Primitive, proper power, and Seifert curves in the boundary of a genus two handlebody

By Lemma 2.1, there exists a cutting disk $D_{A}$ in $H$ such that $\alpha$ meets $\partial D_{A}$ transversely in a single point. Consider the regular neighborhood $N$ of $\partial D_{A}\cup\alpha$ in $\partial H$. Then $N$ is a once-punctured torus which contains $\partial D_{A}$, and whose boundary $\partial N$ bounds a separating disk of $H$. Let $N^{\prime}=\overline{\partial H-N}$. Then it follows by cutting $H$ open along $D_{A}$ that there exists a unique cutting disk $D_{B}$ of $H$ up to isotopy whose boundary $\partial D_{B}$ lies on $N^{\prime}$.

This partition $\{N,N^{\prime}\}$ of $\partial H$ with $\partial D_{A}\subset N$ and $\partial D_{B}\subset N^{\prime}$ gives rise to an R-R diagram of $\alpha$ in which $N$ and $N^{\prime}$ correspond to the $A$-handle and $B$-handle respectively. Since $\alpha$ lies in $N$ and intersects $\partial D_{A}$ in a single point, $\alpha$ has an R-R diagram of the form shown in Figure 1. This completes the proof. ∎

In order to find all possible R-R diagrams of a proper power curve $\beta$, we consider the following cases separately.

1. (1)

   $\beta$ has a 0-connection in at least one of the handles in its R-R diagram.

2. (2)

   $\beta$ has no 0-connections in either handle in its R-R diagram.

The case where $\beta$ has a $0$-connection in one handle gives restriction to other curves in $\partial H$ disjoint from $\beta$. In other words, if $\gamma$ is a simple closed curve in $\partial H$ disjoint from $\beta$, then $\gamma$ must have only bands of connections labeled by $0$ or $1$ in that handle. Therefore we put this case into one type of possible proper power curves, which gives Type I in Theorem 1.2.

Now we assume that $\beta$ has no 0-connections in either handle.

Suppose $\beta$ has only one generator in $\pi_{1}(H)=F(A,B)$. Then up to replacement of $A$ with $A^{-1}$, $B$ with $B^{-1}$, or exchange of $A$ and $B$, we may assume that $[\beta]=B^{s}$ for some $s>1$. Since $\beta$ has no 0-connections, this implies that $\beta$ has no connections in the $A$-handle and only one connection in the $B$-handle. Thus this case gives rise to Type II of a proper power curve in Theorem 3.2 with an R-R diagram of the form shown in Figure 4.

Now suppose that $\beta$ has the two generators $A$ and $B$ in $\pi_{1}(H)$. By Theorem 3.4, we may assume that $\beta=AB^{m_{1}}\cdots AB^{m_{l}}$, where $\{m_{1},\dots,m_{l}\}\subseteq\{s,s+\epsilon\}$ with $\epsilon=\pm 1$ and min$\{s,s+\epsilon\}>0$. This implies that every connection in the $A$-handle is labeled by 1. There are two cases to consider:

• (1)

  $\beta$ has only one band of 1-connections in the $A$-handle,

• (2)

  $\beta$ has two bands of 1-connections in the $A$-handle.

Case (1): Assume that $\beta$ has only one band of 1-connections in the $A$-handle.

Suppose $\{m_{1},\dots,m_{l}\}\subsetneq\{s,s+\epsilon\}$. Without loss of generality, we may assume that $\{m_{1},\dots,m_{l}\}=\{s\}$. If $s>1$, then there must be only one band of $s$-connections in the $B$-handle, in which case $\beta$ is a primitive curve with $[\beta]=AB^{s}$. Therefore $s=1$ and there must be two bands of 1-connections in the $B$-handle. Then with $A$ and $B$ exchanged, this case belongs to Type III of a proper power curve in Theorem 3.2 with an R-R diagram of the form shown in Figure 5.

Suppose $\{m_{1},\dots,m_{l}\}=\{s,s+\epsilon\}$. Then $\beta$ has at least two bands of connections in the $B$-handle. If $\beta$ has two bands of connections in the $B$-handle, $\beta$ must have an R-R diagram of the form shown in Figure 9 and by Lemma 3.5, $\beta$ is not a proper power curve. If $\beta$ has three bands of connections in the $B$-handle, then these three bands of connections should be labeled by 1, 2 and 1 respectively, and $\beta$ has the R-R diagram of the form shown in Figure 11. By Lemma 3.6, $\beta$ is not a proper power curve.

