# Ado’s theorem for groups with dilations?

Ado’s theorem  is equivalent with the following:

Theorem. Let $G$ be a local Lie group. Then there is a real, finite dimensional vector space $V$ and an injective, local group morphism from (a neighbourhood of the neutral element of) $G$ to $GL(V)$, the linear group of $V$.

Any proof I am aware of, (see this post for one proof and relevant links),  mixes the following ingredients:

–  the Lie bracket and the BCH formula,

– either reduction to the nilpotent case or (nonexclusive) use of differential equations,

– the universal enveloping algebra.

WARNING: further I shall not mention the “local” word, in the realm of spaces with dilations everything is local.

We may pass to the following larger frame of spaces with dilations, dilatation structures or emergent algebras:

– locally compact groups with dilations instead of Lie groups

– locally compact conical groups instead of vector spaces

– linearity in the sense of dilation structures instead of usual linearity.

Conjecture:  For any locally compact group with dilations $G$ there is a locally compact conical group $N$ and an injective morphism $\rho: G \rightarrow GL(N)$ such that for every $x \in N$ the map $g \in G \mapsto \rho(g)x$ is differentiable.

In this frame:

– we don’t have the corresponding Lie bracket and BCH formula, see the related problem of the noncommutative BCH formula,

– what nilpotent means is no longer clear (or needed?)

– we don’t have a clear tensor product, therefore we don’t have a correspondent of the universal enveloping algebra.

Nevertheless I think the conjecture is true and actually much easier to prove than Ado’s theorem, because of the weakening of the conclusion.

# Three problems and a disclaimer

In this post I want to summarize the list of problems I am currently thinking about. This is not a list of regular mathematical problems, see the disclaimer on style written at the end of the post.

Here is the list:

1. what is “computing with space“? There is something happening in the brain (of a human or of a fly) which is akin to a computation, but is not a logical computation: vision. I call this “computing with space”. In the head there are a bunch of neurons chirping one to another, that’s all. There is no euclidean geometry, there are no a priori coordinates (or other extensive properties), there are no problems to solve for them neurons, there is  no homunculus and no outer space, only a dynamical network of gates (neurons and their connections). I think that a part of an answer is the idea of emergent algebras (albeit there should be something more than this).  Mathematically, a closely related problem is this: Alice is exploring a unknown space and then sends to Bob enough information so that Bob could “simulate” the space in the lab. See this, or this, or this.

Application: give the smallest hint of a purely relational  model of vision  without using any a priori knowledge of the (euclidean or other) geometry of outer space or any  pre-defined charting of the visual system (don’t give names to neurons, don’t give them “tasks”, they are not engineers).

2. non-commutative Baker-Campbell-Hausdorff formula. From the solution of the Hilbert’s fifth problem we know that any locally compact topological group without small subgroups can be endowed with the structure of a “infinitesimally commutative” normed group with dilations. This is true because  one parameter sub-groups  and Gleason metrics are used to solve the problem.  The BCH formula solves then another problem: from the infinitesimal structure of a (Lie) group (that is the vector space structure of the tangent space at the identity and the maniflod structure of the Lie group) and from supplementary infinitesimal data (that is the Lie bracket), construct the group operation.

The problem of the non-commutative BCH is the following: suppose you are in a normed group with dilations. Then construct the group operation from the infinitesimal data (the conical group structure of the tangent space at identity and the dilation structure) and supplementary data (the halfbracket).

The classical BCH formula corresponds to the choice of the dilation structure coming from the manifold structure of the Lie group.

In the case of a Carnot group (or a conical group), the non-commutative BCH formula should be trivial (i.e. $x y = x \cdot y$, the equivalent of $xy = x+y$ in the case of a commutative Lie group, where by convention we neglect all “exp” and “log” in formulae).

3. give a notion of curvature which is meaningful for sub-riemannian spaces. I propose the pair curvdimension- curvature of a metric profile. There is a connection with problem 1: there is a link between the curvature of the metric profile and the “emergent Reidemeister 3 move” explained in section 6 of the computing with space paper. Indeed, at page 36 there is this figure. Yes, $R^{x}_{\epsilon \mu \lambda} (u,v) w$ is a curvature!

