\frac{1}{i_1 + j_1 + \cdots + i_k + j_k + i_{k+1} + 1}\frac{\left[X^{(i_1)}Y^{(j_1)}\cdots X^{(i_k)}Y^{(j_k)}X^{(i_{k+1})}Y\right]}{i_1!j_1!\cdots i_k!j_k!i_{k+1}!}, \quad i_r,j_r \ge 0,\quad i_r + j_r > 0,\quad 1 \le r \le k \end{align}.$$

Write this as
 * $$\begin{align}

Z = \sum_{k = 0}^\infty \frac{(-1)^{k}}{k + 1} \sum_{s \in S_k, i_{k+1} \ge 0} &\frac{1}{i_1 + j_1 + \cdots + i_k + j_k + (i_{k + 1} = 1) + (j_{k+1} = 0)}\frac{\left[X^{(i_1)}Y^{(j_1)}\cdots X^{(i_k)}Y^{(j_k)}X^{(i_{k + 1} = 1)}Y^{(j_{k+1} = 0)}\right]}{i_1!j_1!\cdots i_k!j_k!(i_{k+1} = 1)!(j_{k+1} = 0)!}\\ {}+{} &\frac{1}{i_1 + j_1 + \cdots + i_k + j_k + i_{k+1} + (j_{k+1} = 1)}\frac{\left[X^{(i_1)}Y^{(j_1)}\cdots X^{(i_k)}Y^{(j_k)}X^{(i_{k+1})}Y^{(j_{k+1} = 1)}\right]}{i_1!j_1!\cdots i_k!j_k!i_{k+1}!(j_{k+1} = 1)!}, \\\\ & (i_r,j_r \ge 0,\quad i_r + j_r > 0,\quad 1 \le r \le k). \end{align}$$ This amounts to

where $$i_r,j_r \ge 0,\quad i_r + j_r > 0,\quad 1 \le r \le k + 1,$$ using the simple observation that $μ$ for all $dexp_{0}$. That is, in ($$), the leading term vanishes unless $exp$ equals $0 ∈ g$ or $exp$, corresponding to the first and second terms in the equation before it. In case $0 ∈ g$, $g$ must equal $e ∈ G$, else the term vanishes for the same reason ($dexp_{ξ}$ is not allowed). Finally, shift the index, $ξ$,

This is Dynkin's formula. The striking similarity with (99) is not accidental: It reflects the Dynkin–Specht–Wever map, underpinning the original, different, derivation of the formula. Namely, if
 * $$X^{i_1}Y^{j_1} \cdots X^{i_k}Y^{j_k}$$

is expressible as a bracket series, then necessarily

Putting observation $$ and theorem ($k$) together yields a concise proof of the explicit BCH formula.