A Stringy Universe II

Now if stringy black holes are all about abstract entanglement, why do string theorists insist on talking about classical landscapes and awful complex analysis? After all, haven't they figured out yet that the Standard Model isn't completely right? On the other hand, then one remembers that the analysis hides a few devils. But it simply isn't acceptable to talk about gravity using such blatant prejudices about geometry, leaving one clear, horrifying option: to rewrite analysis.

Topos theorists have thought quite a bit about this problem. But as simple minded physicists we can always think about it more pragmatically, trying to build up the complex numbers piece by little piece, perhaps starting with small finite fields as truth values for MUB matrix entries. A full complex Hilbert space requires an uncountable number of truth values. In this case, a $1$-ordinal heirarchy of structures isn't going to be enough to reach $\mathbb{C}$, but we must just keep on going somehow (with surreal trees and $n$-ordinals and other goodies). So the complex numbers really ought to be taboo until one studies $\omega$-categories and multicategories, a lesson from both topology and topos theory.

M Theory Lesson 268

Tom Leinster's computation of the Euler characteristic of a (finite) category uses the $n \times n$ incidence matrix, $Z$, of the underlying graph. Let $u$ be the vector $(1,1, \cdots, 1)$ of length $n$. If there exists a vector $a = (a_1,a_2,a_3, \cdots, a_n)$ such that $Z a = u$, the Euler characteristic is given by

$\chi = \sum a_{i}$

Let us try to recover the cardinality of a set from this characteristic, by generalising the set to a connected groupoid on three objects. Our favourite 3 element set will do. Now the equation shows that each hom set in the groupoid must have cardinality $1/3$ for the even weighting to work. Fortunately, this is precisely the cardinality of a group. For it to work for any number of elements $n$, this group should be something like the cyclic group of order $n$. There are $n^{2}$ such hom sets in the groupoid.

Observe how the normalisation factor here has a real effect on the possibilities for hom sets. Without the $1/3$, the vector $a$ would have to be scaled, resulting in an Euler characteristic of only $1$, for any $n$. In other words, when each hom set is the trivial group the information about the cardinality of the set is lost. The simplest possible categorification of the set n therefore uses the cyclic groups.

A Stringy Universe

As the age of blogging rolls on, people seem to be more and more enthusiastic about the prospects of string theory. Today kneemo highlights a new paper by Kallosh. Mottle continues to entertain with links to F theory, for experts only, and of course Woit somehow manages to continuously whine.

Meanwhile, I have been looking again at certain stringy black holes in four dimensions whose entropy is measured by quantities that occur very naturally in the study of entanglement. One may well ask where the $d = 4$ comes from in the quantum information theory, because obviously the messy string theory derivation is quite unimportant compared to these more fundamental considerations.

Well, notice that the three spatial dimensions from $d=4$ matches the number of MUBs for a qubit. Similarly, $d=5$ black holes mysteriously require qutrit states, which have four basic MUBs. Moreover, if one correctly accounts for the fourth roots in the Pauli MUB case, one might guess the dimension should be 6, which happens to be the dimension of the compactified piece in type IIB theory. So instead of ridiculous numbers of dimensions in some arbitrary classical space, we just have dimensions of Hilbert spaces.

Later on I might discuss how one can rewrite this entanglement measure for three qubits in terms of symmetric $3 \times 3$ matrices with entries dependent on only 6 of the 8 amplitudes. Of course, Carl Brannen used similar operators in his paper on the hadron masses, but this paper was rejected due to the unfortunate circumstance that it had almost nothing to do with QCD.

M Theory Lesson 267

Usually when playing with a category of vector spaces over a field $K$, one either puts $K = \mathbb{C}$ or one doesn't worry about the field at all. But there is a situation when finite fields are the only appropriate choice.

Recall that the adjunction from Set to Vect contains a functor $G:$ Vect $\rightarrow$ Set which sends a vector space to its underlying set. Over $\mathbb{C}$ this is clearly an infinite set. So if we wanted to restrict to FinSet, the category of finite sets, there would be no way to maintain the adjunction. On the other hand, with a finite field, although a finite dimensional vector space may be large, it is still finite.

Mathematicians sometimes say that finite fields are a lot like the complex numbers anyway. Without zero, the multiplicative structure is just like the roots of unity in the complex plane. And MUB matrices for finite dimensional Hilbert spaces in Mersenne prime dimensions only require finite fields.

