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?

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    $\begingroup$ This is unlikely. The expected dimension of $X(M)$ is half of the dimension of $X(F)$, where $X$ stands for the character variety. Since there are only countably many compact 3-manifolds $M$, "most" points of $X(F)$ probably not come from restrictions. $\endgroup$ – Moishe Kohan Feb 23 '20 at 12:36
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    $\begingroup$ Hi Charlie--It seems to me that you are asking if there are non-trivial classes in $H_2(BG^\delta)$, or perhaps how to detect such classes. You might like to have a look at Milnor's paper, On the homology of Lie groups made discrete. He cites an argument in a paper of Alperin-Dennis (due to Mather) for the case of G=SL(2,R) that might be relevant. $\endgroup$ – Danny Ruberman Feb 23 '20 at 17:41

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 $U$.

  • $\begingroup$ Sorry, I am pretty unsophisticated about ``computablity''. Where can I learn about the algorithm for coefficients in $\overline{Q}$? $\endgroup$ – Charlie Frohman Feb 23 '20 at 14:52
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    $\begingroup$ @CharlieFrohman: The algorithm, as I said, is silly and totally non-practical: Enumerate all triangulated compact 3-manifolds with boundary $F$. For each of these manifolds, we have a polynomial morphism $r$ as above. For a morphism of affine varieties $r: Y\to X$, there is an algorithm for computation of defining equations and inequalities of $r(Y)$. (I will find a reference if you like.) Then check these on the given point in $X(F)$. If $[\rho]\in r(X(M))$ for some $M$, then this algorithm will terminate and find the manifold $M$. $\endgroup$ – Moishe Kohan Feb 23 '20 at 14:58

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.

  • $\begingroup$ Thanks! I need to read this Milnor paper. $\endgroup$ – Charlie Frohman Feb 24 '20 at 1:01

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.


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