M Theory Lesson 175

By placing each knot crossing in a box, we see 4 output lines for each box, defining two ribbon strands. Thus there are always twice as many extra faces (as squares) on an associated polytope in $\mathbb{R}^{3}$. The associahedron satisfies this condition, as does the deformed octahedron of cubic triality (which has four globule faces). The Euler characteristic defines a sequence of such polytopes via $E = V + F - 2$.

The ribbon diagram for the trefoil knot is the familiar once punctured torus (elliptic curve). Maps relating elliptic curves to the Riemann sphere go back a long way. In particular, the Weierstrass function $P: E(w_{1} , w_{2}) \rightarrow \mathbb{P}^{1}$ is defined via theta functions (for $\tau = \frac{w_{2}}{w_{1}}$) by

$P (z, \tau) = \pi^{2} \theta^{2} (0, \tau) \theta_{10}^{2} (0, \tau) \frac{\theta_{01}^{2} (0, \tau)}{\theta_{11}^{2} (0, \tau)} - \frac{\pi^{2}}{3} (\theta^{4} (0, \tau) + \theta_{10}^{4} (0, \tau))$

Recall that it is the functional relation on $\theta (0, \tau)$ which gives the functional relation for the Riemann zeta function, and these theta functions also define the triality of the j invariant.

Cartoon Fairy

You may have already seen this, but AF definitely needs the link.

Riemann Rainbow II

As it happens, a very friendly bee gave me the data for the new degree 3 $L$ function. Since the data is not mine to divulge, I won't post it here (I'll just have fun playing with the numbers myself for a while). Oh, all right, maybe one little remark: the first positive zero (they are asymmetric about the real axis) is roughly (certainly not exactly) at $3 \pi$, which is about $\frac{2}{3}$ of the first Riemann zeta zero. Recall that the first zero is probably closely related to a low lying energy level for some fundamental physical system.

M Theory Lesson 174

By smoothing the crossing on one boundary disc of the pants (see last lesson) the trefoil turns into two distinct trivalent ribbon vertices. Recall that Bar-Natan's picture of Khovanov homology for the trefoil knot associates a parity cube to all possible smoothings of the three crossings. So once again we start with the Stasheff associahedron and obtain the cube. Moreover, this cube gives us an invariant for the trefoil knot.

For knots with more crossings, Khovanov homology requires higher dimensions, but maybe we can squeeze those into three dimensions by looking at more complicated polytopes with more square faces. For example, recall the 6 crossing knot which we drew on the Klein quartic surface. Can we obtain this knot from the permutohedron, which has six square faces?

Extra, Extra

From Motl's latest post (which discusses this paper):

Conventional physics uses quadratic Lagrangians, two-dimensional worldsheets, second-rank tensors under Yang-Mills groups, commutators between two objects, and similar structures based on the number "2" all the time. We know them quite well.

Still, it looks likely that there exists a whole realm of wisdom that remains mostly hidden in a cloud of mystery ... There exist hints that these largely unknown structures might be based on the number "3" in a similar way as the known theories are based on the number "2". This comment looks extremely vague but there are many reasons to see this prophesy.

Hmmm. Sounds familiar.

M Theory Lesson 173

Let us consider the binary operation on three objects $i$, $j$ and $k$ such that

$i \cdot j = k$
$j \cdot k = i$
$k \cdot i = j$

For example, if the objects were sides of a triangle, the operation might take distinct edges to the edge opposite their common vertex. Note that $k \cdot (j \cdot k) = k \cdot i = j$ so associativity implies that

$j \cdot j = i \cdot i = k \cdot k$

Rules of the form $k \cdot k \cdot k = i$ follow, so one need never encounter more than cubic terms. If the operation were also commutative, then $i \cdot i = k \cdot (j \cdot j) \cdot k = i \cdot k = j$. It follows that $j \cdot j = j$, and in fact all the objects are idempotent. But now there are just too many relations between these objects, so it might be more interesting to drop commutativity and/or associativity. Unfortunately, we can then no longer consider the simple example of an ordinary triangle.

Here is a picture of the three squares on the associahedron pair of pants. When a crossing is marked on each square, there is one path around the faces of the associahedron. By choosing crossings correctly, we can draw a trefoil knot. Note how the picture almost looks like two pieces of ribbon too. By mixing the shown edges with actual polytope edges, one can draw a trivalent ribbon vertex on both the front and back of the pair of pants. Now we can have fun dreaming up new quandle examples using this geometry, associated to the circulant mass operators.

