### M Theory Lesson 154

The three pentagons on the upper half plane (ie. the front of the pair of pants) are marked by boundary pieces from the fundamental domain tesselation of the modular group. The compact pentagon has three edges on the real axis as shown, where the point $\omega$ is the vertex where the three pentagons meet. The real points -1 and 2 define the ends of the larger circles on the ribbon graph. Thus there exist modular transformations mapping points from one pentagon into the others. We would like to consider a triality for the pentagons, such as that given by $(ST)^{3} = 1$, which fixes $\omega$.

Recall also that the three strand braid group $B_3$, which covers the modular group, is the fundamental group for the complement of a trefoil knot in $\mathbb{R}^{3}$. Embedding the Riemann sphere in $\mathbb{R}^{3}$, we can draw the trefoil on the sphere and choose to place the three knot crossings near the points $0$, $1$ and $\infty$, which are the squares of the associahedron. The cell complex dual to this associahedron in $\mathbb{R}^{3}$ turned up in the description of the moduli for the 6-valent ribbon vertex of Mulase et al.

Aside: The critical line of the Riemann zeta function, which is self dual under the functional relation, is the line that divides the hexagon into two pentagons on the front of the pair of pants.

Recall also that the three strand braid group $B_3$, which covers the modular group, is the fundamental group for the complement of a trefoil knot in $\mathbb{R}^{3}$. Embedding the Riemann sphere in $\mathbb{R}^{3}$, we can draw the trefoil on the sphere and choose to place the three knot crossings near the points $0$, $1$ and $\infty$, which are the squares of the associahedron. The cell complex dual to this associahedron in $\mathbb{R}^{3}$ turned up in the description of the moduli for the 6-valent ribbon vertex of Mulase et al.

Aside: The critical line of the Riemann zeta function, which is self dual under the functional relation, is the line that divides the hexagon into two pentagons on the front of the pair of pants.

## 8 Comments:

Kea, you might find this Perimeter Institute lecture of interest, as it sort of intersects things that you and I are doing. The dimension 2^2 = 4 case is the Pauli algebra where the 2+1 = 3 "mutually unbiased bases" are precisely the snuark algebra with its 3 choices of measurement direction x, y and z.

The dimension 3^2 = 9 case is the 3x3 matrix case. The circulant matrices are one of the three choices of base, I believe you talk about a variation of circulant matrices, was it "anti circulant"? that would be another one of the three bases, and there is presumably a third basis set.

Thanks, Carl. I'll check it out.

Gee, Carl, that was enjoyable. I especially liked the bit where he says they are interested in an axiomatic formulation that doesn't necessarily rely on Hilbert spaces. And the story about Alice and the mean king was fun.

For others: Hilbert spaces of prime power dimension have nice properties that other Hilbert spaces don't seem to have, precisely because one can form a field with p^n elements. He discusses the discrete Fourier transform, and dimensions 2 and 4 (2 qubits) are discussed in some detail.

Carl, it's a pity he didn't look at d=3, or they might have noticed that the particle mass ratios arise in this way.

Kea, I haven't done the calculation, but intuitively, I want to say that the 3rd MUB for the 3x3 matrices (besides circulant and anti-circulant) are the diagonals.

That might give another way of writing the generations, one that is more consistent with the way that the guys are playing with E8.

Yes, it will be something simple like that, Carl. My guess would be 0-circulants, since diagonals show up in the first two bases. Maybe we should check out this guy's papers and work it out.

It also fits nicely with these polygons that I'm playing with: when you split three edges of the hexagon (say at 0,1,oo) you create a 9-sided polygon, where the hexagon corresponded to the circulant and anticirculant sets. This 9-gon bounds the three pentagons, or rather the upper half plane.

Kea! I think I've solved the MUB problem in 3 dimensions in a pure density matrix form which will extend to 6 dimensions! If true, that might allow me to write down a 4 basis MUB solution for 6 dimensions, one more than the record in the literature. And it may give hints on how to get a 7 basis solution! Details to come on my blog later tonight (I'm supposed to be working right now).

Oh, I can already sense that it won't extend to 6 dimensions, (cause exp(2i \pi/6) is going to cause problems) however, it will work beautifully in 3 dimensions with expi(2i \pi/3).

It might apply to situations where the three preons are of mixed type, i.e. the quarks. One ends up with 4 basis choices. The circulants, which apply to the leptons, the diagonals, which I would think would apply to the bare preons themselves, and the other two, which should apply to the quarks.

Maybe a better splitting is to have the diagonals be the charged leptons, the circulants be the neutrinos (which take the geometric phase exp(2i \pi/12) where the charged leptons are real), and the two mixed bases would be the up and down quarks.

I'll try to type it up tonight, but I don't know if I'll have connectivity to upload it.

Sounds good, Carl. Thanks.

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