Proof. Notice that all components of \(\partial F\) are parallel essential curves on the torus \(\partial E(K)\). Therefore, we can choose a curve \(c\) transverse to \(\partial F\) that intersects each boundary component exactly once. After fixing orientations of \(c\) and \(F\), we assign a sign to each point of intersection in \(F \cap c\): an intersection is positive if \(c\) is parallel to the normal vector of \(F\) (see Figure 1), and negative otherwise.

Traversing the curve \(c\) and label its successive intersection points with \(F\) by \(p_1, p_2, \dots, p_b\). For each \(i = 1, \dots, b\), the subarc \(c_i\) of \(c\) from \(p_i\) to \(p_{i+1}\) contains no other intersection points with \(F\), where we set \(p_{b+1} = p_1\). If the intersection points \(p_i\) and \(p_{i+1}\) have the same sign, then the arc \(c_i\) connects opposite sides of \(F\); in particular, \(F\) is non-separating. Suppose instead that none of the arcs \(c_i\) connects opposite sides of \(F\). Then, for each \(i\), the points \(p_i\) and \(p_{i+1}\) must have opposite signs (as illustrated in Figure 1). As a consequence, the signs at the endpoints of each arc \(c_i\) cancel, and we obtain \[0 = \sum_{i=1}^b \bigl(\operatorname{sign}(p_i) + \operatorname{sign}(p_{i+1})\bigr) = 2 \sum_{i=1}^b \operatorname{sign}(p_i).\] This implies \(\sum_{i=1}^b (\pm 1) = 0,\) and hence \(b\) must be even, contradicting the assumption that \(b\) is odd.

 ◻

Proof. Observe that \(F\) must have a non-empty boundary. Otherwise, \(F\) would be separating in \(E(K)\).

Fix an orientation of \(F\), which determines a homology class \(w = [F, \partial F] \in H_1(E(K), \partial E(K)) \cong \mathbb{Z}\). Because \(F\) is non-separating, there exists a curve \(c\) in \(E(K)\) that intersects \(F\) only once. We can find \(c\) by taking a curve connecting opposite sides of \(F\) and sliding its ends near \(F\) to coincide in \(F\), forming the loop \(c\). After fixing an orientation on \(c\), it represents an element \(\alpha = [c]\) in \(H_1(E(K))\cong \mathbb{Z}\). Take the cap product \(H^1(E(K)) \times H_1(E(K)) \to H^0(E(K)) \cong \mathbb{Z}\). By Poincaré duality we identify \(H_2(E(K), \partial E(K))\) with \(H^1(E(K))\), and hence obtain the intersection pairing. This defines a bilinear form \(\beta: H_2(E(K), \partial E(K)) \times H_1(E(K)) \to \mathbb{Z}\) that takes any oriented surface \((S, \partial S)\) in \((E(K), \partial E(K))\) and an oriented loop \(\gamma\) in \(E(K)\) and assign \(\beta([S,\partial S], [\gamma])\) as the sum of intersections with signs between \(S\) and \(\gamma\). This is described via transversality.

As \(\beta([F, \partial F], [c]) = 1\), both \([F, \partial F]\) and \([c]\) must be primitive in their respective homology group. As \(H_2(E(K), \partial E(K)) \cong \mathbb{Z}\) is generated by a Seifert surface \([S_0, \partial S_0]\), it must be that \([F, \partial F] = \pm [S_0, \partial S_0]\) in \(H_2(E(K), \partial E(K))\). Now consider the connecting homomorphism \(\delta: H_2(E(K), \partial E(K)) \to H_1(\partial E(K))\). The group \(H_1(\partial E(K))\) is generated by a meridian \(\mu\) (a closed curve that bounds a disk \(D\) in \(S^3\) and intersects \(K\) in a single point) and a longitude \(\lambda\) ( \(= \partial S_0\)). Now \([\partial F] = \delta([F, \partial F]) = \pm \delta([S_0, \partial S_0])= \pm [ \partial S_0] = \pm [\lambda] = 0 [\mu]+ (\pm1)[\lambda]\). This implies that the slope of \(\partial F\) is zero in \(\partial E(K)\). ◻

