View source for Derivative of the exponential map

{{Short description|Formula in Lie group theory}}
[[Image:poincare.jpg|220px|thumb|right|In 1899, Henri Poincaré's investigations into group multiplication in Lie algebraic terms led him to the formulation of the [[Universal enveloping algebra|universal enveloping algebra]].<ref>{{harvnb|Schmid|1982}}</ref>]]
In the theory of [[Lie group]]s, the [[Exponential map (Lie theory)|exponential map]] is a map from the [[Lie algebra]] {{math|'''g'''}} of a Lie group {{math|''G''}} into {{math|''G''}}. In case {{math|''G''}} is a matrix Lie group, the exponential map reduces to the [[Matrix exponential|matrix exponential]]. The exponential map, denoted {{math|exp:'''g''' → ''G''}}, is [[Analytic function|analytic]] and has as such a [[Derivative|derivative]] {{math|{{sfrac|''d''|''dt''}}exp(''X''(''t'')):T'''g''' → T''G''}}, where {{math|''X''(''t'')}} is a {{math|''C''<sup>1</sup>}} [[Path (topology)|path]] in the Lie algebra, and a closely related [[Differential of a function|differential]] {{math|''d''exp:T'''g''' → T''G''}}.<ref name=Rossmann_A>{{harvnb|Rossmann|2002}} Appendix on analytic functions.</ref>

The formula for {{math|''d''exp}} was first proved by Friedrich Schur (1891).<ref>{{harvnb|Schur|1891}}</ref> It was later elaborated by [[Biography:Henri Poincaré|Henri Poincaré]] (1899) in the context of the problem of expressing Lie group multiplication using Lie algebraic terms.<ref>{{harvnb|Poincaré|1899}}</ref> 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&ndash;Campbell&ndash;Hausdorff formula, and it is used frequently in physics<ref>{{harvnb|Suzuki|1985}}</ref>  for example in [[Physics:Quantum field theory|quantum field theory]], as in the [[Physics:Magnus expansion|Magnus expansion]] in [[Physics:Perturbation theory|perturbation theory]], and in [[Physics:Lattice gauge theory|lattice gauge theory]].

Throughout, the notations {{math|exp(''X'')}} and {{math|''e''<sup>''X''</sup>}} 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 {{math|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<ref name=Rossmann_2>{{harvnb|Rossmann|2002}} Theorem 5 Section 1.2</ref>
{{Equation box 1
|indent =:
|equation = 
<math>\frac{d}{dt}e^{X(t)} = e^{X(t)}\frac{1 - e^{-\mathrm{ad}_{X}}}{\mathrm{ad}_{X}}\frac{dX(t)}{dt}.</math>  &nbsp;   &nbsp;   &nbsp;   &nbsp;   &nbsp;   &nbsp;   &nbsp;    {{EquationRef|(1)}}
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;Explanation
{{unordered list
| {{math|1=''X'' = ''X''(''t'')}} is a {{math|''C''<sup>1</sup>}} (continuously differentiable) path in the Lie algebra with derivative {{math|1=''X''′(''t'') = {{sfrac|''dX''(''t'')|''dt''}}}}. The argument {{math|''t''}} is omitted where not needed. 
| {{math|ad<sub>''X''</sub>}} is the linear transformation of the Lie algebra given by {{math|1=ad<sub>''X''</sub>(''Y'') = [''X'', ''Y'']}}. It is the adjoint action of a Lie algebra on itself.
| The fraction {{math|{{sfrac|1 − exp(−ad<sub>''X''</sub>)|ad<sub>''X''</sub>}}}} is given by the power series
{{NumBlk|:|<math>\frac{1 - e^{-\mathrm{ad}_{X}}}{\mathrm{ad}_{X}} = \sum_{k = 0}^\infty \frac{(-1)^k}{(k + 1)!}(\mathrm{ad}_X)^k.</math> |{{EquationRef|2}}}}
derived from the power series of the exponential map of a linear endomorphism, as in matrix exponentiation.<ref name=Rossmann_2/>
| When {{math|''G''}} is a matrix Lie group, all occurrences of the exponential are given by their power series expansion.
| When {{math|''G''}} is ''not'' a matrix Lie group, {{math|{{sfrac|1 − exp(−ad<sub>''X''</sub>)|ad<sub>''X''</sub>}}}} is still given by its power series ({{EquationNote|2}}), while the other two occurrences of {{math|exp}} in the formula, which now are the [[Exponential map (Lie theory)|exponential map in Lie theory]], refer to the time-one [[Flow (mathematics)|flow]] of the [[Lie group#Basic concepts|left invariant]] [[Vector field|vector field]] {{mvar|X}}, i.e. element of the Lie algebra as defined in the general case, on the Lie group {{mvar|G}} viewed as an [[Analytic manifold|analytic manifold]]. This still amounts to exactly the same formula as in the matrix case. Left multiplication of an element of the algebra {{math|'''g'''}} by an element {{math|exp(''X''(''t''))}} of the Lie group is interpreted as applying the differential of the left translation {{math|dL<sub>exp(''X''(''t''))</sub>}}.
| The formula applies to the case where {{math|exp}} is considered as a map on matrix space over {{math|ℝ}} or {{math|'''C'''}}, see [[Matrix exponential|matrix exponential]]. When {{math|1=''G'' = GL(''n'', '''C''')}} or {{math|GL(''n'', '''R''')}}, the notions coincide precisely.
}}

