Dual of the extended beta move. Termination as implosion or blow-out

Thanks Stephen for this comment, it made me think about the dual of the extended beta move and about the elimination of the termination gate.

The extended beta move has been introduced here. It is not, for the moment, part of the tutorial for graphic lambda calculus, but it will be soon.

Meanwhile, let me explore further the duality suggested in the post “Lambda scale in graphic lambda calculus and diagram crossings“.

The duality, although unclear as a precise mathematical object, is described by these three correspondences:


The last correspondence should be understood like it appears in the left hand sides of the “dual” moves from the “Graphic beta move extended …” which I reproduce here:



OK, so what is the dual of the extended beta move, then?  First, let me remind you the extended beta move:


The dual should be this:


If we look at the graph from the right hand side, then we see we can apply a beta move, like this:


Therefore the “dual extended beta” move is equivalent with the curious looking move:


That’s kind of a pruning move, but it is not on the list. Should it be?

We reason like this: by using the extended beta move and the Reidemeister 2 move (for emergent algebras) we arrive at


We may repeat indefinitely this succesion of moves, or, alternatively, we may use the extended beta move with an arbitrary \mu  from the group \Gamma. The intuition behind this is that the gate \bar{\varepsilon} is a dilation gate, which has an input coming from the fan-out gate and the other connected to the output of the gate, therefore, by circulating along this loop, we apply an endless number of dilations of coefficient \varepsilon. At the limit (if such a concept makes sense at this level of generality), either the things blow to infinity (if \varepsilon > 1) or they implode to the value given by the fan-out gate(if \varepsilon < 1), or they circulate forever in a loop (if \varepsilon = 1) . In all cases the graph with only one input and no outputs, which we see in the previous figure, behaves exactly like the termination gate! Hence the mystery move (??) is simply a local pruning move.

Therefore, we may as well eliminate the termination gate and replace it in all the formalism by the gate from the previous figure.


By this replacement the “dual extended beta move” is true, as a consequence of the usual graphic beta move. Moreover, if we do this replacement, then we shall have pure trivalent graphs in the formalism because the ugly univalent termination gate is replaced by a trivalent graph!

What do you think?


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