M Theory Lesson 172

In a post about TWF 261, Baez asks about the trefoil and a knot quandle. The knot quandle looks like

$i \cdot j = k$
$j \cdot k = i$
$k \cdot i = j$

which M theorists will recognise as a three dimensional cyclic rule similar to the logic of mass operators. The braid group $B_3$ (recall that this is the fundamental group of the complement of the trefoil) is associated to this quandle, and this is our favourite group covering the modular group.

Michael Batanin pointed out that Loday had put the trefoil on the Stasheff associahedron. In M Theory we like to put knot crossings on the squares of this $3D$ polytope, because the polytope can be turned into the pair of pants with marked trivalent vertices which we put onto the Riemann sphere (which has a lot to do with the modular group) and recall that the squares end up on the real axis, where we might eventually want branch cuts that can accommodate knot crossings, just as in the Ghrist ribbon templates.

Clearly there is something very fundamental about knots here that we do not really understand. Recall that we also wanted knots on the squares (rather circle boundaries) of the pants so that we could use planar diagrams orthogonally to the Chern-Simons type knots contained in the tubes, and where we could rotate sources and targets on the circles before gluing.

GRB 080319B

AAVSO alert. Stolen from the ccd astronomy blog: an image of this week's record bright GRB 080319B. Thanks to Tommaso Dorigo for the alert. Distance indicated by a significant redshift of 0.94. No prizes for guessing the next question ... has anyone phoned LIGO?

Update: NASA news and images.

Quote of Last Century

From Wigner's The Unreasonable Effectiveness of Mathematics in the Natural Sciences:

It is not the intention of the present discussion to refute the charge that the physicist is a somewhat irresponsible person.

The Dirac Code IV

On extending Rowlands' quaternion units to the seven octonions of the Fano plane, one encounters a 3 Time interpretation, accounting for the three generations via the projected hexagon, as usual. Whereas Rowlands finds 4 choices of sign in the 3 quaternion terms, resulting in a 4 component state, in the octonion case, leaving 3 positive mass terms, there are $2^{4} = 16$ sign choices. But the Fano plane relations suggest a reduction of these degrees of freedom, perhaps to the 12 expected for 3 generations.

It would be interesting to combine this octonionic framework with related $E8$ ideas, although classical groups are not of particular interest in M Theory, except in establishing links with other formalisms.

The Dirac Code III

The Pauli exclusion principle for solutions to Dirac's equation, and nilpotency in general, are summed up by the expression

$D^{2} = 0$

for an operator $D$. Mathematicians really like this expression. It immediately brings to mind the (co)homological dimension raising and lowering operators. And since we really want to do M Theory, why not skip the boring manifold de Rham theory (including equivariant cohomology, for that matter) and go straight to a universal motivic cohomology? After all, our particle states are to be represented by knotty diagrams with interpretations in higher logos logic, so the cohomology would naturally be universal (and we're supposed to be doing Quantum Gravity, dammit).

Dimension shifts are categorical. That's why we try so hard to view cardinalities of sets (such as particle counts) in the context of higher topos axiomatics. Thus we don't even know how to count to 3 until we reach the land of tricategories and their multicategorical analogues. Fortunately, as J. W. Gray showed some time ago, by dimension 3, weak $n$-categories reveal a remarkable surprise: the ability to contain dimension altering operations.

Rowlands talks about simple fermions, for which two spin states form the basis of the quantum logic. The squareness of nilpotency, as opposed to the more general $D^{n} \simeq 0$, may be viewed as a consequence of exclusion in two steps, arising from the spin quantum numbers. A mass analogue therefore suggests the study of $D^{3} \simeq 0$. In the context of generalised cohomology, this asks for an enormous extension of the idea of cohomology itself, which relies on concepts such as duality, as opposed to triality. If the categorical structure was significantly extended with each dimension, such as via a concept of $n$-ordinal category, the nilpotent case should be sufficiently rich to reproduce the cohomology theory in question. And there would be so much more.