To compute the '''differential''' {{math|''d''exp}} of {{math|exp}} at {{math|''X''}}, {{math|''d''exp<sub>''X''</sub>: T'''g'''<sub>''X''</sub> → T''G''<sub>exp(''X'')</sub>}}, the standard recipe<ref name=Rossmann_A/>
: <math>d\exp_XY = \left .\frac{d}{dt}e^{Z(t)}\right|_{t = 0}, Z(0) = X, Z'(0) = Y</math>
is employed. With {{math|1=''Z''(''t'') = ''X'' + ''tY''}} the result<ref name=Rossmann_2/>
{{NumBlk|:|<math>d\exp_XY = e^{X}\frac{1 - e^{-\mathrm{ad}_{X}}}{\mathrm{ad}_{X}}Y</math>|{{EquationRef|3}}}}

follows immediately from {{EquationNote|(1)}}. In particular, {{math|1=''d''exp<sub>0</sub>:T'''g'''<sub>0</sub> → T''G''<sub>exp(0)</sub> = T''G''<sub>''e''</sub>}} is the identity because {{math|T'''g'''<sub>''X''</sub> ≃ '''g'''}} (since {{math|'''g'''}} is a vector space) and {{math|T''G''<sub>''e''</sub> ≃ '''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 {{math|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 {{math|''s''}} of the parametrized expression
:<math>\Gamma(s, t) = e^{-sX(t)}\frac{\partial}{\partial t} e^{sX(t)}</math>
to obtain a first order differential equation for {{math|Γ}} which can then be solved by direct integration in {{mvar|s}}. The solution is then {{math|''e''<sup>''X''</sup> Γ(1, t)}}.

'''Lemma'''

Let {{math|[[Adjoint representation|Ad]]}} denote the [[Adjoint representation|adjoint action]] of the group on its Lie algebra. The action is given by {{math|1=Ad<sub>''A''</sub>''X'' = ''AXA''<sup>−1</sup>}} for {{math|''A'' ∈ ''G'', ''X'' ∈ '''g'''}}. A frequently useful relationship between {{math|Ad}} and {{math|ad}} is given by<ref>{{harvnb|Hall|2015}} Proposition 3.35</ref><ref group=nb>A proof of the identity can be found in [[Baker–Campbell–Hausdorff formula#An important lemma|here]]. The relationship is simply that between a representation of a Lie group and that of its Lie algebra according to the Lie correspondence, since both {{math|Ad}} and {{math|ad}} are representations with {{math|1=ad = ''d''Ad}}.</ref>
{{Equation box 1
|indent =:
|equation = 
<math>\mathrm{Ad}_{e^X} = e^{\mathrm{ad}_X}, ~~X \in \mathfrak{g}~.</math>   &nbsp;   &nbsp;   &nbsp;   &nbsp;   &nbsp;   &nbsp;   &nbsp;    {{EquationRef|(4)}}
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'''Proof'''

