Proof. By remark 3.2, we see that $D \subset \Sigma$ is checkerboard colourable, and we choose the colouring so that every crossing has type $b$. The black surface $B$ is a union of disks and half-twisted bands, and with this choice each band has a left-handed half-twist. (The white surface $W$ is likewise a union of disks and half-twisted bands, and each band has a right-handed half-twist.)

Let $\Gamma _{B}$ be the Tait graph for the black surface; it is a graph in $\Sigma$ with one vertex for each black disk and one edge for each crossing. Choose arbitrarily an orientation for the edges, and label the edge associated with a crossing $x$ of $D$ with its incidence number $\eta _x$. (Since every crossing has type $b$, each edge of $\Gamma _B$ is labelled $-1$.) The black surface $B$ admits a deformation retraction onto $\Gamma _{B}$, hence $H_1(B)\cong H_1(\Gamma _{B})$.

Now consider an element $\gamma$ in $H_1(B) \cong H_1(\Gamma _B),$ which we can view as a sum $\gamma _1 + \cdots + \gamma _k$ of cycles in $\Gamma _B$. We can write each cycle $\gamma _i$ as a sum of oriented edges. If the same edge occurs with opposite signs in two of the cycles, then they can be eliminated without altering the homology class of $\gamma$. Thus, we can assume that no edge occurs with opposite signs in two cycles.

Let $a_i$ be a simple closed curve in $B$ corresponding to $\gamma _i$ for $i=1,\,\ldots,\, k$. We can assume that the curves $a_i$ and $a_j$ intersect transversely for $i \neq j$, and that their intersection points lie within the black regions. This implies that $a_i \cap a_j$ consists of finitely many points, none of which contribute to $G_F(a_i,\,a_j)$ (see figure 3 (right)). Since no edge occurs with opposite signs in the cycles $\gamma _i$ and $\gamma _j$, whenever two curves $a_i$ and $a_j$ cross the same band, they cross it in the same direction.

Since $D$ is cellularly embedded, any nontrivial cycle $\gamma _i$ will have edge set with $\ell _i>0$ edges. Therefore, $\left \bracevert \gamma _i \right \bracevert =\mathcal {G}_B(\gamma _i,\,\gamma _i)=-\ell _i<0$. This step follows by computing it as a sum of local contributions, each of which comes from a band crossing of $a_i$ and is negative. For $i \neq j$, we see that $\mathcal {G}_B(\gamma _i,\,\gamma _j) \leq 0$. This follows by again viewing it as a sum of local contributions, each of which comes from a band crossing of $a_i$ and $a_j$ and is non-positive. (This step uses the condition that whenever the two curves $a_i$ and $a_j$ cross the same band, they cross it in the same direction.) Thus, for $\gamma = \gamma _1+\cdots + \gamma _k,$ we see that

\begin{align*} \left\bracevert \gamma \right\bracevert_B & = \mathcal{G}_B(\gamma, \gamma) = \mathcal{G}_B(\gamma_1 +\cdots+\gamma_k, \gamma_1 +\cdots+\gamma_k), \\ & = \sum_{i=1}^{k} \mathcal{G}_B(\gamma_i,\gamma_i) +2 \sum_{i< j}\mathcal{G}_B(\gamma_i,\gamma_j) <0. \end{align*}

Since this holds for all nontrivial homology classes in $H_1(B)$, it follows that $B$ is negative definite.

A similar argument shows that the white surface $W$ is positive definite.

Proof. The proof is by contradiction, and here we sketch the main idea and refer to the proof of [Reference Boden, Chrisman and Karimi1, theorem 3.1] for further details.

The key step is to show that if $D \subset \Sigma$ is not cellularly embedded, then one of $\mathcal {G}_B$ or $\mathcal {G}_W$ is singular.