Case (2): $\beta$ has two bands of 1-connections in the $A$-handle.

Suppose $\{m_{1},\dots,m_{l}\}\subsetneq\{s,s+\epsilon\}$. Without loss of generality, we may assume that $\{m_{1},\dots,m_{p}\}=\{s\}$. Then if $\beta$ has only one band of connections in the $B$-handle, then the band must be labeled by $s$, in which case Type III of a proper power curve in Theorem 3.2 arises with an R-R diagram of the form shown in Figure 5. If $\beta$ has two bands of connections in the $B$-handle, then $s$ must be 1 and this case yields Type IV of a proper power curve in Theorem 3.2 with an R-R diagram of the form shown in Figure 6.

Suppose $\{m_{1},\dots,m_{l}\}=\{s,s+\epsilon\}$. If there are only two bands of connections in the $B$-handle, then $\beta$ must have R-R diagram of the form shown in Figure 12 and by Lemma 3.7, $\beta$ is not a proper power curve.

If there are three bands of connections in the $B$-handle, then $\beta$ has two types of R-R diagrams as shown in Figure 8. By Lemma 3.8, the R-R diagram of $\beta$ in Figure 8a cannot be a proper power curve. Therefore we have the R-R diagram of $\beta$ in Figure 8b, which gives Type V of a proper power curve in Theorem 3.2.

Thus we complete the proof of Theorem 3.2 ∎

Note that since $\beta$ is a simple closed curve, gcd$(a,b)=1$. Let $b=\rho a+\eta$, where $\rho\geq 0$ and $0\leq\eta<a$ (if $\eta=0$, then $\rho>0$ and $a=1$). Now we record the curve $\beta$ algebraically by starting the $a$ parallel arcs, i.e., the band of width $a$, entering into the $(s+\epsilon)$-connection in the $B$-handle. It follows that $\beta$ is the product of two subwords $AB^{s+\epsilon}(AB^{s})^{\rho}$ and $AB^{s+\epsilon}(AB^{s})^{\rho+1}$ with $|AB^{s+\epsilon}(AB^{s})^{\rho}|=a-\eta$ and $|AB^{s+\epsilon}(AB^{s})^{\rho+1}|=\eta$. Here, for example, $|AB^{s+\epsilon}(AB^{s})^{\rho}|$ denotes the total number of appearances of $AB^{s+\epsilon}(AB^{s})^{\rho}$ in the word of $\beta$ in $\pi_{1}(H)=F(A,B)$.

There is a change of cutting disks of the handlebody $H$, which induces an automorphism of $\pi_{1}(H)$ that takes $A\mapsto AB^{-s}$ and leaves $B$ fixed. Then by this change of cutting disks, $AB^{s+\epsilon}(AB^{s})^{\rho}$ and $AB^{s+\epsilon}(AB^{s})^{\rho+1}$ are sent to $A^{\rho+1}B^{\epsilon}$ and $A^{\rho+2}B^{\epsilon}$ respectively. Therefore the resulting Heegaard diagram of $\beta$ realizes a new R-R diagram of the form in Figure 10, where the positions of the $A$ and $B$-handles are switched. The new R-R diagram has the same form as that in Figure 9 with less number of arcs. One can continue inductively until one of the labels of parallel arcs is 0, in which case $[\beta]=AB^{j}$ for some $j>0$ up to replacement of $A$ with $A^{-1}$, $B$ with $B^{-1}$, or exchange of $A$ and $B$. Such a curve $\beta$ is a primitive curve. ∎

First, note that gcd$(a+b,b+c)=1$ in order for $\beta$ to be a simple closed curve. Since $AB$ and $AB^{2}$ appear in $[\beta]$, by considering $\{AB,AB^{2}\}$ as a generating set of $F(A,B)$ and by Theorem 3.4 one of the two must appear with exponent $1$. However, from the R-R diagram one can see that if one reads $\beta$ from the right-hand edge of the band of width $a$ entering the 1-connection in the $B$-handle, then $AB$ appears twice consecutively. Therefore $AB^{2}$ must have only exponent $1$.

