# CHAPTER 6

## Braids on surfaces

In this chapter, we study the LCS of surface braid groups and their partitioned versions. This may be seen as a generalisation of results from §3, where we studied classical Artin braids, that is, braids on the disc. Contrary to what is usually done in the literature, we choose to consider braids on any surface. In particular, our surfaces may be non-compact, and they may have (countably) infinite genus or boundary components. For braids, which are always compactly supported and can be pushed away from the boundary, this level of generality does not really complicate things. We also do not assume that our surfaces are orientable, because the techniques that we use work very similarly for orientable surfaces and for non-orientable ones.

We first recall what we need from Richards' classification of surfaces (§6.1), then we introduce the tools that we need from the general theory of braids on surfaces (§6.2) and we review presentations of braid groups on compact surfaces (§6.3). We then find ourselves ready to tackle the study of LCS. We do this first for the whole braid group $B_n(S)$: we show that the LCS stops if $n \ge 3$, and we completely compute the Lie ring in this case (§6.4). We then generalise these results to partitioned surface braid groups $B_\lambda(S)$ in §6.5, whose LCS stops if the blocks of the partition have size at least 3, a stability hypothesis under which we can compute the associated Lie ring. Finally, we study the unstable cases in §6.6. There, four cases stand out from the crowd: the braid groups on the sphere $S^2$, on the torus $T^2$, on the Möbius strip $M^2$ and on the projective plane $\mathbb{P}^2$.

### 6.1. Surfaces

Recall that a *surface* is a (second countable) 2-manifold, which we will *not* suppose compact or orientable in general. Such manifolds are well-understood: the classification of surfaces without boundary has been achieved by Richards [Ric63], and the classification of surfaces with boundary, which is more complicated, was completed more than fifteen years later by Brown and Messer [BM79].

**REMARK 6.1.** By a *manifold*, we mean a locally Euclidean, Hausdorff space. Moreover, all our manifolds are assumed *second countable*. For connected manifolds, this is equivalent to either metrisability or paracompactness, and it implies triangulability (see [Rad25] for the latter implication).

For studying braids, we will in fact only need to consider Richards' classification. Indeed, let $S$ be any connected surface (possibly with boundary). If we glue a copy of $\partial S \times [0, \infty)$ to $S$ by identifying $\partial S \times 0$ with $\partial S$, we obtain a surface $S'$ without boundary, and it is easy to show that the canonical injection $S \hookrightarrow S'$ is an isotopy equivalence. Moreover, one can show that an isotopy equivalence induces homotopy equivalences between configuration spaces and, in turn, isomorphisms
between braid groups. In the sequel, we thus identify braids on $S$ with braids on $S'$ (or on $S - \partial S$). For instance, braids on the closed disc are identified with braids on the plane. This holds in particular for braids on one strand, that is, for fundamental groups, whose computation is recalled below in Proposition 6.4.

Let us briefly recall Richards’ construction of all surfaces, up to homeomor-
phism; for a detailed account, the reader is referred to [Ric63], in particular to §5
and §6, especially Theorem 3, therein.

PROPOSITION 6.2. Let $S$ be a connected surface. Then $S$ is homeomorphic to a surface constructed in the following way:

* Consider the Cantor set $K$ embedded in the sphere $S^2$ in the usual manner. Choose some closed subset $X$ of $K$, and remove it from $S^2$.

* Choose a finite or countably infinite sequence of pairwise disjoint closed 2-discs in $S^2 - X$, which has no accumulation point outside of $X$.

* Along each of these discs, perform a connected sum operation with either $T^2$ or $P^2$.

REMARK 6.3. At the second step, one can in fact choose an explicit sequence of discs depending only on $X$ together with the subset $X_{np} \subseteq X$ of accumulation points of the sequence of discs; see [Ric63].

As a direct corollary of Richards' classification, one can compute fundamental
groups of surfaces:

PROPOSITION 6.4. Let $S$ be a connected surface without boundary. Then $\pi_1(S)$ is a free group, except when $S$ is closed. Moreover, it is of finite type if and only if $S$ is obtained (up to homeomorphism) from a closed surface by removing a finite number of points.

REMARK 6.5. We see that closed surfaces (that is, compact surfaces without boundary) are singled out here, as will also be the case later in our study of braid groups; see for instance Proposition 6.15 and Theorem 6.52.

Fundamental groups are in fact almost all homotopy groups of surfaces:

COROLLARY 6.6. Let $S$ be a connected surface without boundary. Then $\pi_{>2}(S)$ is trivial, except when $S$ is the sphere or the projective plane.

PROOF. The universal covering $\tilde{S}$ of $S$ is a simply connected surface without boundary. Proposition 6.4 implies that such a surface must be of finite type, and the explicit computation of fundamental groups of surfaces of finite type shows that $\tilde{S}$ must be homeomorphic either to the sphere or to the plane. If $\tilde{S} \simeq S^2$, then $S$ is compact, and the fibers in the covering must be finite, so $\pi_1(S)$ is finite, and $S$ is either the sphere itself, or the projective plane. In all the other cases, $\pi_{>2}(S) \simeq \pi_{>2}(\tilde{S}) \simeq \pi_{>2}(\mathbb{R}^2) = 0$. $\square$

Another immediate corollary of Richards' classification is the following di-
chotomy:

PROPOSITION 6.7. Let $S$ be a connected surface without boundary. Then either $S$ can be embedded into the sphere $S^2$, or it contains the 1-punctured torus or the Möbius strip as an embedded subsurface.

In our study of braids, this appears as a trichotomy, between the following
cases:
* S is planar, i.e. it embeds into the plane;

* S is the sphere $\mathbb{S}^2$;

* S contains an embedded 1-punctured torus (a *handle*) or an embedded Möbius strip (a *crosscap*).

In this regard, see in particular Proposition 6.19.

**CONVENTION 6.8.** In the usual way, an embedded Möbius strip on a surface will be indicated in our figures by a *crosscap*, that is, a circle drawn on the surface, bounding a disc on which is drawn a cross (see for instance Figure 6.1a). In order to obtain the surface that is meant from the surface on which the crosscaps are drawn, one must, for each crosscap, remove the interior of the corresponding disc, and then glue together opposite points of the remaining circle.

The reader not familiar with this classical representation should check, as a good exercise, that a disc with one crosscap is a Möbius strip, that a sphere with one crosscap is a projective plane (so that adding a crosscap on a surface is the same as taking the connected sum with a projective plane), and that a sphere with two crosscaps is a Klein bottle. They should also keep in mind Dyck's theorem: three crosscaps on a surface is the same as a handle and a crosscap (see for instance [FW99]).

## 6.2. Braids on surfaces: general theory

We gather here some fundamental results in the theory of braids on surfaces. The main tools that we need are Goldberg's theorem (§6.2.2), the Fadell-Neuwirth exact sequences (§6.2.3) and a little calculation showing that the usual pure braid generators become commutators on non-planar surfaces (§6.2.4). Goldberg's theorem [Gol73, Th. 1] says that surface braid groups are generated by braids on the disc, together with braids obtained from loops on the surface; we give a new simple proof of this result, incidentally extending it to possibly non-compact surfaces. As for the Fadell-Neuwirth exact sequences, which are traditionally stated for pure braid groups, since they involve forgetting strands, we generalise them easily to partitioned braid groups on possibly non-compact surfaces, where the projections forget blocks of strands.

**6.2.1. Definitions, notations and conventions.** Let $S$ be a connected surface. Let us consider the configuration space

$$F_n(S) = \{(x_1, \dots, x_n) \in S^n \mid \forall i \neq j, x_i \neq x_j\} \subset S^n.$$

The *braid group on the surface* $S$ on $n$ strings is the fundamental group $B_n(S)$ of the unordered configuration space $C_n(S) = F_n(S)/\mathfrak{S}_n$. When $S$ is the 2-disc $\mathbb{D}$, this group is exactly Artin's braid group, that is $B_n = B_n(\mathbb{D})$.

Let us fix an embedded closed disc $\mathbb{D} \subset S$, together with a base configuration $c = (c_1, \dots, c_n) \in F_n(S)$ of points $c_i \in \mathbb{D}$. Since the assignment $S \mapsto B_n(S)$ is functorial with respect to embeddings of surfaces, we have a (not necessarily injective) group morphism:

$$\varphi: B_n = B_n(\mathbb{D}) \to B_n(S).$$

In the sequel, we omit most mentions of $\varphi$: if $\beta \in B_n$, we still denote by $\beta$ its image in $B_n(S)$ (although we should denote it by $\varphi(\beta)$).
groups:

$$
\begin{equation} \tag{6.1}
\begin{split}
& \pi_1(S - Q_n) \xrightarrow{\iota_*} \mathbf{P}_{n+1}(S) \xrightarrow{p_*} \mathbf{P}_n(S) \\
(6.1) \qquad & \Big\downarrow_{u} \qquad \Big\downarrow_{v} \qquad \Big\downarrow_{w} \\
& \pi_1(S) \hookrightarrow \pi_1(S)^{n+1} \twoheadrightarrow \pi_1(S)^n,
\end{split}
\end{equation}
$$

whose bottom line is obviously exact. The map $p$ is a (locally trivial) fibration by
[FN62, Th. 3] (see also [Bir74, Th. 1.2]), and the first line is part of its exact
sequence in homotopy. Thus it is exact (but $\iota_*$ need not be injective in general).

Let us prove our second statement. For $n = 1$, $\psi_1$ is the canonical isomorphism $\pi_1(S) \cong \mathbf{P}_1(S) = \mathbf{B}_1(S)$, and there is nothing to prove.

Let us now suppose that the conclusion holds for some $n \ge 1$. By applying the Seifert-van Kampen theorem, we see that $\pi_1(S - Q_n) \cong \mathbf{F}_n *_{\mathbb{Z}} \pi_1(S - \mathbb{D})$, where $\mathbf{F}_n = \pi_1(\mathbb{D} - Q_n)$ is free on $n$ generators. Then $\iota_* : \mathbf{F}_n *_{\mathbb{Z}} \pi_1(S - \mathbb{D}) \to \mathbf{P}_n(S)$ identifies with the map induced by $\mathbf{F}_n \hookrightarrow \mathbf{P}_n$ (the kernel of $\mathbf{P}_n \twoheadrightarrow \mathbf{P}_{n-1}$) and $\psi_{n+1} : \pi_1(S - \mathbb{D}) \to \mathbf{P}_n(S)$.

Now let $G$ be the subgroup of $\mathbf{P}_{n+1}(S)$ generated by $\mathbf{P}_{n+1}$ and the images of the $\psi_i$ for $i \le n+1$. It contains the image of $\iota_*$, which is the kernel of $p_*$. Moreover, its image by $p_*$ contains the images of $\psi_1, \dots, \psi_n$, and $\mathbf{P}_n$, hence all of $\mathbf{P}_n(S)$, by the induction hypothesis. As a consequence, $G = \mathbf{P}_{n+1}(S)$, which was the desired conclusion.

Let us prove our third statement. For $n = 1$, $\pi_S$ is the canonical isomorphism $\mathbf{B}_1(S) = \mathbf{P}_1(S) \cong \pi_1(S)$ (inverse to $\psi_1$). Then $\mathbf{P}_n^\circ(S)$ and $\mathbf{P}_n$ are trivial, and there is nothing to prove.

Let us now suppose that the conclusion holds for some $n \ge 1$. Consider the
induced maps between kernels of the vertical morphisms in (6.1). By definition,
the kernels of $v$ and $w$ are respectively $\mathbf{P}_{n+1}^\circ(S)$ and $\mathbf{P}_n^\circ(S)$. Let us denote by $K$
the kernel of $u$. We get induced maps:

$$
K \xrightarrow{\iota_{\#}} \mathbf{P}_{n+1}^{\circ}(S) \xrightarrow{p_{\#}} \mathbf{P}_{n}^{\circ}(S)
$$

such that $p_{\#} \circ \iota_{\#} = 1$. An easy chase in the diagram (or an application of the Snake Lemma) shows that we can lift any element in the kernel of $p_{\#}$ to an element of $K$: the above sequence is exact.

The morphism $u$ identifies with the projection $\mathbf{F}_n *_{\mathbb{Z}} \pi_1(S - \mathbb{D}) \to \{1\} *_{\mathbb{Z}} \pi_1(S - \mathbb{D}) \cong \pi_1(S)$ killing the first factor, hence $K$ identifies with the normal closure of $\mathbf{F}_n$ in $\mathbf{F}_n *_{\mathbb{Z}} \pi_1(S)$. Moreover, $\iota_*$ sends $\mathbf{F}_n = \pi_1(\mathbb{D} - Q_n)$ to a subgroup of $\mathbf{P}_{n+1}$, so the image of $K$ in $\mathbf{P}_{n+1}(S)$ is contained in the normal closure of $\mathbf{P}_{n+1}$.

Now, let $N$ be the normal closure of $\mathbf{P}_{n+1}$ in $\mathbf{P}_{n+1}(S)$. Since $v(\mathbf{P}_{n+1}) = \{1\}$, we have $v(N) = \{1\}$, which means that $N \subseteq \mathbf{P}_{n+1}^\circ(S)$. By the induction hypothesis, $p_\#(N) = \mathbf{P}_n^\circ(S)$. Moreover, $N$ contains the image of $i_\#$, which is the kernel of $p_\#$. Thus $N = \mathbf{P}_{n+1}^\circ(S)$, which was the desired conclusion. $\square$

REMARK 6.11. For a closed, oriented surface, the group $K$ appearing in the proof is exactly the group $F_{n+1}$ from [GMP04, page 227], where they give a precise description of it in this particular case. However, in their paper, they were using the very result that we are recovering and generalising here, quoting [Bir74] for it [GMP04, page 225].
The proof of Proposition 6.10 also works for manifolds in higher dimension, allowing us to recover the classical [Bir69, Th. 1]:

**PROPOSITION 6.12.** For any manifold $M$ of dimension at least 3, the morphism $\pi_M: \mathbf{B}_n(M) \to \pi_1(M) \ltimes \mathfrak{S}_n$ is an isomorphism.

**PROOF.** One can directly check that the proof of Proposition 6.10 works if we replace the surface $S$ with a connected manifold $M$ of any dimension $d \ge 2$ and the disc $\mathbb{D}$ with a $d$-disc $\mathbb{D}^d$. Then $\mathbf{P}_n$ gets replaced with $\mathbf{P}_n(\mathbb{D}^d)$, which is trivial whenever $d \ge 3$ (the configuration space $F_n(\mathbb{D}^d) \cong F_n(\mathbb{R}^d)$ is obtained from $\mathbb{R}^{nd}$ by removing subspaces of codimension $d \ge 3$, so it is simply connected). Thus, the normal closure $\mathbf{P}_n^\circ(M)$ of $\mathbf{P}_n(\mathbb{D}^d)$ is trivial too, and the latter is exactly the kernel of $\pi_M$. $\square$

**REMARK 6.13.** This means that *braid groups on manifolds of dimension at least 3 are exactly wreath products*, whose LCS is studied in Appendix A.4; see in particular Corollaries A.27 and A.29.

**6.2.3. The Fadell-Neuwirth exact sequences.** The locally trivial fibrations used in the proof of Proposition 6.10 are particular instances of the Fadell-Neuwirth fibrations. These induce exact sequences between pure braid groups, that are in fact exact sequences between partitioned braid groups. We now recall how these work, and when these exact sequences split.

**DEFINITION 6.14.** Let $S$ be a surface, let $n \ge 1$ be an integer, and let $\lambda = (n_1, \dots, n_l)$ be a partition of $n$. The corresponding *partitioned surface braid group* is:

$$ \mathbf{B}_{\lambda}(S) := \pi^{-1}(\mathfrak{S}_{\lambda}) = \pi^{-1}(\mathfrak{S}_{n_1} \times \dots \times \mathfrak{S}_{n_l}) \subseteq \mathbf{B}_n(S). $$

There are canonical surjections between partitioned braid groups, obtained by forgetting blocks. For most surfaces, these projections behave exactly as they do for the disc. However, their behaviour for closed surfaces is somewhat trickier, especially when it comes to the sphere and the projective plane. The latter are in fact the only ones whose braid groups admit non-trivial torsion elements, a fact that can be seen as a consequence of their second homotopy group being non-trivial.

**PROPOSITION 6.15 (Fadell-Neuwirth exact sequences).** Let $S$ be a connected surface, let $\mu$ be a partition of an integer $m \ge 1$, $\nu$ be a partition of an integer $n \ge 1$, and let us denote by $\mu\nu$ their concatenation, which is a partition of $m+n$. The following sequence of canonical maps is exact:

$$ \mathbf{B}_{\mu}(S - \{n \text{ pts}\}) \to \mathbf{B}_{\mu\nu}(S) \to \mathbf{B}_{\nu}(S) \to 1. $$

Moreover, except when $S = \mathbb{S}^2$ and $n = 1, 2$ or $S = \mathbb{P}^2$ and $n = 1$, this is in fact a short exact sequence:

$$ 1 \to \mathbf{B}_{\mu}(S - \{n \text{ pts}\}) \to \mathbf{B}_{\mu\nu}(S) \to \mathbf{B}_{\nu}(S) \to 1. $$

Furthermore, if $S$ is not closed, the surjection $\mathbf{B}_{\mu\nu}(S) \to \mathbf{B}_{\nu}(S)$ splits.

**REMARK 6.16.** Recall that base configurations, hence also the $n$ points removed from $S$, must not be on the boundary of $S$, if $S$ has a non-trivial boundary.

**PROOF OF PROPOSITION 6.15.** Let us first recall that in considering braid groups, we consider surfaces up to isotopy equivalence, so we can remove the boundary of $S$ if it is non-trivial, and assume that $\partial S = \emptyset$. Recall that if $\lambda$ is a partition
of $N$, we denote by $C_\lambda(S)$ the configuration space $F_N(S)/\mathcal{S}_\lambda$. Forgetting the first $n$ points induces a map of configuration spaces $C_{\mu\nu}(S) \to C_\nu(S)$, which is a locally trivial fibration with fibres homeomorphic to $C_\mu(S - \{n \text{ pts}\})$, by a slight adaptation of [FN62, Th. 3] (or [Bir74, Th. 1.2]), which works for any manifold (without boundary). Since its fibres are path-connected, part of its long exact sequence of homotopy groups is:

$$ \pi_2(C_\nu(S)) \to \mathbf{B}_\mu(S - \{n \text{ pts}\}) \to \mathbf{B}_{\mu\nu}(S) \to \mathbf{B}_\nu(S) \to 1. $$

The map $F_n(S) \to C_\nu(S)$ is a covering, so that $\pi_2(C_\nu(S)) \cong \pi_2(F_n(S))$, which is trivial except when $S = \mathbb{S}^2$ and $n = 1, 2$ or $S = \mathbb{P}^2$ and $n = 1$. When $S$ is not the sphere or the projective plane, this follows from [Bir74, Prop. 1.3], using the fact that higher homotopy groups of surfaces different from $\mathbb{S}^2$ and $\mathbb{P}^2$ are trivial (see Corollary 6.6). When $S = \mathbb{S}^2$ this is [FvB62, Cor. p. 244] and when $S = \mathbb{P}^2$ it is [vB66, Cor. p. 82].

If $S$ is not closed, then there is an isotopy equivalence between $S$ and a proper subsurface $S'$ of $S$. Then one can choose a configuration of $m$ points in $S - S'$ and add them to each configuration of $n$ points of $S'$, getting a map $F_n(S') \to F_{m+n}(S)$. The induced map $\pi_1(C_\nu(S)) \cong \pi_1(C_\nu(S')) \to \pi_1(C_{\mu\nu}(S))$ is the required section. $\square$

**REMARK 6.17.** A weaker form of the asphericity statement for $\pi_2(F_n(S))$ used in the above proof may be found [FN62, Cor. 2.2]. They ask that the surface be compact, but this is only in order to be able to use the classification of compact surfaces, which we easily replaced by the classification of all surfaces in our proof.

**REMARK 6.18.** When $S$ is not closed, the construction of the splitting in the proof of Proposition 6.15 can also be used to get a morphism $\iota$ from $\mathbf{B}_n(S)$ to $\mathbf{B}_{n+1}(S)$, corresponding to adding an $(n+1)$-st strand near a boundary or an end of $S$. The restriction of this morphism to pure braid groups is a split injection (it is the section in the particular case $\mu = (1)$ and $\nu = (1, \dots, 1)$ in the above proof). Since it also induces an injection from $\mathbf{B}_n(S)/\mathbf{P}_n(S) \cong \mathcal{S}_n$ into $\mathbf{B}_{n+1}(S)/\mathbf{P}_{n+1}(S) \cong \mathcal{S}_{n+1}$, the morphism $\iota$ itself is injective (compare for instance with the proof of Corollary 4.5). Notice that, thanks to the connectedness of $S$, $\iota$ depends on the choices made in its construction only up to conjugation. By contrast, if $S$ is closed, there is no obvious construction of such a map, and in fact it does not exist in general [GG04a, GG10].

**6.2.4. Pure braid generators and commutators.** Some of the results below will hold for all surfaces $S$. However, in order to get more precise results, we need to get more specific and use the classification of surfaces recalled in §6.1. Recall that all the generators $\sigma_i$ of $\mathbf{B}_n$ are identified in $\mathbf{B}_n^{ab} \cong \mathbb{Z}$ (see Example 2.3), hence also in $\mathbf{B}_n(S)^{ab}$. The next proposition deals notably with the order of their common class $\sigma$. The trichotomy that appears here, which comes from Proposition 6.7, will play an important role in all that follows.

**PROPOSITION 6.19.** Let $n \ge 2$. Let us consider the generator $A_{ij}$ of $\mathbf{P}_n$ as an element of $\mathbf{B}_n(S)$.

* If $S$ is planar, then the class $\bar{A}_{ij} \in \mathbf{B}_n(S)^{ab}$ has infinite order.

* If $S \cong \mathbb{S}^2$, then the class $\bar{A}_{ij} \in \mathbf{B}_n(S)^{ab}$ has order $n-1$. However, its class in $\mathbf{P}_n(S)^{ab}$ has infinite order.

* In all the other cases, $A_{ij}$ is the commutator of two elements of $\mathbf{P}_n(S)$.
PROOF. If $S$ is planar: then $S$ can be embedded in a disc. Such an embedding induces a morphism $\mathbf{B}_n(S) \to \mathbf{B}_n(\mathbb{D}) = \mathbf{B}_n$, which in turn induces a morphism from $\mathbf{B}_n(S)^{ab}$ to $\mathbf{B}_n^{ab}$. The latter is infinite cyclic, generated by $\sigma$. Our element $\bar{A}_{ij}$ is sent to $\sigma^2$, hence it cannot be of finite order.

If $S$ is the sphere: then from the usual presentation of $\mathbf{B}_n(\mathbb{S}^2)$ (see for instance Corollary 6.26 below), we get that $\mathbf{B}_n(\mathbb{S}^2)^{ab} \cong \mathbb{Z}/(2(n-1))$, generated by $\sigma$. Again, $\bar{A}_{ij} = \sigma^2$, whose order is $n-1$.

If $S$ cannot be embedded in the sphere: then $S$ contains a handle or a crosscap; see Proposition 6.7 and the remark following it. We can then use the explicit isotopies drawn in Figures 6.1a and 6.1b to show that $A_{ij}$ is a bracket of two pure braids (which are respectively in the image of $\psi_j$ and in the image of $\psi_i$). $\square$

<figure><img src="image_31.png" /><figcaption>(A) Pure braid generator as a commutator on a surface with a crosscap.</figcaption></figure>

<figure><img src="image_32.png" /><figcaption>(B) Pure braid generator as a commutator on a surface with a handle.</figcaption></figure>

FIGURE 6.1. Explicit isotopies expressing a pure braid generator as a commutator. We use them in many different guises throughout the chapter, each time with respect to a different embedding of the crosscap or the handle in our surface, whose image contains exactly two points of the base configuration. For instance, with the notations from Figure 6.2 below, for any choice of $1 \le i < j \le n$ and of $1 \le r \le g$, the first one gives $A_{ij} = [c_j^{(r)}, (c_i^{(r)})^{-1}]$ in $\mathbf{B}_n(\mathcal{N}_g)$.

## 6.3. Presentations of surface braid groups

In order to prove some of the results below, in particular to determine completely the Lie rings of partitioned braid groups in the stable case, we will need to use presentations of braids groups on compact surfaces. The main tool for determining presentations of surface braid groups (including braids on the disc, which are usual Artin braids) are the Fadell-Neuwirth exact sequences; see Proposition 6.15. These were already used in the course of the proof of Proposition 6.10 to obtain generators of these groups. Let us now briefly explain how they may be used in order to determine defining relations on these generators. For $S$ a non-closed surface, these exact sequences give a decomposition of $\mathbf{B}_{1,n-1}(S)$ as a semi-direct product of $\mathbf{B}_{n-1}(S)$ with a free group $F$. Then suppose that one has a set of relations satisfied in $\mathbf{B}_n(S)$, defining a group $G_n$ and a well-defined surjection $\pi: G_n \to \mathbf{B}_n(S)$. One
can consider the subgroup $G_{1,n-1} = \pi^{-1}(\mathbf{B}_{1,n-1}(S))$, use the relations to show that it is of index at most $n$ in $G_n$ (which implies that the induced surjection of $G_n/G_{1,n-1}$ onto $\mathbf{B}_n(S)/\mathbf{B}_{1,n-1}(S)$ is a bijection – but of course not a group morphism, since these quotients do not bear a group structure), and determine a presentation of $G_{1,n-1}$ by using the Reidemeister-Schreier method. Then one shows that $G_{1,n-1}$ decomposes as a semi-direct product of a quotient isomorphic to $G_{n-1}$ with a kernel $K$ generated by a family of elements sent by $\pi$ to a basis of the free group $F$. The latter fact implies that this family must be a free basis of $K$, which means that $\pi: K \to F$ is an isomorphism. By induction, $\pi: G_{n-1} \to \mathbf{B}_{n-1}(S)$ is an isomorphism. Then $\pi: G_{1,n-1} \to \mathbf{B}_{1,n-1}(S)$ must be an isomorphism too. And since the induced surjection of $G_n/G_{1,n-1}$ onto $\mathbf{B}_n(S)/\mathbf{B}_{1,n-1}(S)$ is a bijection, $\pi: G_n \to \mathbf{B}_n(S)$ is an isomorphism (the reader can easily convince themselves that the latter implication works regardless of the existence of a group structure on the quotients).

This method can be used to get presentations of the braid groups of every non-closed surface of finite type; see [Bel04] for instance. It can also be adapted to the case of closed surfaces, replacing semi-direct product decompositions by non-split extensions, with some care for the exceptional cases where this is not even an extension. However, we prefer to deduce the case of closed surfaces from the non-closed one: we give here a direct general argument presenting $\mathbf{B}_n(S)$ as the quotient of $\mathbf{B}_n(S-\text{pt})$ by one explicit relation, by applying the Seifert-van Kampen theorem to configuration spaces; see Proposition 6.25.

**6.3.1. Surfaces with one boundary component.** Presentations of braid groups of compact surfaces with one boundary component can be found in [HOL02, §4] and in [Bel04, Th. 1.1 and A.2]. We re-write them with our own conventions, which we now explain.

Let us denote by $\Sigma_{g,1}$ the orientable connected compact surface of genus $g$ with one boundary component, and by $N_{g,1}$ the non-orientable connected compact surface of genus $g$ with one boundary component. We draw $\Sigma_{g,1}$ as a rectangle with $2g$ handles attached to it, and $N_{g,1}$ as a rectangle with $g$ crosscaps (see Convention 6.8). Our notations for braid generators are detailed in Figure 6.2. Our drawings of braids are to be thought of as seen from above, and the left-to-right direction in products corresponds to the foreground-to-background direction in our drawings. For instance, with the notations of Figure 6.2, we have that $\sigma_1 a_k^{(i)} \sigma_1^{-1} = a_k^{(i+1)}$. As an illustration of these conventions, we draw different representations of the classical pure braid generator $A_{i,i+1}$ in Figure 6.3.

**PROPOSITION 6.20 ([Bel04, Th. 1.1]).** Let $g \ge 0$. A presentation of the braid group on $\Sigma_{g,1}$, generated by $\sigma_1, \dots, \sigma_{n-1}$, $a_1, \dots, a_g$, $b_1, \dots, b_g$, is given by the braid relations for $\sigma_1, \dots, \sigma_{n-1}$, to which are added the following four families of relations (where $x$ and $y$ denote either a or $b$, and $1 \le r, s \le g$):

$$ (6.2) \qquad \left\{ \begin{array}{ll} (BS1) & \sigma_i \stackrel{\leftarrow}{\leftrightarrow} x_r \qquad \text{for all } r \text{ and all } i \ge 2; \\ (BS2) & x_r \stackrel{\leftarrow}{\leftrightarrow} \sigma_1 y_s \sigma_1^{-1} \qquad \text{for } r < s; \\ (BS3) & (\sigma_1 x_r)^2 = (x_r \sigma_1)^2 \qquad \text{for all } r; \\ (BS4) & [\sigma_1 b_r \sigma_1^{-1}, a_r^{-1}] = \sigma_1^2 \qquad \text{for all } r. \end{array} \right. $$

Recall that $g \stackrel{\leftrightarrow}{h}$ means that $g$ commutes with $h$ (Notation 4.1).
<figure><img src="image_33.png" /><figcaption>FIGURE 6.2. **Generators of surface braid groups.** For each generator (except for $\sigma_i$, where the points $i$ and $i+1$ move), only one point of the configuration moves, and the others stay put. We often denote $x_k^{(1)}$ by $x_k$, for $x = a, b, c$.</figcaption></figure>

<figure><img src="image_34.png" /><figcaption>FIGURE 6.3. The classical pure braid generator $A_{i,i+1} = \sigma_i^2$.</figcaption></figure>

Notice that it is easy to check that these relations hold in $\mathbf{B}_n(\Sigma_{g,1})$ by drawing explicit isotopies. See for instance Figure 6.1b for a drawing of (BS4) (which generalises to $[b_s^{(j)}, (a_r^{(j)})^{-1}] = A_{r,s}$ if $r < s$). The translation between Bellingeri's conventions and ours is as follows:

* Our statement is the case $p=1$ of [Bel04, Th. 1.1], whence the absence of the $z_k$, and of the relations involving them.

* Our $\sigma_i$ is his $\sigma_i^{-1}$, our $a_r$ is his $b_r^{-1}$, and our $b_r$ is his $a_r^{-1}$.

* Our (BS1) – (BS4) are his (R1) – (R4), with 2 and 3 exchanged.

**REMARK 6.21.** Although [Bel04, Th. 1.1] is stated for $g \ge 1$, the proof works equally well if $g=0$. In fact, the case $g=0$ of our statement is just the usual presentation of braid groups on the disc.

