# Noncommutative Baker-Campbell-Hausdorff formula: the problem

I come back to a problem alluded in a previous post, where the proof of the Baker-Campbell-Hausdorff formula from this post by Tao is characterized as “commutative”, because of the “radial homogeneity” condition in his Theorem 1 , which forces commutativity.

Now I am going to try to explain this, as well as what the problem of a “noncommutative” BCH formula would be.

Take a Lie group $G$ and identify a neighbourhood of its neutral element with a neighbourhood of the $0$ element of its Lie algebra. This is standard for Carnot groups (connected, simply connected nilpotent groups which admit a one parameter family of contracting automorphisms), where the exponential is bijective, so the identification is global. The advantage of this identification is that we get rid of log’s and exp’s in formulae.

For every $s > 0$ define a deformation of the group operation (which is denoted multiplicatively), by the formula

(1)                $s(x *_{s} y) = (sx) (sy)$

Then we have $x *_{s} y \rightarrow x+y$ as $s \rightarrow 0$.

Denote by $[x,y]$ the Lie bracket of the (Lie algebra of the) group $G$ with initial operation and likewise denote by $[x,y]_{s}$ the Lie bracket of the operation $*_{s}$.

The relation between these brackets is: $[x,y]_{s} = s [x,y]$.

From the Baker-Campbell-Hausdorff formula we get:

$-x + (x *_{s} y) - y = \frac{s}{2} [x,y] + o(s)$,

(for reasons which will be clear later, I am not using the commutativity of addition), therefore

(2)         $\frac{1}{s} ( -x + (x *_{s} y) - y ) \rightarrow \frac{1}{2} [x,y]$       as        $s \rightarrow 0$.

Remark that (2) looks like a valid definition of the Lie bracket which is not related to the group commutator. Moreover, it is a formula where we differentiate only once, so to say. In the usual derivation of the Lie bracket from the group commutator we have to differentiate twice!

Let us now pass to a slightly different context: suppose $G$ is a normed group with dilations (the norm is for simplicity, we can do without; in the case of “usual” Lie groups, taking a norm corresponds to taking a left invariant Riemannian distance on the group).

$G$ is a normed group with dilations if

• it is a normed group, that is there is a norm function defined on $G$ with values in $[0,+\infty)$, denoted by $\|x\|$, such that

$\|x\| = 0$ iff $x = e$ (the neutral element)

$\| x y \| \leq \|x\| + \|y\|$

$\| x^{-1} \| = \| x \|$

– “balls” $\left\{ x \mid \|x\| \leq r \right\}$ are compact in the topology induced by the distance $d(x,y) = \|x^{-1} y\|$,

• and a “multiplication by positive scalars” $(s,x) \in (0,\infty) \times G \mapsto sx \in G$ with the properties:

$s(px) = (sp)x$ , $1x = x$ and $sx \rightarrow e$ as $s \rightarrow 0$; also $s(x^{-1}) = (sx)^{-1}$,

– define $x *_{s} y$ as previously, by the formula (1) (only this time use the multiplication by positive scalars). Then

$x *_{s} y \rightarrow x \cdot y$      as      $s \rightarrow 0$

uniformly with respect to $x, y$ in an arbitrarry closed ball.

$\frac{1}{s} \| sx \| \rightarrow \|x \|_{0}$, uniformly with respect to $x$ in a closed ball, and moreover $\|x\|_{0} = 0$ implies $x = e$.

1. In truth, everything is defined in a neighbourhood of the neutral element, also $G$ has only to be a local group.

2. the operation $x \cdot y$ is a (local) group operation and the function $\|x\|_{0}$ is a norm for this operation, which is also “homogeneous”, in the sense

$\|sx\|_{0} = s \|x\|_{0}$.

Also we have the distributivity property $s(x \cdot y) = (sx) \cdot (sy)$, but generally the dot operation is not commutative.

3. A Lie group with a left invariant Riemannian distance $d$ and with the usual multiplication by scalars (after making the identification of a neighbourhood of the neutral element with a neighbourhood in the Lie algebra) is an example of a normed group with dilations, with the norm $\|x\| = d(e,x)$.

4. Any Carnot group can be endowed with a structure of a group with dilations, by defining the multiplication by positive scalars with the help of its intrinsic dilations. Indeed, take for example a Heisenberg group $G = \mathbb{R}^{3}$ with the operation

$(x_{1}, x_{2}, x_{3}) (y_{1}, y_{2}, y_{3}) = (x_{1} + y_{1}, x_{2} + y_{2}, x_{3} + y_{3} + \frac{1}{2} (x_{1}y_{2} - x_{2} y_{1}))$

multiplication by positive scalars

$s (x_{1},x_{2},x_{3}) = (sx_{1}, sx_{2}, s^{2}x_{3})$

and norm given by

$\| (x_{1}, x_{2}, x_{3}) \|^{2} = (x_{1})^{2} + (x_{2})^{2} + \mid x_{3} \mid$

Then we have $X \cdot Y = XY$, for any $X,Y \in G$ and $\| X\|_{0} = \|X\|$ for any $X \in G$.

Carnot groups are therefore just a noncommutative generalization of vector spaces, with the addition operation $+$ replaced by a noncommutative operation!

5. There are many groups with dilations which are not Carnot groups. For example endow any Lie group with a left invariant sub-riemannian structure and hop, this gives a norm group with dilations structure.

In such a group with dilations the “radial homogeneity” condition of Tao implies that the operation $x \cdot y$ is commutative! (see the references given in this previous post). Indeed, this radial homogeneity is equivalent with the following assertion: for any $s \in (0,1)$ and any $x, y \in G$

$x s( x^{-1} ) = (1-s)x$

which is called elsewhere “barycentric condition”. This condition is false in any noncommutative Carnot group! What it is true is the following: let, in a Carnot group, $x$ be any solution of the equation

$x s( x^{-1} ) = y$

for given $y \in G$ and $s \in (0,1)$. Then

$x = \sum_{k=0}^{\infty} (s^{k}) y$ ,

(so the solution is unique) where the sum is taken with respect to the group operation (noncommutative series).

Problem of the noncommutative BCH formula: In a normed group with dilations, express the group operation $xy$ as a noncommutative series, by using instead of “$+$” the operation “$\cdot$” and by using a definition of the “noncommutative Lie bracket” in the same spirit as (2), that is something related to the asymptotic behaviour of the “approximate bracket”

(3)         $[x,y]_{s} = (s^{-1}) ( x^{-1} \cdot (x *_{s} y) \cdot y^{-1} )$.

Notice that there is NO CHANCE to have a limit like the one in (2), so the problem seems hard also from this point of view.

Also notice that if $G$ is a Carnot group then

$[x,y]_{s} = e$ (that is like it is equal to $o$, remember)

which is normal, if we think about $G$ as being a kind of noncommutative vector space, even of $G$ may be not commutative.

So this noncommutative Lie bracket is not about commutators!