The Dirac Code II

In a new post, Carl Brannen compares Rowlands' nilpotents with the idempotents of the density operator formalism. Rowlands says on slide 22, "this is intriguingly close to twistor algebra", in reference to 4 complex variables arising from a combination of his quaternions ($1$, $I$, $J$, $K$) and multivariate vectors ($i$, $u$, $v$, $w$). This results in 64 possible products of 8 units, which may be generated, for example, by the combinations

$iK$, $uI$, $vI$, $wI$, $1J$

namely 5 in number, as the Dirac gamma matrices. Rowlands then writes the Dirac equation in the form

$[ iK \frac{\partial}{\partial t} + Iu \frac{\partial}{\partial x} + Iv \frac{\partial}{\partial y} + Iw \frac{\partial}{\partial z} + iJ m ] \psi = 0$

thereby associating the quaternion units $I$, $J$ and $K$ with momentum, mass and energy. The nilpotency appears for the amplitude $A$ when trying to interpret $\psi$ as a plane wave solution. See the slides for extensions of these ideas. For example, requiring $iKE + Ip + Jm$ to be nilpotent, we obtain the expression $E^{2} = p^{2} + m^{2}$ of special relativity. It is OK to put $c = 1$ here, because we work in the one time approximation.

From the perspective of M Theory, even novel algebras are merely representative of the meta-algebraic categorical axioms (Rowlands eliminates equations on slide 40), but analogous number theoretic structures, such as those arising from the $\mathbb{F}_{3}$ matrices for the quaternions in a Langlands type context, contain an even richer potential for interpreting operators in a measurement context, where numbers are the inevitable outcome.

The Dirac Code

Thanks to Carl Brannen for links to slides by physicist Peter Rowlands. I thought the name was familiar: had I met him at a conference a few years ago? If so, it was odd that I could not recall somebody working on a nilpotent operator theory related to Brannen's measurement algebra. Ah! That's it! He has also been banned by the arxiv server, without explanation, despite being a qualified physicist and author of a book on the foundations of physics.

Do any of the blacklisted physicists not have an interest in this approach to unification?

Riemann Rainbow

David Corfield brings our attention to an AIM press release about the discovery of a new L function. As the blurb explains, this new function $L(s)$ satisfies a degree 3 symmetric functional relation

$F(s) \equiv \frac{\sqrt{q}}{\pi^{3}} \Gamma (\frac{s}{2} + r_{1}) \Gamma (\frac{s}{2} + r_{2}) \Gamma (\frac{s}{2} + r_{3}) L(s) = F(1 - s)$

for some integer $q$, in contrast to the degree 1 behaviour of the Riemann zeta function (for which $q = 1$). Of course I immediately emailed Michael Rubinstein to ask for a reference on the actual values of these Langlands' parameters, as well as values for the first few known zeroes, which lie on the critical line. I eagerly await a reply, but my server may well be treated as a spam generator. In the meantime, Minhyong Kim has kindly provided helpful comments and links.

M Theory Lesson 171

As one moves up the $n$-ordinal ladder, perhaps by adding levels to a tree (as in the case of the extension of the associahedra to the permutoassociahedra), the spherical polytopes acquire more and more vertices and faces. That is, they begin to better approximate a sphere. In all dimensions, both the cube and the permutohedron tile $\mathbb{R}^{N}$. The translation lattice for the 3 dimensional permutohedron may be generated by the vectors $(1,1,-3)$, $(1,-3,1)$ and $(1,1,-3)$. The associahedra do not share this property, but recall that they instead tile the real points of interesting moduli spaces.

M Theory Lesson 170

Hamilton circuits are drawn on the cube and associahedron. Observe how a step on the parity cube shifts a bit by 1. This circuit is used to define the space filling Gray code by shifting one edge across a boundary. Tony Smith explains how this is related to $E8$ lattices.

M Theory Lesson 169

Thanks to Steven H. Cullinane for this diagram of Hamilton's quaternions, the whirligigs. Recall that this plane over the three element finite field $\mathbb{F}_3$ arises in the $d = 3^{n}$ MUB problem, associated to the Fourier transform of the $p = 3$ mass matrices.

The Dark Side

I'm a bit behind the times down here sometimes. I only just noticed, whilst passing the news stand at the supermarket, that the cover story of the last issue of New Scientist is about the possible non-existence of The Dark Force. I didn't need to open it to know that it mentioned David Wiltshire, but of course not Louise Riofrio, Matti Pitkanen or a whole of host of other quantum gravity researchers who think The Dark Force is absurd.

M Theory Lesson 168

Like Hamilton's dodecahedron, the squashed permutoassociahedron also permits a circuit that passes once through each of the 120 vertices. It helps to paint the squares, pentagons and dodecagons in different colours. Try it!