Using the product rule twice one finds,
:<math>\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}.</math>
Then one observes that
:<math>\frac{\partial\Gamma}{\partial s} = \mathrm{Ad}_{e^{-sX}}X' = e^{-\mathrm{ad}_{sX}}X',</math>
by {{EquationNote|(4)}} above. Integration yields
:<math>\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.</math>
Using the formal power series to expand the exponential, integrating term by term, and finally recognizing ({{EquationNote|2}}),
:<math>\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},</math>
and the result follows. The proof, as presented here, is essentially the one given in {{harvtxt|Rossmann|2002}}. A proof with a more algebraic touch can be found in {{harvtxt|Hall|2015}}.<ref>See also {{harvnb|Tuynman|1995}} from which Hall's proof is taken.</ref>

===Comments on the general case===
The formula in the general case is given by<ref>{{harvnb|Sternberg|2004}} This is equation (1.11).</ref>
:<math>\frac{d}{dt}\exp(C(t)) = \exp(C)\phi(-\mathrm{ad}(C))C~ ',</math>
where<ref group="nb">
It holds that
:<math>\tau(\log z)\phi(-\log z) = 1</math>
for |z − 1| < 1 where
:<math>\tau(w) = \frac{w}{1 - e^{-w}}.</math>
Here, {{math|''τ''}} is the exponential generating function of
:<math>(-1)^k b_k,</math>
where {{math|''b''<sub>''k''</sub>}} are the Bernoulli numbers.
</ref>
:<math>\phi(z) = \frac{e^z - 1}{z} = 1 + \frac{1}{2!}z + \frac{1}{3!}z^2 + \cdots,</math>
which formally reduces to
:<math>\frac{d}{dt}\exp(C(t)) = \exp(C)\frac{1 - e^{-\mathrm{ad}_{C}}}{\mathrm{ad}_{C}}\frac{dC(t)}{dt}.</math>
Here the {{math|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 {{harvtxt|Sternberg|2004}} 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,
:<math>\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}</math>
where each factor owes its place  to the non-commutativity of {{math|''X''(''t'')}} and {{math|''X''´(''t'')}}.

Dividing the unit interval into {{math|''N''}} sections {{math|1=Δ''s'' = {{sfrac|Δ''k''|''N''}}}} ({{math|1=Δ''k'' = 1}} since the sum indices are integers) and letting {{mvar|N}} → ∞, {{math|Δ''k'' → ''dk'', {{sfrac|''k''|''N''}} → ''s''}}, {{math|Σ → ∫}} yields
:<math>\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}</math>

==Applications==
===Local behavior of the exponential map===
The [[Inverse function theorem|inverse function theorem]] together with the derivative of the exponential map provides information about the local behavior of {{math|exp}}. Any {{math|''C''<sup>''k''</sup>, 0 ≤ ''k'' ≤ ∞, ''ω''}} map {{math|''f''}} between vector spaces (here first considering matrix Lie groups) has a {{math|''C''<sup>''k''</sup>}} inverse such that {{math|''f''}} is a {{math|''C''<sup>''k''</sup>}} bijection in an open set around a point {{math|''x''}} in the domain provided {{math|''df''<sub>''x''</sub>}} is invertible. From ({{EquationNote|3}}) it follows that this will happen precisely when

