SU(2) representations of alternating knot groups 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?
 A: 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 $H_0$. 
A: 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