**PROPOSITION 6.22 ([Bel04, Th. A.2])**. Let $g \ge 1$. A presentation of the braid group on $\mathcal{N}_{g,1}$, generated by $\sigma_1, \dots, \sigma_{n-1}$, $c_1, \dots, c_g$ is given by the braid relations for $\sigma_1, \dots, \sigma_{n-1}$, to which are added the following three families of relations (where $1 \le r, s \le g$):

$$ (6.3) \qquad \left\{ \begin{array}{ll} (BN1) & \sigma_i \leftrightarrow c_r \qquad \text{for all } r \text{ and all } i \ge 2; \\ (BN2) & c_r \leftrightarrow \sigma_1 c_s \sigma_1^{-1} \qquad \text{for } r < s; \\ (BN3) & [\sigma_1 c_r \sigma_1^{-1}, c_r^{-1}] = \sigma_1^2 \qquad \text{for all } r. \end{array} \right. $$

Here again, it is easy to check these relations explicitly. See for instance Figure 6.1a for a drawing of (BN3) (which generalises to $[c_s^{(j)}, (c_r^{(j)})^{-1}] = A_{rs}$ if $r < s$). Also, it is the case $p=1$ of Bellingeri's statement so, again, the $z_k$ and the corresponding relations are irrelevant. Moreover, our $\sigma_i$ is his $\sigma_i^{-1}$, our $c_i$ are his $a_i^{-1}$, and the indexation of our relations is the same as his.
REMARK 6.23. The statement of [Bel04, Th. A.2] is for $g \ge 2$, but the proof works equally well if $g = 1$.

In the case of non-orientable surfaces, we will also need the case $p > 1$ of [Bel04, Th. A.2], which will be used twice, in the proofs of Proposition 6.58 and Proposition 6.84. Let us denote by $\mathcal{N}_{g,n+1}$ the non-orientable connected compact surface of genus $g$ with either $n+1$ punctures or $n+1$ boundary components (recall that, up to isotopy, removing a point and removing an open disc are equivalent). The braid group $\mathbf{B}_m(\mathcal{N}_{g,n+1})$ may be seen as a subgroup of $\mathbf{B}_{m+n}(\mathcal{N}_{g,1})$; see Proposition 6.15. Namely, this subgroup is generated by $\sigma_1, \dots, \sigma_{n-1}, c_1, \dots, c_g$ and $z_j := A_{1,m+j}$ for all $1 \le j \le n$.

PROPOSITION 6.24 ([Bel04, Th. A.2]). Let $g \ge 0$ and $m \ge 1$. A presentation of the braid group $\mathbf{B}_m(\mathcal{N}_{g,n+1})$, generated by $\sigma_1, \dots, \sigma_{m-1}, c_1, \dots, c_g, z_1, \dots, z_n$ is given by the relations from Proposition 6.22, together with the following four families of relations:

$$ \left\{ \begin{array}{ll} (BN4) & z_j \rightleftharpoons \sigma_i \quad \text{for all } j \le n \text{ and all } i \in \{2, \dots, m-1\}; \\ (BN5) & c_r \rightleftharpoons \sigma_1 z_j \sigma_1^{-1} \quad \text{for all } j \le n \text{ and all } r \le g; \\ (BN6) & z_i \rightleftharpoons \sigma_1 z_j \sigma_1^{-1} \quad \text{for } i > j; \\ (BN7) & (\sigma_1 z_j)^2 = (z_j \sigma_1)^2 \quad \text{for all } j \le n. \end{array} \right. $$

Once more, these relations are easy to check explicitly. The translation between [Bel04, Th. A.2] and our statement is the same as above, our $z_j$ being the same as Bellingeri's.

### 6.3.2. Closed surfaces.
When $S$ is a closed surface, one needs to add a single relation to a presentation of $\mathbf{B}_n(S - pt)$ to get a presentation of $\mathbf{B}_n(S)$. In fact, this is a very general fact, which does not require any hypothesis on the surface.

PROPOSITION 6.25. Let $S$ be a connected surface and $x \in S$ any point in its interior. The inclusion of $S - x$ into $S$ induces a surjective homomorphism

$$ \mathbf{B}_n(S - x) \to \mathbf{B}_n(S) $$

whose kernel is normally generated by a single element $\beta$. Explicitly, $\beta$ is a braid with $n-1$ trivial strands, whose remaining strand loops once around the puncture $x$.

PROOF. Choose a subdisc $D \subset S$ containing $x$ in its interior, and a metric on $D$. Write $\mathcal{U}_n(S, x)$ for the subspace of $C_n(S) = F_n(S)/\mathfrak{S}_n$ of (unordered) configurations that have a unique closest point in $D$ to $x$ (which may be $x$ itself). Together with $C_n(S - x)$, this forms an open cover

$$ \{\mathcal{U}_n(S, x), C_n(S - x)\} $$

of $C_n(S)$, with intersection $\mathcal{U}_n(S, x) \cap C_n(S - x) = \mathcal{U}_n(S - x, x)$ the space of $n$-point configurations in $S - x$ that have a unique closest point in $D - x$ to $x$. Note that these subspaces of $C_n(S)$ are all path-connected. Let us choose a basepoint for $C_n(S)$ that lies in $\mathcal{U}_n(S - x, x)$. The Seifert-van Kampen theorem then gives us a pushout square of groups:

$$ \begin{tikzcd} \pi_1(\mathcal{U}_n(S - x, x)) \arrow[r] \arrow[d] & \mathbf{B}_n(S - x) \arrow[d] \\ (6.4) \arrow[u] \arrow[r] & \pi_1(\mathcal{U}_n(S, x)) \arrow[r] \arrow[u] \arrow[d] \arrow[d] & \mathbf{B}_n(S). \end{tikzcd} $$
There is a well-defined projection $\mathcal{U}_n(S, x) \to D$ given by remembering just the unique closest point in $D$ to $x$, which restricts to a projection $\mathcal{U}_n(S-x, x) \to D-x$. These are both locally trivial fibrations with fibres canonically homeomorphic to $C_{n-1}(S-x)$. (Over a point $p \in D$, the homeomorphism is induced by the evident homeomorphism between $S-x$ and $S-\tilde{B}_{d(p,x)}(x)$, where $\tilde{B}_r(x)$ denotes the closed ball of radius $r$ in $D$ centred at $x$ and $d(-, -)$ is the metric that we chose on $D$.) The inclusion of the latter into the former is therefore a map of locally trivial fibrations which is the identity on fibres. Considering the induced long exact sequences of homotopy groups, we obtain a map of exact sequences:

(6.5)

$$ \begin{array}{ccc} 1 \longrightarrow \mathbf{B}_{n-1}(S-x) \longrightarrow \pi_1(\mathcal{U}_n(S-x,x)) \longrightarrow \pi_1(D-x) \cong \mathbb{Z} \longrightarrow 1 & & \\ & \xrightarrow{\cong} \Big\downarrow_{\mathcal{U}_n(S,x)}^{\mathcal{U}_n(S,x)} \Big\downarrow_{\mathcal{U}_n(S,x)} & \\ 1 \longrightarrow \mathbf{B}_{n-1}(S-x) \xrightarrow{\cong} \pi_1(\mathcal{U}_n(S,x)) \longrightarrow \pi_1(D) = 1. & & \end{array} $$

The map $r$ obtained from this diagram is a retraction for the upper short exact sequence, whose existence implies that $\pi_1(\mathcal{U}_n(S-x,x))$ is the direct product of $\pi_1(\mathcal{U}_n(S,x))$ and $\mathbb{Z}$. Then, the left-hand vertical map in (6.4) identifies with the projection that forgets the $\mathbb{Z}$ factor. Together with the fact that (6.4) is a pushout square, this implies that the right-hand vertical map in (6.4) is the quotient of $\mathbf{B}_n(S-x)$ by (the subgroup normally generated by) the image of the $\mathbb{Z}$ factor of $\pi_1(\mathcal{U}_n(S-x,x))$ in $\mathbf{B}_n(S-x)$. We may choose for a generator of this $\mathbb{Z}$ factor any element of $\pi_1(\mathcal{U}_n(S-x,x))$ that projects to a generator of $\pi_1(D-x)$, for example the braid described in the proposition. $\square$

Let us make this explicit:

**COROLLARY 6.26.** For all $g \ge 0$, the braid group $\mathbf{B}_n(\Sigma_g)$ is the quotient of $\mathbf{B}_n(\Sigma_{g,1})$ by the relation:

$$ \sigma_1 \cdots \sigma_{n-2} \sigma_{n-1}^2 \sigma_{n-2} \cdots \sigma_1 = \prod_{r=1}^{g} [a_r, b_r^{-1}]. $$

Similarly, the braid group $\mathbf{B}_n(\mathbb{N}_g)$ is the quotient of $\mathbf{B}_n(\mathbb{N}_{g,1})$ by the relation:

$$ \sigma_1 \cdots \sigma_{n-2} \sigma_{n-1}^2 \sigma_{n-2} \cdots \sigma_1 = c_1^2 \cdots c_g^2. $$

We note that we recover as a particular case the usual presentations of the braid group on the sphere (see [FvB62] or [Bir74, Th. 1.11]) and of the projective plane (see [vB66, §III, page 83]). Not having to treat these as exceptional cases is one of the great advantages of the present method.

**PROOF OF COROLLARY 6.26.** This is a direct application of Proposition 6.25, using the fact that $\Sigma_g-\text{pt}$ (resp. $\mathbb{N}_g-\text{pt}$) is isotopy equivalent to $\Sigma_{g,1}$ (resp. to $\mathbb{N}_{g,1}$). We note that $\sigma_1 \cdots \sigma_{n-2} \sigma_{n-1}^2 \sigma_{n-2} \cdots \sigma_1 = A_{12} \cdots A_{1n}$ is the (pure) braid obtained by making the first strand turn once around all the other ones; see Figure 6.4 for the relevant drawings. $\square$

**6.3.3. Partitioned braids on closed surfaces.** The above proof of Proposition 6.25 generalises to partitioned braid groups without much difficulty; see Remark 6.31. However, we prefer to deduce these generalisations directly from
See Figure 6.5 for an example of such choices. The latter relations imply $N' \subseteq N$.

We now show that $N'$ contains all the conjugates of $\beta$ by elements of $\mathbf{B}_n(S-x)$, which implies $N' \supseteq N$. In order to do this, we need only show that it contains $t^{-1}\beta t$ for $t$ in a set of representatives of classes modulo $\mathbf{P}_n(S-x)$: then every element of $\mathbf{B}_n(S-x)$ is of the form $t\alpha$ for some such $t$ and some $\alpha \in \mathbf{P}_n(S-x)$, and $(t\alpha)^{-1}\beta(t\alpha) = \alpha^{-1}(t^{-1}\beta t)\alpha$ must be in $N'$.

Every element $\tau \in \mathfrak{S}_n \cong \mathbf{B}_n(S-x)/\mathbf{P}_n(S-x)$ is the product of an element $\tau'$ fixing 1 with some cycle $\tau_1 \cdots \tau_{i-1}$ (precisely, $i = \tau(1)$). Since $\beta$ commutes with every element of the subgroup $\langle\sigma_2, \dots, \sigma_{n-1}\rangle \subset \mathbf{B}_n(S-x)$, and since this subgroup surjects onto permutations fixing 1, we can choose a lift $t'$ of $\tau'$ commuting with $\beta$, so that the lift $t = t'\sigma_1 \cdots \sigma_{i-1}$ of $\tau$ to $\mathbf{B}_n(S-x)$ satisfies:

$$ \begin{aligned} t^{-1}\beta t &= (\sigma_1 \cdots \sigma_{i-1})^{-1} t'^{-1} \beta t'(\sigma_1 \cdots \sigma_{i-1}) \\ &= (\sigma_1 \cdots \sigma_{i-1})^{-1} \beta (\sigma_1 \cdots \sigma_{i-1}) = \beta_i \in N', \end{aligned} $$

whence our result.

□

<figure><img src="image_35.png" /><figcaption>FIGURE 6.5. Braids $\beta$ and $\beta_1, \dots, \beta_n$ from the proof of the equiv- alence of Propositions 6.25 and 6.27.</figcaption></figure>

It is not difficult to generalise this to any partitioned braid group:

COROLLARY 6.28. Let $S$ be a connected surface, $x \in S$ any point in its interior and $\lambda = (n_1, \dots, n_l)$ be a partition of $n$ of length $l$. The inclusion of $S - x$ into $S$ induces a surjective homomorphism

$$ \mathbf{B}_{\lambda}(S-x) \to \mathbf{B}_{\lambda}(S) $$

whose kernel is normally generated by $l$ elements $\beta_1, \dots, \beta_l$. Explicitly, $\beta_i$ is a braid with $n-1$ trivial strands, except one block in the $i$-th block which loops once around the puncture $x$.

PROOF. We have $\mathbf{P}_n(S-x) \subset \mathbf{B}_{\lambda}(S-x) \subset \mathbf{B}_n(S-x)$. As a consequence, if $N$ is a normal subgroup of $\mathbf{P}_n(S-x)$ which is normal in $\mathbf{B}_n(S-x)$, we have $\mathbf{P}_n(S-x)/N \subset \mathbf{B}_{\lambda}(S-x)/N \subset \mathbf{B}_n(S-x)/N$. Moreover, normal generators for $N$ in $\mathbf{P}_n(S-x)$ are also normal generators for $N$ in $\mathbf{B}_{\lambda}(S-x)$ and, in fact, fewer generators are needed to normally generate $N$ in $\mathbf{B}_{\lambda}(S-x)$. Namely, using the
notations from the previous proof, we have $\beta_{\alpha+1} = \sigma_{\alpha}^{-1}\beta_{\alpha}\sigma_{\alpha}$, so that $\beta_{\alpha+1}$ and $\beta_{\alpha}$ are conjugate in $\mathbf{B}_{\lambda}(S-x)$ whenever $\sigma_{\alpha} \in \mathbf{B}_{\lambda}(S-x)$, which happens when $\alpha$ and $\alpha+1$ are in the same block of $\lambda$. As a consequence, we need to pick only one index $\alpha$ in each block of $\lambda$ in order for the $\beta_{\alpha}$ to normally generate $N$ in $\mathbf{B}_{\lambda}(S-x)$. $\square$

REMARK 6.29. This boils down to considering representatives modulo $\mathbf{B}_{\lambda}(S-x)$ instead of modulo $\mathbf{P}_{n}(S-x)$ (that is, elements of $\mathfrak{S}_{n}/\mathfrak{S}_{\lambda}$ instead of $\mathfrak{S}_{n}$) in the previous reasoning. In fact, one can see that there are straightforward equivalences between all these statements for the different partitions of $n$.

Let us make these statements explicit. We use the usual convention $A_{ji} = A_{ij}$ and $A_{ii} = 1$:

COROLLARY 6.30. For all $g \ge 0$ and any partition $\lambda = (n_1, \dots, n_l)$, the braid group $\mathbf{B}_{\lambda}(\Sigma_g)$ is the quotient of $\mathbf{B}_{\lambda}(\Sigma_{g,1})$ by the relations:

$$A_{\alpha_1} \cdots A_{\alpha_n} = \prod_{r=1}^{g} [a_r^{(\alpha)}, (b_r^{(\alpha)})^{-1}].$$

Similarly, the braid group $\mathbf{B}_{\lambda}(\mathfrak{N}_g)$ is the quotient of $\mathbf{B}_{\lambda}(\mathfrak{N}_{g,1})$ by the relations:

$$A_{\alpha_1} \cdots A_{\alpha_n} = (c_1^{(\alpha)})^2 \cdots (c_g^{(\alpha)})^2.$$

In both cases, $\alpha$ runs through any set of representatives of the blocks of $\lambda$.

REMARK 6.31. Instead of the reasoning above, one could adapt the proof of Proposition 6.25 to partitioned configuration spaces. Precisely, we get an open cover $\{\mathcal{U}_{\lambda}(S, x), C_{\lambda}(S-x)\}$ of $C_{\lambda}(S) = F_n(S)/\mathfrak{S}_{\lambda}$ by an obvious adaptation of the definition of $\mathcal{U}_n(S, x)$. However, this time $\mathcal{U}_{\lambda}(S, x)$ is disconnected, with one path-component for each block of the partition $\lambda$. As a consequence, one needs to apply the Seifert-van Kampen theorem once for each path-component of $\mathcal{U}_{\lambda}(S, x)$, resulting in taking the quotient of the fundamental group by one additional relation for each application of the theorem. In doing so, one needs to be careful about basepoints, since one obviously cannot choose a common basepoint in the different path-components of $\mathcal{U}_{\lambda}(S, x)$.

## 6.4. The lower central series of the whole group

We now turn to the study of the LCS of $\mathbf{B}_n(S)$, which we completely determine for any $n \ge 3$ and any surface $S$. We begin by computing the abelianisation of this group; see Proposition 6.32. Then we study $\mathbf{B}_n(S)/\Gamma_{\infty}$, and we show that when $n \ge 3$, it is nilpotent of class at most 2, which means that the LCS of $\mathbf{B}_n(S)$ stops at most at $\Gamma_3$; see Theorem 6.34 and Corollary 6.36. Finally, we compute the Lie ring, generalising a result of [BGG08]; see Theorem 6.42.

### 6.4.1. The abelianisation.
We first compute the abelianisation of $\mathbf{B}_n(S)$, for any $n$ and any surface $S$. In order to do this, recall that the group morphism $\varphi: \mathbf{B}_n \to \mathbf{B}_n(S)$ induces a map $\mathbf{B}_n^{ab} \to \mathbf{B}_n(S)^{ab}$, and that, since all the $\sigma_i$ are identified in $\mathbf{B}_n^{ab} \cong \mathbb{Z}$, they are so also in $\mathbf{B}_n(S)^{ab}$. We denote by $\sigma$ their common image in $\mathbf{B}_n(S)^{ab}$.

PROPOSITION 6.32. In general, for all $n \ge 2$, we have:

$$\mathbf{B}_n(S)^{ab} \cong \pi_1(S)^{ab} \times \langle \sigma \rangle.$$

Moreover, $\sigma$ is:
• of infinite order if S is planar,

• of order 2(n - 1) if S ≅ S²,

• of order 2 in all the other cases.

PROOF. Consider the short exact sequence $\mathbf{P}_n^\circ(S) \hookrightarrow \mathbf{B}_n(S) \xrightarrow{\pi_S} \pi_1(S) \wr \mathfrak{S}_n$. By Lemma 1.11, it induces an exact sequence of abelian groups:

$$ (\mathbf{P}_n^\circ(S)^{ab})_{\mathbf{B}_n(S)} \longrightarrow \mathbf{B}_n(S)^{ab} \longrightarrow (\pi_1(S) \wr \mathfrak{S}_n)^{ab} \longrightarrow 0. $$

On the one hand, the quotient $(\pi_1(S) \wr \mathfrak{S}_n)^{ab}$ is isomorphic to $\pi_1(S)^{ab} \times \mathbb{Z}/2$ (Lemma A.24). On the other hand, it follows from Proposition 6.10 that $\mathbf{P}_n^\circ(S)$ is generated by $\mathbf{P}_n$ under the action of $\mathbf{B}_n(S)$. As a consequence, the map $\mathbf{P}_n^{ab} \to (\mathbf{P}_n^\circ(S)^{ab})_{\mathbf{B}_n(S)}$ induced by $\varphi$ is surjective. Moreover, it factors through $(\mathbf{P}_n^{ab})_{\mathbf{B}_n} = (\mathbf{P}_n^{ab})_{\mathfrak{S}_n} \cong \mathbb{Z}$. Thus $(\mathbf{P}_n^\circ(S)^{ab})_{\mathbf{B}_n(S)}$ is cyclic, and its image in $\mathbf{B}_n(S)^{ab}$ is generated by $\sigma^2$, which is the image of any pure braid generator.

All this implies that $\mathbf{B}_n(S)^{ab} / \langle \sigma \rangle \cong \pi_1(S)^{ab}$. Moreover, the corresponding projection $\mathbf{B}_n(S)^{ab} \to \pi_1(S)^{ab}$ splits, a splitting being induced by any of the $\psi_i$. As a consequence, $\mathbf{B}_n(S)^{ab}$ identifies with $\pi_1(S)^{ab} \times \langle \sigma \rangle$, and we can use Proposition 6.19 to get a complete calculation. $\square$

**6.4.2. The lower central series.** Let us now turn to the study of the LCS of $\mathbf{B}_n(S)$. Our main tool for studying it is a decomposition theorem (Theorem 6.34 below), whose proof relies on the following:

LEMMA 6.33. Let $n \ge 3$. The image of $\mathbf{B}_n$ in $\mathbf{B}_n(S)/\Gamma_\infty$ is cyclic, and it is central. Namely, it is generated by the common class $\sigma$ of the usual generators $\sigma_i$ of $\mathbf{B}_n$.

PROOF. The morphism $\varphi$ sends $\Gamma_\infty(\mathbf{B}_n)$ to $\Gamma_\infty(\mathbf{B}_n(S))$. We know that $\sigma_i\sigma_j^{-1} \in \Gamma_\infty(\mathbf{B}_n)$ for all $i, j < n$, so $\sigma_i \equiv \sigma_j \pmod{\Gamma_\infty(\mathbf{B}_n(S))}$. Let us denote by $\sigma \in \mathbf{B}_n(S)/\Gamma_\infty$ the common image of the $\sigma_i$. Since the $\sigma_i$ generate $\mathbf{B}_n$, the image of $\mathbf{B}_n$ in $\mathbf{B}_n(S)/\Gamma_\infty$ is the cyclic subgroup generated by $\sigma$. In particular, its elements commute with $\sigma$. Moreover, for all $\gamma \in \pi_1(S - \mathbb{D})$, the braids $\sigma_2$ and $\psi_1(\gamma)$ have disjoint support, hence $\sigma$ also commutes with the image of $\psi_1$. Since the images of $\varphi$ and $\psi_1$ generate $\mathbf{B}_n(S)$ by Proposition 6.10, this means that $\sigma$ is a central element of $\mathbf{B}_n(S)/\Gamma_\infty$. $\square$

We can now state our main decomposition theorem:

THEOREM 6.34. For all $n \ge 3$, there is a central extension:

$$ \langle \sigma^2 \rangle \longrightarrow \mathbf{B}_n(S)/\Gamma_\infty \longrightarrow \pi_1(S)^{ab} \times \mathbb{Z}/2. $$

PROOF. Since $\pi_S$ is surjective, it sends $\Gamma_\infty(\mathbf{B}_n(S))$ onto a normal subgroup of $\pi_1(S) \wr \mathfrak{S}_n$. This normal subgroup is contained in $\Gamma_\infty(\pi_1(S) \wr \mathfrak{S}_n)$ and contains the $\tau_i\tau_j^{-1}$, which are the images of the $\sigma_i\sigma_j^{-1} \in \Gamma_\infty(\mathbf{B}_n(S))$. By Lemma A.26, it is
equal to $\Gamma_2(\pi_1(S) \wr \mathfrak{S}_n)$. Then, we can apply the Nine Lemma to the diagram:

<figure><img src="image_36.png" /></figure>

Then recall that $(\pi_1(S)\wr \mathfrak{S}_n)^{ab} \cong \pi_1(S)^{ab} \times \mathbb{Z}/2$. Thus, we are left with analysing the kernel $\mathbf{P}_n^\circ(S)/\Gamma_\infty(\mathbf{B}_n(S)) \cap \mathbf{P}_n^\circ(S)$ of $\bar{\pi}_S$, which is the image of $\mathbf{P}_n^\circ(S)$ in $\mathbf{B}_n(S)/\Gamma_\infty$. Since $\mathbf{P}_n^\circ(S)$ is the normal closure of $\mathbf{P}_n$ in $\mathbf{B}_n(S)$ by Proposition 6.10, its image in $\mathbf{B}_n(S)/\Gamma_\infty$ is the normal closure of the image of $\mathbf{P}_n$. But $\mathbf{P}_n$ is sent to $\langle\sigma^2\rangle$ which is central (and, in particular, normal) in $\mathbf{B}_n(S)/\Gamma_\infty$, whence the result. $\square$

REMARK 6.35. We also have a central extension:

$$ \langle\sigma\rangle \longrightarrow \mathbf{B}_n(S)/\Gamma_\infty \dashrightarrow \pi_1(S)^{ab}. $$

This slightly different statement tells us slightly different things: it implies that $\sigma$ is central, whereas the statement of the theorem says that $\bar{\sigma}$ is not trivial in $\mathbf{B}_n(S)^{ab}$.

COROLLARY 6.36. For any $n \ge 3$, we have $\Gamma_3(\mathbf{B}_n(S)) = \Gamma_4(\mathbf{B}_n(S))$.

PROOF. Proposition 6.34 implies that $\mathbf{B}_n(S)/\Gamma_\infty$ is 2-nilpotent, which means exactly that its $\Gamma_3$ is trivial. In other words, $\Gamma_3 \subseteq \Gamma_\infty$ for $\mathbf{B}_n(S)$. $\square$

The remaining cases consist in the cases when $n = 1$ and when $n = 2$.

PROPOSITION 6.37. For any connected surface $S$, either $\mathbf{B}_1(S) = \pi_1(S)$ is abelian, which occurs precisely when $S \in \{\mathbb{D} - pt, \mathbb{D}, \mathbb{S}^2, \mathbb{T}^2, \mathbb{P}^2, \mathbb{M}^2\}$ up to isotopy equivalence, or its LCS does not stop.

PROOF. If $S$ is closed and not in $\{\mathbb{S}^2, \mathbb{P}^2, \mathbb{T}^2\}$ then the LCS of $\pi_1(S)$ does not stop. To see this, note that it admits a presentation of the form $\langle a_1, b_1, \dots, a_g, b_g \mid [a_1, b_1] \cdots [a_g, b_g] = 1 \rangle$ or $\langle c_1, \dots, c_g \mid c_1^2 \cdots c_g^2 = 1 \rangle$ for $g \ge 2$, and these both project onto $\mathbb{Z}/2 * \mathbb{Z}/2 = \langle x, y \mid x^2 = y^2 = 1 \rangle$ by sending $a_1, b_1, c_1$ to $x$ and all other generators to $y$; the LCS of $\mathbb{Z}/2 * \mathbb{Z}/2$ does not stop by Proposition A.16. If $S$ is not closed, its fundamental group is free by Proposition 6.4. If it is also not in $\{\mathbb{D} - pt, \mathbb{D}, \mathbb{M}^2\}$ then its fundamental group is moreover non-abelian free, and hence its LCS does not stop by [Mag35] (see also [MKS04, Chap. 5]). $\square$

REMARK 6.38. The fundamental group of a surface is known to always be residually nilpotent, and almost always residually torsion-free nilpotent (the only exceptions to the latter being the projective plane and the Klein bottle). This follows from [MKS04, Chap. 5] for non-closed surfaces, whose fundamental group is free. For closed surfaces, it follows from [Bau10] if $S$ contains a handle: then the Seifert-van Kampen theorem applied to a decomposition $S = \mathbb{T}^2\#S'$ gives a presentation of the form of Theorem 1 therein. The remaining surfaces are the sphere, the projective plane and the Klein bottle, whose LCS are easily computed explicitly (see Proposition A.4 for the last one).

PROPOSITION 6.39. When $S$ is not $\mathbb{D}$, $\mathbb{S}^2$ or $\mathbb{P}^2$, up to isotopy equivalence, the LCS of $\mathbf{B}_2(S)$ does not stop.
**PROOF.** The group $\mathbf{B}_2(S)$ surjects onto $\pi_1(S) \wr \mathfrak{S}_2$. Since $\pi_1(S)^{ab}$ surjects onto $\mathbb{Z}$ except in the three excluded cases of the statement, we can apply Corollary A.29 to see that the LCS of $\pi_1(S) \wr \mathfrak{S}_2$ does not stop. Then the one of $\mathbf{B}_n(S)$ does not either by Lemma 1.1. $\square$

**REMARK 6.40.** For orientable surfaces, $\mathbf{B}_2(S)$ is residually nilpotent [BB09b, Cor. 10]. This result can in fact be extended to non-orientable surfaces, with the same method, using the results quoted in Remark 6.38.

**REMARK 6.41.** Clearly, the LCS of $\mathbf{B}_2(\mathbb{D}) \cong \mathbb{Z}$ and of $\mathbf{B}_2(\mathbb{S}^2) \cong \mathbb{Z}/2$ both stop at $\Gamma_2$. The fact that $\mathbf{B}_2(\mathbb{S}^2) \cong \mathbb{Z}/2$ is the case $n=2$ and $g=0$ of Corollary 6.26. On the other hand, the group $\mathbf{B}_2(\mathbb{P}^2)$ is the dicyclic group of order 16 (Corollary 6.82), which is 3-nilpotent, so its LCS stops at $\Gamma_4$.

**6.4.3. The Lie ring.** We can be more precise about our description of the LCS of surface braid groups. In particular, Corollary 6.36 says that, for $n \ge 3$, the LCS stops at most at $\Gamma_3$, but it does not say when it stops at $\Gamma_2$ (then the associated Lie ring consists only of the abelianisation, which has already been computed in Proposition 6.32), or when it stops at $\Gamma_3$ (in which case the Lie ring is 2-nilpotent but not abelian). We now show that the latter holds only for non-planar orientable surfaces, and we compute precisely $\mathcal{L}(\mathbf{B}_n(S))$, generalising results of [BGG08].

**THEOREM 6.42.** Let $n \ge 3$ be an integer and $S$ be a connected surface. The LCS of $\mathbf{B}_n(S)$:

* stops at $\Gamma_2$ if $S$ is either planar or non-orientable, or if $S \cong \mathbb{S}^2$.

* stops at $\Gamma_3$ in the other cases. Then $\mathcal{L}_2(\mathbf{B}_n(S))$ is cyclic, generated by the common class $\sigma^2$ of the pure braid generators $A_{ij}$.

Moreover, in the second case, $\sigma^2$ is of finite order if and only if $S$ is closed, in which case its order is $n + g - 1$, where $g$ is the genus of $S$.

**PROOF.** Since the LCS of $\mathbf{B}_n(S)$ stops at most at $\Gamma_3$, $\mathcal{L}_2(\mathbf{B}_n(S))$ identifies with $\Gamma_2(\mathbf{B}_n(S)/\Gamma_\infty)$ (the latter being $\Gamma_2/\Gamma_\infty(\mathbf{B}_n(S)) = \Gamma_2/\Gamma_3(\mathbf{B}_n(S))$). Moreover, using the central extension of Theorem 6.34, we see that in $\mathbf{B}_n(S)/\Gamma_\infty$, the subgroup $\Gamma_2$ must be contained in $\langle\sigma^2\rangle$, which implies that it is cyclic, generated by a power of $\sigma^2$.