Assume then that $D$ is not cellularly embedded. Then we can find a non-contractible simple closed curve $\gamma$ disjoint from $D$. Since $\gamma$ and $D$ are disjoint, $\gamma$ must be contained entirely in either one of the black regions or one of the white regions. Without loss of generality, we can assume that $\gamma$ lies in a black region.

We claim there exists a simple closed curve $\alpha$ in $\Sigma$ lying entirely in a black region such that its homology class $[\alpha ] \in H_1(\Sigma )$ is nontrivial. Indeed, if $\gamma$ is non-separating, then we can take $\alpha =\gamma.$ Otherwise, if $\gamma$ is separating, then since $D$ is connected, it must lie in one of the connected components of $\Sigma \smallsetminus \gamma$. Both components have positive genus (since $\gamma$ is non-contractible), and the component disjoint from $D$ is contained entirely in a black region. There exists a simple closed curve $\alpha$ in that component with $[\alpha ]\neq 0$ in $H_1(\Sigma )$.

Consider now the map $H_1(B) \to H_1(\Sigma )$ induced by $B \hookrightarrow \Sigma \times I \to \Sigma$, the composition of inclusion and projection. Since $[\alpha ]$ is nontrivial in $H_1(\Sigma )$, it must also be nontrivial in $H_1(B).$ Further, since $\alpha$ is a simple closed curve, the set $\{[\alpha ]\}$ can be extended to a basis $U$ for $H_1(B)$. Since $\alpha$ lies entirely within one of the black regions, we have $\mathcal {G}_{B}(\alpha,\, \alpha )=\left \bracevert \alpha ^{}_{} \right \bracevert _B=0$. Any other element of $U$ can be represented as a simple closed curve $\beta$ on $B$. Since $\alpha$ lies entirely in a black region, the two curves $\alpha$ and $\beta$ have only intersection points; there are no points where $\alpha$ crosses over or under $\beta$. However, as we have seen, an intersection point contributes 0 to the pairing, and thus it follows that $\mathcal {G}_{B}(\alpha,\, \beta )=0$. This holds for all $\beta \in U$, therefore the Gordon–Litherland pairing $\mathcal {G}_{B}$ is singular.

We now prove that $L$ is non-split. Suppose to the contrary that $L$ is split, and let $D' \subset \Sigma$ be a disconnected diagram for $L$. Notice that $D'$ is not cellularly embedded, and that the checkerboard surfaces need not be connected. By adding small tubes, we can connect them. However, one or both of the resulting spanning surfaces will have singular Gordon–Litherland pairing. This is a contradiction (cf. remark 3.4).

Proof. Let $D$ be a cellularly embedded alternating diagram for $L$. Then $D$ is checkerboard colourable (remark 3.2), and lemma 3.3 implies that the black and white surfaces are definite. In particular, their Gordon–Litherland pairings are non-singular. The conclusion now follows from theorem 3.5.

Proof. In general we have

\begin{align*} \sigma_{W}(L) & = \operatorname{sig}(\mathcal{G}_W)-\mu_W(D), \\ \sigma_{B}(L) & = \operatorname{sig}(\mathcal{G}_B) -\mu_B(D). \end{align*}

Since all crossings have type $b$ and referring to figure 7, we see that

\begin{align*} \mu_W(D)& = \sum_{x \, \text{type I}}-\eta_x = \sum_{x \, \text{type I}}\varepsilon_x= c_{+}(D), \\ \mu_B(D)& = \sum_{x \, \text{type II}}\eta_x= \sum_{x \, \text{type II}}\varepsilon_x={-}c_{-}(D), \end{align*}

where $c_+(D)$ is the number of positive crossings of $D$ and $c_-(D)$ is the number of negative crossings. Hence

(3.4)\begin{equation} \mu_W(D) -\mu_B(D) = c_{+}(D) + c_{-}(D) = c(D). \end{equation}

FIGURE 7. Two type $b$ crossings.