If $b=1$, then $AB^{2}$ appears only once and thus $\beta$ is a primitive curve. Suppose $b>1$. Consider following the band of width $b$ around the R-R diagram. Since $\beta$ is a simple closed curve, this band must split into two subbands at the endpoint of the $(-1)$-connection in the $A$-handle. If the band splits into two subbands belonging to the bands of width $a$ and $b$ respectively, then by tracing arcs of this band we see that $\beta$ has two subwords $AB^{2}(AB)^{n}$ and $AB^{2}(AB)^{n+e}$ for some positive integers $n$ and $e$ with $e>1$. It follows from Theorem 3.4 that $\beta$ cannot be a proper power curve. Similarly for the case where the band splits into two subbands belonging to the bands of width $b$ and $c$ respectively. ∎

We use the argument of hybrid diagrams which are introduced in [K20]. Consider the corresponding hybrid diagram as shown in Figure 13a. Then we drag the vertex A^- together with the edges meeting the vertex A^- over the $s$-connection in the $B$-handle. This corresponds to the change of the cutting disks of $H$ inducing an automorphism of $\pi_{1}(H)$ which takes $A\mapsto AB^{-s}$ and leaves $B$ fixed. The resulting hybrid diagram is shown in Figure 13b.

Transforming the hybrid diagram in Figure 13b back into an R-R diagram, there are three possible cases to consider as follows:

1. (1)

   There is only one band of connections in the $A$-handle;

2. (2)

   There are only two bands of connections in the $A$-handle;

3. (3)

   There are three bands of connections in the $A$-handle.

Note from Figure 13b that since $b+c>0$, all of the labels of bands of connections in the $A$-handle are greater than $0$ and at least one of the bands of connections has label greater than $1$.

(1) Suppose that there is only one band of connections in the $A$-handle. Then $a$ must be 1 and $\beta=B^{\epsilon}A^{b+c+1}$, which is primitive.

(2) Suppose that there are only two bands of connections in the $A$-handle. Let $p$ and $q$ be the labels of the two bands of connections. Since $b+c>0$ and thus at least one of the bands of connections has label greater than $1$, $p\neq q$. Theorem 3.4 forces $|p-q|$ to be $1$. However, by Lemma 3.5 such a curve $\beta$ is a primitive curve.

(3) Suppose that there are three bands of connections in the $A$-handle. Let $p,q,$ and $r$ be the labels of the three bands of connections with $q=p+r$. In order for $\beta$ to be a proper power, $(p,q,r)=(1,2,1)$. However, by Lemma 3.6, $\beta$ is not a proper power curve. ∎

Consider the corresponding hybrid diagram as shown in Figure 15a.

Then we drag the vertex A^- together with the edges meeting the vertex A^- over the $2$-connection in the $B$-handle. This corresponds to the change of the cutting disks of $H$ inducing an automorphism of $\pi_{1}(H)$ which takes $A\mapsto AB^{-2}$ and leaves $B$ fixed. The resulting hybrid diagram is shown in Figure 15b.

Transforming the hybrid diagram in Figure 15b back into an R-R diagram, there are three possible cases to consider as follows:

1. (1)

   There is only one band of connections in the $A$-handle;

2. (2)

   There are only two bands of connections in the $A$-handle;

3. (3)

   There are three bands of connections in the $A$-handle.

Note from Figure 15b that since $b>0$, all of the labels of bands of connections in the $A$-handle are greater than $0$ and at least one of the bands of connections has label greater than $1$.

(1) Suppose that there is only one band of connections in the $A$-handle. Let $p$ be the label of the band of connections. Since there are the $b$ edges connecting A⁺ and A^-, $p>1$. On the other hand, in Figure 14, consider chasing back and forth the outermost arc in the band of width $a$ entering the 1-connection in the $A$-handle. This represents $ABAB\cdots$, which is carried into $AB^{-1}AB^{-1}\cdots$ by the automorphism $A\mapsto AB^{-2}$. This implies that $\beta$ has a 1-connection in the $A$-handle, a contradiction.

