### M Theory Lesson 238

We can think of the braid group $B_{3}$ as the general matrix group over the field with one element, associated to sets as vector spaces. It is also the fundamental group of the complement of the trefoil knot. Recall that the trefoil knot corresponds to the Pauli quandle of operators $\sigma_X$, $\sigma_Y$ and $\sigma_Z$.

This quandle can be thought of as a group ring for a field with one element. Additively, there is only one choice for the coefficients of $\sigma_X$, $\sigma_Y$ and $\sigma_Z$, and so the formal sum $\sigma_X + \sigma_Y + \sigma_Z$ represents the three element set as the union of labelled one element sets. Multiplicatively, the cyclic quandle rules hold, and these are the only rules.

What does it mean to take the fundamental group (or groupoid) not of the trefoil, but of the Pauli quandle? What is the complement of the quandle in MUB space? A truncated braid group of type $B_3$ naturally arises for the $3 \times 3$ operators. Moreover, M theory is very interested in how the Pauli operators interact with this three dimensional case. Somehow M theory doesn't mind that $B_3$ is specialised to truncated knots when considering three objects. After all, the fundamental group is really about maps of a circle into a space, but a circle is what one obtains only after considering (at least) an infinite number of objects.

This quandle can be thought of as a group ring for a field with one element. Additively, there is only one choice for the coefficients of $\sigma_X$, $\sigma_Y$ and $\sigma_Z$, and so the formal sum $\sigma_X + \sigma_Y + \sigma_Z$ represents the three element set as the union of labelled one element sets. Multiplicatively, the cyclic quandle rules hold, and these are the only rules.

What does it mean to take the fundamental group (or groupoid) not of the trefoil, but of the Pauli quandle? What is the complement of the quandle in MUB space? A truncated braid group of type $B_3$ naturally arises for the $3 \times 3$ operators. Moreover, M theory is very interested in how the Pauli operators interact with this three dimensional case. Somehow M theory doesn't mind that $B_3$ is specialised to truncated knots when considering three objects. After all, the fundamental group is really about maps of a circle into a space, but a circle is what one obtains only after considering (at least) an infinite number of objects.

## 3 Comments:

This whole thing about braids makes me wonder if there is a better way of putting what I am doing to handle generations.

In the permutations on 3 elements, I want to raise each element of the permutation group to a complex multiple of a permutation matrix. But to get generations, I have to allow permutations that change phase. Of course it turns out that the phase changes will be cubed roots of unity.

This makes the solutions to the idempotency equation involve cubed roots of unity, and this triples the solutions, more or less, which is what gives generations. Now is there a more elegant way of saying this with braids?

Well, Carl, yes. And by making contact with braids, there are connections a lot more mathematics related to modern QFT. On the other hand, braid diagrams don't in themselves yield interesting numerical invariants. That's what we have to figure out.

To get a java applet to do the IJKRGB arithmetic requires (a) user interface, (b) arithmetic packages for IJKRGB type addition, multiplication, and subtraction, and (c) the code to hook these together. I've got (a) and (b) running. Should finish the other soon, but I'm going on vacation and will be travelling.

Keep thinking about braids, surely there is an easier way to do this.

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