The derivative of the exponential map is given by^[6]

Explanation

To compute the differential dexp of exp at X, dexp_X: Tg_X → TG_exp(X), the standard recipe^[2]

${\displaystyle d\exp _{X}Y=\left.{\frac {d}{dt}}e^{Z(t)}\right|_{t=0},Z(0)=X,Z'(0)=Y}$ 

is employed. With Z(t) = X + tY the result^[6]

${\displaystyle d\exp _{X}Y=e^{X}{\frac {1-e^{-\mathrm {ad} _{X}}}{\mathrm {ad} _{X}}}Y}$

(3)

follows immediately from (1). In particular, dexp₀:Tg₀ → TG_exp(0) = TG_e is the identity because Tg_X ≃ g (since g is a vector space) and TG_e ≃ g.

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 exp is correctly interpreted. See comments on the general case below.

The outline of proof makes use of the technique of differentiation with respect to s of the parametrized expression

${\displaystyle \Gamma (s,t)=e^{-sX(t)}{\frac {\partial }{\partial t}}e^{sX(t)}}$ 

to obtain a first order differential equation for Γ which can then be solved by direct integration in s. The solution is then e^X Γ(1, t).

Lemma

Let Ad denote the adjoint action of the group on its Lie algebra. The action is given by Ad_AX = AXA⁻¹ for A ∈ G, X ∈ g. A frequently useful relationship between Ad and ad is given by^{[7][nb 1]}

Proof

Using the product rule twice one finds,

${\displaystyle {\frac {\partial \Gamma }{\partial s}}=e^{-sX}(-X){\frac {\partial }{\partial t}}e^{sX(t)}+e^{-sX}{\frac {\partial }{\partial t}}\left[X(t)e^{sX(t)}\right]=e^{-sX}{\frac {dX}{dt}}e^{sX}.}$ 

Then one observes that

${\displaystyle {\frac {\partial \Gamma }{\partial s}}=\mathrm {Ad} _{e^{-sX}}X'=e^{-\mathrm {ad} _{sX}}X',}$ 

by (4) above. Integration yields

${\displaystyle \Gamma (1,t)=e^{-X(t)}{\frac {\partial }{\partial t}}e^{X(t)}=\int _{0}^{1}{\frac {\partial \Gamma }{\partial s}}ds=\int _{0}^{1}e^{-\mathrm {ad} _{sX}}X'ds.}$ 

Using the formal power series to expand the exponential, integrating term by term, and finally recognizing (2),

${\displaystyle \Gamma (1,t)=\int _{0}^{1}\sum _{k=0}^{\infty }{\frac {(-1)^{k}s^{k}}{k!}}(\mathrm {ad} _{X})^{k}{\frac {dX}{dt}}ds=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(k+1)!}}(\mathrm {ad} _{X})^{k}{\frac {dX}{dt}}={\frac {1-e^{-\mathrm {ad} _{X}}}{\mathrm {ad} _{X}}}{\frac {dX}{dt}},}$ 

and the result follows. The proof, as presented here, is essentially the one given in Rossmann (2002). A proof with a more algebraic touch can be found in Hall (2015).^[8]

Comments on the general case edit

The formula in the general case is given by^[9]

${\displaystyle {\frac {d}{dt}}\exp(C(t))=\exp(C)\phi (-\mathrm {ad} (C))C~',}$ 

where^{[nb 2]}

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

which formally reduces to

${\displaystyle {\frac {d}{dt}}\exp(C(t))=\exp(C){\frac {1-e^{-\mathrm {ad} _{C}}}{\mathrm {ad} _{C}}}{\frac {dC(t)}{dt}}.}$ 

Here the exp-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 Sternberg (2004) reference.

A direct formal argument edit

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,

${\displaystyle {\begin{aligned}{\frac {d}{dt}}e^{X(t)}&=\lim _{N\to \infty }{\frac {d}{dt}}\left(1+{\frac {X(t)}{N}}\right)^{N}\\&=\lim _{N\to \infty }\sum _{k=1}^{N}\left(1+{\frac {X(t)}{N}}\right)^{N-k}{\frac {1}{N}}{\frac {dX(t)}{dt}}\left(1+{\frac {X(t)}{N}}\right)^{k-1}~,\end{aligned}}}$ 

where each factor owes its place to the non-commutativity of X(t) and X´(t).

