2. Bouquets and braid groups

We denote by $T_a$ the positive Dehn twist along a simple closed curve a in an oriented surface $\Sigma$ . Given two simple closed curves $a,b \subset \Sigma$ that intersect transversally in one point, we obtain the following equality between curves, up to isotopy: $T_a(b)=T_b^{-1}(a)$ . Rewriting this as $T_b T_a(b)=a$ and applying the change of coordinates $T_{T_b T_a(b)}=(T_a T_b)T_a(T_a T_b)^{-1}$ , we obtain that $T_a$ and $T_b$ satisfy the braid relation:

\begin{equation*}T_a T_b T_a=T_b T_a T_b.\end{equation*}

For a more detailed proof, including the reverse implication; see Chapter 3 in [Reference Farb and Margalit6]. More generally, let $a_1,a_2,\ldots,a_n \subset \Sigma$ be a of set of simple closed curves such that their union is $\pi_1$ -injective in $\Sigma$ and such that they are pairwise disjoint, except for pairs with consecutive indices, which intersect transversally in one point. Such a family of curves is called a chain.

For every oriented compact surface $\Sigma$ , the subgroup of the mapping class group of $\Sigma$ generated by the Dehn twists associated with a chain of n curves is isomorphic to the braid group $B_{n+1}$ . This is a consequence of work by Birman and Hilden in [Reference Birman and Hilden4] (see also Chapter 9 in [Reference Farb and Margalit6]). An interpretation of that subgroup as the monodromy group of a plane curve singularity of type $A_n$ was later described in [Reference Perron and Vannier9]; the case of curves intersecting in a general tree-like pattern was solved by Wajnryb in [Reference Wajnryb11]. The $\pi_1$ -injectivity is needed to rule out “false chains,” such as $a,b,\bar{a}$ , where the curves a and $\bar{a}$ cobound an embedded annulus. In that case, the resulting subgroup is isomorphic to the braid group $B_3$ or its quotient $\textrm{SL}(2,{\mathbb Z})$ rather than $B_4$ .

Here is an important relation between bouquets and chains of curves: suppose that the simple closed curves $a_1,a_2,\ldots,a_n \subset \Sigma$ form a $\pi_1$ -injective bouquet, numbered in the anticlockwise direction around the common intersection point. Then the set of transformed curves

\begin{equation*}a_1,T_{a_1}^{-1}(a_2),T_{a_2}^{-1}(a_3)\ldots,T_{a_{n-1}}^{-1}(a_n)\end{equation*}

forms a chain, as shown in Figure 1 for $n=4$ (where the new curves are labeled 1’,2’,3’,4’). Moreover, the Dehn twists along these new curves generate the same subgroup in the mapping class group of $\Sigma$ as the Dehn twists associated with the curves of the initial bouquet. This is another consequence of the equation:

\begin{equation*}T_{T_x^{-1}(y)}=T_x^{-1} T_y T_x.\end{equation*}

Using the result about chains from the last paragraph, we conclude that the Dehn twists associated with the curves of a bouquet generate a subgroup isomorphic to the braid group $B_{n+1}$ .

Figure 1. Bouquet and chain.

As for the second statement of Proposition 1, we need to analyze how the braid and cycle relations among the Dehn twists along the curves $a_1,a_2,\ldots,a_n$ translate into the usual braid and commutation relation among the Dehn twists associated with the transformed curves $a_1,T_{a_1}^{-1}(a_2),T_{a_2}^{-1}(a_3)\ldots,T_{a_{n-1}}^{-1}(a_n)$ .

Let $a,b,c \in \{a_1,a_2,\ldots,a_n\}$ be a triple of curves ordered in the anticlockwise way, and let $x=a,y=T_a^{-1}(b),z=T_b^{-1}(c)$ be the transformed curves. The Dehn twists $T_x,T_y,T_z$ satisfy the two braid relations:

\begin{equation*}T_x T_y T_x=T_y T_x T_y \ , \ T_y T_z T_y=T_z T_y T_z\end{equation*}

and the commutation relation:

\begin{equation*}T_x T_z=T_z T_x.\end{equation*}

Moreover, this is a complete set of relations, again by the result about chains from above. We need to show that these are equivalent to the three braid relations:

\begin{equation*}T_a T_b T_a=T_b T_a T_b \ , \ T_b T_c T_b=T_c T_b T_c \ , \ T_c T_a T_c=T_a T_c T_a\end{equation*}

