"Modern" proof for the Baker-Campbell-Hausdorff formula - MathOverflow

"Modern" proof for the Baker-Campbell-Hausdorff formula

2012-06-10T11:14:03Z

Does someone has a reference to a modern proof of the Baker-Campbell-Hausdorff formula?

All proofs I have ever seen are related only to matrix Lie groups / Lie algebras and are not at all geometric (i.e. depend on indices, bases ect.)

By a 'modern' proof I'm thinking of a proof entirely in terms of differential geometry, i.e. in terms of the tangent bundle on the Lie group manifold or even better in terms of jets.

I'll keep the formulation vague on purpose, to higher my chances to get a good reference. I think the question is pretty clear anyway.

Beyond the plain BCH-equation I would like to get a deeper understanding WHY the commutator (and the linear structure of the Lie algebra) is enough to define the group product locally and what is geometrically going on.

Answer by alvarezpaiva for "Modern" proof for the Baker-Campbell-Hausdorff formula

2012-06-10T11:47:03Z

It seems to me that Baker-Cambell-Hausdorff is just (noncommutative) algebra. A nice closed formula was found by Dynkin On a representation of the series $\log(e^x e^y)$ in non-commuting $x$ and $y$ via their commutators Math. Sb. 25 (67) 1949. The book "Methods of Noncommutative Analysis" by Nazaikinskii, Shatalov, and Sternin has a nice presentation.

There is also an interesting paper of Mosolova: New formula for $\log(e^B e^A)$ in terms of commutators of $A$ and $B$. Math. Notes. 29 1978.

Answer by Theo Johnson-Freyd for "Modern" proof for the Baker-Campbell-Hausdorff formula

2012-06-10T14:24:53Z

In the Lie Theory notes on my website based on the 2008 class by Mark Haiman, section 3.3, we give the following discussion. I should mention that I think of BCH and breaking into two parts. The first is purely algebraic:

Let $U$ be a bialgebra over field $\mathbb k$, and $s$ a formal (commuting) variable. Then we can form the $s$-adic completion $U[\![ s ]\!]$ of $U \otimes_{\mathbb k} \mathbb k[s]$. It is an algebra, of course. It is not a bialgebra in the algebraic sense. On the other hand, we can extend the comultiplication $\Delta : U \to U\otimes U$ by linearity (and continuity – it is $s$-adic-continuous) to $U[\![ s ]\!] \to (U\otimes U)[\![ s ]\!]$, and the map $U[\![ s ]\!] \otimes_{\mathbb k[\![ s ]\!]} U[\![ s ]\!] \to (U\otimes U)[\![ s ]\!]$ realizes the latter as the $s$-adic-completion of the former. So $U[\![ s ]\!]$ is a bialgebra for this $s$-adic-completed tensor product, which I will denote by $\hat\otimes$. Denote this completed comultiplication by $\hat\Delta$. Note that $\hat\Delta$ is continuous in the $s$-adic topology.
Choose $\psi \in U[\![ s ]\!]$ such that $\psi(s=0) = 0$. Then $\psi$ is primitive (i.e. $\hat\Delta(\psi) = \psi \otimes 1 + 1\otimes \psi$ iff $e^\psi$ is grouplike (i.e. $\hat\Delta(e^\psi) = e^\psi \hat\otimes e^\psi$). Note that $e^\psi = \sum_n \frac1{n!}\psi^n$ converges in the $s$-adic topology since $\psi(s=0) = 0$. Proof: By continuity, $\hat\Delta(e^\psi) = e^{\hat\Delta(\psi)}$. As a remark, note that the power series $\psi \in U[\![ s ]\!]$ is primitive iff it is primitive term-by-term.
Everything above continues to work if we use two commuting formal variables $s$ and $t$. Let $\mathfrak f$ denote the free Lie algebra on two generators $x$ and $y$. Then its universal enveloping algebra is the free associative algebra $T$ on the same two generators. (Proof: "free" functors are left adjoint to "forget to vector spaces" and "universal enveloping" is left adjoint to "forget from associative to Lie", and the composition of left adjoints is left adjoint to the corresponding composition of forgetful functors.) We will work in the (completed) bialgebra $T[\![s,t]\!]$. Define the formal series $b(sx,ty)$ by the formula $$ e^{b(sx,ty)} = e^{sx}e^{ty}. $$ Since $x$ and $y$ are primitive and the product of grouplikes is grouplike, it follows from 2. that $b(sx,ty)$ is primitive. Indeed, the coefficient on $s^mt^n$ in $b(sx,ty)$ is primitive.
The primitives in a universal enveloping algebra (in this case, $T$) are precisely the original Lie algebra (in this case, $\mathfrak f$). Thus the coefficient on $s^mt^n$ in $b(sx,ty)$ is a Lie polynomial (composition of brackets) in the noncommuting variables $x,y$. By degree counting, it is homogeneous of degree $m$ in $x$ and homogeneous of degree $n$ in $y$. More generally, $b(sx,ty)$ is a Lie series. If you work to define the correct topology so that you can talk about power series in noncommuting variables, then it makes sense to talk about the series $b(x,y)$.