Lemma 3.3 shows that $W$ is positive definite and $B$ is negative definite, hence

\begin{align*} \sigma_{W}(L)-\sigma_{B}(L) & = \operatorname{sig}(\mathcal{G}_W)-\mu_W(D)-(\operatorname{sig}(\mathcal{G}_B)-\mu_B(D)), \\ & = b_1(W)+b_1(B)- (\mu_W(D) -\mu_B(D)) =2\,g, \end{align*}

where the last step follows from equations (3.3) and (3.4).

Proof. If $D$ is alternating, then lemma 3.3 gives the desired conclusion.

Conversely, suppose $B$ is negative definite and $W$ is positive definite. Let $a_{\pm }$ be the number of type $a$ crossings of $D$ with $\varepsilon _x=\pm 1$, and $b_{\pm }$ be the number of type $b$ crossings of $D$ with $\varepsilon _x=\pm 1$. Then

\[ \mu_{W}(D) = a_{-}-b_{+} \text{ and } \mu_{B}(D) ={-}a_{+}+b_{-}. \]

It follows that

\begin{align*} \mu_{W}(D)-\mu_{B}(D) & = a_{-} - b_{+}-({-}a_{+}+b_{-}), \\ & =a(D)-b(D), \end{align*}

where $a(D)=a_{+}+a_{-}$ is the total number of type $a$ crossings and $b(D)=b_{+}+b_{-}$ is the total number of type $b$ crossings.

Therefore,

(3.5)\begin{equation} |\mu_{W}(D)-\mu_{B}(D)|\leq c(D), \end{equation}

with equality if and only if $a(D)=0$ or $b(D)=0$. In the first case, all crossings have type $b$, and in the second, they all have type $a$. In either case, we see that $D$ is alternating. Thus, it only remains to show that $|\mu _{W}(D)-\mu _{B}(D)|= c(D)$.

Given a spanning surface $F$ for $L$, by the discussion after lemma 2.1, we obtain a new surface $F\#_\tau \Sigma$ by connecting it to a parallel copy of $\Sigma$ by a thin tube $\tau$. The surfaces $F$ and $F\#_\tau \Sigma$ are not $S^{*}$-equivalent.

Since $B$ and $W$ are not $S^{*}$-equivalent, and since a non-split link in $\Sigma \times I$ has exactly two $S^{*}$-equivalence classes of spanning surfaces, it follows that $B$ and $W\#_\tau \Sigma$ must be $S^{*}$-equivalent. By lemma 2.1, $\sigma _{B}(L)=\sigma _{W\#_\tau \Sigma }(L)$. We have $|\sigma _{W}(L)-\sigma _{W\#_\tau \Sigma }(L)|\leq 2\,g$. Hence $|\sigma _{W}(L)-\sigma _{B}(L)|\leq 2\,g$. Further, since $B$ and $W$ are definite surfaces of opposite sign, we see that $|\operatorname {sig}(\mathcal {G}_W)-\operatorname {sig}(\mathcal {G}_B)| = b_{1}(W)+b_{1}(B).$ These two observations, combined with equation (3.3) and inequality (3.5), show that:

\begin{align*} 2\,g & \geq |\sigma_{W}(L)-\sigma_{B}(L)|, \\ & = |\operatorname{sig}(\mathcal{G}_W)-\mu_{W}(D)-(\operatorname{sig}(\mathcal{G}_B)-\mu_{B}(D))|,\\ & \geq |\operatorname{sig}(\mathcal{G}_W)-\operatorname{sig}(\mathcal{G}_B)|-|\mu_{W}(D)-\mu_{B}(D)|,\\ & = b_{1}(W)+b_{1}(B)-|\mu_{W}(D)-\mu_{B}(D)|,\\ & = 2g+c(D)-|\mu_{W}(D)-\mu_{B}(D)|\geq 2\,g. \end{align*}

Therefore we have equality throughout, and it follows that $D$ is alternating.

