Arcadian Functor

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Marni D. Sheppeard

Thursday, November 27, 2008

M Theory Lesson 240

The Tutte polynomial $T(x,y)$ of a graph $G$ with edge set $E$ is defined by simple recursion rules. Let $G/e$ denote $G$ with the edge $e$ contracted, and $G - e$ denote $G$ with the edge $e$ deleted. Then the rules are:

$\bullet T(G) = 1$ if $E$ is empty
$\bullet T(G) = xT(G/e)$ if the deletion of $e$ disconnects $G$
$\bullet T(G) = yT(G - e)$ if $e$ is a loop on a single vertex
$\bullet T(G) = T(G - e) + T(G/e)$ else

For example, the 2-colored trefoil is associated to the triangle graph on three vertices, or its dual graph, the trivalent node. For the triangle, the computation of the Tutte polynomial proceeds as follows. Now let $g$ be the number of vertices in $G$ and $d$ the number of vertices in the dual graph. The writhe of the knot is $w$. In terms of $T$, the Jones polynomial for an alternating knot is given by

$J(t) = (-1)^{w} t^{(g - d + 3w)/4} T(-t, \frac{-1}{t})$

For the trefoil knot, which has a writhe of $+3$, we calculate the Jones polynomial using the Tutte polynomial:

$J(t) = -1 \cdot t^{(3 - 4 + 9)/4} \cdot (t^{2} - \frac{1}{t} - t)$
$= - t^{2} (t^{2} - \frac{1}{t} - t) = t + t^{3} - t^{4}$

Hopefully this polynomial is familiar to M theorists.


Blogger CarlBrannen said...

The polynomial I've been solving for the weak quantum numbers is the idempotency equation t^2 = t, or t(t-1) = 0. But I only get half the solutions, the other half satisfy t(t+1) = 0.

For each elementary fermion, one shows up in the one solution as a particle, and as an anti particle in the other solution. So to write a polynomial that gets all of them, we'd presumably solve t(t^2-1) = 0, or t^3-t = 0.

The funny thing about this is that t^3 = t is an interesting variant of t^3 = t. It brings to mind triality or braids or something like that.

Meanwhile, I've got the code mostly running for an RPN calculator where each "number" is a pair of 3x3 complex matrices (36 real numbers in total), which one thinks of as the IJK matrix and the RGB matrix. I guess I'll put it up on the web and blog it after I get it checked out. The idea is that the entries for this calculator are the "t" values.

November 27, 2008 7:38 PM  
Blogger Kea said...

Carl, t^n = t has come up previously, for example with the Abel sums for trees. For the Jones polynomials, it appears most notably in the writhe term, where there is always a factor of t^{n*3/4}, and application of t^n=t (coming from the (n-1)th root of unity choice for t) reduces this to t^{3/4}. There is another factor of 1/t, making the writhe factor = t^{-1/4}.

This is a really natural occurrence of turning cubed roots into 12th roots, 6th roots into 24th roots etc.

November 28, 2008 11:53 AM  

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