Manifolds Resulting from Gluing Tori

Question

I'm trying to show that if solid tori $T_1, T_2; T_i=S^1 \times D^2$ ,are glued by homeomorphisms between their respective boundaries, then the homeomorphism type of the identification space depends on the choice of homeomorphism up to, I think, isotopy ( Please forgive the rambling; I'm trying to put together a lot of different things from different sources, and I don't yet have a very coherent general picture.) I first thought of Lens spaces, but the gluing here is not done by a homeomorphism.

I have some fuzzy ideas here that I would like to make precise: I know this has to see with Heegard splittings; specifically, this is a genus-1 splitting ( actually, genus-1 gluing ) and the gluing may be determined by a mapping in $SL(2,\mathbb Z)$, which determines the induced map on the top homology , and different induced maps would result in different homeomorphic types on the glued spaces.

I think we can also see this from the perspective of Dehn surgery ( please feel free to correct anything I write here ), where we remove a link $L$ and a tubular 'hood $T(L)$ of $L$ , and then glue another torus. I know then an n-framing is equivalent to removing a solid torus, twisting n times and then regluing. But it's obvious from the post that I don't know how to show that the homeomorphism class of the space glued along $h: \partial T_1 \rightarrow \partial T_2$ depends on $h$.

Thanks, and sorry for the rambling ( not my fault, I was born a rambling man.)

Answer 1

To see that isotopic maps produce homoeomorphic manifolds, consider the manifold given by $$D^2 \times S^1 \cup_f S^1 \times S^1 \times I \cup_g D^2 \times S^1.$$ If $f$ and $g$ are isotopic, then omitting either the first or the second term produces a space homeomorphic to a solid torus rel boundary, so the whole thing is homeomorphic to either gluing.

In the other direction, if you believe $S^2 \times S^1$ and $S^3$ are not homeomorphic, either of these is relatively easy to produce (e.g. $S^3 = \{x^2 + y^2 + z^2 + w^2 = 1\} = \{x^2 + y^2 \le 1/2\} \cup \{z^2+w^2 \le 1/2\}$).

Answer 2

Let me provide here two different objects that can be obtained by gluing two solid tori.

Before start, let me fix a notation that $T_1$, $T_2$ be two solid tori $S^1 \times D^2$ and $f : S^1 \times S^1 \to S^1 \times S^1$ be a map from $\partial T_1$ to $\partial T_2$, along which we glue two solid tori.

First, $S^3$ is obtained when $f$ is a map $(x, y) \mapsto (y, x)$. To see this, let us first delete a meridian disk $D_M$ from $T_1$ and then glue $T_1 \setminus D_M$ and $T_2$ by $f(x, y) = (y, x)$. It results in a 3-ball, then gluing $D_M$ again produces $S^3$.

The other thing we can have is $S^2 \times S^1$, which is realized when $f$ is an identity. Note first that $S^2 \times S^1$ is obtained by identifying two boundary components of $S^2 \times [0, 1]$. Taking a circle $C \subset S^2$, $C \times [0, 1]$ becomes a torus after identifying two boundary components $S^2 \times \{0\}$ and $S^2 \times \{1\}$. We can now drill out this torus from $S^2 \times S^1$, which makes us get two solid tori with boundary $C \times S^1$.