Example
Let \(H_1,H_2,H_3\subset \mathbb P^2_{xyz}\) be three hyperplanes (copies of \(\mathbb P^1)\) meeting at a point \(\pt\) in \(\mathbb P^2\). For example, in coordinates we can choose
- \(H_1 = V(x = 0) = \{[0:t:1]\}\)
- \(H_2 = V(y = 0) = \{[t:0:1]\}\)
- \(H_3 = V(x = y) = \{[t:t:1]\}\)
so that in the affine chart \(z \neq 0\), \(\pt = (0,0)\), \(H_1\) and \(H_2\) are the \(y\)-axis and \(x\)-axis respectively and \(H_3\) is the line \(x = y\). The intersection lattice, ordered with respect to reverse inclusion is then
\begin{equation*} \mathcal L = \hspace{0.5em}where \(\hat 0 = \mathbb P^2\) is the unique minimal element in \(\mathcal L\). Arrows flow small to large. We find the unique building set for \(\mathcal L\), its nested complex and the building algebra.
Building Set Here there is only one possible building set: \[\mathcal G = \{H_1, H_2, H_3, \pt\} = \mathcal L\setminus \{\hat 0\}\] which is the minimal building set since it is precisely the set of irreducibles of \(\mathcal L\) (see [Convex Geometry of Building Sets, text before Lemma 2.2]).
Nested Complex Any subset of \(\mathcal G\) of cardinality at least 2 consisting of pairwise incomparable elements will necessarily have a total join equal to \(\pt\), hence the only nested sets are those either with cardinality less than 2 or with only comparable elements. Of the \(\# 2^{\mathcal G} = 16\) subsets of \(\mathcal G\), we see that \(8\) are nested and \(8\) are not. The nested sets are \[\mathcal N(\mathcal L, \mathcal G) = \overbrace{\Big\{\emptyset, \underset{\times 3}{\{H_i\}}, \{\pt\}\Big\}}^{\text{sets of cardinality < 2}} \cup \overbrace{\Big\{\underset{\times 3}{\{H_i, \pt\}} \Big\}}^{\substack{\text{sets with only} \\ \text{comparable elements}}}\] and the non-nested sets are \[2^\mathcal G \setminus \mathcal N(\mathcal L, \mathcal G) = \Big\{\underset{\times 3}{\{H_i, H_j\}}, \{H_1, H_2, H_3\}, \underset{\times 3}{\{H_i, H_j, \pt\}}, \{H_1, H_2, H_3, \pt\}\Big\}.\] The abstract simplicial complex formed by \(\mathcal N(\mathcal L, \mathcal G)\) has diagramatic representation given by the top half of the lattice above, with edges labeled by set sets \(\{H_i, \pt\}\in \mathcal N(\mathcal L, \mathcal G)\).
Building Algebra The building algebra is a quotient of a polynomial algebra \(\mathbb Z[\{x_G\}_{G\in \mathcal G}]\) by the ideal with one generator of the form \[\prod_{G\in S} x_{G}\] for every subset \(S\subset \mathcal G\) which is not nested (i.e. \(S\not\in \mathcal N(\mathcal L, \mathcal G)\)) and one generator of the form \[\sum_{\substack{G\in \mathcal G,~ A\leq G}} x_G \] for every atom \(A \in \mathcal L\). In our case, we have four elements in our building set corresponding to four generators \(x_1,x_2,x_3,x_4\). Every non-nested subset of \(\mathcal L\) is also a subset of \(\mathcal G\) implying that we have \(8\) generators of the first form, and we have three atoms \(H_1, H_2 \) and \(H_3\) so \(3\) elements of the second form. Thus the ideal generated by products is
\begin{align*} I_1 &= \big(x_1x_2, x_1x_3,x_2x_3, x_1x_2x_3, x_1x_2x_4, x_1x_3x_4, x_2x_3x_4, x_1x_2x_3x_4 \big) \\ &= \big(x_1x_2, ~ x_1x_3, ~ x_2x_3\big) \end{align*}and the ideal generated by sums is
\begin{align*} I_2 &= \big(x_1 + x_4, ~ x_2 + x_4, ~ x_3 + x_4\big). \end{align*}It is not hard to see that \[D(\mathcal L, \mathcal G) := \mathbb Z[x_1,x_2,x_3, x_4]/(I_1 + I_2) \xrightarrow{x_4 \mapsto x}\mathbb Z[x]/(x^2)\] is an isomorphism; the image of \(x_4\) determines the image of every other \(x_i\) due to the relations provided by \(I_1 \) and \(x_4^2 \in I_1 + I_2\) since \[x^2_4 = x_4(x_1 + x_4) - x_1(x_2 + x_4) + x_1x_2.\]