M Theory Lesson 167

At PF, Lawrence B. Crowell taught us about the remarkable invention of non-commutative geometry by the great Hamilton, the inventor of the quaternions. But I do not refer to the quaternions themselves. Rather, as Janet Heine Barnett explains in a beautiful article on the icosian game, in Hamilton's own words:

I have lately been led to the conception of a new system, or rather family of systems, of non-commutative roots of unity, which are entirely distinct from the i j k of quaternions, though having some general analogy thereto.

The basic icosian calculus describes moves through the vertices of a dodecahedron and is generated by three kinds of move, let us say $a$, $b$ and $c$, such that $a^{2} = 1$, $b^{3} = 1$, $c^{5} = 1$ and $c = ab$. Observe the appearance of the rules for the modular group. All these moves apply to the oriented graph and are given by
a. reverse the edge (eg. $ST \mapsto TS$)
b. rotate (say left) around the endpoint (eg. $HG \mapsto BG$)
c. move one edge (to the right) along a pentagon (eg. $BZ \mapsto ZQ$) At least one crazy retired physicist has incorporated this calculus into a spacetime model for the leptons and quarks, in which the $E8$ lattice magically appears out of paired quaternion like (ie. octonion) operations. A triality involving three $E8$s is briefly discussed.

Actually, it was supposedly Hamilton who first considered the complex numbers algebraically as an ordered pair of reals, in a paper entitled, Theory of Conjugate Functions, or Algebraic Couples; with a Preliminary and Elementary Essay on Algebra as the Science of Pure Time. Hamilton's next publication was entitled, On the Propagation of Light in Vacuo. (I almost wish I was 15 again so that I had time to read more.)

Resolving Power

It is quite a while since we looked at Batanin's picture of the three dimensional permutoassociahedron, so here it is! Observe how it resolves each vertex of a permutohedron (labelled by an ordering of four letters) into a pentagon (which represents the Mac Lane bracketings of four letters). This is a 120 vertex polytope.

M Theory Lesson 166

Recall that the third theta function in the j invariant numerator, namely $\theta_{10}(0, \tau)$, transforms under $\tau \mapsto \tau + 1$ to $\theta_{10}(0, \tau)$ multiplied by the 8th root of unity $\textrm{exp}(i \frac{\pi}{4})$. For the j invariant, the 8th root disappears with the power of 8. Perhaps we should think of the power of 8 here as representing 8 vertices of a cube, where 3 cubes are formed from the six squares of a permutohedron, each paired to its opposite. Every vertex of such a cube is obtained by rotating the octahedron dual by $\textrm{exp}(i \frac{\pi}{4})$, as mentioned yesterday.

But why would one try to associate the j invariant with such low dimensional polytopes rather than, say, the $E8$ or Leech lattices? Why not! It is far easier to visualise what is going on in three dimensions, because that happens to be the dimension of our classical space template. And for 6 dimensional twistor type moduli, perhaps we can put 3 and 3 together.

M Theory Lesson 165

In his newest preon model, Carl Brannen (whose diagram I have filched) uses a total of 24 objects, including six squares as shown. In other words, think of the six squares of a permutohedron, given by the squares of a truncated octahedron. These ternary geometry type squares were rotated by 45 degrees from the squares used to specify the usual dual, namely a cube.

Autumn Days

I am thinking of going on a walk soon with some friends down south.

Quote of the Week

From Dynamics of Cats, regarding the WMAP release: If you're a cosmologist, you need to rewrite your proposal. Now.

Aside: according to Mottle (oh, wait, it's in the abstract) the 5 year data strongly suggests 3 (or 4) neutrino species, of mass sum $\leq 0.61 eV$, in agreement with Carl Brannen's prediction of $\Sigma m = 0.0600(40) eV$.

Update: Louise Riofrio has an informative post on the results.

Time Machine

Although it caused quite a stir in the press and on the blogosphere, I didn't take much notice of the Time Machine paper until today, when I realised it was written by Irina Aref'eva and Volovich, who happen to work on p-adic strings and the quantization of the Riemann zeta function. Ultimately, they are simply speculating about new kinds of objects, related to classical causality violation, that may be visible at the LHC, and the catchy title is simply a gimmick without which it is difficult these days to get papers posted on the arxiv.

In this paper, the authors discuss some pretty hairy mathematics, in the physicist's characteristic shockingly hand-wavy manner. To quote:

[this lends] additional support to the proposal that the Beilinson conjectures on the values of L-functions of motives can be interpreted as dealing with the cosmological constant problem ... in section 6 we shall discuss an approach of how to use a Galois group and quantum L-functions instead of SUSY to improve the spectrum.