(2) Suppose that there are only two bands of connections in the $A$-handle. Let $p$ and $q$ be the labels of the two bands of connections. By the similar argument in the proof of Lemma 3.7, $|p-q|=1$. However, by Lemma 3.7 such a curve $\beta$ cannot be a proper power.

(3) Suppose that there are three bands of connections in the $A$-handle. Let $p,q,$ and $r$ be the labels of the three bands of connections with $q=p+r$. In order for $\beta$ to be a proper power, $(p,q,r)=(1,2,1)$. Then by switching the $A$- and $B$-handles, and the signs of the labels, the R-R diagram of $\beta$ has the same form as in Figure 14 with less number of the arcs connecting the $A$- and $B$-handles. So if we can continue to perform the change of cutting disks of $H$, then since the number of the edges is strictly decreasing under the change of cutting disks, this case must eventually belong to the cases (1) and (2). ∎

We take an inner product by $U^{\perp}$ on both sides of $W=xU+yV$ . Then since $U^{\perp}\circ U=0$ and $U^{\perp}\circ V=\pm 1$, the result follows. ∎

We start by showing that if $H[\alpha]$ is Seifert-fibered over $D^{2}$ with two exceptional fibers, then $\alpha$ has an R-R diagram of the claimed form.

The key idea is that $H[\alpha]$, which is defined to be the manifold obtained by adding a 2-handle to $H$ along $\alpha$, induces a genus two Heegaard decomposition of a Seifert-fibered space over $D^{2}$ with two exceptional fibers. However Heegaard decompositions of a Seifert-fibered space over $D^{2}$ are well understood. For instance, Theorem 4.4, of Boileau, Rost and Zieschang, completely describes the genus two Heegaard diagrams of a Seifert-fibered space over $D^{2}$ with two exceptional fibers. Using this result, Theorem 4.5 shows how the Heegaard diagrams described in Theorem 4.4 translate into R-R diagrams. And then Lemma 4.6 adds a finishing detail to the proof of this direction by showing that it is always possible to assume that $n>1$ in Figure 16b.

With the proof of one direction of Theorem 4.2 finished, it remains to show that if $\alpha$ has an R-R diagram with the form of Figure 16a with $n,s>1$, or Figure 16b with $n,s>1$, $a,b>0$, and $\gcd(a,b)=1$, then $H[\alpha]$ is Seifert-fibered over $D^{2}$ with two exceptional fibers.

To see that this is the case, consider Figure 17 in which each of the R-R diagrams of Figure 16 has been augmented with a simple closed curve $\beta$ disjoint from $\alpha$. Then, in each diagram of Figure 17, two parallel copies of $\beta$ bound an essential separating annulus $\mathcal{A}$ in $H$. (Figure 18 illustrates the situation when $s=2$, and $D_{A}$ and $D_{B}$ are cutting disks of $H$ underlying the A-handle and B-handle of the R-R diagram of $\alpha$.)

Cutting $H$ apart along $\mathcal{A}$ yields a genus two handlebody $W$ and a solid torus $V$. Note that $\alpha$ lies in $\partial W$ as a primitive curve in $W$ implying that $W[\alpha]$ is a solid torus, because any component of $D_{B}\cap W$ is a cutting disk $D_{C}$ of $W$ such that an R-R diagram of $\alpha$ with respect to $\{D_{A},D_{C}\}$ of $W$ has the form in Figure 16 with $s$ replaced by $1$, in which case $\alpha$ in Figure 16a intersects $D_{C}$ only once and thus is primitive, and $\alpha$ in Figure16b is primitive by Lemma 3.5. It follows that $H[\alpha]$ is obtained by gluing the two solid tori $W[\alpha]$ and $V$ together along $\mathcal{A}$. So $H[\alpha]$ is Seifert-fibered over $D^{2}$ with $\beta$ as a regular fiber and the cores of $W[\alpha]$ and $V$ as exceptional fibers.

The last step is to compute the indexes of the two exceptional fibers of $H[\alpha]$. It is clear that the annulus $\mathcal{A}$ wraps around the solid torus $V$ $s$ times longitudinally, so the core of $V$ is an exceptional fiber of index $s>1$. For the other index, it follows by computing $\pi_{1}((W[\alpha])[\beta])(=\pi_{1}((W[\beta])[\alpha])=\mathbb{Z}_{n})$ that if $\alpha$ has an R-R diagram with the form of Figure 16a, then the core of $W[\alpha]$ is an exceptional fiber of index $n$.