Dividing the unit interval into N sections Δs = Δk/N (Δk = 1 since the sum indices are integers) and letting N → ∞, Δk → dk, k/N → s, Σ → ∫ yields

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

Local behavior of the exponential map edit

The inverse function theorem together with the derivative of the exponential map provides information about the local behavior of exp. Any C^k, 0 ≤ k ≤ ∞, ω map f between vector spaces (here first considering matrix Lie groups) has a C^k inverse such that f is a C^k bijection in an open set around a point x in the domain provided df_x is invertible. From (3) it follows that this will happen precisely when

${\displaystyle {\frac {1-e^{\mathrm {ad_{X}} }}{\mathrm {ad} _{X}}}}$ 

is invertible. This, in turn, happens when the eigenvalues of this operator are all nonzero. The eigenvalues of 1 − exp(−ad_X)/ad_X are related to those of ad_X as follows. If g is an analytic function of a complex variable expressed in a power series such that g(U) for a matrix U converges, then the eigenvalues of g(U) will be g(λ_ij), where λ_ij are the eigenvalues of U, the double subscript is made clear below.^{[nb 3]} In the present case with g(U) = 1 − exp(−U)/U and U = ad_X, the eigenvalues of 1 − exp(−ad_X)/ad_X are

${\displaystyle {\frac {1-e^{-\lambda _{ij}}}{\lambda _{ij}}},}$ 

where the λ_ij are the eigenvalues of ad_X. Putting 1 − exp(−λ_ij)/λ_ij = 0 one sees that dexp is invertible precisely when

${\displaystyle \lambda _{ij}\neq k2\pi i,k=\pm 1,\pm 2,\ldots .}$ 

The eigenvalues of ad_X are, in turn, related to those of X. Let the eigenvalues of X be λ_i. Fix an ordered basis e_i of the underlying vector space V such that X is lower triangular. Then

${\displaystyle Xe_{i}=\lambda _{i}e_{i}+\cdots ,}$ 

with the remaining terms multiples of e_n with n > i. Let E_ij be the corresponding basis for matrix space, i.e. (E_ij)_kl = δ_ikδ_jl. Order this basis such that E_ij < E_nm if i − j < n − m. One checks that the action of ad_X is given by

${\displaystyle \mathrm {ad} _{X}E_{ij}=(\lambda _{i}-\lambda _{j})E_{ij}+\cdots \equiv \lambda _{ij}E_{ij}+\cdots ,}$ 

with the remaining terms multiples of E_mn > E_ij. This means that ad_X is lower triangular with its eigenvalues λ_ij = λ_i − λ_j on the diagonal. The conclusion is that dexp_X is invertible, hence exp is a local bianalytical bijection around X, when the eigenvalues of X satisfy^{[10][nb 4]}

${\displaystyle \lambda _{i}-\lambda _{j}\neq k2\pi i,\quad k=\pm 1,\pm 2,\ldots ,\quad 1\leq i,j\leq n=\dim V.}$ 

In particular, in the case of matrix Lie groups, it follows, since dexp₀ is invertible, by the inverse function theorem that exp is a bi-analytic bijection in a neighborhood of 0 ∈ g in matrix space. Furthermore, exp, is a bi-analytic bijection from a neighborhood of 0 ∈ g in g to a neighborhood of e ∈ G.^[11] 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_ξ itself is invertible for ξ sufficiently small.^[12]

Derivation of a Baker–Campbell–Hausdorff formula edit

If Z(t) is defined such that

${\displaystyle e^{Z(t)}=e^{X}e^{tY},}$ 

an expression for Z(1) = log( exp X exp Y ), the Baker–Campbell–Hausdorff formula, can be derived from the above formula,

