Injective Split Systems

Abstract

A split system $\mathcal S$ on a finite set X, $|X|\ge 3$, is a set of bipartitions or splits of X which contains all splits of the form $\{x,X-\{x\}\}$, $x \in X$. To any such split system $\mathcal S$ we can associate the Buneman graph $\mathcal B(\mathcal S)$ which is essentially a median graph with leaf-set X that displays the splits in $\mathcal S$. In this paper, we consider properties of injective split systems, that is, split systems $\mathcal S$ with the property that ${{\,\textrm{med}\,}}_{\mathcal B(\mathcal S)}(Y) \ne {{\,\textrm{med}\,}}_{\mathcal B(\mathcal S)}(Y')$ for any 3-subsets $Y,Y'$ in X, where ${{\,\textrm{med}\,}}_{\mathcal B(\mathcal S)}(Y)$ denotes the median in $\mathcal B(\mathcal S)$ of the three elements in Y considered as leaves in $\mathcal B(\mathcal S)$. In particular, we show that for any set X there always exists an injective split system on X, and we also give a characterization for when a split system is injective. We also consider how complex the Buneman graph $\mathcal B(\mathcal S)$ needs to become in order for a split system $\mathcal S$ on X to be injective. We do this by introducing a quantity for |X| which we call the injective dimension for |X|, as well as two related quantities, called the injective 2-split and the rooted-injective dimension. We derive some upper and lower bounds for all three of these dimensions and also prove that some of these bounds are tight. An underlying motivation for studying injective split systems is that they can be used to obtain a natural generalization of symbolic tree maps. An important consequence of our results is that any three-way symbolic map on X can be represented using Buneman graphs.

1 Introduction

Let X be a finite set $|X| \ge 3$. A (three-way) symbolic map (on X) is a map $\delta :{X \atopwithdelims ()3} \rightarrow M$ to some set M of symbols. In [16], a special type of symbolic map was studied, called a symbolic tree map which arises as follows. Let T be a phylogenetic tree with leaf-set X (i.e. an unrooted tree with no vertices of degree two and leaf set X [20]) in which each interior vertex v of T is labelled by some element l(v) in M by some labelling map l. The symbolic tree map $\delta$ associated to T is the map from ${X \atopwithdelims ()3}$ to M that is obtained by setting

$$\begin{aligned} \delta (Y) = l({{\,\textrm{med}\,}}_T(Y)), \quad Y \in {X \atopwithdelims ()3}, \end{aligned}$$

where ${{\,\textrm{med}\,}}_T(Y)$ is the unique interior vertex of T that belongs to the shortest paths between each pair of the three vertices in Y, and ${X \atopwithdelims ()3}$ denotes the set of all 3-subsets of X. For example, for the symbolic tree map $\delta$ associated to the labelled tree in Fig. 1(i), $\delta (\{1,2,3\})=c$, and $\delta (\{2,3,4\})=b$. Symbolic tree maps are closely related to symbolic ultrametrics [6] and also appear in the theory of hypergraph colourings [14]—see [16] for more details, where amongst other results, a characterization of symbolic tree maps is presented. There are also close connections with cograph theory [15] and modular decompositions [7].

Fig. 1

For $X=\{1, \ldots , 5\}$, a phylogenetic tree on X in (i) and a Buneman graph on X in (ii). In (i) the internal vertices are labelled by the elements in $M=\{a,b,c\}$ and in (ii) they are labelled by the elements in $M=\{a,b,\dots ,k\}$

