Theory Update 47

Recall that a knot quandle is a set of relations based on a labeling of arcs on a knot diagram. For the trefoil knot the three arcs $a$, $b$ and $c$ satisfy quandle rules, and these generators may be mapped to the usual braid generators $\sigma_i$.

Twistor Revolution

Mottle tells us about a new talk (which I cannot view at present) by Arkani-Hamed about the latest results from the twistor vanguard. Apparently Arkani-Hamed (who is a very eloquent chap, and taken seriously as a physicist) has decided to rename the Theory of Quantum Gravity in a fashion that pays less respect to stringy jargon. It is now officially about an ultratwistoholographic T theory triality!

Hmm, so that must be why no one has read my 2006 work on twistor triality. I wasn't using big enough words!

Theory Update 46

The braid group on three strands is associated to the spatial complement of the trefoil knot in three dimensional space. Let us draw Loday's trefoil on the tetrahedron associahedron.

Theory Update 45

The missing three pentagons are added by embedding the hexagon in one face of a tetrahedron. This is the tetrahedron associahedron.

Theory Update 44

This post is for kneemo, without whom it would not exist. That is, at least one person knows exactly what I am talking about!

A tetractys has an equal number of internal edges and external edges. Thus we can swap the edge weights for these two sets, obtaining a new tetractys.

This tetractys appears as a set of blue triangles in a diagram that begins with a hexagon. The hexagon has an internal triangle, just like the chorded hexagon that marks a square face on an associahedron polytope in dimension $3$. Dual to this chorded hexagon is a pink diagram, selecting $3$ pentagons and $3$ squares, namely $6$ faces of the associahedron.

Cool Blogging

Some time ago Arcadian Functor made the list of Best 50 Mathematics blogs. Now we must thank Lab Tech for putting us on the Best 50 Astrophysics blogs. Not bad for a blog that nobody reads!

Theory Update 43

Recall that the three dimensional associahedron is obtained as the secondary polytope of a hexagon in the plane, defining vertices in terms of triangulations of the hexagon. In general, counting triangulations of planar point sets is a difficult combinatorial problem. Convex polygons with $d + 3$ sides lead to the associahedron in dimension $d$, whose vertices are enumerated by the Catalan numbers.

The tetractys diagram introduces internal points. How do we count the many triangulations of a tetractys? First observe that there are three natural ways to cut the triangle into square (or rhombus) blocks, as shown below in brown, blue and purple. Each square has two possible chords, leading to $24$ possible triangulations.
But we can mess up the pattern of square blocks to form further triangulations of the tetractys. This tends to create internal triangles around the central vertex. Counting incoming edges at a vertex, we recover the vertex weights of the qutrit paths.

Quote of the Week

... to do number theory as if it were physics, where one looks for conjectures by playing with prime numbers with a computer. For example, a physicist would say that the Riemann hypothesis is amply justified by experiment, because many calculations have been done, and none contradicts it ...

Now in physics, to go from Newtonian physics to relativity theory, to go from relativity theory to quantum mechanics, one adds new axioms. One needs new axioms to understand new fields of human experience. In mathematics one doesn't normally think of doing this. But a physicist would say that the Riemann hypothesis should be taken as a new axiom, because it's so rich and fertile in consequences. Of course, a physicist has to be prepared to throw away a theory and say that even though it looked good, in fact it's contradicted by further experience. Mathematicians don't like to be put in that position.

G. J. Chaitin

Theory Update 42

If the total of path weights on the edges of a tetractys is $27$ (the number of three qutrit paths) then the total of the vertex weights (where each incoming edge counts once) must be $54$, since each edge has two bounding vertices. The number $27$ is the dimension of a nice Jordan algebra, and $54$ is the dimension of a complexification of the algebra. Using dual honeycombs, the vertices become two dimensional regions.

Theory Update 41

The dots and edges of the path table do not give the letters $X$ or $Y$ as arrows, suggesting an alternative view with the qubit sequence in terms of two dimensional arrows.
Note that the $4$ paths of the two qubit set are now given by arrow compositions. A qutrit triangle may be built from these arrows. By fixing the mismatch between the outer $XYZ$ and the inner $YZX$, we can still label the path simplex edges with single letters, as shown.

Ignoring the edge directions, we can form $24$ out of the $27$ length three qutrit paths with this two qutrit diagram. All paths of type $XYZ$ are closed triangles. When using these globule diagrams, the degenerate case of classical (one dimensional) objects is represented by basic strings of $1$-arrows, which is to say, ordinary paths. Qubits and qutrits are naturally higher dimensional objects. Observe now that the ordinary tetractys may be labeled by length two words along the edges, and these edges may be interpreted as globules.

Theory Update 40

When interpreting strands as objects in a category, there is one label for each strand. However, we may fix three points $X$, $Y$ and $Z$, and then single strands are specified by paths such as $XY$ or $YZ$. Putting these two qutrit paths from the path table into a $3 \times 3$ matrix, we may select a circulant permutation as shown by assigning equal weights to the paths $XY$, $YZ$ and $ZX$. Circulants are used to form the Koide mass matrices.