**Planar surfaces:** if $S$ is planar, then the common class $\sigma$ of the $\sigma_i$ in $\mathbf{B}_n(S)^{ab} = (\mathbf{B}_n(S)/\Gamma_\infty)^{ab}$ is of infinite order by Proposition 6.32. Hence $\Gamma_2(\mathbf{B}_n(S)/\Gamma_\infty)$ does not contain any power of $\sigma = \bar{\sigma}_1$. But $\Gamma_2(\mathbf{B}_n(S)/\Gamma_\infty)$ is contained in $\langle\sigma^2\rangle$, so it must be trivial, which means that $\Gamma_2 = \Gamma_\infty$ for $\mathbf{B}_n(S)$.

**The sphere:** if $S = \mathbb{S}^2$, then $\mathbf{B}_n(\mathbb{S}^2)$ is a quotient of $\mathbf{B}_n$ (see Corollary 6.26), which implies that its LCS also stops at $\Gamma_2$.

**Non-orientable surfaces:** the surface $S$ is non-orientable if and only if it contains an embedded Möbius band. Then $\sigma_1^2 = [\sigma_1 c \sigma_1^{-1}, c^{-1}]$ for some $c \in \mathbf{B}_n(S)$: precisely, $c$ is a braid whose first strand goes around the Möbius strip once, that is, through the crosscap; see Figure 6.1a. Since $\sigma_1$ is sent to the central element $\sigma$ of $\mathbf{B}_n(S)/\Gamma_\infty$, this relation implies that $\sigma^2 = [\bar{c}, \bar{c}^{-1}] = 1$. Thus $\langle\sigma^2\rangle$ is trivial, and so is its subgroup $\Gamma_2(\mathbf{B}_n(S)/\Gamma_\infty) \cong \mathcal{L}_2(\mathbf{B}_n(S))$.

**Non-planar orientable surfaces:** if $S$ has a handle, then $\sigma_1^2 = [a^{-1}, \sigma_1 b \sigma_1^{-1}]$ for some $a, b \in \mathbf{B}_n(S)$. Precisely, $a$ and $b$ are braids whose first strands go around a handle; see Figure 6.1b. Hence $\sigma^2 \in \Gamma_2(\mathbf{B}_n(S)/\Gamma_\infty)$, which implies that $\mathcal{L}_2(\mathbf{B}_n(S))$ is generated by $\sigma^2$.
If $S$ is a non-planar compact orientable surface with at least one boundary component, then $S$ can be embedded in some $\Sigma_{g,1}$ for an arbitrary large $g$, by attaching a disc to each boundary component save one. The induced map $\mathcal{L}_2(\mathbf{B}_n(S)) \to \mathcal{L}_2(\mathbf{B}_n(\Sigma_{g,1}))$ sends $\sigma^2$ to $\sigma^2$, and the latter is of infinite order (Proposition 6.43).

If $S$ is a non-compact non-planar orientable surface, let us suppose that $\sigma^2$ is a torsion element in $\mathbf{B}_n(S)/\Gamma_\infty = \mathbf{B}_n(S)/\Gamma_3$. Then for some integer $k$, $\sigma_1^{2k}$ is equal to some product of commutators of length at least 3 in $\mathbf{B}_n(S)$. Such a formula involves only a finite number of braids. Let us choose a representative of each of these isotopy classes. These involve a finite number of paths on the surface. If moreover we choose an isotopy realising the aforementioned equality of braids (using the concatenation of the chosen representatives as a representative of the right-hand side), the image of this isotopy is contained in a compact subsurface $S'$ of $S$. Thus, our formula also holds in $\mathbf{B}_n(S')$: $\sigma_1^{2k}$ is equal to some product of commutators of length at least 3 in there. Then $\sigma^{2k} = 1$ in $\mathbf{B}_n(S')/\Gamma_3$. However, this contradicts the previous case, since $S'$ cannot be closed (nor planar). We conclude that $\sigma^2$ cannot be a torsion element in $\mathbf{B}_n(S)/\Gamma_\infty$; equivalently, $\mathcal{L}_2(\mathbf{B}_n(S)) \cong \mathbb{Z}$.

If $S$ is a closed orientable surface $\Sigma_g$ of genus $g \ge 1$, then $\mathcal{L}_2(\mathbf{B}_n(\Sigma_g)) \cong \mathbb{Z}/(n+g-1)$. This is [BGG08, Th. 1], which can be deduced from Proposition 6.43 below. We will do so in a more general context later; see Proposition 6.56. $\square$

The proof of the following result is inspired from [BGG08]. In fact, it is equivalent to [BGG08, Th. 1] in a quite straightforward way; see Remark 6.44.

**PROPOSITION 6.43.** Let $g \ge 1$, and let $\Sigma_{g,1}$ denote the compact surface of genus $g$ with one boundary component. For every $n \ge 1$, $\mathcal{L}_2(\mathbf{B}_n(\Sigma_{g,1})) \cong \mathbb{Z}$, generated by the common class $\sigma^2$ of the pure braid generators.

**PROOF.** We compute completely $\mathbf{B}_n(\Sigma_{g,1})/\Gamma_3$. Let us consider the quotient of $\mathbf{B}_n(\Sigma_{g,1})$ by the relations $\sigma_i = \sigma_{i+1}$, that is, by the normal closure $N$ of the $\sigma_i\sigma_{i+1}^{-1}$ for $1 \le i \le n-1$. We already know that, modulo $\Gamma_3$, the braid generators $\sigma_i$ have a common class $\sigma$. In particular, $N \subseteq \Gamma_3$, and we will show that $N = \Gamma_3$ by showing that $G := \mathbf{B}_n(\Sigma_{g,1})/N$ is 2-nilpotent. Thanks to Proposition 6.20, we can compute a presentation of this quotient. It is generated by $\sigma$, together with $a_r$ and $b_r$ for $1 \le r \le g$. The braid relations on the $\sigma_i$ become trivial there. (BS1) says that $\sigma$ is central. (BS2) says that the $a_r$ and the $b_s$ commute with one another, except $a_r$ and $b_r$ (for each $r$). Since $\sigma$ is central, (BS3) becomes trivial. Finally, (BS4) can be written as $b_r a_r b_r^{-1} = a_r \sigma^{-2}$. From this presentation, one can see that $G \cong (\mathbb{Z} \times \mathbb{Z}^g) \rtimes \mathbb{Z}^g$, where the three factors are free abelian on $\sigma$, the $a_r$ and the $b_r$ respectively; the action of each $b_r$ is trivial on $\sigma$ and the $a_s$ if $s \ne r$, and $b_r \cdot a_r = a_r - 2\sigma$. Since this group is 2-nilpotent, we have $N = \Gamma_3(\mathbf{B}_n(\Sigma_{g,1}))$, as announced. Moreover, the Lie ring of $G$, which is the Lie ring of $\mathbf{B}_n(\Sigma_{g,1})$ by Corollary 6.36, is easy to compute. Namely, $\mathcal{L}_1(G) = G^{ab} \cong (\mathbb{Z}/2)^2 \times \mathbb{Z}^{2g}$, $\mathcal{L}_2(G) = \Gamma_2(G) \cong \mathbb{Z}$ is generated by $\bar{\sigma}^2$, and the only non-trivial brackets of generators are $[\bar{a}_r, \bar{b}_r] = \bar{\sigma}^2$. $\square$

**REMARK 6.44.** Here we choose to recover [BGG08, Th. 1] from Proposition 6.43. However, one can also deduce Proposition 6.43 from [BGG08, Th. 1]. Namely, one can embed $\Sigma_{g,1}$ into $\Sigma_{g'}$ for any $g' \ge g$, by attaching $\Sigma_{g'-g,1}$ along the boundary component. The induced map $\mathcal{L}_2(\mathbf{B}_n(\Sigma_{g,1})) \to \mathcal{L}_2(\mathbf{B}_n(\Sigma_{g'}))$ sends $\sigma^2$ to
$\sigma^2$, and the latter is of order $n + g' - 1$ by [BGG08, Th. 1]. Thus, by varying $g'$, we see that $\sigma^2$ cannot be of finite order inside $\mathcal{L}_2(\mathbf{B}_n(\Sigma_{g,1}))$.

REMARK 6.45. The group $(\mathbb{Z} \times \mathbb{Z}^g) \times \mathbb{Z}^g$ appearing in the proof of Proposition 6.43, which is then the maximal nilpotent quotient of $\mathbf{B}_n(\Sigma_{g,1})$, has a nice interpretation as a matrix group resembling the Heisenberg group of a symplectic vector space. Namely, it is the subgroup of $\mathrm{GL}_{g+2}(\mathbb{Q})$ given by:

$$ \begin{pmatrix} 1 & \mathbb{Z} & \mathbb{Z} & \cdots & \mathbb{Z} & \frac{1}{2}\mathbb{Z} \\ & 1 & 0 & \cdots & 0 & \mathbb{Z} \\ & & \ddots & \ddots & \vdots & \vdots \\ & & & \ddots & 0 & \mathbb{Z} \\ & & & & 1 & \mathbb{Z} \\ & & & & & 1 \end{pmatrix} $$

REMARK 6.46 (About the Riemann-Hurwitz formula). The result [BGG08, Th. 1] quoted above and recovered below (Proposition 6.56) says that when $S$ is an orientable closed surface, the information encoded in $\mathcal{L}_2(\mathbf{B}_n(S))$ is essentially the genus of $S$, or its Euler characteristic. In fact, one can recover the Riemann-Hurwitz formula for (unramified) coverings of closed orientable surfaces from this computation. Indeed, let $p: \Sigma_h \to \Sigma_g$ be a $k$-sheeted covering. It induces a continuous map $p^*: C_n(\Sigma_g) \to C_{kn}(\Sigma_h)$ between unordered configuration spaces sending a configuration of $n$ points to the configuration of the $kn$ preimages of these points by $p$. The induced map between the fundamental groups sends the pure braid generator $A_{12}$ to the product of the $A_{j,j+1}$ for $j \in p^{-1}(1)$ (if the correct conventions are chosen). Thus the induced map from $\mathcal{L}_2(\mathbf{B}_n(\Sigma_g))$ to $\mathcal{L}_2(\mathbf{B}_{kn}(\Sigma_h))$ sends $\sigma^2$ to $k\sigma^2$. Since $\sigma^2$ is of order $n + g - 1$ in $\mathcal{L}_2(\mathbf{B}_n(\Sigma_g))$, we obtain $(n + g - 1)k\sigma^2 = 0$ in $\mathcal{L}_2(\mathbf{B}_{kn}(\Sigma_h))$. However, $\sigma^2$ is of order $kn+h-1$ there, implying that $kn+h-1$ divides $kn+k(g-1)$. This holds for all $n$. For $n$ big enough, $kn + k(g - 1) < 2(kn + h - 1)$, so the only possibility is that $kn + k(g - 1) = kn + h - 1$, that is, $k(g - 1) = h - 1$, which is equivalent to the Riemann-Hurwitz formula for $p$.

## 6.5. Partitioned braids on surfaces

Let us now study the LCS of partitioned surface braid groups $\mathbf{B}_\lambda(S)$, generalising the results from §6.4, which can be seen as the case of the trivial partition $\lambda = (n)$ of $n$. We follow the same steps: we first compute the abelianisations of $\mathbf{B}_\lambda(S)$ in Proposition 6.47, before studying $\mathbf{B}_\lambda(S)/\Gamma_\infty$ and showing that the LCS of $\mathbf{B}_\lambda(S)$ stops at most at $\Gamma_3$ when the partition $\lambda$ has only blocks of size at least 3, in Theorem 6.50 and Corollary 6.51. Finally, under the latter hypothesis, we compute the associated Lie rings in Theorem 6.52.

**6.5.1. The abelianisation.** Let $\lambda = (n_1, \dots, n_l)$ be a partition of an integer $n$. A computation of $\mathbf{B}_\lambda(S)^{ab}$ can be obtained by a quite straightforward generalisation of the computation of $\mathbf{B}_n(S)^{ab}$ from Proposition 6.32.

Let us first recall that the morphism $\varphi$ from §6.2 induces a map $\mathbf{B}_\lambda^{ab} \to \mathbf{B}_\lambda(S)^{ab}$. Then, from Proposition 3.5, we get that:

* For each $i \le l$ such that $n_i \ge 2$, all the $\sigma_\alpha$ with $\alpha$ and $\alpha + 1$ in the $i$-th block of $\lambda$ have a common image in $\mathbf{B}_\lambda(S)^{ab}$, called $s_i$.
• For each $i < j \le l$, all the $A_{\alpha\beta}$ with $\alpha$ (resp. $\beta$) in the $i$-th (resp. the $j$-th) block of $\lambda$ have a common image in $\mathbf{B}_{\lambda}(S)^{ab}$, called $a_{ij}$ (or $a_{ji}$).

Let us now consider the short exact sequence:

$$ \mathbf{P}_{n}^{\circ}(S) \to \mathbf{B}_{\lambda}(S) \xrightarrow{\pi_{S}} \pi_{1}(S) \to \mathfrak{S}_{\lambda}. $$

We can apply Lemma 1.11 to it, and we get an exact sequence of abelian groups:

$$ (\mathbf{P}_{n}^{\circ}(S)^{ab})_{\mathbf{B}_{\lambda}(S)} \to \mathbf{B}_{\lambda}(S)^{ab} \to (\pi_{1}(S) \to \mathfrak{S}_{\lambda})^{ab} \to 0. $$

On the one hand, the quotient $(\pi_1(S) \to \mathfrak{S}_\lambda)^{ab}$ is isomorphic to the product of the $(\pi_1(S) \to \mathfrak{S}_{n_i})^{ab}$, which is $(\pi_1(S)^{ab})^l \times (\mathbb{Z}/2)^{l'}$, where $l'$ is the number of indices $i \le l$ such that $n_i \ge 2$ (see Corollary A.25). On the other hand, it follows from Proposition 6.10 that $\mathbf{P}_n^\circ(S)$ is generated by $\mathbf{P}_n$ under the action of $\mathbf{P}_n(S)$, which is a subgroup of $\mathbf{B}_\lambda(S)$. As a consequence, the map $\mathbf{P}_n^{ab} \to (\mathbf{P}_n^\circ(S)^{ab})_{\mathbf{B}_\lambda(S)}$ induced by $\varphi$ is surjective. Moreover, it factors through the quotient $(\mathbf{P}_n^{ab})_{\mathbf{B}_\lambda} = (\mathbf{P}_n^{ab})_{\mathfrak{S}_\lambda} \cong \mathbb{Z}^{l(l-1)/2} \times \mathbb{Z}^{l'}. Thus the image of $(\mathbf{P}_n^\circ(S)^{ab})_{\mathbf{B}_\lambda(S)}$ in $\mathbf{B}_\lambda(S)^{ab}$ is generated by the images of the pure braid generators, which are the elements $2s_i$ for $i \le l$ such that $n_i \ge 2$, and $a_{ij}$ for $i < j \le l$.

Now, let $H$ be the subgroup of $\mathbf{B}_\lambda(S)^{ab}$ generated by the $s_i$ and the $a_{ij}$. From the above, we get an isomorphism $\mathbf{B}_\lambda(S)^{ab}/H \cong (\pi_1(S)^{ab})^l$. Moreover, the corresponding projection $\mathbf{B}_\lambda(S)^{ab} \to (\pi_1(S)^{ab})^l$ splits: a splitting is defined by sending $(g_i)_i$ to $\sum \psi_i(g_i)$ where, for each $i$, $\psi_i$ is induced by $\psi_\alpha$ for any $\alpha$ in the $i$-th block of $\lambda$. As a consequence, $\mathbf{B}_\lambda(S)^{ab}$ identifies with $(\pi_1(S)^{ab})^l \times H$, and we can use Proposition 6.19 to get a complete calculation:

**PROPOSITION 6.47.** Let $n \ge 1$ and $\lambda = (n_1, \dots, n_l)$ be a partition of $n$. Then:

$$ \mathbf{B}_{\lambda}(S)^{ab} \cong (\pi_1(S)^{ab})^l \times (\mathbf{B}_{\lambda}^{ab}/R), $$

where $\mathbf{B}_{\lambda}^{ab}$ is free abelian on the $s_i$ and the $a_{ij}$ from Proposition 3.5, and $R$ is:

• trivial if $S$ is planar,

• generated by the relations $2(n_i-1)s_i + \sum_{j \neq i} n_j a_{ij}$ ($i=1, \dots, n$) if $S \cong \mathbb{S}^2$,

• generated by the $2s_i$ and the $a_{ij}$ in all the other cases.

Explicitly, if $l'$ denotes the number of indices $i \le l$ such that $n_i \ge 2$, we have, for $S \not\cong \mathbb{S}^2$:

$$ \mathbf{B}_{\lambda}^{ab}/R \cong \begin{cases} \mathbb{Z}^{l'} \times \mathbb{Z}^{l(l-1)/2} & \text{if } S \text{ is planar,} \\ (\mathbb{Z}/2)^{l'} & \text{if } S \text{ does not embed into the sphere.} \end{cases} $$

**PROOF.** If $S \neq \mathbb{S}^2$, all the relations are direct consequences of Proposition 6.19, and it remains to prove that they are the only ones. In the first case, we can directly use the result of Proposition 3.5: an embedding of $S$ into $\mathbb{R}^2 \cong \mathbb{S}^2 - pt$ induces a retraction $\mathbf{B}_\lambda(S)^{ab} \to \mathbf{B}_\lambda^{ab}$ of the morphism $\mathbf{B}_\lambda^{ab} \to \mathbf{B}_\lambda(S)^{ab}$ induced by $\varphi$, whence the result in this case. In the third one, the same proof as the proof of Proposition 3.5 works: projections onto the factors are given by projections onto the $\mathfrak{S}_{n_i}^{ab} \cong \mathbb{Z}/2$ for $n_i \ge 2$.

If $S = \mathbb{S}^2$, Corollary 6.30 describes $\mathbf{B}_\lambda(\mathbb{S}^2)$ as the quotient of $\mathbf{B}_\lambda$ by the relations $A_{\alpha_1}A_{\alpha_2} \cdots A_{\alpha_n} = 1$ for all $1 \le \alpha \le n$. The abelianisation $\mathbf{B}_\lambda(\mathbb{S}^2)^{ab}$ is then the quotient of $\mathbf{B}_\lambda^{ab}$ (described in Proposition 3.5) by the classes of the above relations. If $\alpha$ is in the $i$-th block of $\lambda$, then the class of $A_{\alpha\beta}$ is either $2s_i$ if $\beta$ is also in the
i-th block (which holds for $n_i - 1$ values of $\beta$), or $a_{ij}$ if $\beta$ is in the j-th block for some $j \neq i$ (which happens for $n_j$ values of $\beta$, for each $j \neq i$). Thus the classes of the $A_{\alpha_1}A_{\alpha_2} \cdots A_{\alpha_n}$ are indeed the relations of the statement. $\square$

**REMARK 6.48.** The relation $2(n_i - 1)s_i + \sum_{j \neq i} n_j a_{ij}$ makes sense even if $n_i = 1$: in this case, with our conventions, there is no generator $s_i$, but this does not matter since its coefficient is $2(n_i - 1) = 0$ (alternatively, one could add a generator $s_i$ and ask that $s_i = 0$).

**6.5.2. The lower central series.** The work done above to show that the LCS of $\mathbf{B}_n(S)$ stops if $n \ge 3$ (see Corollary 6.36) generalises to the partitioned braid group when all the blocks of the partition are of size at least 3. First, here follows the generalisation of Lemma 6.33.

**LEMMA 6.49.** Let $\lambda = (n_1, \dots, n_l)$ be a partition of $n$, with $n_i \ge 3$ for all $i$. The image of $\mathbf{B}_\lambda$ in $\mathbf{B}_\lambda(S)/\Gamma_\infty$ is central. In particular, it is a quotient of $\mathbf{B}_\lambda^{ab}$.

**PROOF.** It follows from Theorem 3.6 that $\Gamma_\infty(\mathbf{B}_\lambda) = \Gamma_2(\mathbf{B}_\lambda)$. As a consequence, the morphism $\varphi: \mathbf{B}_\lambda \to \mathbf{B}_\lambda(S)/\Gamma_\infty$ factors through $\mathbf{B}_\lambda/\Gamma_2 = \mathbf{B}_\lambda^{ab}$. Hence its image is abelian, generated by the images of the $s_i$ and the $a_{ij}$ from Proposition 3.5.

In order to show that it is central, we need to show that these elements commute with generators of $\mathbf{B}_\lambda(S)/\Gamma_\infty$. We deduce from Proposition 6.10 that $\mathbf{B}_\lambda(S)$ is generated by the images of $\varphi$ and of the $\psi_\alpha$. In fact, we can restrict to taking one $\alpha$ in each block of $\lambda$, since $\psi_\alpha$ and $\psi_\beta$ are conjugated by elements of $\mathbf{B}_\lambda$ if $\alpha$ and $\beta$ are in the same block. The $s_i$ and the $a_{ij}$ already commute with each other, so we only need to show that they commute with the images of the selected $\psi_\alpha$. Since the blocks of $\lambda$ have size at least 3, we can find representatives of the $s_i$ and the $a_{ij}$ having disjoint support with elements of all the chosen $\mathrm{Im}(\psi_\alpha)$. Thus, the $s_i$ and the $a_{ij}$ commute with a family of generators of $\mathbf{B}_\lambda(S)/\Gamma_\infty$, which proves our claim. $\square$

We can now generalise our main decomposition theorem (Theorem 6.34) to partitioned braids:

**THEOREM 6.50.** Let $\lambda = (n_1, \dots, n_l)$ be a partition of $n$, with $n_i \ge 3$ for all $i$. There is a central extension:

$$ \langle s_i^2, a_{ij} \rangle_{i,j \le l} \longrightarrow \mathbf{B}_\lambda(S)/\Gamma_\infty \dashrightarrow (\pi_1(S))^{ab} \times \mathbb{Z}/2^l. $$

**PROOF.** The proof is essentially the same as the proof of Theorem 6.34, so we only stress what changes. The element $\sigma_\alpha \sigma_\beta^{-1}$ is in $\Gamma_\infty(\mathbf{B}_n(S))$ only when $\alpha$ and $\beta$ are in the same block of $\lambda$. However, their images $\tau_\alpha \tau_\beta^{-1}$ in $\pi_1(S) \rlap{/}{\mathfrak{S}}_\lambda$ still normally generate $\Gamma_2(\pi_1(S) \rlap{/}{\mathfrak{S}}_\lambda)$, because $\pi_1(S) \rlap{/}{\mathfrak{S}}_\lambda$ is the product of the $\pi_1(S) \rlap{/}{\mathfrak{S}}_{n_i}$, whose $\Gamma_2$ is normally generated by the $\tau_\alpha \tau_\beta^{-1}$ for $\alpha$ and $\beta$ in the i-th block of $\lambda$ (Lemma A.26). Thus, arguing exactly as in the proof of Theorem 6.34, we get a short exact sequence:

$$ \mathbf{P}_n^\circ(S)/\Gamma_\infty(\mathbf{B}_\lambda(S)) \cap \mathbf{P}_n^\circ(S) \longrightarrow \mathbf{B}_\lambda(S)/\Gamma_\infty \xrightarrow{\pi_S} (\pi_1(S) \rlap{/}{\mathfrak{S}}_\lambda)^{ab}. $$

Recall that $(\pi_1(S) \rlap{/}{\mathfrak{S}}_\lambda)^{ab} \cong (\pi_1(S)^{ab} \times \mathbb{Z}/2)^l$. Moreover, the kernel is the normal closure of the image of $\mathbf{P}_n$, but this image is already normal, and even central by Lemma 6.49, generated by the $a_{ij}$ and the squares of the $s_i$. $\square$
Finally, Corollary 6.36 adapts readily to this context:

COROLLARY 6.51. If all the blocks of $\lambda$ have size at least 3, then:

$$ \Gamma_3(\mathbf{B}_{\lambda}(S)) = \Gamma_4(\mathbf{B}_{\lambda}(S)). $$

**6.5.3. The Lie ring.** Here again, in the stable case (that is, when all the blocks of the partition $\lambda$ have size at least 3), we can be more precise about the description of the LCS. Namely, our next goal is to show the following generalisation of Theorem 6.42, whose proof occupies the rest of the present section.

**THEOREM 6.52.** Let $\lambda = (n_1, \dots, n_l)$ be a partition of an integer $n$ with $n_i \ge 3$ for all $i$, and let $S$ be a connected surface. The LCS of $\mathbf{B}_{\lambda}(S)$:

* stops at $\Gamma_2$ if $S$ is planar, if $S \cong \mathbb{S}^2$ or if $l = 1$ and $S$ is non-orientable.

* stops at $\Gamma_3$ in the other cases.

**REMARK 6.53.** In both cases, the Lie ring can be computed completely. Namely, $\mathcal{L}_1(\mathbf{B}_{\lambda}(S)) = \mathbf{B}_{\lambda}(S)^{ab}$ has already been computed in Proposition 6.47. In the first case, no further computation is required. In the second case, $\mathcal{L}_2(\mathbf{B}_{\lambda}(S))$ is described completely in Proposition 6.55 and Proposition 6.56 for orientable surfaces, and in Corollary 6.59 and Corollary 6.61 for non-orientable ones. Moreover, one can easily describe the Lie bracket from the computations given there. Precisely, the only non-trivial brackets come from the computations depicted in Figure 6.1.

The proof of Theorem 6.52 begins with the following observation, which will be of essence in our study of the Lie ring of $\mathbf{B}_{\lambda}(S)$:

**FACT 6.54.** Let $\lambda = (n_1, \dots, n_l)$ be a partition of $n$ with all blocks of size at least 3. Then $\mathcal{L}_2(\mathbf{B}_{\lambda}(S))$ identifies with $\Gamma_2(\mathbf{B}_{\lambda}(S)/\Gamma_{\infty})$, which is included in the subgroup $\langle s_i^2, a_{ij} \rangle_{i,j \le l}$ of $\mathbf{B}_{\lambda}(S)$ from Theorem 6.50. Furthermore, if $S$ has a handle or a crosscap, then $\mathcal{L}_2(\mathbf{B}_{\lambda}(S)) \cong \langle s_i^2, a_{ij} \rangle_{i,j \le l}$.

**PROOF.** We know that $\Gamma_3 = \Gamma_{\infty}$ for $\mathbf{B}_{\lambda}(S)$ by Corollary 6.51. Then, since $\mathcal{L}_2 = \Gamma_2/\Gamma_3$, $\mathcal{L}_2(\mathbf{B}_{\lambda}(S))$ identifies with $\Gamma_2(\mathbf{B}_{\lambda}(S)/\Gamma_{\infty})$. Moreover, in $\mathbf{B}_{\lambda}(S)/\Gamma_{\infty}$, the subgroup $\Gamma_2$ must be contained in the kernel $\langle s_i^2, a_{ij} \rangle_{i,j \le l}$ of the central extension of Theorem 6.50. Now, if $S$ has a handle or a crosscap, then the quotient $(\pi_1(S)^{ab} \times \mathbb{Z}/2)^l$ of $\mathbf{B}_{\lambda}(S)/\Gamma_{\infty}$ identifies with its abelianisation (see Proposition 6.47). As a consequence, the subgroup $\Gamma_2(\mathbf{B}_{\lambda}(S)/\Gamma_{\infty})$ is the whole of $\langle s_i^2, a_{ij} \rangle_{i,j \le l}$. $\square$

**6.5.3.1. Non-planar orientable surfaces.** Let us turn our attention to the case where the surface is not closed.

**PROPOSITION 6.55.** If $S$ is a non-planar orientable surface which is not closed, then $\mathcal{L}_2(\mathbf{B}_{\lambda}(S))$ is free abelian on the $a_{ij}$ and the $s_i^2$.

**PROOF.** We already know that these elements generate $\mathcal{L}_2(\mathbf{B}_{\lambda}(S))$ by Fact 6.54, so we need to show that they are linearly independent. The argument from the proof of Proposition 3.5 works here, using the maps from $\mathcal{L}_2(\mathbf{B}_{\lambda}(S))$ to $\mathcal{L}_2(\mathbf{B}_{n_i}(S)) \cong \mathbb{Z}$ and $\mathcal{L}_2(\mathbf{B}_{n_i+n_j}(S)) \cong \mathbb{Z}$ instead of the maps from $\mathbf{B}_{\lambda}^{ab}$ to $\mathbf{B}_{n_i}^{ab}$ and $\mathbf{B}_{n_i+n_j}^{ab}$. This uses the fact that $\mathcal{L}_2(\mathbf{B}_{n}(S)) \cong \mathbb{Z}$ if $n \ge 3$, from Theorem 6.42. $\square$

The case of closed orientable surfaces is a bit trickier. In fact, we first need to generalise [BGG08, Th. 1]:
PROPOSITION 6.56. Let $g \ge 1$. Then $\mathcal{L}_2(\mathbf{B}_\lambda(\Sigma_g))$ is the quotient of the free abelian group on the $a_{ij}$ and the $s_i^2$ by the relations:

$$ (n_i + g - 1)s_i^2 + \sum_{j \neq i} n_j a_{ij} = 0, \quad \text{for all } 1 \le i \le n. $$

PROOF. Let us consider the subsurface $\Sigma_{g,1}$ of $\Sigma_g$ obtained by removing an open disc. The corresponding embedding of $\Sigma_{g,1}$ into $\Sigma_g$ induces the quotient map described in Corollary 6.30. In particular, $\mathbf{B}_\lambda(\Sigma_g)/\Gamma_3$ is the quotient of $\mathbf{B}_\lambda(\Sigma_{g,1})/\Gamma_3$ by the classes of the relations from Corollary 6.30. Since the $A_{\alpha\beta}$ are in $\Gamma_2(\mathbf{B}_\lambda(\Sigma_{g,1}))$, these relations are between elements of $\Gamma_2$, that is, elements of $\langle s_i^2, a_{ij} \rangle_{i,j \le l}$. In order to write them as relations between the $s_i^2$ and the $a_{ij}$, we need to recall that if $\alpha$ is in the $i$-th block of $\lambda$, for any $r \le g$, we have, in $\mathcal{L}_2(\mathbf{B}_\lambda(S))$:

$$ \overline{[a_r^{(\alpha)}, (b_r^{(\alpha)})^{-1}]} = -\overline{[a_r^{(\alpha)}, b_r^{(\alpha)}]} = -\overline{[\sigma_{\alpha} b_r^{(\alpha)} \sigma_{\alpha}^{-1}, (a_r^{(\alpha)})^{-1}]} = -\overline{\sigma_{\alpha}^2} = -s_i^2, $$

where the third equality comes from Figure 6.1b. Moreover, the class of $A_{\alpha\beta}$ modulo $\Gamma_3(\mathbf{B}_\lambda(S))$ is either $s_i^2$ if $\alpha$ and $\beta$ are both in the $i$-th block of $\lambda$, or $a_{ij}$ if $\alpha$ is in the $i$-th block and $\beta$ is in the $j$-th block for some $j \neq i$. Finally, using additive notations in the central subgroup $\langle s_i^2, a_{ij} \rangle$, the classes of the relations from Corollary 6.30 are:

$$ (n_i - 1)s_i^2 + \sum_{j \neq i} n_j a_{ij} = -gs_i^2 \quad \text{for all } 1 \le i \le n. $$

Moreover, these relators are central in $\mathbf{B}_\lambda(\Sigma_g)/\Gamma_\infty$, so their normal closure is only the subgroup they generate, which implies that $\Gamma_2(\mathbf{B}_\lambda(\Sigma_g)/\Gamma_3)$ is the quotient of $\Gamma_2(\mathbf{B}_\lambda(\Sigma_{g,1})/\Gamma_3)$ by these relations. By Proposition 6.55, the latter is free on the $s_i^2$ and the $a_{ij}$. Whence our claim. $\square$

REMARK 6.57. Let us point out that this computation of $\mathcal{L}_2(\mathbf{B}_\lambda(\Sigma_g))$ is very similar to the computation of $\mathcal{L}_1(\mathbf{B}_\lambda(\mathbb{S}^2))$ in the proof of Proposition 6.47 (note that $s_i^2$ was equal to $2s_i$ there, but here $s_i$ and $s_i^2$ do not live in the same part of the Lie ring). The only difference lies in the degree of our relations with respect to the LCS: if the surface has a handle, then the pure braid generators $A_{\alpha\beta}$ belong to the derived subgroup (see Figure 6.1b).