:<math>\frac{1 - e^{\mathrm{ad_X}}}{\mathrm{ad}_X}</math>

is invertible. This, in turn, happens when the eigenvalues of this operator are all nonzero. The eigenvalues of {{math|{{sfrac|1 − exp(−ad<sub>''X''</sub>)|ad<sub>''X''</sub>}}}} are related to those of {{math|ad<sub>''X''</sub>}} as follows. If {{math|''g''}} is an analytic function of a complex variable expressed in a power series such that {{math|''g''(''U'')}} for a matrix {{math|''U''}} converges, then the eigenvalues of {{math|''g''(''U'')}} will be {{math|''g''(''λ''<sub>''ij''</sub>)}}, where {{math|''λ''<sub>''ij''</sub>}} are the eigenvalues of {{math|''U''}}, the double subscript is made clear below.<ref group=nb>This is seen by choosing a basis for the underlying vector space such that {{math|''U''}} is [[Triangular matrix|triangular]], the eigenvalues being the diagonal elements.  Then {{math|''U''<sup>''k''</sup>}} is triangular with diagonal elements {{math|''λ''<sub>''i''</sub><sup>''k''</sup>}}. It follows that the eigenvalues of {{math|''U''}} are {{math|''f''(''λ''<sub>''i''</sub>)}}. See {{harvnb|Rossmann|2002}}, Lemma 6 in section 1.2.</ref> In the present case with {{math|1=''g''(''U'') = {{sfrac|1 − exp(−''U'')|''U''}}}} and {{math|1=''U'' = ad<sub>''X''</sub>}}, the eigenvalues of {{math|{{sfrac|1 − exp(−ad<sub>''X''</sub>)|ad<sub>''X''</sub>}}}} are
:<math>\frac{1 - e^{-\lambda_{ij}}}{\lambda_{ij}},</math>
where the {{math|''λ''<sub>''ij''</sub>}} are the eigenvalues of {{math|ad<sub>''X''</sub>}}. Putting {{math|1={{sfrac|1 − exp(−''λ''<sub>''ij''</sub>)|''λ''<sub>''ij''</sub>}} = 0}} one sees that {{math|''d''exp}} is invertible precisely when
:<math>\lambda_{ij} \ne k2\pi i, k = \pm1, \pm2, \ldots.</math>

The eigenvalues of {{math|ad<sub>''X''</sub>}} are, in turn, related to those of {{math|''X''}}. Let the eigenvalues of {{math|''X''}} be {{math|''λ''<sub>''i''</sub>}}. Fix an ordered basis {{math|''e''<sub>''i''</sub>}} of the underlying vector space {{math|''V''}} such that {{math|''X''}} is lower triangular. Then 
:<math>Xe_i = \lambda_ie_i + \cdots,</math>
with the remaining terms multiples of {{math|''e''<sub>''n''</sub>}} with {{math|''n'' > ''i''}}. Let {{math|''E''<sub>''ij''</sub>}} be the corresponding basis for matrix space, i.e. {{math|1=(''E''<sub>''ij''</sub>)<sub>''kl''</sub> = ''δ''<sub>''ik''</sub>''δ''<sub>''jl''</sub>}}. Order this basis such that {{math|''E''<sub>''ij''</sub> < ''E''<sub>''nm''</sub>}} if {{math|''i'' − ''j'' < ''n'' − ''m''}}. One checks that the action of {{math|ad<sub>''X''</sub>}} is given by
:<math>\mathrm{ad}_XE_{ij} = (\lambda_i - \lambda_j)E_{ij} + \cdots \equiv \lambda_{ij}E_{ij} + \cdots,</math>
with the remaining terms multiples of {{math|''E''<sub>''mn''</sub> > ''E''<sub>''ij''</sub>}}. This means that {{math|ad<sub>''X''</sub>}} is lower triangular with its eigenvalues {{math|1=''λ''<sub>''ij''</sub> = ''λ''<sub>''i''</sub> − ''λ''<sub>''j''</sub>}} on the diagonal. The conclusion is that {{math|''d''exp<sub>''X''</sub>}} is invertible, hence {{math|exp}} is a local bianalytical bijection around {{math|''X''}}, when the eigenvalues of {{math|''X''}} satisfy<ref>{{harvnb|Rossmann|2002}} Proposition 7, section 1.2.</ref><ref group=nb>Matrices whose eigenvalues {{math|''λ''}} satisfy {{math|{{!}}Im ''λ''{{!}} < ''π''}} are, under the exponential, in bijection with matrices whose eigenvalues {{math|''μ''}} are not on the negative real line or zero. The {{math|''λ''}} and {{math|''μ''}} are related by the complex exponential. See {{harvtxt|Rossmann|2002}} Remark 2c section 1.2.</ref>
:<math>\lambda_i - \lambda_j \ne k2\pi i, \quad k = \pm1, \pm2, \ldots, \quad 1 \le i, j \le n = \dim V.</math>

