# Gluing two 3 manifolds along their boundary

Let $X,Y$ be two compact, smooth, orientable 3 manifolds, each with an incompressible boundary component diffeomorphic to some genus $g$ surface $S_g$. Under an orientation-reversig diffeomorphism $f:S_g \to S_g$, those two manifolds can be glued together to obtain a new smooth, orientable manifold $X \cup_f Y$. I wonder now in how far the diffeomorphism type of this result depends on the choice of $f$. Using a collar, one can show that if $f$ and $g$ are two isotopic diffeomorphisms of $S_g$, then the corresponding gluings are diffeomorphic . Is this also a necessary condition ?

Some thoughts I have made so far: 1) If both X and Y are irreducible, then so is $X \cup_f Y$, and if $X \cup_f Y$ still has some boundary component, it must be an aspherical manifold. Hence, all important Information is contained in the fundamental group. Is it true that the homeomorphism type of aspherical 3 - manifolds obtained this way is already determined by their fundamental group ?

2) Also, I wonder if its true that the isotopy type of two orientation-reversing diffeomorphisms on a surface $S_g$ is determined by their action on the fundamental group. Edit: This is probably false, since mapping class groups of surfaces are very distinct from the corresponding fundamental group.

Any help is appreciated.

Edit: I apologize for missing and/or inaccurately placed capital letters. I wrote this question yesterday evening on my phone, and it was impossible to fix all the mistakes made by auto-correct (I am writing on a german phone).

Edit 2: I have also updated this question, according to what has already been solved by the answers and what is still open.

• 3-manifolds with boundary are never acyclic. Perhaps you mean irreducible? – ThiKu Nov 7 '15 at 1:08
• On the other hand it is true that the homotopy and even homeomorphism type of irreducible 3-manifolds is determined by their fundamental group. – ThiKu Nov 7 '15 at 1:10
• I meant aspherical. True, I forgot that classifying spaces are unique up to homotopy equivalence – Berni Waterman Nov 7 '15 at 1:13
• But why is even the homeomorphism type determined by the fundamental group ? – Berni Waterman Nov 7 '15 at 1:14
• I know Borel conjecture, but it does not really apply here, since our manifold is not closed. – Berni Waterman Nov 7 '15 at 1:29

The original question:

Using a collar, one can show that if $f$ and $g$ are two isotopic diffeomorphisms of $S_g$, then the corresponding gluings are diffeomorphic. Is this also a necessary condition?

No, it is not. Suppose we have glued to obtain $M = X \cup_f Y$. Suppose that $X$ admits a self-homeomorphism $\Phi$. Define $\phi = \Phi|\partial X$. Then the map $g = \phi \circ f$ gives the manifold $N = X \cup_g Y$ and this is homeomorphic to $M$.

Now, it is simple to find such $\Phi$ if $X$ has compressible boundary - namely we can do a Dehn twist on a disk. You've ruled that out. But we can still find examples by twisting along an essential properly embedded annulus in $X$. For example, if $X$ is a twisted $I$-bundle over a non-orientable surface.

If you further assume that $X$ is "acylindrical" then there are still examples, but they are harder to find. We can build a hyperbolic manifold $X$ which has a self-homeomorphism $\Phi$ of finite order (eg Thurston's knotted Y).

If you further assume that $X$ (and $Y$) has no symmetries, then examples should still exist, but they will be very hard to find. Basically, we need to find a manifold $M$ that contains homeomorphic surfaces $S$ and $S'$, but where there is no homeomorphism of $M$ taking $S$ to $S'$. We then need to "get lucky" and find that $M - n(S)$ and $M - n(S')$ are homeomorphic and win. One way to do this is by a search through one of the many censuses of closed three-manifolds (eg snappy or regina). Another way that should work is to think deeply about hyperbolic three-manifolds with "corners".

Ad (2): the mapping class group of surfaces is isomorphic to the outer automorphism group of its fundamental group - this is a theorem of Baer-Dehn-Nielsen. In particular there is even a finite set of closed loops such that an element of the mapping class group is determined by its action on this set.

Ad (1): irreducible, orientable 3-manifolds (with the exception of $S^2\times S^1$) have $\pi_2=0$ and - assuming $\mid \pi_1\mid=\infty$ - it is an exercise in algebraic topology to prove that they are then aspherical. It is then a classical fact that the homotopy type is determined by $\pi_1$. From the answer to 3-manifolds with isomorphic fundamental groups one sees that even the homeomorphism type is determined.

Ad the original question, whether the gluing map $f$ is determined up to isotopy by the homeomorphism type of $X\cup_fY$, I do not know. (I think I remember there are $3$-manifolds with non-equivalent fiberings, although I don't know a reference. In any case, this only yields examples with disconnected boundary.)

• You said that your answer only deals with closed, orientable 3-manifolds. Recall that mine is not closed. I am also pretty sure that only non-compact, irreducible 3 manifolds are automatically aspherical ( take S^3 as a counterexample). – Berni Waterman Nov 7 '15 at 8:16
• I forgot the assumption on infinite fundamental group, which is now added. Obviously that is implied by your assumptions (gluing along incompressible boundary). – ThiKu Nov 7 '15 at 8:42
• If you still have remaining boundary components after gluing, then homotopy type does not necessarily determine homeomorphism type. This has been discussed in Johannson's book "Homotopy equivalences of 3-manifolds with boundaries", an up-to-date account is section 2.2 in uni-regensburg.de/Fakultaeten/nat_Fak_I/friedl/papers/… – ThiKu Nov 7 '15 at 8:46
• Although in general, the homeorphism type of aspherical 3-manifolds-with-boundary might not be determined by their fundamental group, we know that in our case, we have two such manifolds $A$, $B$ with the additional property that each remaining boundary component of $A$ can be paired up with a diffeomorphic boundary component of $B$. Maybe this helps... – Berni Waterman Nov 7 '15 at 9:09
• Addendum: For the 3-manifolds with non-equivalent fiberings see mathoverflow.net/questions/241822/… – ThiKu Jul 22 '16 at 14:24