# When does one obtain different 3-manifolds by pasting two tori?

Consider a compact solid torus $T$ and a diffeomorphic copy of it $T' \subset T$ embedded in the interior of $T$ in such a way that it makes two turns around the central circle of $T$.

I would like to consider the compact boundaryless 3-manifolds $M_f$ obtained from the difference $M = T \setminus \mathring{T'}$ of $T$ and the interior of $T'$ by gluing the two torus boundaries $T_1$ and $T_2$ of $M$ with a diffeomorphism $f$.

If two diffeomorphisms $f,f'$ are isotopic then $M_f$ and $M_{f'}$ are diffeomorphic.

Question: Is this the only way one can have $M_f$ diffeomorphic to $M_{f'}$? If not, is there a characterization of when one obtains diffeomorphic manifolds? Is there a way to know what Thurston geometry the manifold $M_f$ will have?

I've spent some time trying to calculate the fundamental group of the resulting manifolds via Van Kampen's theorem but I didn't get very far.

Let me address your third question, about the Thurston geometries. What I have to say is covered in standard references such as Peter Scott's article "The geometries of 3-manifolds".

Thurston's result about geometries of compact 3-manifolds requires one to first decompose the 3-manifold canonically into pieces; only after doing that will the pieces have one of eight geometries. Your manifold is already prime, so only the decomposition along incompressible tori is needed. Every prime manifold can be cut along a unique minimal collection of incompressible tori (unique up to isotopy, that is) so that the complementary pieces are either Seifert fibered or atoroidal.

In your case, the starting manifold $M$ is indeed Seifert fibered, in fact it is locally trivially fibered over the 3-punctured sphere, with circle fibers. One can work out from your description of $M$ how the circle fibers restrict to the two boundary tori, for example the circle fibers on the outer boundary torus form a foliation by circles of slope $(1,2)$ running parallel to the inner torus. The geometry of $M$ is $\mathbb{H}^2 \times \mathbb{R}$, because the 3-punctured sphere embeds in $M$ transverse to the fibers so that the monodromy map is (up to isotopy) of finite order $2$.

After you glue the two boundary tori to obtain the manifold $M_f$, it may or may not be Seifert fibered, depending on whether the gluing map $f$ matches up the circle fibers of the inner boundary with the circle fibers of the outer boundary (up to isotopy).

If $f$ does not match up the inner and outer boundary fibers then $M_f$ is not Seifert fibered, and it is not atoroidal, so it has no geometry itself; you will have to recut first along the glued torus, to get the Seifert fibered manifold $M$. In this case $M_f$ is a "graph manifold", which simply means a prime manifold such that the pieces of its torus decomposition are all Seifert fibered.

If $f$ does match up the inner and outer boundary fibers then $M_f$ is Seifert fibered and its geometry has one of the two Seifert fibered geometries over $\mathbb{H}^2$, either $\mathbb{H}^2 \times \mathbb{R}$ or $PSL(2,\mathbb{R})$: if $f$ matches up the outer meridian curve to the inner meridian curve then $M_f$ has $\mathbb{H}^2 \times \mathbb{R}$ geometry; otherwise $M_f$ has $PSL(2,\mathbb{R})$ geometry.

• Thanks for this very interesting and surprising answer! I never expected $SL(2,\mathbb{R})$ to show up. – Pablo Lessa Apr 21 '13 at 11:58

Here's a way to get different gluing maps giving the same glued-up manifold.

There's an annulus embedded in M that meets each boundary in a closed curve (on the outer torus $T_1$ this closed curve runs twice around the longitudinal direction). There's a self-diffeomorphism of $M$ supported near the annulus that restricts to a Dehn twist $D_1$ on the closed curve in $T_1$ and to a Dehn twist $D_2$ on the closed curve on $T_2$. It follows that if $f : T_1 \rightarrow T_2$ is a gluing map, then for $f' = D_2 \circ f \circ D_1^{-1}$ we have $M_f = M_{f'}$.

Let $M_1, M_2$ be two oriented manifolds with boundary $\newcommand{\pa}{\partial}$ and suppose that $\Phi,\Psi:\pa M_1\to\pa M_2$ are two orientation reversing diffeomorphisms. We get two manifolds

$$X_\Phi=M_1\cup_\Phi M_2, \;\; X_\Psi= M_1\cup_\psi M_2.$$

If $\Psi^{-1}\circ \Phi: \pa M_1\to \pa M_1$ extends to a diffeomorphism of $M_1$ then $X_\Phi$ and $X_\Psi$ are diffeomorphic.

The answer to your question lies in Kirby calculus. Your manifold seems to have a rather reasonable surgery description, but Kirby calculus is not my cup of tea.

• Hey Liviu! Thanks for the answer. I'll look into it (Kirby Calculus that is). – Pablo Lessa Apr 21 '13 at 12:01