Arcadian Functor

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Marni D. Sheppeard

Thursday, February 07, 2008

Mutual Unbias II

While Carl has already launched an attack on the unsolved problem of finding a 7 MUB set for $d=6$, I have just begun mulling over the MUB literature.

This recent paper, by Monique Combescure, studies MUBs for prime dimensions, and shows that in this case Fourier matrices and circulants are sufficient to construct a full MUB set. There is evidence that a set of $d+1$ MUBs may be impossible if $d$ is not a prime power. For example, finite projective planes with $d+1$ points on a line are associated to MUBs in dimension $d$.

An example of a finite projective plane with 3 points on a line is the Fano plane, the non-zero points of $\mathbb{F}_{2}^{3}$, which we saw was related to a three qubit triality via the Hamming code. Thus MUBs provide a connection between Koide mass matrices and E8 type trialities, perhaps explaining the current enthusiasm for Lisi's work.

16 Comments:

Anonymous Tony Smith said...

Kea, this is related of course to the discussion going on over at physics forums to which you provided a link with the word "enthusiasm".

Thanks for the reference to the Combescure paper at 0710.5642 in which she said
"... It is known that the maximum number of MUB’s in any dimension d is d + 1, and that this number is attained if d = p^m, p being a prime number. ...".

Do I understand it correctly as follows:

For prime 7 there are up to 8 of the 7d MUBs which might correspond to the 8 different E8 lattices which are based on the 7 imaginary octonion basis elements,
and
for prime 23 there are up to 24 of the 23d MUBs which might correspond to 23 types of Leech lattice structures.

???

If so, those might be keys as to how to build explicit examples of MUBs related to basis-things for Algebraic Quantum Field Theoretical generalized hyperfinite II1 von Neumann factors,
which can be built from classical Lagrangian structures of Garrett-type E8 physics.

In other words, maybe this is a key of how to go from classical Lagrangians so useful in the Standard Model to nice Algebraic Quantum Field Theory.

If you are still going to have a meeting in NZ, that would be fun for a workshop.

Tony Smith

February 07, 2008 4:56 PM  
Anonymous Tony Smith said...

Sorry - when I said
"... might correspond to 23 types of Leech lattice structures ..."

I think I should have said
"... might correspond to 23+1 = 24 types of Leech lattice structures ..."

Tony Smith

February 07, 2008 5:00 PM  
Blogger Kea said...

You clearly understand perfectly well! Yes, it would be a great topic for a workshop: Moonshine in Godzone. Do you think anybody would fund it? I told my boss at the restaurant that I was through to round 2 of the fqxi grants, but he was relieved when I added that these grants won't be decided for some months to come.

February 07, 2008 5:38 PM  
Blogger Matti Pitkanen said...

I checked my little argument for the construction of MUBs in the general N-dimensional case as a tensor product of MUBs for prime power factors of N in the representation N-D Hilbert space as a tensor product of prime factors. To my best understanding the overlaps equal to 1/N trivially. Could anyone tell why this does not work?

Prime decomposition of Hilbert spaces seems to be a crucial element of quantum information processing and number theory would become part of Hilbert space theory. What additional axiomatic input this could mean? This might relate also to the fact that spins 0,1/2,1 with Hilbert spaces dimension 1,2,3 for massless/massive quanta are fundamental.

February 07, 2008 5:59 PM  
Blogger CarlBrannen said...

Matti, I wonder if this is what you did:

There are 3 MUBs for 2-d, and 4 MUBs for 3-d, so there should be 12 MUBs for 6-d? Bad arithmetic, taking the tensor product doesn't increase the number of MUBs.

Instead, if you take the 2+1 = 3 MUBs for 2-d and three of the 3+1 = 4 MUBs for 3-d, then you end up with three MUBs for 6-d which is exactly the best that anyone's ever found.

But this method is not as good as the primitive root method listed in the literature. For example, using the above method to find the 4-d case you'd only get 3 of the 5 MUBs because you only began with 3 for the two 2-d cases.

I tried working on the problem in density operator formalism. I wrote the 6x6 = 36-dimensional operator space for 6-Hilbert as a tensor on the 3x3 and 2x2 operator spaces for the 2 and 3-Hilbert. Another way of saying this is that I wrote the "Dirac bilinears" as products of the form A^a B^b C^c D^d where A and C are good square roots of unity taken from the 2-d case, and B and D are good cube roots of unity taken from the 3-d case, in each case computed as a tensor product with unity; and a,c are 0,1, and b,d are 0,1,2.

