I'm struggling to understand a statement in Sec 2.2 of "A simplicial complex of Nichols algebras" by Cuntz and Lentner. Simplicial arrangements are sets of hyperplanes in real vector spaces satisfying various properties, and arise in the study of Weyl groupoids. They are defined as:
- A simplicial arrangement is a finite set $A=\{H_1,\dots,H_n\}$ of hyperplanes in $V=\mathbb{R}^r$ such that every chamber $K$ (i.e. connected component in $V\backslash \cup_{H\in A}H$) is an open simplicial cone, meaning $\exists \alpha_1^\vee,\dots,\alpha_r^\vee\in V$ s.t.
$$K=\{\sum_{i=1}^r\lambda_i \alpha_i^\vee|\ \lambda_i>0\ \forall i\}=:\langle \alpha_1^\vee,\dots,\alpha_r^\vee \rangle_{>0}$$
Now for each hyperplane $H_i$ we can choose $x_i\in V^*$ s.t. $H=\ker(x_i)$. Let $R:=\{\pm x_1,\dots,\pm x_n\}$. It is then stated that for each chamber $K$ there is a unique choice of basis $\{\alpha_1^\vee,\dots,\alpha_r^\vee\}$ of $V$ such that $K=\langle \alpha_1^\vee,\dots,\alpha_r^\vee \rangle_{>0}$ and the dual basis $\{\alpha_1,\dots,\alpha_r\}$ for $V^*$ is a subset of $R$.
I'm trying to verify this statement with a simple example, but running into difficulties. Let $V=\mathbb{R}^2$, and take hyperplanes $H_1,H_2$ (i.e. just lines) given by the span of vectors $v_1=(1,0),v_2=(1,1)$ respectively. Then all the chambers can be given as $$K_1=\langle v_1,v_2 \rangle_{>0},\ K_2=\langle -v_1,v_2 \rangle_{>0},\ K_3=\langle v_1,-v_2 \rangle_{>0},\ K_4=\langle -v_1,-v_2 \rangle_{>0}$$
So by choosing $\alpha_i^\vee=\pm v_i$ with signs depending on chamber chosen, I believe it then follows that this is a simplicial arrangement? Any pair from $\{\pm v_1,\pm v_2\}$ is basis as well, but considering the corresponding dual bases, these have kernels on the hyperplanes/lines orthogonal to $H_1$ or $H_2$, which do not coincide with $H_1,H_2$, and therefore these dual basis vectors cannot be in $R$, so something isn't working here.
Perhaps my example is not a simplicial arrangement for some reason, I expect I've overlooked something quite trivial.
Edit based on answer: So now if $\triangle^K$ denotes the dual basis $\{\alpha_1,\dots,\alpha_r\}$ of $V^*$ corresponding to chamber $K$, then we have “crystallographic arrangement" if for all chambers $K$, $R\subset \sum_{\alpha\in \triangle^K} \mathbb{Z}\alpha$. If I now understand things better my example should be crystallographic too, because, for instance $\triangle^{K_2}=\{-f_2,f_1\}$, and indeed the elements of $R=\{\pm f_1,\pm f_2\}$ are $\mathbb{Z}$-linear combinations of $f_1,-f_2$. And the other chambers follow similarly.
Best Answer
Your example is a simplicial arrangement, but you are mixing up $V$ and $V^*$ or have a wrong notion of dual bases in mind. In your example in $\mathbb R^2$ you have for $i=1,2$ $$ H_i = \langle v_i \rangle = \operatorname{ker}(f_i) $$ where $v_1=(1,0)$, $v_2=(1,1)$, $f_1(x,y)=y$ and $f_2(x,y)=x-y$. So $R = \{\pm f_1,\pm f_2\}$.
Now consider some chamber, say $K_2 = \langle -v_1, v_2\rangle_{>0}$. The basis $\{-v_1, v_2\}$ of $\mathbb R^2$ has dual basis $\{-f_2, f_1\}$ of $(\mathbb R^2)^*$ and indeed both $-f_2$ and $f_1$ are elements of $R$.