In [16] it was asked how results on symbolic tree maps might be extended to Buneman graphs [10] (see also [7, p.8]), as these graphs provide a natural way to generalize phylogenetic trees and have useful applications in evolutionary biology [2]. More specifically, given a split system (on X), i. e.  a set $\mathcal S$ of bipartitions or splits of X that contains all splits of the form $\{\{x\}, X-\{x\}\}$, $x \in X$, then the Buneman graph $\mathcal B(\mathcal S)$ on X associated to $\mathcal S$ is essentially a median graph with leaf-set X (see Sect. 2 for more details). The fact that $\mathcal B(\mathcal S)$ is a median graph implies that for any 3-subset Y of X, there exists a unique vertex ${{\,\textrm{med}\,}}_{\mathcal B(\mathcal S)}(Y)$ in $\mathcal B(\mathcal S)$ (or median), that lies on shortest paths between any pair of elements in Y. Since every phylogenetic tree is a Buneman graph, the notion of a symbolic tree map naturally generalises by considering labelling maps $\delta$ that can be represented by labelling the internal vertices of some Buneman graph $\mathcal B(\mathcal S)$, and, for any 3-subset Y of X, taking $\delta (Y)$ to be the label of ${{\,\textrm{med}\,}}_{\mathcal B(\mathcal S)}(Y)$. For example, for the map $\delta$ associated to the interior vertex-labelled Buneman graph depicted in Fig. 1(ii), $\delta (\{1,2,3\})=k$, and $\delta (\{3,4,5\})=f$.

It is therefore of interest to understand under what circumstances we can represent for a split system $\mathcal S$ on X a symbolic map $\delta$ on X by labelling the vertices of some Buneman graph $\mathcal B(\mathcal S)$ on X and vertex set V. In other words, we want to find some labelling map $l :V-X\rightarrow M$ such that $\delta (\{x,y,z\}) = l({{\,\textrm{med}\,}}_{\mathcal B(\mathcal S)}(Y))$ for all $Y \in {X \atopwithdelims ()3}$. Clearly this is the case if there is some split system $\mathcal S$ on X such that

$$\begin{aligned} {{\,\textrm{med}\,}}_{\mathcal B(\mathcal S)}(Y) \ne {{\,\textrm{med}\,}}_{\mathcal B(\mathcal S)}(Y') \text{ for } \text{ all } \text{ distinct } Y, Y' \in {X \atopwithdelims ()3}, \end{aligned}$$

(1)

since then we can just label the vertex ${{\,\textrm{med}\,}}_{\mathcal B(\mathcal S)}(Y)$ by $\delta (Y)$ for every 3-subset Y of X. For example, the Buneman graph depicted in Fig. 1(ii) enjoys Property (1), whereas the phylogenetic tree T (which is a Buneman graph for the split system obtained by deleting all seven edges in turn) in Fig. 1(i) does not since, for example, ${{\,\textrm{med}\,}}_T(\{1,5,3\})={{\,\textrm{med}\,}}_T(\{1,5,4\})$. Motivated by these considerations, we call a split system $\mathcal S$ injective if Property (1) holds. In this paper we shall focus on understanding such split systems, in particular presenting some results concerning their properties. We now briefly summarize them.

In the next section, we begin by presenting some preliminaries concerning Buneman graphs and also briefly explain how they arise in evolutionary applications. In Sect. 3 we then prove that for any finite set X with $|X|\ge 3$, there always exists some injective split system on X. In particular, we show that the split system on X which contains all those splits $\{A,B\}$ of X with $\min \{|A|,|B|\}| \le 2$, and the split system that is obtained by deleting any pair of edges in a cycle with vertex set X are both injective (Theorem 1). In particular, as mentioned above, it follows that any symbolic map $\delta$ on a set X can be represented by some Buneman graph.

In Sect. 4, we provide a characterization of injective split systems (Theorem 2). This characterization is obtained by considering how the restriction of a split system on X to small subsets of X partitions these subsets. In particular, it implies that it can be decided if a split system $\mathcal S$ on X is injective or not by considering the restriction of $\mathcal S$ to subsets of X with size at most 6.

In general, since we can always represent a symbolic map by some Buneman graph, we would like to find representations that are as simple as possible. Since for any split system $\mathcal S$ the Buneman graph $B(\mathcal S)$ is an isometric subgraph of an $|\mathcal S|$-cube in which the convex hull of any isometric cycle of length k is a k-cube, $k\ge 3$, a natural measure for the complexity of a split system $\mathcal S$ is the dimension of the largest isometric k-cube in $B(\mathcal S)$. We call this quantity the dimension of $\mathcal S$; for example, the split systems in Fig. 1(i) and (ii) have dimension 1 and 2, respectively.

