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Suppose that $K$ is an $\textit{alternating}$ knot in $S^3$, and let $R_0$ be the space of homomorphisms from $\pi_1(S^3 - K)\to SU(2)$ which send meridians to trace free matrices. Denote the subset of $R_0$ consisting of metabelian representations as $R_m \subset R_0$.

Question: when $K$ is prime, is there any reason to think that $R_0$ retracts to $R_m$? Could at least $H_*(R_0;\mathbb{Z})=H_*(R_m;\mathbb{Z})$? Does anyone know of anything about such questions in the literature?

Might there be similar analyses for more general $SU(2)$ representations of knot groups?

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Do you mean deformation retract (this is suggested by your $H_*$ question): generically the space of trace free reps (mod conjugation) is finite and so there would obviously be a retract. – Paul Mar 18 '10 at 3:13
It seems that $R_m$ is the set of binary dihedral reps, which is a finite set up to conjugation. IIRC it has |p(-1)-1|/2 points where p(t) is the alex. poly (see Klassen TAMS). One can probably find examples with $H_1(R_0)$ is non-zero if $S^3-K$ contains incompressible tori by bending the rep along the torus, but your assumption that K is alternating and prime presumably prevents this. For torus knots, $R_m=R_0$. – Paul Mar 18 '10 at 19:09
Ah, that's interesting. Yes, there are definitely non-alternating examples, I think. Maybe using hyperbolicity for the alternating (non-torus) knots could give some leverage here? Indeed I did mean deformation retrace; I'm really just interested in the homology (for now). – Sam Lewallen Mar 22 '10 at 23:06
up vote 4 down vote accepted

I doubt there is such a retraction. Such representations are representations of the $\pi$-orbifold, obtained by killing the square of the meridian (at least if one quotients by $\pm Id$). If one takes a Montesinos knot, these orbifolds are Seifert fibered, and such representations should factor through the quotient orbifold. Such oribfolds can have several non-metabelian isolated representations into $SO(3)$, so I expect that the answer is no even on $H_0$.

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Ah, this is interesting, thanks! But many of the more interesting montesinos knots are non-alternating, right (e.g. if the cover is a Brieskorn homology sphere, the knot won't be alternating). Is there any possibility that the isolated non-metabelians only occur for alternating knots? I'm writing this comment just after reading your response, so I'm sorry if I've asked anything dumb... I'll think about it more soon – Sam Lewallen Mar 22 '10 at 23:09
I mean "only occur for NON-alternating knots" – Sam Lewallen Mar 22 '10 at 23:09
I don't think it should matter whether the Montesinos knot is alternating or not, it should only depend on the base orbifold. However, what I haven't checked is which $SO(3)$ reps lift to $SU(2)$ reps. The simplest non-trivial example would be the $(2,3,7)$-pretzel. This maps to a $(2,3,7)$ triangle orbifold, which is arithmetic. Thus, the orbifold fundamental group has a Galois conjugate which is in $O(3)$. However, one can actually get this to lie in $SO(3)$. Since $Isom(\mathbb{H}^2) < PSL(2, \mathbb{C})$, one can take the Galois conjugate to lie in $PSU(2)=SO(3)$. – Ian Agol Mar 26 '10 at 4:36

This looks like a good place to start from (if you haven't already read it)

MR2488756 (2009m:57024) Nagasato, Fumikazu . Finiteness of a section of the ${\rm SL}(2,\Bbb C)$-character variety of the knot group. Kobe J. Math. 24 (2007), no. 2, 125--136.

This paper shows that for any knot, there are only finitely many irreducible metabelian characters in the ${\rm SL}(2,{\bf C})$ character variety. It is also shown that the number of conjugacy classes is given by a simple formula involving the Alexander polynomial. In the context of two bridge knots, there are inequalities involving the $A$-polynomial of the knot. Results of this nature were previously obtained by X. S. Lin [Acta Math. Sin. (Engl. Ser.) 17 (2001), no. 3, 361--380; MR1852950 (2003f:57018)] for ${\rm SU}(2)$ representations.

This paper can be found at

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