2.1. Rigid expansions

In [Reference Hernández Hernández6], we gave a general description for the concept of a rigid subgraph and rigid expansions of abstract simplicial graphs. In this paper, we give the definitions focused on the case of the curve graph.

Given a curve α on $S_{g,n}$, we define the set $\mathrm{adj}(\alpha)$ as the set of all curves on S that are disjoint from α. Note that $\alpha \notin \mathrm{adj}(\alpha)$. The notation comes from the set of vertices adjacent to the vertex corresponding to the curve α.

We say a curve $\alpha \in \mathcal{C}(S_{g,n})$ is uniquely determined by $A \subset \mathcal{C}(S_{g,n})$, denoted by $\alpha = \langle A \rangle$, if α is the unique curve on $S_{g,n}$ that is disjoint from every element in A, i.e.

\begin{equation*} \{\alpha\} = \bigcap_{\gamma \in A} \mathrm{adj}(\gamma). \end{equation*}

See Figure 2 for examples.

Figure 2.

In both examples, the blue curve is uniquely determined by the set of red curves.

Let $Y \lt \mathcal{C}(S_{g,n})$ be a induced subgraph of $\mathcal{C}(S_{g,n})$, and let V(Y) denote its vertex set. We define the first rigid expansion of Y, denoted by Y ¹ as the subgraph of $\mathcal{C}(S_{g,n})$ induced by the following vertices:

\begin{equation*} V(Y^{1}) := V(Y) \cup \{\alpha \in \mathcal{C}(S_{g,n}) : \alpha = \langle A \rangle,\text{for some } A \subset V(Y)\}. \end{equation*}

We then define the $(n+1)$th rigid expansion as $(Y^{n})^{1}$, and Y ^ω as the induced subgraph of $\mathcal{C}(S_{g,n})$ induced by the following vertex set:

\begin{equation*} V(Y^{\omega}) := \bigcup _{n \geqslant 1} V(Y^{n}). \end{equation*}

Recall that a graph morphism between simplicial graphs is a map between the corresponding vertex sets that also maps edges to edges.

The following theorem is Theorem B in [Reference Hernández Hernández6] for the special case of the curve graph, using the terminology presented above:

Now, given a curve α on $S_{g,n}$, we define the star of α, denoted by $\mathrm{star}(\alpha)$, as the subgraph whose vertex set is $\{\alpha\} \cup \mathrm{adj}(\alpha)$, and the edges are those from $\mathcal{C}(S_{g,n})$ that have α as one of its endpoints, i.e. the edges incident to α.

A locally injective map $\varphi: X \rightarrow \mathcal{C}(S_{g,n})$ is a simplicial map whose restriction to $\mathrm{star}(\alpha)$ is injective for every $\alpha \in X$.

Now, a induced subgraph $X \lt \mathcal{C}(S_{g,n})$ is called rigid if every locally injective map from X to $\mathcal{C}(S_{g,n})$ is induced by an automorphism of $\mathcal{C}(S_{g,n})$ (again, if $(g,n) \neq (1,2)$ it becomes a homeomorphism of $S_{g,n}$).

Note that not every supergraph of a rigid subgraph is rigid (see Proposition 3.2 in [Reference Aramayona and Leininger2]). However, any rigid expansion of a rigid subgraph is rigid (see Proposition 3.5 in [Reference Aramayona and Leininger2]).

The following corollary is Corollary C in [Reference Hernández Hernández6] for the special case of the curve graph, and it comes as a direct consequence of Theorem 1 and the definition of rigidity.

3.1. The basis of $\mathscr{Y}(S)$

For $k\geqslant 2$, let $O = \{\alpha_{0}, \ldots, \alpha_{k}\}$ be a set of outer curves. We say O is an outer chain if $i(\alpha_{i},\alpha_{j}) = 2$ if $|i-j|=1$ and α_i is disjoint from α_j otherwise. This notation comes as an analogue for a chain in higher genus. We say O is a closed outer chain if (counting the subindices modulo k + 1) $i(\alpha_{i},\alpha_{j}) = 2$ if $|i-j|=1$ and α_i is disjoint from α_j otherwise. See Figure 3 for examples. In both cases, we set the length of O as its cardinality.

Figure 3.

On the left, an example of an outer chain. On the right, an example of a closed outer chain.

Note that a closed outer chain of maximal length has cardinality n, and its regular neighbourhood is homeomorphic to a 2n-punctured annulus, of which n punctures are punctures of the surface S, while the other n punctures correspond to non-essential curves.

Let $\mathscr{O} = \{\alpha_{0}, \ldots, \alpha_{n-1}\}$ be a closed outer chain of maximal length. While these curves seem to be a good candidate for $\mathscr{Y}(S)$, they are not enough since their iterated rigid expansions eventually stabilize. So, we have to add some auxiliary curves.

3.2. Auxiliary curves

To define the auxiliary curves, we must first number the punctures of S in such a way that $\alpha_{i} \in \mathscr{O}$ bounds the ith and $(i+1)$th punctures (with indices modulo n). See Figure 4 for an example.

Figure 4.

The ith and $(i+1)$th puncture along with the curve α_i.

Now, let α be an outer curve on S. We define the arc corresponding to α to be the unique arc disjoint from α whose endpoints are the punctures bounded by α. Similarly, let a be an arc on S with different endpoints; we call the outer curve corresponding to a, to the unique outer curve that bounds precisely the endpoints of a and is disjoint from a. See Figure 5 for examples.

Figure 5.

The blue and purple arcs have the green and orange curves as their (respective) corresponding outer curves.

Note that there is a bijective correspondence between the set of all outer curves on S and the set of all arcs on S with two distinct endpoints.

Let $A = \{a_{i}\}$ be the set of arcs a_i corresponding to $\alpha_{i} \in \mathscr{O}$. Let also H ₁ and H ₂ be two identical n-gons with numbered vertices such that S can be identified with the result of gluing H ₁ and H ₂ together by their sides (and eliminating the vertices), and the labelling of the vertices in H ₁ and H ₂ coincides with the labelling of the punctures on S. Moreover, we particularly choose the identification so that the (glued together) sides of H ₁ and H ₂ are identified with the elements of A. See Figure 6.

Figure 6.

The n-gons H ₁ and H ₂ (left), glued together to obtain the surface S (right).

For $0 \leqslant i \lt j \lt n$, let $b_{i,j}$ be the arc on S joining the ith and jth punctures of S that is identified with the diagonal on H ₁ joining the ith and jth vertices. Similarly, for $0 \leqslant i \lt j \lt n$, let $c_{i,j}$ be the arc on S joining the ith and jth punctures of S that is identified with the diagonal on H ₂ joining the ith and jth vertices. See Figure 7 for an example.

Figure 7.

The arcs $b_{1,5}$ and $c_{1,6}$ that induce the curves $\beta_{1,5}$ and $\gamma_{1,6}$ respectively.

Then, let $\mathscr{D}_{1} = \{\beta_{i,j}\}$ be the set of outer curves $\beta_{i,j}$ corresponding to the arc $b_{i,j}$, and let $\mathscr{D}_{2} = \{\gamma_{i,j}\}$ be the set of outer curves $\gamma_{i,j}$ corresponding to the arc $c_{i,j}$.

Finally, we define:

\begin{equation*} \mathscr{Y}(S) := \mathscr{O} \cup \mathscr{D}_{1} \cup \mathscr{D}_{2}. \end{equation*}

3.3. Proof of Theorem A

For the proof of Theorem A, we first recall some definitions and properties around outer curves on S and half-twists.

Let α be a curve on S. We denote by η_α the (left) half-twist along α (see Figure 8); recall that this is defined if and only if α is an outer curve. Also recall that there is exactly one half-twist along α if $S \ncong S_{0,4}$.

Figure 8.

On the left, the outer curve α and the curve β; on the right, the curve $\eta_{\alpha}(\beta)$.

Let O be a set of outer curves and A be a set of curves on S. We denote by $\eta_{O}(A)$ the following set

\begin{equation*} \eta_{O}(A) := \bigcup_{\varepsilon \in O} \eta_{\varepsilon}(A). \end{equation*}

Let $\mathscr{G} := \{\eta_{\alpha_{i}}^{\pm 1}: \alpha_{i} \in \mathscr{O}\}$. It is a well-known fact (see Theorem 4.9 in [Reference Farb and Margalit4]) that $\mathscr{G}$ is a symmetric generating set of $\mathrm{Mod}(S)$. With this, if A is a set of curves, we denote by $\mathscr{G} \cdot A$ the following set $\mathscr{G} \cdot A := \eta_{\mathscr{O}}^{\pm 1}(A)$.

Now, similarly to the proofs in [Reference Hernández Hernández7], we state a key lemma for the proof of Theorem A. The proof of this lemma is discussed and given after the proof of Theorem A (see § 3.5 and 3.6).

Proof of Theorem A

We divide this proof into two parts: first we prove that all outer curves on S are elements in $\mathscr{Y}(S)^{\omega}$; then we argue that any other curve on S is uniquely determined by a finite number of outer curves and therefore we obtain the desired result.

First part: Let β be an outer curve on S. Since all outer curves on S have the same topological type, then there exists a element $\alpha_{i} \in \mathscr{O}$ and a homeomorphism $h \in \mathrm{Mod}(S)$ such that $h(\alpha_{i}) = \beta$. Given that $\mathscr{G}$ is a symmetric generating set of $\mathrm{Mod}(S)$, then there exist elements $f_{1}, \ldots, f_{k} \in \mathscr{G}$ and powers $n_{1}, \ldots, n_{k} \geqslant 0$ such that

\begin{equation*} f_{1}^{n_{1}} \circ \cdots \circ f_{k}^{n_{k}} (\alpha_{i}) = \beta. \end{equation*}

We claim that by an iterative use of Lemma 1, this implies that $\beta \in \mathscr{Y}(S)^{3(n_{1} + \cdots + n_{k})}$: By Lemma 1, $f_{k}(\alpha_{i}) \in f_{k}(\mathscr{Y}(S)) \subset \mathscr{Y}(S)^{3}$. Since f_k is a homeomorphism, the same proof for Lemma 1 works for $f_{k}(\mathscr{Y}(S))$, thus $\mathscr{G} \cdot f_{k}(\mathscr{Y}(S)) \subset f_{k}(\mathscr{Y}(S))^{3} \subset \mathscr{Y}(S)^{6}$. Applying this argument iteratively, we obtained (as claimed above) that $\beta \in \mathscr{Y}(S)^{3(n_{1} + \cdots + n_{k})}$.

Thus, every outer curve on S is an element of $\mathscr{Y}(S)^{\omega}$.

Second part: Let γ be a curve on S that is not an outer curve. Since there are only finitely many topological types of such curves, which depend on how they separate the punctures of S, up to homeomorphism γ is as in the example on Figure 9. Then, there exists two (finite) outer chains, namely A and B, that uniquely determine γ (see Figure 9 for an example). Given that by the first part of this proof $A \cup B \subset \mathscr{Y}(S)^{k}$ for some $k \geqslant 0$, we have that $\gamma \in \mathscr{Y}(S)^{k+1}$. Therefore, every curve on S that is not an outer curve is also an element of $\mathscr{Y}(S)^{\omega}$.

Figure 9.

The set A (red curves) and the set B (green curves) uniquely determining the curve γ (in blue).

As for the proof of Lemma 1, we first prove the lemma for the complexity 2 case in § 3.4; then we prove them for complexity at least 3 in more generality. The reason for this division is that in the complexity 2 we ‘do not have enough space’ for the general method to work.

For the complexity at least 3 case, we divide the proof into the following claims:

Claim 1: $\eta_{\mathscr{O}}^{\pm 1}(\mathscr{O}) \subset \mathscr{Y}(S)$.