Proof. A detailed proof can be found in [Reference Carter, Silver and Williams6], so we only sketch the argument. Since $\Sigma$ has genus $g \geq 1$, the universal cover of $\Sigma \times I$ is ${{\mathbb {R}}}^{2}\times I$, which is irreducible, then by [Reference Hatcher9, proposition 1.6] it follows that $\Sigma \times I$ is irreducible. Any embedded 2-sphere in $X_L \subset \Sigma \times I$ must bound a 3-ball $Y \subset \Sigma \times I$, and the hypotheses ensure that $Y\subset X_L$. This completes the argument.

Proof. If $F'$ is $S^{*}$-equivalent to $F$, then lemma 2.1 implies that $\sigma _F(L)=\sigma _{F'}(L).$ If, in addition, $e(F)=e(F')$, then it follows that $\operatorname {sig}(\mathcal {G}_F)= \operatorname {sig}(\mathcal {G}_F').$

Now suppose $F$ is definite. Then we have

\[ b_1(F) = |\operatorname{sig}(\mathcal{G}_F)| = |\operatorname{sig}(\mathcal{G}_{F'})| \leq b_1(F'), \]

which shows the first claim.

If, in addition, $b_1(F')=b_1(F),$ then we have $b_1(F') = |\operatorname {sig}(\mathcal {G}_{F'})|,$ hence $F'$ must also be definite. Since $\operatorname {sig}(\mathcal {G}_{F}) = \operatorname {sig}(\mathcal {G}_{F'}),$ it follows that $F$ and $F'$ must have the same sign.

Proof. Suppose to the contrary that $F$ is compressible. Let $F'$ be the surface obtained from $F$ by a compression. Then $F'$ is $S^{*}$-equivalent to $F$ and $b_1(F') < b_1(F).$ Further, $e(F') =e(F)$. However, this is impossible, for it would contradict lemma 4.2 if $F$ is definite.

Proof. Any surface $S$ with genus $g(S)< g(\Sigma )$ has $[S]= 0$ in $H_2(\Sigma \times I)$. This proves the first statement.

Assume now that $g(S)= g(\Sigma )$. If $S$ is not incompressible, there exists a non-contractible simple closed curve $\gamma$ in $S$ which bounds a disk in $\Sigma \times I$. Let $S'$ be the surface obtained by cutting $S$ along $\gamma$ and gluing in two disks along the newly created boundary components. Then $[S'] =[S]$ in $H_2(\Sigma \times I).$ If $\gamma$ is non-separating on $S$, then $g(S') = g(S)-1 < g(\Sigma )$ and $[S']=0$, a contradiction. Otherwise, if $\gamma$ is separating on $S$, then $S'$ is a disjoint union $S_1' \sqcup S'_2$ of two surfaces of positive genus satisfying $g(S')=g(S_1') + g(S'_2) =g(S)$. Hence $0< g(S_i')< g(S)$ for $i=1,\,2$. It follows that $[S'_1] =0$ and $[S_2']=0$ in $H_2(\Sigma \times I).$ Thus $[S'] = [S_1'] + [S_2'] = 0,$ again a contradiction. This proves the second statement and completes the proof.

Proof. Suppose $P$ and $N$ are positive and negative definite surfaces, respectively, and $\gamma$ is a simple closed curve contained in $P\cap N\cap X_L$. Let $\nu (\gamma )$ be a small regular neighbourhood of $\gamma$ in $\Sigma \times I$, which contains no other intersection of $P$ and $N$. Clearly $\nu (\gamma )$ is a $D^{2}$ bundle over $\gamma$, and since $\Sigma \times I$ is orientable, so is $\nu (\gamma )$ and it follows that $\nu (\gamma )\approx S^{1}\times D^{2}$ is a trivial bundle. The annuli $P \cap \nu (\gamma )$ and $N \cap \nu (\gamma )$ intersect only in $\gamma$, so the framing $\left \bracevert \gamma \right \bracevert _{P}$ of $\gamma$ in $P$ is equal to the framing $\left \bracevert \gamma \right \bracevert _{N}$ of $\gamma$ in $N$. Since $P$ and $N$ are positive and negative definite, respectively, we have

\[ \left\bracevert\gamma \right\bracevert_{N}\leq 0 \leq \left\bracevert\gamma \right\bracevert_{P}. \]