Finally, suppose the R-R diagram of $\alpha$ has the form of Figure 16b. In this case, Lemma 4.1 can be used to compute the index of the second exceptional fiber formed by the core of $W[\alpha]$.

Abelianizing $\pi_{1}(W)$, we have: $[\vec{\alpha}]$ = $(n(a+b)+b,a+b)$ and $[\vec{\beta}\,]$ = $(0,1)$. By Lemma 4.1:

in $H_{1}(W[\alpha])$. Thus the regular fiber $\beta$ wraps around the solid torus $W[\alpha]$ longitudinally $n(a+b)+b>1$ times, so the core of $W[\alpha]$ is an exceptional fiber of index $n(a+b)+b$. ∎

First, observe that the curves $\alpha$ in Figures 19, 20, and 21 represent $s^{p}t^{-q}$, $W_{p,\nu}(u^{-1},t^{q})$, and $W_{q,\omega}(v^{-1},s^{p})$ respectively in $\pi_{1}(H)$.

Next, consider the diagram of Figure 19, and let $C_{A}$ and $C_{B}$ be cores of the A-handle and B-handle of $H$. Then $C_{A}$ and $C_{B}$ are the exceptional fibers of the Seifert-fibration of $H[\alpha]$. Let $N(C_{A})$ and $N(C_{B})$ be closed regular neighborhoods of $C_{A}$ and $C_{B}$ respectively in $H$, and let $M_{A}$ and $M_{B}$ be meridional disks of $N(C_{A})$ and $N(C_{B})$ respectively.

Observe that the pair of dotted curves $\beta$, $\gamma_{s}$ on the A-handle of Figure 19 can be considered to lie on $\partial N(C_{A})$, while the pair of dotted curves $\beta^{\prime}$, $\gamma_{t}$ on the B-handle of Figure 19 can be considered to lie on $\partial N(C_{B})$. Also observe that as indicated in Remark 4.3, the curves $\beta$ and $\beta^{\prime}$ represent regular fibers of the Seifert-fibration of $H[\alpha]$. Then $\partial M_{A}$ = $(\beta^{\nu}\gamma_{s}^{p})^{\pm 1}$ in $\pi_{1}(\partial N(C_{A}))$, while $\partial M_{B}$ = $(\beta^{\prime\omega}\gamma_{t}^{q})^{\pm 1}$ in $\pi_{1}(\partial N(C_{B}))$. So $C_{A}$ and $C_{B}$ are exceptional fibers of types $\nu/p$ and $\omega/q$ in the Seifert-fibration of $H[\alpha]$.

This leaves the diagrams of Figures 20 and 21. Since these diagrams are similar, we will only consider Figure 20 in detail. To start, note that the configuration of the curves $\beta$ and $\gamma_{t}$ on the B-handle of Figure 20 is identical to that of $\beta^{\prime}$ and $\gamma_{t}$ on the B-handle of Figure 19. Since $\beta$ is again a regular fiber in the Seifert-fibration of $H[\alpha]$ when $\alpha$ has an R-R diagram on $\partial H$ with the form of Figure 20, the core of the B-handle $C_{B}$ of $H$ is again an exceptional fiber of type $\omega/q$. The other exceptional fiber that exists in $H[\alpha]$ when $\alpha$ has an R-R diagram on $\partial H$ with the form of Figure 20, arises in a slightly different way.

As in Figure 18, two parallel copies of the regular fiber $\beta$ in Figure 20 bound an essential separating annulus $\mathcal{A}$ in $H$. Cutting $H$ open along $\mathcal{A}$ cuts $H$ into a genus two handlebody $W$ and a solid torus $V$ which has $C_{B}$ as its core. The curve $\alpha$ lies on $\partial W$, and the R-R diagram of $\alpha$ on $\partial W$ appears in Figure 22. Since $\alpha$ is primitive in $W$, $W[\alpha]$ is a solid torus $V^{\prime}$. Let $C_{V^{\prime}}$ be the core of $V^{\prime}$. Then $C_{V^{\prime}}$ is the second exceptional fiber of the Seifert-fibration of $H[\alpha]$.