Claim 2: $\eta_{\mathscr{O}}^{\pm 1}(\mathscr{D}_{1}) \subset \mathscr{Y}(S)^{2}$.

Claim 3: $\eta_{\mathscr{O}}^{\pm 1}(\mathscr{D}_{2}) \subset \mathscr{Y}(S)^{3}$.

3.4. Complexity 2 case

Let S be a five-punctured sphere, $\alpha \in \mathscr{O}$ and $\delta \in \mathscr{Y}(S)$. Note that if α and δ are disjoint, we have that $\eta_{\alpha}^{\pm 1}(\delta) = \delta$. Thus, we focus solely on the cases when they intersect.

We can easily check that

\begin{equation*} \eta_{\alpha_{i+1}}(\alpha_{i}) = \gamma_{i,i+2} \in \mathscr{Y}(S), \end{equation*}

\begin{equation*} \eta_{\alpha_{i-1}}(\alpha_{i}) = \beta_{i,i+2} \in \mathscr{Y}(S). \end{equation*}

This and Remark 1 imply that

\begin{equation*} \eta_{\alpha_{i+1}}^{-1}(\alpha_{i}) = \eta_{\alpha_{i}}(\alpha_{i+1}) \in \mathscr{Y}(S), \end{equation*}

\begin{equation*} \eta_{\alpha_{i-1}}^{-1}(\alpha_{i}) = \eta_{\alpha_{i}}(\alpha_{i-1}) \in \mathscr{Y}(S). \end{equation*}

Hence, $\eta_{\mathscr{O}}^{\pm 1}(\mathscr{O}) \subset \mathscr{Y}(S)$.

Since S is a five-punctured sphere, then every $\beta \in \mathscr{D}_{1}$ is of the form either $\beta_{i,i+2}$ or $\beta_{i,i+3}$ for some $0 \leqslant i \lt n$ (with the indices modulo 5).

If $\beta = \beta_{i,i+3}$, we have that

\begin{equation*} \eta_{\alpha_{i-1}}(\beta_{i,i+3}) = \alpha_{i-2} \in \mathscr{Y}(S), \end{equation*}

\begin{equation*} \eta_{\alpha_{i}}(\beta_{i,i+3}) = \langle \{\beta_{i,i+2}, \gamma_{i-1,i+2}\} \rangle \in \mathscr{Y}(S)^{1}. \end{equation*}

See Figure 10 for an example.

Figure 10.

On the left, the curves $\gamma_{i-1,i+2}$ (orange) and $\beta_{i,i+2}$; on the right, the curve they uniquely determine, $\eta_{\alpha_{i}}(\beta_{i,i+3})$, in green.

If $\beta = \beta_{i,i+2}$, on one hand, we have that

\begin{equation*}\eta_{\alpha_{i-1}}(\beta_{i,i+2}) = \beta_{i-1,i+2} \in \mathscr{Y}(S).\end{equation*}

On the other hand, for $\eta_{\alpha_{i}}(\beta_{i,i+2})$, we need an auxiliary curve, which we define as follows (using indices modulo 5):

\begin{equation*}\delta = \langle \{\beta_{i-1,i+2}, \gamma_{i-1,i+1}\} \rangle.\end{equation*}

See Figure 11. Then, we have that

\begin{equation*}\eta_{\alpha_{i}}(\beta_{i,i+2}) = \langle\{\alpha_{i-2}, \delta\}\rangle \in \mathscr{Y}(S)^{2}.\end{equation*}

Figure 11.

On the left, the curves $\beta_{i-1,i+2}$ (blue) and $\gamma_{i-1,i+1}$ (green); on the right, the auxiliary curve δ (orange) which they uniquely determine.

See Figure 12.

Figure 12.

On the left, the curves $\alpha_{i-2}$ (blue) and the auxiliary curve δ (orange); on the right, $\eta_{\alpha_{i}}(\beta_{i,i+2})$, which they uniquely determine.

Given that the rest of the cases (with $\alpha_{i+1}$, $\alpha_{i+2}$ and $\alpha_{i+3}$) are analogous to the cases above, we get that $\eta_{\mathscr{O}}(\mathscr{D}_{1}) \subset \mathscr{Y}(S)^{2}$.

Again, since S is a five-punctured sphere, then every $\gamma \in \mathscr{D}_{2}$ is of the form either $\gamma_{i,i+2}$ or $\gamma_{i,i+3}$ for some $0 \leqslant i \lt n$ (with the indices modulo 5).

If $\gamma = \gamma_{i,i+2}$, we have that

\begin{equation*}\eta_{\alpha_{i}}(\gamma_{i,i+2}) = \alpha_{i+1} \in \mathscr{Y}(S),\end{equation*}

and that the case for $\eta_{\alpha_{i-1}}(\gamma_{i,i+2})$ is analogous to the case for $\eta_{\alpha_{i}}(\beta_{i,i+3})$; hence

\begin{equation*}\eta_{\alpha_{i-1}}(\gamma_{i,i+2}) \in \mathscr{Y}(S)^{1}.\end{equation*}

If $\gamma = \gamma_{i,i+3}$, we have that

\begin{equation*}\eta_{\alpha_{i}}(\gamma_{i,i+3}) = \gamma_{i+1,i+3} \in \mathscr{Y}(S),\end{equation*}

and that the case for $\eta_{\alpha_{i-1}}(\gamma_{i,i+3})$ is analogous to the case for $\eta_{\alpha_{i}}(\beta_{i,i+2})$; hence

\begin{equation*}\eta_{\alpha_{i-1}}(\gamma_{i,i+3}) \in \mathscr{Y}(S)^{2}.\end{equation*}

Analogously to the cases for $\mathscr{D}_{1}$, the cases above imply that $\eta_{\mathscr{O}}(\mathscr{D}_{2}) \subset \mathscr{Y}(S)^{2}$.

Finally, the cases for $\eta_{\mathscr{O}}^{-1}(\mathscr{D}_{1})$ and $\eta_{\mathscr{O}}^{-1}(\mathscr{D}_{2})$ mirror the cases for $\eta_{\mathscr{O}}(\mathscr{D}_{2})$ and $\eta_{\mathscr{O}}(\mathscr{D}_{1})$, respectively. Therefore, $\eta_{\mathscr{O}}^{\pm 1}(\mathscr{Y}(S)) \subset \mathscr{Y}(S)^{2}$.

3.5. Proof of Claim 1: $\eta_{\mathscr{O}}^{\pm 1}(\mathscr{O}) \subset \mathscr{Y}(S)$

Let $\alpha_{i}, \alpha_{j} \in \mathscr{O}$. Note that if α_i is disjoint from α_j, then $\eta_{\alpha_{i}}^{\pm 1}(\alpha_{j}) = \alpha_{j} \in \mathscr{Y}(S)$. So, we only focus on the cases where α_i intersects α_j, i.e. $|i-j| = 1$.

For the proof of Claim 1, let $0 \leqslant i \lt n$, then (by the same arguments as above) we have that (using indices modulo n):

\begin{equation*}\eta_{\alpha_{i+1}}(\alpha_{i}) = \gamma_{i,i+2}.\end{equation*}

Also, we have

\begin{equation*}\eta_{\alpha_{i+1}}^{-1}(\alpha_{i}) = \beta_{i,i+2}.\end{equation*}

With this and Remark 1, we have that $\eta_{\mathscr{O}}^{\pm 1}(\mathscr{O}) \subset \mathscr{Y}(S)$.

3.6. Proof of Claim 2: $\eta_{\mathscr{O}}^{\pm 1}(\mathscr{D}_{1}) \subset \mathscr{Y}(S)^{2}$

Let $\beta_{i,j} \in \mathscr{D}_{1}$ and $\alpha_{k} \in \mathscr{O}$. Note that if α_k is disjoint from $\beta_{i,j}$ then $\eta_{\alpha_{k}}^{\pm 1}(\beta_{i,j}) = \beta_{i,j} \in \mathscr{Y}(S)$. So, we only focus on the cases where α_k intersects $\beta_{i,j}$.

Now, we divide the proof of this claim into three cases according to the auxiliary curves needed for the proof; that said, the auxiliary curves are easier to define according to the triple $(i,j,k)$:

First case: Suppose that one of the following is true: [ $j \neq i +2$ and $i \neq j+2$] (both modulo n), or [ $j = i+2$ and $k \neq i, i+1$], or [ $i = j+2$ and $k \neq j,j+1$]. According to which of these is true, to ease the definition of the auxiliary curve $\beta_{i,j}$, we have the following subcases:

1. (i) Suppose $j \neq i +2$ and $i \neq j+2$ (both modulo n).

2. (ii) Suppose $j = i+2$ and $k \neq i, i+1$.

3. (iii) Suppose $i = j+2$ and $k \neq j,j+1$.

Then, there exists non-empty sets $O \subset \mathscr{O}$ and $D \subset \mathscr{D}_{2}$ such that $\beta_{i,j} = \langle O \cup D \rangle$ and α_k is disjoint from every element in D. Thus, $\eta_{\alpha_{k}}^{\pm 1}(\beta_{i,j}) = \langle \eta_{\alpha_{k}}^{\pm 1}(O) \cup \eta_{\alpha_{k}}^{\pm 1}(D) \rangle = \langle \eta_{\alpha_{k}}^{\pm 1}(O) \cup D\rangle$. From this, by Claim 1, we have that $\eta_{\alpha_{k}}^{\pm 1}(\beta_{i,j}) \in \mathscr{Y}(S)^{1}$.

The existence of these sets for each subcase is inferred from the following examples (see Figure 13):

1. (i) If $j \neq i+2$ and $i \neq j+2$ (both modulo n), and if $k=i-1$ or k = i, we set $O = \{\alpha_{i+1}, \ldots, \alpha_{j-2}\} \cup \{\alpha_{j+1}, \ldots, \alpha_{i-2}\}$ and $D = \{\gamma_{j-1,j+1}\}$. The case for $k = j-1$ or k = j is analogous.

2. (ii) If $j = i+2$ and $k \neq i, i+1$, we set $O = \{\alpha_{j+1}, \ldots, \alpha_{i-2}\}$ and $D = \{\gamma_{i-1,i+1}\}$.

3. (iii) The case for $i = j+2$ and $k \neq j,j+1$ is analogous to the subcase (ii).

Second case: Suppose $j = i+2$ and k = i (the case $i = j+2$ and k = j is completely analogous). Then $\eta_{\alpha_{i}}^{-1}(\beta_{i,i+2}) = \alpha_{i+1}$.

Figure 13.

On top, the curves corresponding to the subcase (i) and the curve $\beta_{i,j}$; on the bottom, the curves corresponding to the subcase (ii). On the right, the curves $\beta_{i,j}$ they uniquely determine.

For $\eta_{\alpha_{i}}(\beta_{i,i+2})$, we need an auxiliary curve, which we define as follows (recall the indices are modulo n):

\begin{equation*}\delta := \langle \{\alpha_{i+4}, \ldots \alpha_{i-2}\} \cup \{\beta_{i+2,i+4}\} \cup \{\gamma_{i-1,i+1}\}\rangle.\end{equation*}

See Figure 14 for an example.

Figure 14.

On top, the curves needed to uniquely determine the auxiliary curve δ. On the bottom, the curves needed to uniquely determine $\eta_{\alpha_{i}}(\beta_{i,i+2})$ in the second case.

Then, we have that

\begin{equation*}\eta_{\alpha_{i}}(\beta_{i,i+2}) = \langle \{\alpha_{i+3}, \ldots, \alpha_{i-2}\} \cup \{\delta\}\rangle \in \mathscr{Y}(S)^{2}.\end{equation*}

See Figure 14 for an example.