${\displaystyle \exp(-Z(t)){\frac {d}{dt}}\exp(Z(t))={\frac {1-e^{-\mathrm {ad} _{Z}}}{\mathrm {ad} _{Z}}}Z'(t).}$ 

Its left-hand side is easy to see to equal Y. Thus,

${\displaystyle Y={\frac {1-e^{-\mathrm {ad} _{Z}}}{\mathrm {ad} _{Z}}}Z'(t),}$ 

and hence, formally,^[13][14]

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

However, using the relationship between Ad and ad given by (4), it is straightforward to further see that

${\displaystyle e^{\mathrm {ad} _{Z}}=e^{\mathrm {ad} _{X}}e^{t\mathrm {ad} _{Y}}}$ 

and hence

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

Putting this into the form of an integral in t from 0 to 1 yields,

${\displaystyle Z(1)=\log(\exp X\exp Y)=X+\left(\int _{0}^{1}\psi \left(e^{\operatorname {ad} _{X}}~e^{t\,{\text{ad}}_{Y}}\right)\,dt\right)\,Y,}$ 

an integral formula for Z(1) that is more tractable in practice than the explicit Dynkin's series formula due to the simplicity of the series expansion of ψ. Note this expression consists of X+Y and nested commutators thereof with X or Y. A textbook proof along these lines can be found in Hall (2015) and Miller (1972).

Derivation of Dynkin's series formula edit

Dynkin's formula mentioned may also be derived analogously, starting from the parametric extension

${\displaystyle e^{Z(t)}=e^{tX}e^{tY},}$ 

whence

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

so that, using the above general formula,

${\displaystyle Z'={\frac {\mathrm {ad} _{Z}}{1-e^{-\mathrm {ad} _{Z}}}}~\left(e^{-t\,\mathrm {ad} _{Y}}X+Y\right)={\frac {\mathrm {ad} _{Z}}{e^{\mathrm {ad} _{Z}}-1}}~\left(X+e^{t\,\mathrm {ad} _{X}}Y\right).}$ 

Since, however,

${\displaystyle {\begin{aligned}\mathrm {ad_{Z}} &=\log \left(\exp \left(\mathrm {ad} _{Z}\right)\right)=\log \left(1+\left(\exp \left(\mathrm {ad} _{Z}\right)-1\right)\right)\\&=\sum \limits _{n=1}^{\infty }{\frac {(-)^{n+1}}{n}}(\exp(\mathrm {ad} _{Z})-1)^{n}~,\quad \|\mathrm {ad} _{Z}\|<\log 2~~,\end{aligned}}}$ 

the last step by virtue of the Mercator series expansion, it follows that

${\displaystyle Z'=\sum \limits _{n=1}^{\infty }{\frac {(-)^{n-1}}{n}}\left(e^{\mathrm {ad} _{Z}}-1\right)^{n-1}~\left(X+e^{t\,\mathrm {ad} _{X}}Y\right)~,}$

(5)

and, thus, integrating,

${\displaystyle Z(1)=\int _{0}^{1}dt~{\frac {dZ(t)}{dt}}=\sum _{n=1}^{\infty }{\frac {(-)^{n-1}}{n}}\int _{0}^{1}dt~\left(e^{t\,\mathrm {ad} _{X}}e^{t\mathrm {ad} _{Y}}-1\right)^{n-1}~\left(X+e^{t\,\mathrm {ad} _{X}}Y\right).}$ 

It is at this point evident that the qualitative statement of the BCH formula holds, namely Z lies in the Lie algebra generated by X, Y and is expressible as a series in repeated brackets (A). For each k, terms for each partition thereof are organized inside the integral ∫dt t^k−1. The resulting Dynkin's formula is then

For a similar proof with detailed series expansions, see Rossmann (2002).

