Given a compact oriented aspherical $3$--manifold $M$ with torus boundary $\partial M\simeq T^2$ (*e.g.* a knot complement), the condition that the images in $\pi\_1 M$ of basis $x,y\in \pi\_1 T^2$ under the inclusion generate the boundary of a compact oriented $3$-manifold with fundamental group $\pi\_1 M$ (the $3$-manifold $M$ itself, to be precise) is algebraically encoded by the vanishing of their [Pontryagin product](http://en.wikipedia.org/wiki/Pontryagin_product) $\langle x,y\rangle\in H\_2 (\pi\_1 M)$, which you can think of as the fundamental class of the torus under the inclusion. Recall that the Pontryagin product sends two commuting elements of a group to an element of $H\_2$ of the group.



> 
> **Question:** Given a compact oriented aspherical $3$--manifold $M$ with boundary a closed surface $\partial M \simeq \Sigma$ of genus $g\geq 2$, how can I express the statement that $\Sigma$ bounds a $3$--manifold (the $3$-manifold $M$) on the level of fundamental groups? What condition must images under the inclusion of a basis for $\pi\_1 \Sigma$ satisfy? How can I express the fundamental class of $\Sigma$ in terms of elements of $\pi\_1$, if I know what $\Sigma$ is topologically?
> 


I'm trying to identify some homomorphs of the fundamental group of a knotted theta, with peripheral data, and the answer to this question would, I presume, be something all such homomorphs would have to satisfy.



What is a higher genus analogue of the Pontryagin product?
An obvious necessary condition is that the pair $(M,\Sigma)$ satisfies Lefschetz duality. Since $M$ and $\Sigma$ are aspherical, this can be reformulated in terms of group cohomology.


Let $\Gamma=\pi\_1(M)$ and $S=\pi\_1(\Sigma)$. Assume that the inclusion $\Sigma\subset M$ is $\pi\_1$-injective, so that $S$ is a subgroup of $\Gamma$. Then the pair $(\Gamma,S)$ is a *Poincaré duality pair* of dimension $3$. See Definition 4.6 of [this survey by Davis](http://www.math.osu.edu/~davis.12/pdgroup.pdf), which refers to papers of Bieri and Eckmann. 


This should place some pretty serious restrictions on how the homology of $\Sigma$ sits in the homology of $M$, from which you might be able to extract a useful answer to your question. 


