# What group is this? (Parallel transport in spaces with dilations, II)

I continue from Parallel transport in spaces with dilations, I.   Recall that we have a set $X$ , which could be see as the complete directed graph $X^{2}$. By a construction using binary decorated trees, with leaves in $X$, we obtain first a set of finite trees $FinT(X)$, then we put an equivalence relation $\sim$ on this set, namely two finite trees $A$ and $B$ are close $A \sim B$ if $A \bullet B$ is a finite tree. The class of finite points $PoinT(X)$ is formed by the equivalence classes $[A]$ of finite trees $A$  with respect to the closeness relation $\sim$.

Notice that the equality relation is $\leftrightarrow$ , in this world.  This equality relation is generated by the “oriented Reidemeister moves”  R1a and R2a, which appear also as moves in graphic lambda calculus. (By the way, this construction can be made in graphic lambda calculus, which has the moves R1a and R2a. In this way we obtain a higher level of abstraction, because in the process we eliminate the set $X$. Graphic lambda calculus does not need variables. More about this at a future time.) If you are not comfortable with this equality relation than you can just factorize with it and replace it by equality.

It is clear that to any “point” $x \in X$ is associated a finite point $[x] \in PoinT(X)$. Immediate questions jump into the mind:

• (Q1)  Is the function $x \in X \mapsto [x] \in PoinT(X)$ injective? Otherwise said, can you prove that if $x \not = y$ then $x \bullet y$ is not a finite tree?
• (Q2)  What is the cardinality of $PoinT(X)$? Say, if $X$ is finite is then  $PoinT(X)$ infinite ?

Along with these questions, a third one is almost immediate. To any two finite trees $A$ and $B$ is associated the function $[AB] : [B] \rightarrow [A]$  defined by

$[AB](C) = A \circ (B \bullet C)$ .

The function is well defined: for any $C \in [B]$ we have $B \bullet C \in FinT(X)$, by definition. Therefore $[AB](C) \in [A]$, because $A \bullet \left( [AB](C) \right) \leftrightarrow B \bullet C$ .

Consider now the groupoid $ParaT(X)$ with the set of objects $PoinT(X)$ and the set of arrows generated by the arrows $[AB]$ from $[B]$ to $[A]$.  The third question is:

• (Q3)  What is the isotropy group of a finite point $[A]$   (in particular $[x]$ ) in this groupoid? Call this isotropy group $IsoT(X)$ and remark that because the groupoid $ParaT(X)$ is connected, it follows that the isotropy groupoid does not depend on the object (finite point), in particular is the same at any point $x \in X$ (seen of course as $[x] \in PoinT(X)$ ).

In a future post I shall explain the answers to these questions, which I think they are the following:

• Q1:  yes.
• Q2: infinite.
• Q3: a kind of free nilpotent group.

But feel free to contradict me, or to propose solutions. Of course, I shall cite any valuable contribution, even if it appears in a blog  (via +Graham Steel).

# Parallel transport in spaces with dilations, I

I intended to call this series of posts “What group is this?”, but I switched to this more precise, albeit more bland name. In this first post of the series I take again, in more generality, the construction explained in the post   Towards geometric Plünnecke graphs.

The construction starts in the same way, almost. After I give this first part of the construction, an interpretation in term sof groupoids is provided.  We consider only the moves  R1a and R2a, like in the post “A roadmap to computing with space“:

(The names “R1a”, “R2a” come from the names of oriented Reidemeister moves, see arXiv:0908.3127  by M. Polyak.)
Definition 1. The moves R1a, R2a  act on the set of binary trees  $T(X)$ with nodes decorated with two colours (black and white) and leaves decorated with elements of a set of “variable names”  $X$ which has at least two elements.  I shall denote by  $A, B, C$ … such trees and by  $x, y, z, u, v, w$ … elements of  $X$.

The edges of the trees are oriented upward. We admit  $X$ to be a subset of  $T(X)$, thinking about  $x \in X$ as an edge pointing upwards which is also a leaf decorated with $x$.

The moves are local, i.e. they can be used for any portion of a tree from  $T(X)$ which looks like one of the patterns from the moves, with the understanding that the rest of the respective tree is left unchanged.