Disclaimer on style. I am not a problem solver, in the sense that I don’t usually like to find the solution of an already formulated problem. Instead, what I do like to do is to understand some phenomenon and prove something about it in the simplest way possible.  When thinking about a subject, I like to polish the partial understanding I have by renouncing to use any “impure” tools, that is any (mathematical) fact which is strange to the subject. I know that this is not the usual way of doing the job, but sometimes less is more.

# Baker-Campbell-Hausdorff polynomials and Menelaus theorem

This is a continuation of the previous post on the noncommutative BCH formula. For the “Menelaus theorem” part see this post.

Everything is related to “noncommutative techniques” for approximate groups, which hopefully will apply sometimes in the future to real combinatorial problems, like the Tao’ project presented here, and also to the problem of understanding curvature (in non-riemannian settings), see a hint here, and finally to the problem of higher order differential calculus in sub-riemannian geometry, see this for a comment on this blog.

Remark: as everything this days can be retrieved on the net, if you find in this blog something worthy to include in a published paper, then don’t be shy and mention this. I believe strongly in fair practices relating to this new age of scientific collaboration opened by the www, even if in the past too often ideas which I communicated freely were taken in published papers without attribution. Hey, I am happy to help! but unfortunately I have an ego too (not only an ergobrain, as any living creature).

For the moment we stay in a Lie group , with the convention to take the exponential equal to identity, i.e. to consider that the group operation can be written in terms of Lie brackets according to the BCH formula:

$x y = x + y + \frac{1}{2} [x,y] + \frac{1}{12}[x,[x,y]] - \frac{1}{12}[y,[y,x]]+...$

For any $\varepsilon \in (0,1]$ we define

$x \cdot_{\varepsilon} y = \varepsilon^{-1} ((\varepsilon x) (\varepsilon y))$

and we remark that $x \cdot_{\varepsilon} y \rightarrow x+y$ uniformly with respect to $x,y$ in a compact neighbourhood of the neutral element $e=0$. The BCH formula for the operation labeled with $\varepsilon$ is the following

$x \cdot_{\varepsilon} y = x + y + \frac{\varepsilon}{2} [x,y] + \frac{\varepsilon^{2}}{12}[x,[x,y]] - \frac{\varepsilon^{2}}{12}[y,[y,x]]+...$

$BCH^{0}_{\varepsilon} (x,y) = x \cdot_{\varepsilon} y$

and $BCH^{0}_{0}(x,y) = \lim_{\varepsilon \rightarrow 0} BCH^{0}_{\varepsilon}(x,y) = x + y$.

Define the “linearized dilation$\delta^{x}_{\varepsilon} y = x + \varepsilon (-x+y)$ (written like this on purpose, without using the commutativity of the “+” operation; due to limitations of my knowledge to use latex in this environment, I am shying away to put a bar over this dilation, to emphasize that it is different from the “group dilation”, equal to $x (\varepsilon(x^{-1}y))$).

Consider the family of $\beta > 0$ such that there is an uniform limit w.r.t. $x,y$ in compact set of the expression

$\delta_{\varepsilon^{-\beta}}^{BCH^{0}_{\varepsilon}(x,y)} BCH^{0}_{0}(x,y)$

and remark that this family has a maximum $\beta = 1$. Call this maximum $\alpha_{0}$ and define

$BCH^{1}_{\varepsilon}(x,y) = \delta_{\varepsilon^{-\alpha_{1}}}^{BCH^{0}_{\varepsilon}(x,y)} BCH^{0}_{0}(x,y)$

and $BCH^{1}_{0}(x,y) = \lim_{\varepsilon \rightarrow 0} BCH^{1}_{\varepsilon}(x,y)$.

Let us compute $BCH^{1}_{0}(x,y)$:

$BCH^{1}_{0}(x,y) = x + y + \frac{1}{2}[x,y]$

and also remark that

$BCH^{1}_{\varepsilon}(x,y) = x+y + \varepsilon^{-1} ( -(x+y) + (x \cdot_{\varepsilon} y))$.

We recognize in the right hand side an expression which is a relative of what I have called in the previous post an “approximate bracket”, relations (2) and (3). A better name for it is a halfbracket.

