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Let $M$ be an orientable cusped hyperbolic 3-manifold. Let $\rho \in \mathrm{Hom}(\pi_1(M),\mathrm{SL}_2(\mathbb{C}))$ be an irreducible representation.

Is $\rho$ scheme reduced ?

What can prevent $\rho$ from being a smooth point of the representation variety ?

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  • $\begingroup$ Perhaps contact Misha Kapovich (math.ucdavis.edu/~kapovich) with this question? He'll know, or know who to point you to. $\endgroup$
    – Sam Nead
    Commented Mar 18, 2014 at 21:42
  • $\begingroup$ In particular, see his answer to this question mathoverflow.net/questions/123198/… $\endgroup$
    – Peter Samuelson
    Commented Mar 18, 2014 at 21:53
  • $\begingroup$ In the question I am interested in the representation itself not the character but in case of hyperbolic manifold with n cusps we have an analytic fibration $R^{\text{irr}}(M)\to X^{\text{irr}}(M)$ of the irreducible representation over the irreducible characters so is it possible that $\rho$ is not scheme reduced (resp. smooth) if $\chi_{\rho}$ is not scheme reduced (resp. smooth) ? $\endgroup$
    – christian
    Commented Mar 19, 2014 at 21:03
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    $\begingroup$ You should rewrite the question instead of reinterpreting the question in the comments. For the question you are now asking the answer is: $X^{irr}$ is smooth (reduced) at $[\rho]$ iff $Hom$ is smooth (reduced) at $\rho$. This has nothing to do with hyperbolicity. $\endgroup$
    – Misha
    Commented Mar 20, 2014 at 10:32

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I am turning Misha's comment into an answer, so this is not left unanswered under this tag.

The current question is:

Is it possible that irreducible $\rho$ is not smooth (reduced) if $\chi_\rho$ is not smooth (reduced)?

Here is Misha's answer (which I agree with):

$\mathfrak{X}^{irr}(\pi_1(M), \mathrm{SL}(2,\mathbb{C}))$ is smooth (reduced) at $\chi_\rho$ if and only if $\mathrm{Hom}^{irr}(\pi_1(M),\mathrm{SL}(2,\mathbb{C}))$ is smooth (reduced) at $\rho$.

Here is a comment from me:

It is possible for singular points in $\mathrm{Hom}(\Gamma,G)$ to be smooth in $\mathfrak{X}(\Gamma,G)=\mathrm{Hom}(\Gamma, G)//G$. Take for example a genus 2 surface group $\Gamma$ with $G=\mathrm{SU}(2)$, then $\mathrm{Hom}(\Gamma, G)/G\cong\mathbb{CP}^3$ but there are singularities in the $\mathrm{Hom}(\Gamma, G)$ (coming from reducible representations). Worse, even for free groups there are Lie groups $G$ of arbitrarily large rank with the property that $\mathfrak{X}(F_2,G)$ has smooth reducibles and singular irreducibles for a rank 2 free group $F_2$. See here and here for more information.

One should also read Misha's great answer to a related question here.

Tangentially related are the answers here.

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