This method gives the right degrees of freedom for the operators in the 6-Hilbert. And it does allow one to write down basis states. However, they're not unbiased. I didn't check to see if it was possible to use this method to divide the degrees of freedom into commuting groups (which was the method I used on the Dirac algebra and also, I guess, on the qutrit cast).

Maybe there's a more complicated tensor product method here...

Also, I've got a new post talking about quantum numbers and symmetry breaking in MUBs

February 08, 2008 1:08 AM  
Blogger Matti Pitkanen said...

Carl,

thank you for demonstrating my mistake. I had replaced for N+1 with N from beginning.

I see now the reason why the argument fails. Moduli squared equal to 1/N for inner product of tensor product only if all tensor factors correspond to different MUB. One must select sub-set of all possible tensor product basis with this property for each factor. The basis associated with the smallest prime power in the decomposition must be different for all MUBs and only p_min^n_i basis are obtained in this manner.

February 08, 2008 3:03 AM  
Anonymous Tony Smith said...

Kea said that Carl was working on finding 7 MUB for the 6d MUB problem.
Based on the hope that sphere-packing lattices might correspond to MUB, here are some thoughts:

As Coxeter (see his papers "Integral Cayley Numbers" (reprinted in his book "The Beauty of Geometry Twelve Essays")and "Regular and Semi-Regular Polytopes III" (reprinted in the book "Kaleidoscopes" edited by Sherk et al) has said,
if you look at the Octonions with basis
{1, i,j,k, e, ie,je,ke}
and fix one of the 480 distinct Octonion Product multiplication rules,
you see that
there are 7 independent E8 lattices that form integral domains (and there is a natural 8th, but it does not close under the same multiplication rule as the 7).

If you look for 7d MUB you are looking for 7-dim E7 sublattices of those E8 lattices
then, according to Conway and Sloane in their book "Sphere Packings, Lattices, and Groups" at pages 124-125,
"...The vectors in E8 perpendicular to any minimal vector v in E8 form the lattice E7 ...[one of the]... possible coordinate systems .. leads to
E6 = { (x1, ..., x8) in E8 such that x6 = x7 = x8 } ...".

Although I don't see where Conway and Sloane explicitly say so, it seems to me that you would get 7 independent E7 lattices from the 7 independent E8 lattices,
and so if you include the 8th E8 lattice you might get 7+1 = 8 MUB for 7d MUB for prime number 7.

If you look for 6d MUB you are looking for 6-dim E6 sublattices of those E8 lattices then, according to Conway and Sloane in their book "Sphere Packings, Lattices, and Groups" at pages 125-127,
"... The vectors in E8 prerpendicular to any A2-sublattice V in E8 for the lattice E6 ...
[one of the]... possible coordinate systems .. leads to
E6 = { (x1, ..., x8) in E8 such that x6 = x7 = x8 } ...
E6 also has a simple construction as a three-dimensional complex lattice, a lattice over the Eisenstein integers ...".

Again, it seems to me that, from the 7 independent E8 lattices, you would get 7 independent E6 lattices,
and that might give you the 7 MUB of the 6d MUB problem.

The relation of E6 to the Eisenstein integers and 3x3 complex matrices leads me to think about going from 2x3 = 6 to 8x3 = 24 and the Leech lattice and 23d and 24d MUB.
Conway and Sloane say at pages 480 and 508
"... We investigate the points in 24-dimensional space that are at maximal distance from the Leech lattice, i.e. the "deep holes" in that lattice. ... there are 23 inequivalent types of deep hole, one for each of the 23 even unimodular 24-dimensional lattices found by Niemeier. ...
We .. present 23 constructions for the Leech lattice, one for each class of hole or Niemeier lattice. Two of these are the usual constructions of the Leech lattice from the Golay codes over F2 and F3. ...".

It may be that the 23 Niemeier lattices roughly correspond to the 7 independent E8 lattices, and that the 24th (Leech lattice) corresponds to the 8th E8 lattice described above,
and so if you include the 24th Leech lattice you might get 23+1 = 24 MUB for 23d MUB for prime number 23.


