Handle body $H_g$ of genus $g$ as a CW complex

Question

Often is the torus (as a $2$-dimensional manifold) given as an example of a CW-complex (see here, for example, for the genus $g$ manifold). However, one can ask how to decompose the $3$-manifold (handle body, $H_g$) of genus $g$, which is bounded by the $2$-dimensional surface of genus $g$, $M_g$. It seems that it is possible to just glue a single $3$-cell to $M_g$ in order to create $H_g$, but I wasn't able to describe it fully.

Is this true that attaching a single $3$-cell will suffice? How can I describe in more details the CW-structure of $H_g$?

Lee Mosher · Answer 1 · 2023-05-29T00:28:40.937

2

Attaching a single 3-cell does not suffice. You see why by starting with the fact that $H_1(M_g)$ has rank $2g$, and then use the Mayer-Vietoris sequence to verify that after attaching a single 3-cell the rank of $H_1$ is unchanged. However, the rank of $H_1(H_g)$ is equal to $g < 2g$.

You can, however, get a CW structure on $M_g$ like this. Start with the usual family of simple closed curves $a_1, b_1, ..., a_g, b_g \subset M_g$ which form a homology basis for $H_1(M_g)$. Now attach $g$ 2-cells, along the curves $a_1,...,a_g$, and now one can attach a single $3$-cell to get $H_g$.

edited May 29 '23 at 00:28

answered May 29 '23 at 00:22

Lee Mosher

120,280

Thank you! Is there a way to present this explanation homology-free? I am struggling to visualize it – Robert May 29 '23 at 00:32
1

That's what's so great about homology, quick easy proofs. It would be good to learn to use it! – Lee Mosher May 29 '23 at 00:37
Indeed, but I have not reached it yet, and I am expected to solve these kinds of problems without it. – Robert May 29 '23 at 00:39
I'm curious from where this expectation arises. Regardless, I doubt there's a proof without algebraic topology. – Lee Mosher May 29 '23 at 00:40
Not all professors solve themselves the problems they give to students. – Robert May 29 '23 at 00:43

Handle body $H_g$ of genus $g$ as a CW complex

1 Answers1