In particular, in the case of matrix Lie groups, it follows, since {{math|''d''exp<sub>0</sub>}} is invertible, by the [[Inverse function theorem|inverse function theorem]] that {{math|exp}} is a bi-analytic bijection in a neighborhood of {{math|0 ∈ '''g'''}} in matrix space. Furthermore, {{math|exp}}, is a bi-analytic bijection from a neighborhood of {{math|0 ∈ '''g'''}} in {{math|'''g'''}} to a neighborhood of {{math|''e'' ∈ ''G''}}.<ref>{{harvnb|Hall|2015}} Corollary 3.44.</ref> 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|implicit function theorem]] that {{math|''d''exp<sub>''ξ''</sub>}} itself is invertible for {{math|''ξ''}} sufficiently small.<ref>{{harvnb|Sternberg|2004}} Section 1.6.</ref>

=== Derivation of a Baker–Campbell–Hausdorff formula ===
{{Main|Baker–Campbell–Hausdorff formula}}
If {{math|''Z''(''t'')}} is defined such that 
:<math>e^{Z(t)} = e^{X} e^{tY},</math>
an expression for {{math|1=''Z''(1) = log( exp ''X'' exp ''Y'' )}}, the [[Baker–Campbell–Hausdorff formula]], can be derived from the above formula,
:<math>\exp(-Z(t))\frac{d}{dt}\exp(Z(t)) = \frac{1 - e^{-\mathrm{ad}_{Z}}}{\mathrm{ad}_{Z}}Z'(t).</math>

Its left-hand side is easy to see to equal ''Y''. Thus,
:<math>Y = \frac{1 - e^{-\mathrm{ad}_{Z}}}{\mathrm{ad}_{Z}}Z'(t),</math>
and hence, formally,<ref>{{harvnb|Hall|2015}}Section 5.5.</ref><ref>{{harvnb|Sternberg|2004}} Section 1.2.</ref>
:<math>
  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.
</math>

However, using the relationship between {{math|Ad}} and {{math|ad}} given by {{EquationNote|(4)}}, it is straightforward to further see that 
:<math> e^{\mathrm{ad}_{Z}} = e^{\mathrm{ad}_{X}} e^{t\mathrm{ad}_{Y}} </math>
and hence 
:<math>Z'(t) = \psi\left(e^{\mathrm{ad}_{X}} e^{t\mathrm{ad}_{Y}}\right)Y.</math>
Putting this into the form of an integral in ''t'' from 0 to 1 yields,
:<math>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, </math>
an [[Baker–Campbell–Hausdorff formula#Selected tractable cases|integral formula]] for {{math|''Z''(1)}} that is more tractable in practice than the explicit [[Baker–Campbell–Hausdorff formula#An explicit Baker–Campbell–Hausdorff formula|Dynkin's series formula]] due to the simplicity of the series expansion of {{math|''ψ''}}. Note this expression consists of {{math|''X+Y''}} and nested commutators thereof with {{mvar|X}} or {{mvar|Y}}. A textbook proof along these lines can be found in {{harvtxt|Hall|2015}}  and {{harvtxt|Miller|1972}}.

