Let $M$ be a compact 4manifold with finite fundamental group such that the prime factorization of its boundary $\partial M$ is connected and has no aspherical factors. Assuming $\partial M$ is incompressible in $M$, then $\partial M$ is elliptic by the elliptization theorem. Barring making this assumption, are there other conditions that guarantee $\partial M$ is elliptic? By elliptic I mean a spherical space form. [I've edited this question because I earlier mistakenly used the terminology spherical.]

$\begingroup$ What convention are you using for a 4manifold having an incompressible boundary? $\endgroup$ – Ryan Budney Dec 18 '16 at 7:54

$\begingroup$ By incompressible I meant that the homomorphism between fundamental groups induced by the inclusion map $i:\partial M \rightarrow M$ is injective. $\endgroup$ – Topology Student Dec 18 '16 at 14:28

1$\begingroup$ I'm not sure I understand the question; which hypotheses do you want to delete and which do you want to retain. Also, why does $\partial M$ being incompressible and also a sum of spherical manifolds imply that there's only one summand? $\endgroup$ – Danny Ruberman Dec 18 '16 at 21:11

$\begingroup$ I am interested in keeping the $M$ has $\pi_1(M) < \infty$ and $\partial M$ is a connected sum of elliptic manifolds and $S^2 \times S^1$'s conditions. I would like to know what some sufficient conditions for $\partial M$ to be an elliptic manifold are. The assumption that $\partial M$ is incompressible, which with $\pi_1(M) < \infty$ implies $\pi_1(\partial M) < \infty$, seems too strong as it supersedes the other condition on $\partial M$. $\endgroup$ – Topology Student Dec 19 '16 at 23:07

$\begingroup$ Perhaps you could provide some context for the question; does it come from some particular circumstance? There is a 3manifold that is a connected sum as described. Do you know something else? Are you given some information about a 4manifold that it bounds? $\endgroup$ – Danny Ruberman Dec 20 '16 at 15:28
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If $M$ admits a locally smooth circle action that restricts to an action on $\partial M$, then $\partial M$ will be spherical.