Therefore, $\left \bracevert \gamma \right \bracevert _{P}=\left \bracevert \gamma \right \bracevert _{N}=0$. It follows that $\gamma$ is null-homologous in both $P$ and $N$. Thus, $\gamma$ is separating on $P$ and $N$. Let $P'$ and $N'$ be orientable subsurfaces of $P$ and $N$, respectively, with $\partial P'=\partial N'=\gamma$. By choosing $\gamma$ an innermost curve on $N$, we can arrange that $N'$ is disjoint from $P.$ By lemma 4.6, $P'$ is positive definite and $N'$ is negative definite.

Set $V=P' \cup _\gamma N'$. Since $P'$ and $N'$ are both orientable, $V$ is a closed orientable surface, and $\frac {1}{2}e(P',\,\gamma )=0=\frac {1}{2}e(N',\,\gamma )$. Therefore, $\sigma _{P'}(\gamma ) = \operatorname {sig}(\mathcal {G}_{P'})$ and $\sigma _{N'}(\gamma )=\operatorname {sig}(\mathcal {G}_{N'})$.

Suppose firstly that $P'$ and $N'$ are $S^{*}$-equivalent. Then

\[ 0\leq b_{1}(P') = \operatorname{sig}(\mathcal{G}_{P'}) = \sigma_{P'}(\gamma) = \sigma_{N'}(\gamma) = \operatorname{sig}(\mathcal{G}_{N'})={-}b_{1}(N')\leq 0. \]

Therefore, $P'$ and $N'$ are disks in this case.

Suppose now that $P'$ and $N'$ are not $S^{*}$-equivalent. Then $P'$ and $N'\#_\tau \Sigma$ are $S^{*}$-equivalent, and so $\operatorname {sig}(\mathcal {G}_{P'}) = \operatorname {sig}(\mathcal {G}_{N' \#_\tau \Sigma })$. Therefore, if $g=\text {genus}(\Sigma ),$ we see that

\[ 2\,g \geq \operatorname{sig}(\mathcal{G}_{N' \#_\tau \Sigma}) - \operatorname{sig}(\mathcal{G}_{N'}) = \operatorname{sig}(\mathcal{G}_{P'}) - \operatorname{sig}(\mathcal{G}_{N'}) = b_1(P') + b_1(N'). \]

In particular, $V$ is a closed, orientable surface with $\text {genus}(V) \leq g$. In fact, its genus must be exactly $g$, since otherwise we would have $[V] = 0$ in $H_2(\Sigma ;{{\mathbb {Z}}}_2)$, which would imply that $P'$ and $N'$ are $S^{*}$-equivalent.

Lemma 4.4 applies and shows that $V$ is incompressible in $\Sigma \times I.$ Therefore, by [Reference Waldhausen19, corollary 3.2], it follows that $V$ is isotopic to $\Sigma \times \{t_0\}$ for some $0< t_0<1$ (see remark 4.5). Since $g(V)\geq 1$, there is a non-separating simple closed curve $\alpha$ in either $P'$ or $N'$. Assume $\alpha \subset P'$. Then $[\alpha ]\neq 0 \in H_1(P')$, so $\left \bracevert \alpha ^{}_{} \right \bracevert _{P'}>0$. On the other hand, since $\alpha \subset P'\subset V$ and $V$ is isotopic to $\Sigma \times \{t_0\}$, the framing of $\alpha$ in $P'$ is the same as the framing of $\alpha$ in $V$, so $\left \bracevert \alpha ^{}_{} \right \bracevert _{P'}=0$. This is a contradiction, therefore, $P'$ and $N'$ are $S^{*}$-equivalent. It follows that $P'$ and $N'$ are both disks, and $V$ is a 2-sphere. By proposition 4.1, $X_L$ is irreducible, thus, $V$ bounds a 3-ball. Using this ball, we can set-up an isotopy which separates $P'$ and $N'$, and makes the number of components of $P \cap N \cap X_L$ smaller. This is a contradiction, and the result follows.