Tony Smith

PS - As Coxeter points out, the 7 independent integral domain E8 lattices correspond to the 7 imaginary octonion basis elements.

What does category theory say about a thing (such as an E8 lattice)
that corresponds to
a component element of the thing (such as the octonion imaginary basis element corresponding to that lattce) ?

How does category theory characterize naturally reflexive things such as Lisp-type computer languages ?

February 08, 2008 4:08 AM  
Anonymous Tony Smith said...

Sorry - where I said in my previous comment

If you look for 7d MUB you are looking for 7-dim E7 sublattices of those E8 lattices
then, according to Conway and Sloane in their book "Sphere Packings, Lattices, and Groups" at pages 124-125,
"...The vectors in E8 perpendicular to any minimal vector v in E8 form the lattice E7 ...[one of the]... possible coordinate systems .. leads to
E6 = { (x1, ..., x8) in E8 such that x6 = x7 = x8 } ...".

I should have said

If you look for 7d MUB you are looking for 7-dim E7 sublattices of those E8 lattices
then, according to Conway and Sloane in their book "Sphere Packings, Lattices, and Groups" at pages 124-125,
"...The vectors in E8 perpendicular to any minimal vector v in E8 form the lattice E7 ...[one of the]... possible coordinate systems .. leads to
E7 = { (x1, ..., x8) in E8 such that x7 = x8 } ...".


My apologies for the typo due to cut-and-paste-without-thought.

Tony Smith

February 08, 2008 5:41 AM  
Blogger CarlBrannen said...

Tony, does that mean that the solution has to be non associative? That would make a lot of sense to me. The prime powers have primitive elements and consequently are associative because one can write p^n p^m = p^(n+m), I think.

I guess it should be mentioned (but has not yet), that the J and M that I used to split the "Dirac bilinears" for 3-Hilbert correspond, respectively, to the "generation" operator and the "color" operator.

That is, the circulant matrices (which give the Koide formula when you correct the phase angle, a topic I owe a blog post upon) are eigenstates of J, while the other four bases are eigenstates of M or various combinations of J and M.

So the association of the 1-plet with the leptons, and the 3-plet with the quarks is very natural here.

By the way, Matti, your tensor observation is exactly the method used to give the best lower bound for the MUBs of composite dimension Hilbert spaces.

February 08, 2008 6:23 AM  
Anonymous Tony Smith said...

carlbrannen asks "... does that [ octonionic product structure of E8 lattices ] mean that the solution has to be non associative? ...".

Not necessarily.
The vertices of an E8 lattice close under octonionic multiplication, which is nonassociative.
If you look at the 240 vertices nearest the origin, they close under nonassociative octonionic multiplication (which is why they can be regarded as unit octonions),
but
they also represent the root vectors of the E8 Lie algebra, and (along with 8 Cartan subalgebra elements) also close under the Lie algebra bracket product, which is how Garrett's model sees them.

Further, using Garrett's preferred breakdown of the E8 Lie algebra as
E8 = 120-dim adjoint Spin(16) + 128-dim half-spinor of Spin (16)
and
seeing that adjoint Spin(16) is the Lie algebra of the bivectors of the Cl(16) Clifford algebra
and that the half-spinor of Spin(16) is half of the spinors of the Cl(16) Clifford algebra,

you can represent Garrett's model in terms of the Cl(16) Clifford algebra which uses the associative Clifford product.

If you use real Clifford 8-periodicity to see Cl(16) as the tensor product Cl(16) = Cl(8) (x) Cl(8)
you can represent Garrett's model in terms of tensor products of Cl(8),
and
if you take the completion of the union of all the tensor products of Cl(8)
i.e., take the union of all things
like Cl(8) (x) ...N times... (x) Cl(8)
and then complete that huge union,
you should (in my opinion) get a generalization of the hyperfinite II1 von Neumann factor algebra
which would then be an Algebraic Quantum Field Theory structure that describes Garrett's model as a quantized model,
for which the stuff we have been talking about might lead to construction of nice MUB structure.

Since that algebra uses the associative Clifford product, it would not have the problems that have been found in earlier attempts to formulate octonionic-type Hilbert space-type things larger than J(3,O).

In short, the nonassociative octonions underly the Algebraic Quantum Field Theory solution,
but do not impose nonassociativity on the parts of the structure where it is not helpful.