In Sect. 5, we investigate the notion of the injective dimension ID(n) which we define to be the smallest dimension of any injective split system on a set of size n, $n \ge 3$. In particular, as well as giving the values of ${{\,\textrm{ID}\,}}(n)$ for all $n \le 8$, we show that ${{\,\textrm{ID}\,}}(n) \le \lfloor \frac{n}{2} \rfloor$, and that ${{\,\textrm{ID}\,}}(n) \ge 3$ for all $n \ge 8$ (Theorem 3). As an immediate corollary to this result it follows that to represent arbitrary symbolic maps on sets X of size 6 or more using Buneman graphs, Buneman graphs that contain 3-cubes are required.

We continue by considering two variants of the injective dimension. The first variant, ${{\,\textrm{ID}\,}}_2(n)$, is considered in Sect. 6 and is given by restricting the definition of ${{\,\textrm{ID}\,}}(n)$ to split systems $\mathcal S$ for which every split $\{A,B\} \in \mathcal S$ has $\min \{|A|,|B|\}| \le 2$. We show that for all $n \ge 5$, $\lfloor \frac{n}{2} \rfloor \le {{\,\textrm{ID}\,}}_2(n) \le n-3$ (Theorem 4) which implies that ${{\,\textrm{ID}\,}}_2(5)=2$. The second variant, ${{\,\textrm{ID}\,}}^r(n)$, is considered in Sect. 7, and is defined by modifying the definition of injectivity as follows: We say that a split system $\mathcal S$ on X is rooted-injective relative to some $r \in X$ if

$$\begin{aligned} {{\,\textrm{med}\,}}_{\mathcal B(\mathcal S)}(Z\cup \{r\}) \ne {{\,\textrm{med}\,}}_{\mathcal B(\mathcal S)}(Z' \cup \{r\}) \text{ for } \text{ all } \text{ distinct } Z, Z' \in {X \atopwithdelims ()2}. \end{aligned}$$

The quantity ${{\,\textrm{ID}\,}}^r(n)$ is given in an analogous way to ${{\,\textrm{ID}\,}}(n)$ by taking the minimum over rooted-injective splits systems relative to r. Using a recent result from [7] concerning rooted median graphs, we show that, in contrast to ${{\,\textrm{ID}\,}}_2(n)$, ${{\,\textrm{ID}\,}}^r(n)=2$ for all $n \ge 4$. We conclude in Sect. 8 with a discussion of some open problems.

2 Preliminaries

2.1 Graphs and Median Graphs

We consider undirected graphs $G=(V,E)$ whose vertex sets V are finite with $|V| \ge 2$, and whose edge sets E are contained in $\left( {\begin{array}{c}V\\ 2\end{array}}\right)$, i.e., graphs without loops and multiple edges. A leaf in such a graph is a vertex with degree one. A cycle is a connected graph in which every vertex has degree two. The length of a cycle C is the number of edges or, equivalently, the number of vertices in C. A connected graph that does not contain a cycle is called a tree.

If G is connected then we denote by $d_{G}(v,w)$ the length of a shortest path between two vertices v and w of G. Note that $d_{G}(v,w)=0$ if and only if $v=w$. A connected subgraph $G'$ of G is called isometric if $d_{G'}(v,w) = d_{G}(v,w)$, for all vertices v and w in $G'$.

A vertex x in G is called a median of three vertices $u,v,w\in V$ if $d_{G}(u,x)+d_{G}(x,v)=d_{G}(u,v)$, $d_{G}(v,x)+d_{G}(x,w)=d_{G}(v,w)$ and $d_{G}(u,x)+d_{G}(x,w)=d_{G}(u,w)$. A connected graph is called a median graph if any three of its vertices have a unique median [18]. In other words, G is a median graph if for all vertices u, v, and w in G, there is a unique vertex that belongs to shortest paths between each pair of u, v and w. We denote the unique median of three vertices u, v and w in a median graph G by ${{\,\textrm{med}\,}}_{G}(u,v,w)$. Median graphs have several interesting characterizations and properties, see e.g. [19]. For example, a connected graph G is a median graph if and only if the convex hull¹ of any isometric cycle of G is a hypercube (see e.g. [17]).

