is a lower order ideal in ${\mathscr{T}}_B$ for every choice of base chamber $B\in{\mathscr{T}}\cap{\mathscr{L}}_{\geq C}$.

The weighted complex {#section_weighted_complex}
--------------------

Let $\lambda\in V$. We consider the following subset of the face poset ${\mathscr{L}}$: $${\mathscr{L}}_{\lambda}=\{C\in{\mathscr{L}}:  C \subseteq \overline{H_\lambda^+}\}.$$ More generally, if $C_0$ is a fixed face of ${\mathscr{L}}$, we also consider the intersection $${\mathscr{L}}_{\lambda,\geq C_0}={\mathscr{L}}_\lambda\cap{\mathscr{L}}_{\geq C_0} .$$

(See [@OM Section 4.5] for definitions.) If $\lambda\not=0$, then the hyperplane arrangement $\{H_\lambda\}\cup\{H_{\alpha_e} : e \in E\}$ defines an affine oriented matroid with distinguished hyperplane $H_\lambda$. If $H_\lambda$ is in general position relative to the $H_{\alpha_e}$, that is, if $\lambda$ is not in the span of any family $(\alpha_e)_{e \in F}$ for $|F| \leq \dim(V)-1$, then ${\mathscr{L}}_{\lambda}$ coincides with the bounded complex of this affine oriented matroid. In general, ${\mathscr{L}}_{\lambda}$ is larger.

The basic properties of the subsets ${\mathscr{L}}_{\lambda}$ and ${\mathscr{L}}_{\lambda,\geq C_0}$ are given in the following proposition.

The following two statements hold:

-   For a fixed face $C_0$ of ${\mathscr{L}}$ the set ${\mathscr{L}}_{\lambda,\geq C_0}$ is a lower order ideal in ${\mathscr{L}}_{\geq C_0}$.

-   Let $C\in{\mathscr{L}}_{\lambda}$. Then there exists $T\in{\mathscr{T}}\cap{\mathscr{L}}_{\lambda}$ such that $C\leq T$.

\[prop\_weighted\_complex\_basic\]

It suffices to prove (i) when $C_0$ is the minimal face of ${\mathscr{L}}$. Let $C,D\in{\mathscr{L}}$ such that $C \leq 