Third case: Suppose $j = i+2$ and $k = i+1$ (the case $i = j+2$ and $k =j+1$ is completely analogous). Then $\eta_{\alpha_{i+1}}(\beta_{i,i+2}) = \alpha_{i}$.

Now, for $\eta_{\alpha_{i+1}}^{-1}(\beta_{i,i+2})$, we need again an auxiliary curve, which we define as follows (recall the indices are modulo n):

\begin{equation*}\delta := \langle \{\alpha_{i+3}, \ldots, \alpha_{i-3}\} \cup \{\beta_{i-2,i}\} \cup \{\gamma_{i+1,i+3}\}\rangle.\end{equation*}

See Figure 15 for an example.

Then, we have that

\begin{equation*}\eta_{\alpha_{i+1}}^{-1}(\beta_{i,i+2}) = \langle \{\alpha_{i+3}, \ldots, \alpha_{i-2}\} \cup \{\delta\}\rangle \in \mathscr{Y}(S)^{2}.\end{equation*}

See Figure 15 for an example.

Figure 15.

On top, the curves needed to uniquely determine the auxiliary curve δ. On the bottom, the curves needed to uniquely determine $\eta_{\alpha_{i+1}}^{-1}(\beta_{i,i+2})$ in the second case.

Thus, $\eta_{\mathscr{O}}(\mathscr{D}_{1}) \subset \mathscr{Y}(S)^{2}$.

3.7. Proof of Claim 3: $\eta_{\mathscr{O}}^{\pm 1}(\mathscr{D}_{2}) \subset \mathscr{Y}(S)^{3}$

The proof of this claim follows from Claim 1 and Claim 2: If $\gamma \in \mathscr{D}_{2}$, then there exist sets $O \subset \mathscr{O}$ and $D \subset \mathscr{D}_{1}$ such that $\gamma = \langle O \cup D \rangle$. Thus, for all $\alpha_{i} \in \mathscr{O}$, $\eta_{\alpha_{i}}^{\pm 1}(\gamma) = \langle \eta_{\alpha_{i}}^{\pm 1}(O) \cup \eta_{\alpha_{i}}^{\pm 1}(D)\rangle$. By Claims 1 and 2, this implies that $\eta_{\alpha_{i}}(\gamma) \in \mathscr{Y}(S)^{3}$.

The existence of these sets can be easily inferred from the following examples (see Figure 16):

• • If $j \neq i+2$ and $i \neq j+2$ (both modulo n), we set $O = \{\alpha_{i+1}, \ldots, \alpha_{j-2}\} \cup \{\alpha_{j+1}, \ldots, \alpha_{i-2}\}$ and $D = \{\beta_{i+1,j+1}\}$.

• • If $j = i+2$ (modulo n), we set $O = \{\alpha_{j+1}, \ldots, \alpha_{i-2}\}$ and $D = \{\beta_{i+1,j+1}\}$.

The rest of the cases are analogous to one of the above.

Figure 16.

On the left, the sets O (red) and D (blue); on the right, the curve $\gamma_{i,j}$ they uniquely determine.

3.8. Proof of Theorem B

For the purposes of this work, we define the set $\mathfrak{X}(S)$ as the set $\mathfrak{X}_{A}$ from Section 4 in [Reference Aramayona and Leininger2], which was proved to be rigid (see Theorem 4.2 in [Reference Aramayona and Leininger2]).

According to the notation above, we define $\mathfrak{X}(S)$ as follows:

\begin{equation*}\mathfrak{X}(S) := \mathscr{O}^{1} \cup \mathscr{G}\cdot\mathscr{O}.\end{equation*}

Proof of Theorem B

To prove Theorem B, we only need to prove that $\mathscr{Y}(S) \subset \mathfrak{X}(S)^{1}$, since this would imply that (by Theorem A) $\mathcal{C}(S) = \mathscr{Y}(S)^{\omega} \subset \mathfrak{X}(S)^{\omega}$.

For $\gamma \in \mathscr{O}$, it is obvious that $\gamma \in \mathfrak{X}(S)$. Thus, we suppose that $\gamma \in \mathscr{D}_{1} \cup \mathscr{D}_{2}$.

If $\gamma \in \mathscr{D}_{1}$, there exist $0 \leqslant i \lt j \lt n$ such that $\gamma = \beta_{i,j}$. Thus, we have two possible cases:

1. (1) Suppose that $j \neq i+2$ (modulo n). Then (see Figure 17)

   \begin{equation*}\beta_{i,j} = \langle \{\alpha_{i+1}, \ldots, \alpha_{j-2}\} \cup \{\alpha_{j+1}, \ldots \alpha_{i-2}\} \cup \{\eta_{\alpha_{i}}(\alpha_{i-1})\}\rangle.\end{equation*}

2. (2) Suppose that $j = i+2$ (modulo n). Then, by definition of a half-twist, we have that $\beta_{i,j} = \eta_{\alpha_{i+1}}^{-1}(\alpha_{i})$.

Figure 17.

On the left, the curves needed to uniquely determine the curve $\beta_{i,j}$ on the right.

If $\gamma \in \mathscr{D}_{2}$, there exist $0 \leqslant i \lt j \lt n$ such that $\gamma = \gamma_{i,j}$. Thus, we have two possible cases, which are analogous to (1) and (2) above, simply substituting $\eta_{\alpha_{i}}(\alpha_{i-1})$ for $\eta_{\alpha_{i}}^{-1}(\alpha_{i-1})$ in case (1) and $\eta_{\alpha_{i+1}}^{-1}(\alpha_{i})$ for $\eta_{\alpha_{i+1}}(\alpha_{i})$ in case (2).

Then, $\mathscr{Y}(S) \subset \mathfrak{X}(S)^{1}$, and therefore $\mathfrak{X}(S)^{\omega} = \mathcal{C}(S)$.

4.1. The foundation of $\mathscr{Y}(S)$

Let $\mathscr{C} := \{\alpha_{0}^{0}, \ldots, \alpha_{0}^{n-1}, \alpha_{1}\}$ be a set of curves such that $\alpha_{0}^{i}$ is disjoint from $\alpha_{0}^{j}$ if i ≠ j, and $i(\alpha_{1},\alpha_{0}^{i}) = 1$ for all $0 \leqslant i \lt n$. See Figure 18.

Using $\mathscr{C}$, we number the punctures in S in the following way: the ith puncture is in the annulus bounded by $\alpha_{0}^{i-1}$ and $\alpha_{0}^{i}$. In Figure 18, this means that the ith puncture is on the left of $\alpha_{0}^{i}$.

Figure 18.

The curve α ₁ in red, and the curves $\alpha_{0}^{0}, \alpha_{0}^{1}, \alpha_{0}^{2}$ in blue.

With this, we can define for each $0 \leqslant i \lt n$ the following curve (see Figure 19):

\begin{equation*}\beta_{i} := \langle \mathscr{C} \backslash \{\alpha_{0}^{i}\}\rangle.\end{equation*}

Figure 19.

On the left, the curves needed to uniquely determine the curve β ₁ on the right.

Now, we define $\mathscr{D} := \{\beta_{i}\}_{i = 0}^{n-1}$. Note that $\mathscr{D}$ is a closed outer chain.

Ideally we would like $\mathscr{Y}(S)$ to be equal to the union of $\mathscr{C}$ and $\mathscr{D}$; however, $(\mathscr{C} \cup \mathscr{D})^{2} = (\mathscr{C} \cup \mathscr{D})^{1}$. So, we need some auxiliary curves.

4.2. Auxiliary curves

To define the auxiliary curves, first we must recall (from § 3.3) that if α is an outer curve on S, then η_α denotes the half-twist along α.

For each $0 \leqslant i \lt n$, we now define the following curves (with indices modulo n):

\begin{equation*}\varepsilon_{i}^{+} := \eta_{\beta_{i}}(\beta_{i-1}) \qquad\qquad \varepsilon_{i}^{-} := \eta_{\beta_{i}}^{-1}(\beta_{i-1}).\end{equation*}

See Figure 20.

Figure 20.

On the left the curve $\varepsilon_{i}^{+}$; on the right the curve $\varepsilon_{i}^{-}$.

Thus, we can define the sets $\mathscr{A}^{+} := \{\varepsilon_{i}^{+}\}_{i = 0}^{n-1}$, $\mathscr{A}^{-} := \{\varepsilon_{i}^{-}\}_{i = 0}^{n-1}$, and $\mathscr{A} := \mathscr{A}^{+} \cup \mathscr{A}^{-}$.

Finally, we define

\begin{equation*}\mathscr{Y}(S) := \mathscr{C} \cup \mathscr{D} \cup \mathscr{A}.\end{equation*}

4.3. Proof of Theorem A

For the proof of Theorem A, we first recall some definitions and properties of Dehn twists along curves on S.

Let α be a curve on S; we denote by τ_α the (left) Dehn twist along α. Similarly to Remark 1, we have the following property.

Similarly to the notation around half-twists, if C and A are sets of curves on S, we denote by $\tau_{C}(A)$ the following set

\begin{equation*}\tau_{C}(A) = \bigcup_{\gamma \in C} \tau_{\gamma}(A).\end{equation*}

Now, let $\mathscr{G} := \{\tau_{\alpha}^{\pm 1}: \alpha \in \mathscr{C}\} \cup \{\eta_{\beta}^{\pm 1}: \beta \in \mathscr{D}\}$. It is a well-known fact (Theorem 4.14 in [Reference Farb and Margalit4]) that $\mathscr{G}$ is a symmetric generating set of $\mathrm{Mod}(S)$ (often called the Humphries–Lickorish generating set). As in the previous section, if A is a set of curves on S, we denote by $\mathscr{G} \cdot A$ the following set

\begin{equation*} \begin{array}{rcl} \mathscr{G} \cdot A & := & \tau_{\mathscr{C}}^{\pm 1}(A) \cup \eta_{\mathscr{D}}^{\pm 1}(A)\\ & = & \left(\bigcup_{\gamma \in \mathscr{C}} \tau_{\gamma}^{\pm 1}(A)\right) \cup \left(\bigcup_{\varepsilon \in \mathscr{D}} \eta_{\varepsilon}(A)\right). \end{array} \end{equation*}

As in the previous section, we state a key lemma for the proof of Theorem A, leaving its proof to the following subsections (see § 4.4, 4.5, 4.6 and 4.7).

Proof of Theorem A

We divide the proof into two parts: first we prove that all outer curves and non-separating curves on S are elements in $\mathscr{Y}(S)^{\omega}$; then we argue that any curve on S is uniquely determined by a finite set of non-separating curves and outer curves, thus obtaining the desired result.

First part: Let γ be either an outer curve or a non-separating curve on S. Recall that all outer curves on S have the same topological type, and it is the same case for all non-separating curves. Then, there exists a curve $\delta \in \mathscr{Y}(S)$ and a homeomorphism $h \in \mathrm{Mod}(S)$ such that $h(\delta) = \gamma$. Since $\mathscr{G}$ is a symmetric generating set of $\mathrm{Mod}(S)$, there exist elements $f_{1}, \ldots, f_{k} \in \mathscr{G}$ and powers $n_{1}, \ldots, n_{k} \geqslant 0$ such that

\begin{equation*}f_{1}^{n_{1}} \circ \cdots \circ f_{k}^{n_{k}}(\delta) = \gamma.\end{equation*}

By the same argument as in the proof of Theorem A for the case of genus zero, an iterated use of Lemma 2 for $\mathscr{Y}(S)$ and its translates by powers of $\mathscr{G}$ implies that $\gamma \in \mathscr{Y}(S)^{6(n_{1} + \cdots + n_{k})}$. Thus, every outer curve and every non-separating curve on S is an element of $\mathscr{Y}(S)^{\omega}$.