By the spectrum they are referring to their analysis, inspired by the non trivial zeroes of the Riemann zeta function, which correspond to $m^{2}$ values in Klein-Gordon operators. In other words, the zeta function is defined not on numbers, but as a pseudodifferential operator. The Hypothesis says that the zeta field is given by a sum of such Klein-Gordon Lagrangians.

Note that in M Theory, we prefer to replace $\Lambda$ with the heirarchy of Planck scales, but this idea is basically present in their work.

M Theory Lesson 164

Let's do some really basic algebra. On Friday we wondered whether or not the inverses of Riemann zeroes might be related to that damned number. Note that inverse eigenvalues naturally occur for inverse matrices, since

$A^{-1} A v = v = A^{-1} (\lambda v)$

where $\lambda$ is an eigenvalue for $A$. If $A$ (assumed complex) is both Hermitian and unitary, it satisfies $A^{\dagger} = A^{-1} = A$, from which it follows that $A^{2} = I$. What are the solutions to this equation?

In the $2 \times 2$ case one quickly finds that either $A = I$ or $A = \sigma_{x}$. For the $3 \times 3$ case, relations of the form

$a_{11}^{2} - a_{22}^{2} - a_{33}^{2} - 2 a_{23} \overline{a_{23}} = 1$

suggest setting off diagonal elements to be real, since diagonal elements are already real. So if $a_{12} = 1$, it immediately follows that $A$ is completely specified by the circulant matrix

0 1 0
1 0 0
0 0 1

The remaining possibilities are left to the reader. Now note that if $A = A^{-1}$ then the real eigenvalues come in pairs $(\lambda , \lambda^{-1})$, with $\lambda = \pm 1$ considered a double eigenvalue. As a factor of the characteristic polynomial, these 2 roots give

$x^{2} - x(\lambda + \frac{1}{\lambda}) + 1$

which in the $2 \times 2$ case is precisely the statement that $\textrm{det} (A) = 1$ (the constant term) and $\textrm{tr} (A) = \lambda + \frac{1}{\lambda}$.

History Meme

Tommaso Dorigo has tagged me with the history meme, but my charity towards pyramid schemes only extends to rules 1 and 2, and I omit links to other blogs in the post. Besides, I don't actually know enough people to pester. So then, seven weird and/or random things about an historical figure...

The idea of a favourite never really makes sense to me. How does one compare the merits of a Marie Curie to a Cyrus the Great? Anyway, I settled on the warrior Artemisia, the Queen of Halicarnassus:

1. most famous for deceptively ramming a friendly Persian ship, which sunk along with its crew, in order to escape after the Persian defeat at Salamis, where Artemisia commanded 5 ships
2. thinking that the sunken ship was Greek, Xerxes remarked: my men fought like women and my women fought like men
3. she supposedly fought in the previous Persian victory near Cape Artemisium, at the same time as the Persian victory on land at Thermopylae, at which point the Greek situation wasn't looking good
4. it was Artemisia that advised Xerxes to return to Asia, and it was her that had warned Xerxes of the Greek strength at sea
5. she is said to have put out the eyes of her beloved but neglectful Dardanus
6. as punishment she died by taking the lover's leap
7. the later namesake, Queen Artemisia II, built a great tomb for her brother (and husband) Mausolus, who died in 353 BC (this is where the word mausoleum comes from)

Day At Work

I was about to commiserate with Tommaso regarding the state of bureaucracy in Italy, having just received a telegram giving me a time for an interview there, which of course I am completely unable to attend, my swimming abilities not extending to global circumnavigation. Now, I've never actually received a telegram before. It reminded me of my favourite childhood Agatha Christie stories, of adventurous women wandering in the African jungle in between afternoon tea and biscuits and telegrams.

But then I recalled that my email address has changed a number of times since I sent off the application, and I had to change my cellphone number because I was being harrassed, and I don't have have any other phone numbers or means of contact. Anyway, the state of bureaucracy in Italy may well be dismal, but I am in no position to judge it.

Meanwhile, I am attempting to deal with the forms for a local postdoctoral fellowship. I sent off a draft to the university research office and it was returned with a large number of suggested corrections, most notably that I should cut down the technical summary to 300 words or less. The remaining 20 or so pages are for detailing the benefits of the proposal for the country's economic, cultural, industrial and educational welfare. This is going to take me some time to figure out.