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I have a question about connected sums in 3-dimensional spaces in relation to normal surfaces. I was reading the paper https://arxiv.org/abs/math/9712269 by Joel Hass, which at one point outlines a proof of "Knesers Theorem", which states that a triangulated $3$-dimensional manifold (without boundary) $M$ can be decomposed non-trivially along $2$-spheres (in terms of the connected sum) up to a number of times bounded by a constant $k(M)$ dependent on $M$ and its triangulation.

For the proof, the notion of a normal surface was introduced, which is a surface embedded in $M$ which is a disjoint union of normal triangles or normal quadrilaterals, where the latter are intersecting a tetrahedron of $M$ in 3 or 4 edges and faces.

At Page 7 in the proof, the notion of a "punctured ball" is introduced, which is just defined as "a ball with some open balls removed". At Page 8 in the proof it is mentioned, that if we use a separating (normal) $2$-sphere $S$ to decompose $M$ into a connected sum and $S$ bounds a punctured ball, then one of the summands in the connected sum is trivial, meaning its homeomorphic to $S^3$. I can't get my head around the following things:

  1. Do we have specific extra conditions regarding the definition of a punctured ball? For example that we only allow for finitely many open balls cut out or that they have to be completely enclosed in the main ball?
  2. I can't really see why a $2$-sphere bounding a punctured ball leads to a $S^3$-summand. Is it a simple trick I don't see (maybe having to do with normal $2$-sphere) or is this maybe done by calculation of homology groups? I think I am missing something important here.

I appreciate any kind of help and ideas, thanks in advance for your time!

Mathdealer
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1 Answers1

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  1. Yes, there are finitely many and they have disjoint closures.

  2. The summands (by definition of the connect sum) are gotten by gluing 3-balls to each 2-sphere in a connected region in the complement of the 2-spheres, This results in a 3-manifold with no boundary for each component.

If one of the summands is a punctured 3-ball then gluing in 3-balls to each boundary 2-sphere (ie filling back in the punctures) results in a 3-sphere, which is a trivial summand.

Joel
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