Proof. For $g(\Sigma )=0$, the result follows from [Reference Greene8], therefore we assume $g(\Sigma )\geq 1$.

Suppose $L$ is non-split and $D$ is an alternating diagram for $L$. Since $L$ has minimal genus, $D$ is cellularly embedded. Further, $D$ is checkerboard colourable by remark 3.2, and lemma 3.3 implies that the checkerboard surfaces $W$ and $B$ are positive and negative definite, respectively. This proves one direction, and it remains to prove the other.

Suppose then that $P$ and $N$ are two definite spanning surfaces for $L$, with $P$ positive definite and $N$ negative definite.

Let $X_L=\Sigma \times I\smallsetminus \operatorname {int}(\nu (L))$ be the exterior of $L$. We write $\partial X_L = \partial _1 X_L \cup \cdots \cup \partial _m X_L$ according to the components of the link $L = K_1 \cup \cdots \cup K_m$. Clearly, each $\partial _i X_L$ is a torus. Assume further that, $P$ and $N$ intersect transversely in $X_L$ such that the number of components of $P \cap N \cap X_L$ is minimized.

For $i=1,\,\ldots,\, m$ set $\lambda ^{P}_i=P\cap \partial _i X_L$ and $\lambda ^{N}_i=N\cap \partial _i X_L$. Thus $\lambda ^{P}_i$ and $\lambda ^{N}_i$ intersect transversely in $\partial _i X_L$. We further set $\lambda ^{P} = \bigcup _i \lambda ^{P}_i$ and $\lambda ^{N} = \bigcup _i \lambda ^{N}_i.$

By lemma 4.7, we can assume that $P\cap N\cap X_L$ does not contain any closed components. Thus $P\cap N\cap X_L$ is a union of arc components which we call double arcs. Each double arc connects a pair of distinct points in $\lambda ^{P}\cap \lambda ^{N}$. Thus $\lambda ^{P}\cap \lambda ^{N}$ consists of an even number of points, equal to twice the number of double arcs. Since $P$ is positive definite and $N$ is negative definite, the number of points in $\lambda ^{P}_i\cap \lambda ^{N}_i$ is equal to the difference in framings $\tfrac {1}{2}\left \bracevert [K_i]\right \bracevert _{P}-\tfrac {1}{2}\left \bracevert [K_i]\right \bracevert _{N}$. Summing over the components, we get that

\[ \sum_{i=1}^{m} \left( \tfrac{1}{2}\left\bracevert [K_i]\right\bracevert_{P}-\tfrac{1}{2} \left\bracevert [K_i]\right\bracevert_{N} \right) = e(N)-e(P). \]

The number of arc components in $P \cap N \cap X_L$ is therefore equal to $\frac {1}{2}(e(N)-e(P))$.

An orientation on $X_L$ induces one on $\partial X_L$, and an orientation on each $K_i$ induces ones on $\lambda ^{P}_i$ and $\lambda ^{N}_i$. This defines a sign $\varepsilon _x \in \{\pm 1\}$ for each point $x\in \lambda ^{P}_i\cap \lambda ^{N}_i$. The sign $\varepsilon _x$ is positive if the orientation of $\lambda ^{P}_i$ followed by the orientation of $\lambda ^{N}_i$ at $x$ agree with the orientation of $\partial _i X_L$, and it is negative otherwise. A key point is that every point in $\lambda ^{P}_i\cap \lambda ^{N}_i$ has the same sign; this follows from the fact that

\[ \#(\lambda^{P}_i\cap \lambda^{N}_i)=\tfrac{1}{2}\left\bracevert [K_i]\right\bracevert_{P}-\tfrac{1}{2} \left\bracevert [K_i]\right\bracevert_{N}=\left|{\sum \varepsilon_{x}}\right|, \]

where the sum on the right is taken over all $x \in \lambda ^{P}_i\cap \lambda ^{N}_i.$

