# Saturated Torus in a Seifert fibered manifold

Let $M$ be a Seifert fibered 3-manifold and $B$ be its orbit-manifold. If we consider a simple closed curve in $B$ and take its pre-image under the projection map, then its a saturated torus in $M$. I'm stuck in a claim in Jaco's book that says that if the simple closed curve bounds a disk in $B$ and the disk contains at least 2 exceptional points then the saturated torus is incompressible in $M$. What is special about 2? What happens if the no.is 1? Any idea would be very helpful.

• If the number is $1$ then the saturated torus bounds a solid torus, and so it is compressible. Nov 17, 2016 at 3:30
• Thank you Professor @LeeMosher Nov 17, 2016 at 4:19

First of all: the statement as phrased above is incorrect. It can happen that your simple closed curve $\gamma \subset B$ divides $B$ into two disks $D_1$ and $D_2$ with $D_1$ containing at most one singular point of the Seifert fibration $p:M \to B$, and $D_2$ more than two. In this case the torus $T=p^{-1}(\gamma)$ is compressible (as Lee Mosher said because $p^{-1}(D_1)$ is a solid torus) despite $\gamma$ bounds a disk containing more than two singular points.

The correct way to phrase what you said is the following.

Denote by $p:M \to B$ the Seifert projection. Suppose that $\gamma \subset B$ is a simple closed curve and that none of the components of $B- \gamma$ is a disk containing at most one singular point of the projection $p$. Then the torus $T=p^{-1}(\gamma)$ is incompressible.

Cutting along $T$ we can reduce the proof of the previous statement to the one of the following claim.

A Seifert manifold with boundary $M$ is $\partial$-icompressible if and only if its orbit manifold $B$ is different from a disk with at most one singular point.

The proof goes as follows. As I already explained here, $\pi_1(M)$ has a presentation with generators $a_1, b_1 \dots a_g , b_g, c_1 \dots c_m, d_1 \dots d_k, t$ and relations: $$[a_1, b_1] \dots [a_g, b_g]= d_1 \dots d_k \cdot c_1 \dots c_m \ ,$$ $$\ [t, a_i]=[t, b_i]=[t, c_i]=[t, d_i]=1 \ ,$$ $$c_i^{p_i} \cdot t^{q_i} =1 \ \ \ \ \ \ \ \ \ \ \ \ i=1, \dots ,m.$$ Here $t$ denotes (the homotopy class of) a regular fiber of M and $d_1 \dots d_k$ the simple close curves in $\partial M$ corresponding to the boundary components of $M$. Notice that $t$ has infinite order in $\pi_1(M)$ (you can find this basically in any expository paper about Seifert manifolds).

Suppose by contradiction that there is an essential curve $\mu\subset p^{-1}(d_i)$ that bounds a disk $D$ properly embedded in $M$. Write $\mu= d_i^\alpha t^\beta$. Since $t$ has infinite order and $1=\partial D=\mu$, in $\pi_1(M)$ we have the relation $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1=d_i^\alpha t^\beta \ \ \ \ \ \ \ \ \text{ with }\alpha \not=0$$ and we can conclude that $d_i$ has finite order in $\pi_1(B)=\pi_1(M)/<t, c_1, c\dots c_m>$. The only case in which this can happen is when $B$ is a disk $D^2$ and $d=d_i$ corresponds to its only boundary component. In this case $\pi_1(M)$ has a presentation with generators $c_1 \dots c_m, d, t$ and relations: $$d= (c_1 \dots c_m)^{-1}, \ \ \ \ c_i^{p_i} \cdot t^{q_i} =1, \ \ \ \ [t, c_i]=1 \ .$$ Now, the relation $d= (c_1 \dots c_m)^{-1}$ tell us that in $$\pi_1(M)/<t> \ \simeq \ \mathbb{Z}/p_1 \mathbb{Z} * \dots * \mathbb{Z}/p_m \mathbb{Z}$$ the product of the generators has finite order, a contradiction.

• by 'Seifert manifold with boundary M is $\partial$ incompressible', do you mean that $\partial M$ is $\partial$-incompressible? Nov 19, 2016 at 6:05
• That there is no essential curve $\mu \subset \partial M$ that bounds a properly embedded disk $D \subset M$. Nov 19, 2016 at 6:15