Let $F$ be a compact oriented surface and $\rho:\pi\_1(F)\rightarrow SL\_2\mathbb{C}$ be a representation. Does there exist a compact oriented three-manifold $M$ with $\partial M=F$ and a homomorphism $\tilde{\rho}:\pi\_1(M)\rightarrow SL\_2\mathbb{C}$ so that the restriction of $\tilde{\rho}$ to $\pi\_1(F)$ is equal to $\rho$?


If not is there an obstruction that allows you to identify the representations that do extend?


For instance let $BSL\_2\mathbb{C}^\delta$ denote the classifying space of $SL\_2\mathbb{C}$ as a discrete group. Corresponding to $\rho:\pi\_1(F)\rightarrow SL\_2\mathbb{C}$ is a continuous map $f:F\rightarrow BSL\_2\mathbb{C}^\delta$. If $\rho$ extends over a three-manifold $M$ then the homology class represented by $f\_\*[F]$ is zero. Is there a computable way to detect this?



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](https://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem). Then $[\rho]=f^{-1}(p)$ does not lie in $U$. 