Second part: Let γ be a separating curve on S that is not an outer curve. The topological type of γ is then determined the punctures it bounds. Thus, up to homeomorphism, γ is as in the example on Figure 21 (right). Then, there exist finite sets N and O of non-separating and outer curves, respectively, that uniquely determine γ. See Figure 21 for an example. The first part of this proof implies that there exists $k \geqslant 0$ such that $N \cup O \subset \mathscr{Y}(S)^{k}$. Thus, $\gamma \in \mathscr{Y}(S)^{k+1}$ and therefore every curve on S that is neither an outer curve nor a non-separating curve is also an element of $\mathscr{Y}(S)^{\omega}$.

Figure 21.

On the left, the set N (red) and the set O (green); on the right the non-outer separating curve they uniquely determine.

We now proceed to the proof of Lemma 2. Similarly to the previous section, we prove Lemma 2 by proving the following claims:

Claim 1: $\tau_{\mathscr{C}}^{\pm 1}(\mathscr{C} \cup \mathscr{D}) \subset \mathscr{Y}(S)^{3}$.

Claim 2: $\tau_{\mathscr{C}}^{\pm 1}(\mathscr{A}) \subset \mathscr{Y}(S)^{6}$.

Claim 3: $\eta_{\mathscr{D}}^{\pm 1}(\mathscr{C} \cup \mathscr{D}) \subset \mathscr{Y}(S)^{1}$.

Claim 4: $\eta_{\mathscr{D}}^{\pm 1}(\mathscr{A}) \subset \mathscr{Y}(S)^{3}$.

4.4. Proof of Claim 1: $\tau_{\mathscr{C}}^{\pm 1}(\mathscr{C} \cup \mathscr{D}) \subset \mathscr{Y}(S)^{3}$

We divide this proof into two parts:

First part $\tau_{\mathscr{C}}^{\pm 1}(\mathscr{D}) \subset \mathscr{Y}(S)^{2}$: Note that $\tau_{\alpha}(\beta) = \beta$ if α and β are disjoint from each other. So, we only need to prove that $\tau_{\alpha_{0}^{i}}^{\pm 1}(\beta_{i}) \in \mathscr{Y}(S)^{2}$; to prove this, we need some auxiliary curves.

For each $0 \leqslant i \lt n$, we define the following curves (recall the subindices are modulo n)

\begin{equation*}\gamma_{i}^{+} := \langle\{\beta_{i+1}, \ldots, \beta_{i-2}\} \cup \{\varepsilon_{i}^{+}\} \cup \{\alpha_{1}\}\rangle \in \mathscr{Y}(S)^{1},\end{equation*}

\begin{equation*}\gamma_{i}^{-} := \langle \{\beta_{i+1}, \ldots, \beta_{i-2}\} \cup \{\varepsilon_{i}^{-}\} \cup \{\alpha_{1}\}\rangle \in \mathscr{Y}(S)^{1}.\end{equation*}

See Figure 22 for examples.

Figure 22.

On the left, the curves needed to uniquely determine the curves $\gamma_{i}^{+}$ and $\gamma_{i}^{-}$.

Now, we have the following (see Figure 23):

\begin{equation*}\tau_{\alpha_{0}^{i}}(\beta_{i}) = \langle \{\gamma_{i}^{-},\gamma_{i+1}^{+}\} \cup \{\alpha_{0}^{i+1}, \ldots, \alpha_{0}^{i-1}\}\rangle \in \mathscr{Y}(S)^{2},\end{equation*}

\begin{equation*}\tau_{\alpha_{0}^{i}}^{-1}(\beta_{i}) = \langle \{\gamma_{i}^{+}, \gamma_{i+1}^{-}\} \cup \{\alpha_{0}^{i+1}, \ldots, \alpha_{0}^{i-1}\}\rangle \in \mathscr{Y}(S)^{2}.\end{equation*}

Figure 23.

On the left, the curves needed to uniquely determine the curves on the right $\tau_{\alpha_{0}^{i}}(\beta_{i})$ (top) and $\tau_{\alpha_{0}^{i}}^{-1}(\beta_{i})$ (bottom).

Thus, $\tau_{\mathscr{C}}^{\pm 1}(\mathscr{D}) \subset \mathscr{Y}(S)^{2}$.

Second part $\tau_{\mathscr{C}}^{\pm 1}(\mathscr{C}) \subset \mathscr{Y}(S)^{3}$: Recalling Remark 2 and that $\tau_{\alpha}(\beta) = \beta$ if α and β are disjoint from each other, we only need to prove that $\tau_{\alpha_{0}^{i}}^{\pm1}(\alpha_{1}) \in \mathscr{Y}(S)^{3}$ for each $0 \leqslant i \lt n$.

Note that $\alpha_{1} = \langle \mathscr{D} \rangle$ (see Figure 24), thus $\tau_{\alpha_{0}^{i}}^{\pm1}(\alpha_{1}) = \langle \tau_{\alpha_{0}^{i}}^{\pm1}(\mathscr{D})\rangle$. This implies that $\tau_{\alpha_{0}^{i}}^{\pm1}(\alpha_{1}) \in \mathscr{Y}(S)^{3}$ for each $0 \leqslant i \lt n$.

Figure 24.

On the left, the set $\mathscr{D}$; on the right the curve α ₁.

Therefore, $\tau_{\mathscr{C}}^{\pm 1}(\mathscr{C} \cup \mathscr{D}) \subset \mathscr{Y}(S)^{3}$.

4.5. Proof of Claim 2: $\tau_{\mathscr{C}}^{\pm 1}(\mathscr{A}) \subset \mathscr{Y}(S)^{6}$

As was done in the previous subsection, we only need to prove that for each $0 \leqslant i \lt n$ we have that $\tau_{\alpha_{0}^{i-1}}^{\pm 1}(\varepsilon_{i}^{\pm}), \tau_{\alpha_{0}^{i}}^{\pm 1}(\varepsilon_{i}^{\pm}) \in \mathscr{Y}(S)^{6}$, since any other curve in $\mathscr{C}$ is disjoint from both $\varepsilon_{i}^{+}$ and $\varepsilon_{i}^{-}$.

To do this, we divide the proof into two parts, according to the number of punctures in S.

First part, n = 3: Since we do not have enough punctures for this part, we uniquely determine $\varepsilon_{i}^{\pm}$ by using the curves $\gamma_{i}^{\pm}$ from § 4.4 twice.

In the case for $\alpha_{0}^{i-1}$ recall and check that:

\begin{equation*}\gamma_{i+1}^{\pm} := \langle \{\beta_{i+2}\} \cup \{\varepsilon_{i+1}^{\pm}\} \cup \{\alpha_{1}\} \rangle \,\text{(see Figure 22)}.\end{equation*}

\begin{equation*}\gamma_{i}^{\pm} = \langle \{\beta_{i+1}\} \cup \{\alpha_{1}\} \cup \{\gamma_{i+1}^{\mp}\} \rangle \,\text{(see Figure 25 top)}.\end{equation*}

\begin{equation*}\varepsilon_{i}^{\pm} = \langle \{\alpha_{0}^{i-2}\} \cup \{\alpha_{1}\} \cup \{\gamma_{i}^{\pm}\} \rangle \,\text{(see Figure 25 bottom)}.\end{equation*}

Figure 25.

On the left, the curves needed to uniquely determine the curves $\gamma_{i}^{+}$ (top) and $\varepsilon_{i}^{+}$ (bottom) on the right.

Thus, by Claim 1, we have that

\begin{equation*}\begin{array}{rl} \tau_{\alpha_{0}^{i-1}}^{\pm 1}(\gamma_{i+1}^{\pm}) & = \langle \tau_{\alpha_{0}^{i-1}}^{\pm 1}(\{\beta_{i+2}\}) \cup \tau_{\alpha_{0}^{i-1}}^{\pm 1}(\{\varepsilon_{i+1}^{\pm}\}) \cup \tau_{\alpha_{0}^{i-1}}^{\pm 1}(\{\alpha_{1}\}) \rangle\\ & = \langle \tau_{\alpha_{0}^{i-1}}^{\pm 1}(\{\beta_{i+2}\}) \cup \{\varepsilon_{i+1}^{\pm}\} \cup \tau_{\alpha_{0}^{i-1}}^{\pm 1}(\{\alpha_{1}\}) \rangle \in \mathscr{Y}(S)^{4}, \end{array}\end{equation*}

hence

\begin{equation*}\tau_{\alpha_{0}^{i-1}}^{\pm 1}(\gamma_{i}^{\pm}) = \langle \tau_{\alpha_{0}^{i-1}}^{\pm 1}(\{\beta_{i+1}\}) \cup \tau_{\alpha_{0}^{i-1}}^{\pm 1}(\{\alpha_{1}\}) \cup \tau_{\alpha_{0}^{i-1}}^{\pm 1}(\{\gamma_{i+1}^{\mp})\} \rangle \in \mathscr{Y}(S)^{5},\end{equation*}

and finally

\begin{equation*}\tau_{\alpha_{0}^{i-1}}^{\pm 1}(\varepsilon_{i}^{\pm}) = \langle \tau_{\alpha_{0}^{i-1}}^{\pm 1}(\{\alpha_{0}^{i-2}\}) \cup \tau_{\alpha_{0}^{i-1}}^{\pm 1}(\{\alpha_{1}\}) \cup \tau_{\alpha_{0}^{i-1}}^{\pm 1}(\{\gamma_{i}^{\pm}\}) \rangle \in \mathscr{Y}(S)^{6}.\end{equation*}

The case for $\alpha_{0}^{i}$ is completely analogous, substituting $\gamma_{i+1}^{\pm}$ by $\gamma_{i-1}^{\pm}$.

Second part, $n \geqslant 4$: We proceed similarly to the first case in § 3.6 to prove that $\tau_{\alpha}(\varepsilon_{i}^{\pm}) \in \mathscr{Y}(S)^{4}$ with $\alpha = \alpha_{0}^{i-1}, \alpha_{0}^{i}$. We can uniquely determine $\varepsilon_{i}^{\pm}$ as follows: We define the set

\begin{equation*}C = \{\alpha_{0}^{i+1}, \ldots, \alpha_{0}^{i-2}\} \cup \{\alpha_{1}\},\end{equation*}

and the curve

\begin{equation*}\varepsilon^{\pm} = \left\{ \begin{array}{cl} \varepsilon_{i+1}^{\pm} & \text{for the case}\ \alpha_{0}^{i-1}, \\ \varepsilon_{i-1}^{\pm} & \text{for the case}\ \alpha_{0}^{i}. \end{array}\right.\end{equation*}

Then we check that (see Figure 26)

\begin{equation*}\varepsilon_{i}^{\pm} = \langle C \cup \{\varepsilon^{\mp}\}\rangle.\end{equation*}

Figure 26.

On the left, the curves needed to uniquely determine $\varepsilon_{i}^{+}$ on the right, for the cases $\alpha_{0}^{i-1}$ (top) and $\alpha_{0}^{i}$ (bottom).