The Even Prime

The question in dimension 2 is, when defining the circulant MUB operator why do we need the complex number $i$? Why couldn't we just take the circulant $C = (1,-1)$? Well, $C$ has the obvious problem that both eigenvector columns are the same, making it useless as a basis. But let's step back and think a moment about a general circulant of the form $(1, \omega)$, where we don't know exactly what $\omega$ should be. Then conjugating on the diagonal $\sigma_{Z}$, which lists the usual spin eigenvalues, results in Now checking the independence of eigenvectors in the relation tells us that $\omega^{2} \neq 1$, namely that $-1$ is not an option. Then the eigenvalue equation tells us that $(\overline{\omega}^{2} - 1) \overline{\omega} = \omega - \overline{\omega}$, which is to say that $\overline{\omega}^{3} = \omega$. So we simply must have that $\omega = i$, if it's an ordinary number of some sort.

Oxford Life IV

Strangely enough, the frequent obstacles of construction work and hordes of tourists here reminds me a lot of Christchurch NZ. But in other ways, England is really very different. One is supposed to tip in restaurants (sacrebleu), an average bathroom is about the size of a bathtub, and people seem to have trouble understanding that in the southern summer there are places where it is 13 hours ahead of GMT.

Meanwhile, my passport has finally wandered off into the bowels of the Home Office together with my original PhD certificate, which is nicely soiled and crumpled after its three month holiday in Wellington. Unsurprisingly, I had to fill out a long form containing almost exactly the same information they have already been given several times previously, such as the fact that I harbour no terrorist intentions or am not in any other way a person of dubious character!

M Theory Lesson 266

Thinking once again about knotted and coloured string networks, it seems that trivalent vertices on three colours are a lot like a (Tutte) dual representation of a Pauli quandle for a trefoil, where each arc of the knot is a different colour.

These networks also allow crossings, such that the undercrossing preserves the colour. As a quandle rule, such colourings correspond to the action $b \circ a = a$. An example of a quandle that obeys this simple law is one generated by a single invertible operator $M$, such that $M^{n} = 1$. The quandle operation (as is usual for a group) is conjugation, and this acts trivially on $M^{k}$ because we have only powers of $M$ to play with. Without the trivalent vertices, this quandle generates $n$ separate link components, because the arcs at a crossing never mix.

Seminar Heaven III

Lurie decided to give four lectures instead of one, kicking off today with yet another introduction to the Baez-Dolan cobordism hypothesis. There are also seminars to attend here in the Comlab. Meanwhile Physics is running a series this week on how the LHC works, but unfortunately there's no way I'll make those, and it's not long now until this conference.

Unfortunately, computer science, physics and mathematics all lie next to each other on campus. I would be much fitter if they didn't.

Oxford Life III

Spring is in the air here, with daffodils lining many a walkway. I am slowly settling in, having enjoyed my first college lunch last week, during which I glimpsed the old main hall with large portraits of its founder, Elizabeth I, and other notable patrons. A number of friendly academics managed to demolish venison, veges, fruit, dessert and coffee (in the senior common room, as they say) and then walk back to the office, all in the space of a lunch hour! It is a little embarrassing discussing the sorry state of the economy in such surroundings.

M Theory Lesson 265

There is a nice series of papers by Godsil et al on combinatorics associated to MUBs. Consider the simple qubit Combescure matrix This is the only choice for the eigenvectors of the Pauli matrix $\sigma_{Y}$ that satisfies the following properties. The Schur multiplication of two matrices simply defines entries by the product of matching entries from the two components $A$ and $B$. That is $M_{ij} = A_{ij} B_{ij}$. Under this product, an inverse for $R_2$ is found relative to the democratic matrix (the Schur identity): An invertible matrix in this sense is type II if $M (M^{-1})^{T} = n I$, where $I$ is the ordinary identity matrix. This works for $R_{2}$, although only $R_{2}^{8} = I$.

A type II matrix is a spin model, in the sense of Jones, if all vectors of the form

$Me_{i} \circ M^{-1}e_{j}$

(for $e_{i}$ the standard basis vectors) are eigenvectors for $M$. One checks that this holds for $R_2$. Note that the Fourier (Hadamard) operator $F_{2}$, although a type II matrix, is not a spin model matrix.

