5
$\begingroup$

I would like to ask questions about the definition of the Heegaard splitting. The following are the facts I know.

  1. A Heegaard splitting says that any 3-manifold is built up from two handlebodies and a homeomorphism between boundaries of the handlebodies.

  2. If $f$ and $g$ are isotopic such homeomorphisms, the 3-manifolds obtained are homeomorphic.

This is the fact what I know and want to prove it. But I don't know how to prove the second part.

  1. How do I show that two isotopic homeomorphisms of boundaries of handlebodies produce the homeomorphic 3-manifolds?

  2. Also, more generally, let $M$ and $M'$ be 3-manifolds with boundary. Suppose that $A\subset \partial M$ and $B \subset \partial M'$ are homeomorphic sub manifolds. Let $f:A \to B$ be a homeomorphism from $A$ to $B$. We glue $M$ and $M'$ via $f$. Does the homeomorphism class of the resulting manifold depend only on the isotopy class of the homeomorphism $f$?

  3. Does the answer of the previous questions depend on what 3-manifolds I want to consider? Like, smooth, topological, piece-wise linear etc.

Edit: I am not familiar with ''collar'' in the comment below. I appreciate if one can explain more detail. I also want to know if collar exists for any type of manifolds.

$\endgroup$
  • 6
    $\begingroup$ I believe what you do is choose a collar of the boundary and perform the isotopy along the collar. $\endgroup$ – Tim kinsella Nov 5 '14 at 20:33
  • 2
    $\begingroup$ In the smooth category this is a quite general result: gluing manifolds along isotopic diffeomorphisms gives diffeomorphic manifolds. Have a look at Hirsch, Differential Topology, chapter 8 sections 1 and 2, especially theorem 2.3 $\endgroup$ – Lor Nov 5 '14 at 20:59
  • $\begingroup$ Same in the topological (and PL) category; works in all dimensions too. $\endgroup$ – Moishe Kohan Nov 7 '14 at 19:51
  • $\begingroup$ For the concept of a collar, see Lee's Introduction to Smooth Manifolds. $\endgroup$ – Tim kinsella Nov 7 '14 at 23:33
0
$\begingroup$

The comments above and the references they contains are good answers in my opinion.

Let me just add a picture of how to build an explicit homeo, given two isotopic gluings. As Tim kinsella says, use collars and the cylinder given by the isotopy.

enter image description here

I hope this very bad picture can help.

The first row of the picture tells you how to build a manifold inserting between $M_1$ and $M_2$ a cylinder $\partial M_1\times [0,1]$. This is the cylinder of the isotopy of the two gluings $f_1$ and $f_2$. Say that you glue on the left via $f_1$ and on the right via $f_2$. Call this manifold $\hat M$

The two bottom part of the picture says that you can "insert" the cylinder of the isotopy in a collar of $\partial M_1$ as well as in a collar of $\partial M_2$.

So the second line is the result of gluing $M_1$ and $M_2$ via $f_2$ and the third is the result of gluing $M_1$ and $M_2$ via $f_1$.

Both are homeo to $\hat M$.

$\endgroup$
  • $\begingroup$ Thank you for a good explanation. Could you answer the question part 2 as well? I am not sure if I can use the method above for restricted collar to submanifold in the boundary. $\endgroup$ – Snow Nov 12 '14 at 15:53
  • $\begingroup$ as far as you have collars (you need collars of $\partial A$ in $\partial M$), then yes the same construction works. (b.t.w. what do you mean exactly by submanifold of $\partial M$? is just an open subset of $\partial M$? or a compact sub-manifold with boundary of $\partial M$, or a connected component of $\partial M$? also, I guess $B\subset \partial M'$, right?) In low dimension (1,2, and 3) there are no problems of existence of collars (in the case $A$ is a compact submanifold of $\partial M$) and the three categories: Top, Pl, Smooth coincide. Problems can arise in higer dimension. $\endgroup$ – user126154 Nov 12 '14 at 16:32
  • $\begingroup$ Yes $B \subset \partial M$. I fixed it. By submanifold of $\partial M$, I meant an embedding A into $\partial M$. Does it make sense? $\endgroup$ – Snow Nov 12 '14 at 20:02
  • $\begingroup$ What I confuse in Question 2 is that if if $A \subset \partial M$ is not all of $\partial M$, it looks like for me the isotopy cannot be absorbed in $M$ nicely? $\endgroup$ – Snow Nov 12 '14 at 20:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.