6.5.3.2. Non-orientable surfaces. For non-orientable surfaces, we already know that the $s_i^2$ vanish in the quotient by $\Gamma_\infty$, as in the proof of Theorem 6.42. As a consequence, $\Gamma_2(\mathbf{B}_n(S)/\Gamma_\infty)$ is generated by the $a_{ij}$. The following proposition is an analogue of [BGG17, Prop. 3.7] in the non-orientable case:

PROPOSITION 6.58. Let $g \ge 0$ and $m, n \ge 3$. We have:

$$ \mathbf{B}_{m,n}(\mathbb{N}_{g,1})/\Gamma_\infty \cong (\mathbb{Z}/2)^2 \times (\mathbb{Z} \times \mathbb{Z}^g) \times \mathbb{Z}^g, $$

where the factors are respectively generated by $s_1$ and $s_2$, $a_{12}$, $c'_r$ and $c_r$. The action is given by $c_r c'_r c_r^{-1} = c'_r a_{12}$ (for all $r \le g$) and all the other pairs of generators commuting. In particular, this group is 2-nilpotent, and $\mathcal{L}_2(\mathbf{B}_{m,n}(\mathbb{N}_{g,1}))$ is infinite cyclic, generated by $a_{12}$.

PROOF. Recall that we have a split extension (from Proposition 6.15):

$$ \mathbf{B}_m(\mathbb{N}_{g,1+n}) \hookrightarrow \mathbf{B}_{m,n}(\mathbb{N}_{g,1}) \twoheadrightarrow \mathbf{B}_n(\mathbb{N}_{g,1}). $$
Thus, we can get a presentation of $\mathbf{B}_{m,n}(\mathcal{N}_{g,1})$ from the presentation of the quotient described in Proposition 6.22 and the presentation of the kernel from Proposition 6.24, using the method of Appendix B. The generators of this presentation are the $\sigma_i$ for $i \neq m$, the $c_r := c_r^{(1)}$, the $c'_r := c_r^{(m+1)}$ and the $z_j = A_{1,m+j}$ (where our conventions are those from §6.3).

We use this to get a presentation of the quotient of $\mathbf{B}_{m,n}(\mathcal{N}_{g,1})$ by the normal closure $N$ of the $\sigma_i\sigma_{i+1}^{-1}$ for $i < m+n$ and $i \notin \{m-1, m\}$, together with the $z_jz_{j+1}^{-1}$ for $j < n$. This quotient is generated by the common class $s_1$ (resp. $s_2$, resp. $a_{12}$) of $\sigma_1, \dots, \sigma_{m-1}$, the common class $s_2$ of $\sigma_{m+1}, \dots, \sigma_{m-1}$ and the common class $a_{12}$ of $z_1, \dots, z_n$, together with the $c_r$ and the $c'_r$ for $r \le g$. They are subject to the following relations:

* Relations coming from those of $\mathbf{B}_n(\mathcal{N}_{g,1})$: $s_2$ commutes with the $c'_r$ (BN1), the $c'_r$ commute with one another (BN2) and $s_2^2 = 1$ (BN3).

* Relations coming from those of $\mathbf{B}_m(\mathcal{N}_{g,1+n})$: $s_1$ commutes with the $c_r$ (BN1), the $c_r$ commute with one another (BN2), $s_1^2 = 1$ (BN3), $a_{12}$ commutes with $s_1$ (BN4), and with the $c_r$ (BN5). (BN6) and (BN7) become trivial.

* Relations describing the action by conjugation of $s_2$ and the $c'_r$ on $s_1$, $a_{12}$ and the $c_r$. This action is easily seen to be trivial in most cases, since most of the pairs of elements involved come from elements having disjoint support in $\mathbf{B}_{m,n}(\mathcal{N}_{g,1})$, hence they commute. Namely, this holds for $c_s$ and $c'_r$ when $r < s$: using the fact that $a_{12}$ commutes with $c_s$, we see that $c_s$ is the class of $z_1c_r^{(1)}z_1^{-1}$, whose support is disjoint from the support of $c_r^{(m+1)}$ (up to isotopy). This is again true for $a_{12}$ and the $c'_r$, since $s_2$ commutes with the $c'_r$: $c'_r$ is the class of $c_r^{(m+2)} = \sigma_{m+1}c_r^{(m+1)}\sigma_{m+1}^{-1}$, whose support is disjoint from that of $z_1$. Finally, the only pair of generators under scrutiny for which this does not hold are $c_r$ and $c'_r$ (for $r \le g$). But for them, the situation is the one from Figure 6.1a: $[(c'_r)^{-1}, c_r] = \bar{A}_{1,m+1} = a_{12}$.

These relations can be summed up as:

* All the generators commute pairwise, except $c_r$ and $c'_r$ for $r \le g$,

* $s_1^2 = s_2^2 = 1$,

* $c_r c'_r c_r^{-1} = c'_r a_{12}$ (for $r = 1, \dots, g$).

This is a presentation of the group described in the statement. Its commutator subgroup is the factor $\mathbb{Z}$ of the decomposition, which is infinite cyclic, generated by $a_{12}$. It is also central, hence the group is 2-nilpotent. Since this group is $\mathbf{B}_{m,n}(\mathcal{N}_{g,1})/N$, $N$ contains $\Gamma_3(\mathbf{B}_{m,n}(\mathcal{N}_{g,1}))$. But the elements normally generating $N$ are in $\Gamma_\infty(\mathbf{B}_{m,n}(\mathcal{N}_{g,1}))$; see Lemma 6.49 and its proof. Thus $N = \Gamma_\infty(\mathbf{B}_{m,n}(\mathcal{N}_{g,1}))$, and the Lie ring of the quotient $\mathbf{B}_{m,n}(\mathcal{N}_{g,1})/N$ identifies with the Lie ring of $\mathbf{B}_{m,n}(\mathcal{N}_{g,1})$. In particular, since its $\Gamma_3$ is trivial, its $\mathcal{L}_2$ coincides with its $\Gamma_2$, which is infinite cyclic, generated by $a_{12}$. $\square$

**COROLLARY 6.59.** Let $S$ be a non-orientable surface that is not closed. Let $\lambda = (n_1, \dots, n_l)$ be a partition whose blocks are of size at least 3. Then $\mathcal{L}_2(\mathbf{B}_\lambda(S))$ is free abelian on the $a_{ij}$ for $1 \le i < j \le l$.

**PROOF.** We first show that if $m, n \ge 3$, $\mathcal{L}_2(\mathbf{B}_{m,n}(S))$ is infinite cyclic, generated by $a_{12}$. This is true for $S = \mathcal{N}_{g,1}$ by Proposition 6.58. Then we can follow the same method as in the proof of Theorem 6.42, to which the reader is referred for
more details. Namely, we already know that $\mathcal{L}_2(\mathbf{B}_{m,n}(S))$ is generated by $a_{12}$, and we need to show that it has infinite order. If $S$ is compact, we can embed $S$ into some $\mathcal{N}_{g,1}$, and the image of $a_{12}$ by the corresponding morphism from $\mathcal{L}_2(\mathbf{B}_{m,n}(S))$ to $\mathcal{L}_2(\mathbf{B}_{m,n}(\mathcal{N}_{g,1}))$ has infinite order, whence our claim in this case. Then, if $S$ is not compact, a relation saying that $a_{12}$ has finite order in $\mathbf{B}_{m,n}(S)/\Gamma_3$ would hold in a compact subsurface, which is impossible by the previous case.

From this, we can deduce the result for $\lambda$ having more than two blocks, reasoning as in the proofs of Propositions 3.5 and 6.55. Indeed, each canonical map $\mathcal{L}_2(\mathbf{B}_{\lambda}(S)) \to \mathcal{L}_2(\mathbf{B}_{n_i,n_j}(S)) \cong \mathbb{Z}$ kills all the $a_{kl}$, except $a_{ij}$, which is sent to a generator of the target. Thus the $a_{ij}$ must be linearly independent. $\square$

**PROPOSITION 6.60.** Let $g \ge 0$ and $m, n \ge 3$. Then $\mathcal{L}_2(\mathbf{B}_{m,n}(\mathcal{N}_g))$ is cyclic of order 2, generated by $a_{12}$.

**PROOF.** Let us consider $G := \mathbf{B}_{m,n}(\mathcal{N}_g)/\Gamma_3$. This group is the quotient of $\mathbf{B}_{m,n}(\mathcal{N}_{g,1})/\Gamma_3 = \mathbf{B}_{m,n}(\mathcal{N}_g)/\Gamma_\infty \cong (\mathbb{Z}/2)^2 \times (\mathbb{Z} \times \mathbb{Z}^g) \times \mathbb{Z}^g$ described in Proposition 6.58 by the images of the boundary relations from Corollary 6.30. Before considering these relations, we can already remark that the commutator subgroup of $\mathbf{B}_{m,n}(\mathcal{N}_{g,1})/\Gamma_3$ is cyclic generated by $a_{12}$, so the same holds for $G$. Since $\Gamma_2(G)$ identifies with $\mathcal{L}_2(\mathbf{B}_{m,n}(\mathcal{N}_g))$, we only need to show that $a_{12}$ has order 2 in $G$ to prove our statement.

Recall that $A_{\alpha\beta}$ is sent to $s_1^2 = s_2^2 = 1$ if $\alpha$ and $\beta$ are in the same block of the partition $(m, n)$, and to $a_{12}$ if they are not. As a consequence, these relations are:

$$c_1^2 \cdots c_g^2 = a_{12}^m \quad \text{and} \quad (c'_1)^2 \cdots (c'_g)^2 = a_{12}^n.$$

Since $c_1$ commutes with all the other generators except $c'_1$, and $[c_1, c'_1] = a_{12}$, by applying the commutator with $c_1$ to the second relation, we get $a_{12}^2 = 1$. We can thus consider the intermediate quotient by the central element $a_{12}^2$, and see $G$ as a quotient of $(\mathbb{Z}/2)^2 \times ((\mathbb{Z}/2) \times \mathbb{Z}^g) \times \mathbb{Z}^g$. Modulo $a_{12}^2$, the above relations (hence the structure of $G = \mathbf{B}_{m,n}(\mathcal{N}_g)/\Gamma_3$) depend only on the parity of $m$ and $n$. Moreover, we now know that $a_{12}^2 = 1$ in $G$, so we are left with showing that $a_{12}$ is not trivial in $G$.

Suppose first that $m$ and $n$ are both even. Then the relations become $c_1^2 \cdots c_g^2 = (c'_1)^2 \cdots (c'_g)^2 = 1$. Notice that after the quotient by $a_{12}^2$, all the $c_r^2$ and the $(c'_r)^2$ have become central, so both $c_1^2 \cdots c_g^2$ and $(c'_1)^2 \cdots (c'_g)^2$ are central in $(\mathbb{Z}/2)^2 \times ((\mathbb{Z}/2) \times \mathbb{Z}^g) \times \mathbb{Z}^g$. Thus, if $A$ denotes the abelian group $\mathbb{Z}^g/(2, 2, \dots, 2)$ (which is also $\pi_1(\mathcal{N}_g)^{ab}$), we see that $G \cong (\mathbb{Z}/2)^2 \times ((\mathbb{Z}/2) \times A) \times A$, whose commutator subgroup is cyclic of order 2, generated by $a_{12}$.

If $m$ and $n$ have different parities, we can assume (by symmetry) that $m$ is even and $n$ is odd. Then the relations become $c_1^2 \cdots c_g^2 = 1$ and $(c'_1)^2 \cdots (c'_g)^2 = a_{12}$. In $(\mathbb{Z}/2)^2 \times ((\mathbb{Z}/2) \times \mathbb{Z}^g) \times \mathbb{Z}^g$, both $c_1^2 \cdots c_g^2$ and $(c'_1)^2 \cdots (c'_g)^2 a_{12}^{-1}$ are central, so they general cyclic normal subgroups. Thus, if we denote by $\tilde{A}$ the quotient $\mathbb{Z}^g/(4, 4, \dots, 4)$ of $(\mathbb{Z}/2) \times \mathbb{Z}^g$ by $(1, 2, \dots, 2)$, we have $G \cong (\mathbb{Z}/2)^2 \times \tilde{A} \times A$, whose commutator subgroup is cyclic of order 2, generated by $a_{12}$ (the latter identifies with the class of $(2, 2, \dots, 2)$ in $\tilde{A}$).

If $m$ and $n$ are both odd, then the relations become $c_1^2 \cdots c_g^2 = (c'_1)^2 \cdots (c'_g)^2 = a_{12}$. In this case, there is no obvious semi-direct product decomposition of $G$ where $a_{12}$ is clearly non-trivial, so we need another argument to show that $a_{12} \ne 1$. If $g = 1$, one can see that $G \cong (\mathbb{Z}/2)^2 \times Q_8$, where $c_1$ is sent to $i$, $c'_1$ is sent to $j$,
and $a_{12}$ identifies with the central element $-1$ of the quaternion group $Q_8$. Indeed, one can easily check that the above correspondence defines a morphism from $G$ to $(\mathbb{Z}/2)^2 \times Q_8$, and use the presentation $Q_8 = \langle i, j \mid i^2 = i^2 = (ij)^2 \rangle$ to construct its inverse. For $g \ge 1$, we can find a similar quotient of $G$, by considering the quotient $(Q_8)^g/H$, where $H$ is the hyperplane of $\mathbb{Z}((Q_8)^g) \cong (\mathbb{Z}/2)^g$ defined by the vanishing of the sum of the coordinates. Since $H$ is central in $(Q_8)^g$, it is normal. Moreover, $(Q_8)^g$ decomposes as a central extension of $(\mathbb{Z}/2)^{2g}$ by $(\mathbb{Z}/2)^g$, which induces a central extension:

$$ (\mathbb{Z}/2)^g/H \cong \mathbb{Z}/2 \longrightarrow (Q_8)^g/H \longrightarrow (\mathbb{Z}/2)^{2g}. $$

Using the presentation of $G$, we can see that there is a well-defined projection from $G$ onto $(Q_8)^g/H$ sending $c_r$ to $(1, \dots, 1, i, 1, \dots, 1)$, $c'_r$ to $(1, \dots, 1, j, 1, \dots, 1)$ (where the non-trivial coordinate is the $r$-th one in both cases), and $s_1$ and $s_2$ to 1. This projection sends $a_{12} = [c_1, c'_1]$ to the generator of the centre $(\mathbb{Z}/2)^g/H \cong \mathbb{Z}/2$. Thus $a_{12}$ is again not trivial in $G$, whence our claim. $\square$

**COROLLARY 6.61.** Let $g \ge 0$ and let $\lambda = (n_1, \dots, n_l)$ be a partition whose blocks are of size at least 3. Then $\mathcal{L}_2(\mathbf{B}_\lambda(\mathcal{N}_g)) \cong (\mathbb{Z}/2)^{l(l-1)/2}$ is the free elementary abelian 2-group on the $a_{ij}$ for $1 \le i < j \le l$.

**PROOF.** The proof is again the same as the proofs of Propositions 3.5 and 6.55, using the canonical maps $\mathcal{L}_2(\mathbf{B}_\lambda(\mathcal{N}_g)) \to \mathcal{L}_2(\mathbf{B}_{n_i,n_j}(\mathcal{N}_g)) \cong \mathbb{Z}/2$, where the latter is generated by the image of $a_{ij}$, by Proposition 6.60. $\square$

We can now finish the proof of the main result of this section.

**PROOF OF THEOREM 6.52.** Under our hypotheses, $\mathcal{L}_2(\mathbf{B}_\lambda(S))$ identifies with $\Gamma_2(\mathbf{B}_\lambda(S)/\Gamma_\infty) \subseteq \langle s_i^2, a_{ij} \rangle_{i,j \le l}$ (see Fact 6.54).

**Planar surfaces:** if $S$ is planar, then the canonical projection from $\mathbf{B}_\lambda(S)/\Gamma_\infty$ to $\mathbf{B}_\lambda(S)^{ab}$ sends the $s_i$ and the $a_{ij}$ to their counterparts in $\mathbf{B}_\lambda(S)^{ab}$. The latter is a linearly independent family by Proposition 6.47, so the restriction $\langle s_i^2, a_{ij} \rangle_{i,j \le l} \to \mathbf{B}_\lambda(S)^{ab}$ must be injective. But its kernel is $\Gamma_2(\mathbf{B}_\lambda(S)/\Gamma_\infty) = \mathcal{L}_2(\mathbf{B}_\lambda(S))$, which must then be trivial.

**The sphere:** if $S = \mathbb{S}^2$, then $\mathbf{B}_\lambda(\mathbb{S}^2)$ is a quotient of $\mathbf{B}_\lambda$ (see Corollary 6.30), which implies that its LCS also stops at $\Gamma_2$.

**Non-planar orientable surfaces:** Proposition 6.55 deals with most of them; the remaining ones are closed orientable surfaces, for which Proposition 6.56 is the relevant statement.

**Non-orientable surfaces:** these are dealt with in Corollary 6.59, except for the closed ones, whose Lie ring is studied separately, in Corollary 6.61. $\square$

## 6.6. Partitions with small blocks

Now that we have a complete description of the LCS of $\mathbf{B}_\lambda(S)$ in the stable case, that is, when the blocks of the partition $\lambda$ have size at least 3 (§6.5), we turn our attention to the cases where $\lambda$ does have blocks of size 1 or 2. Then we ask ourselves: under this assumption, when does the LCS of $\mathbf{B}_\lambda(S)$ stop? For most surfaces, it does not; see Proposition 6.62. In fact, there are only six surfaces to which this result does not apply. One of them is the disc, for which an answer has already been given in Chapter 3. Another one is the cylinder, whose case can easily be deduced from the case of the disc. Four surfaces remain: the torus $\mathbb{T}^2$ (§6.6.2),
the Möbius strip $M^2$ (§6.6.3), the sphere $S^2$ (§6.6.4) and the projective plane $\mathbb{P}^2$ (§6.6.5).

**6.6.1. The generic cases.** As a direct corollary of Propositions 6.37 and 6.39, we get the following result:

**PROPOSITION 6.62.** Let $\lambda = (n_1, \dots, n_l)$ be a partition of an integer $n \ge 1$.

* If $\lambda$ has at least one block of size 1 and $\pi_1(S)$ is not abelian (that is, if we suppose that $S \notin \{\mathbb{D}-pt, \mathbb{D}, S^2, T^2, \mathbb{P}^2, M^2\}$ up to isotopy equivalence), then the LCS of $\mathbf{B}_\lambda(S)$ does not stop.

* If $\lambda$ has at least one block of size 2 and $\pi_1(S)^{ab}$ is not finite (that is, if we suppose that $S \notin \{\mathbb{D}, S^2, \mathbb{P}^2\}$ up to isotopy equivalence), then the LCS of $\mathbf{B}_\lambda(S)$ does not stop.

**PROOF.** In the first case, there is a surjection $\mathbf{B}_\lambda(S) \to \mathbf{B}_1(S) \cong \pi_1(S)$, and in the second one, a surjection $\mathbf{B}_\lambda(S) \to \mathbf{B}_2(S)$. Propositions 6.37 and 6.39 say that, under our hypotheses, the LCS of the quotient does not stop in either case. The result then follows from Lemma 1.1. $\square$

Thus the question of whether the LCS of $\mathbf{B}_\lambda(S)$ stops has been answered for every partition, except for the six surfaces $\mathbb{D}-pt, \mathbb{D}, S^2, T^2, \mathbb{P}^2$ and $M^2$. In fact, $\mathbf{B}_\lambda(\mathbb{D}) = \mathbf{B}_\lambda$ has already been considered in Chapter 3; see Theorem 3.6. Also, since $\mathbf{B}_1(\mathbb{D}) = \{1\}$, we have an isomorphism $\mathbf{B}_\lambda(\mathbb{D}-pt) \cong \mathbf{B}_{1,\lambda}(\mathbb{D})$ for every partition $\lambda$ by Proposition 6.15, so we can deduce the remaining answer for $\mathbb{D}-pt$ from the answer for the disc. Namely, Lemma 3.8 and Proposition 3.11 imply:

**LEMMA 6.63.** If $\lambda$ has at least two blocks of size 1, then the LCS of $\mathbf{B}_\lambda(\mathbb{D}-pt)$ does not stop. If $\lambda = (1, n_2, \dots, n_l)$ where every $n_i \ge 3$, then its LCS stops at $\Gamma_2$.

Therefore, we are left with four remaining cases: the torus, the Möbius strip, the sphere and the projective plane.

**6.6.2. Partitioned braids on the torus.** We know that the LCS of $\mathbf{B}_\lambda(\mathbb{T}^2)$ stops if $\lambda$ has only blocks of size at least 3, and that it does not if there is at least one block of size 2. The remaining cases are dealt with using the following generalisation of [BGG08, Lem. 17]:

**PROPOSITION 6.64.** There is an isomorphism $\mathbf{B}_{1,\mu}(\mathbb{T}^2) \cong \mathbf{B}_\mu(\mathbb{T}^2 - pt) \times \mathbb{Z}^2$, for any partition $\mu$.

**PROOF.** Proposition 6.15 gives a short exact sequence:

$$ \mathbf{B}_{\mu}(\mathbb{T}^2 - pt) \to \mathbf{B}_{1,\mu}(\mathbb{T}^2) \to \pi_1(\mathbb{T}^2) \cong \mathbb{Z}^2. $$

Let $n \ge 2$ such that $\mu$ is a partition of $n-1$. The centre of $\mathbf{B}_n(\mathbb{T}^2)$ is generated by two braids $\alpha$ and $\beta$ corresponding to rotating all the punctures along each factor of $S^1 \times S^1$ [PR00, Prop. 4.2]. These are pure braids, so they are in the subgroup $\mathbf{B}_{1,\mu}(\mathbb{T}^2)$ of $\mathbf{B}_n(S)$. The above projection (forgetting all strands but one) maps these two elements to a basis of $\pi_1(\mathbb{T}^2) \cong \mathbb{Z}^2$. As a consequence, it restricts to an isomorphism $\langle \alpha, \beta \rangle \cong \mathbb{Z}^2$. Thus the above short exact sequence splits, and the corresponding action of $\mathbb{Z}^2$ is trivial ($\alpha$ and $\beta$ being central), hence it is in fact a direct product. $\square$

It is then an easy task to finish proving the following summary of our results for partitioned torus braids:
THEOREM 6.65. Let $\lambda$ be a partition of an integer $n \ge 1$. The LCS of $\mathbf{B}_\lambda(\mathbb{T}^2)$:

* does not stop if $\lambda$ has at least two blocks of size 1 or one block of size 2.

* stops at $\Gamma_3$ in all the other cases, except for $\mathbf{B}_1(\mathbb{T}^2) \cong \mathbb{Z}^2$.

PROOF. If $\lambda$ has at least two blocks of size 1, then $\mathbf{B}_\lambda(\mathbb{T}^2)$ surjects onto $\mathbf{B}_{1,1}(\mathbb{T}^2)$, which is isomorphic to $\mathbf{B}_1(\mathbb{T}^2 - pt) \times \mathbb{Z}^2 \cong \mathbf{F}_2 \times \mathbb{Z}^2$ by Proposition 6.64. The LCS of $\mathbf{F}_2$ does not stop, whence the result in this case, by Lemma 1.1.

If $\lambda$ has exactly one block of size 1 and no block of size 2, then Proposition 6.64 gives an isomorphism $\mathbf{B}_\lambda(\mathbb{T}^2) \cong \mathbf{B}_\mu(\mathbb{T}^2 - pt) \times \mathbb{Z}^2$ where $\mu$ has only blocks of size at least 3. Then Corollary 6.51 implies that the LCS stops at most at $\Gamma_3$ in this case. In fact, if $\mu$ is non-trivial, then the LCS of $\mathbf{B}_\mu(\mathbb{T}^2 - pt)$ stops exactly at $\Gamma_3$; see Theorem 6.52.

The other cases have already been treated as part of Corollary 6.51, Theorem 6.52 and Proposition 6.62. $\square$

### 6.6.3. Partitioned braids on the Möbius strip.
As in the case of the torus, we know that the LCS of $\mathbf{B}_\lambda(\mathbb{M}^2)$ stops if $\lambda$ has only blocks of size at least 3, and that it does not if there is at least one block of size 2. The only remaining cases are the ones when $\lambda$ has some blocks of size 1, the other ones being of size at least 3. We begin by showing that the LCS does not stop when there are at least two blocks of size 1. In this case, $\mathbf{B}_\lambda(\mathbb{M}^2)$ surjects onto $\mathbf{B}_{1,1}(\mathbb{M}^2) = \mathbf{P}_2(\mathbb{M}^2)$ so, by Lemma 1.1, this case follows from the following study of $\mathbf{P}_2(\mathbb{M}^2)$:

LEMMA 6.66. A presentation of the pure braid group $\mathbf{P}_2(\mathbb{M}^2)$ is given by generators $\gamma_1, \gamma_2$ and $A$ subject to the relations

$$ \left\{ \begin{array}{l} \gamma_1 A \gamma_1^{-1} = \gamma_2^{-1} A^{-1} \gamma_2 \\ \gamma_1 \gamma_2 \gamma_1^{-1} = \gamma_2^{-1} A^{-1} \gamma_2^2. \end{array} \right. $$

Recall that $\mathbb{M}^2$ is the surface $\mathcal{N}_{1,1}$ from §6.3. Here, we have changed the notations from §6.3 to lighter ones, more suited to the case with only one crosscap $(\gamma_i := c_1^{(i)})$ and only two strands ($A := A_{12}$).

PROOF. Proposition 6.15 gives a decomposition $\mathbf{P}_2(\mathbb{M}^2) \cong \pi_1(\mathbb{M}^2 - pt) \rtimes \pi_1(\mathbb{M}^2) \cong \mathbf{F}_2 \rtimes \mathbb{Z}$, where the projection onto $\pi_1(\mathbb{M}^2) \cong \mathbb{Z}$ is given by forgetting one strand (say, the first one). Then the factor $\mathbb{Z}$ is generated by $\gamma_1$, and a free basis of the factor $\mathbf{F}_2$ is given by $\gamma_2$ and $A$ (with the notations introduced just before the proof). Moreover, the action of $\gamma_1$ by conjugation on $\langle A, \gamma_2 \rangle$ is not difficult to compute. Namely, we have $\gamma_1 A \gamma_1^{-1} = \gamma_2^{-1} A^{-1} \gamma_2$; see Figure 6.6. Then, from Figure 6.1a, we get that $[\gamma_2, \gamma_1^{-1}] = A$, which is equivalent to $\gamma_1 \gamma_2 \gamma_1^{-1} = \gamma_1 A \gamma_1^{-1} \gamma_2$. Using the previous relation, the latter equals $\gamma_2^{-1} A^{-1} \gamma_2^2$. These relations determine $\mathbf{P}_2(\mathbb{M}^2)$, since the group $G$ that they define decomposes as $\langle A, \gamma_2 \rangle \rtimes \langle \gamma_1 \rangle$, and the obvious projection of $G$ onto $\mathbf{P}_2(\mathbb{M}^2)$ must be an isomorphism on both the kernel and the quotient, hence it must be an isomorphism. $\square$

COROLLARY 6.67. The LCS of $\mathbf{P}_2(\mathbb{M}^2)$ does not stop.

