Lattice structure in the root poset 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.
 A: 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.
$\tilde{A_2}$.">
A: 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.

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 $y$.
