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We know from Morse theory that smooth manifold(with or without boundary) is a handlebody. However, I found a paper "Three-dimensional manifolds with boundary of nonnegative Ricci curvature" by Ananov, N. G.(2-AOS2); Burago, Yu. D.(2-AOS2); Zalgaller, V. A.(2-AOS2). They proved

Every 3-dimensional compact Riemannian manifold M (∂M≠∅) with nonnegative (at all points and in all directions) Ricci curvature (Ric≥0) and nonnegative mean curvature of the boundary (H≥0), which is positive at least at one point of ∂M, is diffeomorphic to a handle-body.

Why this need a proof? I am pretty sure that I missed something, but what is it? One thing I know is they proved the boundary is connected, which can be viewed as a splitting theorem.

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    $\begingroup$ It sounds like different notions of "handlebody" are being used in your two usages of the term. You might want to check which assumptions everyone is using. Your 1st occurance contradicts the standard notion of handlebody in 3-manifold theory and should probably be stated as "has a handle decomposition". The 2nd occurrence is likely more close to a traditional 3-manifold theoretic notion of handlebody -- for example, go to the Wikipedia page for Heegaard Splitting. $\endgroup$
    – Ryan Budney
    Commented Nov 28, 2014 at 3:00
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    $\begingroup$ I looked at the wikipedia's page of Handlebody, which is the 1st occurance. This is what meant in their paper "Μ is homeomorphic to a ball or to a ball with a finite number of solid handles". So this means different things? or simply means no handle can be attached along $S^1\times I$? $\endgroup$
    – user60933
    Commented Nov 28, 2014 at 3:06
  • $\begingroup$ Then your 1st sentence is incorrect. $\mathbb RP^3$ is not a handlebody, in any definition of the term. $\endgroup$
    – Ryan Budney
    Commented Nov 28, 2014 at 3:13
  • $\begingroup$ What if I add a condition "orientable". The 2nd defenition allow nonorientability in their paper, I didn't copy it. $\endgroup$
    – user60933
    Commented Nov 28, 2014 at 3:15
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    $\begingroup$ $\mathbb RP^3$ is orientable. $\endgroup$
    – Ryan Budney
    Commented Nov 28, 2014 at 4:53

2 Answers 2

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3-dimensional handlebodies are obtained by gluing 1-handles to a 3-ball. It is true, that every compact, orientable 3-manifold has a handle decomposition, but this needs not just 1-handles but also 2- and (in the closed case) 3-handles.

In fact, what you get from a handle decomposition (with w.l.o.g. just one 0- and one 3-handle) is the so-called Heegaard splitting, that is a decomposition of a closed, orientable 3-manifold into two handlebodies: the union of a 3-ball with the 1-handles is one of them, the other comes from the 2- and 3-handles.

For 3-manifolds with boundary, the handle decomposition provides you with a Heegaard splitting into one handlebody and one compression body. Here, the compression body is constructed by attaching 1-handles to $\partial M\times I$.

The relevant Wikipedia article is https://en.wikipedia.org/wiki/Heegaard_splitting

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There are (at least) two different senses of the term "handlebody".

When speaking of manifolds of general dimension ("n-dimensional"), "handlebody" usually means "has a handle decomposition". So, for example, every smooth manifold has the structure of a handlebody in this sense.

For 3-manifolds, "handlebody" usually means "built out of a 0-handle and some 1-handles".

The Wikipedia article on handlebodies discusses both of these senses.

(Ryan Budney's first comment on the original question says more or less the same thing, but I thought it was worth putting in an answer.)

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  • $\begingroup$ I'm fairly certain the Kosinski textbook has a further intermediate notion of handlebody which reduces to the 3-manifold one in dimension 3. $\endgroup$
    – Ryan Budney
    Commented Nov 28, 2014 at 5:22
  • $\begingroup$ OK, I'll change "two" to "at least two". But if I understand the original question correctly, the two meanings I list are the relevant ones here. $\endgroup$
    – Kevin Walker
    Commented Nov 28, 2014 at 5:28

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