and the following version of the cycle relation, due to our choice of numbering:

\begin{equation*}T_b T_a T_c T_b=T_c T_b T_a T_c,\end{equation*}

Deriving these relations from the braid relations among $T_x,T_y,T_z$ is an easy task, using the expressions:

\begin{equation*}\begin{split}T_a &= T_x \\[3pt]T_b &= T_a T_y T_a^{-1}=T_x T_y T_x^{-1} \\[3pt]T_c &= T_b T_z T_b^{-1}=T_x T_y T_x^{-1} T_z T_x T_y^{-1} T_x^{-1}=T_x T_y T_z T_y^{-1} T_x^{-1}.\end{split}\end{equation*}

Indeed, after an identification of $T_x,T_y,T_z$ with the standard braid generators $\sigma_1,\sigma_2,\sigma_3 \in B_4$ , the four relations among $T_a,T_b,T_c$ admit a pictorial proof:

\begin{align*}T_a T_b T_a&=\sigma_1^2 \sigma_2=T_b T_a T_b,\\[3pt]T_b T_c T_b&=\sigma_1 \sigma_2^2 \sigma_3 \sigma_1^{-1}=T_c T_b T_c,\\[3pt]T_a T_c T_a&={\sigma_1^2 \sigma_2 \sigma_3 \sigma_2^{-1}}=T_c T_a T_c,\\[3pt]T_b T_a T_c T_b&=\sigma_1^2 \sigma_2 \sigma_3=T_c T_b T_a T_c.\end{align*}

For the reverse direction, we express

\begin{equation*}\begin{split}T_x & = T_a \\[3pt]T_y &= T_a^{-1} T_b T_a \\[3pt]T_z &= T_b^{-1} T_c T_b, \\\end{split}\end{equation*}

use the shortcuts $a=T_a,b=T_b,c=T_c$ in order to save space and derive

\begin{equation*}T_y T_x T_y=a^{-1}baaa^{-1}ba=baa=aa^{-1}baa=T_x T_y T_x.\end{equation*}

Here, we used the braid relation $bab=aba$ . The second braid relation is a bit trickier:

\begin{equation*}\begin{split}T_z T_y T_z&=b^{-1}cba^{-1}bab^{-1}cb=cbc^{-1}a^{-1}bacbc^{-1}=cbc^{-1}a^{-1}cba,\\[3pt]T_y T_z T_y&=a^{-1}bab^{-1}cba^{-1}ba=a^{-1}bacbc^{-1}a^{-1}ba=a^{-1}cbba.\end{split}\end{equation*}

Here, we used a version of the braid relation, $b^{-1}cb=cbc^{-1}$ , as well as the cycle relation $bacb=cbac$ . The equality $T_z T_y T_z=T_y T_z T_y$ is thus equivalent to

\begin{equation*}acbc^{-1}a^{-1}c=cb.\end{equation*}

Thanks to the cycle relation $bacb=cbac$ , the left-hand side is equal to

\begin{equation*}b^{-1}cbacc^{-1}a^{-1}c=b^{-1}cbc=cb.\end{equation*}

Finally, here is the commutation relation:

\begin{equation*}T_x T_z=ab^{-1}cb=acbc^{-1}=b^{-1}bacbc^{-1}=b^{-1}cbacc^{-1}=b^{-1}cba=T_z T_x.\end{equation*}

A similar derivation of the equivalence of these two group presentations can be found in Section 2 of [Reference Baader and Lönne2], where the cycle relation is used to define an invariant of positive braids. Applying the above procedure to all triples of curves among $a_1,a_2,\ldots,a_n$ , we obtain a complete set of relations, as stated in Proposition 1.

3. Triple bouquets

In this section, we prove that whenever three simple closed curves a, b and c in an oriented closed surface $\Sigma$ satisfy pairwise braid relations and a cycle relation, then the set of curves a, b, c forms a bouquet or are all isotopic. Note that this settles Theorem 1 for the case $n = 3$ , since the converse follows from previous considerations. More concretely, in Section 2, it was shown that the cycle relation follows algebraically from $T_xT_z = T_z T_x$ , where $x = a$ and $z = T_b^{-1}(c)$ .