We denote by $A \leftrightarrow B$ the fact that the $A$ can be transformed into $B$ by a finite sequence of moves.
_________________
Definition 2. The class of finite trees   $FinT(X) \subset T(X)$ is the smallest subset of  $T(X)$ with the  properties:

•   $X \subset FinT(X)$,
• if $A, B \in FinT(X)$ then  $A \circ B \in FinT(X)$  , where $A \circ B$ is the tree

• if $A, B, C \in FinT(X)$ then  $Sum(A,B,C) \in FinT(X)$ and $Dif(A,B,C) \in FinT(X)$, where  $Sum(A,B,C)$ is the tree

and $Dif(A,B,C)$ is the tree

• if $A \in FinT(X)$ and we can pass from  $A$ to  $B$ (i.e. $A \leftrightarrow B$ )  by one of the moves then  $B \in FinT(X)$.

_________________

Definition 3. Two graphs   $A, B \in FinT(X)$  are close, denoted by  $A \sim B$, if there is   $C \in FinT(X)$ such that  $B$ can be moved into   $A \circ C$.

_________________

Notice that $A \leftrightarrow B$ then $A \sim B$.
Proposition 1. The closeness relation is an equivalence.

Proof.  I start with the remark that $A \sim B$ if and only if $A \bullet B \in FinT(X)$, where $A \bullet B$ is the tree

Indeed, $A \sim B$ if there is $C \in FinT(X)$ such that $B \leftrightarrow A \circ C$. Then

which proves that $A \bullet B \in FinT(X)$.  Then  $A \sim A$ for any $A \in FinT(X)$, because $A \leftrightarrow A \bullet A$, therefore $A \bullet A \in FinT(X)$. Suppose now that  $A \sim B$. Then $A \bullet B \in FinT(X)$. Notice that $B \bullet A \leftrightarrow Dif(A, A \bullet B, A)$, by the following sequence of moves:

But $Dif(A, A \bullet B, A) \in FinT(X)$, from the hypothesis. Therefore $B \bullet A \in FinT(X)$, which is equivalent with $B \sim A$.

Finally, suppose that $A \sim B$, $B \sim C$. Then $B \sim A$ by the previous reasoning. Then there are $A', C' \in FinT(X)$ such that $A \leftrightarrow B \circ A'$ and $C \leftrightarrow B \circ C'$. It follows that $A \bullet C \leftrightarrow Dif(B, A', C')$, therefore $A \bullet C \in FinT(X)$, which proves that $A \sim C$.

_________________
Definition 4. The class of finite points  of $T(X)$ is  $PoinT(X)$  is the set of equivalence classes w.r.t  $\sim$.

_________________

Same construction, with groupoids.  We may see $\leftrightarrow$ as being an equivalence relation. Let  $T_{0}(X)$  be the set of equivalence classes w.r.t $\leftrightarrow$. We can define on  $T_{0}(X)$ the operations $(A,B) \mapsto A \circ B$ and $(A,B) \mapsto A \bullet B$  (because the moves R1a, R2a are local). Then  $(T_{0}(X), \circ, \bullet)$ is the free left idempotent right quasigroup generated by the set $X$.

Idempotent right quasigroups are the focus of the article arXiv:0907.1520, where emergent algebras are introduced as deformations of such objects. An idempotent right quasigroup $(M, \circ, \bullet)$ is a non-empty set endowed with two operations, such that

•   (idempotence) $x \circ x = x \bullet x = x$ for any $x \in M$,
•   (right quasigroup) $x \bullet (x \circ y) = x \circ (x \bullet y) = y$ for any $x, y \in M$

Let $T_{0}(X)^{2}$ be the trivial (pair) groupoid over $T_{0}(X)$. This is the groupoid with objects which are elements of  $T_{0}(X)$ and arrows of the form  $(A,B) \in T_{0}(X) \times T_{0}(X)$. Equivalently, we see $T_{0}(X)^{2}$ to be the set of it’s arrows, we identify objects with their identity arrows (in this case we identify $A \in Ob T_{0}(X)^{2}$ with it’s identity arrow $(A,A) \in T_{0}(X)^{2}$). Seen like this, the trivial groupoid  $T_{0}(X)^{2}$ is just the set  $T_{0}(X) \times T_{0}(X)$, with the partially defined operation (composition of arrows)

$(A,B) (B,C) = (A,C)$

and with the unary inverse operation

$(A,B)^{-1} = (B,A)$ .

