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Suppose there is a knot $K\subset M$, where $M$ is a closed 3-manifold. What's the relation between $\chi(\pi_{1}(M-K))$ and $\chi(\pi_{1}(M))$?

Note: $\chi(G)$ means the $\text{SL}_{2}(C)$-representation variety of $G$ modulo conjugation.

$\chi(\pi_{1}(M))$ is a subset of $\chi(\pi_{1}(M-K))$. Is it the subvariety defined by the equation: $Tr(\rho(m))=2$, where $m$ is the meridian?

It's not obvious because even if $[\rho]\in\chi (\pi_{1}(M-K) )$ satisfy $Tr(\rho(m))=2$, $\rho(m)$ may not be $Id$, so $\rho$ may not extend to $\pi_{1}(M)$.

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    $\begingroup$ Trace 2 does not imply identity, it could be also parabolic and such examples are easy to construct. But on the level of rep. variety the relation is very simple: The rep. variety for M is given by the equation that meridian maps to the identity. $\endgroup$
    – Misha
    Commented Mar 23, 2013 at 22:18
  • $\begingroup$ Even the case $M=S^3$ is worth looking at. $\endgroup$
    – Liviu Nicolaescu
    Commented Mar 24, 2013 at 11:29
  • $\begingroup$ Dear Lin: I think you should wait a bit before accepting answers. There are other people on MO besides me who know thing or two about character varieties of 3-manifold groups, e.g. Nathan Dunfield, Adam Sikora and Kevin Walker. They could give more enlightening answers than mine. $\endgroup$
    – Misha
    Commented Mar 24, 2013 at 18:55

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I do not think there is a particularly nice answer to this. Below is a suggestion on how you could proceed.

Consider representations $\rho$ of a finitely generated group $\pi$ to $Sl(2, C)$. First of all, the condition that $Tr(\rho(g))=2$ is necessary but not sufficient for $\rho(g)=1$. You can see this by looking at unipotent matrices in $Sl(2, C)$. If $\rho$ maps $g$ to the identity, then for every element $h$ in the normal closure $H$ of $g$ in the ambient group $\pi$, you have $Tr(\rho(h))=2$. One can show that this condition is necessary and sufficient for $\rho(g)=1$. By the Nullstellensatz, it suffices to check this condition only for finitely many elements $h$ in the normal closure $H$.

Now, it should be possible to find an explicit set of such elements of $H$. Doing so is a worthwhile task, I do not think anybody computed this set, it will depend on the word which represents $g$ in terms of generators of $\pi$. If you manage to do this, you can then apply the answer to the special case of generalized knot groups as in your question where $g=m$. Even in the case of knots in the sphere the answer is very likely to be very ugly, and will depend on the knot diagram. However, this is the best I can suggest.

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  • $\begingroup$ Thank you very much! It's very helpful. I think this is not an obvious question. I was a little bit misled by the $SU(2)$ case, which is very different. $\endgroup$
    – Lin Jianfeng
    Commented Mar 24, 2013 at 19:40

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