PROOF. Let us consider the quotient of $\mathbf{P}_2(\mathbb{M}^2) = \mathbf{F}_2 \rtimes \mathbb{Z}$ by $\Gamma_2(\mathbf{F}_2)$. It is $\mathbb{Z}^2 \rtimes \mathbb{Z}$, where the generator $\bar{\gamma}_1$ of $\mathbb{Z}$ acts via the involution $\tau$ sending $\bar{A}$ to $-\bar{A}$ and $\bar{\gamma}_2$ to $\bar{\gamma}_2 - \bar{A}$. Then $V := \mathrm{Im}(\tau - 1)$ is $\mathbb{Z} \cdot \bar{A}$. By Proposition A.10, for $i \ge 2$, we have $\Gamma_i(\mathbb{Z}^2 \rtimes \mathbb{Z}) = 2^{i-2}V$, so the LCS of $\mathbb{Z}^2 \rtimes \mathbb{Z}$ does not stop. Thus the LCS of $\mathbf{P}_2(\mathbb{M}^2)$ does not either. $\square$
<figure><img src="image_37.png" /><figcaption>FIGURE 6.6. Relation in $\mathbf{P}_2(\mathbb{M}^2)$: $\gamma_1 A \gamma_1^{-1} = \gamma_2^{-1} A^{-1} \gamma_2$.</figcaption></figure>

The answer for the remaining cases are consequences of the following result:

**PROPOSITION 6.68.** For every $m \ge 3$, the LCS of $\mathbf{B}_{1,m}(\mathbb{M}^2)$ does not stop.

**PROOF.** We can use Proposition 6.15 to get a decomposition:

$$ \mathbf{B}_{1,m}(\mathbb{M}^2) \cong \mathbf{B}_m(\mathbb{M}^2 - pt) \rtimes \pi_1(\mathbb{M}^2), $$

where $\pi_1(\mathbb{M}^2) \cong \mathbb{Z}$ is identified with the subgroup of $\mathbf{B}_{1,m}(\mathbb{M}^2)$ generated by $\gamma_1$ (as above, we denote $c_1^{(i)}$ by $\gamma_i$). We know that $\Gamma_\infty(\mathbf{B}_{1,m}(\mathbb{M}^2))$ contains $\Gamma_\infty(\mathbf{B}_m(\mathbb{M}^2 - pt))$. The latter is fully invariant in $\mathbf{B}_m(\mathbb{M}^2 - pt)$, hence normal in $\mathbf{B}_{1,m}(\mathbb{M}^2)$, and we can consider the quotient

$$ G := \mathbf{B}_{1,m}(\mathbb{M}^2)/\Gamma_\infty(\mathbf{B}_m(\mathbb{M}^2 - pt)) \cong (\mathbf{B}_m(\mathbb{M}^2 - pt)/\Gamma_\infty) \rtimes \mathbb{Z}. $$

Moreover, Lemma 6.33 and Remark 6.35 give a central extension:

$$ \langle\sigma\rangle \dashrightarrow \mathbf{B}_m(\mathbb{M}^2 - pt)/\Gamma_\infty \dashrightarrow \pi_1(\mathbb{M}^2 - pt)^{ab}. $$

The element $\sigma$ is the common class of the usual generators $\sigma_\alpha$ of $\mathbf{B}_m$. Since these commute with $\gamma_1$ in $\mathbf{B}_{1,m}(\mathbb{M}^2)$, $\sigma$ is central not only in $\mathbf{B}_m(\mathbb{M}^2 - pt)/\Gamma_\infty$, but in fact in $G$. In particular, $\langle\sigma\rangle$ is normal in $G$, and the quotient decomposes as:

$$ G/\sigma \cong [(\mathbf{B}_\mu(\mathbb{M}^2 - pt)/\Gamma_\infty)/\sigma] \rtimes \pi_1(\mathbb{M}^2) \cong \pi_1(\mathbb{M}^2 - pt)^{ab} \rtimes \pi_1(\mathbb{M}^2) \cong \mathbb{Z}^2 \rtimes \mathbb{Z}. $$

A basis of the $\mathbb{Z}^2$ factor is given by $\bar{\gamma}_2$ and $A = \bar{A}_{12}$. Consider the morphism $\mathbf{P}_2(\mathbb{M}^2) \to \mathbf{B}_{1,m}(\mathbb{M}^2)$ corresponding to adding $m-1$ trivial strands to the second block (constructed as in the proof of Proposition 6.15). Composing with the projection, we get a morphism from $\mathbf{P}_2(\mathbb{M}^2) = \mathbf{F}_2 \times \mathbb{Z}$ to $G/\sigma = \mathbb{Z}^2 \times \mathbb{Z}$. From the explicit description of both of these groups, one easily sees that it has to induce an isomorphism $\mathbf{P}_2(\mathbb{M}^2)/\Gamma_2(\mathbf{F}_2) \cong G/\sigma$. But the LCS of $\mathbf{P}_2(\mathbb{M}^2)/\Gamma_2(\mathbf{F}_2)$ does not stop (see the proof of Corollary 6.67), so we have found a quotient of $\mathbf{B}_{1,m}(\mathbb{M}^2)$ whose LCS does not stop, whence the result. $\square$

Let us sum up our results about the LCS of $\mathbf{B}_\lambda(\mathbb{M}^2)$:

**THEOREM 6.69.** Let $\lambda = (n_1, \dots, n_l)$ be a partition of an integer $n \ge 1$. The LCS of $\mathbf{B}_\lambda(\mathbb{M}^2)$:

* does not stop if $\lambda$ has at least one block of size 1 or 2, with the exception of $\mathbf{B}_1(\mathbb{M}^2) \cong \mathbb{Z}$.

* stops in all the other cases, at $\Gamma_3$ if $l \ge 2$ and at $\Gamma_2$ if $l = 1$.

**PROOF.** The first statement follows from Proposition 6.62 (for blocks of size 2) and from Corollary 6.67 and Proposition 6.68 (for blocks of size 1). The second one is part of the general results of Corollary 6.51 and Theorem 6.52. $\square$
**6.6.4. Partitioned braids on the sphere.** For a partition $\lambda = (n_1, \dots, n_l)$, the inclusion of the disc into the sphere induces a surjection $\mathbf{B}_{\lambda} \to \mathbf{B}_{\lambda}(\mathbb{S}^2)$ by Corollary 6.30. We can thus apply Lemma 1.1 to deduce that the LCS of $\mathbf{B}_{\lambda}(\mathbb{S}^2)$ stops whenever the one of $\mathbf{B}_{\lambda}$ does. Namely, by Theorem 3.6, it stops at $\Gamma_2$ if $n_i \ge 3$ for all $i$, save at most two indices for which $n_i = 1$. Since the abelianisation of $\mathbf{B}_{\lambda}(\mathbb{S}^2)$ has already been computed in Proposition 6.47, we have a complete description of the LCS in these cases.

The above argument can be extended somewhat if we remark that, because of Proposition 6.15, $\mathbf{B}_1(\mathbb{S}^2) = \pi_1(\mathbb{S}^2) = \{1\}$ is the cokernel of the canonical morphism $\mathbf{B}_{\mu}(\mathbb{D}) \to \mathbf{B}_{1,\mu}(\mathbb{S}^2)$ (for any partition $\mu$), which is thus surjective. As a consequence, the LCS of $\mathbf{B}_{\lambda}(\mathbb{S}^2)$ also stops at $\Gamma_2$ when $\lambda$ has three blocks of size 1, the other ones being of size at least 3. This proves the first half of the following theorem. The second half is proven in the remainder of this subsection in several different cases, which are synthesised into a proof of the theorem at the end of the subsection.

**THEOREM 6.70.** Let $n \ge 1$ be an integer, let $\lambda = (n_1, \dots, n_l)$ be a partition of $n$. The LCS of $\mathbf{B}_{\lambda}(\mathbb{S}^2)$:

* stops at $\Gamma_2$ if $n_i \ge 3$ for all $i$, save at most three indices for which $n_i = 1$.

* does not stop in all the other cases, except for $\mathbf{B}_2(\mathbb{S}^2) \cong \mathbb{Z}/2$, $\mathbf{B}_{2,1}(\mathbb{S}^2) \cong \mathbb{Z}/4$ and the $\mathbf{B}_{2,m}(\mathbb{S}^2)$ with $m \ge 3$, whose LCS stop either at $\Gamma_{v_2(m)+1}$ or at $\Gamma_{v_2(m)+2}$ when $m$ is even, where $v_2$ is the 2-adic valuation, and either at $\Gamma_2$ or at $\Gamma_3$ when $m$ is odd.

In the case of $\mathbf{B}_{2,m}(\mathbb{S}^2)$ with $m \ge 3$, where the answer given above is ambiguous between two consecutive possibilities, we conjecture that its LCS stops at $\Gamma_{v_2(m)+2}$ for all $m \ge 3$; see Remark 6.74.

**6.6.4.1.** *Blocks of size 1.* We need to consider the case of the pure braid group on four strands:

**LEMMA 6.71.** *The LCS of $\mathbf{P}_4(\mathbb{S}^2)$ does not stop.*

**PROOF.** We sketch a proof of the decomposition of $\mathbf{P}_4(\mathbb{S}^2)$ from [GG04c, Th. 4]. Recall that we have a short exact sequence by Proposition 6.15:

$$ \mathbf{P}_1(\mathbb{S}^2 - \{3 \text{ pts}\}) \hookrightarrow \mathbf{P}_4(\mathbb{S}^2) \to \mathbf{P}_3(\mathbb{S}^2). $$

It is known that $\mathbf{P}_3(\mathbb{S}^2) \cong \mathbb{Z}/2$. In fact, an isomorphism between $\pi_1(\mathrm{SO}_3(\mathbb{R}))$ and $\mathbf{P}_3(\mathbb{S}^2)$ is induced by $\varphi \mapsto (\varphi(e_1), \varphi(e_2), \varphi(e_3))$ from $\mathrm{SO}_3(\mathbb{R})$ to $\mathrm{F}_3(\mathbb{S}^2)$. Moreover, a splitting of the above short exact sequence is given by sending the generator of $\mathbf{P}_3(\mathbb{S}^2)$ to the full twist. Since the latter is central, $\mathbf{P}_4(\mathbb{S}^2)$ is the direct product of $\mathbb{Z}/2$ with $\pi_1(\mathbb{S}^2 - \{3 \text{ pts}\}) \cong \mathbf{F}_2$. Thus $\mathbf{P}_4(\mathbb{S}^2)$ is residually nilpotent (but not nilpotent), whence the result. $\square$

**6.6.4.2.** *Blocks of size 2.* Let us begin with the case where the partition has at least three blocks:

**PROPOSITION 6.72.** Let $\lambda$ be a partition of $n$ with at least three blocks, and at least one block of size 2. Then the LCS of $\mathbf{B}_{\lambda}(\mathbb{S}^2)$ does not stop.

**PROOF.** Let $\mu$ be any partition. Let us consider the quotient of $\mathbf{B}_{2,\mu}(\mathbb{S}^2)$ by $\Gamma_2(\mathbf{B}_{1,1,\mu}(\mathbb{S}^2))$, which is (clearly) an extension:

$$ (6.6) \qquad \mathbf{B}_{1,1,\mu}(\mathbb{S}^2)^{ab} \longrightarrow G = \mathbf{B}_{2,\mu}(\mathbb{S}^2)/\Gamma_2(\mathbf{B}_{1,1,\mu}(\mathbb{S}^2)) \longrightarrow \mathfrak{S}_2. $$
Let us fix $(n_1, \dots, n_l) := (1, 1, \mu)$. From a presentation of the kernel (from Proposition 6.32) and a presentation of the quotient, using the method from Appendix B, we can write down a presentation of $G$. Precisely, $G$ admits the presentation with generators $s, s_i$ (for $1 \le i \le l$), and $a_{ij}$ (for $1 \le i < j \le l$), subject to the following relations:

$$ \left\{ \begin{array}{ll} (1) & s^2 = a_{12}, \\ (2) & s_i = 1 \quad \text{if } n_i = 1, \\ (3) & [s_i, s_j] = [s_i, a_{pq}] = [a_{pq}, a_{uv}] = 1 \quad \forall i, j, p, q, u, v, \\ (4) & [s, s_i] = 1 \quad \forall i \ge 1, \\ (5) & [s, a_{ij}] = 1 \quad \forall j > i \ge 3, \\ (6) & sa_{1j}s^{-1} = a_{2j} \text{ and } sa_{2j}s^{-1} = a_{1j} \quad \forall j \ge 3, \\ (7) & 2(n_i - 1)s_i + \sum_{j \ne i} n_j a_{ij} = 0 \quad \forall i \ge 1, \end{array} \right. $$

where the last relation uses additive notation in the abelian subgroup generated by the $s_i$ and the $a_{ij}$.

Consider the subgroup $H$ of $G$ generated by $s^2$ together with the $s_i$ and the $a_{ij}$ for $i, j \ge 3$. One can easily see that it is central in $G$. Moreover, we can deduce from the presentation of $G$ a presentation of $G/H$, which turns out to be very simple. Indeed, it is generated by $s$, the $a_{1i}$ and the $a_{2i}$ for $i \ge 3$. Moreover, since $n_1 = n_2 = 1$, for $i \ge 3$ the last relation becomes $a_{1i} + a_{2i} = 0$, which means that $a_{1i}$ and $a_{2i}$ are inverse to each other. For $i \in \{1, 2\}$, we find the relations $\sum n_j a_{1j} = 0$ and $\sum n_j a_{2j} = 0$, which are equivalent to each other modulo the previous relations. Thus, if we denote by $a_i$ the element $a_{1i} = a_{2i}^{-1}$ of $G/H$, we get that $G/H$ is generated by $a_3, a_4, \dots, a_l$ and $s$, subject to the relations:

$$ \left\{ \begin{array}{ll} (1) & s^2 = 1 \\ (3) & [a_i, a_j] = 1 \quad \forall i, j, \\ (6) & sa_j s^{-1} = a_j^{-1} \quad \forall j \ge 3, \\ (7) & \sum n_j a_j = 0. \end{array} \right. $$

Finally, $G/H \cong A \rtimes (\mathbb{Z}/2)$, where $A$ is the quotient of the free abelian group on $a_3, a_4, \dots, a_l$ by the single relation $\sum n_j a_j = 0$, and $\mathbb{Z}/2$ acts on $A$ via $-id$. We can use Corollary A.8 to compute the LCS of $G/H$. Namely, since $A \cong \mathbb{Z}^{l-3} \times \mathbb{Z}/\gcd(n_i)$, it does not stop whenever $l > 3$. Thus, we have found a quotient of $\mathbf{B}_{2,\mu}(\mathbb{S}^2)$ having a non-stopping LCS whenever $\mu$ has at least two blocks. $\square$

If $\lambda$ has only two blocks, we first assume that the other block is large enough:

**PROPOSITION 6.73.** Let $m \ge 3$ be an integer. The LCS of $\mathbf{B}_{2,m}(\mathbb{S}^2)$ stops at $\Gamma_{v_2(m)+1}$ or at $\Gamma_{v_2(m)+2}$ when $m$ is even, where $v_2$ is the 2-adic valuation, and at $\Gamma_2$ or at $\Gamma_3$ when $m$ is odd.

**PROOF.** We have seen above that the LCS of $\mathbf{B}_{1,1,m}(\mathbb{S}^2)$ (which is a quotient of $\mathbf{B}_{1,1,m}$) stops at $\Gamma_2$ if $m \ge 3$. Then, as in the proof of Proposition 3.12, we have that $\Gamma_\infty(\mathbf{B}_{2,m}(\mathbb{S}^2))$ contains $\Gamma_\infty(\mathbf{B}_{1,1,m}(\mathbb{S}^2)) = \Gamma_2(\mathbf{B}_{1,1,m}(\mathbb{S}^2))$. Thus it is enough to understand when the LCS of $G := \mathbf{B}_{2,m}(\mathbb{S}^2)/\Gamma_2(\mathbf{B}_{1,1,m}(\mathbb{S}^2))$ stops. In order to do this, we use the calculations already done in the course of the proof of Proposition 6.72: the group $G$ is a central extension of $(\mathbb{Z}/m) \times (\mathbb{Z}/2)$, where $\mathbb{Z}/2$ acts via $-id$. Thanks to Corollary A.8, we know that the LCS of the latter is
given by $\Gamma_i = 2^{i-1}\mathbb{Z}/m$ for $i \ge 2$. Hence it stops at $\Gamma_2$ if $m$ is odd and at $\Gamma_{v_2(m)+1}$ if $m$ is even. Thus, the result will follow from Corollary 1.6, if we show that the kernel $H$ of the central extension $H \hookrightarrow G \to (\mathbb{Z}/m) \times (\mathbb{Z}/2)$ injects into $G^{ab}$. We can compute $G^{ab}$ using the presentation of $G$ from the proof of Proposition 6.72, or using Proposition 6.47 and the fact that $G^{ab} \cong \mathbf{B}_{2,m}(\mathbb{S}^2)^{ab}$ (by definition of $G$). Either way, we find that it is generated by three elements $s (= s_1)$, $s_3$ and $a$ (the latter being the common image of $a_{13}$ and $a_{23} = sa_{13}s^{-1}$ in $G^{ab}$), subject to the relations:

$$ \begin{cases} 2s + ma = 0; \\ 2(m-1)s_3 + 2a = 0. \end{cases} $$

In other words, $G^{ab} = \mathbb{Z}^3/R$ where, if $(s, s_3, a)$ denotes a basis of $\mathbb{Z}^3$, $R$ is generated by $2s + ma$ and $2(m-1)s_3 + 2a$. The image $Q$ of $H$ in $G^{ab}$ is generated by $2s$ and $s_3$. An easy calculation shows that, in $\mathbb{Z}^3$, $\langle 2s, s_3 \rangle \cap R$ is generated by $r_m := 2s + m(m-1)s_3$ if $m$ is even, and $r_m := 4s + 2m(m-1)s_3$ if $m$ is odd. This means that, as an abelian group, $Q$ is presented as the quotient of the free abelian group $\mathbb{Z}^2$ over $2s$ and $s_3$ by the relation $r_m = 0$. But this relation already holds in $H$. Indeed, in $\mathbf{B}_{1,1,m}(\mathbb{S}^2)^{ab} \subset G$, we have $s^2 = a_{12} - ma_{13} = -ma_{23}$ (see the proof of Proposition 6.72), so if we multiply the relation $2(m-1)s_3 + a_{13} + a_{23} = 0$ by $m/2$ if $m$ is even (resp. by $m$ if $m$ is odd), we obtain $m(m-1)s_3 = s^2$ (resp. $2m(m-1)s_3 = 2s^2$), hence $r_m = 0$ in $H \subset G$. Finally, the projection $H \to Q$ must be an isomorphism (one can construct an inverse to it using the presentation of $Q$), whence our conclusion. $\square$

**REMARK 6.74.** It seems difficult to decide theoretically between the two possibilities, although our experimental calculations [DPS22] using GAP [Gro22] and the package NQ [Nic96] suggest that the LCS of $G$ (and hence also the one of $\mathbf{B}_{2,m}(\mathbb{S}^2)$) always stops at $\Gamma_{v_2(m)+2}$ (for both even and odd $m$); we conjecture that this is the case.

Finally, let us show that the LCS of $\mathbf{B}_{2,2}(\mathbb{S}^2)$ does not stop, using Lemma 1.1. Namely, we are looking for a quotient of $\mathbf{B}_{2,2}(\mathbb{S}^2)$ whose LCS does not stop. As a manageable non-abelian quotient, one can think of $\mathbf{B}_{2,2}(\mathbb{S}^2)/\Gamma_2(\mathbf{P}_4(\mathbb{S}^2))$, which is an extension of $\mathfrak{S}_2 \times \mathfrak{S}_2$ by $\mathbf{P}_4(\mathbb{S}^2)^{ab}$. In fact, in order to make it even more manageable, we take a further quotient, turning it into a split extension to which the methods of Appendix A apply.

**PROPOSITION 6.75.** *The LCS of $\mathbf{B}_{2,2}(\mathbb{S}^2)$ does not stop.*

**PROOF.** Recall that $\Gamma_2(\mathbf{P}_4(\mathbb{S}^2))$ is fully invariant in $\mathbf{P}_4(\mathbb{S}^2)$, hence normal in $\mathbf{B}_{2,2}(\mathbb{S}^2)$, so the quotient $G = \mathbf{B}_{2,2}(\mathbb{S}^2)/\Gamma_2(\mathbf{P}_4(\mathbb{S}^2))$ is a well-defined extension of $\mathfrak{S}_2 \times \mathfrak{S}_2$ by $\mathbf{P}_4(\mathbb{S}^2)^{ab}$. The latter it is the abelian group generated by $a_{ij}$ for $1 \le i < j \le 4$, subject to the four relations $\sum_{j \ne i} a_{ij} = 0$ (for each $i \le 4$); this classical computation is part of Proposition 6.32.

The action of $\mathbf{B}_{2,2}(\mathbb{S}^2)$ on $\mathbf{P}_4(\mathbb{S}^2)^{ab}$ induced by conjugation factors through $\mathbf{B}_{2,2}(\mathbb{S}^2)/\mathbf{P}_4(\mathbb{S}^2) \cong \mathfrak{S}_2 \times \mathfrak{S}_2$. This action is by permutation of the indices of the generators $a_{ij}$ of $\mathbf{P}_4(\mathbb{S}^2)^{ab}$. In particular, it fixes $a_{12}$ and $a_{34}$, which then generate a central subgroup $H$ of $G$. Let us consider the quotient by this central subgroup, that is, the quotient of $\mathbf{B}_{2,2}(\mathbb{S}^2)$ by its subgroup $\tilde{H} = \langle A_{12}, A_{34}, \Gamma_2(\mathbf{P}_4(\mathbb{S}^2)) \rangle$. This
quotient is an extension:

$$ \mathbf{P}_4(\mathbb{S}^2)^{ab}/H \to G/H \to \mathfrak{S}_2 \times \mathfrak{S}_2. $$

By definition of $\tilde{H}$, we have $\sigma_1^2 = \sigma_3^2 = 1$ in $\mathbf{B}_{2,2}(\mathbb{S}^2)/\tilde{H} = G/H$, so this short exact sequence splits. Moreover, $\mathbf{P}_4(\mathbb{S}^2)^{ab}/H$ is the quotient of $\mathbf{P}_4(\mathbb{S}^2)^{ab}$ by the relations $a_{12} = a_{34} = 0$, modulo which the relations defining $\mathbf{P}_4(\mathbb{S}^2)^{ab}$ become $a_{13} = -a_{14} = a_{13} = -a_{23}$. Thus $\mathbf{P}_4(\mathbb{S}^2)^{ab}/H \cong \mathbb{Z}$, and $G/H \cong \mathbb{Z} \times (\mathfrak{S}_2)^2$, where both transpositions act via a sign. This action factors through the signature $\varepsilon: \mathfrak{S}_2 \times \mathfrak{S}_2 \to \mathbb{Z}/2$, so that $\Gamma_*^{\mathfrak{S}_2 \times \mathfrak{S}_2}(\mathbb{Z}) = \Gamma_*^{\mathbb{Z}/2}(\mathbb{Z})$; see §A.1 for the definition of relative LCS. From Lemma A.4 (whose proof could alternatively be repeated in this situation), we get that $\Gamma_i(\mathbb{Z} \times (\mathfrak{S}_2)^2) = 2^{i-1}\mathbb{Z}$ for $i \ge 2$, so the LCS of $\mathbf{B}_{2,2}/\tilde{H} = G/H$ does not stop. $\square$

**PROOF OF THEOREM 6.70.** The first point of the theorem was proven just before its statement. We explain how to deduce the second point from the results above.

If $\lambda$ has at least four blocks of size 1, then $\mathbf{B}_\lambda(\mathbb{S}^2)$ surjects onto the pure braid group $\mathbf{P}_4(\mathbb{S}^2)$ (by forgetting the other blocks). Using Lemma 1.1, the result in this case follows from Lemma 6.71.

The remaining cases are the ones where there is at least one block of size 2. If the partition has at least three blocks, the result follows from Proposition 6.72. If the partition has exactly two blocks, and if the size of the other block is at least 2, then we can apply either Proposition 6.73 or Proposition 6.75.

Finally, let us compute $\mathbf{B}_{2,1}(\mathbb{S}^2)$ and $\mathbf{B}_2(\mathbb{S}^2)$. As recalled in the proof of Lemma 6.71, $\mathbf{P}_3(\mathbb{S}^2) \cong \pi_1(\mathrm{SO}_3(\mathbb{R})) \cong \mathbb{Z}/2$, so that $\mathbf{B}_{2,1}(\mathbb{S}^2)$ is an extension of $\mathfrak{S}_2 \cong \mathbb{Z}/2$ by $\mathbb{Z}/2$. Thus, it must be isomorphic to $(\mathbb{Z}/2)^2$ or to $\mathbb{Z}/4$. In order to decide between the two, one can use the computation of $\mathbf{B}_{2,1}^{ab}$ from Proposition 6.32, and find that $\mathbf{B}_{2,1}(\mathbb{S}^2) \cong \mathbb{Z}/4$. As for $\mathbf{B}_2(\mathbb{S}^2)$, we use Proposition 6.15 to get an exact sequence $\pi_1(\mathbb{S}^2 - \{pt\}) \to \mathbf{P}_2(\mathbb{S}^2) \to \pi_1(\mathbb{S}^2)$, which implies that $\mathbf{P}_2(\mathbb{S}^2) = 1$, whence $\mathbf{B}_2(\mathbb{S}^2) \cong \mathfrak{S}_2$. $\square$

**6.6.5. Partitioned braids on the projective plane.** As in §6.3.2, we see the projective plane $\mathbb{P}^2 = \mathcal{N}_1$ as a sphere with one crosscap, which is the quotient of the Möbius strip $\mathbb{M}^2 = \mathcal{N}_{1,1}$ by its boundary. This allows us to use the conventions of Figure 6.2b (with $g = 1$) to describe braids on the projective plane. We only modify slightly our notation, as we did before for braids on the Möbius strip:

**NOTATION 6.76.** We use the notation of §6.3 for braids on non-orientable surfaces, but since there is only one crosscap, we denote $c_1^{(\alpha)}$ by $\gamma_\alpha$.

We now prove the following theorem, which describes when the LCS of $\mathbf{B}_\lambda(\mathbb{P}^2)$ stops, except for $\mathbf{B}_{2,m}(\mathbb{P}^2)$ with $m \ge 3$ (for this case, see Conjecture 6.94):

**THEOREM 6.77.** Let $n \ge 1$ be an integer, let $\lambda = (n_1, \dots, n_l)$ be a partition of $n$. The LCS of $\mathbf{B}_\lambda(\mathbb{P}^2)$:

* stops at $\Gamma_2$ if $l = 1$, except for $\mathbf{B}_2(\mathbb{P}^2)$, which is (strictly) 3-nilpotent.

* stops at $\Gamma_3$ if $l \ge 2$ and $n_i \ge 3$ for all $i$.

* does not stop in all the other cases where $l \ge 3$.
Moreover, the LCS of $\mathbf{B}_{2,2}(\mathbb{P}^2)$ and of $\mathbf{B}_{1,2}(\mathbb{P}^2)$ do not stop, the LCS of $\mathbf{B}_{1,1}(\mathbb{P}^2)$ stops at $\Gamma_3$ and, for $m \ge 3$, the LCS of $\mathbf{B}_{1,m}(\mathbb{P}^2)$ stops at $\Gamma_{v_2(m)+2}$ or at $\Gamma_{v_2(m)+3}$ when $m$ is even, where $v_2$ is the 2-adic valuation, and at $\Gamma_3$ or $\Gamma_4$ when $m$ is odd.

In the case of $\mathbf{B}_{1,m}(\mathbb{P}^2)$ with $m \ge 3$, where the answer given above is ambiguous between two consecutive possibilities, we conjecture that its LCS stops at $\Gamma_{v_2(m)+3}$ for all $m \ge 3$; see Remark 6.81.

The proof splits into many different cases, which we consider individually below; they are synthesised into a proof of Theorem 6.77 immediately after Remark 6.93.

6.6.5.1. *Blocks of size 1.* In order to study partitioned braids with blocks of size 1 on the projective plane, we need to know what the Fadell-Neuwirth exact sequence becomes in the exceptional case $n=1$ of Proposition 6.15.

**PROPOSITION 6.78.** For any partition $\mu$ of an integer $m$, we have a short exact sequence:

$$ \mathbf{B}_{\mu}(\mathbb{M}^2)/\xi^2 \dashrightarrow \mathbf{B}_{1,\mu}(\mathbb{P}^2) \dashrightarrow \pi_1(\mathbb{P}^2) \cong \mathbb{Z}/2, $$

where $\xi$ is the central element of $\mathbf{B}_{\mu}(\mathbb{M}^2)$ given by all the strands going once along $\partial\mathbb{M}^2$.

**SKETCH OF PROOF.** One needs to show that $\xi^2$ is the image of a generator of $\pi_2(\mathbb{P}^2)$ by the connecting morphism in the long exact sequence of the proof of Proposition 6.15. A generator of $\pi_2(\mathbb{P}^2)$ is given by the canonical projection $\mathbb{S}^2 \to \mathbb{P}^2$, which can be lifted to a map from the disc $\mathbb{D}$ to $F_{m+1}(\mathbb{P}^2)$ as follows. Let $\pi$ denote the projection of $\mathbb{D}$ onto $\mathbb{S}^2 = \mathbb{D}/\partial\mathbb{D}$ sending $\partial\mathbb{D}$ to the south pole $P = -N$. Let $\rho: \mathbb{D} \to SO_3(\mathbb{R})$ be the unique continuous function sending each $x \in \text{Int}(\mathbb{D})$ to the rotation of axis $N \times \pi(x)$ and angle $(N, \pi(x))$ (so that $\rho(x)(N) = \pi(x)$). Let $\text{ev}: SO_3(\mathbb{R}) \to F_{m+1}(\mathbb{P}^2)$ be evaluation at a base configuration whose first element is $\pm N$. Then $\text{ev} \circ \rho$ is the required lift. Moreover, when $x$ goes once around $\partial\mathbb{D}$, then $\rho(x)$ goes twice around the circle of rotations of angle $\pi$ with axis orthogonal to $N$. Since $\mathbb{P}^2$ is to be thought of as the quotient of the Möbius strip by its boundary (which goes to $\pm N$), this circle of rotations evaluates to the element $\xi$ described in the statement. $\square$

**PROPOSITION 6.79.** Let $\mu$ be a partition having at least two blocks. Then the LCS of $\mathbf{B}_{1,\mu}(\mathbb{P}^2)$ does not stop.

