I continue from Parallel transport in spaces with dilations, I. Recall that we have a set , which could be see as the complete directed graph . By a construction using binary decorated trees, with leaves in , we obtain first a set of finite trees , then we put an equivalence relation on this set, namely two finite trees and are close if is a finite tree. The class of finite points is formed by the equivalence classes of finite trees with respect to the closeness relation .

Notice that the equality relation is , 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 . 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” is associated a finite point . Immediate questions jump into the mind:

- (Q1) Is the function injective? Otherwise said, can you prove that if then is not a finite tree?
- (Q2) What is the cardinality of ? Say, if is finite is then infinite ?

Along with these questions, a third one is almost immediate. To any two finite trees and is associated the function defined by

.

The function is well defined: for any we have , by definition. Therefore , because .

Consider now the groupoid with the set of objects and the set of arrows generated by the arrows from to . The third question is:

- (Q3) What is the isotropy group of a finite point (in particular ) in this groupoid? Call this isotropy group and remark that because the groupoid is connected, it follows that the isotropy groupoid does not depend on the object (finite point), in particular is the same at any point (seen of course as ).

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).

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Have you considered the effect of dilations on a pair of finite graphs, A and B, that are linked together such that for at least one cycle in A there is an edge of cycle in B that penetrates it? It seems that this would prevent isotropy under most conditions.

These finite graphs are in fact trees, therefore they don’t have cycles (not even if you eliminate variable names by using fan-out gates; in this case you get “diamonds”, like in the construction which associates to any lambda term a graph). Isotropy, a good question though, for example the Heisenberg group has the (conformally) symplectic group as group of isotropy, in a sense (i.e. if you think “direction” as related to dilations). Is it isotropic? [Note: the isotropy group from this post means the group of arrows in the respective groupoid which have as source and target a fixed object. The isotropy group of the Heisenberg group is the group of linear (commuting with dilations) transformations which preserve the neutral element of the group.]

I agree. I am trying to apply your theory to networks that contain cycles… As far as I can tell, linking a pair of networks as I described would act to limit the range of the dilation of the networks to a lower bound such that the usual definition of isotropy might be weakened.

Thanks, I shall think about this, although for the moment I don’t see how to express this linking in graph theoretic terms, but surely is possible.