# Relations between relations in the positive braid monoid

The positive braid monoid on $$n$$ strands is the monoid with generators $$s_1$$, $$s_2$$, ..., $$s_{n-1}$$ and relations $$s_i s_{i+1} s_i = s_{i+1} s_i s_{i+1} \qquad s_i s_j = s_j s_i \text{for}\ |i-j| \geq 2.\qquad (\ast)$$ For any element $$\beta$$ in the positive braid monoid, let $$\Gamma(\beta)$$ be the graph whose vertices are words for $$\beta$$, and whose edges correspond to using a single braid relation from the list $$(\ast)$$. For example, $$\Gamma(s_1 s_1 s_2 s_1)$$ looks like $$s_1 s_1 s_2 s_1 \longleftrightarrow s_1 s_2 s_1 s_2 \longleftrightarrow s_2 s_1 s_2 s_2.$$

This graph is a tree, but it is easy to find $$\beta$$ where this graph has cycles.

Main Question: What is a list of cycles which generates $$\pi_1(\Gamma(\beta))$$?

Here is the list of cycles which I conjecture works.

(1) If a word $$w$$ for $$\beta$$ can have two braid relations applied to it in non-overlapping positions, than we can apply the relations in either order, making a $$4$$-cycle in $$\Gamma(\beta)$$.

(2) If a word for $$\beta$$ contains a substring $$s_p$$, $$s_q$$, $$s_r$$, where $$s_p$$, $$s_q$$ and $$s_r$$ all commute, then we can glue in a hexagon, as in the figure below:

(3) If a word for $$\beta$$ contains a substring $$s_i s_j s_{j+1} s_j$$ where $$s_i$$ commutes with $$s_i$$, $$s_j$$, then we can glue in an octagon, as in the figure below:

(4) If a word for $$\beta$$ contains a reduced word for the longest element in $$S_4$$, then we can glue in a $$14$$-gon, as shown in the figure below.

Question: Does this list of cycles generate? If not, can we always generate $$\pi_1(\Gamma(\beta))$$ with a list of cycles like this, coming from subwords of some bounded length?

• Is this a duplicate of mathoverflow.net/questions/432075 ? Feb 17 at 20:12
• Yup, I do. I'm trying to understand these "weaves" that Cassals, Gorsky and company work with, which involve very nonreduced words. Feb 17 at 22:55
• If it turned out that cycles supported on reduced subwords (like all the examples I have listed) generate $\pi_1$, that would be very good news for me. Feb 17 at 22:56
• I skimmed the Tits paper doi.org/10.1007/978-1-4612-5648-9_35 . The result for reduced words is Proposition 4, and I believe that it is the same list of generators that I have conjectured: I believe that Tits' (a) is non-overlapping braid relation and (b) is moves supported on a rank $3$ subgroup which, in type $A$, means $A_1^3$, $A_2 \times A_1$ and $A_3$, which are the types I listed. Feb 18 at 13:28
• Hi David, Ben and I also stumbled across this question and answered it (in some form) here: arxiv.org/abs/1405.4928. There is also a discussion in Ronan's book on buildings if I recall correctly. There is also a higher-categorical question (relations between relations between relations...) which should have a nice answer in terms of finite parabolic subgroups of certain ranks. (E.g. the finite parabolics of rank 3 give rise to the "relations between relations" that you see.) Feb 18 at 23:33

I found a published reference! This is the main result of:

Fukushi, Takeo, On a braid monoid analogue of a theorem of Tits., SUT J. Math. 47, No. 1, 45-53 (2011). ZBL1235.20036.

I'll keep my write up below.

My conjectured list generates $$\pi_1$$. This proof is basically the one in Lemma 3 and Proposition 4 of Tits' paper that Sam Hopkins referenced in the comments, plus some facts from Garside theory.