**PROOF.** We use the extension from Proposition 6.78:

$$ \mathbf{B}_{\mu}(\mathbb{M}^2)/\xi^2 \dashrightarrow \mathbf{B}_{1,\mu}(\mathbb{P}^2) \dashrightarrow \pi_1(\mathbb{P}^2) \cong \mathbb{Z}/2. $$

We denote the partition $\lambda := (1, \mu)$ of the integer $n$ by $(n_1, \dots, n_l)$, and we denote by $l'$ the number of indices $i$ such that $n_i \ge 2$. Using notations from the §6.3 (changed to $\gamma_i := c_1^{(i)}$), we have that the quotient in the previous extension is generated by the class of $\gamma_1$, and we can write $\xi$ as the following product of commuting braids:

$$ \xi = \prod_{i=2}^{n} \gamma_i^2 (A_{i,i+1} \cdots A_{i,n})^{-1}. $$

Let us consider the quotient $G$ of $\mathbf{B}_{1,\mu}(\mathbb{P}^2)$ by $\Gamma_2(\mathbf{B}_{\mu}(\mathbb{M}^2)/\xi^2)$. It is an extension of $\mathbb{Z}/2$ by $\mathbf{B}_{\mu}(\mathbb{M}^2)^{ab}/2\xi$. Recall from Proposition 6.47 that:

$$ \mathbf{B}_{\mu}(\mathbb{M}^2)^{ab} \cong \mathbb{Z}^{l-1} \times (\mathbb{Z}/2)^{l'}. $$
A basis of the first factor is given by $c_2, \dots, c_l$, where each $c_i$ is the common class of the $\gamma_\alpha$ with $\alpha$ in the $i$-th block of $\lambda$. A $(\mathbb{Z}/2)$-basis of the second factor is given by the elements $s_i$ described in Proposition 3.5. The images of the $s_i$ in $G$ commute with the class $c_1$ of $\gamma_1$, since they have lifts with disjoint support. As a consequence, they are central not only in $\mathbf{B}_\mu(\mathbb{M}^2)^{ab}/2\bar{\xi}$, but in $G$. Let them generate the subgroup $A$ of $G$, and consider $G/A$. There is an extension:

$$ \mathbb{Z}^{l-1}/2\bar{\xi} \to G/A \to \mathbb{Z}/2, $$

where $\bar{\xi} = 2 \sum_{i \ge 2} n_i c_i$. This extension is not split, but we can quotient further to get a split extension of abelian groups. Namely, we need to kill the element $c_1^2$ of $G$. In order to understand it, we use the relations from Corollary 6.30: $\gamma_\alpha^2 = A_{\alpha 1} \cdots A_{\alpha n}$ in $\mathbf{B}_{1,\mu}(\mathbb{P}^2)$. In $G/A$, where the $A_{\alpha\beta}$ are killed if $\alpha, \beta \ge 2$, these relations give $\bar{A}_{1\alpha} = \bar{\gamma}_\alpha^2$ if $\alpha \ge 2$ and

$$ c_1^2 = \bar{\gamma}_1^2 = \bar{A}_{12} \cdots \bar{A}_{1n} = \bar{\gamma}_2^2 \cdots \bar{\gamma}_n^2 = 2 \sum_{i \ge 2} n_i c_i = \bar{\xi}. $$

In particular, $\bar{\xi}$ commutes with $c_1$, which implies that it is central in $G/A$. The quotient $H$ of $G/A$ by $\bar{\xi}$ is thus a semi-direct product:

$$ H \cong (\mathbb{Z}^{l-1}/\bar{\xi}) \times \mathbb{Z}/2. $$

Finally, we need to compute the action of $c_1$ by conjugation on the other $c_i$. Recall that $[\gamma_\alpha, \gamma_1^{-1}] = A_{1,\alpha}$ in $\mathbf{B}_{1,\mu}(\mathbb{P}^2)$ (see Figure 6.1a), and $\bar{A}_{1,\alpha} = \bar{\gamma}_\alpha^2$ in $H$, so that $[c_i, c_1^{-1}] = c_i^2$ in $H$, whence $c_1^{-1}c_1^{-1}c_1 = c_i$, which implies that $c_1$ acts via $-id$ on $\langle c_2, \dots, c_l \rangle = \mathbb{Z}^{l-1}/\bar{\xi} \cong \mathbb{Z}^{l-2} \times (\mathbb{Z}/2)$. Since we have assumed that $l \ge 3$, the LCS of $H = (\mathbb{Z}^{l-2} \times (\mathbb{Z}/2)) \times (\mathbb{Z}/2)$ does not stop; see Corollary A.6. $\square$

**PROPOSITION 6.80.** For $m \ge 3$, the LCS of $\mathbf{B}_{1,m}(\mathbb{P}^2)$ stops either at $\Gamma_{v_2(m)+2}$ or at $\Gamma_{v_2(m)+3}$ when $m$ is even, where $v_2$ is the 2-adic valuation, and either at $\Gamma_3$ or at $\Gamma_4$ when $m$ is odd.

**PROOF.** Again, we use the extension from Proposition 6.78:

$$ \mathbf{B}_m(\mathbb{M}^2)/\xi^2 \to \mathbf{B}_{1,m}(\mathbb{P}^2) \to \pi_1(\mathbb{P}^2) \cong \mathbb{Z}/2. $$

Recall that, since $m \ge 3$, the LCS of $\mathbf{B}_m(\mathbb{M}^2)$ stops at $\Gamma_2$; see Theorem 6.42. Moreover, $\mathbf{B}_m(\mathbb{M}^2)/\Gamma_\infty = \mathbf{B}_m(\mathbb{M}^2)^{ab} \cong \mathbb{Z} \times \mathbb{Z}/2$, where the first factor is generated by the common class $\gamma$ of $\gamma_2, \dots, \gamma_{m+1}$ and the second factor is generated by the common class $\sigma$ of the $\sigma_i$; see Proposition 6.32. The image of the central element $\xi$ of $\mathbf{B}_m(\mathbb{M}^2)^{ab}$ is then:

$$ \bar{\xi} = \sum_{i=2}^{m+1} \frac{\gamma_i^2 (A_{i,i+1} \cdots A_{i,m+1})^{-1}}{2} = 2m\gamma - 2 \cdot \frac{m(m+1)}{2} \sigma = 2m\gamma, $$

so that $\mathbf{B}_m(\mathbb{M}^2)/(\xi^2\Gamma_2) \cong \mathbb{Z}/4m \times \mathbb{Z}/2$. Let us consider the quotient $G$ of $\mathbf{B}_{1,m}(\mathbb{P}^2)$ by the image of $\Gamma_2(\mathbf{B}_m(\mathbb{M}^2))$. The image of $\Gamma_2(\mathbf{B}_m(\mathbb{M}^2)) = \Gamma_\infty(\mathbf{B}_m(\mathbb{M}^2))$ is inside $\Gamma_\infty(\mathbf{B}_{1,m}(\mathbb{P}^2))$, so that $G$ and $\mathbf{B}_{1,m}(\mathbb{P}^2)$ have the same associated Lie ring. Now, $G$ is an extension:

$$ \mathbb{Z}/4m \times \mathbb{Z}/2 \to G \to \mathbb{Z}/2. $$

We can already see that $G$ is finite, and deduce that its LCS stops, so that $\Gamma_*(\mathbf{B}_{1,m}(\mathbb{P}^2))$ stops too. In order to be more precise, let us recall that $G = \langle \gamma, \sigma, \gamma_1 \rangle$,
where the class of $\gamma_1$ generates the quotient in the previous extension. Notice that $\gamma_1^2$ is in the kernel, so it commutes with $\gamma$ and $\sigma$. Since it obviously commutes with $\gamma_1$, it is central in $G$. Moreover, as in the previous proof, the boundary relations in $\mathbf{B}_{1,m}(\mathbb{P}^2)$ give:

$$ \gamma = \overline{\gamma_i^2} = \overline{A_{i,1} \cdots A_{i,m+1}} = \overline{A_{i,1}} \quad \text{for } i \ge 2 \text{ and } \gamma_1^2 = \overline{A_{1,2} \cdots A_{1,m+1}} = \gamma^{2m} = \overline{\xi}. $$

Thus, the quotient of $G$ by its central subgroup $A = \langle \gamma^{2m} \rangle$ is an extension of $\mathbb{Z}/2$ by $(\mathbb{Z}/2m) \times \mathbb{Z}/2$. Since $\gamma_1^2 = 1$ in $G/A$, this extension splits as a semi-direct product:

$$ G/A \cong ((\mathbb{Z}/2m) \times \mathbb{Z}/2) \rtimes \mathbb{Z}/2. $$

The element $\gamma_1$ of $G/A$ commutes with $\sigma$, and, using once again the isotopy from Figure 6.1a:

$$ [\gamma, \gamma_1^{-1}] = [\overline{\gamma_2, \gamma_1^{-1}}] = \overline{A_{12}} = \gamma^2, $$

which implies that $\gamma_1^{-1}\gamma^{-1}\gamma_1 = \gamma$. This means that $\mathbb{Z}/2$ acts trivially on the $\mathbb{Z}/2$ factor and via $-id$ on the $\mathbb{Z}/2m$ factor. This explicit description allows us to compute completely the LCS of $G/A \cong \mathbb{Z}/2 \times (\mathbb{Z}/2m \times \mathbb{Z}/2)$ using Corollary A.8. Precisely, $\Gamma_i(G/A) = 2^{i-1}\mathbb{Z}/(2m)$ for $i \ge 2$. Hence it stops at $\Gamma_3$ if $m$ is odd and at $\Gamma_{v_2(m)+2}$ if $m$ is even. Finally, since $A$ is cyclic of order 2, we can apply Lemma 1.2 (with $l=1$) to conclude that the LCS of $G$ (whence the one of $\mathbf{B}_{1,m}(\mathbb{P}^2)$) stops at $\Gamma_{v_2(m)+2}$ or at $\Gamma_{v_2(m)+3}$ when $m$ is even and at $\Gamma_3$ or $\Gamma_4$ when $m$ is odd. $\square$

**REMARK 6.81.** Similarly to the situation of $\mathbf{B}_{2,m}(\mathbb{S}^2)$ (see Remark 6.74), it seems difficult to decide theoretically between the two possibilities in Proposition 6.80. However, based on experimental calculations [DPS22] using GAP [Gro22] and the package NQ [Nic96], we conjecture that the LCS of $\mathbf{B}_{1,m}(\mathbb{P}^2)$ always stops at $\Gamma_{v_2(m)+3}$ (for both even and odd $m$).

We are left with two cases to consider where there is a block of size 1, namely $\mathbf{B}_{1,1}(\mathbb{P}^2)$ and $\mathbf{B}_{1,2}(\mathbb{P}^2)$. The group $\mathbf{B}_{1,1}(\mathbb{P}^2) = \mathbf{P}_2(\mathbb{P}^2)$ is isomorphic to the quaternion group $Q_8$ (see Corollary 6.82 below), which is 2-nilpotent, so its LCS stops at $\Gamma_3$. Notice that this means that the conclusion of Proposition 6.80 is correct also for $m=1$, since that proposition would assert that the LCS of $\mathbf{B}_{1,1}(\mathbb{P}^2)$ stops at $\Gamma_3$ or $\Gamma_4$. This fact is also compatible with our conjecture in Remark 6.81. In contrast, we will show that the LCS of $\mathbf{B}_{1,2}(\mathbb{P}^2)$ does not stop (Proposition 6.88). The study of this latter case is postponed, and will be part of our study of $\mathbf{B}_{2,m}(\mathbb{P}^2)$, of which a presentation will be computed in Proposition 6.84 for every $m \ge 1$.

6.6.5.2. *Blocks of size 2*. We now study the case where there is at least one block with exactly two strands.

Proposition 6.78 can be used to recover the following classical calculations from [vB66, p. 87]:

**COROLLARY 6.82.** *The pure braid group $\mathbf{P}_2(\mathbb{P}^2)$ is isomorphic to the quaternion group $Q_8$ (which is 2-nilpotent), and $\mathbf{B}_2(\mathbb{P}^2)$ to the dicyclic group of order 16 (which is 3-nilpotent). Precisely, a presentation of the latter is:*

$$ \mathbf{B}_2(\mathbb{P}^2) = \langle \sigma_1, \gamma_1 \mid \gamma_1^2 = [\sigma_1 \gamma_1 \sigma_1^{-1}, \gamma_1^{-1}] = \sigma_1^2 \rangle, $$
where $\sigma_1$ and $\gamma_1$ are the elements of $\mathbf{B}_2(\mathbb{P}^2)$ defined above (Notation 6.76).

**PROOF.** Recall that the dicyclic group of order 16 can be defined by the presentation:

$$ \mathrm{Dic}_{16} = \langle s, x, y \mid sx = ys, x^2 = y^2 = (xy)^2 = s^2 \rangle, $$
and that it contains the subgroup $Q_8$ of quaternions as the index-2 subgroup generated by $x$ and $y$. Notice that, modulo the other relations, $(xy)^2 = s^2$ is equivalent to $s^2x^{-1}yxs^2 = s^2$, which is equivalent to $[y, x^{-1}] = s^2$ by passing to the inverses. We can also use the first relation to eliminate $y = sxs^{-1}$. As a consequence, we also have:

$$ \mathrm{Dic}_{16} = \langle s, x \mid x^2 = [sxs^{-1}, x^{-1}] = s^2 \rangle. $$

It is easy to check that the elements $\sigma_1, \gamma_1$ and $\gamma_2$ of $\mathbf{B}_2(\mathbb{P}^2)$ satisfy the above relations, so that $s \mapsto \sigma_1$ and $x \mapsto \gamma_1$ define a morphism $\varphi$ from $\mathrm{Dic}_{16}$ to $\mathbf{B}_2(\mathbb{P}^2)$. Indeed, the relation $\gamma_1^2 = \sigma_1^2$ is one of the boundary relations from Corollary 6.26, and an isotopy witnessing the last one is drawn in Figure 6.1a. Note that $\varphi$ sends the element $y = sxs^{-1}$ to $\sigma_1\gamma_1\sigma_1^{-1} = \gamma_2$.

By taking $m = n = 1$, the short exact sequence of Proposition 6.78 specialises to:

$$ \pi_1(\mathbb{M}^2)/\xi^2 \cong \mathbb{Z}/4 \longrightarrow \mathbf{P}_2(\mathbb{P}^2) \longrightarrow \pi_1(\mathbb{P}^2) \cong \mathbb{Z}/2. $$

Indeed, $\mathbf{B}_1(\mathbb{M}^2) = \pi_1(\mathbb{M}^2)$ is isomorphic to $\mathbb{Z}$, and the element $\xi$, which is a loop parallel to the boundary of the Möbius strip, is the square of a generator. From this, we deduce that $\mathbf{P}_2(\mathbb{P}^2)$ has eight elements, and that it is generated by $\gamma_1$ (the image of a generator of $\pi_1(\mathbb{M}^2)$) and $\gamma_2$ (a lift of the generator of $\pi_1(\mathbb{P}^2)$). As a consequence, $\varphi$ must induce an isomorphism between $Q_8 = \langle x, y \rangle \subset \mathrm{Dic}_{16}$ and $\mathbf{P}_2(\mathbb{P}^2)$. Then, we can use the usual extension

$$ \mathbf{P}_2(\mathbb{P}^2) \longrightarrow \mathbf{B}_2(\mathbb{P}^2) \longrightarrow \mathfrak{S}_2 $$

to deduce that $\sigma_1, \gamma_1$ and $\gamma_2$ generate $\mathbf{B}_2(\mathbb{P}^2)$, and that $\mathbf{B}_2(\mathbb{P}^2)$ has sixteen elements. Hence $\varphi$ is an isomorphism. $\square$

We first deal with the case where there are at least three blocks:

**PROPOSITION 6.83.** Let $\mu$ be a partition having at least two blocks. Then the LCS of $\mathbf{B}_{2,\mu}(\mathbb{P}^2)$ does not stop.

**PROOF.** We are looking for a quotient whose LCS can be computed, and does not stop. Let us consider the Fadell-Neuwirth extension from Proposition 6.15:

$$ \mathbf{B}_{\mu}(\mathbb{M}^2 - \{pt\}) \longrightarrow \mathbf{B}_{2,\mu}(\mathbb{P}^2) \longrightarrow \mathbf{B}_2(\mathbb{P}^2). $$

We use notations similar to the ones from the proof of Proposition 6.79: we denote the partition $\lambda := (1, 1, \mu)$ of the integer $n$ by $(n_1, \dots, n_l)$, and we denote by $l'$ the number of indices $i$ such that $n_i \ge 2$. We also use Notation 6.76 for braids on the projective plane.

In order to get a more manageable extension, we first take the quotient $G$ of $\mathbf{B}_{2,\mu}(\mathbb{P}^2)$ by $\Gamma_2(\mathbf{B}_{\mu}(\mathbb{M}^2 - \{pt\}))$, getting an extension:

$$ \mathbf{B}_{\mu}(\mathbb{M}^2 - \{pt\})^{ab} \longrightarrow G \longrightarrow \mathbf{B}_2(\mathbb{P}^2). $$

The kernel $\mathbf{B}_{\mu}(\mathbb{M}^2 - \{pt\})^{ab}$ is computed in Proposition 6.47: it is the product $(\pi_1(\mathbb{M}^2 - \{pt\})^{ab})^{l-2} \times (\mathbb{Z}/2)^{l'}$, where the first factor is generated by the classes $c_i$ of $\gamma_{\alpha}$, $a_{1i}$ of $A_{1\alpha}$ and $a_{2i}$ of $A_{2\alpha}$ (for $\alpha$ in the $i$-th block, with $i \ge 3$), subject to the relations $a_{1i} + a_{2i} = 2c_i$ (for each $i$, we get a copy of $\pi_1(\mathbb{M}^2 - \{pt\})^{ab} \cong \mathbb{Z}^2$), and the second one is generated by the classes $s_i$ of the $\sigma_{\alpha}$ (for $\alpha$ and $\alpha + 1$ in the $i$-th
block of $\lambda$, which is possible only if $n_i \ge 2$). The group $G$ is generated by these elements, together with the classes of $\sigma_1$ and $\gamma_1$ (whose images generate $\mathbf{B}_2(\mathbb{P}^2)$).

The elements $s_i$ commute with the other elements of $\mathbf{B}_\mu(\mathbb{M}^2 - \{pt\})^{ab}$ (which is abelian), but also with $\sigma_1$ and $\gamma_1$ (for reasons of support). Hence they are central elements of $G$. Let us denote by $A \cong (\mathbb{Z}/2)^l$ the central subgroup they generate, and let us consider the extension:

$$ \mathbb{Z}^{2(l-2)} \to G/A \to \mathbf{B}_2(\mathbb{P}^2). $$

Corollary 6.82 gives a presentation of the quotient, namely:

$$ \mathbf{B}_2(\mathbb{P}^2) = \langle \sigma_1, \gamma_1 \mid \gamma_1^2 = [\sigma_1 \gamma_1 \sigma_1^{-1}, \gamma_1^{-1}] = \sigma_1^2 \rangle. $$

We now compute a presentation of $G/A$, using the method from Appendix B. Generators are $c_i$, $a_{1i}$ and $a_{2i}$ (for $3 \le i \le l$), together with $\sigma_1$ and $\gamma_1$. Relations defining the kernel are the ones saying that the $c_i$, the $a_{1i}$ and the $a_{2i}$ commute with each other, together with $a_{1i} + a_{2i} = 2c_i$ (for each $i \ge 3$). The latter could be used to eliminate $a_{2i} = 2c_i - a_{1i}$. However, we will choose not to do so here, and to give a redundant, but more tractable presentation of $G/A$. Relations lifting the above presentation of the quotient are:

$$ \left\{ \begin{array}{l} \gamma_1^2 = \sigma_1^2 \cdot a_{13}^{n_3} \cdots a_{1l}^{n_l}, \\ [\sigma_1 \gamma_1 \sigma_1^{-1}, \gamma_1^{-1}] = \sigma_1^2. \end{array} \right. $$

Indeed, these hold in $\mathbf{B}_{2,\mu}(\mathbb{P}^2) \subset \mathbf{B}_n(\mathbb{P}^2)$: the first one is one of the boundary relations from Corollary 6.26 and the second one is the one pictured in Figure 6.1a. Moreover, these are clearly lifts of the relations defining $\mathbf{B}_2(\mathbb{P}^2)$.

We are left with understanding how $\sigma_1$ and $\gamma_1$ (whence also $\gamma_2 = \sigma_1 \gamma_1 \sigma_1^{-1}$) act by conjugation on the $c_i$, the $a_{1i}$ and the $a_{2i}$. We claim that the following relations hold in $G/A$:

$$ \left\{ \begin{array}{l} \sigma_1 a_{1i} = a_{2i} \sigma_1, \quad \sigma_1 a_{2i} = a_{1i} \sigma_1 \quad \text{and} \quad \sigma_1 \stackrel{\cdot}{\rightleftharpoons} c_i \\ \gamma_1 a_{ki} \gamma_1^{-1} = (-1)^{\delta_{1k}} a_{ki} \quad \text{and} \quad \gamma_1 c_i \gamma_1^{-1} = c_i - a_{1i}, \end{array} \right. $$

where we use additive notations in the (abelian) subgroup generated by the $c_i$, the $a_{1i}$ and the $a_{2i}$. These are images of relations holding in $\mathbf{B}_{2,\mu}(\mathbb{P}^2)$, which can be proved by drawing explicit isotopies. Precisely, the first one comes from $\sigma_1 A_{1i} \sigma_1^{-1} = A_{2i}$, the second one from $\sigma_1 A_{2i} \sigma_1^{-1} = A_{2i}^{-1} A_{1i} A_{2i}$ and the third one from the commutation of $\sigma_1$ with all the $\gamma_\alpha$ if $\alpha \ge 3$. The other relation comes from $\gamma_1$ commuting with $A_{2i}$, and from $[\gamma_i, \gamma_1^{-1}] = A_{1i}$ (Figure 6.1a), that is, $\gamma_1^{-1} \gamma_i \gamma_1 = A_{1i}^{-1} \gamma_i$. Notice that the relation involving $\gamma_1 a_{1i} \gamma_1^{-1}$ can be deduced from the other two, using $a_{1i} = 2c_i - a_{2i}$.

We now have a presentation of $G/A$. In order to get a simpler quotient, we quotient further by $\sigma_1^2$ and $\gamma_1^2$. That is, we add the relations $\sigma_1^2 = \gamma_1^2 = 1$ to the previous presentation. The result is a split extension:

$$ \mathbb{Z}^{2(l-2)} / \left( \sum_{i=3}^{l} n_i a_{1i} = \sum_{i=3}^{l} n_i a_{2i} = 0 \right) \rtimes W_2. $$

If we quotient further by $\sum n_i c_i$ (which is central in the above semi-direct product, since it is fixed by the action of $W_2$), we obtain a semi-direct product $M \rtimes W_2$, where $M$ has an explicit workable description. In fact, as a $W_2$-representation, $M \cong \Lambda \otimes B$, where $\Lambda$ is the canonical representation of $W_2$ defined in §A.2.4 and $B$ is the quotient
of $\mathbb{Z}^{l-2}$ by the vector $(n_3, \dots, n_l)$, seen as a trivial $W_2$-representation. Precisely, if $e_i$ ($i=3, \dots, l$) is the generating family of $B$ obtained from the canonical basis of $\mathbb{Z}^2$, an isomorphism $\Lambda \otimes B \cong M$ is given, with the notations from Remark A.15, by $a \otimes e_i \mapsto a_{2i}$, $b \otimes e_i \mapsto a_{1i}$ and $c \otimes e_i \mapsto c_i$ (with notations from Remark A.15).

We finally use our hypothesis: since $l \ge 4$, the rank of $B$ is not 0, so it surjects onto $\mathbb{Z}$. Thus, $M$ surjects onto $\Lambda \otimes \mathbb{Z} = \Lambda$ (as a $W_2$-representation), and $M \times W_2$ surjects onto $\Lambda \times W_2$, whose LCS (computed in Proposition A.12) does not stop. $\square$

Now, we are left with the case where there are precisely two blocks, one of which has exactly two strands. First of all, we give an explicit presentation of the associated group:

**PROPOSITION 6.84.** Let $m \ge 1$ be an integer. The group $\mathbf{B}_{2,m}(\mathbb{P}^2)$ admits the presentation with generators $\sigma_1, \sigma_3, \sigma_4, \dots, \sigma_{m+1}, \gamma_1, \gamma_3$ and $A_{23}$, subject to the following relations:

$$
\left\{
\begin{array}{lcl}
(\mathbb{P}R1) & \sigma_4, \dots, \sigma_{m+1} & \stackrel{\gamma_3, A_{23};}{\rightleftharpoons} \\
(\mathbb{P}R2) & A_{23} & \stackrel{\sigma_3\gamma_3\sigma_3^{-1};}{\rightleftharpoons} \\
(\mathbb{P}R3) & (\sigma_3 A_{23})^2 & = (A_{23}\sigma_3)^2; \\
(\mathbb{P}R4) & [\sigma_i\gamma_i\sigma_i^{-1}, \gamma_i^{-1}] & = \sigma_i^2 \quad \text{for } i \in \{1,3\}; \\
(\mathbb{P}R5) & \gamma_1^2 & = \sigma_1 \left( \prod_{k=2}^{m+1} (\sigma_k \cdots \sigma_3) A_{23}(\sigma_k \cdots \sigma_3)^{-1} \right) \sigma_1; \\
(\mathbb{P}R6) & \sigma_1 & \stackrel{\sigma_3, \sigma_4, \dots, \sigma_{m+1}, \gamma_3;}{\rightleftharpoons} \\
(\mathbb{P}R7) & \gamma_1 & \stackrel{\sigma_3, \sigma_4, \dots, \sigma_{m+1}, A_{23};}{\rightleftharpoons} \\
(\mathbb{P}R8) & \gamma_3^2 & = (\sigma_1^{-1} A_{23}\sigma_1) A_{23} \cdot \prod_{k=3}^{m+1} (\sigma_k \cdots \sigma_4) \sigma_3^2 (\sigma_k \cdots \sigma_4)^{-1}; \\
(\mathbb{P}R9) & [\gamma_3, \gamma_1^{-1}] & = \sigma_1^{-1} A_{23}\sigma_1.
\end{array}
\right.
$$

As above, the choice of names for the generators is coherent with their geometric interpretation (see Notation 6.76 and Figure 6.2b).

**REMARK 6.85.** This set of generators is not minimal, although it is close to being minimal. Indeed, we can eliminate $A_{23}$ using (PR9). The set of generators thus obtained is then minimal, at least for $m=1,2$, since the classes of the generators are a $\mathbb{Z}/2$-basis of the abelianisation in this case. However, this elimination would render the relations much less tractable; the above seems a better compromise between the number of generators and readability of the relations.

**PROOF OF PROPOSITION 6.84.** Let us begin by checking that these relations hold for the usual elements $\sigma_1, \sigma_3, \sigma_4, \dots, \sigma_{m+1}, \gamma_1, \gamma_3$ and $A_{23}$ of $\mathbf{B}_{2,m}(\mathbb{P}^2)$. Let us first remark that we can express other usual elements in terms of these elements. Namely, $A_{13} = \sigma_1^{-1}A_{23}\sigma_1$ and, if $k \in \{3, \dots, m+2\}$, $\gamma_k$ (resp. $A_{2k}, A_{1k}$) is obtained from $\gamma_3$ (resp. $A_{23}, A_{13}$) by conjugation by $\sigma_k \cdots \sigma_4$ on the left (by convention, this product equals 1 if $k=3$). Then one can see that (PR5) and (PR8) are the boundary relations from Corollary 6.26, corresponding to the strands 1 and 3 (recall that $A_{12} = \sigma_1^2$ and that $\sigma_1$ commutes with $\sigma_3, \dots, \sigma_{m+1}$). All the other relations can be checked by drawing explicit isotopies. Precisely, the relations (PR1), (PR2), (PR6) and (PR7) are commutation relations between elements having disjoint support; (PR4) and (PR9) are instances of the relation drawn in Figure 6.1a; (PR3) can either be proved by drawing an explicit isotopy, or it can be deduced from the
similar relation in $\mathbf{B}_{1,2}$ (see Lemma 3.16), by considering the morphism from $\mathbf{B}_{1,2}$ to $\mathbf{B}_{2,m}(\mathbb{P}^2)$ induced by a well-chosen embedding of $\mathbb{D}$ into $\mathbb{P}^2$.

In order to show that these relations describe the group, we now apply the methods of Appendix B to the Fadell-Neuwirth extension from Proposition 6.15:

$$ \mathbf{B}_m(\mathbb{P}^2 - \{2 \text{ pts}\}) \to \mathbf{B}_{2,m}(\mathbb{P}^2) \to \mathbf{B}_2(\mathbb{P}^2). $$

Since $\mathbb{P}^2 - \{2 \text{ pts}\} = \mathcal{N}_{1,2}$, a presentation of the kernel is given by Proposition 6.24, for $n=1$. In order for it to identify with the right subgroup of $\mathbf{B}_{2,m}(\mathbb{P}^2)$ (corresponding to braids on the strands $3, \dots, m+2$), indices are shifted by 2 (so that, for instance, $c_1$ becomes $\gamma_3$), and we take $z_1 = A_{23}$; this last choice changes the presentation a little bit, but one can easily figure out the necessary modifications. Thus, $(BN1)$ and $(BN4)$ give $(PR1)$; $(BN2)$ and $(BN6)$ are empty; $(BN3)$ is the case $i=3$ of $(PR4)$; $(BN5)$ is $(PR2)$; finally, $(BN7)$ is the case $i=3$ of $(PR3)$.

A presentation of the quotient is given by Corollary 6.82 and its proof:

$$ \mathbf{B}_2(\mathbb{P}^2) = \langle \sigma_1, \gamma_1, \gamma_2 \mid \gamma_1^2 = [\sigma_1 \gamma_1 \sigma_1^{-1}, \gamma_1^{-1}] = \sigma_1^2 \rangle. $$

The elements $\gamma_1, \sigma_1 \in \mathbf{B}_{2,m}(\mathbb{P}^2)$ are lifts of the elements $\gamma_1, \sigma_1 \in \mathbf{B}_2(\mathbb{P}^2)$. The second relation holds without change for these lifts, giving the case $i=1$ of $(PR4)$. The relation $\gamma_1^2 = \sigma_1^2$ lifts to the boundary relation associated with the first strand, which is $(PR5)$.

