# rigidity of representations of dehn fillings on the figure eight knot

Consider a Dehn filling of the figure eight knot with coprime parameters (p,q) denoted by M. I am interested in the representation space of M in $SU_2$.

I wonder wether all such representations are infinitesimally isolated or said differently $H^1(M,Ad_\rho)=0$ for all non central $\rho:\pi_1(M)\to SU_2$.

The representation space of the figure eight knot is described (except for the 2 isolated solutions) by the curve $$F(x,y)=cos(x)+1-cos(4y)+cos(2y)=0$$ where $x$ and $y$ are the angles of the meridian and the longitude respectively.

The question is equivalent to saying that the line of equation $px+qy=0$ meets transversely the curve $F(x,y)=0$. This is clear for geometric reasons if $|p/q|<\sqrt{20}$ and can be proven for $|p/q|>\sqrt{20}$ by showing that the polynomial $$\Phi=X^{2p}-X^{p+4q}+X^{p+2q}+2X^p+X^{p-2q}-X^{p-4q}+1$$ has simple roots except $-1$ if $p$ is odd and $\pm i$ if $4|p$.

This is a simple test which I ran with my computer for all $p<200$: the answer is yes. But I don't find a general argument. Any help is welcome!

Julien -- what exactly is $Ad_\rho$? Is it $\rho$ composed with the adjoint repesentation? – algori Apr 21 '11 at 15:35
@algori: yes. Then $H^1(\pi; ad_\rho)$ is identified (by Weil) with the zariski tangent space of the space of conjugacy classes of representations at $\rho$. So Julien is wondering if all non-central reps are "smooth" 0-diml varieties. Since the Figure 8 alexander polynomial has no roots on the unit circle, the usual way to find non-smooth isolated non-central reps of surgeries doesn't apply. (e.g. the corresponding answer is no for the trefoil) – Paul Apr 21 '11 at 16:44
thanks for your comments. I confirm the answer of Paul, Ad$_\rho$ is $\rho$ composed with the Adjoint representation. I think we know that the reps are isolated by conjugation because if not, it would mean that the curve F=0 would contain segments of the line $px+qy=0$ which is certainly no true, but I still cannot show that there are no tangency. – Julien Marché Apr 28 '11 at 9:29
@Julien: I know, it is old question, but I do find it interesting. However, there is something odd with what you wrote: It seems that you know how to prove your conjecture for all values of p, q except in the case $|p/q|^2=20$. However, the equality cannot happen since $p, q$ are integers. Maybe 20 should be replaced with something else? On general grounds, it follows from the work of Kerchkoff and Hodgson that representation of fundamental group of a Dehn filling of a hyperbolic manifold which belongs to the "main component" is infinitesimally rigid, except for a controlled... – Misha Mar 24 '13 at 18:37