The Garside theory part:

I'll write $$B_n^+$$ for the positive braid monoid, $$S_n$$ for the symmetric group. For any subset $$J$$ of $$[n-1]$$, I write $$(w_0)_J$$ for the longest element in the corresponding parabolic subgroup of $$S_n$$ and $$\Delta_J$$ for the element of $$B_n^+$$ given by any reduced word for $$(w_0)_J$$.

We equip $$B_n^+$$ with right weak order, meaning that $$u \preceq w$$ if we can write $$w = uv$$ for $$u$$ and $$v$$ in $$B_n^+$$.

Theorem: $$B_n^+$$ is a meet semilattice with respect to $$\preceq$$, i.e., for any $$x$$ and $$y$$ in $$B_n^+$$, there is an element $$x \wedge y$$ of $$B_n^+$$ such that $$z \preceq x$$, $$y$$ if and only if $$z \preceq x \wedge y$$.

I found several sources attributing this to [Garside], but I couldn't find it there on a quick read. A proof can be found in [Epstein et al, Corollary 9.3.7].

We call $$s_i$$ a left descent of $$w \in B_n^+$$ if $$s_i \preceq w$$, in other words, if there is a word for $$w$$ starting with $$s_i$$.

Corollary: Let $$J \subset [n-1]$$ and let $$w \in B_n^+$$. Suppose that, for all $$i \in J$$, the braid generator $$s_i$$ is a left descent of $$w$$. Then $$\Delta_J \preceq w$$.

For $$|J|=2$$, this is Theorem H in [Garside]; I will need it for $$|J|=2$$ and $$|J|=3$$.

Proof Let $$z = w \wedge \Delta_J$$. Then we have $$s_i \preceq z$$ for all $$i \in J$$, and $$z \preceq \Delta_J$$. But the interval below $$\Delta_J$$ is isomorphic to the weak order on the Coxeter group $$(S_n)_J$$, and the only element in this weak order which is above $$s_i$$ for all $$i \in J$$ is $$(w_0)_J$$. So $$z = \Delta_J$$ and $$w \succeq \Delta_J$$. $$\square$$.

The part due to Tits:

I will need Lemma 3 from [Tits]. I think that Lemma 3 is missing a key word, which I have inserted in bold.

Lemma: Let $$X$$ and $$X'$$ be connected graphs and let $$\phi : X \to X'$$ be a surjective morphism of graphs, meaning that, for each edge $$(a,b)$$ of $$X$$, either $$\phi(a) = \phi(b)$$ or else $$(\phi(a), \phi(b))$$ is an edge of $$X'$$. Let $$C$$ be a collection of cycles in $$X$$. Suppose the following:

(1) For each vertex $$a$$ of $$X$$, the graph $$\phi^{-1}(x)$$ is connected, and those cycles of $$C$$ lying in $$\phi^{-1}(a)$$ generate $$\pi_1(\phi^{-1}(a))$$.

(2) For any edge $$(a', b')$$ in $$X'$$ and any two edges $$e_1= (a_1, b_1)$$, $$e_2 =(a_2, b_2)$$ in $$X$$ with $$\phi(a_1) = \phi(a_2) = a'$$ and $$\phi(b_1) = \phi(b_2) = b'$$, there is a cycle in the subgroup generated by $$C$$ of the form $$e_1 f e_2^{-1} g$$ where $$f$$ is a path $$b_1 \leadsto b_2$$ in $$\phi^{-1}(b')$$ and $$g$$ is a path $$a_2 \leadsto a_1$$ in $$\phi^{-1}(a')$$.

Then $$C$$ generates $$\pi_1(X)$$ if and only if $$\phi(C)$$ generates $$\pi_1(X')$$.

Using this lemma of Tits, I prove the result in almost exactly the same way that Tits proves his Proposition 4:

Proof of main result We induct on the length of $$\beta$$; the base case $$\ell(\beta) = 0$$ is clear. Let $$J$$ be the set of left descents of $$\beta$$; let $$K$$ be the complete graph on vertex set $$J$$, and define a map $$\phi : \Gamma(\beta) \longrightarrow K$$ sending a word $$w$$ for $$\beta$$ to its first letter. We now need to verify several things, in order to conclude by the Lemma.

