Arcadian Functor

occasional meanderings in physics' brave new world

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Location: New Zealand

Marni D. Sheppeard

Friday, November 28, 2008

M Theory Lesson 242

Another way to write the Tutte polynomial for the square is since any tree with $n$ edges has Tutte polynomial $x^{n}$. In fact, for any polygon we have $T_{n} = T_{n - 1} + \overline{T_{n-1}}$. What does this duality mean? Observe that a triangle plus a trivalent node can make a tetrahedron, so the square can be thought of as representing the four faces of a tetrahedron in three dimensions, which is the dimension of the MUB operators of $T_{4}$.

It follows that a polygon with $n$ sides represents the $n$ faces of an $n$-simplex. The Tutte terms then match cells of the simplex: for $n = 4$, $x^{3}$ is a 3-cell, $x^{2}$ is a 2-cell and $x$ is a single edge. The oddball $y$ term comes from the loop on a single vertex.

Since the dual star graph puts three vertices in the region outside the trefoil knot on the plane, it marks three points in the plane, which we will call 0, 1 and $\infty$ as usual. This is like drawing a trefoil on an associahedron, the dual cell complex of which divides $\mathbb{R}^{3}$ into 9 regions, three of which are the square faces of the trefoil knot crossings. When the associahedron is drawn on the pair of pants, the central vertex of the trivalent node is marked with the number $\omega$. Thus the Tutte map $t \mapsto T(t)$ for $t = \omega$ (at $n = 4$) is a symmetry of the associahedron, which swaps sides of the pair of pants, or equivalently, acts via complex conjugation on the Belyi diagram for the j invariant.

In this paper the authors show that the Jones polynomial for any link can be obtained from a ribbon graph (dessins) diagram for planar link projections via a (three variable) generalisation of the Tutte polynomial, called the Bollobas-Riordan-Tutte polynomial.

M Theory Lesson 241

For polygon graphs with $n$ sides one quickly finds that the Tutte polynomial $T_{n}$ is given by

$T_{n} = x^{n-1} + T_{n-1} = x^{n-1} + x^{n-2} + \cdots + x + y$

For $x = -t$ and $y = -1/t$ this becomes

$T_{n} = (-1)^{n-1}t^{n-1} + (-1)^{n-2}t^{n-2} + \cdots - t - \frac{1}{t}$

With the choice $y = 1/x$ the expansion looks similar in form to the infinite Fourier expansion of a function such as the j invariant $J(q) = j(q) - 744$, although $J(q)$ has positive integer coefficients. Naturally, in M theory we would like to associate the polygon graphs with MUB cycles, generalising the Pauli MUB triangle.

Note that $T_{4}$ at a cubed root of unity $t = \omega$ spits out the cubed root $\omega^{2}$. Similarly, $T_{5}$ at a fourth root of unity ($t = i$) is equal to $i$. The Jones invariant for the trefoil equals $-3$ at $t = -1$. The Tutte component, $T_{3} = x^{2} + x + 1/x$, contributes the $3$, because here $x = -t = 1$. Similarly, $T_{n} = n$ whenever $x = y = 1$. That is, setting $t = -1$ is one way to express the usual ordinals $n$ as dimensions of MUB spaces.

Thursday, November 27, 2008

M Theory Lesson 240







Hopefully this polynomial is familiar to M theorists.

Spring is Here

Photo of lupins, courtesy of my colleague Fraser Gunn, who also takes excellent astronomical photographs.

Wednesday, November 26, 2008

Moving North IV



Sunday, November 23, 2008

M Theory Lesson 239

Truncating the braid group $B_{3}$ on three strands at a cubed root of unity is a little bizarre, because it leads to equivalences which do not preserve the number of loops in a link. Most notably, since each generator satisfies a rule $\sigma^{3} = 1$, there is an identity taking the trefoil knot on two strands to two straight strands. However, we knew that the Jones polynomial for the trefoil knot,

$J(t) = t + t^{3} - t^{4}$

evaluated at a cubed root of unity, is equal to 1. This is also the value for the unknot. There are 24 elements in the truncated group. More general torus knots have polynomials that may be normalised with respect to the trefoil polynomial, which is another way of setting $J$ to 1.

Aside: For the even simpler case of $B_{2}$, forcing the double crossing to be the identity sends the Hopf link polynomial to $2i$ at $t = -1$, or to zero at $t = i$.

Thursday, November 20, 2008

M Theory Lesson 238

We can think of the braid group $B_{3}$ as the general matrix group over the field with one element, associated to sets as vector spaces. It is also the fundamental group of the complement of the trefoil knot. Recall that the trefoil knot corresponds to the Pauli quandle of operators $\sigma_X$, $\sigma_Y$ and $\sigma_Z$.

This quandle can be thought of as a group ring for a field with one element. Additively, there is only one choice for the coefficients of $\sigma_X$, $\sigma_Y$ and $\sigma_Z$, and so the formal sum $\sigma_X + \sigma_Y + \sigma_Z$ represents the three element set as the union of labelled one element sets. Multiplicatively, the cyclic quandle rules hold, and these are the only rules.

What does it mean to take the fundamental group (or groupoid) not of the trefoil, but of the Pauli quandle? What is the complement of the quandle in MUB space? A truncated braid group of type $B_3$ naturally arises for the $3 \times 3$ operators. Moreover, M theory is very interested in how the Pauli operators interact with this three dimensional case. Somehow M theory doesn't mind that $B_3$ is specialised to truncated knots when considering three objects. After all, the fundamental group is really about maps of a circle into a space, but a circle is what one obtains only after considering (at least) an infinite number of objects.

