In the weak Bruhat order, is every element bounded by a power of a Coxeter element? Let $(W,S)$ be a Coxeter System with $|W| = \infty$. Let $c = s_1\ldots s_{|S|}$ for some total ordering of $S$, a Coxeter element of $W$.
Is it true that for all $w \in W$, there exists a $k\in \mathbb{N}$ such that $w \le c^k$ in the weak Bruhat Order?
I would guess that this should not hold true in general but maybe something close to this is? Are there any sources that considers this?
Thanks!
 A: This is certainly true for the strong Bruhat order. This paper proves that if $(W,S)$ is infinite and irreducible (I'll leave you to check what happens when passing to the reducible case) then writing $\tilde{c}=(s_1,\dots,s_{|S|})$ for the reduced word representing the element $c$, the word $\tilde{c}^{\ast k}$ ($k$ copies of $\tilde{c}$ concatenated to form a single word of length $k|S|$) is a reduced expression for the element $c^k$.
Now let $w$ be and arbitrary element of $W$, and suppose $\tilde{w}=(s_{i_1},\dots, s_{i_d})$ is a reduced word representing $w$. Then $\tilde{w}$ is a subword of $\tilde{c}^{\ast d}$, which is clear by choosing $s_{i_1}$ from the first copy of $\tilde{c}$, $s_{i_2}$ from the second copy, and so on. Now apply the subword property of the strong Bruhat order.
This property now fails in general for the weak Bruhat order. Consider the case $(W,S)$ is the free Coxeter group of rank $|S|\ge3$, then $\tilde{c}^{\ast k}$ is not only a reduced word, it is the unique reduced word representing $c^k$ (since by Tits' solution to the word problem, and other reduced expression can be found by applying a relation which doesn't change the length of the word, but in this case $W$ doesn't contain any such relations). Clearly most elements in $W$ are not represented by reduced word (again unique) which is a prefix or suffix of $\tilde{c}^{\ast k}$ for any $k$. Here by "most" I mean $$\lim_{d\to\infty}\frac{|\{w\in W\mid l(w)\le d,\;w\le_Lc^k\;\textrm{for some}\;k\}|}{|\{w\in W\mid l(w)\le d\}|}=\lim_{d\to\infty}\frac{O(d)}{O(|S|^d)}=0.$$
Edit: What happens more generally?
The question asked by OP in the comments is how to get an intuition for this, and what can we say in general about for which groups this works. To get an intuition I like to think about Bruhat order geometrically in the context of the Coxeter complex of $(W,S)$. First however I will introduce some notation. $\le$ will be the strong Bruhat order, and $\le_L$ will be the weak left Bruhat order (I will ignore the right order because a) it is equivalent to the left order under $w\mapsto w^{-1}$, and b) it is geometrically a lot less nice. Now for $w\in W$, write $$[e,w]=\{u\in W\mid u\le w\}\;\textrm{and}\;[e,w]_L=\{u\in W\mid u\le_L w\}$$ for the order ideals of $w$ with respect to the strong order and weak order respectively. Denote by $R$ the set of all reflections in $(W,S)$, ie the set of all conjugates of the generators $S$. For $w\in W$ the set of inversions of $w$ is $\mathrm{inv}(w)=\{r\in R\mid l(rw)<l(w)\}$ (in Bjorner and Breti this is call $T_L(w)$ but I prefer this notation which comes from thinking about type $A$ Coxeter groups in terms of permutations). Finally the left descent set of $w$ is $D_L(w)=\textrm{inv}(w)\cap S$, and is the set of all generators which appear as the first letter in some reduced word representing $w$.
Now we can ask a more general question: given $(W,S)$, when is there an element $w_0\in W$ such that the sequence of sets $\{[e,w_0^k]\}_{k=0}^\infty$ or $\{[e,w_0^k]_L\}_{k=0}^\infty$ exhausts $W$, and when $w_0$ exists, can we classify all $w_0$ having this property?
I will focus on the case $(W,S)$ is infinite and irreducible, although much of what I'll say will hold in general.
Geometric intuition for string and weak order ideals
Before tackling this question I want to be able to draw pictures of order ideals. I will draw these in the Coxeter complex $X$ of $(W,S)$ and I'll assume familiarity with the ideas of chambers, galleries, walls, and the correspondence between chambers and elements of $W$, and between (minimal) galleries and (reduced) words. For an excellent exposition of these ideas consult the first three chapters of Buildings by K Brown.
The set of reflections $R$ corresponds to the set of walls $\mathcal{H}$ in $X$, and the set of inversions $\textrm{inv}(w)$ to the set of walls which separate the chamber labelled $w$, which by abuse of language I will identify with $w$ itself, from (the chamber labelled) $e$. The left descent set is the set of these inversion walls which bound $e$. The following is quite easy to prove:
Lemma 1: The set $[e,w]$ is invariant under the action of the finite special subgroup $W_{D_L(w)}$.
The two types of order ideals admit nice geometric descriptions as well. The set $[e,w]$ is the smallest set of chambers containing $w$ which has the following property:

For any $u\in[e,w]$ and any $r\in\textrm{inv}(u)$, the chamber $u'$ which is the reflection of $u$ in the wall $H_r$ corresponding to $r$ is also in $[e,w]$ (this chamber is labelled $ur$ and has $l(ur)<l(u)$).

The set $[e,w]_L$ is the union of all minimal galleries joining $e$ to $w$. It also has another geometric description. For $r\in R-\textrm{inv}(w)$ let $H_r^+$ be the half space (thought of as a union of chambers) with respect to the wall $H_r$ which contains both $e$ and $w$. Then $$[e,w]_L=\bigcap_{r\in R-\textrm{inv}(w)}H_r^+,$$ in particular this is a convex set.
Example 2: The Coxeter complex for the Coxeter group of type $\tilde{B}_2$ is the tiling of the Euclidean plane by $45-45-90$ triangles. Below is the set of chambers corresponding to the strong order ideal for $w=s_1s_2s_1s_2s_3s_1s_2s_3s_1$. Note that $D_L(w)=\{s_1,s_2\}$ and $[e,w]$ is invariant under the action of the dihedral group generated by this set (cf Lemma 1). Outlined in blue is the weak order ideal $[e,w]_L$, which satisfies both geometric criteria given above, and is also clearly a subset of $[e,w]$. In the right hand picture I have drawn $\textrm{inv}(w)$.

The question for the strong order
The picture above is quite typical, and in particular you may notice that $[e,w]$ looks like a ball in the Coxeter complex. You might imagine that as $k$ increases, $[e,w^k]$ approximates larger and larger balls, and so exhausts $W$. Indeed, I think this is the case once the obvious problematic choices of $w$ are excluded.
Conjecture 3: Let $(W,S)$ be an infinite irreducible Coxeter group, and let $w\in W$ have infinite order and not lie in any conjugate of a proper special subgroup (ie not lie in any proper parabolic subgroup). Then $\{[e,w^k]\}_{k=0}^\infty$ exhausts $W$.
I have not made any serious effort to prove this, it's based on my intuition, but as this is not a peer-reviewed paper I'm happy to throw caution to the wind and call it a conjecture. This conjecture would follow by the argument I gave in my original answer if you proved the following property of $w$:

As $k\to\infty$, the number of occurrences of each $s\in S$ in some (equivalently any) reduced word for $w^k$ is unbounded.

This holds true, for example, if $w$ satisfies the conditions in conjecture (which in particular imply that every element of $S$ appear in every reduced expression for $w$), and $D_L(w)\cap D_L(w^{-1})=\emptyset$ and $$W_{D_L(w)\cup D_L(w^{-1})}=W_{D_L(w)}\ast W_{D_L(w^{-1})}.$$ This condition guarantees there is no cancellation between consecutive copies of $w$ in $w^k$, however it is much stronger than it needs to be, I'm sure.
The question for the weak order
If $\{[e,w^k]_L\}_{k=0}^\infty$ is to exhaust $W$, then in particular $S\subset W$ must be a subset of $\bigcup_{k=0}^N [e,w^k]_L$ for some $N$.  In finite Coxeter groups, there are elements (eg Coxeter elements) which have the longest element as a power, and if $w^k=w_0$ is the longest element we have $S=D_L(w_0)\subset[e,w_0]_L$. Outside of this situation, we always have $[e,w^k]_L\cap S=D_L(w^k)\subsetneq S$ (cf Example 2), so for $\bigcup_{k=0}^N [e,w^k]_L$ to contain $S$, there must be distinct powers of $w$ contributing.
Of course even if this is achieved somehow, $w$ must still have infinite order in the hopes of the weak ideals exhausting $W$.
Conjecture 4: Let $(W,S)$ be an infinite irreducible Coxeter group, and let $w\in W$ have infinite order, then $\bigcup_{k=0}^\infty D_L(w^k)\subsetneq S$. In particular there is no $w$ such that $\{[e,w^k]_L\}_{k=0}^\infty$ exhausts $W$.
Again I have given no serious thought to proving this. Even if this is not true, I strongly suspect this is true for most $w$, in the same sense of "most" in my original answer.
