Baker-Campbell-Hausdorff formula

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In mathematics, the Baker-Campbell-Hausdorff formula is the solution to

for non-commuting x and y. It is named for H. F. Baker, John Edward Campbell, and Felix Hausdorff. It was first noted in print by Campbell, elaborated by Poincare and Baker, and systematized by Hausdorff. The formula below was introduced by Eugene Dynkin. <ref>Rossmann, Wolf: "Lie Groups: An Introduction through Linear Groups". Oxford University Press, 2002.</ref>

Specifically, let G be a simply-connected Lie group with Lie algebra <math>\mathfrak g</math>. Let

<math>\exp : \mathfrak g\rightarrow G </math>

be the exponential map, defining

<math>Z = X * Y = \log(\exp X\exp Y),\qquad X,Y,Z\in\mathfrak g.</math>

The general formula is given by:

<math>X * Y =

\sum_{n>0}\frac {(-1)^{n-1}}{n} </math>

<math>\sum_{ \begin{matrix} & {r_i + s_i > 0}

                     \\ & {1\le i \le n} \end{matrix}}

\frac{(\sum_{i=1}^n (r_i+s_i))^{-1}}{r_1!s_1!\cdots r_n!s_n!}

\times(\mbox{ad} X)^{r_1}(\mbox{ad} Y)^{s_1}\cdots (\mbox{ad} X)^{r_n}(\mbox{ad} Y)^{s_n - 1}Y.

</math>

Here ad(A)B = [A,B] is the adjoint endomorphism. In terms in the sum where <math>s_n = 0</math>, the last three factors should be interpreted as <math>(\mbox{ad} X)^{r_n - 1} X</math>.

The first few terms are well-known, with all higher-order terms involving [X,Y] and commutator nestings thereof (thus in the Lie Algebra):

<math>X*Y = \,\!</math>	<math>X + Y + \frac {1}{2}[X,Y] + \frac {1}{12}[X,[X,Y]] - \frac {1}{12}[Y,[X,Y]] </math>
	<math>- \frac {1}{48}[Y,[X,[X,Y]]] - \frac{1}{48}[X,[Y,[X,Y]]] + \cdots</math>

There is no expression in closed form for an arbitrary Lie algebra, though there are exceptional tractable cases, as well as efficient algorithms for working out the expansion in applications.

For example, note that if [X,Y] vanishes, then the above formula manifestly reduces to X +Y. If the commutator [X,Y] is a constant (central), then all but the first three terms on the right-hand-side of the above vanish. If the commutator is =s Y, the formula reduces to Z = X+ Ys/(1−exp(−s)). There are numerous such well-known expressions applied routinely in physics (cf Magnus).

For a matrix Lie group <math>G \sub \mbox{GL}(n,\mathbb{R})</math> the Lie algebra is the tangent space of the identity I, and the commutator is simply [X, Y] = XY − YX; the exponential map is the standard exponential map of matrices,

<math>\mbox{exp}\ X = e^X = \sum_{n=0}^{\infty}{\frac{X^n}{n!}}.</math>

When we solve for Z in

we obtain a simpler formula:

<math> Z =

\sum_{n>0} \frac{(-1)^{n-1}}{n} \sum_{\begin{matrix} &{r_i+s_i>0}

                    \\ & {1\le i\le n}\end{matrix}}

\frac{X^{r_1}Y^{s_1}\cdots X^{r_n}Y^{s_n}}{r_1!s_1!\cdots r_n!s_n!}.</math>

We note that the first, second, third and fourth order terms are:

<math>z_1 = X + Y\,\!</math>

<math>z_2 = \frac{1}{2} (XY - YX)</math>

<math>z_3 = \frac{1}{12} (X^2Y + XY^2 - 2XYX + Y^2X + YX^2 - 2YXY)</math>

<math>z_4 = \frac{1}{24} (X^2Y^2 - 2XYXY - Y^2X^2 + 2YXYX).</math>

1 Integral version
2 See also
3 References
4 External links

[edit] Integral version

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An integral version of the formula exists as well. If two n×n complex matrices X and Y with norms sufficiently small, then

<math>\log (e^X e^Y ) = \int^1_0 f(e^{t \, \mbox{ad}_X} e^{t \, \mbox{ad}_Y})(X) \, dt

+ \int^1_0 f(e^{t \, \mbox{ad}_X} e^{t \, \mbox{ad}_Y})e^{t \, \mbox{ad}_X}(Y) \, dt</math>

where the function f is defined as

<math>f(A) = \sum_{n=0}^\infty a_n(A - I)^n </math>

<math>f(z) = \sum_{n=0}^\infty a_n(z - 1)^n </math>

[edit] See also

Dyson series

[edit] References

L. Corwin & F.P Greenleaf (1990) Representation of nilpotent Lie groups and their applications, Part 1: Basic theory and examples, ISBN 0-521-36034-X.