===Derivation of Dynkin's series formula ===
[[Image:Eugene Dynkin 2003.jpg|220px|thumb|right|[[Biography:Eugene Dynkin|Eugene Dynkin]] at home in 2003. In 1947 Dynkin proved the explicit BCH series formula.<ref name=Dynkin>{{harvnb|Dynkin|1947}}</ref> Poincaré, Baker, Campbell and Hausdorff were mostly concerned with the ''existence'' of a bracket series, which suffices in many applications, for instance, in proving central results in the Lie correspondence.<ref>{{harvnb|Rossmann|2002}} Chapter 2.</ref><ref>{{harvnb|Hall|2015}} Chapter 5.</ref> Photo courtesy of the Dynkin Collection.]]

Dynkin's formula mentioned may also be derived analogously, starting from the parametric extension
:<math>e^{Z(t)} = e^{tX} e^{tY},</math>
whence 
:<math>e^{-Z(t)} \frac{de^{Z(t)}}{dt} = e^{-t \, \mathrm{ad}_{Y}}X + Y ~,</math>
so that, using the above general formula,
:<math>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) .</math>

Since, however, 
:<math>\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{(-)^{n+1}}{n} (\exp(\mathrm{ad}_Z) - 1)^n ~, \quad \|\mathrm{ad}_Z\| < \log 2   ~~,
\end{align}</math>
the last step by virtue of the [[Mercator series]] expansion, it follows that
{{NumBlk|:|<math>Z' = \sum\limits^{\infty}_{n=1} \frac{(-)^{n-1}}{n} \left(e^{\mathrm{ad}_Z} - 1\right)^{n-1} ~ \left(X + e^{t \, \mathrm{ad}_{X}}Y\right)~,</math>|{{EquationRef|5}}}}
and, thus, integrating,
:<math>Z(1) = \int^1 _0 dt ~\frac{dZ(t)}{dt} = \sum^{\infty}_{n=1} \frac{(-)^{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) .</math>

It is at this point evident that the qualitative statement of the BCH formula holds, namely {{math|''Z''}} lies in the Lie algebra generated by {{math|''X'', ''Y''}} and is expressible as a series in repeated brackets {{EquationRef|(A)}}. For each {{mvar|k}}, terms for each partition thereof are organized inside the integral {{math|∫''dt'' ''t''<sup>''k''−1</sup>}}. The resulting Dynkin's formula is then
{{Equation box 1
|indent =
|equation = <math>Z = \sum_{k = 1}^\infty \frac{(-1)^{k-1}}{k} \sum_{s \in S_{k}} \frac{1}{i_1 + j_1 + \cdots + i_k + j_k}\frac{[X^{(i_1)}Y^{(j_1)}\cdots X^{(i_k)}Y^{(j_k)}]}{i_1!j_1!\cdots i_k!j_k!}, \quad i_r,j_r \ge 0,\quad i_r + j_r > 0,\quad 1 \le r \le k.</math>
|cellpadding= 6
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}}      
For a similar proof with detailed series expansions, see {{harvtxt|Rossmann|2002}}.

===Combinatoric details===
Change the summation index in ({{EquationNote|5}}) to {{math|1=''k'' = ''n'' − 1}} and expand
{{NumBlk|:|<math>\frac{dZ}{dt} = \sum_{k = 0}^\infty \frac{(-1)^{k}}{k + 1}\left\{\left(e^{\mathrm{ad}_{tX}}e^{\mathrm{ad}_{tY}} - 1\right)^kX + \left(e^{\mathrm{ad}_{tX}}e^{\mathrm{ad}_{tY}} - 1\right)^ke^{\mathrm{ad}_{tX}}Y\right\}</math>|{{EquationRef|97}}}}
in a power series. To handle the series expansions simply, consider first {{math|1=''Z'' = log(''e''<sup>''X''</sup>''e''<sup>''Y''</sup>)}}. The {{math|log}}-series and the {{math|exp}}-series are given by
:<math>\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!}</math>
respectively. Combining these one obtains
{{NumBlk|:|<math>\log\left(e^X e^Y\right) = \sum_{k = 1}^\infty \frac{(-1)^{k + 1}}{k}{\left(e^Xe^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 \ge 0, i + j > 1}^\infty \frac{X^iY^j}{i!j!} \right)^k.</math>|{{EquationRef|98}}}}
This becomes
{{Equation box 1
|indent =
|equation = 
<math>Z = \log\left(e^X e^Y\right) = \sum_{k = 1}^\infty \frac{(-1)^{k + 1}}{k} \sum_{s \in S_k} \frac{X^{i_1}Y^{j_1}\cdots X^{i_k}Y^{j_k}}{i_1!j_1!\cdots i_k!j_k!}, \quad i_r, j_r \ge 0, \quad i_r + j_r > 0, \quad 1 \le r \le k,</math>&nbsp; &nbsp; &nbsp; &nbsp; {{EquationRef|(99)}}
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where {{math|''S''<sub>''k''</sub>}} is the set of all sequences {{math|1=''s'' = (''i''<sub>1</sub>, ''j''<sub>1</sub>, ..., ''i''<sub>''k''</sub>, ''j''<sub>''k''</sub>)}} of length {{math|2''k''}} subject to the conditions in {{EquationNote|(99)}}.