Remark that the function  $F: T_{0}(X)^{2} \rightarrow T_{0}(X)^{2}$  defined by  $F(B,A) = (A \circ B, A)$  is a bijection of the set of arrows and moreover

•   it preserves the objects $F(A,A) = (A,A)$,
• the inverse has the expression  $F^{-1}(B,A) = (A \bullet B, A)$.

Define the groupoid  $F \sharp T_{0}(X)^{2}$ by declaring $F$ to be an isomorphism of groupoids. This means  $F \sharp T_{0}(X)^{2}$ to be  the set of arrows  $T_{0}(X)\times T_{0}(X)$, with the partially defined composition of arrows given by

$(B,A) * (D,C) = F^{-1} \left( F(B,A) F(D,C)) \right)$
for any pair of arrows $(B,A), (D,C)$ such that $F(B,A)$ can be composed in  $T_{0}(X)^{2}$ with $F(D,C)$, and unary inverse operation given by

$(B,A)^{-1,*} = F^{-1} \left( \left( F(B,A) \right)^{-1} \right)$  .

The groupoid  $F \sharp T_{0}(X)^{2}$ has then the composition operation

$(B, C \circ D) * (D,C) = (Sum(C,D,B), C)$ ,

the unary inverse operation

$(B,A)^{-1,*} = (Dif(A,B,A), A \circ B)$

and the set of objects $Ob(F \sharp T_{0}(X)^{2}) = T_{0}(X)$ .

Consider the set  $X^{2} = X \times X$, seen as a subset of arrows of the groupoid   $F \sharp T_{0}(X)^{2}$ .

The class of finite trees $FinT(X)$ appears in the following way. First define  $Fin_{0}T(X)$ to be the set of equivalence classes w.r.t  $\leftrightarrow$ of elements in $FinT(X)$.

Remark that $\left( Fin_{0} T(X)\right)^{2}$ is a sub-groupoid of $F \sharp T_{0}(X)^{2}$, which moreover it contains $X^{2}$ and is closed w.r.t. the application of $F$, seen this time as a function (which is not a morphism) from $F \sharp T_{0}(X)^{2}$ to itself. In fact $Fin_{0} T(X)$ is the smallest subset of $T_{0}(X)$ with this property. Let’s give to the groupoid $\left( Fin_{0} T(X)\right)^{2}$ the name   $\langle X^{2} \rangle$, seen as a sub-groupoid of  $F \sharp T_{0}(X)^{2}$ .

Moreover  $F\left( \langle X^{2} \rangle \right)$ is a sub-groupoid of the trivial groupoid  $T_{0}(X)^{2}$, with set of objects  $Fin_{0}T(X)$. But sub-groupoids of the trivial groupoid are the same thing as equivalence relations. In this particular case $(A,B) \in F\left( \langle X^{2} \rangle \right)$ if and only if  $A, B \in Fin_{0}T(X)$ and $A \sim B$.

Next time you’ll see some groups (which are associated to parallel transport in dilation structures) which are in some sense universal, but I don’t know (yet) what structure they have. “What group is this?” I shall ask next time.

________

Do you remark at which stage of this construction the map becomes the territory, thus creating points out of abstract nonsense?

To get a sense of this, replace the set of arrows $X^{2}$ with a graph with nodes in $X$.

# Geometric Ruzsa inequality on groupoids and deformations

This is a continuation of  Geometric Ruzsa triangle inequalities and metric spaces with dilations .  Proposition 1 from that post may be applied to groupoids. Let’s see what we get.

Definition 1. A groupoid is a set $G$, whose elements are called arrows, together with a partially defined composition operation

$(g,h) \in G^{(2)} \subset G \times G \mapsto gh \in G$

and a unary “inverse” operation:

$g \in G \mapsto g^{-1} \in G$

which satisfy the following:

•  (associativity of arrow composition) if $(a,b) \in G^{(2)}$ and $(b,c) \in G^{(2)}$  then $(a, bc) \in G^{(2)}$ and  $(ab, c) \in G^{(2)}$ and moreover  we have $a(bc) = (ab)c$,
•  (inverses and objects)   $(a,a^{-1}) \in G^{(2)}$ and $(a^{-1}, a) \in G^{(2)}$  ; for any $a \in G$ we define the origin of the arrow $a$ to be  $\alpha(a) = a^{-1} a$ and  the target of $a$ to be  $\omega(a) = a a^{-1}$;  origins and targets of arrows form the set of objects of the groupoid $Ob(G)$,
• (inverses again) if $(a,b) \in G^{(2)}$ then $a b b^{-1} = a$  $a^{-1} a b = b$.