Doh. OK, so that braid doesn't go with that particular circulant, but you get the idea.

Interlude

For eight years now, in between waitressing jobs, I have been questioning the ideas of stringers and loopies, at countless conferences and seminars at research universities, and in private conversations and emails. Along the way I obtained a PhD on my own quantum gravity research. In the decade before that, I spent year after year studying mathematical physics, at interesting places around the world. In the decade before that I worked, amongst many other things, as a professional experimental physicist. Now the experimental results pointing away from traditional stringy foundations are rolling out at breakneck speed, and I continue to write this blog.

As far as I can tell, after all that, not one professional physicist values my opinions on quantum gravity. In fact, people with backgrounds in mathematics, computer science, philosophy, chemistry (and God knows what else) have sternly offered their advice to me, as if their university status and gender is itself a sufficient measure of their superior knowledge about physics. Well, it is not for us to judge. We shall let Nature do that.

Theory Update 39

Now we could use a Fun representation of braids, as discussed in this paper, where the Fourier type transform uses both ordinary matrix multiplication and a permutation transformation $P$ to recover the Z boson diagonal.
This takes a flat neutrino diagram in $B_3$ and creates twists in copies of $B_2$. As we have seen, quarks and charged leptons (in $B_6$ perhaps) are also needed to define the transform.

Theory Update 38

In Khovanov homology the bit values, $0$ and $1$, are associated to smoothings of a knot crossing. The trefoil knot in the braid group $B_2$ has three crossings. Let us label these three bits $X$, $Y$ and $Z$. A Bilson-Thompson particle braid is an element of the braid group $B_3$, although its ribbon strands may be either twisted or untwisted. Two crossings on a ribbon strand represent a $1/3$ down quark charge.

Theory Update 37

Let us recall the parity cube as it appears in Bar-Natan's account of the Khovanov invariant. His image shows all eight possible smoothings of the trefoil knot, which has three crossings. Each crossing is resolved into uncrossed strands in one of two distinct ways.

Since quarks also fit onto the central six vertices of a three qubit cube, their fractional charge is associated to the bit sum of a qubit word. The association of charge and braid strands is also familiar to us from the Bilson-Thompson braid spectrum. Some prefer to call these diagrams octupi.

In The News V

This year's FQXi grants have been announced! Apparently, only members of the Old Boys' club are qualified to think about The Nature of Time. Of course, none of them seem to know the first thing about Quantum Gravity, but since when does it matter what you know? As usual, there are two, nice, well behaved token females, just for good form.

Riemann Revisited

Marco Frasca brings up one of our favourite old subjects: the unsolved Riemann hypothesis! Quantum arithmetic is not commutative, because three qubits are not the same thing as two qutrits.

In particular, $8$ is not equal to $9$, much as there are two distinct ways to create the number $57$ from two $3 \times 3$ matrices. As every child knows, one cannot fit a square peg in a triangular hole.

In The News IV

Oh, look. Ed Witten is now working on knots and categorified knot polynomials. Heh, Mottle, maybe you should set him straight.

Another Thought

Around two years ago the gossip of the day, amongst professionals, was about how a certain supersymmetry guru had managed to get guarantees regarding a new building for physics written into his contract. Let's just acknowledge that some researchers are highly valued. Meanwhile, new physics anomalies, all unpredicted by Susy, are being released at an alarming rate.

They say that 2011 is the last year of a world. Let's hope so. Then, it seems there is little I can do at present. With only a few CDs, I have a tendency to listen to Verdi's Dies Irae on my laptop, over and over again. That is some comfort, I suppose.

A Thought

The modern stringers' megalith stands proud amongst the gargantuan forests of modern mathematics. How could they possibly be wrong? Clearly it is not enough to reformulate the Standard Model using other methods, to predict quantitative results in agreement with experiment, or to show that string theory predictions are false. This leaves only one alternative: to derive the string theories as a relatively uninteresting subset of the true M Theory. How far are we from this goal? Almost everything we know about classical spaces is lurking, somewhere in the stringers' toolbox. But this is not an obstacle to be ignored, since in order to derive General Relativity, any decent new theory must reconsider our understanding of manifolds. No attempt that relies blithely on a priori classical number fields has any chance of success ... oh, wait: that covers just about everything the stringers have done! Oh, well, too bad. Everyone can make mistakes.

Dimuon Asymmetry

Jure Zupan gives a PI talk on the D0 anomaly.

Theory Update 36

A while back we played with Terence Tao's honeycomb pictures. Now we see that a one hexagon honeycomb is dual to the tetractys diagram for qutrits.
The Tao honeycomb represents three sets of eigenvalues for $3 \times 3$ matrices, with a zero sum. These eigenvalues are now indexed by tetractys edges, such as the edge $XXX$ goes to $XXY$.

In The News III

Planck at ESA has been releasing early sky map results. Behold dust!