We may continue indefinitely this recipe. Namely for any natural number $i\geq 1$ we first define the maximal number $\alpha_{i}$ among all $\beta > 0$ with the property that the (uniform) limit exists

$\lim_{\varepsilon \rightarrow 0} \delta_{\varepsilon^{-\beta}}^{BCH^{i}_{\varepsilon}(x,y)} BCH^{i}_{0}(x,y)$

Generically we shall find $\alpha_{i} = 1$. We define then

$BCH^{i+1}_{\varepsilon}(x,y) = \delta_{\varepsilon^{-\alpha_{i}}}^{BCH^{i}_{\varepsilon}(x,y)} BCH^{i}_{0}(x,y)$

and $BCH^{i+1}_{0}(x,y) = \lim_{\varepsilon \rightarrow 0} BCH^{i+1}_{\varepsilon}(x,y)$.

It is time to use Menelaus theorem. Take a natural number $N > 0$. We may write (pretending we don’t know that all $\alpha_{i} = 1$, for $i = 0, ... N$):

$x \cdot_{\varepsilon} y = BCH^{0}_{\varepsilon}(x,y) = \delta^{BCH^{0}_{0}(x,y)}_{\varepsilon^{\alpha_{0}}} \delta^{BCH^{1}_{0}(x,y)}_{\varepsilon^{\alpha_{1}}} ... \delta^{BCH^{N}_{0}(x,y)}_{\varepsilon^{\alpha_{N}}} BCH^{N+1}_{\varepsilon}(x,y)$

Let us denote $\alpha_{0} + ... + \alpha_{N} = \gamma_{N}$ and introduce the BCH polynomial $PBCH^{N}(x,y)(\mu)$ (the variable of the polynomial is $\mu$), defined by: $PBCH^{N}(x,y)(\mu)$ is the unique element of the group with the property that for any other element $z$ (close enough to the neutral element) we have

$\delta^{BCH^{0}_{0}(x,y)}_{\mu^{\alpha_{0}}} \delta^{BCH^{1}_{0}(x,y)}_{\mu^{\alpha_{1}}} ... \delta^{BCH^{N}_{0}(x,y)}_{\mu^{\alpha_{N}}} z = \delta^{PBCH^{N}(x,y)(\mu)}_{\mu^{\gamma_{N}}} z$

Such an element exists and it is unique due to (Artin’ version of the) Menelaus theorem.

Remark that $PBCH^{N}(x,y)(\mu)$ is not a true polynomial in $\mu$, but it is a rational function of $\mu$ which is a polynomial up to terms of order $\mu^{\gamma_{N}}$. A straightforward computation shows that the BCH polynomial (up to terms of the mentioned order) is a truncation of the BCH formula up to terms containing $N-1$ brackets, when we take $\mu =1$.

It looks contorted, but written this way it works verbatim for normed groups with dilations! There are several things which are different in detail. These are:

1. the coefficients $\alpha_{i}$ are not equal to $1$, in general. Moreover, I can prove that the $\alpha_{i}$ exist (as a maximum of numbers $\beta$ such that …) for a sub-riemannian Lie group, that is for a Lie group endowed with a left-invariant dilation structure, by using the classical BCH formula, but I don’t think that one can prove the existence of these numbers for a general group with dilations! Remark that the numbers $\alpha_{i}$ are defined in a similar way as Hausdorff dimension is!

2. one has to define noncommutative polynomials, i.e. polynomials in the frame of Carnot groups (at least). This can be done, it has been sketched in a previous paper of mine, Tangent bundles to sub-riemannian groups, section 6.

UPDATE: (30.10.2011) See the post of Tao

Associativity of the Baker-Campbell-Hausdorff formula

where a (trained) eye may see the appearance of several ingredients, in the particular commutative case, of the mechanism of definition of the BCH formula.

The associativity is rephrased, in a well known way,  in proposition 2 as a commutativity of say left and  right actions. From there signs of commutativity (unconsciously assumed) appear:  the obvious first are the “radial  homogeneity  identities”, but already at this stage a lot of familiar  machinery is put in place and the following is more and more heavy of  the same. I can only wonder:  is this  all necessary? My guess is: not. Because for starters, as explained here and in previous posts, Lie algebras are of a commutative blend, like the BCH formula. And (local, well known from the beginning) groups are not.

# 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!

# Planar rooted trees and Baker-Campbell-Hausdorff formula

Today on arXiv was posted the paper

with the abstract

We introduce the combinatorial notion of posetted trees and we use it in order to write an explicit expression of the Baker-Campbell-Hausdorff formula.

The paper may be relevant (check also the bibliography!) for the subject of writing “finitary“, “noncommutative” BCH formulae, from self-similarity arguments using dilations.