8
$\begingroup$

Let me give a worked-out example: The following cubic planar non-simple graph

$\hskip2.3in$enter image description here

has the adjacency matrix $A=\pmatrix{0&3\\3&0}$. The graph has three faces, so the rank of $G$ is $\chi(G)=2$. The reciprocal of Ihara's $\zeta$ function can be evaluated $$ \frac{1}{\zeta_G(u)}={(1-u^2)^{\chi(G)-1}\det(I - Au + 2u^2I)}\\ ={(1-u^2) (4u^4-5u^2+1)} $$

EDIT: Then $\zeta_G(u)=\frac{(1-u^2)^{-1} }{(4u^4-5u^2+1)}=\prod_p (1-u^{L(p)})$ with the product running over prime paths $p$ and $L(p)$ being their lenghts. The $-1$ in the numerator's exponent $(1-u^2)^{-1}$ is due to $|V|-|E|=2-3=-1$...

Looking at the orientation of the edges around the vertices, it is obvious that left and right are oriented opposite.

Now blow up every edge like in a ribbon or fat graph. Including the change of orientation the resulting graph looks like:

$\hskip1.7in$enter image description here

where I stuck to the convention that I flipped every fat graph edge in the same direction. The resulting knot is a trefoil, which has the following polynomial invariants:

The Alexander polynomial of the trefoil knot is $ \Delta (t)=t-1+t^{-1},$ and the Conway polynomial is $ \nabla (z)=z^{2}+1.$ The Jones polynomial is $ V(q)=q^{-1}+q^{-3}-q^{-4}.$..

None of the latter mentioned matches Ihara's $\zeta$. My list is not complete, but

Is there a relation between Ihara's $\zeta$ and any knot polynomial?

EDIT:

Further, the bicubic planar graphs can be related to Riemann surfaces (see here and references therein). Is there a relation between the Riemann surfaces and the knot?

How does the topology of the graphs' Riemann surface relate to its knot representation?

  • 0
    Do you know how to show here the $\zeta_G(s)$ obtained from the prime cycles and the Euler product is the same as what you wrote [in term of the adjacency matrix](https://en.wikipedia.org/wiki/Ihara_zeta_function#Ihara.27s_formula) ? – 2017-01-23
  • 0
    @user1952009 I think that is exactly what Ihara has proven. I tried to come from [non-backtracking paths](http://math.stackexchange.com/q/177722/19341) via Chebyshev polynmomials and think I got pretty close [here](http://math.stackexchange.com/q/1580353/19341), but I'm still looking for answers... – 2017-01-24
  • 0
    It's not clear to me because $4u^4-5u^2+1 = (2u-1)(2u+1)(u+1)(u-1)$ so your $\zeta_G(u)$ is not an Euler product generating the number of cycles of a given length. Did you try normalizing the adjacency matrix ? – 2017-01-24
  • 0
    @user1952009 I added something to explain that. Which matrix norm would you recommend and why? How could normalization help? – 2017-01-25
  • 0
    Sideremark for me: $-\log((1-x^2)(4x^4-5x^2+1))=12\frac {x^2} 2+36 \frac{x^4}4+132\frac{x^6}6+\cdots$ $12$ means: select $g,y,r$ or its inverse for the first step, so that's $6$. Then you have two other colors to go back: $6\cdot 2=12$ ... – 2017-04-10
  • 0
    Re-posted at MO: https://mathoverflow.net/q/275112/11856... – 2017-07-10
  • 1
    I don't think we need to close this before the MO one was answered. – 2017-07-10

0 Answers 0