Let $M$ be the meridional disk of $V^{\prime}$, and note that the curves $\beta$ and $\gamma_{u}$ of Figure 22 lie on $\partial W$ and $\beta$ and $\gamma_{u}$ form a basis for $\pi_{1}(\partial W[\alpha])$. The next step is to obtain an expression for $\partial M$ in $\pi_{1}(\partial W[\alpha])$ in terms of the basis $\beta$, $\gamma_{u}$ of $\pi_{1}(\partial W[\alpha])$.

We can do this by using Lemma 4.1. Abelianizing $\pi_{1}(W)$, we have: $[\vec{\alpha}]$ = $(-n(a+b)-a,a+b)$ = $(-p,\nu)$, $[\vec{\gamma}_{u}]$ = $(-1,0)$, and $[\vec{\beta}\,]$ = $(0,1)$. By Lemma 4.1:

in $H_{1}(W[\alpha])$, where $\delta=\pm 1$. It follows that $\partial M$ = $(\beta^{\nu}\gamma_{u}^{p})^{\pm 1}$ in $\pi_{1}(\partial W[R])$. So $C_{V^{\prime}}$ is an exceptional fiber of type $\nu/p$ in the Seifert-fibration of $H[\alpha]$ when $\alpha$ has an R-R diagram with the form of Figure 20.

Similarly, one sees that if $\alpha$ has an R-R diagram with the form of Figure 21, then $H[\alpha]$ is also Seifert-fibered over $D^{2}$ with two exceptional fibers of types $\nu/p$ and $\omega/q$. ∎

Suppose a simple closed curve $\alpha$ in the boundary of a genus two handlebody $H$ has an R-R diagram with the form of Figure 20 with $n=1$. For the form of Figure 21, the similar argument can apply. Note that since $n=1$, $p=2a+b$ and $\nu=a+b$. It is easy to see that the underlying Heegaard diagram of $\alpha$ on $\partial H$ does not have minimal complexity. Thus in order to have minimal complexity, we can perform a change of cutting disks of $H$, i.e., replace the cutting disk $D_{B}$ of $H$ which underlies the $B$-handle with a new cutting disk $D^{\prime}_{B}$ by bandsumming $D_{B}$ with the cutting disk $D_{A}$ in the $A$-handle along the arc of $\alpha$.

Specifically, for the weights $a$ and $b$ in the R-R diagram, since gcd$(a,b)=1$, we can let $b=\rho a+r$, where $\rho\geq 0$ and $0\leq r<a$. However we may assume $r>0$, otherwise $a=1$ and thus $p=2+b$ and $\nu=1+b$, which implies that $\nu\equiv-1\mod p$ and it follows from Theorem 4.4 that this Heegaard decomposition is homeomorphic to $HD_{0}$.

Now we record $\alpha$ by starting the $a$ parallel arcs entering into the $-2$-connection in the $A$-handle. It follows from the R-R diagram that $\alpha$ is the product of two subwords $A^{-2}B^{q}(A^{-1}B^{q})^{\rho}$ and $A^{-2}B^{q}(A^{-1}B^{q})^{\rho+1}$ with $|A^{-2}B^{q}(A^{-1}B^{q})^{\rho}|=a-r$ and $|A^{-2}B^{q}(A^{-1}B^{q})^{\rho+1}|=r$. We perform a change of cutting disks of the handlebody $H$ which induces an automorphism of $\pi_{1}(H)$ that takes $A^{-1}\mapsto A^{-1}B^{-q}$. Then by this change of cutting disks, $A^{-2}B^{q}(A^{-1}B^{q})^{\rho}$ and $A^{-2}B^{q}(A^{-1}B^{q})^{\rho+1}$ are sent to $A^{-1}B^{-q}(A^{-1})^{\rho+1}$ and $A^{-1}B^{-q}(A^{-1})^{\rho+2}$. Therefore the resulting Heegaard diagram of $\alpha$ has minimal complexity and realizes a new R-R diagram of the form in Figure 24, which is the same form as in Figure 20 with $n>1$. ∎