We are left with finding relations describing the action of $\sigma_1$ and $\gamma_1$ (or of $\sigma_1^{-1}$ and $\gamma_1^{-1}$ – see Remark B.4) by conjugation on the other generators. The commutation relations $(PR6)$ and $(PR7)$ describe most of this action. Only two elements still need to be expressed in terms of the generators $\sigma_3, \dots, \sigma_{m+1}, \gamma_3$ and $A_{23}$ of the kernel, namely $\sigma_1^{-1}A_{23}\sigma_1$ and $\gamma_1^{-1}\gamma_3\gamma_1$. The boundary relation $(PR8)$ deals with $\sigma_1^{-1}A_{23}\sigma_1$. Finally, $(PR9)$ deals with $\gamma_1^{-1}\gamma_3\gamma_1$: it says that $\gamma_1^{-1}\gamma_3\gamma_1 = \gamma_3\sigma_1^{-1}A_{23}\sigma_1$, and the right-hand side can be expressed in terms of the generators of the kernel using the previous relation $(PR8)$. This finishes the proof that the above relations are the ones obtained using the method of Appendix B, whence the result. $\square$

Let us now consider the case $m=1$. In this case, there is no $\sigma_i$ for $i \ge 3$, so the presentation is much simpler; in fact, in the extension of the proof, the kernel is just $\pi_1(\mathbb{P}^2 - \{2 \text{ pts}\})$, which is free on $\gamma_3$ and $A_{23}$. Thus, $(PR1)$, $(PR2)$ and $(PR3)$ are empty in this case, and there is no case $i=3$ in $(PR4)$. The other relations reduce to:

COROLLARY 6.86. The group $\mathbf{B}_{2,1}(\mathbb{P}^2)$ has a presentation with 4 generators $\sigma_1$, $\gamma_1$, $\gamma_3$ and $A_{23}$ and 6 relations (indexed as above):

$$ \left\{ \begin{array}{l} (4) \quad [\sigma_1 \gamma_1 \sigma_1^{-1}, \gamma_1^{-1}] = \sigma_1^2; \\ (5) \quad \gamma_1^2 = \sigma_1 A_{23} \sigma_1; \\ (6) \quad \sigma_1 \stackrel{\leftrightarrow}{\gamma_3}; \end{array} \right. $$

$$ \left\{ \begin{array}{l} (7) \quad \gamma_1 \stackrel{\leftrightarrow}{\to} A_{23}; \\ (8) \quad \gamma_3^2 = \sigma_1^{-1} A_{23} \sigma_1 A_{23}; \\ (9) \quad [\gamma_3, \gamma_1^{-1}] = \sigma_1^{-1} A_{23} \sigma_1. \end{array} \right. $$

REMARK 6.87. This presentation is smaller than van Buskirk’s presentation from [vB66, Lem. p. 84], which has 6 generators and 13 relations. Note that in our language, his generators are as follows, where the right-hand side of each equation
uses our notation and the left-hand side uses his:

$$
\left\{
\begin{array}{l}
\sigma_2 = \sigma_1, \\
a_2 = \sigma_1^{-1} A_{23} \sigma_1, \\
a_3 = A_{23},
\end{array}
\right.
\qquad
\left\{
\begin{array}{l}
\rho_1 = \gamma_3, \\
\rho_2 = \gamma_1 \sigma_1^{-1} A_{23}^{-1} \sigma_1, \\
\rho_3 = \sigma_1^{-1} \gamma_1 \sigma_1.
\end{array}
\right.
$$

PROPOSITION 6.88. *The LCS of* $\mathbf{B}_{2,1}(\mathbb{P}^2)$ *does not stop.*

PROOF. Let $G$ be the quotient of $\mathbf{B}_{2,1}(\mathbb{P}^2)$ by the single relation $\sigma_1\gamma_1\sigma_1^{-1} = \gamma_1^{-1}$. Let us consider the presentation of $G$ given by the presentation of Corollary 6.86, to which this relation is added. Then relation (4) becomes $\sigma_1^2 = 1$, that is, $\sigma_1^{-1} = \sigma_1$. Relation (5) becomes $A_{23} = \sigma_1\gamma_1^2\sigma_1 = \gamma_1^{-2}$, hence (7) becomes redundant. Relation (8) is then equivalent to $\gamma_3^2 = 1$. Relation (9) becomes $\gamma_3\gamma_1^{-1}\gamma_3^{-1} = \gamma_1$. If we add to this relation (6), which says that $\sigma_1$ commutes with $\gamma_3$, we get a presentation of $\mathbb{Z} \times (\mathbb{Z}/2)^2$ (where both elements of a basis of $(\mathbb{Z}/2)^2$ act via $-id$). Thus, $G \cong \mathbb{Z} \times (\mathbb{Z}/2)^2$, whose LCS does not stop (one can either compute it with the method of Appendix A, or quotient further by $\sigma_1 = \gamma_3$ to get $\mathbb{Z} \times (\mathbb{Z}/2)$ as a quotient and apply Corollary A.8). $\square$

REMARK 6.89. Even if the imposed relation looks very much like a relation defining the infinite dihedral group $\mathbb{Z} \times (\mathbb{Z}/2)$, it is not at all clear a priori why adding this one relation should work. Much experimentation has been needed before ending up here.

Next, let us consider the case $m = 2$. In this case, there is no $\sigma_i$ for $i \ge 4$. Thus, $(\mathbb{P}R1)$ is empty, and the boundary relations $(\mathbb{P}R5)$ and $(\mathbb{P}R8)$ become much simpler. Let us spell out the result in this case:

COROLLARY 6.90. *The group* $\mathbf{B}_{2,2}(\mathbb{P}^2)$ *has a presentation with 5 generators* $\sigma_1$, $\gamma_1$, $\gamma_3$ *and* $A_{23}$ *and 8 relations (indexed as above):

$$
\left\{
\begin{array}{l}
(2) \quad A_{23} \stackrel{\leftarrow}{\leftrightarrow} \sigma_3 \gamma_3 \sigma_3^{-1}; \\
(3) \quad (\sigma_3 A_{23})^2 = (A_{23} \sigma_3)^2; \\
(4) \quad [\sigma_i \gamma_i \sigma_i^{-1}, \gamma_i^{-1}] = \sigma_i^2 \quad \text{for } i \in \{1, 3\}; \\
(5) \quad \gamma_1^2 = \sigma_1 A_{23} \sigma_3 A_{23} \sigma_3^{-1} \sigma_1;
\end{array}
\right.
\qquad
\left\{
\begin{array}{l}
(6) \quad \sigma_1 \stackrel{\leftarrow}{\leftrightarrow} \sigma_3, \gamma_3; \\
(7) \quad \gamma_1 \stackrel{\leftarrow}{\leftrightarrow} \sigma_3, A_{23}; \\
(8) \quad \gamma_3^2 = \sigma_1^{-1} A_{23} \sigma_1 A_{23} \sigma_3^2; \\
(9) \quad [\gamma_3, \gamma_1^{-1}] = \sigma_1^{-1} A_{23} \sigma_1.
\end{array}
\right.
$$

PROPOSITION 6.91. *The LCS of* $\mathbf{B}_{2,2}(\mathbb{P}^2)$ *does not stop.*

PROOF. Let us consider, as above, the projection $p: \mathbf{B}_{2,2}(\mathbb{P}^2) \to \mathbf{B}_{2}(\mathbb{P}^2) \cong \mathrm{Dic}_{16}$ induced by forgetting the last two strands. Recall that Corollary 6.82 gives a presentation of this quotient:

$$
\mathbf{B}_2(\mathbb{P}^2) = \langle \sigma_1, \gamma_1 \mid \gamma_1^2 = [\sigma_1 \gamma_1 \sigma_1^{-1}, \gamma_1^{-1}] = \sigma_1^2 \rangle.
$$

Since the second relation is already true in $\mathbf{B}_{2,2}(\mathbb{P}^2)$ (it is the case $i=1$ of relation (4) in Corollary 6.90), the projection $p$ becomes split if we impose the relation $\sigma_1^2 = \gamma_1^2$. We will in fact consider the quotient $G$ of $\mathbf{B}_{2,2}(\mathbb{P}^2)$ by the two relations:

$$
\left\{
\begin{array}{l}
(Q1) \quad \sigma_1^2 = \gamma_1^2; \\
(Q2) \quad \sigma_3 \gamma_3 \sigma_3^{-1} = \gamma_3^{-1}.
\end{array}
\right.
$$

Since the second relator also sits in the kernel of $p$, we get an induced split projection
$\bar{p}: G \to \mathbf{B}_2(\mathbb{P}^2)$.
Let us consider the presentation of $G$ given by the presentation of Corol-
lary 6.90, together with the two relations $(Q1)$ and $(Q2)$. Modulo $(Q1)$, relation
(5) becomes $\sigma_3 A_{23} \sigma_3^{-1} = A_{23}^{-1}$. Modulo $(Q2)$, relation (2) says that $A_{23}$ commutes
with $\gamma_3$, and the case $i=3$ of relation (4) gives $\sigma_3^2 = 1$. At this point, let us remark
that the relations obtained so far say that the subgroup $\langle \gamma_3, A_{23}, \sigma_3 \rangle$ is a quotient
of $\mathbb{Z}^2 \times (\mathbb{Z}/2)$, where the action of $\mathbb{Z}/2$ is by $-id$.

Continuing our investigation, we remark that relation (3) is a consequence
of the previous relation (both sides of it are killed modulo these). Relation (8)
becomes $\sigma_1^{-1}A_{23}\sigma_1 = \gamma_3^2 A_{23}^{-1}$ and (remembering that $\sigma_3^2 = 1$) relation (9) becomes
$\gamma_1^{-1}\gamma_3\gamma_1 = A_{23}\gamma_3^{-1}$. If we add relations (6) and (7) without change, we get a
presentation of $G$, which will allow us to describe it in an explicit way.

Let us first consider the subgroup $A := \langle \gamma_3, A_{23} \rangle$ (which is abelian by relation (2)). Relations (Q2) and (2) imply that it is stable under conjugation by $\sigma_3$ (which equals $\sigma^{-1}$, by (4)). Relations (6) and (8) imply that it is stable under conjugation by $\sigma_i^{-1}$, and also by $\sigma_i$ (as one sees by conjugating (8) by $\sigma_i$, taking (6) into account). In the same way, relations (7) and (9) imply that it is stable under conjugation by $\gamma_1^{\pm 1}$. Finally, all this implies that it is normal in $G$. Moreover, the presentation of $G/A$ that we get from the one of $G$ clearly gives $G/A \cong \text{Dic}_{16} \times (\mathbb{Z}/2)$ and, in fact, the relations defining $G/A$ were already true in $G$ for the generators $\sigma_1, \gamma_1$ and $\sigma_3$, so $G \to G/A$ splits, and finally:

$$
G \cong A \times (\mathrm{Dic}_{16} \times (\mathbb{Z}/2)).
$$

We are left with understanding $A$ (which is a quotient of $\mathbb{Z}^2$) and the action of $K := \mathrm{Dic}_{16} \times (\mathbb{Z}/2)$ on it. In order to do this, let us consider the relations describing the action of $K$ on $A$, namely:

$$
\left\{
\begin{array}{l}
(Q2) \quad \sigma_3 \gamma_3 \sigma_3^{-1} = \gamma_3^{-1}; \\
(5) \quad \sigma_3 A_{23} \sigma_3^{-1} = A_{23}^{-1}; \\
(6) \quad \sigma_1 \gamma_3 \sigma_1^{-1} = \gamma_3; \\
(8) \quad \sigma_1 A_{23} \sigma_1^{-1} = A_{23}^{-1} \gamma_3^2; \\
(9) \quad \gamma_1 \gamma_3 \gamma_1^{-1} = \gamma_3^{-1} A_{23}; \\
(7) \quad \gamma_1 A_{23} \gamma_1^{-1} = A_{23}.
\end{array}
\right.
$$

We remark that these already define an action of $K$ on $\mathbb{Z}^2$, which is exactly the
action on $\mathbb{Z}^2 = \Gamma$ considered at the end of §A.2.4. Precisely, with the identifications
$\sigma_1 \mapsto \sigma$, $\gamma_1 \mapsto \gamma$ and $\sigma_3 \mapsto \tau$ for generators of $K$, we get an equivariant map $\Gamma \to A$
sending $a$ to $A_{23}$ and $c$ to $\gamma_3$ (with the notations of Remark A.15). This induces a
surjective morphism from $\Gamma \times_K \text{ onto } A \times_K G$. It is then easy to check that all
the relations defining $G$ are in fact already true in $\Gamma \times_K \text{, which allows us to define}$
a converse isomorphism } G \cong \mathbb{Z}^2 \times_K \text{.}
$$

Finally, Proposition A.12, together with the equality $\Gamma_*^K(\Lambda) = \Gamma_*^{W_2}(\Lambda)$ from the end of §A.2.4, gives us a complete description of the LCS of $G$ which, in particular, does not stop. $\square$

REMARK 6.92. The projection onto $\mathrm{Dic}_{16} \times (\mathbb{Z}/2)$ in the proof can be seen
as coming from the geometry. Precisely, it is the factorisation through $G$ of the
projection

$$
q: \mathbf{B}_{2,2}(\mathbb{P}^2) \to \mathbf{B}_2(\mathbb{P}^2) \times \mathfrak{S}_2 \cong \mathrm{Dic}_{16} \times (\mathbb{Z}/2)
$$
whose first factor forgets the last two strands and whose second factor forgets the first two strands and then applies the usual projection $\pi: \mathbf{B}_2(\mathbb{P}^2) \to \mathcal{S}_2$.

REMARK 6.93. This quotient looks very much like the one from the proof of Proposition 6.88, and the same remark applies (see Remark 6.89). Namely, (Q1) is a natural relation to impose (making the extension split), whereas it is much less clear why quotienting by (Q2) (which is the same relation as in the aforementioned proof, up to re-indexing the strands) should work.

We may now complete the proof of Theorem 6.77.

PROOF OF THEOREM 6.77. The first two statements are part of the general results of Corollary 6.51 and Theorem 6.52, except for $\mathbf{B}_1(\mathbb{P}^2) \cong \mathbb{Z}/2$ and $\mathbf{B}_2(\mathbb{P}^2)$, which is the dicyclic group of order 16 (Corollary 6.82). The third statement combines Propositions 6.79 (if $\lambda$ has blocks of size 1) and 6.83 (if $\lambda$ has blocks of size 2). The fourth statement combines Propositions 6.80, 6.88 and 6.91, together with the fact that $\mathbf{B}_{1,1}(\mathbb{P}^2) = \mathbf{P}_2(\mathbb{P}^2)$ is the quaternion group $Q_8$ (Corollary 6.82). $\square$

The remaining cases to consider are $\mathbf{B}_{2,m}(\mathbb{P}^2)$ for $m \ge 3$. These are the only examples of partitioned surface braid groups for which we have not been able to answer the question of whether their LCS stop. We can still say something about these LCS, using the proof of Proposition 6.83. Recall that in this proof, the hypothesis on the number of blocks of $\mu$ was not used until the end. Moreover, in the case $\mu = (m)$ and $m \ge 3$, Proposition 6.42 applies, implying that the first quotient $G$ is the quotient by $\Gamma_\infty(\mathbf{B}_m(\mathbb{M}^2 - \{pt\}))$. The latter must be contained in $\Gamma_\infty(\mathbf{B}_{2,m}(\mathbb{P}^2))$ so, in order to understand the LCS of $\mathbf{B}_{2,m}(\mathbb{P}^2)$, we only need to understand the LCS of $G$. Then, since the central subgroup $A = \langle s_3 \rangle$ (where $s_3$ is the class of $\sigma_3$) is cyclic of order 2, we can apply Lemma 1.2 to see that the LCS of $G$ stops if and only if the one of $G/A$ does (with possibly one more step). We have the same presentation of $G/A$ as in the proof of Proposition 6.83, from its decomposition as an extension of $\mathbf{B}_2(\mathbb{P}^2) = \mathrm{Dic}_{16}$ by $\mathbb{Z}^2$. Note that the action of $\mathrm{Dic}_{16}$ on the (abelian) kernel in this extension is exactly the one on $\Lambda$ from §A.2.4, which is through the quotient $\mathrm{Dic}_{16} \to W_2$. Precisely, as in the previous proof, $a = a_{23}$ and $c = c_3$ identify with the basis of $\Lambda$ from Remark A.15. However, this extension is not split, so computing its LCS seems tricky. We can try to make it split, by considering the quotient by the relation $\sigma_1^2 = \gamma_1^2$ (which is equivalent to $a^m = 1$), but then we also kill $c^{2m}$, getting the finite quotient:

$$ (G/A)/\sigma_1^2 \gamma_1^{-2} \cong (\Lambda/\langle ma, 2mc \rangle) \times \mathrm{Dic}_{16} \cong (\mathbb{Z}/m \times \mathbb{Z}/2m) \times \mathrm{Dic}_{16}. $$

In fact, using the notation of Proposition A.12, we have $\langle ma, 2mc \rangle = mV \subset \Lambda$. Thus we can deduce from Proposition A.12 a computation of

$$ \Gamma_*^{\mathrm{Dic}_{16}}(\mathbb{Z}/m \times \mathbb{Z}/2m) = \Gamma_*^{W_2}(\Lambda/mV). $$

Namely, if $m = 2^\nu m'$, with $m'$ odd, we have that $2^\nu V$ contains $mV$, and that $2^{\nu+1}V$ equals $2^\nu V$ modulo $mV$, so $\Gamma_*^{W_2}(\Lambda/mV)$ stops at $\Gamma_{2\nu+2}^{W_2}(\Lambda/mV) = 2^\nu V/mV$. Finally, the LCS of the above quotient stops at $\Gamma_k$, where $k = 2v_2(m) + 2$ if $m$ is even and $k = 4$ is $m$ is odd (in the latter case, note that the relative LCS stops at the second step, but $\mathrm{Dic}_{16}$ is 3-nilpotent).

This gives a lower bound for the step at which the LCS of $\mathbf{B}_{2,m}(\mathbb{P}^2)$ stops: it cannot stop before $\Gamma_k$ for $k = \max\{4, 2v_2(m) + 2\}$. However, this lower bound is far from optimal: our experimental calculations [DPS22] using GAP [Gro22] and
the package NQ [Nic96] show that, for all $m \le 1024$, the LCS of $G/A$, and hence those of $G$ and of $\mathbf{B}_{2,m}(\mathbb{P}^2)$, do not stop before $\Gamma_{100}$ (we also verified this for all $m = 2^\nu$ with $\nu \le 23$). We thus conjecture:

CONJECTURE 6.94. If $m \ge 3$, the LCS of $\mathbf{B}_{2,m}(\mathbb{P}^2)$ does not stop.
<figure><img src="image_38.png" /></figure>
# APPENDIX A

## Some calculations of lower central series

This appendix is devoted to computing the LCS of some combinatorially-defined groups. These include notably the Klein group $Z \times Z$, the free products $Z/2 * Z/2$ and $Z * Z/2$, the Artin group of type $B_2$ and wreath products. Our main tool here is the decomposition of the LCS of a semi-direct product into a semi-direct product of filtrations, which we recall first.

### A.1. Relative lower central series

In order to obtain actual computations, we need to recall some material from [Dar21, §3] about the LCS of a semi-direct product.

**DEFINITION A.1.** Let $G$ be a group, of which $H$ is a normal subgroup. We define the relative lower central series $\Gamma_*^G(H)$ by:

$$ \begin{cases} \Gamma_1^G(H) := H, \\ \Gamma_{k+1}^G(H) := [G, \Gamma_k^G(H)]. \end{cases} $$

If $G$ is the semi-direct product of $H$ with a group $K$, we write $\Gamma_*^K(H)$ for $\Gamma_{H \times K}^*(H)$ (which does not cause any confusion: if $H$ is a normal subgroup of a group $G$, then $\Gamma_*^G(H) = \Gamma_{H \times G}^*(H)$, for the semi-direct product associated to the conjugation action of $G$ on $H$). It was shown in [Dar21] that in this case:

$$ \Gamma_*(H \times K) = \Gamma_*^K(H) \times \Gamma_*(K). $$

Moreover, the filtration $\Gamma_*^K(H) = H \cap \Gamma_*(H \times K)$ does have the property that $[\Gamma_i^K(H), \Gamma_j^K(H)] \subseteq \Gamma_{i+j}^K(H)$ (for all $i, j \ge 1$), which allows one to define an associated graded Lie ring $\mathcal{L}^K(H)$ (with brackets induced by commutators, as in §1.2). Then, the Lie ring of $H \times K$ decomposes into a semi-direct product of Lie rings:

$$ \mathcal{L}(H \times K) = \mathcal{L}^K(H) \times \mathcal{L}(K). $$

This is in fact a generalisation of Lemma 1.12, which is the degree-one part (one can check that $\mathcal{L}_1^K(H) = (H^{ab})_K$).

We can devise an analogue of Lemma 2.6 in this context, which gives a criterion for the relative LCS to stop:

**LEMMA A.2.** Let a group $K$ act on a group $H$. Let the set $S_K$ generate $K^{ab}$ and let the set $S_H$ generate $(H^{ab})_K$. Suppose that, for each pair $(s, t) \in S_H^2$ (resp. each pair $(s, t) \in S_H \times S_K$), we can find representatives $\tilde{s}, \tilde{t} \in H$ (resp. $\tilde{s} \in H$ and $\tilde{t} \in K$) of $s$ and $t$ such that $\tilde{s}$ and $\tilde{t}$ commute in $H$ (resp. in $H \times K$). Then $\Gamma_2^K(H) = \Gamma_3^K(H)$, which means that $\Gamma_*^K(H)$ stops at $\Gamma_2^K(H)$, and:

$$ \mathcal{L}(H \times K) \cong (H^{ab})_K \times \mathcal{L}(K), $$

where the first factor is concentrated in degree one.
PROOF. On the one hand, by definition of the relative LCS, an element of $\mathcal{L}_2^K(H)$ is a sum of brackets in $\mathcal{L}(G \times K)$, either of two elements of $\mathcal{L}_1^K(H)$, or of an element of $\mathcal{L}_1(K)$ with an element of $\mathcal{L}_1^K(H)$. On the other hand, the relation $[\tilde{s}, \tilde{t}] = 1$ in $H \times K$ readily implies that $[s, t] = 0$ in $\mathcal{L}(H \times K) \cong \mathcal{L}^K(H) \times \mathcal{L}(K)$. Since $S_H$ linearly generates $\mathcal{L}_1^K(H) = (H^{ab})_K$ and $S_K$ linearly generates $\mathcal{L}_1(K) = K^{ab}$, we infer that under our hypothesis, all elements of $\mathcal{L}_2^K(H)$ are trivial, which means that $\Gamma_2^K(H) = \Gamma_3^K(H)$. Moreover, from the definition of $\Gamma_*^K(H)$, this obviously implies that $\Gamma_i^K(H) = \Gamma_{i+1}^K(H)$ for all $i \ge 2$. The statement about Lie rings is then just a reformulation of the decomposition $\mathcal{L}(H \times K) \cong \mathcal{L}^K(H) \times \mathcal{L}(K)$ taking into account these conclusions. $\square$

## A.2. Semi-direct products of abelian groups

Let a group $G$ act on an abelian group $A$. Then the LCS of $A \times G$ can be computed using linear algebra. Indeed, we have:

$$ \Gamma_{k+1}^G(A) := [A \times G, \Gamma_k^G(A)] = [G, \Gamma_k^G(A)]. $$

These are commutators in $A \times G$, given, for $g \in G$ and $a \in A$, by:

$$ [g, a] = g \cdot a - a = (g - id)(a). $$

As a consequence, $\Gamma_{k+1}^G(A)$ is the subgroup of $A$ generated by the $(g - id)(\Gamma_k^G(A))$ (for $g \in G$), which can be computed by studying the endomorphisms $g - id$ of $A$.

We now study several instances of this situation. We begin by computing the LCS of the Klein group $Z \times Z$ (which is the fundamental group of the Klein bottle). We then generalise this calculation to any semi-direct product of an abelian group by $Z$ acting by $-id$. This can be generalised again, to the case of an action of $Z$ by an involution. Finally, we compute the LCS of $\Lambda \times W_2$, where $W_2 \cong (Z/2) \rtimes S_2$ is the Coxeter group of type $B_2$ (or $C_2$), acting on the lattice $\Lambda$ generated by its root system.

### A.2.1. The Klein group.
There are two distinct automorphisms of $Z$ (that is, $\pm id$), whence only one non-trivial action of $Z$ on $Z$. Thus the following definition makes sense:

**DEFINITION A.3.** The Klein group is the semi-direct product $K = Z \times Z$.

Let us denote by $x$ (resp. by $t$) the element $(1, 0)$ (resp. $(0, 1)$) of $Z \times Z$. A presentation of $K$ is given by

$$ K = \langle x, t \mid txt^{-1} = x^{-1} \rangle. $$

The LCS of $K$ decomposes as $\Gamma_*(K) = \Gamma_*^{\mathbb{Z}}(\mathbb{Z}) \times \Gamma_*(\mathbb{Z})$. Thus, in order to understand it, we need to understand the filtration $\Gamma_*^{\mathbb{Z}}(\mathbb{Z})$.

**PROPOSITION A.4.** The LCS of the Klein group is $\Gamma_i(\mathbb{Z} \times \mathbb{Z}) = (2^{i-1}\mathbb{Z}) \times \{1\}$ for $i \ge 2$. In other words, $\Gamma_i^{\mathbb{Z}}(\mathbb{Z}) = 2^{i-1}\mathbb{Z}$. In particular, $\mathbb{Z} \times \mathbb{Z}$ is residually nilpotent.

**PROOF.** This follows from the formula $[x^{2^j}, t] = x^{2^j}(tx^{-2^j}t^{-1}) = x^{2^{j+1}}$, by induction on $i$. $\square$

**COROLLARY A.5.** The (graded) Lie ring of the Klein group identifies with $(\mathbb{Z}/2)[X] \times \mathbb{Z}$, where the polynomial ring $(\mathbb{Z}/2)[X]$ is seen as a graded abelian Lie ring (where $X^i$ is of degree $i$), and the generator $T$ of $\mathbb{Z}$ (of degree 1) acts via $[X^i, T] = X^{i+1}$.
PROOF. From Proposition A.4, we get a decomposition $\mathcal{L}(K) = \mathcal{L}(2^{*-1}\mathbb{Z}) \rtimes \mathcal{L}(\mathbb{Z})$. Since $\mathbb{Z}$ is abelian, the two factors are abelian Lie rings. The result follows, by calling $X^i$ the class of $x^{2^{i-1}}$ and $T$ the class of $t$. The formula for brackets comes from $[x^{2^j}, t] = x^{2^{j+1}}$. $\square$

**A.2.2. Generalised Klein groups.** Let $A$ be any abelian group and let $\mathbb{Z}$ act on $A$ via the powers of $-id_A$. The corresponding semi-direct product $K_A = A \rtimes \mathbb{Z}$ is a generalisation of the Klein group $K = K_{\mathbb{Z}}$. We can generalise the above results to this context:

PROPOSITION A.6. The LCS of $K_A$ is given by $\Gamma_i(A \rtimes \mathbb{Z}) = (2^{i-1}A) \rtimes \{1\}$ for $i \ge 2$. In other words, for all $i \ge 2$, $\Gamma_i^{\mathbb{Z}}(A) = 2^{i-1}A$. In particular, for any free abelian group $A$, $K_A$ is residually nilpotent.

PROOF. Let $t$ denote the generator of $\mathbb{Z}$. For all $a$ in $A$, we have $[a, t] = a - t \cdot a = 2a$ in $A \rtimes \mathbb{Z}$. Hence $[2^j A, t] = 2^{j+1} A$, from which the calculation of the LCS follows. Then $K_A$ is residually nilpotent if and only if the intersection of the $2^j A$ is trivial, which is true for instance when $A$ is finitely generated or when $A$ is free abelian. $\square$

COROLLARY A.7. Let us consider the graded abelian Lie ring $\mathcal{L}(2^{*-1}A) = \bigoplus 2^{i-1}A/2^i A$ (where the sum is taken over $i \ge 1$). The (graded) Lie ring of $K_A$ identifies with $\mathcal{L}(2^*A) \rtimes \mathbb{Z}$, where the generator $T$ of $\mathbb{Z}$ acts via the degree-one map induced by $a \mapsto 2a$.

PROOF. From Proposition A.6, we get a decomposition $\mathcal{L}(K) = \mathcal{L}(2^{*-1}A) \rtimes \mathcal{L}(\mathbb{Z})$. Since $\mathbb{Z}$ and $A$ are abelian, the two factors are abelian Lie rings. The result follows, since brackets with $T$ come from commutators with $t$, given by $[a, t] = 2a$ in $A \rtimes \mathbb{Z}$. $\square$

Since $t^2$ acts trivially on $A$, it is a central element of $K_A$ (in fact, one easily sees that it generates the centre of $K_A$ if $A$ is not trivial). Thus we can consider $A \rtimes (\mathbb{Z}/2)$ (where $\mathbb{Z}/2$ acts on $A$ via $-id_A$) as a quotient of $K_A$, which behaves in much the same way:

COROLLARY A.8. Consider the group $A \rtimes (\mathbb{Z}/2)$, where $\mathbb{Z}/2$ acts on the abelian group $A$ via $-id_A$. We have $\Gamma_i(A \rtimes (\mathbb{Z}/2)) = (2^{i-1}A) \rtimes \{1\}$ for all $i \ge 2$. In particular, for any finitely generated abelian group $A$, $A \rtimes (\mathbb{Z}/2)$ is residually nilpotent. Moreover, the (graded) Lie ring of this group identifies with $\mathcal{L}(2^{*-1}A) \rtimes (\mathbb{Z}/2)$, where the generator $T$ of $\mathbb{Z}/2$ acts via the degree-one map induced by $a \mapsto 2a$.

Finally, let us spell out the particular case where $A \cong \mathbb{Z}^n$ is free abelian on some basis $x_1, \dots, x_n$. We then denote $K_A$ by $K_n$, for short.

COROLLARY A.9. The (graded) Lie ring of $K_n$ identifies with $(\mathbb{Z}/2)^n[X] \rtimes \mathbb{Z}$, where the polynomial ring $(\mathbb{Z}/2)^n[X]$ is seen as an abelian Lie ring (where $X^i$ is of degree $i$), and the generator $T$ of $\mathbb{Z}$ (of degree 1) acts via $[v \cdot X^i, T] = v \cdot X^{i+1}$, for $v \in (\mathbb{Z}/2)^n$.

PROOF. This is just a matter of identifying $\mathcal{L}(2^{*-1}\mathbb{Z}^n)$ with $(\mathbb{Z}/2)^n[X]$, by calling $v \cdot X^i$ the class of the sum of the $2^{i-1}x_k$ such that $v_k = 1$. $\square$
**A.2.3. A further generalisation.** Let $A$ be an abelian group and let $\mathbb{Z}$ act on $A$ via the powers of some involution $\tau$ (for instance, $\tau$ could exchange two isomorphic direct factors of $A$). Let us denote by $K_\tau$ the corresponding semi-direct product $A \times \mathbb{Z}$. We have:

$$((\tau - 1) + 2)(\tau - 1) = \tau^2 - 1 = 0,$$

which means that $\tau - 1$ acts via multiplication by $-2$ on $V := \text{Im}(\tau - 1)$.

**PROPOSITION A.10.** The LCS of $K_\tau$ is given by $\Gamma_i(A \times \mathbb{Z}) = (2^{i-2}V) \times \{1\}$ for $i \ge 2$. In other words, for all $i \ge 2$, $\Gamma_i^{\mathbb{Z}}(A) = 2^{i-2}V$. In particular, for any free abelian group $A$, $K_\tau$ is residually nilpotent.

**PROOF.** For all $a \in A$, we have $[a, t] = a - \tau(a) = (1 - \tau)(a)$. This implies that $\Gamma_2^{\mathbb{Z}}(A) = \text{Im}(\tau - 1) = V$. Then for all $v$ in $V$, we have $[v, t] = v - \tau(v) = 2v$ in $A \times \mathbb{Z}$, and the rest of the proof is similar to that of Proposition A.6. $\square$

Corollary A.7 generalises immediately to this context:

**COROLLARY A.11.** The (graded) Lie ring of $K_\tau$ identifies with $(A/V \oplus V/2V \oplus 2V/4V \oplus \cdots) \times \mathbb{Z}$, where $A/V$ and $\mathbb{Z}$ are in degree 1, $2^{i-2}V/2^{i-1}V$ is in degree $i$ and the generator $T$ of $\mathbb{Z}$ acts via the degree-one map induced by $1 - \tau$ (which coincides with $v \mapsto 2v$ on $V$).

The reader can also easily write a generalisation of Corollary A.8 to this context, by factoring the action of $\mathbb{Z}$ through $\mathbb{Z}/2$.