2.2 Buneman Graphs

From now on, we let X be a finite set with $|X|\ge 3$. A split (of X) is a bipartition $A|B=B|A$ of X into two non-empty subsets, that is, $A,B\subset X$, $A\cap B=\emptyset$ and $A\cup B=X$. For simplicity, we write $a_1\ldots a_k|b_1\ldots b_l$ or $a_1\ldots a_k|\overline{a_1\ldots a_k}$ for a split A|B if $A=\{a_1,\ldots , a_k\}$ and $B=\{b_1,\ldots , b_l\}$, for some $k,l\ge 1$. We call the sets A and B the parts of the split A|B. If $S=A|B$ is such that $|A|<|B|$ then we call A the small part of S. The size of a split A|B is defined as $\min \{|A|,|B|\}$, and if a split S has size r we call S an r-split. A split A|B of X is called trivial if it has size 1 or, equivalently, if A|B is of the form $x|{\overline{x}}$ for some $x\in X$. For a split $S=A|B$ of X, we let S(x) denote the part of S that contains x. We say that S separates two elements x and y in X if $S(x)\not =S(y)$. From now on we shall assume that all split systems on X contain all trivial splits on X.

Following [10], we define for a split system $\mathcal S$ on X, the Buneman graph $\mathcal B(\mathcal S)$ (on X) to be the graph with vertex set consisting of all maps $\phi : \mathcal S \rightarrow \mathcal P(X)$ (where $\mathcal {P}(X)$ denotes the power-set of X) satisfying the following two conditions:

• (B1) For all $S\in \mathcal S$, $\phi (S) \in S$.

• (B2) For all $S, S' \in \mathcal S$ distinct, $\phi (S) \cap \phi (S') \ne \emptyset$.

Two vertices $\phi$ and $\phi '$ in $\mathcal B(\mathcal S)$ are joined by an edge if there is a unique split $S \in \mathcal S$ such that $\phi (S) \ne \phi '(S)$. For example, the graphs in Fig. 1(i) and (ii) are Buneman graphs on $X=\{1,\ldots , 5\}$ for the split systems

$$\begin{aligned} \mathcal S_1=\{15|234,24|135\} \cup \{x|{\overline{x}} :\, x \in X\} \end{aligned}$$

and

$$\begin{aligned} \mathcal S_2=\{15|234,24|135, 12|345, 34|125,35|124\} \cup \{x|{\overline{x}} :\, x \in X\}, \end{aligned}$$

respectively.

Note that Buneman graphs have applications in evolutionary biology, especially to the analysis of human mitochondrial DNA [2]. More specifically, in mitochondrial applications DNA sequences are typically aligned and then recoded into binary sequences to form a 0/1 matrix whose rows are indexed by a set X of taxa. Columns in this matrix are then deleted if all of their entries are the same or have missing entries, and repeats of columns are also removed. It follows that each column i in the resulting binary matrix M corresponds to a different split of X, where the parts of the split are given by checking whether the row for each element in X has 0 or a 1 in column i. A median network is then constructed for this matrix by repeatedly taking the median sequence for triples of binary sequences starting with the sequences in M until no new sequences are obtained, where the median sequence for each triple is given by taking the majority rule sequence for the three sequences. It can be shown (see e.g [9, Theorem 4.9]) that this median network is isomorphic to the Buneman graph of the set of splits of X corresponding to M. In this way, we see that vertices in Buneman graphs can be thought of as maps on split systems as above, or as binary sequences; in this paper we will use the former terminology as this is more convenient for our arguments. Note that Buneman graphs can also be defined in terms of certain set families—see e.g. [4].

2.3 Properties of Buneman Graphs

We now summarise some relevant properties of the Buneman graph (for proofs of these facts see e.g. [12, Chapter 4]; see also [5] using different notation).

• (S1) For all $x \in X$, the map $\phi _x: \mathcal S \rightarrow \mathcal P(X)$ given by putting $\phi _x(S)=S(x)$, for all $S \in \mathcal S$, is a leaf in $\mathcal B(\mathcal S)$.

