What is a higher genus analogue of the Pontryagin product?

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 $\langle x,y\rangle\in H_2 (\pi_1 M)$, 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.

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Isn't $\pi_1(M)$ non-abelian in general? If so, how do you define the Pontryagin product on its homology? It seems you are just looking at the image of the fundamental class of $\partial M$ under inclusion into $M$, is this true? – Mark Grant Aug 3 '12 at 9:33
Question edited to clarify this point. – Daniel Moskovich Aug 5 '12 at 9:46
I don't really understand your question: if $M$ is e.g. a knot complement, then $H_2(\pi_1(M))=0$, so the Pontryagin product vanishes for any immersed torus. So this doesn't characterize the peripheral torus. – Ian Agol Aug 6 '12 at 17:03
Agol: True- for the fundamental group itself, it's not interesting. The interest comes when one looks at homomorphs G of pi_1(M). These might have non-trivian H_2, but still the image of the Pontryagin product of the images of x and y under the homomorphism will vanish (I think this is noted in "Homomorphs of Knot Groups" by Johnson, where the significance of this condition is discussed- Johnson-Livingston says if you add a condition, this characterizes knot homomorphs). I'm asking about the analogous statement for a higher genus surface. But you're right- it isn't clearly stated. – Daniel Moskovich Aug 6 '12 at 18:04

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, 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.

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Thank you for this!! – Daniel Moskovich Aug 6 '12 at 18:17

Is there such a statement? I don't think so. As long as your manifold supports homotopy equivalences that do not restrict to homotopy equivalences on the boundary you have this problem. So manifolds with compressible boundary are a problem like high genus handelbodies - the handelbody mapping class group is very different from the outer automorphism group of a free group. So there is no such result for these manifolds. You'll want a more restrictive class of three manifolds to get a statement.

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Thanks for this! This is useful. I'm willing to restrict to knotted theta complements in $S^3$. Do you know whether one can write down, at least in this case, at least a necessary condition for "&Sigma; is a boundary" (what the vanishing of the Pontryagin product says in the torus case) in terms of the basis for &Sigma; inside pi_1 of the manifold? – Daniel Moskovich Aug 6 '12 at 4:31
There certainly isn't such a statement for the trivial theta curve, since the complement is a handlebody. – Ryan Budney Aug 6 '12 at 23:35