Proof. We take the path systems given by \[\begin{align} \gamma_0&:& \langle \frac{-1}{p}\rangle &\longrightarrow \frac{x_0}{x_0+y_0} \langle \frac{-1}{p}\rangle + \frac{y_0}{x_0+y_0}\langle \frac{-1}{p-1}\rangle \\ \gamma_1&:& \langle \frac{1}{p}\rangle &\longrightarrow \frac{x_1}{x_1+y_1} \langle \frac{1}{p}\rangle + \frac{y_1}{x_1+y_1} \langle \frac{0}{1}\rangle \\ \gamma_2&: &\langle \frac{1}{q}\rangle &\longrightarrow \frac{x_2}{x_2+y_2} \langle \frac{1}{q}\rangle + \frac{y_2}{x_2+y_2} \langle \frac{0}{1}\rangle \\ \end{align}\]

We write ending points of the edgepaths in \((a,b,c)\) coordinates,

\[\begin{align} \text{End}(\gamma_0)&:& (x_0+y_0, x_0(p-1)+y_0(p-2), -x_0-y_0) \\ \text{End}(\gamma_1)&:& (x_1+y_1, x_1(p-1), x_1) \\ \text{End}(\gamma_2)&:& (x_2+y_2, x_2(q-1), x_2) \\ \end{align}\]

Since each \(a\)-coordinate equals the number of sheets, they must be all equal. By condition E3, all \(b\) coordinates need to be equal, and the sum of all \(c\) coordinates must be zero. These observations translate into the following equations: \[\begin{align} x_0 + y_0 &= x_1 + y_1 = x_2+ y_2\\ x_0(p-1) + y_0(p-2) &= x_1(p-1) =x_2(q-1)\\ x_1+x_2 &= x_0+y_0 \end{align}\]

By elementary number-theoretic arguments, one verifies that all solutions are multiples of the solution: \[\begin{align} x_0= \frac{a+b-a^2}{g} &, &y_0 = \frac{a^2}{g} &, &x_1 =\frac{b}{g} &, &y_1 = \frac{a}{g} &, & x_2 =\frac{a}{g} &, &y_2 = \frac{b}{g}\\ \end{align}\]

where \(g = gcd(a,b)\) (recall that \(a =p-1\) and \(b=q-1\)). Here, we need the condition \(q \geq a^2 -a + 1\) to make \(x_0\) non-negative. So, the three edgepaths have a fractional longitude given by: \[\begin{align} |\gamma_0| = \frac{y_0}{x_0+y_0} = \frac{a^2}{a+b} &, & |\gamma_1| = \frac{a}{a+b}&,& |\gamma_2| = \frac{b}{a+b} \end{align}\]

Then the number of sheets has to be an integer \(s\) such that \(s|\gamma_i|\) is an integer for \(i=0, 1,2\). This implies that \(s\) is a multiple of \((a+b)/gcd(a,b)\). This means that to construct a connected surface, it is enough to consider when the number of sheets is \(s = (a+b)/gcd(a,b)\). Now, the twist \(\tau(\gamma_i)\), in this case, is twice the length of each edgepath with a sign (negative if it goes downwards and positive if it goes upwards in \(\mathcal{D}\)):

\[\begin{align} \tau(\gamma_0) = \frac{2a^2}{a+b} &, & \tau(\gamma_1) = -\frac{2a}{a+b}&,& \tau(\gamma_2) = -\frac{2b}{a+b} \end{align}\]

The total twist is: \[\tau(\gamma_0, \gamma_1, \gamma_2) = \frac{2a^2-2a -2b}{a+b}\]

And by [@HO] we know that the actual slope of the surface \(slope(F)\) satisfy that \(\displaystyle slope(F) \equiv \tau(\gamma_0, \gamma_1, \gamma_2) \equiv \frac{2a^2}{a+b}\pmod{1}\). This means that the denominator of \(slope(F)\), as an irreducible fraction, is equal to \((a+b)/gcd(a+b, 2a^2)\). Finally, the number of boundary components must be: \[|\partial F|=sheets(F)/den(\partial F) = \frac{(a+b)/gcd(a,b)}{(a+b)/gcd(a+b, 2a^2)} = \frac{gcd(a+b, 2a^2)}{gcd(a,b)}.\]

This determines the slope of the candidate. It remains to be proven that it is incompressible and orientable. For incompressibility, we use [@HO], which we state below.

The mentioned cycle of \(r\)-values can be computed from the last edge of each edgepath \(\gamma_i\) as follows. If \(\langle p/q, u/v\rangle\) is the final edge of \(\gamma_i\) (going from \(p/q\) towards \(u/v\)), \(r_i = q - v\) is the \(r\)-value corresponding to \(\gamma_i\). Then, \((r_1, r_2, \dots, r_n)\) is the cycle of \(r\)-values for the final edges. In our case, the final edges (the only edge per edgepath) are \(\langle1/p, 1/p-1\rangle\), \(\langle 1/p, 0/1 \rangle\) and \(\langle 1/q, 0/1 \rangle\). So the cycle of \(r\)-values is \((1, p-1, q-1)\), which is none of the exceptional cases from Corollary 1. So, our candidate surface is incompressible.