Now substitute {{math|(''e''<sup>''X''</sup>''e''<sup>''Y''</sup> − 1)}} for {{math|(''e''<sup>ad<sub>''tX''</sub></sup>''e''<sup>ad<sub>''tY''</sub></sup> − 1)}} in the [[Sides of an equation|LHS]] of ({{EquationNote|98}}). Equation {{EquationNote|(99)}} then gives
:<math>\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}</math>
or, with a switch of notation, see An explicit Baker&ndash;Campbell&ndash;Hausdorff formula,
:<math>\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}.</math>

Note that the summation index for the rightmost {{math|''e''<sup>ad<sub>''tX''</sub></sup>}} in the second term  in ({{EquationNote|97}}) is denoted {{math|''i''<sub>''k'' + 1</sub>}}, but is ''not'' an element of a sequence {{math|''s'' ∈ ''S''<sub>''k''</sub>}}. Now integrate {{math|1=''Z'' = ''Z''(1) = ∫{{sfrac|''dZ''|''dt''}}''dt''}}, using {{math|1=''Z''(0) = 0}},
:<math>\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}.</math>

Write this as 
:<math>\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}</math>
This amounts to
{{NumBlk|:|<math>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}!}, </math>|{{EquationRef|100}}}}
where <math>i_r,j_r \ge 0,\quad i_r + j_r > 0,\quad 1 \le r \le k + 1,</math> using the simple observation that {{math|1=[''T'', ''T''] = 0}} for all {{math|''T''}}. That is, in ({{EquationNote|100}}), the leading term vanishes unless {{math|''j''<sub>''k'' + 1</sub>}} equals {{math|0}} or {{math|1}}, corresponding to the first and second terms in the equation before it. In case {{math|1=''j''<sub>''k'' + 1</sub> = 0}}, {{math|''i''<sub>''k'' + 1</sub>}} must equal {{math|1}}, else the term vanishes for the same reason ({{math|1=''i''<sub>''k'' + 1</sub> = 0}} is not allowed). Finally, shift the index, {{math|''k'' → ''k'' − 1}},
{{Equation box 1
|indent =
|equation =<math>Z = \log e^Xe^Y = \sum_{k = 1}^\infty \frac{(-1)^{k-1}}{k} \sum_{s \in S_{k}} \frac{1}{i_1 + j_1 + \cdots + i_k + 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!},~ i_r,j_r \ge 0,~ i_r + j_r > 0,~ 1 \le r \le k.</math>
|cellpadding= 6
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This is Dynkin's formula. The striking similarity with (99) is not accidental: It reflects the '''Dynkin&ndash;Specht&ndash;Wever map''', underpinning the original, different, derivation of the formula.<ref name=Dynkin/> Namely, ''if''
:<math>X^{i_1}Y^{j_1} \cdots X^{i_k}Y^{j_k}</math>
is expressible as a bracket series, then necessarily<ref>{{harvnb|Sternberg|2004}} Chapter 1.12.2.</ref>
{{NumBlk|:|<math>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}.</math>|{{EquationRef|B}}}}
Putting observation {{EquationNote|(A)}} and theorem ({{EquationNote|B}}) together yields a concise proof of the explicit BCH formula.