Tony Smith

February 08, 2008 10:04 AM  
Blogger CarlBrannen said...

Tony, that all is slightly over my head and I trust you guys to figure it out.

I wonder if one could take the MUB for CL(16) and do something cool with it. If I recall, he had 240 roots used. Therea are 17 MUBs for CL(16), and that gives 17 x 16 = 256+16 states. However, 16 of these are the diagonal states that one would associate with naked preons. If you could figure out how to get rid of another 16 states you'd have the same 240 number.

One of the echoes going on here is that in Garrett's stuff, particle interactions come from vector addition laws. In the MUB, one gets particle interactions between any pair of states from different bases. Hmmmm.

February 08, 2008 6:53 PM  
Anonymous Tony Smith said...

CarlBrannen refers to "... the MUB for CL(16) ...".

Since Cl(16) has 2^16 = 256x256 = 65,536 dimensions,
and since 65,536+1 = 65,537 is prime
(a Fermat prime, like 257 and 17 and 5 and 3)
so
should there be 65,537+1 = 65,538 MUB related to Cl(16) ?

Tony Smith

February 09, 2008 6:16 PM  
Blogger CarlBrannen said...

No, Tony, by Cl(16) I meant the Clifford algebra that is represented by 16x16 complex matrices. I guess that I mean instead C(6,0).

By the way, the method I used to write down MUBs for 2,3, and 4-d Hilbert space doesn't seem to generalize for 2^n-d Hilbert space. At least last night I managed to convince myself that I probably couldn't write down a set of generating elements to make a MUB for 2^4-d Hilbert space using C(6,0).

The argument that it is impossible comes from looking at all possible sets of 15 C(6,0) bilinears that commute in terms of how many bilinears they use from each (non scalar) blade. For each such commuting set, one gets an integer vector in Z^8 whose terms are non negative and add to 15. The 8 dimensions correspond to blades 1 (vectors) thru 8 (psuedoscalar).

For a division of the bilinears to exist, the sum of these integer vectors has to give the blade dimension count for C(6,0) which happens to be (8, 28, 56, 70, 56, 28, 8, 1) as in Pascal's triangle, but ignoring the scalar.

As an example, as far as I can see there is only one blade dimension vector that can hit the psuedoscalar (8-blade). It is (0, 4, 0, 6, 0, 4, 0, 1). Continuing this process, one ends up with a list of blade dimension vectors that cannot add so as to divide up the 255 degrees of freedom into 17 groups of 15.

February 09, 2008 8:52 PM  
Anonymous Tony Smith said...

carlbrannen said "... by Cl(16) I [Carl] meant the Clifford algebra that is represented by 16x16 complex matrices. I guess that I mean instead C(6,0) ...".

No.
A real Clifford algebra Cl(R;N) over an N-dim real vector space has dimension 2^N.
A complex Clifford algebra Cl(C;N) over an N-dim complex vector space has dimension 2 x 2^N.

If you use the signature convention that
(p,q) means p - signs and q + signs
then
real Cl(R;6,0) has 2^6 = 64 = 8x8 dimensions and is the 4x4 quaternion matrix algebra M(4,Q)
while
real Cl(R;0,6) also has 2^6 = 64 = 8x8 dimensions but, since signature makes a difference in the real case, is the 8x8 real matrix algebra M(8,R).

A real Clifford algebra that is the 16x16 real matrix algebra is Cl(R;0,8) which has 2^8 = 256 dimensions and the graded structure
1+8+28+56+70+56+28+8+1

You can complexify that to get a complex Clifford algebra
C x Cl(R;0,8) = Cl(C;8) = 16x16 complex matrices =
= M(16,C)
(since signature doesn't matter with complex numbers the complex Clifford algebra notation just uses 8 instead of (0,8)).

However, there are real Clifford algebras that are the 16x16 complex matrix algebra M(16,C)
such as for example Cl(R;2,7).

Tony Smith

February 10, 2008 5:16 AM  
Blogger phil said...

Hi, all. I found another way to explore possibilities for MUBs using Fourier transforms and phase factors. Hope this helps you in some way.

February 11, 2008 1:18 AM  
Blogger CarlBrannen said...

Actually Tony, it's C(8,0). Your formulas are for the dimensionality of the operator algebra, not the dimensionality of the vectors.

February 11, 2008 11:58 PM  

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