• (S2) Let $S=A|B\in {\mathcal {S}}$. Then the removal of all edges $\{\phi ,\phi '\}$ in $\mathcal B(\mathcal S)$ with $\phi (S)\ne \phi '(S)$ disconnects $\mathcal B(\mathcal S)$ into precisely two connected components, one of which contains the leaves $\phi _a$, $a \in A$ and the other the leaves $\phi _b$, $b \in B$.

• (S3) $\mathcal B(\mathcal S)$ is a median graph.

• (S4) $\mathcal B(\mathcal S)$ is an isometric subgraph of the $|\mathcal S|$-dimensional hypercube consisting of all those maps $\phi : \mathcal S \rightarrow \mathcal P(X)$ that only satisfy Property (B1) in the definition of the Buneman graph (with edge set defined in the analogous same way).

• (S5) For any three vertices $\phi _1, \phi _2,\phi _3$ in $\mathcal B(\mathcal S)$, the median of $\phi _1, \phi _2$ and $\phi _3$ in $\mathcal B(\mathcal S)$ is the map that assigns to each split $S \in \mathcal S$ the part of S of multiplicity two or more in the multiset $\{\phi _1(S),\phi _2(S),\phi _3(S)\}$ (see also [11, p. 1905, Equ. (1)]).

Suppose that ${\mathcal S}$ is a split system on X. In light of Property (S1), we shall consider X as being the leaf-set of $\mathcal B(\mathcal S)$, since each $x \in X$ corresponds to the map $\phi _x$ in $\mathcal B(\mathcal S)$. As an example for (S2), consider the tree in Fig. 1(i). Removing the edge associated to the split 15|234 disconnects the tree into two trees with leaf sets $\{1,5\}$ and $\{2,3,4\}$, respectively. In this way, we see that $\mathcal B(\mathcal S_1)$ displays each of the splits in $\mathcal S_1$.

Note that by Property (S3) and the fact mentioned at the end Sect. 2.1, the convex hull of any isometric cycle in $\mathcal B(\mathcal S)$ is a hypercube. In light of this, we define the dimension $\dim (\mathcal S)$ of a split system $\mathcal S$ to be the dimension of the largest hypercube contained in $\mathcal B(\mathcal S)$ in case $\mathcal B(\mathcal S)$ is not a phylogenetic tree and one otherwise. This dimension can be characterized in terms of splits as follows. Suppose $S=A|B$ and $T=C|D$ are two splits in $\mathcal S$. Then S and T are called incompatible if $S\not =T$ and $A\cap C$, $A \cap D$, $B\cap C$ and $B \cap D$ are all non-empty; otherwise S and T are called compatible. Calling a set $\mathcal S$ of splits incompatible if any two splits in $\mathcal S$ are incompatible, then $\dim (\mathcal S)$ is equal to the maximum size of an incompatible subset of $\mathcal S$ (see e.g. [8, p. 445]). If $\mathcal B(\mathcal S)$ contains a cycle then it must contain a hypercube of dimension two or more. Hence, a split system $\mathcal S$ on X is 1-dimensional if and only if $\mathcal B(\mathcal S)$ is a phylogenetic tree on X (in which case it has $|\mathcal S|+1$ vertices and |X| leaves), a fact which also holds if and only if every pair of splits in $\mathcal S$ is compatible (see e.g. [10]). In particular, as mentioned in the introduction, it follows that any phylogenetic tree is a Buneman graph of some split system, and that any two distinct splits in this split system must be compatible.

3 Two Families of Injective Split Systems

Let $\mathcal S$ be a split system on X. For $Y =\{x,y,z\} \in {X \atopwithdelims ()3}$, we let $\phi _Y = \phi _{xyz}=med_{\mathcal B(\mathcal S)}(Y)$ denote the median of $\phi _x,\phi _y,\phi _z$ in $\mathcal B({\mathcal S})$, which exists by Property (S3). In this notation, $\mathcal S$ is injective if for all $Y,Y' \in {X \atopwithdelims ()3}$ distinct, we have $\phi _Y \ne \phi _{Y'}$. Note that if $|X|=3$, then there is only one split system $\mathcal S$ on X (the one that contains only trivial splits), and that $\mathcal S$ is injective, since $|\left( {\begin{array}{c}X\\ 3\end{array}}\right) |=1$. In this section, we show that for every set X with $|X|\ge 4$ there exists an injective split system on X. To do this, we shall present two infinite families of injective split systems.