____________________

The definition is a bit unnecessary restrictive in the sense that I take groupoids to have sets of arrows and sets of objects. Of course there exist larger groupoids, but for the purpose of this post we don’t need them.

The most familiar examples of groupoids are:

• the trivial groupoid associated to a non-empty set $X$ is $G = X \times X$, with composition $(x,y) (y,z) = (x,z)$ and inverse $(x,y)^{-1} = (y,x)$. It is straightforward to notice that $\alpha(x,y) = (y,y)$ and $\omega(x,y) = (x,x)$, which is a way to say that the set of objects can be identified with $X$ and the origin of the arrow $(x,y)$ is $y$ and the target of $(x,y)$ is $x$.
• any group $G$ is a groupoid,  with the arrow operation being the group multiplication and the inverse being the group inverse. Let $e$ be the neutral element of the group $G$. Then for any “arrow$$g \in G$ we have $\alpha(g) = \omega(g) = e$, therefore this groupoid has only one object, $e$. The converse is true, namely groupoids with only one object are groups. • take a group $G$ which acts at left on the set $X$ , with the action $(g,x) \in G \times X \mapsto gx \in X$ such that $g(hx) = (gh)x$ and $ex = x$. Then $G \times X$ is a groupoid with operation $(h, gx) (g,x) = (hg, x)$ and inverse $(g,x)^{-1} = (g^{-1}, gx)$. We have $\alpha(g,x) = (e,x)$, which can be identified with $x \in X$, and $\omega(g,x) = (e,gx)$, which can be identified with $gx \in X$. This groupoid has therefore $X$ as the set of objects. For the relations between groupoids and dilation structures see arXiv:1107.2823 . The case of the trivial groupoid, which will be relevant soon, has been discussed in the post The origin of emergent algebras (part III). ____________________ The following operation is well defined for any pair of arrows $(g,h) \in G$ with $\alpha(g) = \alpha(h)$ : $\Delta(g,h) = h g^{-1}$ Let $A, B, C \subset G$ be three subsets of a groupoid $G$ with the property that there exists an object $e \in Ob(G)$ such that for any arrow $g \in A \cup B \cup C$ we have $\alpha(g) = e$. We can define the sets $\Delta(C,A)$$\Delta(B,C)$ and $\Delta(B,A)$ . Let us define now the hard functions $f: \Delta(C,A) \rightarrow C$ and $g: \Delta(C,A) \rightarrow A$ with the property: for any $z \in \Delta(C,A)$ we have (1) $\Delta(f(z), g(z)) = z$ (The name “hard functions” comes from the fact that $\Delta$ should be seen as an easy operation, while the decomposition (1) of an arrow into a “product” of another two arrows should be seen as hard.) The following is a corollary of Proposition 1 from the post Geometric Ruzsa triangle inequalities and metric spaces with dilations: Corollary 1. The function $i: \Delta(C,A) \times B \rightarrow \Delta(B,C) \times \Delta(B,A)$ defined by $i(z,b) = (f(z) b^{-1} , g(z) b^{-1})$ is injective. In particular, if the sets $A, B, C$ are finite then $\mid \Delta(C,A) \mid \mid B \mid \leq \mid \Delta(B,C) \mid \mid \Delta(B,A) \mid$ . ____________________ Proof. With the hypothesis that all arrows from the three sets have the same origin, we notice that $\Delta$ satisfies the conditions 1, 2 from Proposition 1, that is 1. $\Delta( \Delta(b,c), \Delta(b,a)) = \Delta(c,a)$ 2. the function $b \mapsto \Delta(b,a)$ is injective. As a consequence, the proof of Proposition 1 may be applied verbatim. For the convenience of the readers, I rewrite the proof as a recipe about how to recover $(z, b)$ from $i(z,b)$. The following figure is useful. We have $f(z) b^{-1}$ and $g(z) b^{-1}$ and we want to recover $z$ and $b$. We use (1) and property 1 of $\Delta$ in order to recover $z$. With $z$ comes $f(z)$. From $f(z)$ and $f(z) b^{-1}$ we recover $b$, via the property 2 of the operation $\Delta$. That’s it. ____________________ There are now some interesting things to mention. Fact 1. The proof of Proposition 2 from the Geometric Ruzsa post is related to this. Indeed, in order to properly understand what is happening, please read again The origin of emergent algebras (part III) . There you’ll see that a metric space with dilations can be seen as a family of defirmations of the trivial groupoid. In the following I took one of the figures from the “origin III” post and modified it a bit. Under the deformation of arrows given by $\delta_{\varepsilon}(y,x) = (\delta^{x}_{\varepsilon} y , x)$ the operation $\Delta((z,e)(y,e))$ becomes the red arrow $(\Delta^{e}_{\varepsilon}(z,y), \delta^{e}_{\varepsilon} z)$ The operation acting on points (not arrows of the trivial groupoid) which appears in Proposition 2 is $\Delta^{e}_{\varepsilon}(z,y)$, but Proposition 2 does not come straightforward from Corollary 1 from this post. That is because in Proposition 2 we use only targets of arrows, so the information at our disposal is less than the one from Corrolary 1. This is supplemented by the separation hypothesis of Proposition 2. This works like this. If we deform the operation $\Delta$ on the trivial groupoid by using dilations, then we mess the first image of this post, because the deformation keeps the origins of arrows but it does not keep the targets. So we could apply the Corollary 1 proof directly to the deformed groupoid, but the information available to us consists only in targets of the relevant arrow and not the origins. That is why we use the separation hypotheses in order to “move” all unknown arrow to others which have the same target, but origin now in $e$. The proof then proceeds as previously. In this way, we obtain a statement about algebraic operations (like additions, see Fact 2.) from the trivial groupoid operation. Fact 2. It is not mentioned in the “geometric Ruzsa” post, but the geometric Ruzsa inequality contains the classical inequality, as well as it’s extension to Carnot groups. Indeed, it is enough to apply it for particular dilation structures, like the one of a real vectorspace, or the one of a Carnot group. Fact 3. Let’s see what Corollary 1 says in the particular case of a trivial groupoid. In this case the operation $\Delta$ is trivial $\Delta((a,e), (c,e)) = (a,c)$ and the “hard functions$ are trivial as well

