# Prime decomposition of 3-manifolds

Let $H_g$ be a three dimensional handlebody bounded by a genus $g$ surface.

Let $M_g$ be a manifold obtained by gluing two copies of $H_g$ via an orientation reversing homeomorphism of the surface of $H_g$.

I would like to know what is a prime decomposition of the manifold $M_g$.

When $g=1$, we have $M_1$ is homeomorphic to $S^2 \times S^1$ and this is a prime decomposition.

What's the decomposition of $M_2$? Is it a connected sum of two $S^2 \times S^1$?

I appreciate any help. Thank you in advance.

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What's the fundamental group? –  user641 Oct 12 '12 at 22:03

Note that in the case $g=1$ you don't always get $S^1 \times S^2$, but you may obtain also $S^3$ or the lens spaces, depending on the homeomorphism you choose for the gluing. The point is that the torus has a lot of non isotopic homeomorphism. The same is true for higher $g$ as well.
What I'm suggesting is that a priori the decomposition will depend on the chosen gluing...I'm not aware of any kind of independence result. As you can see in the $g=1$ case, if your gluing fixes the two generators of $\pi_1 (M)$ then you get $S^1 \times S^2$ which is the prime decomposition of itself (being prime); if your gluing swaps them, then you get $S^3$ which is the prime decomposition of itself (being prime). So you get two different decompositions of two different manifolds. "The manifold obtained gluing two copies of $H_1$" is an ill-posed term, and so is "the decomposition of the manifold obtained gluing two copies of $H_1$".
In general, you have to specify which is the gluing $\varphi \in Homeo (\partial H_g)$ you are performing, at least modulo isotopy of $\partial H_g$ (since isotopic homeomorphisms give homeomorphic manifolds $M_g$).