We begin with a simple but useful lemma.

Lemma 1

Let $\mathcal S$ be a split system on a set X, $|X| \ge 3$, and let $x,y,z \in X$ distinct. Then $\phi _{xyz}$ is the (unique) map in $\mathcal B(\mathcal S)$ that assigns to each split $S \in \mathcal S$ the part $A\in S$ for which $|A \cap \{x,y,z\}| \ge 2$.

Proof

Let $S\in \mathcal S$. Then $\phi _v(S)=S(v)$, for all $v\in \{x,y,z\}$. By Property (S5), $\phi _{xyz}(S)$ is the part of S that appears twice (or more) in the multiset $\{S(x),S(y),S(z)\}$, that is, the part of S that contains (at least) two elements of $\{x,y,z\}$. $\square$

Now, a split system $\mathcal S$ on X is called circular [1] if there exists a labelling $x_1, \ldots , x_n$, $n=|X|$, of the elements of X such that all splits of $\mathcal S$ are of the form $x_i x_{i+1} \ldots x_j|\overline{x_i x_{i+1} \ldots x_j}$, some $1 \le i \le j \le n$. If $\mathcal S$ is a circular split system on X and there is no circular split system $\mathcal S'$ on X such that $\mathcal S \subsetneq \mathcal S'$, then we say that $\mathcal S$ is a maximal circular split system on X (for $n=4,5,6$, see Fig. 2 for a representation of such a split system in terms of the associated Buneman graph). Note that a maximal circular split system on X has size ${|X| \atopwithdelims ()2}$ [1, Section 3].

We now use Lemma 1 to show that there exist families of split systems that are injective.

Theorem 1

Let $\mathcal S$ be a split system on X, $|X| \ge 4$. Then:

• (i) If $\mathcal S$ contains all 2-splits of X, then $\mathcal S$ is injective.

• (ii) If $\mathcal S$ is maximal circular, then $\mathcal S$ is injective.

Proof

For both (i) and (ii), let $Y=\{x,y,z\}$ and $Y'$ denote two distinct subsets of X of size 3. Assume without loss of generality that $x\notin Y'$.

• (i) By Lemma 1, $\phi _Y$ is the unique map $\mathcal B(\mathcal S)$ that assigns to each split $S \in \mathcal S$ the part A of S such that $|A \cap Y| \ge 2$. It follows that for $S=xy|{\overline{xy}}$ (which is an element of $\mathcal S$ as it has size two), $\phi _{xyz}(S)=\{x,y\}$. Since $x\not \in Y'$, we obtain $\phi _{Y'}(S)= X-\{x,y\}$. Consequently, $\phi _Y \ne \phi _{Y'}$.

• (ii) Put $X=\{x_1,\ldots , x_n\}$, $n\ge 4$. Then there exist $i,j,k\in \{1,\ldots , n\}$ with $i<j<k$ ($\textrm{mod}\ n$) such that $x=x_i$, $y=x_j$ and $z=x_k$. With respect to the circular ordering of X induced by $\mathcal S$ it follows that one of the four sets $\{x=x_i,x_{i+1}, \ldots , y=x_j\}$, $\{y=x_j,x_{j+1}, \ldots , x=x_i\}$, $\{x=x_i,x_{i+1}, \ldots , z=x_k\}$ and $\{z=x_k,x_{k+1},\ldots , x=x_i\}$ must contain at most one element of $Y'$. Let A be such a set. Since $\mathcal S$ is maximal circular by assumption, it follows that the split $S=A | X-A$ is contained in $\mathcal S$. By Lemma 1, $\phi _Y(S)=A\not =X-A=\phi _{Y'}(S)$. Hence, $\phi _Y \ne \phi _{Y'}$.$\square$