Combinatoric details edit

Change the summation index in (5) to k = n − 1 and expand

${\displaystyle {\frac {dZ}{dt}}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k+1}}\left\{\left(e^{\mathrm {ad} _{tX}}e^{\mathrm {ad} _{tY}}-1\right)^{k}X+\left(e^{\mathrm {ad} _{tX}}e^{\mathrm {ad} _{tY}}-1\right)^{k}e^{\mathrm {ad} _{tX}}Y\right\}}$

(97)

in a power series. To handle the series expansions simply, consider first Z = log(e^Xe^Y). The log-series and the exp-series are given by

${\displaystyle \log(A)=\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k}}{(A-I)}^{k},\quad {\text{and}}\quad e^{X}=\sum _{k=0}^{\infty }{\frac {X^{k}}{k!}}}$ 

respectively. Combining these one obtains

${\displaystyle \log \left(e^{X}e^{Y}\right)=\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k}}{\left(e^{X}e^{Y}-I\right)}^{k}=\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k}}\left({\sum _{i=0}^{\infty }{\frac {X^{i}}{i!}}\sum _{j=0}^{\infty }{\frac {Y^{j}}{j!}}-I}\right)^{k}=\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k}}\left(\sum _{i,j\geq 0,i+j>1}^{\infty }{\frac {X^{i}Y^{j}}{i!j!}}\right)^{k}.}$

(98)

This becomes

where S_k is the set of all sequences s = (i₁, j₁, ..., i_k, j_k) of length 2k subject to the conditions in (99).

Now substitute (e^Xe^Y − 1) for (e^ad_tXe^ad_tY − 1) in the LHS of (98). Equation (99) then gives

${\displaystyle {\begin{aligned}{\frac {dZ}{dt}}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k+1}}\sum _{s\in S_{k},i_{k+1}\geq 0}&t^{i_{1}+j_{1}+\cdots +i_{k}+j_{k}}{\frac {{\mathrm {ad} _{X}}^{i_{1}}{\mathrm {ad} _{Y}}^{j_{1}}\cdots {\mathrm {ad} _{X}}^{i_{k}}{\mathrm {ad} _{Y}}^{j_{k}}}{i_{1}!j_{1}!\cdots i_{k}!j_{k}!}}X\\{}+{}&t^{i_{1}+j_{1}+\cdots +i_{k}+j_{k}+i_{k+1}}{\frac {{\mathrm {ad} _{X}}^{i_{1}}{\mathrm {ad} _{Y}}^{j_{1}}\cdots {\mathrm {ad} _{X}}^{i_{k}}{\mathrm {ad} _{Y}}^{j_{k}}X^{i_{k+1}}}{i_{1}!j_{1}!\cdots i_{k}!j_{k}!i_{k+1}!}}Y,\quad i_{r},j_{r}\geq 0,\quad i_{r}+j_{r}>0,\quad 1\leq r\leq k,\end{aligned}}}$ 

or, with a switch of notation, see An explicit Baker–Campbell–Hausdorff formula,

${\displaystyle {\begin{aligned}{\frac {dZ}{dt}}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k+1}}\sum _{s\in S_{k},i_{k+1}\geq 0}&t^{i_{1}+j_{1}+\cdots +i_{k}+j_{k}}{\frac {\left[X^{(i_{1})}Y^{(j_{1})}\cdots X^{(i_{k})}Y^{(j_{k})}X\right]}{i_{1}!j_{1}!\cdots i_{k}!j_{k}!}}\\{}+{}&t^{i_{1}+j_{1}+\cdots +i_{k}+j_{k}+i_{k+1}}{\frac {\left[X^{(i_{1})}Y^{(j_{1})}\cdots X^{(i_{k})}Y^{(j_{k})}X^{(i_{k+1})}Y\right]}{i_{1}!j_{1}!\cdots i_{k}!j_{k}!i_{k+1}!}},\quad i_{r},j_{r}\geq 0,\quad i_{r}+j_{r}>0,\quad 1\leq r\leq k\end{aligned}}.}$ 

Note that the summation index for the rightmost e^ad_tX in the second term in (97) is denoted i_{k + 1}, but is not an element of a sequence s ∈ S_k. Now integrate Z = Z(1) = ∫dZ/dtdt, using Z(0) = 0,