As explained in §2 of [Reference Howie10], there are two kinds of double arcs; one is called a standard double arc and the other is called a parallel double arc. A double arc of $P \cap N$ with endpoints $x$ and $y$ is standard if $\varepsilon _x = \varepsilon _y$, and it is parallel if $\varepsilon _x = -\varepsilon _y.$ By the previous observations, we see that every double arc is standard. Each double arc extends to give an arc $a$ in $P\cap N$ with $\partial a=a\cap L$. Since the double arc is standard, there is a neighbourhood $V$ of $a$ modelled on the intersecting checkerboard surfaces in a standard neighbourhood of a crossing in a link diagram. See figures 8 and 9.

FIGURE 8. A standard crossing (left) placed on an octahedron (right).

FIGURE 9. The two ruled surfaces near a crossing. The heavily dashed vertical lines depict the standard arc of intersection of the two surfaces.

We can choose local coordinates near the crossings so that the arcs of the link lie in the $xy$ plane except at the crossings. At each crossing we place an octahedron which intersects the $xy$ plane in the square with vertices $p=(1,\,-1,\,0)$, $q=(-1,\,1,\,0)$, $r=(-1,\,-1,\,0)$, and $s=(1,\,1,\,0)$. We assume that the over-crossing arc connecting $p$ to $q$ is given by

\[ \beta(t)=(1-2t,-1+2t,\text{min}(2t,2-2t))\text{ for } 0\leq t\leq 1, \]

and the under-crossing arc connecting $r$ to $s$ is given by

\[ \gamma(t)=({-}1+2t,-1+2t,\text{max}({-}2t,-2+2t))\text{ for } 0\leq t\leq 1. \]

In the standard crossing, the black surface is parametrized by

\[ B(s,t)=\beta(t)(1-s)+s\cdot\gamma(t)\text{ for } 0\leq s,t\leq 1, \]

and the white surface is parametrized by

\[ W(s,t)=\beta(t)(1-s)+s\cdot\gamma(1-t)\text{ for } 0\leq s,t\leq 1. \]

Notice that, at the crossing, the black surface contains a left half-twist, whereas the white surface contains a right half-twist. The black and white surfaces intersect in the vertical arc $(0,\,0,\,1-2t),$ for $0\leq t \leq 1,$ which connects $(0,\,0,\,1)$ to $(0,\,0,\,-1)$.

Thus, any standard arc $a$ has a neighbourhood $V\subset \Sigma \times I$ such that $a$ is vertical and the projection $p\colon \Sigma \times I\to \Sigma$ maps $(P\cup N\smallsetminus a)\cap V$ homeomorphically onto a once-punctured disk in $\Sigma$.

Let $A$ denote the union of all the double arcs in $P\cap N\cap X_L$. Lemma 4.7 implies that $P\cap N\cap X_L$ does not contain any simple closed curves. Thus it follows that $P\cup N\smallsetminus A$ is a two-dimensional manifold. Furthermore, collapsing the standard models of each of the double arcs $a$ in $A$ down, we see that $\nu (P\cup N)$ is homeomorphic to $\nu (S)$ for some connected surface $S$ embedded in $\Sigma \times I$. We can identify $\nu (S) \approx S\times I$ in such a way that the double arcs are all mapped to distinct points under projection $S\times I\to S$.

Set $c=\frac {1}{2}(e(N)-e(P))$, which is equal to the number of arc components in $P\cap N \cap X_L.$ Therefore, $P \cap N = L \cup A$ and has Euler characteristic $-c.$

Claim: $\chi (S)=\chi (\Sigma )$.

