Menelaus theorem by way of Reidemeister move 3

(Half of) Menelaus theorem is equivalent with a theorem by Emil Artin, from his excellent book Geometric Algebra, Interscience Publishers (1957), saying that the inverse semigroup generated by dilations (in an affine space) is composed by dilations and translations. More specifically, if \varepsilon, \mu > 0 are such that \varepsilon \mu < 1 then the composition of two dilations, one of coefficient \varepsilon and the other of coefficient \mu, is a dilation of coefficient \varepsilon \mu.

Artin contributed also to braid theory, so it may be a nice idea to give a proof of Artin interpretation of Menelaus theorem by using Reidemeister moves.

This post is related to previous ones, especially these three:

Noncommutative Baker-Campbell-Hausdorff formula

A difference which makes four differences, in two ways

Rigidity of algebraic structure: principle of common cause

which I shall use as references for what a normed group with dilations, conical group and associated decorations of tangle diagrams are.

Let’s start! I use a representation of dilations as decorations of an oriented tangle diagram.

For any \varepsilon > 0, dilations of coefficient \varepsilon and \varepsilon^{-1} provide two operations which give to the space (say X) the structure of an idempotent right quasigroup, which is equivalent to saying that decorations of the tangle diagrams by these rules are stable to the Reidemeister moves of type I and II.

A particular example of a space with dilations is a normed group with dilations, where the dilations are left-invariant.

If the decorations that we make are also stable with respect to the Reidemeister move 3, then it can be proved that definitely the space with dilations which I use has to be a conical group! What is a conical group? It is a non-commutative vector space, in particular it could be a real vector space or the Heisenberg group, or a Carnot group and so on. Read the previous posts about this.

Graphically, the Reidemeister move 3 is this sliding movement:

of CC' under the crossing AA'-BB' (remark also how the decorations of crossings \varepsilon and \mu switch places).
Further on I shall suppose that we use for decorations a conical group, with distance function denoted by d. Think about a real vector space with distance given by an euclidean norm, but don’t forget that in fact we don’t need to be so restrictive.

Take now two strings and twist them one around the other, an infinity of times, then pass a third string under the first two, then decorate everything as in the following figure

We can slide twice the red string (the one which is under) by using the Reidemeister move 3. The decorations do not change. If you want to see what is the expression of z' as a function of x,y then we easily write that

z' = \delta^{x}_{\varepsilon} \delta^{y}_{\mu} z = \delta^{x_{1}}_{\varepsilon} \delta^{y_{1}}_{\mu} z

where x_{1}. y_{1} are obtained from x,y according to the rules of decorations.

We may repeat n times the double slide movement and we get that

z' = \delta^{x_{n}}_{\varepsilon} \delta^{y_{n}}_{\mu} z

If we prove that the sequences x_{n}, y_{n} converge both to some point $w$, then by passing to the limit in the previous equality we would get that

z' = \delta^{w}_{\varepsilon} \delta^{w}_{\mu} z = \delta^{w}_{\varepsilon \mu} z

which is the conclusion of Artin’s result! Otherwise said, if we slide the red string under all the twisted pair of strings, then the outcome is the same as passing under only one string, decorated with $w$, with the crossing decorated with \varepsilon \mu.

The only thing left is to prove that indeed the sequences converge. But this is easy: we prove that the recurrence relation between x_{n+1}, y_{n+1} and x_{n},y_{n} is a contraction. See the figure:

Well, that is all.

UPDATE: I was in fact motivated to draw the figures and explain all this after seeing this very nice post of Tao, where an elementary proof of a famous result is given, by using “elementary” graphical means.

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