Sunday, November 16, 2008

Moving North III


Friday, November 14, 2008

M Theory Lesson 237

The tribimaximal mixing matrix $T$, which is easily expressed in circulant form, is Fourier transformed to a block matrix which has exact entries

$\frac{2 \sqrt{2} - 1}{2 \sqrt{6}} \pm \frac{\sqrt{3}}{2 \sqrt{6}} i$
$\frac{\sqrt{3} + \sqrt{6}}{2 \sqrt{6}} \pm \frac{1 - \sqrt{2}}{2 \sqrt{6}} i$
$\frac{1 + \sqrt{2}}{\sqrt{6}} + \frac{1 - \sqrt{2}}{\sqrt{6}} i$

Numerically, this transform takes the form Observe that the norm of both the $GL(1)$ component and the $GL(2)$ component is 1, as in the case of the CKM transform. This reduces the parameterisation to four real parameters. The two dimensional component $A$ does not belong to $SU(2)$, but $A^{\dagger} A$ is a braided circulant, namely where the imaginary part should be exactly $2/3$, but my rounding was lazy. The real part is exactly $\sqrt{3}^{-1}$, and the determinant of $A^{\dagger} A$ is $2/9$ (a number that M theorists will recognise).

Thursday, November 13, 2008

Quote of the Month

Aaron Bergman

Moving North II

In two weeks I leave my waitressing job, and the view, and in December I will head down south to visit friends. It seems that soon I will have an airline ticket to Europe in my pockets. Just like magic!

Wednesday, November 12, 2008

A Day at Work

Two of my colleagues at the Astro Cafe, Chris from Arizona and Morpheus from Hong Kong, weren't quite sure what to make of the seasonal appearance of curious lambs on the way to work, but they captured some lovely shots.

Sunday, November 09, 2008

Talk of the Week II


Friday, November 07, 2008

Picoseconds II

For any power law mass triple $(n^{2},n,1)$, such as $(4,2,1)$, the Fourier 2-cycle $F^{2} = (31)$ acts as a T duality operation, inverting scales. That is,

$F^{2}: (4,2,1) \mapsto (1,2,4) = (\frac{1}{4}, \frac{1}{2}, 1)$

where one can keep the total mass scale the same through a T duality coupling of $n^2 = 4$. Recall also (from mid July, Tommaso) that the vector $(1,4,2)$ appeared in the eigenvalue equation for a full six dimensional standard model operator, with the eigenvalue an elementary 3-cycle, using mod 7 arithmetic.


Don't forget to catch this talk by Rivasseau.

Thursday, November 06, 2008

Picoseconds I






Aside: See also posts by Matti Pitkanen and Carl Brannen.

Moving North I

As expected, my comments about being blacklisted from the arxiv were deleted from a popular blog (this time The Cafe) and of course AF is not listed on respectable blogrolls. Nonetheless, it appears that blogging may have helped me obtain a postdoctoral research job somewhere in the Northern Hemisphere. I would love to tell you all about it, but not until everything is official and my flight is booked. Perhaps this is a brave new world, after all.

Wednesday, November 05, 2008

M Theory Lesson 236

On any diagonal matrix of (square root) mass eigenvalues, one instance of the squared Fourier transform acts as a simple permutation, and so clearly the fourth power of the transform is the identity. In other words, the discrete Fourier transform is like a square root of a basic 2-cycle or Pauli operator $\sigma_{X}$. The other choices for $F$ involve braiding elements.

What would a square root of a braid crossing look like? Geometrically, considering the element of $B_{2}$ as a map between bars with two points, the square root is, instead of a rotation of $\pi$ for the bar, a rotation of $\pi/2$. This configuration lines up the points on the bottom bar so that the strands appear to come together in a diagram that usually represents Hopf algebraic multiplication in a category, only now the points are still separated in the third dimension.

Sunday, November 02, 2008

Talk of the Week

From PIRSA, a talk by Glenn Starkman with the title

Saturday, November 01, 2008

Faster Than Light

BackReaction has a post on special relativity which includes a nice $2 \times 2$ circulant matrix $\Lambda = (\gamma , - \gamma \beta )$, where $\gamma$ and $\beta$ are what you think they are.

Let's see how relativity likes playing with numbers. First observe that since the eigenvalues of a 2d circulant $(A,B)$ are always $A \pm B$, the eigenvalues of $\Lambda$ are given by

$\lambda \in \{ \frac{\sqrt{1 - v/c}}{\sqrt{1 + v/c}} , \frac{\sqrt{1 + v/c}}{\sqrt{1 - v/c}} \}$

and $\lambda \rightarrow 1$ as $c \rightarrow \infty$. Alternatively, we could have simplified $\lambda$ to

$\lambda = \frac{1 \pm v/c}{\sqrt{1 - (v/c)^{2}}}$

Now for $v/c = \sqrt{3}$, which is the speed of an observed preon, the two expressions for the eigenvalues lead to the basic numerical identity

$\sqrt{2} = (\sqrt{3} - 1) \cdot (\frac{\sqrt{3} + 1}{\sqrt{3} - 1})^{1/2} = ((\sqrt{3} + 1)(\sqrt{3} - 1))^{1/2}$

which is easily verified, even on my lousy calculator. This expression for $\sqrt{2}$ just amounts to the difference of squares $3 - 1$, but it's cute. Note also that the same expression applies for any pair $(n, n+1)$ of finite ordinals. That is, all square roots $\sqrt{n}$ are simply expressible in terms of $\sqrt{n + 1}$. For the ordinal $3$, or any $n$ such that $n+1$ is a square, it turns out to be a tautology, because $2 = \sqrt{4}$ trivialises the expression.