The orientability of the candidate surface is discussed in Theorem 10. ◻

Proof. Candidate surfaces are constructed by gluing embedded surfaces \(S_i\) inside each tangle \(p_i/q_i\); we denote by \(B_i\) the ball containing the tangle. Here, \(p_0/q_0 = -1/p\), \(p_1/q_1 = 1/p\), and \(p_2/q_2 = 1/q\). The edgepath \(\gamma_i\) encodes the surfaces \(S_i \subset B_i\), which are generally disconnected. In our case, the edgepaths \(\gamma_0\), \(\gamma_1\), and \(\gamma_2\) are described in the proof of Theorem 9.

One can check that the surfaces \(S_0\), \(S_1\), and \(S_2\) are all disconnected. Our strategy to prove orientability is to orient each component of \(S_i\) so that they induce opposite orientations on \(B_i \cap B_j\). This ensures that the chosen orientations extend to the entire surface, thereby proving orientability.

We need first to recall how each \(S_i\) is constructed. We start by considering a collar neighborhood \(C = S^2 \times I\) of the \(\partial B_i\) inside \(B_i\) where \(S\times \{1\} = \partial B_i\). Put the first part of the arcs of the tangle parallel to \(I\) following the product structure of \(C\), and the tangled part of the arcs resting over the sphere \(S^2 \times \{0\}\) (as in Figure 3). Now take \(s\) parallel copies of slope \(p_i/q_i\) over \(S^2 \times \{0\}\); over the neighborhood of the \(p_i/q_i\) arcs part of the Montesinos’s decomposition of the original knot \(K\) (see Figure 3). Product these parallel copies with the \([0, \epsilon]\) inside \(C\), creating in this way a surface \(S'_i\) with \(2s\) disk components contained in \(B_i\).

To construct \(S_i\) from \(S'_i\), we attach saddle bands according to the edgepath \(\gamma_i\) and transverse to the product structure of the collar \(C = S^2 \times I\), starting from level \(S^2 \times \epsilon\) and moving upwards toward \(S^2 \times 1 = \partial B_i\).

For the case that concern us, the edgepath \(\gamma_i\) prescribe a total of \(y_i\) saddle of the same type for \(i=0,1,2\) . For example, for \(\gamma_1\), we need to attach \(y_1 = \tfrac{a}{g}\) saddles, each transforming a pair of arcs of slope \(1/p\) into a pair of arcs of slope \(0\). So, we start at level \(S^2 \times \epsilon\) with \(s = a/g + b/g\) parallel copies of a pair of arcs of slope \(1/p\) and we end up at level \(S^2 \times 1\) with \(a/g\) of slope \(1/p\) plus \(b/g\) of slope \(0\) . This is achieved using the saddle move illustrated in Figure 4.

Now, to fix an orientation for \(S_i\) we will first fix one for \(S'_i\); which has \(2^{2s}\) possibilities (two per component of \(S'_i\)). We can picture these orientations by their induced orientation in \(\partial S'_i\) on \(S^2 \times \epsilon\); which are \(s\)-parallel copies of directed pairs of arcs of slope \(p_i/q_i\). But not every orientation can be extended to an orientation of \(S_i\). We need to make sure that the saddle preserve that orientation.

For a pair of arcs \(\alpha\) and \(\beta\) of slope \(p_i/q_i\), the two arcs must be oriented so that the saddle being use connects opposite sides of \(\alpha\) and \(\beta\). For the specific case of saddle changing the slope from \(1/p\) to \(0\), the saddle preserve orientation if the arcs \(\alpha\) and \(\beta\) are directed as shown in Figure 4, or by reversing both simultaneously. Fix any of thoses orientation on \(\alpha\) and \(\beta\).