Fig. 2

For $X=\{1,\dots ,n\}$ with $n=4, 5,6$ and the induced natural ordering of X, the respective Buneman graphs on X of the associated maximal circular split systems on X where (i) is $n=4$, (ii) is $n=5$, and (iii) is $n=6$. In all cases, leaves are indicated in terms of the elements in X. Vertices that are of the form $\phi _{xyz}$, some $x,y,z \in X$, are indicated as unfilled circles and all other non-leaf vertices are indicated as filled circles

In view of Theorem 1 (ii), it is interesting to understand if maximal circular split systems admit proper subsets that are also injective. As it turns out, the answer is no in general, as we show in our next result.

Proposition 1

Let $\mathcal S$ be a circular split system on X with $|X|\ge 4$ and let $\mathcal S'$ denote a split system on X that is contained in $\mathcal S$ as a proper subset. Then $\mathcal S'$ is not injective.

Proof

Let $S_0$ be a non-trivial split in $\mathcal S-\mathcal S'$. We show that there exists two subsets Y and Z of $X=\{1,\ldots , n\}$ distinct such that $\phi _Y(S)=\phi _Z(S)$ for all $S \in \mathcal S-\{S_0\}$. In particular, $\phi _Y(S)=\phi _Z(S)$ for all $S \in \mathcal S'$, so $\mathcal S'$ is not injective.

Assume that $\mathcal S$ is circular for the natural ordering of X. Without loss of generality, we may assume that $S_0=1\ldots k|k+1 \ldots n$, some $2 \le k \le \frac{n}{2}$. Consider the sets $Y=\{n,1,k\}$ and $Z=\{n,1,k+1\}$. Let $S\in \mathcal S-\{S_0\}$. If $S(n)=S(1)$ then, by Lemma 1, $\phi _Y(S)=S(n)=\phi _Z(S)$. If $S(n)\not =S(1)$ then S must be of the form $1\dots \ell | \ell +1\dots n$, some $1\le \ell \le n-1$. Since $S\ne S_0$, we have $\ell \ne k$. Hence, $S(k)=S(k+1)$. Moreover, since $S(1)\not =S(n)$ either 1 or n must be contained in S(k). We can then apply Lemma 1 again to conclude that $\phi _Y(S)=\phi _Z(S)$ which completes the proof. $\square$

We remark that a similar result to Proposition 1 does not necessarily hold for non-circular split systems even if they are injective. For example, Theorem 1(i) implies that the split system $\mathcal S$ on $X=\{1, \ldots , n\}$, $n \ge 5$, that consists precisely of all trivial splits and 2-splits on X is injective. Let $\mathcal S^*$ denote the split system containing all splits of $\mathcal S$ except those of the form $1x|{\overline{1x}}$, $x \in X-\{1\}$. Then, $\mathcal S^*$ is injective. To see this, consider the proof of Theorem 1(i). Then, up to potentially having to relabel the elements of Y and $Y'$, the elements x and y can always be chosen to be different from 1. Hence, the split $S=xy|{\overline{xy}}$ such that $\phi _Y(S) \ne \phi _{Y'}(S)$ can always be chosen in such a way that $S\in \mathcal S^*$. As a consequence, it follows that for all $Y, Y' \in {X \atopwithdelims ()3}$ distinct, there exists a split S of $\mathcal S^*$ such that $\phi _Y(S) \ne \phi _{Y'}(S)$ which implies that $\mathcal S^*$ is injective.

4 Characterization of Injective Split Systems and Dicing

In this section, we characterize injective split systems (Theorem 2). To this end, we shall consider the restriction of a split system on X to subsets of X which is defined as follows. Given a split system $\mathcal S$ on X, and a subset $Y \subseteq X$ with $|Y|\ge 3$ then we define the restriction $\mathcal S|_Y$ of $\mathcal S$ to Y as the set of splits $S|_Y$ restricted to Y, that is,

$$\begin{aligned} \mathcal S|_Y = \{ S|_Y= A\cap Y| B \cap Y :\, A|B \in \mathcal S \}. \end{aligned}$$

Note that $\mathcal S|_Y$ is in fact a split system on Y since $\mathcal S|_Y$ contains all trivial splits on Y. We begin by proving a useful lemma concerning such restrictions.