Suppose a, b, c are curves satisfying the three braid relations:

\begin{align*}T_aT_bT_a &= T_bT_aT_b \\[3pt] T_aT_cT_a & =T_cT_aT_c \\[3pt] T_bT_cT_b& =T_cT_bT_c\end{align*}

and the cycle relation $T_b T_a T_c T_b = T_c T_b T_a T_c$ . Using the relations, one checks that if two curves are isotopic, then $T_a=T_b=T_c$ , so all three curves are isotopic. Thus, from here on, we consider a, b, and c to be pairwise non-isotopic. In particular, by the braid relations, a, b, c have pairwise intersection number one. Hence, after an isotopy, they admit a tubular neighborhood either as shown to the left of Figure 2, in which case we write $a<b<c<a$ , or mirrored, in which case we write $a < c < b < a$ ; compare Remark 1 below. Letting $x=a, z = T_b^{-1}(c)$ we have that $T_x$ and $T_z$ commute, under the exact same reasoning as in Section 2, where $T_xT_z = T_zT_x$ is deduced purely algebraically from the braid relations and the cycle relation. This means that x and z in Figure 2 have disjoint representatives in their isotopy classes. Hence, x and z bound a bigon B, since their number of intersections is not minimal.

Figure 2. The curves $x = a$ and $z = T_b^{-1}(c)$ intersect twice.

Notice that this allows us to exclude the case $a < c < b <a$ , since in this case, the curves x and $z = T_b^{-1}(c)$ have intersection number two and hence $T_x$ and $T_z$ do not commute. This is because the union $x \cup z$ bounds two regions, both of whose boundaries are polygonal of length four that intersect themselves in two corners, and in particular the two regions are not bigons.

We can now assume that $a < b < c < a$ . There are two possibilities for the position of B, indicated by the two dotted regions in Figure 3. In the first case, on the left, it is obvious that a, b, c form a bouquet. The second case, on the right, seems slightly more challenging. However, note that the two surfaces that are obtained by filling in the dotted regions are actually diffeomorphic via an orientation preserving diffeomorphism preserving all three curves a, b, c individually as sets. One example of such a diffeomorphism is as follows. Cut up Figure 3 along the dashed lines, as well as along the edges on the drawn boundary that are identified with their opposites, to obtain three X-shaped regions after six cuts. Rotating each of those regions by 180 degrees preserves all identifications and maps the edges of the dotted triangle on the left to the edges of the dotted triangle on the right. Extending this to the dotted regions yields the desired diffeomorphism.

Figure 3. Possible bigons.

It turns out that there are no further cases: the set $x \cup z$ bounds four regions in a small neighborhood of $x \cup z$ , two of which we have now considered. The other two cannot possibly be the boundary of a bigon, since they have homotopically nontrivial boundary, evidenced by the fact that the boundary curves intersect b precisely once.

4. General bouquets

In this section, we prove Theorem 1 by induction on the number of curves n. It is beneficial to be careful about the cyclic order of curves. For a bouquet given as the union of n simple closed curves $c_1,c_2,\ldots,c_n$ in an oriented surface $\Sigma$ , we write

\begin{equation*}c_1<c_2<\cdots <c_n<c_1\end{equation*}

if $c_{i+1}$ occurs next (counter-clockwise) to $c_{i}$ for all $i\in{\mathbb Z}/n{\mathbb Z}$ . For an example with $n=4$ and $c_1<c_2<c_3<c_4<c_1$ ; see the left-hand side of Figure 1. If a set of simple closed curves $c_1,c_2,\ldots,c_n$ forms a bouquet, we write $c_1<c_2<\cdots <c_n<c_1$ if the corresponding bouquet is as above. In other words, by the definition from the introduction, $c_1<c_2<\cdots <c_n<c_1$ means that $c_1,c_2,\ldots,c_n$ form an oriented bouquet.

More generally, let a, b, c be simple closed curves in $\Sigma$ that have pairwise intersection number one. Having $a<b<c<a$ and $a<c<b<a$ , respectively, can be defined as in Section 3. And for bouquets of three curves the notions agree.

Analyzing the case of three curves (as in Section 3) while keeping track of the cyclic order leads to the following proposition, which we use to prove Theorem 1.

As an aside, we note that $c_{n+1}$ being distinct from $c_i$ , up to isotopy, for $i\leq n$ is implied without being assumed.