In The News II

Gosh, one can hardly keep up with the headlines. Now DZero give us new bounds on higher mass W bosons. We eagerly await LHC updates on, for instance, the post SM dimuon asymmetry and also the new anomaly from CDF, although the LHC will presumably take some time sorting out these results.

In The News

OK, so everyone knows this already:

$\bullet$ The LHC continues to restrict the possibilities for Susy.
$\bullet$ The Tevatron will shut down in October. Presumably, they have finally figured out that fairy fields do not belong in Science.
$\bullet$ LIGO improves limits on astrophysical gravity waves. Who believes in them, anyway?
$\bullet$ At NASA, Fermi has photos of antimatter from thunder storms.
$\bullet$ A dwarf galaxy hides a supermassive black hole!

Theory Update 35

Since it is so much fun making chorded hexagons look like cubes, let us also make pentagons look like prisms!

Theory Update 34

Today's view of triality is the $21$ edges of our favourite polytope: the Stasheff associahedron in three dimensions! At each trivalent vertex there is a triplet $(A,B,C)$ of edges. The vertex signs are then defined by the cycling of the letters at the vertex. An arbitrary source and target vertex are marked.

Dear Google

Dear Google

I am a law abiding, non-American Wikileaks supporter. I realise that in the event your government orders you to hand over all my personal details, that there is not much I can do about it. I am not afraid to die, but a few people would be sad if my blog were inadvertently deleted by an unscrupulous government string theory supporter. Please protect my personal details. Please protect my blog.

Thanks
Kea

New Physics from CDF

Cool. Yet more to think about! Stay tuned.

Theory Update 33

Mottle calls it a twistor minirevolution. To stringers, this is the idea that one can completely reformulate the Standard Model without in any way affecting the physical basis for ordinary string theory. (I have personally heard some twistor mathematicians laugh at such nonsense, but what would they know?) Anyway, it seems that most stringers are rather unkeen to investigate ternary extensions of basic twistor geometry.

Thanks to kneemo's excellent advice I have lately been reading two wonderful papers (by Corinne Manogue and collaborators) on an octonion analogue of the complex group with which one begins to study twistors, namely $SL(2,C)$. The 2009 paper defines a group $SL(3,O)$ over the octonions. This behaves like the group $E_6$, which contains many subgroups of interest to stringers, such as three copies of $SO(9,1)$, acting on a $10$ dimensional Lorentzian space. This group is like the octonion group $SL(2,O)$, built from complex octonion matrices. We could also make $3 \times 3$ matrices with $2 \times 2$ blocks from $SL(2,O)$.

Back to the CKM

Let us rephase the $R_2$ factors for the CKM matrix, so that each factor is a nice matrix of determinant $\pm 1$. That is, one parameter appears in a factor of the form
The correct phase for the triplet of such factors is

where the numerator and denominator each contribute a phase of $\pi/12 + x$ for $x = 0.003$ a small parameter. In this form, the triple product is more clearly associated to triality. Note that the numerator of $\phi$ cancels the phasing of each factor, so we have simply modified the complex matrix by the phase $\pi/12 + x$.

Theory Update 32

The PeSla has been pondering the higher dimensional analogues of Pascal's triangle for the binomial coefficients, which come from qubit words. In general, these are called Pascal's simplices, at least by Eric Rowland. As we have seen, triangles correspond to qutrit words, and tetrahedra to ... um ... qutets?

A Professorship

The world's most prestigious academic post is Cambridge's Lucasian Chair. The current holder of this professorship is string theorist Michael Green, who took over from Hawking in 2009, when Hawking reached the retirement age of $67$. When Isaac Newton held the chair, he used the terms of the founding will to argue that he should be exempt from taking holy orders, then compulsory. King Charles II agreed, and exempted all future holders of the Lucasian Chair from dealings with the church.

Now given Michael Green's birth date of 1946, the professorship will once again become vacant in 2013, only two years away. Despite its remarkable secular origins, the post has never been held by a woman. Sounds nice to me.

Theory Update 31

Now we can't have the neutrinos all crammed into a point, so let us expand the path diagram to the $81$ dimensional $4$ qutrit space. Now the outer hexagon has nine edges, or six of varying length ($1$ or $2$). The vertices of this hexagon still form a $24$ dimensional space.

Theory Update 30

Now we can put the $9$ edge hexagon into the tetractys of commuting three qutrit paths, by placing the three chords on an outside edge.
This leaves room for the neutrinos and the charged leptons. That is, the $10$ vertices of the tetractys expand the $8$ vertices of an ordinary parity cube.

The charged lepton words ($XXX$, $YYY$ and $ZZZ$) should all have three charged strands for a total charge of $3$. Recall the embedding of the two qutrit paths into the tetractys. Usually we place the two qutrit hexagon in a corner of the tetractys, but we can choose instead to map it to the quark hexagon as shown. Then the path counts on the original hexagon, namely $1$ (for $XX$) or $2$ (for $XY$ and $YX$), are attached to the quarks as charges.