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Let $M^3$ and $N^3$ be compact orientable manifolds with connected boundary. When we perform the connected sum operation, will the resulting manifold $X = M \#N$ have a connected or disconnected boundary?

For instance, if $M$ and $N$ are two solid tori, their connected sum will be a genus $2$ handlebody, and hence have a connected boundary, or will it be another $3$-manifold whose boundary has $2$ connected components?

Eduardo Longa
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  • If $M$ and $N$ are disks $D^3$ then isn't their connected sum a cylinder $S^2 \times [0, 1]$, with boundary $S^2 \sqcup S^2$? – Qiaochu Yuan Dec 09 '20 at 02:23
  • Aren't you doing the connected sum of the boundaries of the solid tori? What do you mean by connected sum of manifolds with boundary? – Ted Shifrin Dec 09 '20 at 05:46
  • I am trying to understant Theorem 1.1 in this beautiful paper: https://arxiv.org/pdf/1903.11772.pdf – Eduardo Longa Dec 09 '20 at 05:50

1 Answers1

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The trouble is that for manifolds with connected boundary $M_1, M_2$ there are two notions of connected sum:

  1. The ordinary connected sum, denoted $M=M_1\# M_2$ where you take out open balls $B_1, B_2$ from the interiors of $M_1, M_2$ and glue $M_i-B_i$ via a homeomorphism of the new boundary spheres $\partial B_1, \partial B_2$. Then, clearly, $M$ has two boundary components, homeomorphic to $\partial M_i, i=1, 2$.

  2. The boundary connected sum, denoted $N=M_1\#_{\partial} M_2$, where you pick two closed (tame) disks $D_i\subset \partial M_i$ and glue $M_1, M_2$ via a homeomorphism $D_1\to D_2$. Then $N$ will have connected boundary, homeomorphic to $\partial M_1 \# \partial M_2$.

(In both cases, there are some issues with the choice of gluing homeomorphisms, I will ignore these. Separately, there are issues with the category in which everything is done: TOP, PL or DIFF, in dimension 3 this does not matter.)

Thus, the answer to your question is: It depends on your notion of connected sum.

Moishe Kohan
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