Thus, by Claim 1, we have that

\begin{equation*}\begin{array}{rl} \tau_{\alpha_{0}^{i-1}}(\varepsilon_{i}^{\pm}) & = \langle \tau_{\alpha_{0}^{i-1}}(C) \cup \tau_{\alpha_{0}^{i-1}}(\{\varepsilon_{i+1}^{\mp}\})\rangle \\ & = \langle \tau_{\alpha_{0}^{i-1}}(C) \cup \{\varepsilon_{i+1}^{\mp}\}\rangle \in \mathscr{Y}(S)^{4}, \end{array} \end{equation*}

\begin{equation*}\begin{array}{rl} \tau_{\alpha_{0}^{i}}(\varepsilon_{i}^{\pm}) & = \langle \tau_{\alpha_{0}^{i}}(C) \cup \tau_{\alpha_{0}^{i}}(\{\varepsilon_{j}\})\rangle \\ & = \langle \tau_{\alpha_{0}^{i}}(C) \cup \{\varepsilon_{j}\}\rangle \in \mathscr{Y}(S)^{4}. \end{array} \end{equation*}

The cases for $\tau_{\alpha_{0}^{i-1}}^{-1}(\varepsilon_{i}^{\pm}), \tau_{\alpha_{0}^{i}}^{-1}(\varepsilon_{i}^{\pm}) \in \mathscr{Y}(S)^{4}$ are analogous.

Therefore $\tau_{\mathscr{C}}^{\pm 1}(\mathscr{A}) \subset \mathscr{Y}(S)^{6}$.

4.6. Proof of Claim 3: $\eta_{\mathscr{D}}^{\pm 1}(\mathscr{C} \cup \mathscr{D}) \subset \mathscr{Y}(S)^{1}$

Recall that $\varepsilon_{i}^{\pm} = \eta_{\beta_{i}}^{\pm1}(\beta_{i-1})$. Using this and Remark 1, we have that $\eta_{\mathscr{D}}^{\pm 1}(\mathscr{D}) \subset \mathscr{Y}(S)$.

Now, similarly to the previous subsections, we only need to prove that $\eta_{\beta_{i}}^{\pm 1}(\alpha_{0}^{i}) \in \mathscr{Y}(S)^{1}$. This can be done as follows: We define a set $B \subset \mathscr{D}$ as

\begin{equation*} B = \left\{\begin{array}{cl} \varnothing & \text{if}\ n = 3, \\ \{\beta_{i+2}, \ldots, \beta_{i-2}\} & \text{if}\ n \geqslant 4. \end{array}\right.\end{equation*}

Then we have (see Figure 27)

\begin{equation*}\eta_{\beta_{i}}(\alpha_{0}^{i}) = \langle \{\varepsilon_{i}^{+},\varepsilon_{i+1}^{-}\} \cup B \cup \{\alpha_{0}^{i+1}, \ldots, \alpha_{0}^{i-1}\}\rangle,\end{equation*}

\begin{equation*}\eta_{\beta_{i}}^{-1}(\alpha_{0}^{i}) = \langle \{\varepsilon_{i}^{-},\varepsilon_{i+1}^{+}\} \cup B \cup \{\alpha_{0}^{i+1}, \ldots, \alpha_{0}^{i-1}\}\rangle.\end{equation*}

Figure 27.

On the left, the curves needed to uniquely determine the curves $\eta_{\beta_{i}}(\alpha_{0}^{i})$ (top) and $\eta_{\beta_{i}}^{-1}(\alpha_{0}^{i})$ (bottom) on the right.

Therefore, $\eta_{\mathscr{D}}^{\pm 1}(\mathscr{C} \cup \mathscr{D}) \subset \mathscr{Y}(S)^{1}$.

4.7. Proof of Claim 4: $\eta_{\mathscr{D}}^{\pm 1}(\mathscr{A}) \subset \mathscr{Y}(S)^{3}$

For this claim, similarly to the previous subsections, we only need to prove for $\varepsilon = \varepsilon_{i}^{+}, \varepsilon_{i}^{-}$ that $\eta_{\beta_{i-2}}^{\pm 1}(\varepsilon), \eta_{\beta_{i-1}}^{\pm 1}(\varepsilon), \eta_{\beta_{i}}^{\pm 1}(\varepsilon), \eta_{\beta_{i+1}}^{\pm 1}(\varepsilon) \in \mathscr{Y}(S)^{3}$.

First note that by definition $\eta_{\beta_{i-1}}(\varepsilon_{i}^{+}) = \eta_{\beta_{i-1}}^{-1}(\varepsilon_{i}^{-}) = \beta_{i} \in \mathscr{Y}(S)$ and $\eta_{\beta_{i}}^{-1}(\varepsilon_{i}^{+}) = \eta_{\beta_{i}}(\varepsilon_{i}^{-}) = \beta_{i-1} \in \mathscr{Y}(S)$.

Also note that we can uniquely determine $\varepsilon_{i}^{\pm}$ as in § 4.5 as follows (see Figure 28):

\begin{equation*}\varepsilon_{i}^{\pm} = \langle \{\alpha_{1}\} \cup \{\alpha_{0}^{i+1}, \ldots, \alpha_{0}^{i-2}\} \cup \{\gamma_{i}^{\pm}\} \rangle,\end{equation*}

hence, by Claim 3 and using that $\eta_{\beta_{i-2}}^{\pm 1}(\gamma_{i}^{+}) = \eta_{\beta_{i+1}}^{\pm 1}(\gamma_{i}^{+}) = \gamma_{i}^{+}$ and $\eta_{\beta_{i-2}}^{\pm 1}(\gamma_{i}^{-}) = \eta_{\beta_{i+1}}^{\pm 1}(\gamma_{i}^{-}) = \gamma_{i}^{-}$, we get that $\eta_{\beta_{i-2}}^{\pm 1}(\varepsilon_{i}^{+}), \eta_{\beta_{i-2}}^{\pm 1}(\varepsilon_{i}^{-}), \eta_{\beta_{i+1}}^{\pm 1}(\varepsilon_{i}^{+}), \eta_{\beta_{i+1}}^{\pm 1}(\varepsilon_{i}^{-}) \in \mathscr{Y}(S)^{2}.$

Figure 28.

On the left, the curve needed to uniquely determine the curve $\varepsilon_{i}^{+}$ on the right. The case for $\varepsilon_{i}^{-}$ is analogous.

Thus we only need to prove that $\eta_{\beta_{i-1}}^{-1}(\varepsilon_{i}^{+})$, $\eta_{\beta_{i}}(\varepsilon_{i}^{+})$, $\eta_{\beta_{i-1}}(\varepsilon_{i}^{-})$, $\eta_{\beta_{i}}^{-1}(\varepsilon_{i}^{-}) \in \mathscr{Y}(S)^{3}$.

All of these cases can be uniquely determined by an auxiliary curve $\delta \in \mathscr{Y}(S)^{2}$ (a different one for each case) and the set $\mathscr{C} \backslash \{\alpha_{0}^{i-1}, \alpha_{0}^{i}\}$. See Figure 29 for an example.

Figure 29.

An example of the curve $\eta_{\beta_{i-1}}^{-1}(\varepsilon_{i}^{+})$ (right) being uniquely determined by the set $\mathscr{C} \backslash \{\alpha_{0}^{i-1}, \alpha_{0}^{i}\}$ and the curve δ (left).

Figure 30.

On the left, the curves needed to uniquely determine the curve δ on the right.

The subsets of $\mathscr{Y}(S)^{1}$ used to uniquely determine δ are different for each case and again we make use of the curves of the form $\gamma_{j}^{\pm}$ from the first part of § 4.4, but they are completely analogous to each other and can be inferred from the following examples and Figure 30:

• • If n = 3, then we use the following curve to obtain $\eta_{\beta_{i-1}}^{-1}(\varepsilon_{i}^{+})$:

  \begin{equation*}\delta := \langle \{\alpha_{1}\} \cup \{\varepsilon_{i-1}^{-}\} \cup \{\gamma_{i+1}^{-}\}\rangle.\end{equation*}

• • If $n \geqslant 4$, then we use the following curve to obtain $\eta_{\beta_{i-1}}^{-1}(\varepsilon_{i}^{+})$:

  \begin{equation*}\delta := \langle \{\alpha_{1}\} \cup \{\varepsilon_{i-1}^{-}\} \cup \{\beta_{i+1}, \ldots, \beta_{i-3}\} \cup \{\gamma_{i+1}^{-}\}\rangle.\end{equation*}

Then, $\eta_{\mathscr{D}}^{\pm 1}(\mathscr{A}) \subset \mathscr{Y}(S)^{3}.$

4.8. Proof of Theorem B

For this proof, we need the curves $\gamma_{i}^{\pm}$ from § 4.4 and the following auxiliary curves.

If n = 3, for each $0 \leqslant i \lt n$, we define the following curves $\gamma_{i,i+1}^{+} := \gamma_{i+2}^{-}$ and $\gamma_{i,i+1}^{-} := \gamma_{i+2}^{+}$.

If $n \geqslant 4$, for each $0 \leqslant i \lt n$, we define the following curves (see Figure 31):

\begin{equation*}\gamma_{i,i+1}^{+} := \langle \{\alpha_{1}\} \cup \{\gamma_{i}^{+}, \gamma_{i+1}^{+}, \gamma_{i+2}^{-}\} \cup \{\beta_{i+2}, \ldots, \beta_{i-2}\}\rangle,\end{equation*}

\begin{equation*}\gamma_{i,i+1}^{-} := \langle \{\alpha_{1}\} \cup \{\gamma_{i}^{-}, \gamma_{i+1}^{-}, \gamma_{i+2}^{-}\} \cup \{\beta_{i+2}, \ldots, \beta_{i-2}\}\rangle.\end{equation*}

Figure 31.

On the left, the curves needed to uniquely determine the curves $\gamma_{i,i+1}^{+}$ (top) and $\gamma_{i,i+1}^{-}$ (bottom) on the right.

According to the notation above, we can define $\mathfrak{X}(S)$ as follows:

\begin{equation*}\mathfrak{X}(S) := (\mathscr{C} \cup \mathscr{D})^{1} \cup \{\gamma_{i}^{\pm}\}_{i = 0}^{n-1} \cup \{\gamma_{i,i+1}^{\pm}\}_{i = 0}^{n-1}.\end{equation*}

Note that the set $\mathfrak{X}(S)$ defined above is the set $\mathfrak{X}_{1}$ from Section 6 in [Reference Aramayona and Leininger2], which was proved to be rigid (see Proposition 6.2 in [Reference Aramayona and Leininger2]).

Proof of Theorem B

For this proof, we proceed as in § 3.8, so we only need to prove that $\mathscr{Y}(S) \subset \mathfrak{X}(S)^{1}$.

Given that $\mathscr{C} \cup \mathscr{D} \subset \mathfrak{X}(S)$, we only need to prove that $\mathscr{A} \subset \mathfrak{X}(S)^{1}$. For this, as in § 4.7 note that for each $0 \leqslant i \lt n$ we have that (see Figure 28):

\begin{equation*}\varepsilon_{i}^{+} = \langle \{\gamma_{i}^{+}\} \cup (\mathscr{C} \backslash \{\alpha_{i-1}, \alpha_{i}\})\rangle,\end{equation*}

\begin{equation*}\varepsilon_{i}^{-} = \langle \{\gamma_{i}^{-}\} \cup (\mathscr{C} \backslash \{\alpha_{i-1}, \alpha_{i}\})\rangle.\end{equation*}

Thus, $\mathscr{Y}(S) \subset \mathfrak{X}(S)^{1}$, and therefore $\mathfrak{X}(S)^{\omega} = \mathcal{C}(S)$.

5.1. The definition of $\mathscr{Y}(S)$

Let $C = \{\gamma_{1}, \ldots, \gamma_{k}\}$ be a set of curves for k > 1. We say C is a chain if $i(\gamma_{i},\gamma_{j}) = 1$ for $|i-j| =1$, and γ_i is disjoint from γ_j otherwise. Note the similarities with an outer chain (defined in § 3.1). We define the length of C as its cardinality. See Figure 32 for an example.