Next, consider the \(s\) parallel copies \(\alpha_1,\ldots,\alpha_s\) of \(\alpha\) and \(\beta_1,\ldots,\beta_s\) of \(\beta\) in \(\partial S'_i \subset S \times {\epsilon}\), numbered from leftmost to rightmost (see Figure 5). Each arc \(\alpha_j\) may be oriented either consistently with \(\alpha\) or with the opposite orientation. We encode this choice by a vector

\[\mathbf{x}=(x_1,\ldots,x_s)\in\{\pm1\}^s,\]

and similarly we encode the orientations of the arcs \(\beta_i\) by

\[\mathbf{y}=(y_1,\ldots,y_s)\in\{\pm1\}^s.\]

Orientability is preserved after attaching the saddles provided that the orientation choices on the parallel arcs \(\alpha_j\) and \(\beta_j\) agree both with the one of \(\alpha\) and \(\beta\) or its reverse, that is, provided that \(\mathbf{x}=\mathbf{y}\). A convenient choice satisfying this condition is to alternate signs, for instance \[\mathbf{x}=\mathbf{y}=(+1,-1,+1,\ldots,+1,-1).\]

After performing \(\tfrac{a}{g}\) saddle moves, we obtain an induced orientation on \(\partial B\), shown schematically in Figure 6. In this configuration, the orientation data are given by \[\mathbf{x_1}=(-1,+1,-1,+1,\ldots,-1), \qquad \mathbf{y_1}=(+1,-1,+1,-1,\ldots,+1),\] where \(|\mathbf{x_1}|=\tfrac{a}{g}\) and \(|\mathbf{y_1}|=\tfrac{b}{g}\). Here \(|\cdot|\) denotes the length of the vector.

Notice that \(\mathbf{x_1}\) starts in \(-1\) and ends in \(-1\) because \(a/g\) is odd, and the same applies for \(b/g\).

Cutting the diagram along the \(z\)–axis (indicated by the dotted red line in Figure 6) produces two disks, or equivalently two half-planes, which we denote by \(D_1^L\) and \(D_1^R\) (see Figure 7).

An analogous analysis applies to \(\gamma_2\). To obtain compatible orientations, we reverse all signs, namely \[x_2=(-1,+1,-1,+1,\ldots,-1), \qquad y_2=(+1,-1,+1,-1,\ldots,+1),\] yielding the configuration shown in Figure 8. At this stage we have constructed two oriented surfaces \(F_1\) and \(F_2\) which match perfectly along \[F_1 \cap F_2 \subset D_1^R = D_2^L = B_1 \cap B_2.\]

Finally, we apply the same analysis to \(\gamma_0\), using the saddle move from slope \(-1/p\) to \(-1/(p-1)\) shown in Figure 9. Choosing alternating orientations for the parallel copies of \(\alpha_0\) and \(\beta_0\) ensures that orientability is preserved after \(\tfrac{a^2}{g}\) saddle moves, yielding the induced orientations on each half of \(\partial B_0\) depicted in Figure 10. .

This completes the construction and proves that the resulting surface remains orientable throughout the process. ◻

Proof. We apply Theorem 9 using \(a =n\) and \(b = 2n^2q-n\). Observe first that \(gcd(a,b) = n\) and \(gcd(a+b, 2a^2) = gcd(2n^2q, 2n^2)=2n^2\). So, the number of boundary components is the quotient \(gcd(a+b, 2a^2)/gcd(a,b) = 2n^2/n = 2n\), as claimed. And the denominator is the quotient \((a+b)/gcd(a+b, 2a^2) = 2n^2q/2n^2 = q\), which concludes the proof of our statement. ◻

Proof. We employ a construction analogous to that in Theorem 9, applying the algorithm of Hatcher and Oertel [@HO]. Consider the pretzel knot \(K = P(p_1, p_2, p_3)\) with parameters \(p_1 = -(n+1)\), \(p_2 = n+2\), and \(p_3 = n(2nq+1)+2\). We construct a candidate surface using a system of edgepaths on the diagram \(\mathcal{D}\) for each tangle.

Similar to the proof of Theorem 9, we set up the edgepath system equations (E1)–(E4). Let \(x_i, y_i\) be the coordinates determining the edgepaths in the respective tangles. By solving the consistency equations for the number of sheets and the vertical coordinates (Condition E3), we obtain a solution space generated by a minimal integer solution.

Substituting the specific values of \(p_1, p_2, p_3\) given in the theorem statement into the formula for the number of boundary components \(|\partial F|\) and the boundary slope derived from the total twist \(\tau\), we verify the following:

1. The number of boundary components \(|\partial F|\) simplifies to \(2n\).

2. The denominator of the boundary slope simplifies to \(q\).

Furthermore, since the cycle of \(r\)-values for the final edges avoids the exceptional cases in Corollary 1, the surface is incompressible. The orientability is ensured by the parity of the parameters, as \(n\) is odd, which complements the even case in Theorem 11. Thus, the triple \((g, 2n, q)\) is realized for some genus \(g\). ◻