Proof of Proposition 2. As a consequence of the bigon criterion, we can and do isotope all the $c_i$ to achieve that they intersect pairwise transversely and the following holds. The $c_1,c_2,\ldots,c_n$ intersect in the same point p, and $c_i$ and $c_{n+1}$ realize their intersection number and are in general position (their intersections are pairwise different and different from p) for all $i\leq n$ ; see Figure 4(a). We note that, due to (ii’), the curves $c_1$ , $c_n$ , and $c_{n+1}$ do intersect as depicted in Figure 4(a), rather than with the opposite cyclic order; see analysis of the cyclic order at the end of Section 3. We also note that $c_i$ and $c_{n+1}$ intersect at most once since they satisfy the braid relation (i’).

Figure 4. (a) A neighborhood of $c_1\cup c_n\cup c_{n+1}$ (gray), for $2\leq i\leq n-1$ the intersections between $c_i$ and $c_{n+1}$ are not drawn. (b) That neighborhood union the triangle $\Delta$ (dotted). (c) The region C and its intersection with the $c_i$ .

By the argument in Section 3, (i’) and (ii’) imply that the triple of curves $a=c_1$ , $b=c_{n}$ , and $c=c_{n+1}$ forms a bouquet. More precisely, up to an orientation preserving diffeomorphism, we have that a regular neighborhood of $a\cup b\cup c$ union a triangle $\Delta$ is embedded in $\Sigma$ as depicted in Figure 4(b).

Denote by C the connected component of $\Sigma\setminus (a \cup b \cup c)$ containing $\Delta$ . By the assumption on $c_1,c_2,\ldots,c_n$ forming an oriented bouquet, the triangle C has nonempty intersections with all $c_i$ for $1\leq i\leq n$ . Hence, each $c_i$ intersects C in an interval with its end points on $\partial C$ : one at p and the other one in the interior of the interval $c_{n+1}\cap \partial C$ ; see Figure 4(c). Thus, after isotoping $c_{n+1}$ across C, we conclude that $c_1,c_2,\ldots,c_n, c_{n+1}$ form an oriented bouquet.

Proof of Theorem 1. For the case of $n=2$ , recall from the first paragraph of the introduction that two non-isotopic simple closed curves can be isotoped to intersect once transversally if and only if the corresponding Dehn twists satisfy the braid relation. For $n\geq 3$ , we induct on n. The base case (three curves) was treated in Section 3. For the induction step, we assume as the induction hyphothesis that Theorem 1 holds for a fixed $n\geq3$ .

The only if statement follows from Proposition 1. For the if statement, consider $c_1,c_2,c_n,c_{n+1}$ in $\Sigma$ with corresponding positive Dehn twists $T_i$ along them satisfying (i) and (ii) and not all $T_i$ are equal. By cyclically relabeling the curves if needed, we may and do assume that not all $T_1, T_2,\cdots,T_n$ are equal. By the induction hypothesis, $c_1,c_2,\ldots,c_n$ form an oriented bouquet, that is, $c_1<c_2\cdots<c_n<c_1$ .

We consider a pair of consecutive curves b,a in this bouquet; that means, $a=c_{i+1}$ and $b=c_{i}$ for $1\leq i \leq n-1$ or $a=c_1$ and $b=c_n$ . There is at least one choice of b, a such that the cyclic order of $a,b,c_{n+1}$ (as defined in Remark 1) is $a<b<c_{n+1}<a$ . Indeed, assume we have $c_{i+1}<c_{n+1}<c_i<c_{i+1}$ for all $1\leq i\leq n-1$ , then one checks (using $c_1<c_2\ldots<c_n$ ) that $c_1<c_{n}<c_{n+1}<c_1$ .

To conclude, we cyclically relabel $c_1,c_2,\ldots,c_n$ such that $c_1<c_{n}<c_{n+1}<c_1$ . Hence, by Remark 1 the cycle relation for $c_1$ , $c_n$ , and $c_{n+1}$ provided by (ii) is

\begin{equation*}T_{n} T_1 T_{n+1} T_{n} = T_{n+1} T_{n} T_1 T_{n+1}\end{equation*}

or one of its cyclic permutations. Thus, $c_1,c_2,\ldots,c_n, c_{n+1}$ form an oriented bouquet by Proposition 2. This concludes the induction step.

5. An explicit criterion

From the proof of Theorem 1, one notices that we did not use all cycle relations as provided by the assumption (ii). Only linearly many cycle relations (in terms of number of curves) are needed. Indeed, inductive application of Proposition 2 yields the following.