The second part is manifold-theoretic:

Let $G$ denote a Lie group (whose underlying manifold is given analytic structure, and such that multiplication is an analytic map), and $\mathfrak g$ its Lie algebra. Identify the universal enveloping algebra $U\mathfrak g$ with the left-invariant differential operators on $G$. Consider the stalk $\mathcal C(G)_e$ of analytic functions defined in a neighborhood of the identity. Then $u\in U\mathfrak g = 0$ iff $u$ annihilates $\mathcal C(G)_e$.
Recall that we have a map $\exp: \mathfrak g \to G$ (defined, for example, by flowing along the left-invariant vector on $G$ field defined by $x\in \mathfrak g$). It is an analytic isomorphism near the identity, and so near the identity we have an inverse map $\log : G \to \mathfrak g$. By passing to smaller neighborhoods of the identity as needed, we can define $\beta(x,y) = \log(\exp x\exp y)$. It is a $\mathfrak g$-valued analytic function on a neighborhood of $(0,0)\in \mathfrak g \times \mathfrak g$.
Choose $f\in \mathcal C(G)_e$ and $x,y\in \mathfrak g$. Then $e^{sx} \in U\mathfrak g[\![s]\!]$ makes is $\mathbb k[\![s]\!]$-valued differential operator on $\mathcal C(G)_e$, and $e^{sx} f$ is the Taylor expansion in $s$ of $f(\exp sx)$. Similarly, $e^{sx}e^{ty}f$ computes the two-variable Taylor expansion of $f(\exp sx \exp ty)$. Let $\tilde \beta$ denote the Taylor expansion of $\beta(sx,ty)$. Then $$ e^{\tilde \beta(sx,ty)}f = f(\exp(\tilde\beta(sx,ty))) = f(\exp sx \exp ty) = e^{sx}e^{ty} f = e^{b(sx,ty)} f. $$ By 1., we must have $\tilde\beta(sx,ty) = b(sx,ty)$ as formal power series.
But $\tilde\beta$ is the Taylor expansion of the analytic function $\beta$. Thus in a small enough neighborhood of the origin, it converges. By definition, $\tilde \beta = b $ is the Baker–Campbell–Hausdorff series.

Answer by Dror Bar-Natan for "Modern" proof for the Baker-Campbell-Hausdorff formula

2012-06-10T14:56:03Z

A videotaped proof that $\log e^xe^y$ is a Lie series in $x$ and $y$ is at http://katlas.math.toronto.edu/drorbn/dbnvp/wClips-120418.php, starting around minute 10 and ending near the end of the lecture. You may want to stare at the blackboard shots on the right column before viewing the video.

That proof is very short; it takes an hour mostly because all relevant background is presented. A key ingredient that takes the most time is the computation of the differential of the exponential function, which is not obvious in the non-commutative case, and is worth knowing anyway.

The proof uses the "Euler Trick", which again, is useful elsewhere and worth knowing anyway.

Answer by Candela for "Modern" proof for the Baker-Campbell-Hausdorff formula

2012-06-10T15:29:25Z

Perhaps you will find what you are looking for in the book "Topics in noncommutative algebra : the theorem of Campbell, Baker, Hausdorff and Dynkin" by A. Bonfiglioli and R. Fulci.

Answer by Fran Burstall for "Modern" proof for the Baker-Campbell-Hausdorff formula

2012-06-10T16:03:37Z

You will find two proofs in chapter 1 of Shlomo Sternberg's excellent notes on Lie algebras. The first is purely differential geometric in the style you seem to be looking for.

Answer by Jose Arnaldo Dris for "Modern" proof for the Baker-Campbell-Hausdorff formula

2012-06-10T18:11:48Z

You might be interested in some of Terence Tao's posts regarding the Baker-Campbell-Hausdorff formula. Here are the hyperlinks:

254A, Notes 1: Lie groups, Lie algebras, and the Baker-Campbell-Hausdorff formula

Associativity of the Baker-Campbell-Hausdorff formula

The C^{1,1} Baker-Campbell-Hausdorff formula

Notes on local groups

Answer by Igor Rivin for "Modern" proof for the Baker-Campbell-Hausdorff formula

2012-06-10T18:38:47Z

There is a nice prof by Dynkin in @article {MR0021940, AUTHOR = {Dynkin, E. B.}, TITLE = {Calculation of the coefficients in the {C}ampbell-{H}ausdorff formula}, JOURNAL = {Doklady Akad. Nauk SSSR (N.S.)}, VOLUME = {57}, YEAR = {1947}, PAGES = {323--326}, MRCLASS = {20.0X}, MRNUMBER = {0021940 (9,132d)}, MRREVIEWER = {L. Zippin}, }