${\displaystyle {\begin{aligned}Z=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k+1}}\sum _{s\in S_{k},i_{k+1}\geq 0}&{\frac {1}{i_{1}+j_{1}+\cdots +i_{k}+j_{k}+1}}{\frac {\left[X^{(i_{1})}Y^{(j_{1})}\cdots X^{(i_{k})}Y^{(j_{k})}X\right]}{i_{1}!j_{1}!\cdots i_{k}!j_{k}!}}\\{}+{}&{\frac {1}{i_{1}+j_{1}+\cdots +i_{k}+j_{k}+i_{k+1}+1}}{\frac {\left[X^{(i_{1})}Y^{(j_{1})}\cdots X^{(i_{k})}Y^{(j_{k})}X^{(i_{k+1})}Y\right]}{i_{1}!j_{1}!\cdots i_{k}!j_{k}!i_{k+1}!}},\quad i_{r},j_{r}\geq 0,\quad i_{r}+j_{r}>0,\quad 1\leq r\leq k\end{aligned}}.}$ 

Write this as

${\displaystyle {\begin{aligned}Z=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k+1}}\sum _{s\in S_{k},i_{k+1}\geq 0}&{\frac {1}{i_{1}+j_{1}+\cdots +i_{k}+j_{k}+(i_{k+1}=1)+(j_{k+1}=0)}}{\frac {\left[X^{(i_{1})}Y^{(j_{1})}\cdots X^{(i_{k})}Y^{(j_{k})}X^{(i_{k+1}=1)}Y^{(j_{k+1}=0)}\right]}{i_{1}!j_{1}!\cdots i_{k}!j_{k}!(i_{k+1}=1)!(j_{k+1}=0)!}}\\{}+{}&{\frac {1}{i_{1}+j_{1}+\cdots +i_{k}+j_{k}+i_{k+1}+(j_{k+1}=1)}}{\frac {\left[X^{(i_{1})}Y^{(j_{1})}\cdots X^{(i_{k})}Y^{(j_{k})}X^{(i_{k+1})}Y^{(j_{k+1}=1)}\right]}{i_{1}!j_{1}!\cdots i_{k}!j_{k}!i_{k+1}!(j_{k+1}=1)!}},\\\\&(i_{r},j_{r}\geq 0,\quad i_{r}+j_{r}>0,\quad 1\leq r\leq k).\end{aligned}}}$ 

This amounts to

${\displaystyle Z=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k+1}}\sum _{s\in S_{k+1}}{\frac {1}{i_{1}+j_{1}+\cdots +i_{k}+j_{k}+i_{k+1}+j_{k+1}}}{\frac {\left[X^{(i_{1})}Y^{(j_{1})}\cdots X^{(i_{k})}Y^{(j_{k})}X^{(i_{k+1})}Y^{(j_{k+1})}\right]}{i_{1}!j_{1}!\cdots i_{k}!j_{k}!i_{k+1}!j_{k+1}!}},}$

(100)

where ${\displaystyle i_{r},j_{r}\geq 0,\quad i_{r}+j_{r}>0,\quad 1\leq r\leq k+1,}$  using the simple observation that [T, T] = 0 for all T. That is, in (100), the leading term vanishes unless j_{k + 1} equals 0 or 1, corresponding to the first and second terms in the equation before it. In case j_{k + 1} = 0, i_{k + 1} must equal 1, else the term vanishes for the same reason (i_{k + 1} = 0 is not allowed). Finally, shift the index, k → k − 1,

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.^[15] Namely, if

${\displaystyle X^{i_{1}}Y^{j_{1}}\cdots X^{i_{k}}Y^{j_{k}}}$ 

is expressible as a bracket series, then necessarily^[18]

${\displaystyle X^{i_{1}}Y^{j_{1}}\cdots X^{i_{k}}Y^{j_{k}}={\frac {\left[X^{(i_{1})}Y^{(j_{1})}\cdots X^{(i_{k})}Y^{(j_{k})}\right]}{i_{1}+j_{1}+\cdots +i_{k}+j_{k}}}.}$

(B)

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

• Matrix logarithm

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