Figure 32.

On the left, a chain of length 4; on the right, a chain of length 3 and its corresponding bounding pair.

Note that if a chain C has length 2j, any (open) regular neighbourhood of C is homeomorphic to $S_{j,1}$, while if C has length $2j+1$, then any (open) regular neighbourhood of C is homeomorphic to $S_{j,2}$.

Given a chain C of odd length, the bounding pair associated to C is defined as the set containing the boundary curves of a closed regular neighbourhood of C. See Figure 32 for an example.

Let $\mathscr{C}_{0} := \{\alpha_{1}, \ldots, \alpha_{5}\}$ be the chain depicted on the left in Figure 33, and let $\mathscr{C}_{f} := \{\alpha_{0}^{0}, \ldots, \alpha_{0}^{n}\}$ be the multicurve also depicted on the right in Figure 33. Note that for all $0 \leqslant i \leqslant n$ we have that $i(\alpha_{0}^{i},\alpha_{j}) = 1$ if $j = 1, 5$, and $\alpha_{0}^{i}$ is disjoint from α_j otherwise. Then, we define the set $\mathscr{C} := \mathscr{C}_{0} \cup \mathscr{C}_{f}$.

Figure 33.

The curves in $\mathscr{C}_{0}$ on the left and the curves in $\mathscr{C}_{f}$ on the right.

Figure 34.

On the top left, the subsurfaces $\Sigma_{o}^{+}$ (punctured pair of pants at the front) and $\Sigma_{o}^{-}$ (pair of pants at the back); on the top right, the subsurfaces $S_{0}^{+}$ (pair of pants at the top right) and $S_{0}^{-}$ (punctured pair of pants at the bottom); on the bottom, the subsurfaces $S_{i}^{+}$ (punctured pair of pants at the top) and $S_{i}^{-}$ (possibly punctured pair of pants at the bottom).

We can then fix several subsurfaces of S using $\mathscr{C}$ as follows (see Figure 34):

• • Let $\Sigma_{o}^{+}$ be the closed subsurface of S bounded by $\{\alpha_{1},\alpha_{3}, \alpha_{5}\}$ that is not compact, and $\Sigma_{o}^{-}$ be the compact subsurface of S bounded by $\{\alpha_{1},\alpha_{3}, \alpha_{5}\}$.

• • Define $S_{0}^{+}$ as the compact subsurface of S bounded by $\{\alpha_{0}^{0}, \alpha_{2}, \alpha_{4}\}$, and $S_{0}^{-}$ as the non-compact subsurface of S bounded by $\{\alpha_{0}^{0}, \alpha_{2}, \alpha_{4}\}$.

• • For each $0 \lt i \leqslant n$, let $S_{i}^{+}$ be the subsurface of S bounded by $\{\alpha_{0}^{i}, \alpha_{2}, \alpha_{4}\}$ that contains $S_{0}^{+}$. Let $S_{0}^{-}$ be the subsurface of S bounded by $\{\alpha_{0}^{i}, \alpha_{2}, \alpha_{4}\}$ that does not contain $S_{0}^{+}$. Note that all $S_{i}^{+}$ and $S_{0}^{+}$ are closed subsets of S.

Let $C\subset \mathscr{C}$ be a chain of odd length such that $|C \cap \mathscr{C}_{f}| \leqslant 1$, and let B(C) be the bounding pair associated to C. Then, note that the elements of B(C) are either contained in $\mathrm{int}(S_{i}^{+}) \cup \mathrm{int}(S_{i}^{-})$ for some i or contained in $\mathrm{int}(\Sigma_{o}^{+}) \cup \mathrm{int}(\Sigma_{o}^{-})$. We denote by $\beta_{C}^{+}$ the element of B(C) contained in either $\mathrm{int}(S_{i}^{+})$ for some i or $\Sigma_{o}^{+}$; we denote by $\beta_{C}^{-}$ the element of B(C) contained in either $\mathrm{int}(S_{i}^{-})$ for some i or $\Sigma_{o}^{-}$. See Figure 35.

Figure 35.

On the left, the set $C = \{\alpha_{1},\alpha_{2},\alpha_{3}\}$ and the curves $\beta_{C}^{+}$ (front) and $\beta_{C}^{-}$ (back); on the right, the set $C= \{\alpha_{2},\alpha_{3},\alpha_{4}\}$ and the curves $\beta_{C}^{+}$ (top) and $\beta_{C}^{-}$ (bottom).

We define the set $\displaystyle \mathscr{B} := \{\beta_{C}^{+},\beta_{C}^{-} : C \subset \mathscr{C}$ is a chain such that $|C \cap \mathscr{C}_{f}| \leqslant 1\}$.

Note that since we are considering only chains that contain at most one element of $\mathscr{C}_{f}$, then every curve in $\mathscr{B}$ is non-separating.

Finally, we define

\begin{equation*}\mathscr{Y}(S) := \mathscr{C} \cup \mathscr{B}.\end{equation*}

5.2. Auxiliary curves for Theorem A

Unlike the previous sections, here we use auxiliary curves for the key lemma and the proof of Theorem A, instead of using them for the definition of $\mathscr{Y}(S)$.

For each $0 \lt i \lt n$, we define the following outer curves (see Figure 36):

\begin{equation*}\delta_{i} = \langle \mathscr{C} \backslash \{\alpha_{0}^{i}\}\rangle \in \mathscr{C}^{1}.\end{equation*}

Then, we define the following curve (see Figure 36):

\begin{equation*}\Delta = \langle \mathscr{C}_{0} \cup \{\delta_{j}\}_{j=1}^{n-1}\rangle \in \mathscr{C}^{2}.\end{equation*}

Figure 36.

On the left, the curves needed to uniquely determine the curves δ ₁ (top) and Δ (bottom).

With this, we can define the following set

\begin{equation*}\mathscr{D} := \{\delta_{i}: 0 \lt i \lt n\} \cup \{\Delta\} \subset \mathscr{C}^{2}.\end{equation*}

Note that if n = 1, then $\mathscr{D} = \varnothing$, and if n = 2 then $\mathscr{D} = \{\delta_{1}\}$.

5.3. Proof of Theorem A

Similarly to the previous sections, let $\mathscr{G} := \{\tau_{\alpha}^{\pm 1} : \alpha \in (\mathscr{C} \backslash \{\alpha_{5}\})\} \cup \{\eta_{\delta_{i}}^{\pm 1} : 0 \lt i \lt n\}$. It is a well-known fact (see Corollary 4.15 in [Reference Farb and Margalit4]) that $\mathscr{G}$ is a symmetric generating set of $\mathrm{Mod}(S)$ (often called the Humphries–Lickorish generating set). As before, if A is a set of curves on S, we denote by $\mathscr{G} \cdot A$ the following set $\mathscr{G} \cdot A := \tau_{(\mathscr{C} \backslash \{\alpha_{5}\})}^{\pm 1}(A) \cup \eta_{\{\delta_{i} : 0 \lt i \lt n\}}^{\pm 1}(A)$.

Now, we state the key lemma of this section, whose proof we leave to the following subsections (see § 5.5, 5.6, 5.7 and 5.8)

Recalling Proposition 3.7 in [Reference Hernández Hernández7] (restated for our purposes), we obtain an immediate consequence.

Figure 37.

On the left, the curves needed to uniquely determine the non-outer separating curves on the right.

As before, we divide the proof of Lemma 3 into four claims:

Claim 1: $\tau_{(\mathscr{C} \backslash \{\alpha_{5}\})}^{\pm 1}(\mathscr{C}) \subset \mathscr{Y}(S)^{8}$.

Claim 2: $\tau_{(\mathscr{C} \backslash \{\alpha_{5}\})}^{\pm 1}(\mathscr{B}) \subset \mathscr{Y}(S)^{11}$.

Claim 3: $\eta_{\{\delta_{i} : 0 \lt i \lt n\}}^{\pm 1}(\mathscr{C}) \subset \mathscr{Y}(S)^{3}$.

Claim 4: $\eta_{\{\delta_{i} : 0 \lt i \lt n\}}^{\pm 1}(\mathscr{B}) \subset \mathscr{Y}(S)^{6}$.

In the following section, we define the auxiliary curves used to prove these claims.

5.4. Auxiliary curves for the claims

There are two types of auxiliary curves that are used throughout this section.

The first type of curves (which are only defined for surfaces with $n \geqslant 2$ punctures) are defined only for each $k = 2,4$ as follows (see Figure 38):

\begin{equation*}\varepsilon_{k} := \langle \{\alpha_{0}^{i}\}_{i = 1}^{n-1} \cup (\mathscr{C}_{0} \backslash \{\alpha_{k}\})\rangle \in \mathscr{C}^{1}.\end{equation*}

Figure 38.

On the left, the curves needed to uniquely determine the curves ɛ ₂ (top) and ɛ ₄, on the right.

To define the second type of curve we must first define for each $0 \leqslant j \leqslant n$ the following chain:

\begin{equation*}C_{j} := \{\alpha_{0}^{j},\alpha_{1},\alpha_{2}\}.\end{equation*}

Now, the second type of curve is defined for i = 0 as follows:

\begin{equation*}\zeta^{0} := \langle \{\alpha_{1},\alpha_{3}, \alpha_{5}\} \cup \{\beta_{C_{j}}^{+}: 0 \lt j \leqslant n\}\rangle,\end{equation*}

for each $0 \lt i \lt n+1$ as follows:

\begin{equation*}\zeta^{i} = \langle \{\alpha_{1},\alpha_{3}, \alpha_{4}, \alpha_{5}\} \cup \{\beta_{C_{j}}^{+}: j \lt i\} \cup \{\beta_{C_{k}}^{-}: i \leqslant k\}\rangle,\end{equation*}

and finally for $i = n+1$ as follows:

\begin{equation*}\zeta^{n+1} := \langle \{\alpha_{1},\alpha_{3}, \alpha_{5}\} \cup \{\beta_{C_{j}}^{-}: 0 \leqslant j \lt n\}\rangle.\end{equation*}

See Figure 39 for examples.

Figure 39.

On the left, the curves needed to uniquely determine the curves ζ ⁰ (top), ζ ¹ (upper middle), ζ ³ (lower middle) and $\zeta^{n+1}$, on the right.

Note that for all $0 \leqslant i \leqslant n+1$ we have that $\zeta^{i} \in \mathscr{Y}(S)^{1}$.

Finally, we define the sets

\begin{equation*}\mathscr{E} := \left\{\begin{array}{cl} \mathscr{D} & \text{if}\ n = 1, \\ \mathscr{D} \cup \{\varepsilon_{2}, \varepsilon_{4}\} & \text{otherwise,} \end{array} \right.\end{equation*}

(thus $\mathscr{E} \subset \mathscr{C}^{2}$), and

\begin{equation*}\mathscr{Z} := \{\zeta^{j}\}_{j=0}^{n+1}\end{equation*}

(thus $\mathscr{Z} \subset \mathscr{Y}(S)^{1}$).

5.5. Proof of Claim 1: $\tau_{(\mathscr{C} \backslash \{\alpha_{5}\})}^{\pm 1}(\mathscr{C}) \subset \mathscr{Y}(S)^{8}$

Given that S has genus 2, the strategy used in [Reference Hernández Hernández7] for genus at least 3 (find two ‘standard’ cases of the claim and then use the action of a subgroup of $\mathrm{Mod}(S)$) is actually more complicated than listing exactly each case of $\tau_{\alpha}^{\pm 1}(\alpha^{\prime})$ for $\alpha, \alpha^{\prime} \in \mathscr{C}$ (there is too little space in a genus 2 surface). Also, using the fact that $\tau_{\alpha}^{\pm 1}(\alpha^{\prime}) = \alpha^{\prime}$ if and only if $i(\alpha,\alpha^{\prime}) = 0$, and recalling Remark 2, we only need to prove that $\tau_{\alpha_{i}}^{\pm 1}(\alpha_{i-1}) \in \mathscr{Y}(S)^{k}$ for $0 \leqslant i \leqslant 5$ (modulo 6), where $\alpha_{0} = \alpha_{0}^{i}$ for each case of $0 \leqslant i \leqslant n$ and $k \leqslant 8$.