$f(a,c) = (c,e)$ and $g(a,c) =(a,e)$

The conclusion of the Corollary 1 is trivial as well, because $\mid \Delta(C,A) \mid = \mid C \mid \mid A \mid$ (and so on …) therefore the conclusion is

$\mid C \mid \mid A \mid \mid B \mid \leq \mid B \mid^{2} \mid A \mid \mid C \mid$

However, by the magic of deformations provided by dilations structures, from this uninteresting “trivial groupoid Ruzsa inequality” we get the more interesting original one!

# The origin of emergent algebras (part III)

I continue from the post “The origin of emergent algebras (part II)“.   Recall that in the first post on this subject  is mentioned a paragraph by Bellaiche, where he explains that the structure of the tangent bundle is “concealed in dilations”. This is now clear, I hope, but there is something left to do. Bellaiche also mentions the classical construction of the tangent bundle made by Connes, where the tangent bundle of the space $X$  appears as a completion of the trivial pair groupoid $X \times X$.  What I want to explain now is how the dilations interact with the trivial groupoid to give the tangent bundle, in a sort of generalization of Connes construction. See arXiv:1107.2823 for all details.

_________________________

1. The trivial pair groupoid over a set $X$ is the set $X \times X$ with  the partially defined operation:

$(x,u) (u,v) = (x,v)$

This partially defined operation is the composition of arrows in the groupoid which has $X$ as the set of objects and $X \times X$ as the set of arrows.  The source of the arrow $(x,y)$ is $y = \alpha(x,y)$, the target of that arrow is $x = \omega(x,y)$.  (This may seem strange, but we have to define the source and target like this in order to  see $(x,u) (u,v) = (x,v)$ as composition of arrows in a groupoid).

The inverse of the arrow $(x,y)$ is $(y,x) = (x,y)^{-1}$.

