Relations between relations in the positive braid monoid

Question

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 $\pi_1(\Gamma(\beta))$?

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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?

Answer 1

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.

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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.

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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 $w \succeq \Delta_J$. $\square$.

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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 $\beta'$.) $\square$.

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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.