Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

2 Answers

up vote 3 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 $H_0$.

share|improve this answer
    
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
add comment

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 www.math.titech.ac.jp/~fukky/metabelian.pdf

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.