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Recently, A. Carlotto and C. Li proved a complete topological classification of those compact, connected and orientable $3$-manifolds with boundary which support Riemannian metrics of positive scalar curvature and mean-convex boundary. Namely, if $M^3$ is such a manifold, then there exist integers $A, B, C, D \geq 0$ such that $M$ is diffeomorphic to a connected sum of the form \begin{align*} P_{\gamma_1} \# \cdots \# P_{\gamma_A} \# \mathbb{S}^3/ {\Gamma_1} \# \cdots \# \mathbb{S}^3 / {\Gamma_B} \# \left( \#_{i=1}^C \mathbb{S}^2 \times \mathbb{S}^1 \right) \setminus \left( \sqcup_{i=1}^D B_i^3 \right), \end{align*} where $P_{\gamma_i}$, $i \leq A$, are genus $\gamma_i$ handebodies; $\Gamma_i$, $i \leq B$, are finite subgroups of $SO(4)$ acting freely on $\mathbb{S}^3$, $B_i^3$, $i \leq D$, are disjoint $3$-balls in the interior.

My question: Can we classify, in terms of $A,B, C, D$, the $3$-manifolds $M$ of the form above in which any smoothly embedded $2$-sphere in the interior separates $M$?

For instance, if $(A,B,C,D) = (1, 0, 0, 0)$, then this holds. Indeed, if $M = P_{\gamma_1}$, then $H_2(M) = 0$, so the connecting homomorphism $H_2(M, \partial M;\mathbb{Z}) \to H_1(\partial M; \mathbb{Z})$ is injective. Since an embedded $2$-sphere has no boundary, it lies in the kernel of this map, and thus equals to $0$ in $H_2(M, \partial M; \mathbb{Z})$. This means it separates $M$.

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  • $\begingroup$ Every embedded 2-sphere will separate M if and only if C=0, since the handlebody summands and spherical summands are irreducible. $\endgroup$
    – Josh Howie
    Sep 21, 2020 at 0:08
  • $\begingroup$ @JoshHowie deleting balls and making connected sums preserve irreducibility? $\endgroup$ Sep 21, 2020 at 1:04
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    $\begingroup$ No it doesn't preserve irreducibility, but removing 3-balls which are disjoint from any 2-sphere we are interested in, will not change whether the 2-sphere will separate. I have written an answer with more details. $\endgroup$
    – Josh Howie
    Sep 21, 2020 at 1:06
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    $\begingroup$ That result seems obvious, given results in the literature. From a technique of Bray (see also Miao), the double of the manifold admits a positive scalar curvature metric, which is symmetric with respect to reflection. Then Coda-Marques' classification of 3-manifolds with positive scalar curvature and the understanding of fixed points of involutions on them (which comes from the orbifold theorem) completes the picture. See: mathoverflow.net/a/368937/1345 $\endgroup$
    – Ian Agol
    Sep 21, 2020 at 2:22

1 Answer 1

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Every embedded 2-sphere will separate $M$ if and only if $C=0$.

Proof: Suppose $C=0$, and let $\Sigma$ be 2-sphere embedded in $M$. Let $\{S_j\}$ be a collection of 2-spheres which decompose $M$ into prime summands. Look at the intersection of $\Sigma$ with $\{S_j\}$. Let $\Delta$ be an innermost disk on some $S_k$. We can surger $\Sigma$ along $\Delta$, which will decompose $\Sigma$ into two 2-spheres. Repeating this process until there are no intersections with $\{S_j\}$, $\Sigma$ is decomposed into a collection of embedded $2$-spheres which we call $\Sigma'$. Then $\Sigma$ will be non-separating in $M$ if and only if some component of $\Sigma'$ is non-separating in $M$.

Each component of $\Sigma'$ is contained within a single prime summand of $M$. We then cut $M$ along $\{S_j\}$, and glue $3$-balls onto each $2$-sphere boundary component of the resulting $3$-manifolds. It is well-known that handlebodies and closed spherical $3$-manifolds are irreducible, which means that every embedded $2$-sphere bounds a $3$-ball. Thus each component of $\Sigma'$ is separating in its respective prime summand and hence in $M$. Removing the $D$ $3$-balls from $M$ which are disjoint from $\Sigma$ and $\Sigma'$ does not affect whether $\Sigma$ is separating. Therefore $\Sigma$ is separating in $M$.

Conversely, if $C\neq 0$, then we can find a non-separating 2-sphere $\Sigma''$ in some $S^2\times S^1$ component which is disjoint from each $B_i$ and each $S_j$. Furthermore the dual $1$-sphere to $\Sigma''$ is disjoint from each $B_i$ and each $S_j$. Therefore $\Sigma''$ is non-separating in $M$.

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  • $\begingroup$ What do you mean by dual 1-sphere? $\endgroup$ Sep 21, 2020 at 2:25
  • $\begingroup$ A surface in a 3-manifold is non-separating if there exists a circle which intersects that surface transversely exactly once. $\endgroup$
    – Josh Howie
    Sep 21, 2020 at 2:33
  • $\begingroup$ Yes, this is true. I just didn’t understand how do you conclude that such a curve exists for $\Sigma’’$. $\endgroup$ Sep 21, 2020 at 2:36
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    $\begingroup$ Such a circle exists in $S^2\times S^1$, and can be isotoped to be disjoint from any collection of balls, so the circle still exists in $M$. $\endgroup$
    – Josh Howie
    Sep 21, 2020 at 2:40
  • $\begingroup$ Perfect, thank you for the detailed answer. $\endgroup$ Sep 21, 2020 at 2:41

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