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Fix $x,y,z\in \mathbb{C}^*$ and let $M=S^1\times S^1\times S^1$ with $\rho:\pi_1(M)\to \operatorname{SL}_2(\mathbb{C})$ mapping the three generators to diagonal matrices with entries $(x,x^{-1})$, $(y,y^{-1})$, $(z,z^{-1})$ respectively. The question is can you construct a (smooth) 4-manifold $W$ bounding $M$ with a representation $\tilde{\rho}:\pi_1(W)\to \operatorname{SL}_2(\mathbb{C})$ extending $\rho$?

Comments:

1) Such a 4-manifold exists by a non-constructive argument based on the fact that the map $H_3(\mathbb{C}^*,\mathbb{Z})\to H_3(\operatorname{SL}_2(\mathbb{C}),\mathbb{Z})$ induced by the inclusion of diagonal matrices in $\operatorname{SL}_2(\mathbb{C})$ is almost 0.

2) I have a homological argument saying that the extension $\tilde{\rho}$ cannot be abelian.

3) By standard arguments, the same manifold should work for $(x,y,z)$ in a Zariski open subset of $\mathbb{C}^*\times\mathbb{C}^*\times \mathbb{C}^*$.

4) The general motivation is to illustrate topologically the low-dimensional homology of linear groups...

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    $\begingroup$ While you surely know this: A similar result fails in one dimension lower, see here. $\endgroup$
    – Moishe Kohan
    Commented May 29, 2020 at 16:59
  • $\begingroup$ Thanks for your comment. Although I indeed knew it (I wrote a paper on the topic with L. Liechti see arxiv.org/abs/1903.11418), I was not aware of the discussion in MO. $\endgroup$
    – Julien Marché
    Commented May 31, 2020 at 13:10

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