The configuration space $C_n(M)$ of $n$ particles in some connected graph $M$ (thought of as the topological realisation of a one-dimensional CW-complex) is $$M^n \backslash \{ (x_1, \ldots, x_n) \mid x_i=x_j \ \text{for some} \ i \neq j\},$$ and the corresponding unordered configuration space $UC_n(M)$ is the quotient $C_n(M)/ \mathfrak{S}_n$ where $\mathfrak{S}_n$ acts freely on $C_n(M)$ by permuting the coordinates. The braid group $B_n(M)$ is the fundamental group of $UC_n(M)$ (which is connected if $M$ is connected itself).

Is it true that $B_{m}(M)$ admits an injective homomorphism into $B_n(M)$ if $m \leq n$? Is it true at least if $m \leq 2$?

It is worth noticing that the question is not trivial if we replace the graph $M$ with a sphere according to this question.

  • $\begingroup$ My instinct is to say "no" because the standard embedding of the standard braid group depends on basically pushing out the other strands away from the necessary space to produce a braid for each group element $\endgroup$ Aug 1, 2017 at 6:41
  • $\begingroup$ Does "$B_m(M)$ embeds into $B_n(M)$" have to be understood as "there exists an injective group homomorphism $B_m(M)\to B_n(M)$"? $\endgroup$
    – YCor
    Aug 1, 2017 at 7:25
  • $\begingroup$ @YCor: Absolutely. $\endgroup$
    – Seirios
    Aug 1, 2017 at 7:29
  • 3
    $\begingroup$ If this is true you get some very silly embeddings. For example, $B_m(S^1) \cong \mathbb{Z}$ generated by cyclically rotating the $m$ points, so any embedding $B_m(S^1) \to B_n(S^1)$ has to do something slightly strange. More generally, I think that if you carefully think about graphs homeomorphic to $S^1 \vee S^1$ and choose the number of points carefully you can show that there can't be such an embedding of groups that is functorial in graph embeddings. $\endgroup$
    – dvitek
    Aug 1, 2017 at 8:49

2 Answers 2


I think the answer is yes if $M$ has a leaf. Subdivide $M$ sufficiently so that Abram's locally CAT(0) model $X_n(M)$ of $B_n(M)$ (called the reduced braid group, $RB_n(M)$ is this paper of Crisp and Wiest) can be used. We can map $X_n(M)$ to $X_{n+1}(M')$ by mapping a configuration $(p_1, \dots, p_n)$ to $(p_1, \dots, p_n, p_{n+1})$. Here $M$ is the subdivided graph and $M'$ is $M$ with an additional subdivided edge glued onto the degree one vertex of the distinguished leaf of $M$. The point $p_{n+1}$ is fixed and corresponds to the degree one vertex of the distinguished (extended) leaf of $M'$. The map $X_n(M) \to X_{n+1}(M')$ is a cubical map and it seems that the image of the link of any vertex in $X_n(M)$ is a full subcomplex of the link of the image point in $X_{n+1}(M')$. So, this map is $\pi_1$-injective. But, $X_{n+1}(M')$ is isomorphic to $X_{n+1}(M)$, so we have an injective homomorphism $B_n(M) \to B_{n+1}(M)$.

Here's some additional detail/background, though Crisp & Wiest (and Abrams in his thesis) say it much better if you read their work. Given a subdivided graph $M$, the carrier of a point $x$ in $M$ is the closed cell (vertex or edge) containing it. The space $Y_n(M)$ consists of configurations of $n$ ordered points in $M$ such that the carriers of points are mutually disjoint. $X_n(M)$ is the quotient under the action of the symmetric group. The cubical structure comes from the fact that a configuration of $n$ points where exactly $k$ of these points have edges as their carriers corresponds to a $k$-cube.


When $M$ is the 2-sphere, then there is no an injective group homomorphism from $B_{n}(S^2)$ to $B_{m}(S^2)$ if $m$ is greater than $n$. Indeed, $\sigma_1\sigma_2\dots \sigma_{n-1}^2\dots\sigma_2\sigma_1=1$ is trivial in $B_{n}(S^2)$ but not a relation in $B_{m}(S^2)$ so there is no group homomorphism from $B_{n}(S^2)$ to $B_{m}(S^2)$.

  • 3
    $\begingroup$ You're just proving that there's no group homomorphism with a certain assignment on generators. "So there is no group homomorphism" is false: you have the trivial homomorphism, and possibly you have a few others (e.g., with cyclic image? or even larger?). The question (which was indeed a bit unclear but clarified by the OP as an answer to my request) is whether there is an injective group homomorphism. $\endgroup$
    – YCor
    Feb 26, 2018 at 13:25

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