Does every $SL_2\mathbb{C}$ representation of a closed oriented surface extend over a compact oriented three-manifold?

Here is an argument that "most" points in the $SL(2, {\mathbb C})$-character variety $X(F)$ of the surface $F$ do not correspond to representations extendible to 3-manifold groups (as in the question).

Let $M$ be a compact oriented 3-manifold with $\partial M=F$. We then have the "restriction morphism" of $SL(2, {\mathbb C})$-character varieties $$ r: X(M)\to X(F). $$ The image of $r(X(M))$ is "formally Lagrangian" (more precisely, Lagrangian on the scheme-theoretic level) with respect to the standard complex symplectic structure on $X(F)$, see

A. Sikora, Character varieties. Trans. Amer. Math. Soc. 364 (2012), no. 10, 5173–5208.

In particular, $\dim r(X(M))\le \frac{1}{2} \dim X(F)$. Since there are only countably many 3-manifolds $M$ as above, the union $$ U=\bigcup_{M} r(X(M)) \subset X(F) $$ has empty interior (in the Euclidean topology). Thus, "most" points in $X(F)$ do not belong to $U$. I do not know how to detect non-membership in this union algorithmically. Since you are working over the complex numbers, you have to specify what computability even means. For instance, you can restrict to $\overline{{\mathbb Q}}$-points of the character variety (i.e. equivalence classes of representations to $SL(2, \overline{{\mathbb Q}})$); then at least one can use the classical notion of computability and your question is well-defined in this setting. (There is a silly algorithm which terminates for points in $U(\overline{{\mathbb Q}})$.). I do not even know if the membership problem in $U(\overline{{\mathbb Q}})$ is decidable.

PS. This entire discussion feels related to the proof of Theorem 1.3 in

N. Dunfield, W. Thurston, Finite covers of random 3-manifolds. Invent. Math. 166 (2006), no. 3, 457–521.

Edit. Here is one way to find explicit examples of non-extendible representations $\rho$, i.e. such that $[\rho]$ does not belong to $U$ (motivated by Ian Agol's answer in the case of genus $1$). I will use the fact that the variety $X(F)$ is ${\mathbb Q}$-rational, see for instance Theorem 2 in

A. Rapinchuk, V. Benyash-Krivetz, V. Chernousov, Representation varieties of the fundamental groups of compact orientable surfaces. Israel J. Math. 93 (1996), 29–71.

In other words, there exists a birational isomorphism defined over ${\mathbb Q}$, $f: X(F)\to {\mathbb C}^{6g-6}$.

Hence, instead of $X(F)$ we can essentially work in ${\mathbb C}^{6g-6}$ (with its standard rational structure).

Now, take a point $p=(z_1,...,z_{6g-6})$ in ${\mathbb C}^{6g-6}$ whose coordinates generate a field of transcendence degree $>3g-3$. (Such $p$ necessarily belongs to the image of $f$ and $f^{-1}(p)$ is a singleton.) One can find such tuples $p$, for instance, using the Lindemann–Weierstrass theorem. Then $[\rho]=f^{-1}(p)$ does not lie in $U$.

In addition to Moishe Kohan's geometric argument, there's also a bordism-theoretic proof.

$\newcommand{\BDel}{B\mathrm{SL}_2(\mathbb C)^\delta}$ Let $\Omega_*^{\mathrm{SO}}(-)$ denote oriented bordism as a generalized homology theory. Your question is equivalent to asking whether $\Omega_2^{\mathrm{SO}}(\BDel) = 0$.

We can compute this with the Atiyah-Hirzebruch spectral sequence, which has signature $$ E^2_{p,q} = H_p(\BDel, \Omega_q^{\mathrm{SO}}(\mathrm{pt})) \Longrightarrow \Omega^{\mathrm{SO}}_{p+q}(\BDel). $$ $\Omega_q^{\mathrm{SO}}(\mathrm{pt}) = 0$ for $q = 1,2,3$, so in the range $p+q < 4$, this spectral sequence collapses, implying $\Omega_2^{\mathrm{SO}}(\BDel) \cong H_2(\BDel; \mathbb Z)$. Now, as suggested by Danny Ruberman's comment, Milnor's “On the homology of Lie groups made discrete” points out that $H_2(\BDel; \mathbb Z)$ surjects onto an uncountable $\mathbb Q$-vector space, hence is in particular nontrivial, and therefore not every $\mathrm{SL}_2(\mathbb C)$-representation of a closed, oriented surface extends to a compact $3$-manifold.

Unfortunately, this approach is difficult to make explicit for a given representation of a surface group without a better understanding of the homology of $\mathrm{SL}_2(\mathbb C)$ as a discrete group.

This is an extended comment on the algorithmic question. As Moshe points out, the lack of realization as a boundary follows from the Baire category theorem. On the other hand, how does one recognize when an element is not in this infinite countable union of subspaces?

Let's consider the genus 1 case $F=T^2$. Then a rep $\rho:\pi_1(F)\to SL_2(\mathbb{C})$ is determined by two eigenvalues of generators $(\mu, \lambda)$ (assuming the representation is not unipotent). In turn if $F=\partial M$, the boundary of a 3-manifold, then there is an associated $A$-polynomial $A(x,y) \in \mathbb{Z}[x,y,x^{-1},y^{-1}]$ such that $A(\mu,\lambda)=0$. I suspect that $[\rho]=0\in H_2(SL_2(\mathbb{C})^\delta)$ iff $[\rho]$ extends to a representation of a 3-manifold, but I haven't checked this. In any case, one sees that $\mu, \lambda$ are algebraically related in the bounding case. However conversely, if $\mu,\lambda$ satisfy an algebraic relation, it's not clear to me that this implies that $[\rho]=0$, since in general A-polynomials satisfy some non-trivial conditions. I suspect there could be a formulation in terms of algebraic K-theory, but I don't know enough about this.

One may also ask if $H_2(SL_2(\mathbb{C})^\delta)$ is generated by representations of $T^2$? I suspect this might be true. It's not hard to see that a representation of a closed surface of genus $>2$ is cobordant to a sum of representations of genus 2 surfaces (since the commutator map in $SL_2(\mathbb{C})$ is onto). Then I think that genus 2 reps. may be cobordant to a pair of genus 1 reps. (at least the numerology works out, but I haven't checked it). Then one could ask for when a sum of genus 1 reps. is homologically trivial? In turn, this should be realized by a zero of an A-variety. But I'm not sure how one recognizes such points.