The map $$\Gamma(\beta) \to K$$ is surjective. Surjectivity on vertives is obvious. For each edge $$(i,j)$$ of $$K$$, since $$i$$ and $$j$$ are both in $$J$$, we have $$s_i \preceq \beta$$ and $$s_j \preceq \beta$$. By the Corollary, this implies that $$\Delta_{\{ i,j \}} \preceq \beta$$, so there are words for $$\beta$$ of the form $$(s_i s_j \cdots) u$$ and $$(s_j s_i \cdots) u$$. Applying a braid operation to relate these two words gives us an edge of $$\Gamma(\beta)$$ lying above $$(i,j)$$. $$\square$$

The cycles $$\phi(C)$$ generate $$\pi_1(K)$$. Clearly, $$\pi_1(K)$$ is generated by its cycles of length $$3$$. For any $$i$$, $$j$$, $$k \in J$$, if $$s_i$$, $$s_j$$ and $$s_k$$ are all left descents of $$\beta$$, there there is a word for $$\beta$$ of the form $$\Delta_{\{ i,j,k \}} u$$. Using different reduced words for $$\Delta$$, we get one of the three pictures I drew above, so in each case the $$3$$-cycle $$(i \leadsto j \leadsto k \leadsto i)$$ is in the image of a cycle from $$\Gamma(\beta)$$. $$\square$$

For each vertex $$i$$ of $$K$$, the fundamental group of $$\phi^{-1}(i)$$ is generated by the cycles from $$C$$ lying in $$\phi^{-1}(i)$$.

Write $$\beta = s_i \beta'$$. The preimage of $$i$$ is precisely the graph $$\Gamma(\beta')$$, and the claim follows by induction.

Given an edge $$(i, j)$$ in $$K$$, and two edges $$e_1=(u_1, v_1)$$ and $$e_2=(u_2, v_2)$$ lying above $$(i,j)$$ there are paths as required.

Abbreviate $$\Delta_{\{i,j\}}$$ to $$\delta$$ and put $$\beta = \delta \beta'$$.

For $$r \in \{ 1,2 \}$$, $$u_r$$ must be joined to $$v_r$$ by a single braid relation which switches the first letter from $$i$$ to $$j$$. That means that $$u_r$$ and $$v_r$$ must be of the forms $$(s_i s_j \cdots) x_r$$ and $$(s_j s_i \cdots ) x_r$$, with $$x_r$$ a word for $$\beta'$$. Now, $$\Gamma(\beta')$$ is connected, so we can find a path $$x_1 \leadsto x_2$$ in $$\Gamma(\beta')$$; let $$f$$ be the induced path $$(s_j s_i \cdots ) x_1 \leadsto (s_j s_i \cdots) x_2$$ and let $$g$$ be the induced path $$(s_i s_j \cdots) x_2 \leadsto (s_i s_j \cdots) x_1$$. Then we can fill in $$e_1 f e_2 g^{-1}$$ using a collection of $$4$$ cycles involving non-overlapping braid moves. (For each $$4$$-cycle, one braid move is in the prefix $$\delta$$ and the other is in the suffix $$\beta'$$.) $$\square$$.

As should be evident, I have written this in a way which will generalize to other Artin groups, but I am confused about the state of Garside theory for other Artin groups, and I only care about braids right now, so I'll stop here.

[Epstein et al] Epstein, David B. A.; Cannon, James W.; Holt, Derek F.; Levy, Silvio V. F.; Paterson, Michael S.; Thurston, William P., Word processing in groups, Boston, MA etc.: Jones and Bartlett Publishers. xi, 330 p. (1992). ZBL0764.20017.

[Garside] Garside, F. A., The braid group and other groups, Q. J. Math., Oxf. II. Ser. 20, 235-254 (1969). ZBL0194.03303.

[Tits] Tits, Jacques, A local approach to buildings, The geometric vein, The Coxeter Festschr., 519-547 (1982). ZBL0496.51001.