Oxford Life II

Although it's not really my business, I decided to phone the Home Office, just to see if somebody would talk to me. After the inevitable time on hold, I finally spoke to a lady who told me that I should have emailed (not phoned) another division of the Borders Agency, because there was nothing on their system about visas that had been issued overseas. Whilst on hold, the voice message explained that there were no personal appointments available in March, except possibly, under certain circumstances, in Glasgow.

She then explained that issues with visas should be referred to the issuing authority, that is to say, Wellington. Since I am no longer in New Zealand, and my visa stamp expires in only 2 weeks, I expressed some doubt that the Wellington people would be able to help, especially given their achievement record to date, which resulted in the job being given to another country. I wonder, if I posted my passport to a random person at the Home Office in London, how long would it take them to figure out where it was?

Seminar Heaven II

Today Michael Hopkins gave an elementary introduction to TFTs from a topologist's perspective, before outlining a theorem for $\omega$ $n$-categories (ie. with groupoid like arrows above dimension $n$) related to the classification of TFTs. He stressed the importance of category theory for tackling this problem. Lurie will also probably speak about this subject next week. The interesting construction is a choice of subcategory chain

$C^{fd} \rightarrow C^{f} \rightarrow C$

where $C$ is any suitable symmetric monoidal category at the target of the (generalised) TFT functor. The category $C^f$ ($f$ for finite) is the collection of all arrows that have both left and right adjoints, and $C^{fd}$ (for fully dualisable) is the category where objects have duals in a suitable sense. In other words, the categories they study generalise categories such as FinVect (finite dimensional vector spaces) to the infinite dimensional path space realm that topologists love.

Oxford Life

In the land of Monty Python, the Home Office still hasn't seen the error of its ways, and is procrastinating about issuing me with another work permit so that I can apply to extend my visa to a date after the expiry of the work period. Apparently, this problem is not unusual here.

Meanwhile, the University has efficiently issued cards, keys, computing accounts and numerous other useful things to help me get some research done. So I have been wondering a little about how our phenomenological MUBs might relate to, or extend, certain dagger symmetric monoidal structures studied by Coecke et al.

Aside: Apparently the referees did not like Carl's paper because it uses quantum information theory and not QCD, and they know that quantum information theory could not possibly be used to derive such patterns between particle masses.

Keeping Up

It seems that the whole CDF collaboration has taken to writing posts for Tommaso Dorigo with a series of amazing pieces of news, including Higgs mass exclusions and a new hadron with mass 4144 MeV.

Mersenne MUBs

Consider the MUB matrices in prime dimensions $p$. If we demand that the entries form a finite field of $p + 1$ elements (that is, including $0$) it follows that $p$ must be a Mersenne prime.

It is not known if there are an infinite number of such primes or not, but Euler showed that any perfect number $n$ may be written in the form

$n = \frac{1}{2} M_i (M_i + 1)$

for some Mersenne prime $M_i$. In particular, the perfect number $6$ comes from the Mersenne prime $3$.

Seminar Heaven

Apparently, it is possible here to fill an entire work schedule with seminars in computer science, physics, mathematics and philosophy. One could also fill every evening with concerts, lectures and theatre. Happily, MUB physics is interesting enough to keep me entertained.

At the end of my first day, on Thursday, I went to a string theory seminar on gravitino dark matter, which only accelerated the onset of 4pm jetlag crash. On Friday Eugenia Cheng was down from Sheffield to speak about an inductive definition of weak $n$-categories using terminal coalgebras. This is very interesting since it involves both the category Set and the category Top of topological spaces; the latter in order to supply an operad for the construction. Everybody in the group went to the seminar lunch.

Later in the afternoon, there was a seminar entitled Example of a 2-category at the Mathematical Institute, but most non mathematicians were forewarned by the word Langlands in the abstract, and indeed one of the questions at the end was: er, so what does this have to do with categories?

The mornings are still very chilly, but the weather has been quite pleasant so far, and the 20 minute walk to work passes some spectacular historical buildings.

From Oxford

True to form, the Wellington English sent my passport, with no email acknowledgement, to Christchurch, with a visa stamp that expires on April 4, 2009. So I threw everything in my case in 10 minutes, hopped into Kerie's car with Kerie, and headed for Christchurch, six hours away by road. I then took a short flight to Auckland, followed by a very, very long flight to London via Hong Kong, arriving only a few hours ago. Am now enjoying the sights, staying at Oriel College, and will have to figure out where the Computing Lab is tomorrow.