The addition of arrows (i.e. composition) is

$add [(x,u), (u,v)] = (x,v)$

and it is defined for pairs of arrows $((x,u), (u,v))$ such that $\alpha(x,u) = \omega(u,v)$.

It will be useful further to introduce the difference of two arrows:

$dif [(u,x), (v,x)] = (u,x) \, (v,x)^{-1} = (u,x) (x,v) = (u,v)$

The difference is a partially defined operation which  makes sense (is defined for) for pairs of arrows with the same source.

_________________________

2. Dilations from the groupoid point of view.  Let $\Gamma = (0,+\infty)$ be the multiplicative group of strictly positive reals.  For any $\varepsilon \in \Gamma$ we define the dilation of arrows of coefficient $\varepsilon$ to be:

$\delta_{\varepsilon} (x,y) = (\delta_{\varepsilon}^{y} x , y)$

where $\delta_{\varepsilon}^{y} x$ is the intrinsic dilation of coefficient $\varepsilon$, of the point $x$ with respect to the basepoint $y$.  Again, as in the case of the definition of source and target of an arrow, here the definition of the dilation of arrows is turned on its head with respect to the writing convention from left to right.

For simplicity we don’t care about the domain of definition of the intrinsic dilations, or about the domain of definition of the dilation of arrows. Enough is to say that the dilation of arrows is defined for small enough arrows. What could that mean ?It is simple: think about the distance function $d: X \times X \rightarrow [0,+\infty)$ as if it is defined on the trivial groupoid. Then, the distance function appears as a kind of a norm on the set of arrows of the groupoid. The norm of the arrow $(x,y)$ is simply $d(x,y)$  (first time remarked by Lawvere). Then, a short enough arrow is one with a small norm.

Dilations of arrows have the following properties:

• they preserve the source of arrows:  $\alpha (\delta_{\varepsilon} (x,y)) = y = \alpha(x,y)$,
• upon the identification $x \equiv (x,x)$ of objects with their respective identity arrows, any dilation of arrows preserves the objects: $\delta_{\varepsilon}(x,x) = (x,x)$, as a consequence of the fact that $\delta_{\varepsilon}^{x}x = x$,
• finally, they form a one-parameter group: $\delta_{\varepsilon} ( \delta_{\mu} (x,y)) = \delta_{\varepsilon \mu}(x,y)$, because of the relation $\delta^{x}_{\varepsilon} \delta^{x}_{\mu} y = \delta^{x}_{\varepsilon \mu} y$.

_________________________

3. Deforming the trivial groupoid.  I want to use the trivial groupoid in order to explain the recipe for the approximate sum, given by the red dots in the figure

which appears also in the last post. With the notations from emergent algebras we see that $E = \Sigma^{x}_{\varepsilon} (y,z)$.  There is an equivalent construction for the approximate difference of two points, with respect to a basepoint:

Again with the notations from emergent algebras we see that $E' = \Delta^{x}_{\varepsilon}(z,y)$.

By using the trivial groupoid and its deformations by dilations of arrows, it  is easier to explain the origin of the approximate difference. That is because the difference of two arrows is defined on pairs of arrows which have the same source and because dilations of arrows preserve the sources of arrows.

Take two arrows with the same source, say $g = (y,x)$ and $h = (z,x)$. Then $g_{\varepsilon} = \delta_{\varepsilon}(y,x)$ and $h_{\varepsilon} = \delta_{\varepsilon}(z,x)$ have the same source. We can define then the difference of those deformed arrows:

$dif(g_{\varepsilon}, h_{\varepsilon}) = (\delta^{x}_{\varepsilon} y , \delta^{x}_{\varepsilon} z)$

We define now the deformed difference $dif_{\varepsilon}$, which is defined on pairs of arrows with the same source, by the relation

$\delta_{\varepsilon} dif_{\varepsilon} (g,h) = dif(g_{\varepsilon}, h_{\varepsilon})$

A short computation gives

$dif_{\varepsilon} [ (z,x) , (y,x) ] = \, ( \Delta^{x}_{\varepsilon}(z,y) , \delta_{\varepsilon}^{x} z)$

or, with the notations from the last figure

$dif_{\varepsilon} [ (z,x) , (y,x) ] = \, (E', D')$

Finally we can pass to the limit with $\varepsilon$, as in the tangent bundle construction by Connes.