The general strategy to prove that $\tau_{\alpha_{i}}(\alpha_{i-1}) \in \mathscr{Y}(S)^{8}$ (and $\tau_{\alpha_{i}}^{- 1}(\alpha_{i-1}) \in \mathscr{Y}(S)^{8}$) is the following:

• • For n = 1, we first prove that $\tau^{\pm 1}_{\alpha_{5}}(\alpha_{4}) \in \mathscr{Y}(S)^{2}$. We do this is a similar fashion as in [Reference Hernández Hernández7]. Then, for i < 5, we use elements in $\mathscr{Y}(S)$ to fill the complement of the one-holed torus induced by α_i and $\alpha_{i-1}$, and then use $\tau^{\mp 1}_{\alpha_{i+1}}(\alpha_{i})$ to uniquely determine $\tau^{\pm 1}_{\alpha_{i}}(\alpha_{i-1})$. This yields that $\tau^{\pm 1}_{\alpha_{i}}(\alpha_{i-1}) \in \mathscr{Y}(S)^{7}$ for all i.

• • For $n \geqslant 2$, the strategy is almost the same with the exception of needing particular auxiliary curves and $\mathscr{D}$, hence getting that $\tau^{\pm 1}_{\alpha_{5}}(\alpha_{4}) \in \mathscr{Y}(S)^{3}$. This causes that in the end $\tau^{\pm 1}_{\alpha_{i}}(\alpha_{i-1}) \in \mathscr{Y}(S)^{8}$ for all i.

Remark 3. Note that, since $\mathscr{E} \subset \mathscr{C}^{2}$, due to Proposition 1, we have that

\begin{equation*}\tau_{(\mathscr{C} \backslash \{\alpha_{5}\})}^{\pm 1}(\mathscr{E}) \subset \mathscr{Y}(S)^{10}.\end{equation*}

Now, for the purpose of facilitating the exposition to the reader, we illustrate the case n = 1 in § 5.5.1 and in § 5.5.2 we treat the case $n\geqslant 2$, introducing the auxiliary curves and instructing the necessary changes from the previous case.

5.5.1. Case n = 1

First, let $\gamma^{\pm}$ be the following curves (see Figure 40):

\begin{equation*}\gamma^{+} = \langle \{\alpha_{0}^{0}, \alpha_{0}^{1}, \alpha_{3}\} \cup \{\beta_{\{\alpha_{5}, \alpha_{0}^{1}, \alpha_{1}\}}^{+}\} \rangle \in \mathscr{Y}(S)^{1},\end{equation*}

\begin{equation*}\gamma^{-} = \langle \{\alpha_{0}^{0}, \alpha_{0}^{1}, \alpha_{3}\} \cup \{\beta_{\{\alpha_{5}, \alpha_{0}^{0}, \alpha_{1}\}}^{+}\} \rangle \in \mathscr{Y}(S)^{1}.\end{equation*}

Figure 40.

On the left, the curves needed to uniquely determine the curves $\gamma^{+}$ (top) and γ ⁻ (bottom), on the right.

With these curves, we get (see Figure 41)

\begin{equation*}\tau^{\pm 1}_{\alpha_{5}}(\alpha_{4}) = \langle \{\alpha_{1}, \alpha_{2}\} \cup \{\beta_{\{\alpha_{3},\alpha_{4},\alpha_{5}\}}^{+}\} \cup \{\gamma^{\pm}\} \rangle \in \mathscr{Y}(S)^{2}.\end{equation*}

Figure 41.

On the left, the curves needed to uniquely determine the curve $\tau_{\alpha_{5}}(\alpha_{4})$, on the right.

For $\tau^{\pm 1}_{\alpha_{i}}(\alpha_{i-1})$ with i < 5, let A be as in Table 1, and we get

\begin{equation*}\tau^{\pm 1}_{\alpha_{i}}(\alpha_{i-1}) = \langle A \cup \{\tau^{\mp 1}_{\alpha_{i+1}}(\alpha_{i})\} \rangle \in \mathscr{Y}(S)^{j},\end{equation*}

Table 1.

The set A for uniquely determining $\tau^{\pm 1}_{\alpha_{i}}(\alpha_{i-1})$ for n = 1.

where $i + j = 7$ and $\alpha_{0} = \alpha_{0}^{m}$ for each $m = 0, 1$. See Figure 42 for examples.

Figure 42.

On the left, the curves needed to uniquely determine the curves $\tau^{-1}_{\alpha_{4}}(\alpha_{3})$ (top) and $\tau_{\alpha_{3}}(\alpha_{2})$ (bottom), on the right.

5.5.2. Case $n \geqslant 2$

For this case, we first define the following auxiliary curves. For let $\Delta_{1,n} = \Delta \in \mathscr{Y}(S)^{2}$ (see § 5.2); then, for $1 \lt i \lt n$, we define $\Delta_{1,i}$ and $\Delta_{i,n}$ as in Table 2 (see Figure 43).

Table 2.

The auxiliary curves $\Delta_{1,i}$ and $\Delta_{i,n}$.

Note that $\mathscr{D}_{1,i}, \mathscr{D}_{i,n} \subset \mathscr{Y}(S)^{2}$.

With these curves, we define the sets

\begin{equation*}\mathscr{D}_{1,i} := \left\{ \begin{array}{cl} \varnothing & \text{if } i = 1, \\ \{\delta_{1}\} & \text{if } i = 2,\\ \{\delta_{1}, \ldots, \delta_{i-1}\} \cup \{\Delta_{1,i}\} & \text{if } i \geqslant 3. \end{array}\right.\end{equation*}

\begin{equation*}\mathscr{D}_{i,n} := \left\{ \begin{array}{cl} \varnothing & \text{if } i = n, \\ \{\delta_{i-1}\} & \text{if } i = n-1,\\ \{\delta_{i}, \ldots, \delta_{n-1}\} \cup \{\Delta_{i,n}\} & \text{if } i \leqslant n-2. \end{array}\right.\end{equation*}

Using these sets, we define the following auxiliary curves (see Figure 44)

\begin{equation*}\beta_{i}^{+} = \langle (\mathscr{C} \backslash \{\alpha_{0}^{0}, \ldots, \alpha_{0}^{i-1}, \alpha_{2}, \alpha_{4}\}) \cup \mathscr{D}_{1,i} \rangle \in \mathscr{Y}(S)^{3},\end{equation*}

\begin{equation*}\beta_{i}^{-} = \langle (\mathscr{C} \backslash \{\alpha_{0}^{i}, \ldots, \alpha_{0}^{n}, \alpha_{2}, \alpha_{4}\}) \cup \mathscr{D}_{i,n} \rangle \in \mathscr{Y}(S)^{3}.\end{equation*}

Now we can prove the claim for this case. Note that we need only substitute the set $\{\alpha_{0}^{0},\alpha_{0}^{1}\}$ for $\mathscr{C}_{f}$ in the unique determination of the curves $\gamma^{\pm}$ of Case n = 1, to obtain analogous curves (again denoted by $\gamma^{\pm}$) for Case $n \geqslant 2$. Then, for the unique determination of $\tau_{\alpha_{5}}^{\pm 1}(\alpha_{4})$, we proceed as in Case n = 1, adding the set $\mathscr{D}$ and using these new $\gamma^{\pm}$. This implies that $\tau_{\alpha_{5}}^{\pm 1}(\alpha_{4}) \in \mathscr{Y}(S)^{3}$

Similarly, to uniquely determine $\tau_{\alpha_{4}}^{\pm 1}(\alpha_{3})$ and $\tau_{\alpha_{3}}^{\pm 1}(\alpha_{2})$, we also substitute in the set A (see Table 1) the set $\{\alpha_{0}^{0}, \alpha_{0}^{1}\}$ for $\mathscr{C}_{f}$. Thus, $\tau_{\alpha_{4}}^{\pm 1}(\alpha_{3}) \in \mathscr{Y}(S)^{4}$ and $\tau_{\alpha_{3}}^{\pm 1}(\alpha_{2}) \in \mathscr{Y}(S)^{5}$. For $\tau_{\alpha_{2}}^{\pm 1}(\alpha_{1})$, we add in A the set $\mathscr{D}$ and get $\tau_{\alpha_{2}}^{\pm 1}(\alpha_{1}) \in \mathscr{Y}(S)^{6}$.

Figure 43.

On the left, the curves needed to uniquely determine the curves $\Delta_{1,3}$ (top) and $\Delta_{2,4}$ (bottom).

Figure 44.

On the left, the curves needed to uniquely determine $\beta_{2}^{+}$, on the right.

For $\tau_{\alpha_{1}}^{\pm 1}(\alpha_{0}^{0})$ and $\tau_{\alpha_{1}}^{\pm 1}(\alpha_{0}^{n})$, we substitute in A the curves $\beta_{C}^{+}$ for $\beta_{1}^{-}$ and $\beta_{n}^{+}$, respectively, and add the set $\mathscr{D}$. Hence, $\tau_{\alpha_{1}}^{\pm 1}(\alpha_{0}^{0}), \tau_{\alpha_{1}}^{\pm 1}(\alpha_{0}^{n}) \in \mathscr{Y}(S)^{7}$. The curves $\tau_{\alpha_{1}}^{\pm 1}(\alpha_{0}^{i})$ for $1 \leqslant i \leqslant n-1$ are a little trickier: we substitute $\beta_{C}^{+}$ for the set $\{\beta_{i}^{+}, \beta_{i+1}^{-}\}$ and add the sets $\mathscr{D}_{1,i}$ and $\mathscr{D}_{i+1,n}$. See Figure 45 for an example. Then, $\tau_{\alpha_{1}}^{\pm 1}(\alpha_{0}^{i}) \in \mathscr{Y}(S)^{7}$ for all $0 \leqslant i \leqslant n$.

Figure 45.

On the left, the curves needed to uniquely determine $\tau_{\alpha_{1}}(\alpha_{0}^{2})$, on the right.

Proceeding as above, for the curves $\tau_{\alpha_{0}^{0}}^{\pm 1}(\alpha_{5})$ and $\tau_{\alpha_{0}^{n}}^{\pm 1}(\alpha_{5})$, we substitute for $\beta_{1}^{-}$ and $\beta_{n}^{+}$, respectively, while adding $\mathscr{D}$ to A for all of them, and for the curves $\tau_{\alpha_{0}^{i}}^{\pm 1}(\alpha_{5})$ (with $0 \lt i \lt n)$ we substitute in A the curve $\beta_{C}^{+}$ for the set $\{\beta_{i}^{+}, \beta_{i+1}^{-}\}$ and add the sets $\mathscr{D}_{1,i}$ and $\mathscr{D}_{i+1,n}$. See Figure 46. Therefore, $\tau_{\alpha_{0}^{i}}^{\pm 1}(\alpha_{5}) \in \mathscr{Y}(S)^{8}$ for all $0 \leqslant i \leqslant n$.

Figure 46.

On the left, the curves needed to uniquely determine $\tau_{\alpha_{0}^{2}}(\alpha_{5})$, on the right.

