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let $M$ be a closed ortientable irreducible 3-mfd, let $T$ be a non-separating torus in $M$, we cut $M$ along $T$ and glue two solid tori along the two boundary tori, we get a new closed 3-mfd $M_1$ (we need $M_1$ also is irreducible). Now If there exists nonseperable torus $T_1$ in $M_1$, we go on the above process, we get a new closed 3-mfd $M_2$ ...

My question is, whether you can find a $M$, choose suitable $T_i$, glue solid tori suitablely, the process will go infinitely?

or can you prove that it is impossible to find such an example? (for example, from $M$---> $M_1$, some invariant of 3-mfds decrease strictly).

After Kevin's example, I added the condition "$M_1$ also is irreducible". This condition is natural in the original field (for this question): 3-mfd with Anosov flow.

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Notice that $M$ is Haken ($T$ is obviously incompressible), maybe hierarchy of $M$ works to give a prove that it is impossible to find such an example... – Bin Yu Mar 9 '12 at 11:29
All you need is to observe what happens to rank of H^1 after this operation. – Misha Mar 9 '12 at 13:47
@Misha, that is in some sense the first idea in my head. I just hope homology work to decrease the first Betti number under the surgery. But let us have a look at an example: mapping torus with monodromy map (2,1;1,1)---- we call this mfd $M$, $\beta_1(M)=1$. If we cut $M$ along the torus fiber, we obtain $T^2\times I$, then glue two solid tori suitablely, you can get $M_1 =S^1\times S^2$. $\beta_1(M_1)=\beta_1(M)=1$ ! Maybe, you have another way to use homology... – Bin Yu Mar 9 '12 at 14:12
I agree that $M$ must be Haken, but $T$ need not be incompressible. Consider the $S^2$ in $S^2\times S^1$ and then trivially increase the genus of this separating 2-sphere. (I'm assuming that by "nonseperable" you mean simply non-separating.) – Kevin Walker Mar 9 '12 at 14:53
@Kevin, Yes nonseperable =non-separating. – Bin Yu Mar 9 '12 at 15:42
up vote 8 down vote accepted

The answer is "no", although it seems that a homological argument is not enough as Kevin's and Bin's examples show. I describe an argument which uses geometrization.

There is a quantity which decreases strictly at each operation. It is crucial to suppose that both $M$ and $M_1$ are irreducible. The quantity for an irreducible manifold $M$ is the triple $(\|M\|, n(M), s(M))$ of real numbers, where

  • $\|M\|$ is Gromov's norm (i.e. the sum of the volumes of the hyperbolic pieces in the JSJ decomposition of $M$)
  • $n(M)$ is the number of tori in the JSJ decomposition of $M$
  • $s(M)$ is the sum of the $-\chi(S)$ for each Seifert piece of the decomposition with some base surface $S$.

    Triples $(\|M\|, n(M), s(M))$ are ordered lexicographically. I prove below that the surgery you describe (cut along a non-separating torus and glue two solid tori) indeed strictly decreases this quantity, supposing that the resulting manifold $M_1$ is still irreducible (this hypothesis is important). The result then follows because Gromov norms of 3-manifolds form a well-ordered set.

    Let $T$ be the torus you cut. If it is adjacent to at least one hyerbolic piece, the filled manifold $M_2$ has strictly smaller Gromov norm thanks to Thurston's Dehn filling theorem. It remains to consider the case $T$ is adjacent to two (possibly coinciding) Seifert pieces and Gromov norm does not decrease. If $T$ is a torus of the JSJ then $n(M)$ decreases. If $T$ is a torus inside a Seifert piece, then $s(M)$ decreases. You use here the following fact: the hypothesis that $M_2$ is irreducible ensures that the meridian of your Dehn filled tori are not fiber-parallel, and hence only add some (possibly non-singular) fiber at the adjacent Seifert pieces. Therefore the JSJ of the new manifold is easily controlled by the JSJ of the old manifold.

    It is possible that this argument extends to the relaxed case where you only suppose $T$ to be incompressible (and $M$, $M_1$ are any manifolds).

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    @Bruno, so beautiful argument! Thank you very much. Just one point which I don't understand, can you explain to me more detail (reference is also welcome)? : the fact, the hypothesis that $M_2$ is irreducible ensures that the meridian of your Dehn filled tori are not fiber-parallel. – Bin Yu Mar 10 '12 at 14:44
    If the meridian is fiber-parallel, then it creates a sphere in $M_2$. To create a sphere you only need to choose an arc on the base surface connecting two points of the filled boundary: if the arc is non-trivial, the sphere is typically non-trivial and hence $M_2$ is reducible. A good reference for playing with Seifert manifolds is the book of Matveev-Fomenko… – Bruno Martelli Mar 10 '12 at 14:55
    @Bruno, Now I see. thanks a lot. – Bin Yu Mar 10 '12 at 15:11

    Start with $S^2\times \{1\}$ inside $S^2\times S^1$. Increase the genus of this non-separating 2-sphere to obtain a non-separating torus $T\subset S^2\times S^1$. Cutting along $T$ yields $S^2\times I$ with a 1-handle attached to each boundary component. It is possible to glue solid tori to the two boundary components to obtain $S^2\times S^1 \# S^2\times S^1$. This process can be continued indefinitely, always increasing first Betti number.

    Perhaps you want to require that $T$ is incompressible?

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    @Kevin, Nice. But, sorry, I omit something: I need $M_1$ still is irreducible (so non-separating torus is still incompressible)! – Bin Yu Mar 9 '12 at 15:37

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