Proof of the Claim. Since $\nu (S)=\nu (P\cup N)$, we have that

(4.1)\begin{equation} \begin{aligned} \chi(S) & =\chi(\nu(S))=\chi(\nu(P\cup N))=\chi(P\cup N), \\ & =\chi(P)+\chi(N)-\chi(P\cap N), \\ & =(1-b_{1}(P))+ (1-b_{1}(N))+c, \\ & =2-(b_1(P)+b_{1}(N)-c). \end{aligned} \end{equation}

On the other hand, computing the signature of $L$ using $P$ and $N$, we see that

\begin{align*} \sigma_{P}(L)& =\operatorname{sig}(\mathcal{G}_P)+\tfrac{1}{2}e(P, L),\\ \sigma_{N}(L)& =\operatorname{sig}(\mathcal{G}_N)+\tfrac{1}{2}e(N, L). \end{align*}

Thus, by equation (2.1), we have

(4.2)\begin{equation} \begin{aligned} \sigma_{P}(L)-\sigma_N(L) & =\operatorname{sig}(\mathcal{G}_P)-\operatorname{sig}(\mathcal{G}_N)+\tfrac{1}{2}(e(P,L)-e(N,L)),\\ & =\operatorname{sig}(\mathcal{G}_P)-\operatorname{sig}(\mathcal{G}_N)+\tfrac{1}{2}(e(P)-e(N)),\\ & =b_1(P)+b_{1}(N)-c. \end{aligned} \end{equation}

Substituting equation (4.2) into equation (4.1), we see that

(4.3)\begin{equation} \chi(S)=2-(\sigma_{P}(L)-\sigma_N(L)). \end{equation}

We will now show that $P$ and $N$ are not $S^{*}$-equivalent. Suppose to the contrary that $P$ and $N$ are $S^{*}$-equivalent. Then $\sigma _{P}(L)=\sigma _N(L)$, and the equations above give that $\chi (S)=2$. But since $\Sigma \times I$ is irreducible, any 2-sphere bounds a 3-ball. Therefore, the link $L$, which lies on $S$, must be contained in a 3-ball. This however contradicts our assumption that $L \subset \Sigma \times I$ has minimal genus $g(\Sigma )\geq 1.$ Thus, $P$ and $N$ cannot be $S^{*}$-equivalent.

Since $N$ is not $S^{*}$-equivalent to $P$, it follows that $N$ must be $S^{*}$-equivalent to $P\#_{\tau }\Sigma$. Thus

\[ |\sigma_{P}(L)-\sigma_N(L)| =|\sigma_{P}(L)-\sigma_{P\#_{\tau}\Sigma}(L)| \leq 2g(\Sigma). \]

By equation (4.3), we see that

\[ \chi(S) \geq 2 - 2g(\Sigma) = \chi(\Sigma). \]

If this inequality were strict, i.e., if $\chi (S) > \chi (\Sigma )$, then $g(S) < g(\Sigma )$ and it would follow that $[S]=0$ in $H_2(\Sigma \times I; {{\mathbb {Z}}}/2)$. However, that would imply that $P$ and $N$ are $S^{*}$-equivalent. Since they are not, we must have that $\chi (S)=\chi (\Sigma ),$ and this completes the proof of the claim.

Lemma 4.4 applies to show that $S$ is incompressible. As explained in remark 4.5, by [Reference Waldhausen19], this implies that $S$ is isotopic to $\Sigma \times \{t_0\}$ for some $0< t_0<1$. Under the isotopy, $L$ is isotopic to a link $L'$ that lies in $\Sigma \times I'$, where $I'$ is a closed interval containing $t_0$ in its interior.

Projecting $L'$ along $\nu (S)\approx S\times I\to S$, gives a diagram $D$ for $L'$ which by the claim has genus $g(S)=g(\Sigma )$. Furthermore, the checkerboard surfaces of $D$ on $S$ are isotopic relative the boundary to $P$ and $N$. Proposition 3.8 now implies that $D$ is alternating. This finishes the proof.