5.6. Proof of Claim 2: $\tau_{(\mathscr{C} \backslash \{\alpha_{5}\})}^{\pm 1}(\mathscr{B}) \subset \mathscr{Y}(S)^{11}$

Similarly to § 3.6 (first case), 3.7 and 4.5, the proof of this claim follows from the results in § 5.5. Let $\beta \in \mathscr{B}$ and $\alpha \in \mathscr{C} \backslash \{\alpha_{5}\}$. Since S has genus 2, we can uniquely determine β using sets $C \subset \mathscr{C}$ and $E \subset \mathscr{E}$. Then we have $\tau_{\alpha}^{\pm 1}(\beta) = \langle \tau_{\alpha}^{\pm 1}(C \cup E)\rangle$, which by the results in § 5.5 and Remark 3 is an element in $\mathscr{Y}(S)^{11}$.

Figure 47.

On the left, the curves needed to uniquely determine the curves $\beta_{C}^{+}$ for $C=\{\alpha_{1},\alpha_{2},\alpha_{3}\}$ (top), $C=\{\alpha_{0}^{2},\alpha_{1},\alpha_{1},\alpha_{2}\}$ (upper middle), $C=\{\alpha_{5},\alpha_{0}^{2},\alpha_{1}\}$ (lower middle) and $\beta_{C}^{-}$ for $C=\{\alpha_{4},\alpha_{5},\alpha_{0}^{2}\}$ (bottom), on the right.

The existence of these sets can be easily inferred from the following examples (see Figure 47):

• • For $C = \{\alpha_{1},\alpha_{2},\alpha_{3}\}$: Let $D = \varnothing$ if n = 1 and $D= \{\delta_{j}\}_{j=1}^{n-1} \cup \{\varepsilon_{4}\}$ if $n \geqslant 2$. Then we have

  \begin{equation*}\beta_{C}^{+} = \langle \{\alpha_{1},\alpha_{2},\alpha_{3}\} \cup \{\alpha_{5}\} \cup D \rangle,\end{equation*}

  \begin{equation*}\beta_{C}^{-} = \alpha_{5}.\end{equation*}

• • For $C = \{\alpha_{0}^{i},\alpha_{1},\alpha_{2}\}$ with $0 \lt i \lt n$:

  \begin{equation*}\beta_{C}^{+} = \langle \{\alpha_{0}^{i},\alpha_{1},\alpha_{2}\} \cup \{\alpha_{0}^{j}: i \lt j\} \cup \{\alpha_{4}\} \cup \{\delta_{j}: j \lt i\} \cup \{\varepsilon_{4}\}\rangle,\end{equation*}

  \begin{equation*}\beta_{C}^{-} = \langle \{\alpha_{0}^{i},\alpha_{1},\alpha_{2}\} \cup \{\alpha_{0}^{j}: j \lt i\} \cup \{\alpha_{4}\} \cup \{\delta_{j}: i \lt j\} \cup \{\varepsilon_{4}\}\rangle.\end{equation*}

• • For $C = \{\alpha_{0}^{n},\alpha_{1},\alpha_{2}\}$:

  \begin{equation*}\beta_{C}^{+} = \langle \{\alpha_{0}^{n},\alpha_{1},\alpha_{2}\} \cup \{\alpha_{4}\} \cup \mathscr{D}\rangle,\end{equation*}

  \begin{equation*}\beta_{C}^{-} = \alpha_{4}.\end{equation*}

• • For $C = \{\alpha_{5},\alpha_{0}^{i},\alpha_{1}\}$ with $0 \lt i \lt n$:

  \begin{equation*}\beta_{C}^{+} = \langle \{\alpha_{5},\alpha_{0}^{i},\alpha_{1}\} \cup \{\alpha_{3}\} \cup \{\delta_{j}: j \neq i\} \cup \{\varepsilon_{2},\varepsilon_{4}\}\rangle,\end{equation*}

  \begin{equation*}\beta_{C}^{-} = \alpha_{3}.\end{equation*}

• • For $C = \{\alpha_{5},\alpha_{0}^{0},\alpha_{1}\}$:

  \begin{equation*}\beta_{C}^{+} = \langle \{\alpha_{5},\alpha_{0}^{0},\alpha_{1}\} \cup \{\alpha_{3}\} \cup \mathscr{D} \rangle,\end{equation*}

  \begin{equation*}\beta_{C}^{-} = \alpha_{3}.\end{equation*}

• • For $C = \{\alpha_{4},\alpha_{5},\alpha_{0}^{i}\}$ with $0 \lt i \lt n$:

  \begin{equation*}\beta_{C}^{+} = \langle \{\alpha_{4},\alpha_{5},\alpha_{0}^{i}\} \cup \{\alpha_{0}^{j}: i \lt j\} \cup \{\alpha_{2}\} \cup \{\delta_{j}: j \lt i\} \cup \{\varepsilon_{2}\}\rangle,\end{equation*}

  \begin{equation*}\beta_{C}^{-} = \langle \{\alpha_{4},\alpha_{5},\alpha_{0}^{i}\} \cup \{\alpha_{0}^{j}: j \lt i\} \cup \{\alpha_{2}\} \cup \{\delta_{j}: i \lt j\} \cup \{\varepsilon_{2}\}\rangle.\end{equation*}

• • For $C = \{\alpha_{4},\alpha_{5},\alpha_{0}^{n}\}$:

  \begin{equation*}\beta_{C}^{+} = \langle \{\alpha_{4},\alpha_{5},\alpha_{0}^{n}\} \cup \{\alpha_{2}\} \cup \mathscr{D}\rangle,\end{equation*}

  \begin{equation*}\beta_{C}^{-} = \alpha_{2}.\end{equation*}

• • For $C = \{\alpha_{3},\alpha_{4},\alpha_{5}\}$:

  \begin{equation*}\beta_{C}^{+} = \langle \{\alpha_{3},\alpha_{4},\alpha_{5}\} \cup \{\alpha_{1}\} \cup \{\delta_{j}\}_{j=1}^{n-1} \cup \{\varepsilon_{2}\}\rangle,\end{equation*}

  \begin{equation*}\beta_{C}^{-} = \alpha_{1}.\end{equation*}

5.7. Proof of Claim 3: $\eta_{\{\delta_{i} : 0 \lt i \lt n\}}^{\pm 1}(\mathscr{C}) \subset \mathscr{Y}(S)^{3}$

In this subsection, we assume that $n \geqslant 2$ (otherwise there would be no outer curves).

To prove this claim, first note that we only need to prove that $\eta_{\delta_{i}}^{\pm 1}(\alpha_{0}^{i}) \in \mathscr{Y}(S)^{3}$ (all other curves in $\mathscr{C}$ are disjoint from δ_i). To do so, we define the following auxiliary curves.

For each $1 \leqslant i \leqslant n$, we define the following curves (see Figure 48):

\begin{equation*}\gamma_{+}^{i} = \langle \{\alpha_{0}^{j}: j \lt i\} \cup \{\alpha_{1},\alpha_{2}\} \cup \{\alpha_{4}\} \cup \{\zeta^{k}: i \lt k \leqslant n\}\rangle,\end{equation*}

\begin{equation*}\gamma_{-}^{i} = \langle \{\alpha_{0}^{j}: i \lt j\} \cup \{\alpha_{1}, \alpha_{2}\} \cup \{\alpha_{4}\} \cup \{\zeta^{k} : 1 \leqslant k \leqslant i\} \rangle.\end{equation*}

Note that for all $1 \leqslant i \leqslant n$ we have that $\gamma_{+}^{i}, \gamma_{-}^{i} \in \mathscr{Y}(S)^{2}$.

Figure 48.

On the left, the curves needed to uniquely determine the curves $\gamma_{+}^{2}$ (top) and $\gamma_{-}^{2}$ (bottom), on the right.

Finally, for each $1 \leqslant i \lt n$, we have that (see Figure 49):

\begin{equation*}\eta_{\delta_{i}}(\alpha_{i}) = \langle (\mathscr{C} \backslash \{\alpha_{0}^{i}, \alpha_{1}, \alpha_{5}\}) \cup \{\gamma_{+}^{i}\}\rangle,\end{equation*}

\begin{equation*}\eta_{\delta_{i}}^{-1}(\alpha_{i}) = \langle (\mathscr{C} \backslash \{\alpha_{0}^{i}, \alpha_{1}, \alpha_{5}\}) \cup \{\gamma_{-}^{i}\}\rangle.\end{equation*}

Therefore, $\eta_{\{\delta_{i} : 0 \lt i \lt n\}}^{\pm 1}(\mathscr{C}) \subset \mathscr{Y}(S)^{3}$

Figure 49.

On the left, the curves needed to uniquely determine $\eta_{\delta_{i}}(\alpha_{2})$, on the right.

Remark 4. Again (as in § 5.5), since $\mathscr{E} \subset \mathscr{C}^{2}$, we have that

\begin{equation*}\eta_{\{\delta_{i} : 0 \lt i \lt n\}}^{\pm 1}(\mathscr{E}) \subset \mathscr{Y}(S)^{5}.\end{equation*}

5.8. Proof of Claim 4: $\eta_{\{\delta_{i} : 0 \lt i \lt n\}}^{\pm 1}(\mathscr{B}) \subset \mathscr{Y}(S)^{6}$

In this subsection, we assume that $n \geqslant 2$ (otherwise there would be no outer curves).

Analogously to § 5.6, this claim follows from the results in § 5.7: Let $\beta \in \mathscr{B}$ and $\delta \in \{\delta_{i} : 0 \lt i \lt n\}$; since $\beta = \langle C \cup E\rangle$ with $C \subset \mathscr{C}$ and $E \subset \mathscr{E}$ (as in § 5.6), we have that $\eta_{\delta}^{\pm 1}(\beta) = \langle \eta_{\delta}^{\pm 1}(C \cup E)\rangle$, which by the results in § 5.7 and Remark 4 is contained in $\mathscr{Y}(S)^{6}$.

For the existence of these sets, we refer the reader to the examples and figures in § 5.6.

5.9. Proof of Theorem B

Similarly to § 3.8 and 4.8, here we define the set $\mathfrak{X}(S)$ as the set $\mathfrak{X}_{1}$ from Section 5 in [Reference Aramayona and Leininger2], which was proved to be rigid (see Lemma 5.2 in [Reference Aramayona and Leininger2]).

Let X be the set defined as $\mathscr{C}$ union the set of all boundary curves of regular neighbourhoods of sets A with one of the following forms:

1. (1) $A = \{\alpha_{0}^{i},\alpha_{0}^{j},\alpha_{k}\}$ where $0 \leqslant i \leqslant j \leqslant n$ and $k = 1, 5$.

2. (2) $A = \{\alpha_{0}^{i}, \alpha_{0}^{j}, \alpha_{k}, \alpha_{k+1}\}$ where $0 \leqslant i \leqslant j \leqslant n$ and $k = 1, 4$.

3. (3) $A = \{\alpha_{i}, \alpha_{i+1}, \ldots, \alpha_{j}\}$, where $I = \{i, i+1, \ldots, j\}$ is an integer subinterval of $\{0, \ldots, 5\}$ modulo 6 (We assume that α ₀ denotes $\alpha_{0}^{k}$ for some k). If the length of the interval I is odd, then we require that both i and j are even.

Then, we define $\mathfrak{X}(S)$ as the union of X ¹ and the set of all boundary curves of regular neighbourhoods of sets $A = \{\alpha_{i}, \alpha_{i+1}, \ldots, \alpha_{j}\}$, where $0 \lt i \leqslant j \leqslant 3$, and both i and j are odd.

Finally, note that $\mathscr{Y}(S) \subset \mathfrak{X}(S)$, which immediately implies Theorem B.