[Math] Software for Computing Baker-Campbell-Hausdorff

dg.differential-geometrylie-groupsmathematical-softwaremp.mathematical-physics

Does anyone have a recommendation for software which can efficiently calculate the Baker-Campbell-Hausdorff series in classical Lie algebras?

Right now, I have a problem which boils down to understanding Baker-Campbell-Hausdorff with respect to a basis in su(2), and this seems like the kind of thing Sage or Mathematica should be able to handle. However, I haven't had to use computer algebra packages for Lie theory before, so I would love to be pointed in the right direction.

Many thanks,

Jesse

Best Answer

There is a quite comprehensive package for Lie algebras in Maple. It is developed by Ian Anderson (from Utah State not Jethro Tull).

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Differential Geometry – Does the Baker-Campbell-Hausdorff Formula Hold for Vector Fields on a Compact Manifold?

To answer your first question, the composition of two time-1 flows won't necessarily be another time-1 flow.

One way to see this is to note that when a time-1 flow $\phi_X$ has a periodic point $P$ (period > 1), then $P$ can't be hyperbolic since it lies on a closed orbit of the flow for $X$. (The eigenvector of $D\phi_X$ tangent to this orbit has corresponding eigenvalue 1.)

Now, take a flow on $S^2$ whose time-1 map rotates the sphere, switching the north and south poles. Take a second flow for which both poles are hyperbolic attracting fixed points. Composing the two time-1 maps gives you a new diffeomorphism with hyperbolic points of period 2.

[Math] “Modern” proof for the Baker-Campbell-Hausdorff formula

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.

Best Answer

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[Math] “Modern” proof for the Baker-Campbell-Hausdorff formula

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