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Let $W$ be a Coxeter group with simple generators $s_1$, $s_2$, ..., $s_r$. Let $\Phi^+$ be the corresponding positive root system, with $\alpha_i$ the positive root corresponding to $s_i$. Bjorner and Brenti, Combinatorics of Coxeter Groups, Chapter Four define the root poset to be a partial order on $\Phi$ as follows:

If $\beta \in \Phi$ and $s_i \beta - \beta \in \mathbb{R}_{>0} \alpha_i$, then $\beta < s_i \beta$. The root poset is then the transitive closure of this relation.

Bjorner and Brenti, Exercise 4.15 asks:

Is the positive root poset $(\Phi^+, \leq)$, with a bottom element appended, a meet-semilattice?

I can't find the answer to this exercise. Can someone help?

What I would actually like to know is:

Are intervals $[\beta, \gamma]$ in the root poset lattices?

Below, some bibliographic notes:

  • I cheated slightly above: Bjorner and Brenti, as well as the sources below, actually only order $\Phi^+$, not $\Phi$. But I see no reason not to extend the order to the negative roots.

  • This poset was introduced in Henrik Eriksson's PhD thesis and, independently, by Brink and Howlett:

Brink, Brigitte; Howlett, Robert B., A finiteness property and an automatic structure for Coxeter groups, Math. Ann. 296, No. 1, 179-190 (1993). ZBL0793.20036.

  • This poset is not the same as defining $\beta \leq \gamma$ if $\gamma - \beta$ is in the positive span of the $\alpha_i$; a condition which is also sometimes called the root poset.

  • Let $\beta$ be a positive root and $t$ the corresponding reflection. Then $s_i \beta - \beta \in \mathbb{R}_{>0} \alpha_i$ if and only if $s_i$ is an inversion of $s_i t s_i$. Thus, we can define this relation in a purely Coxeter theoretic way, without mentioning root systems.

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  • $\begingroup$ What's an easy example where the two orders on the root system differ? $\endgroup$
    – LSpice
    May 21, 2020 at 15:29
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    $\begingroup$ $B_2$. I'll take my simple roots to be $(-1,1)$ and $(1,0)$ (so the Euclidean form is the standard one). The partial order I want is two chains: $(-1,-1) < (1,-1) < (-1,1) < (1,1)$ and $(0,-1) < (-1,0) < (1,0) < (0,1)$. Just asking that $\gamma -\beta$ be in the span of the positive roots gives $(-1,-1) < (0,-1) < [ (1,-1),\ (-1,0) ] < [ (-1,1),\ (1,0) ] < (0,1) < (1,1)$ where the square brackets surround incomparable pairs. Each incomparable pair is comparable to every element not in the pair. (Doing posets without diagrams is tough!) @LSpice $\endgroup$ May 21, 2020 at 15:42
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    $\begingroup$ This order seems related to the order defining the existence of non-zero homomorphisms between Verma modules; see en.wikipedia.org/wiki/… and also this previous MO question: mathoverflow.net/questions/338226/…. $\endgroup$ May 21, 2020 at 16:49
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    $\begingroup$ It's worth pointing out that Björner and Brenti's definition of the "root poset" is probably not as widespread as the other definition (mentioned near the bottom of David's question). The other definition is the one that shows up in "antichains in the root poset" (AKA "nonnesting partitions") in Coxeter-Catalan combinatorics. $\endgroup$ May 22, 2020 at 16:21
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    $\begingroup$ Another point related to @NathanReading's remark, and also the question about when these two orders differ, is that for say an irreducible crystallographic root system, the 'other' (more usual) root poset $\Phi^+$ is connected; whereas for the one David Speyer is considering, in general there will be two connected components corresponding to the long and short roots (which lie in different Weyl group orbits). $\endgroup$ May 22, 2020 at 16:54

2 Answers 2

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The root poset for $\tilde{A_2}$ is shown in Figure 4.5 in the same reference and copied below. One can check that the elements labelled $112$ and $221$ have both $100$ and $010$ as maximal common lower bounds, so meets don't exist in general. If you take an isomorphic example higher up in this same poset you can get an interval which is not a lattice.

The root poset for type <span class=$\tilde{A_2}$.">

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John Stembridge recently pointed me towards his very nice paper Quasi-Minuscule Quotients and Reduced Words for Reflections, which gives a lot of insight into the root poset.

Here is what I understand from his paper. Let $t$ be a reflection and let $\beta$ be the corresponding root. Then $\phi: w \mapsto - w \beta$ is a surjective poset map from the weak order interval $[e,t]$ to the interval $[-\beta, \beta]$ in the root poset. Of course, $[e,t]$ is a lattice, so, if $\phi$ is a poset isomorphism, then $[-\beta,\beta]$ is a lattice as well.

It turns out that $\phi$ is an isomorphism if and only if it is a bijection. Theorem 2.6 in Stembridge gives a necessary and sufficient condition for $\phi$ to be a bijection, but one important thing to note is that this always occurs if the Dynkin diagram is a forest (in particular, in all finite types and all affine types other than $\tilde{A}$).

I'm not sure if anyone but me cares about this old question, but I worked through the first example of a non-lattice in $\tilde{A}_2$ from this perspective. Let $t = (s_1 s_2 s_3)^2 s_1 (s_1 s_2 s_3)^{-2}$. I have drawn the interval $[e,t]$ in the image below: $[e,t]$ are the labels on the triangles inside the diamond region. In the bottom of the diamond, I have labeled each triangle with a reduced word; in the top, the words got too long so I just put black dots. The edges of the Hasse diagram are dual to the triangular tiling; moving down the page is going down in the lattice.

enter image description here

The root corresponding to $t$ is $\beta:=4 \alpha_1 + 3 \alpha_2 + 3 \alpha_3$. The map from $[e,t]$ to $[-\beta, \beta]$ identifies $(s_1 s_2 s_3)^2$ with $(s_1 s_3 s_2)^2$ (both indicated with light shading) and identifies $(s_1 s_2 s_3)^2 s_1$ with $(s_1 s_3 s_2)^2 s_1$ (dark shading). If we had chosen a larger reflection, we would have gotten a larger diamond $[e,t]$, and then $[-\beta, \beta]$ would be obtained by quotienting this diamond by a one dimensional group of translations so that the poset again had width $3$.

Now, look at the elements $x$ and $y$ (in red). The meet $x \wedge y$ in $\tilde{A}_2$ is $s_1 s_2 s_3 s_1 s_2 s_1$ (marked in black). However, $x$ dominates one of the dark shaded regions and $y$ dominates the other one so, in the quotient $[-\beta, \beta]$, the dark shaded element is a second, incomparable, lower bound for the images of $x$ and $y$.

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