**A.2.4. More actions on abelian groups.** Let us consider the group defined by the presentation $\langle\sigma, \gamma | \sigma^2 = \gamma^2 = (\sigma\gamma)^4 = 1\rangle$, which is the Coxeter group of type $B_2$, also denoted by $W_2 = (\mathbb{Z}/2) \rtimes \mathfrak{S}_2$ in the rest of the memoir. It acts on $\mathbb{R}^2$ in the usual way: $\gamma$ acts by $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ and $\sigma$ by $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. This action preserves the lattice $\Lambda := \mathbb{Z} \cdot (0, 1) \oplus \mathbb{Z} \cdot (\frac{1}{2}, \frac{1}{2})$ (which is generated by roots). It also preserves the lattice $V = \mathbb{Z}^2$, which is of index 2 in $\Lambda$.

**PROPOSITION A.12.** The filtration $\Gamma_*^{W_2}(\Lambda)$ on $\Lambda$ is given by:

$$\Lambda \supset V \supset 2\Lambda \supset 2V \supset 4\Lambda \supset \cdots.$$

In particular, $\Lambda \times W_2$ is residually nilpotent, but not nilpotent.

**PROOF.** One can easily write down explicitly the eight matrices for the actions of elements of $W_2$ (which are the invertible monomial matrices in $\text{GL}_2(\mathbb{Z})$). Recall that, for every $g \in W_2$, $g - \text{id}$ is the commutator by $g$ in $\Lambda \times W_2$. It is then easy to check that the $g - \text{id}$ send $\Lambda$ to $V$ (resp. $V$ to $2\Lambda$) and that the $(g - \text{id})(\Lambda)$ (resp. the $(g - \text{id})(V)$) generate $V$ (resp. $2\Lambda$), whence the result. $\square$

**REMARK A.13.** One can look at the proof in a geometric way, by seeing each element $g \cdot v - v$ as the difference between two vertices of a square centred at 0.

**REMARK A.14.** One can compute completely the associated Lie ring, which is a semi-direct product of the abelian Lie ring $(\mathbb{Z}/2)[X]$ by the mod 2 Heisenberg Lie ring $\mathcal{L}(W_2) \cong \mathfrak{n}_3(\mathbb{Z}/2)$.

**REMARK A.15.** Let us use the notations $a := (0, 1)$, $b := (1, 0)$ and $c := (\frac{1}{2}, \frac{1}{2})$. Then $\Lambda$ can be described abstractly as the abelian group generated by $a$, $b$ and $c$, modulo the relation $a + b = 2c$. Moreover, the action of $\sigma$ fixes $c$ and exchanges
a and b, and the action of $\gamma$ is via $a \mapsto a$, $b \mapsto -b$ and $c \mapsto c - b$. In particular, since $b = 2c - a$ and $c - b = a - c$, in the basis $(a, c)$ of $\Lambda$, $\sigma$ acts by $\begin{pmatrix} -1 & 0 \\ 2 & 1 \end{pmatrix}$ and $\gamma$ by $\begin{pmatrix} 1 & -1 \\ 0 & -1 \end{pmatrix}$. Notice that, in this basis, $V$ is just the subgroup of elements whose second coordinate is even.

In the course of the proof of Proposition 6.91, we encounter a slight variation on the above action of $W_2$ on $\Lambda \cong \mathbb{Z}^2$. Namely, we can construct an action of the group $K = \text{Dic}_{16} \times (\mathbb{Z}/2)$ (where $\text{Dic}_{16}$ is the dicyclic group of order 16, cf. Corollary 6.82, which is an index-2 central extension of $W_2$) on $\Lambda$ by making $\text{Dic}_{16} = \langle\sigma, \gamma\rangle$ act through its quotient $W_2$ (which is the quotient by $\sigma^2$) and making $\mathbb{Z}/2 = \langle\tau\rangle$ act by $-\text{id}$. Notice that there is already an element in $W_2$ acting by $-\text{id}$, namely the central element $(\sigma\gamma)^2$ of $W_2$, so this action is in fact through the quotient $K \to W_2$ sending respectively $\sigma, \gamma$ and $\tau$ to $\sigma, \gamma$ and $(\sigma\gamma)^2$. In particular, this implies that we have:

$$ \Gamma_{*}^{K}(\Lambda) = \Gamma_{*}^{W_2}(\Lambda). $$

### A.3. Free products

**A.3.1. Two examples.** Consider the simplest free product of two groups, which is the infinite dihedral group $\mathbb{Z}/2 * \mathbb{Z}/2 = \langle x, y \mid x^2 = y^2 = 1 \rangle$. We can determine its LCS from its description as a semi-direct product:

**PROPOSITION A.16.** *There is an isomorphism:*

$$ \mathbb{Z}/2 * \mathbb{Z}/2 \cong \mathbb{Z} \times (\mathbb{Z}/2). $$

As a consequence, this group is residually nilpotent, and its Lie ring is $(\mathbb{Z}/2)[X] \rtimes (\mathbb{Z}/2)$, where both factors are abelian Lie rings, and the generator $T$ of $\mathbb{Z}/2$ acts by $[X^i, T] = X^{i+1}$.

**PROOF.** We know a presentation of each of these groups, namely $\mathbb{Z}/2 * \mathbb{Z}/2 = \langle x, y \mid x^2 = y^2 = 1 \rangle$ and $\mathbb{Z} \times (\mathbb{Z}/2) = \langle a, t \mid \text{tat}^{-1} = a^{-1}, t^2 = 1 \rangle$. It is then easy to check that the assignments $x \mapsto t$, $y \mapsto ta$ and $t \mapsto x$, $a \mapsto xy$ define morphisms inverse to each other. The rest is an application of Corollary A.8 with $A = \mathbb{Z}$. $\square$

Consider now the free product $\mathbb{Z} * (\mathbb{Z}/2) = \langle x, y \mid y^2 = 1 \rangle$. We have a similar decomposition into a semi-direct product:

**PROPOSITION A.17.** *There is an isomorphism:*

$$ \mathbb{Z} * (\mathbb{Z}/2) \cong \mathbf{F}_2 \times (\mathbb{Z}/2), $$

where the action of the generator $t$ of $\mathbb{Z}/2$ is given by exchanging the two elements $a$ and $b$ of a basis of the free group $\mathbf{F}_2$.

**PROOF.** Again, we know a presentation of each of these groups, namely $\mathbb{Z} * (\mathbb{Z}/2) = \langle x, y \mid y^2 = 1 \rangle$ and $\mathbf{F}_2 \times (\mathbb{Z}/2) = \langle a, b, t \mid \text{tat}^{-1} = b, \text{tbt}^{-1} = a, t^2 = 1 \rangle$. It is then easy to check that the assignments $x \mapsto tb$, $y \mapsto t$ and $t \mapsto y$, $a \mapsto xy$, $b \mapsto yx$ define morphisms inverse to each other. One may alternatively observe that the Tietze transformation removing the generator $a$ turns the second presentation into the first. $\square$

**REMARK A.18.** The group $\mathbb{Z} * (\mathbb{Z}/2)$ is isomorphic to $\mathbf{wB}_2$ (or $\mathbf{vB}_2$) and the above isomorphism can be identified with $\mathbf{wB}_2 \cong \mathbf{wP}_2 \rtimes \mathfrak{S}_2$, together with $\mathbf{wP}_2 \cong \mathbf{F}_2$ (with basis $(\chi_{12}, \chi_{21})$).
The LCS of $Z^*(Z/2)$ is much more difficult to compute than the one of $Z/2*Z/2$ above. The reader can find a presentation of the associated Lie ring in [Lab77], where the methods used are somewhat different from ours. Our methods could be adapted to recover this result, together with the residual nilpotence of the group, but we will not do so here. We only give a proof of the following:

**PROPOSITION A.19.** *The group $Z^*(Z/2)$ is residually nilpotent, but not nilpotent. Its Lie ring has only 2-torsion elements, except in degree one.*

**PROOF.** Note that the group $Z^*(Z/2)$ surjects onto $Z/2*Z/2$, whose LCS does not stop, by Proposition A.16. Notice also that the statement about torsion has already been proven in Example 2.9. As a consequence, we only need to prove that $Z^*(Z/2)$ is residually nilpotent, which is the difficult part of the statement. We can prove it using a kind of Magnus expansion. Namely, let us consider the associative algebra of non-commutative formal power series $A := F_2\langle\langle X, Y \rangle\rangle/(Y^2 = 1)$. We get a morphism $\Phi$ from $Z^*(Z/2)$ to the group $A^\times$ of units of $A$ by sending $x$ to $1+X$ and $y$ to $1+Y$. It is injective, by the usual argument: if $x^{a_1}y^{\epsilon_1} \cdots x^{a_l}y^{\epsilon_l}x^{a_{l+1}}$ is a non-trivial reduced expression of some non-trivial element $g \in Z^*(Z/2)$ (with $\epsilon_i = \pm 1$, $a_i \in Z$, and $a_1$ and $a_{l+1}$ possibly trivial), then, by writing $a_i = 2^{b_i}(2c_i+1)$, using $(1+X)^{2^k} = 1+X^{2^k}$ and $(1+T)^\alpha = 1+\alpha T + \cdots$, we see that the coefficient of the monomial $X^{2^{b_1}}YX^{2^{b_2}}Y \cdots YX^{2^{b_{l+1}}}$ in $\Phi(g)$ is not trivial, hence $\Phi(g) \neq 1$.

Now let us denote by $(X, Y)$ the ideal generated by $X$ and $Y$ in $A$, and by $A_k^\times$ the subgroup $1+(X, Y)^k$ of $A^\times$ (for $k \ge 1$). It is easy to see that $[A_1^\times, A_k^\times] \subset A_{k+1}^\times$ for all $k \ge 1$. As a consequence, for all $k \ge 1$, we have $\Gamma_k(A_1^\times) \subseteq A_k^\times$. Since the intersection of the $1+(X, Y)^k$ is obviously trivial, $A_1^\times$ is residually nilpotent, whence also $Z^*(Z/2)$, which is isomorphic to one of its subgroups. $\square$

**REMARK A.20.** As mentioned at the beginning of the proof above, if one wishes only to see that the LCS of $Z^*(Z/2)$ does not stop, one need only apply Lemma 1.1 to the obvious projection $Z^*(Z/2) \twoheadrightarrow (Z/2) * (Z/2)$ and one can then deduce the required result from the much simpler explicit computation of the LCS of the infinite dihedral group $(Z/2) * (Z/2) \cong Z \times (Z/2)$ done in Proposition A.16.

**A.3.2. An Artin group.** Consider the Artin group of type $B_2$ (which is also $B_{1,2}$ – see Lemma 3.16), that is:

$$G := \langle \sigma, x \mid (\sigma x)^2 = (x \sigma)^2 \rangle.$$

Let $\delta := \sigma x$. Then $(\sigma x)^2 = (x \sigma)^2$ is equivalent to $\delta^2 = \sigma^{-1}\delta^2\sigma$, so that:

$$G = \langle \sigma, \delta \mid \delta^2 \sigma = \sigma \delta^2 \rangle.$$

The element $\delta^2$ commutes with the generators $\delta$ and $\sigma$, hence it is central in $G$. Since $G^{ab}$ is free on the classes of $\delta$ and $\sigma$ (as is obvious from the presentation), $\delta^2$ is of infinite order. Moreover, the above relation clearly becomes trivial when $\delta^2$ is killed, so that:

$$G/\delta^2 = \langle \sigma, \delta \mid \delta^2 = 1 \rangle \cong Z^*(Z/2).$$

We thus have a central extension:

$$Z \stackrel{\sim}{\longrightarrow} G \longrightarrow Z^*(Z/2).$$

**PROPOSITION A.21.** *The group $G = \langle \sigma, x \mid (\sigma x)^2 = (x \sigma)^2 \rangle$ is residually nilpotent (but not nilpotent).*
**PROOF.** We have observed above that the central subgroup $\langle\delta^2\rangle$ injects into $G^{ab}$, i.e., we have $\langle\delta^2\rangle \cap \Gamma_2(G) = \{1\}$. Thus, we can deduce the (strict) residual nilpotence of $G$ from the fact that $\mathbb{Z} \ast (\mathbb{Z}/2)$ is (strictly) residually nilpotent (see Proposition A.19) by applying Proposition 1.5. $\square$

**REMARK A.22.** The weaker fact that the LCS of $G$ does not stop can also be deduced more directly from the reasoning of Remark A.20.

The decomposition of $G$ into a central extension can also be used to describe its Lie ring. Namely, it can be obtained from the Lie ring of $\mathbb{Z} \ast (\mathbb{Z}/2)$ described in [Lab77]; see §A.3.1.

**PROPOSITION A.23.** The Lie ring of $G = \langle\sigma, x \mid (\sigma x)^2 = (x\sigma)^2\rangle$ is a central extension of $\mathcal{L}(\mathbb{Z} \ast (\mathbb{Z}/2))$ by $\mathbb{Z}$, concentrated in degree one. Precisely, $2\sigma\bar{x}$ is central in $\mathcal{L}(G)$, and $\mathcal{L}(G)/(2\sigma\bar{x}) \cong \mathcal{L}(\mathbb{Z} \ast (\mathbb{Z}/2))$.

**PROOF.** We have seen that $\langle\delta^2\rangle$ injects into $G^{ab}$, which means that $\langle\delta^2\rangle \cap \Gamma_2(G) = \{1\}$. As a consequence, the projections $\pi: \Gamma_k(G) \to \Gamma_k(G/\delta^2)$ have trivial kernels for $k \ge 2$, which means that they are isomorphisms. Thus, the canonical morphism from $\mathcal{L}(G)$ to $\mathcal{L}(G/\delta^2)$ is an isomorphism in degree at least 2. In degree one, it identifies with the projection of $G^{ab} \cong \mathbb{Z}^2$ onto $(G/\delta^2)^{ab} \cong \mathbb{Z} \times (\mathbb{Z}/2)$, whose kernel is generated by $\bar{\delta}^2 = 2\bar{\delta}$. Moreover, since $\delta^2$ is central in $G$, its class in $\mathcal{L}(G)$ must be central, whence our result. $\square$

## A.4. Wreath products

This section is devoted to the study of the LCS of $G\rtimes\mathfrak{S}_\lambda = G^n\rtimes\mathfrak{S}_\lambda$, where $G$ is any group, $\lambda = (n_1, \dots, n_l)$ is a partition of the integer $n$ and $\mathfrak{S}_\lambda \subseteq \mathfrak{S}_n$ acts on $G^n$ by permuting the factors. By Proposition 6.12, wreath products of this form are precisely the (partitioned) braid groups on manifolds of dimensions at least 3. We first compute the abelianisation (Corollary A.25), then we show that the LCS stops under some stability condition (Corollary A.27). Finally, we look at the unstable cases, whose LCS we compute if $G$ is abelian (§A.4.3).

Let us remark that, in order to understand the LCS of $G\rtimes\mathfrak{S}_\lambda$, we only need to study the LCS of $G\rtimes\mathfrak{S}_n$. Indeed:

$$G\rtimes\mathfrak{S}_\lambda \cong \prod_{i=1}^l G\rtimes\mathfrak{S}_{n_i}.$$

**A.4.1. Abelianisations.** Lemma 1.12 allows us to compute abelianisations of wreath products:

**LEMMA A.24.** For any integer $n \ge 2$, we have $(G\rtimes\mathfrak{S}_n)^{ab} \cong G^{ab} \times (\mathbb{Z}/2)$.

**PROOF.** It is a direct consequence of Lemma 1.12, applied to $G\rtimes\mathfrak{S}_\lambda = G^n\rtimes\mathfrak{S}_\lambda$: $((G\rtimes\mathfrak{S}_n)^{ab}) = ((G^n)^{ab})\mathfrak{S}_n \times \mathfrak{S}_n^{ab} = ((G^{ab})^n)\mathfrak{S}_n \times (\mathbb{Z}/2) = G^{ab} \times (\mathbb{Z}/2)$. $\square$

**COROLLARY A.25.** For any partition $\lambda = (n_1, \dots, n_l)$ of $n$, if $l'$ denotes the number of indices $i \le l$ such that $n_i \ge 2$, we have $(G\rtimes\mathfrak{S}_\lambda)^{ab} \cong (G^{ab})^{l'} \times (\mathbb{Z}/2)^{l'}$.

**PROOF.** $G\rtimes\mathfrak{S}_\lambda$ decomposes as the direct product of the $G\rtimes\mathfrak{S}_{n_i}$, and $(G\rtimes\mathfrak{S}_{n_i})^{ab}$ identifies with $G^{ab}$ when $n_i = 1$, and with $G^{ab} \times (\mathbb{Z}/2)$ if $n_i \ge 2$. $\square$
**A.4.2. The stable case.** We now use a disjoint support argument to show that there is a stable behaviour for the LCS of $G \wr \mathfrak{S}_\lambda$, occurring as soon as $n_i \ge 3$ for every $i \le l$.

Recall that the usual generators $\tau_i$ of $\mathfrak{S}_n$ are conjugate to each other, hence $\mathfrak{S}_n^{ab} \cong \mathbb{Z}/2$ is generated by their common class $\tau$, and $\Gamma_2 = \Gamma_\infty$ for $\mathfrak{S}_n$.

**PROPOSITION A.26.** Let $G$ be a group. If $n \ge 3$, the $\tau_i \tau_j^{-1}$ normally generate $\Gamma_2(G \wr \mathfrak{S}_n)$, and $(G \wr \mathfrak{S}_n)^{ab} \cong G^{ab} \times \mathbb{Z}/2$. Moreover, the LCS of $G \wr \mathfrak{S}_n$ stops at $\Gamma_2$.

**PROOF.** Let $N$ be the subgroup of $G \wr \mathfrak{S}_n$ normally generated by the $\tau_i \tau_j^{-1}$. These are in $\Gamma_2(\mathfrak{S}_n)$, whence also in $\Gamma_2(G \wr \mathfrak{S}_n)$, hence the latter contains $N$. In order to show the converse inclusion, we need to show that $(G \wr \mathfrak{S}_n)/N$ is abelian. For any $g \in G$, let us denote by $\bar{g}$ the class of $(g, 1, \dots, 1) \in G^n$ modulo $N$. We now show that the $\bar{g}$, together with $\tau$, generate $(G \wr \mathfrak{S}_n)/N$, and we use a disjoint support argument to show that they commute with one another.

First, let us remark that $\bar{g}$ commutes with $\tau$. This comes from the fact that $\tau_2$ acts trivially on $(g, 1, \dots, 1)$, thus commutes with it in $G \wr \mathfrak{S}_n$. From this, we deduce that $\bar{g}$ is also the class of $\sigma(g, 1, \dots, 1)\sigma^{-1} = (1, \dots, 1, g, 1, \dots, 1)$ for any $\sigma \in \mathfrak{S}_n$ (whose class modulo $N$ is a power of $\tau$). In particular, for all $g, h \in G$, $(g, 1, \dots, 1)$ commutes with $(1, h, \dots, 1)$, hence $\bar{g}$ commutes with $\bar{h}$.

Now, every $(g_1, \dots, g_n) \in G^n$ is the product of the $(1, \dots, 1, g_i, 1, \dots, 1)$, so that all the elements $(1, \dots, 1, g, 1, \dots, 1)$ (for all $g \in G$ and any choice of position), together with the $\tau_i$, generate $G \wr \mathfrak{S}_n$. This implies that their classes $\bar{g}$, together with $\tau$, generate $(G \wr \mathfrak{S}_n)/N$. Since these generators commute with one another, this ends the proof that $N = \Gamma_2(G \wr \mathfrak{S}_n)$.

The rest of the statement is a direct application of Lemmas A.24 and A.2. $\square$

**COROLLARY A.27.** Let $G$ be a group, $n \ge 3$ be an integer and $\lambda = (n_1, \dots, n_l)$ be a partition of $n$ with $n_i \ge 3$ for all $i$. Then the $\tau_\alpha \tau_\beta^{-1}$ for $\alpha$ and $\beta$ in the same block of $\lambda$ normally generate $\Gamma_2(G \wr \mathfrak{S}_\lambda)$, and $(G \wr \mathfrak{S}_\lambda)^{ab} \cong (G^{ab} \times \mathbb{Z}/2)^l$. Moreover, the LCS of $G \wr \mathfrak{S}_\lambda$ stops at $\Gamma_2$.

**PROOF.** Apply Proposition A.26 to each factor of $G \wr \mathfrak{S}_\lambda \cong \prod_{i=1}^l G \wr \mathfrak{S}_{n_i}$. $\square$

**A.4.3. Unstable cases.** Since $G \wr \mathfrak{S}_1 = G$, the only case left in our study of $G \wr \mathfrak{S}_\lambda$ is the case of $G \wr \mathfrak{S}_2$, which can be quite complicated. We treat the case where $G$ is an abelian group, which we denote by $A$.

**PROPOSITION A.28.** Let us denote by $\delta A$ the subgroup $\{(a, -a) \mid a \in A\}$ of $A^2$. For all $i \ge 2$, we have $\Gamma_i(A \wr \mathfrak{S}_2) = 2^{i-2}(\delta A)$. Moreover, the Lie algebra decomposes as:

$$ \mathcal{L}(A \wr \mathfrak{S}_2) \cong (A \oplus A/2A \oplus 2A/4A \oplus \dots) \rtimes (\mathbb{Z}/2), $$
where $A$ and $\mathbb{Z}/2$ are in degree 1 and each factor of the form $2^{i-2}A/2^{i-1}A$ is in degree $i$. The Lie ring $A \oplus A/2A \oplus \dots$ is abelian and the generator $T$ of $\mathbb{Z}/2$ acts on it via the degree-one map:

$$ \left\{ \begin{array}{ll} a \mapsto \bar{a} & \text{in degree 1,} \\ \bar{a} \mapsto 2\bar{a} & \text{in degree at least 2.} \end{array} \right. $$

**PROOF.** This is a straightforward application of Proposition A.10 and Corollary A.11 (adapted to an action of $\mathfrak{S}_2 \cong \mathbb{Z}/2$ instead of $\mathbb{Z}$). Namely, $V = \delta A$ is the
subspace of $A^2$ on which $\mathfrak{S}_2$ acts by $-id$. Moreover, $A^2/V \cong A$ (via $(\overline{a,0}) \leftrightarrow a$) and
the map induced by $1 - \tau$ identifies with the one described in our statement. $\square$

COROLLARY A.29. Let $G$ be a group and $\lambda$ be a partition with at least one block of size 2. Suppose that the filtration $2^*G^{ab}$ of $G^{ab}$ does not stop. Then the LCS of $G \wr \mathfrak{S}_\lambda$ does not stop.

PROOF. The group $G^{ab} \wr \mathfrak{S}_\lambda$ is a quotient of $G \wr \mathfrak{S}_\lambda$, whose LCS does not stop by Proposition A.28. Thus, our statement follows from Lemma 1.1. $\square$

REMARK A.30. If $G^{ab}$ is finitely generated, the condition in Corollary A.29 holds if and only if it is infinite. In general, the condition is equivalent to $2^i A \neq \{0\}$ for all $i \ge 1$, where $A$ is the quotient of $G^{ab}$ by its maximal 2-divisible subgroup.

A.4.4. About general wreath products. Let us indicate briefly how the work done above generalises to study the LCS of $G \wr K = G^X \ltimes K$, where $G$ is a group, $K$ is a group acting on a finite set $X$ and $G^X$ denotes the group of functions $X \to G$. Notice first that Lemma A.24 generalises easily to $(G \wr K)^{ab} \cong (G^{ab})^{X/K} \times K^{ab}$.

Then, by looking closely at the proof of Proposition A.26, one can devise a stability hypothesis that ensures that Lemma A.2 can be applied, so that $L(G \wr K) \cong (G^{ab})^{X/K} \times L(K)$. In fact, one needs to assume that $K$ acts on $X$ without fixed points, and that we can find a set $S$ generating $K$ such that for all $s \in S$, each $K$-orbit of $X$ has a fixed point under the action of $s$. This means that each orbit must be large enough, but also large enough with respect to the supports of the generators of $K$. This is satisfied for instance when $K$ is a subgroup of $\mathfrak{S}_n$ generated by transpositions (which means exactly that $K$ is conjugate to $\mathfrak{S}_\lambda$ for some partition $\lambda$ of $n$), and all its orbits are of size at least 3.
<figure><img src="image_39.png" /></figure>
# APPENDIX B

## Presentation of an extension

Here we recall the classical construction of a presentation of a group extension from a presentation of the quotient and a presentation of the kernel, together with some knowledge of the structure of the extension (see also [HEO05, §2.4.3]). We then apply this construction to show that 2-nilpotent groups whose abelianisation is free are determined by their Lie ring.

Let $G$ be a group, which is an extension of a quotient $K$ by a normal subgroup $H$. Suppose that presentations of $H$ and $K$ are known, namely $H = \langle X|R \rangle$ and $K = \langle Y|S \rangle$, where $R$ is a subset of the free group $F[X]$ (resp. $S \subset F[Y]$). For each $y$ in $Y$, let us fix a lift $\tilde{y}$ of the corresponding generator of $K$ to an element of $G$. Then a presentation of $G$ is given by

$$G = \langle X \sqcup Y | R \cup \tilde{S} \cup T \rangle,$$

where $\tilde{S}$ and $T$ are obtained as follows:

* Each $s \in S$ is a word in the elements of $Y$ and their inverses. If we replace each $y$ in $s$ by its chosen lift $\tilde{y}$, we get an element $\tilde{s}$ of $G$, which is in fact in $H$, since its projection to $K$ is trivial by construction. Each element of $H$ is represented by a word on the elements of $X$ and their inverses, so we can choose some $w_s \in F[X]$ representing $\tilde{s}$. Then $\tilde{S}$ is the following set of relations:

$$\tilde{S} := \{sw_s^{-1} \mid s \in S\} \subset F[X \sqcup Y].$$

* For each $y \in Y$ and each $x \in X$, the element $\tilde{y}x\tilde{y}^{-1}$ is an element of $H$, which can be represented by a word $w_{x,y} \in F[X]$. Then we define:

$$T := \{yxy^{-1}w_{x,y}^{-1} \mid x \in X, y \in Y\} \subset F[X \sqcup Y].$$

**REMARK B.1.** If the presentations of $H$ and $K$ are finite, this construction gives a finite presentation of $G$.

**REMARK B.2 (Split extensions).** When the extension splits (that is, when $G$ is a semi-direct product of $K$ by $H$), one usually chooses the lifts of generators of $K$ to be their images under a fixed section. Then the presentation obtained is somewhat simpler, since the relations in $S$ hold in $G$ (that is, $\tilde{S} = S$).

**PROPOSITION B.3.** *The above presentation is indeed a presentation of the extension $G$.*

**PROOF.** Let $G_0$ be the group defined by the above presentation. By construction, the assignments $x \mapsto x \in H \subset G$ and $y \mapsto \tilde{y}$ induce a well-defined morphism $\pi$ from $G_0$ to $G$. Let $H_0$ be the subgroup of $G_0$ generated by $X$. The morphism $\pi$ restricts to a morphism $\pi_H: H_0 \to H$. Since the relations $R$ are satisfied in $H_0$, we can construct an inverse to $\pi_H$: it is an isomorphism.
The relations $T$ ensure that $H_0$ is stable under left conjugation by the $y \in G_0$. Moreover, for all $h \in H$, $\pi_H(yhy^{-1}) = \pi(y)\pi(h)\pi(y)^{-1} = \tilde{y}\pi_H(h)\tilde{y}^{-1}$. Since $H$ is normal in $G$, left conjugation by $\tilde{y}$ is an automorphism of $H$. Since $\pi_H$ is an isomorphism, left conjugation by $y \in G_0$ must be an automorphism of $H_0$, which implies that $H_0$ is stable under left conjugation by $y^{-1}$. Finally, $H_0$ is stable under left conjugation by $y^{\pm 1}$, and also (clearly) by $x^{\pm 1}$, so it is normal in $G_0$.

As a consequence, $\pi$ induces a morphism of extensions:

$$ \begin{array}{ccc} H_0 & \to & G_0 \\ \big\downarrow_{\pi_H} & & \big\downarrow_{\pi} \\ H & \to & G \end{array} \quad \Rightarrow \quad \begin{array}{c} G_0/H_0 \\ \big\downarrow_{\pi} \\ K. \end{array} $$

The relations $R$ and $T$ become trivial in $G_0/H_0$, and $\tilde{S}$ reduces to $S$ there, so this quotient admits the presentation $\langle Y|S \rangle$. This implies that $\bar{\pi}$ is an isomorphism. Since we already know that $\pi_H$ is an isomorphism, the Five Lemma allows us to conclude that $\pi$ is an isomorphism. $\square$

**REMARK B.4.** We can replace some of the generators $y$ by their inverses before doing this construction, so we can choose $T$ to encode either left conjugation by $y$, or right conjugation by $\tilde{y}$, for each $y$.

**COROLLARY B.5.** If $G$ is a 2-nilpotent group whose abelianisation is free abelian, then $G$ is determined (up to isomorphism) by its associated Lie ring.

**PROOF.** We construct a presentation of $G$ which depends only on the structure of $\mathcal{L}(G)$ (and on some choices not involving elements of $G$). The group $G$ is an extension of $G^{ab} = \mathcal{L}_1(G)$ by $\Gamma_2(G) = \mathcal{L}_2(G)$, to which we can apply the previous construction. Since $G^{ab}$ is free abelian on some set $Y$, a presentation of this group is given by generators $Y$ and relations $\{[y, z] \mid y, z \in Y\}$. Let $\langle X|R \rangle$ be a presentation of the group $\mathcal{L}_2(G)$. Then $G$ admits the presentation with generators $X \sqcup Y$ and relations $R \cup \tilde{S} \cup T$, constructed as above. We need to show that $S$ and $T$ can be recovered from calculations in $\mathcal{L}(G)$ alone.

* Let $s \in S$, that is, $s = [y, z]$ for some $y, z \in Y$. Then, by definition of $\mathcal{L}(G)$, $\tilde{s} = [\tilde{y}, \tilde{z}]$ is the element of $\Gamma_2(G) = \mathcal{L}_2(G)$ given by the bracket of $y, z \in \mathcal{L}_1(G)$.

* Since $[G, \Gamma_2(G)] = \{1\}$, the set $T$ consists of the relations $[y, x]$, for $x \in R$ and $y \in S$.

Thus the above presentation of $G$ can be obtained from the data of $\mathcal{L}(G)$, as claimed. $\square$

**REMARK B.6.** Corollary B.5 is not true in general if the abelianisation of $G$ is not free. For example, the dihedral group $D_8$ of order 8 and the quaternion group $Q_8$ are not isomorphic, but they are 2-nilpotent groups whose Lie rings are isomorphic. Indeed, in both cases, we have $\mathcal{L}_1G = (\mathbb{Z}/2)^2$ and $\mathcal{L}_2G = \mathbb{Z}/2$ and the Lie structure is fully determined by saying that whenever $a$ and $b$ are two distinct non-trivial elements of $\mathcal{L}_1G$, then $[a, b]$ is non-trivial in $\mathcal{L}_2G$.