==See also==
*Matrix logarithm

==Remarks==
{{reflist|group=nb}}

==Notes==
{{reflist}}

==References==
*{{citation|last=Dynkin|first=Eugene Borisovich|year=1947|language=Russian|title=Вычисление коэффициентов в формуле Campbell–Hausdorff|trans-title=Calculation of the coefficients in the Campbell–Hausdorff formula|journal=Doklady Akademii Nauk SSSR|volume=57|pages=323–326}} ;  translation from  [https://books.google.com/books?id=D9ZF5O_JH2gC&dq=Dynkin+Yushkevich++Campbell&pg=PA31       Google books].
*{{Citation| last=Hall|first=Brian C.|title=Lie groups, Lie algebras, and representations: An elementary introduction|edition=2nd|series=Graduate Texts in Mathematics|volume=222|publisher=Springer|year=2015|isbn=978-3319134666}}
*{{citation|last=Miller| first=Wllard| title=Symmetry Groups and their Applications|publisher=Academic Press| year=1972| isbn=0-12-497460-0}}
*{{citation|last=Poincaré|first=H.|year=1899|title=Sur les groupes continus|journal=Cambridge Philos. Trans.|volume=18|pages=220–55|}}
*{{citation|last=Rossmann|first= Wulf|title=Lie Groups – An Introduction Through Linear Groups|publisher=Oxford Science Publications|year=2002|series=Oxford Graduate Texts in Mathematics|isbn=0-19-859683-9}}
*{{citation|last=Schur|first=F.|year=1891|title=Zur Theorie der endlichen Transformationsgruppen|journal=Abh. Math. Sem. Univ. Hamburg|volume=4|pages=15–32|}}
*{{cite journal|last=Suzuki |first=Masuo| year=1985|title=Decomposition formulas of exponential operators and Lie exponentials with some applications to quantum mechanics and statistical physics|journal=Journal of Mathematical Physics| volume=26| issue=4|pages=601–612|doi=10.1063/1.526596| bibcode=1985JMP....26..601S}}
*{{citation|author=Tuynman|year=1995|title=The derivation of the exponential map of matrices|journal=Amer. Math. Monthly|volume=102|issue=9|pages=818–819|doi=10.2307/2974511|jstor=2974511}}
*Veltman, M, [[Biography:Gerard 't Hooft|'t Hooft, G]] & [[Biography:Bernard de Wit|de Wit, B]] (2007). "Lie Groups in Physics", [http://www.staff.science.uu.nl/~hooft101/lectures/lieg07.pdf online lectures].
*{{cite journal|doi=10.1063/1.1705306 | author =Wilcox, R. M.  | title=Exponential Operators and Parameter Differentiation in Quantum Physics | journal=Journal of Mathematical Physics | volume=8 | pages=962–982 | year=1967|issue=4| bibcode =1967JMP.....8..962W  }}

==External links==
*{{citation|last=Sternberg|first=Shlomo|year=2004|url=http://www.math.harvard.edu/~shlomo/docs/lie_algebras.pdf|title=Lie Algebras}}
*{{citation|last=Schmid|first=Wilfried|year=1982|url=https://www.ams.org/journals/bull/1982-06-02/S0273-0979-1982-14972-2/S0273-0979-1982-14972-2.pdf|title=Poincaré and Lie groups|journal=Bull. Amer. Math. Soc.|volume=6|issue=2|pages=175–186|doi=10.1090/s0273-0979-1982-14972-2|doi-access=free}}

[[Category:Mathematical physics]]
[[Category:Matrix theory]]
[[Category:Lie groups]]
[[Category:Exponentials]]

{{Sourceattribution|Derivative of the exponential map}}