Lemma 2

Suppose that S is a split on X with $|X|\ge 4$, and that x, y, z, p are distinct elements of X. Then the following holds for $Y=\{x,y,z,p\}$.

• (i) $\phi _{xyz}(S) \ne \phi _{xyp}(S)$ if and only if $S|_Y\in \{xz|yp,yz|xp\}$. In particular, $S|_Y\not =xy|pz$.

• (ii) If $|X|\ge 5$ and $q\in X-Y$ then $\phi _{xyz}(S) \ne \phi _{xpq}(S)$ if and only if $S|_{Y\cup \{q\}}$ is one of the splits yz|xpq, pq|xyz, xy|zpq, xz|ypq, xp|yzq or xq|yzp.

• (iii) If $|X|\ge 6$ and $q,r\in X-Y$ distinct then $\phi _{xyz}(S) \ne \phi _{pqr}(S)$ if and only if $S|_{Y\cup \{q,r\}}$ is a 3-split or it is a 2-split of $Y \cup \{q,r\}$ whose part of size 2 is contained in $\{x,y,z\}$ or $\{p,q,r\}$.

Proof

Let $S=A|B$.

To see Assertion (i) observe that, by Lemma 1, we have $\phi _{xyz}(S) \ne \phi _{xyp}(S)$ if and only if one of A and B, say A, contains at least two elements of $\{x,y,z\}$ while B contains at least two elements of $\{x,y,p\}$. Since $A\cap B=\emptyset$, this is only possible if and only if $z\in A$ and $p\in B$ while either $x\in A$ and $y\in B$ or $y\in A$ and $x\in B$. The latter is equivalent to $S|_Y\in \{xz|yp,yz|xp\}$ which, in particular, implies that $S|_Y\ne xy|pz$. Hence, Assertion (i) must hold.

To see Assertion (ii), observe that, by Lemma 1, $\phi _{xyz}(S) \ne \phi _{xpq}(S)$ if and only if one of A and B, say A, contains at least two elements of $\{x,y,z\}$ and B contains at least two elements of $\{x,p,q\}$. As is easy to see, this is the case if and only if $S|_{Y'}$ is not a trivial split on $Y'=Y\cup \{q\}$ and one of $S(y)=S(z)$ or $S(p)=S(q)$ holds. Consideration of all ten non-trivial splits on $Y'$ shows that $S|_{Y'}$ must be one of yz|xpq, pq|xyz, xy|zpq, xz|ypq, xp|yzq or xq|yzp. Hence, Assertion (ii) must hold.

To see Assertion (iii), observe that, by Lemma 1, $\phi _{xyz}(S) \ne \phi _{pqr}(S)$ holds if and only if one of A and B, say A, contains at least two elements of $\{x,y,z\}$ and B contains at least two elements of $\{p,q,r\}$. Put $Y'=Y\cup \{q,r\}$, $A'= A\cap Y'$ and $B'=B\cap Y'$. Since $A\cap B=\emptyset$ it follows that $S|_{Y'}$ must be a 2- or 3-split and that if $S|_{Y'}$ is a 2-split, its part of size 2 is contained in $\{x,y,z\}$ or $\{p,q,r\}$.

Conversely, put $A=\{x,y,z\}$ and $B=\{p,q,r\}$ again. If $S|_{Y'} = A'|B'$ is a 3-split on $Y'=Y\cup \{q,r\}$ then, clearly, $|A'\cap \{x,y,z\}|\ge 2$ and $|B'\cap \{p,q,r\}|\ge 2$. Since $A'\subseteq A$ and $B'\subseteq B$, we obtain $\phi _{xyz}(S) \ne \phi _{pqr}(S)$. Furthermore, if $S|_{Y} = A'|B'$ is a 2-split such that the part of size 2 is contained in A or B, then the other part must be of size 4 and must contain B or A. Consequently, $\phi _{xyz}(S) \ne \phi _{pqr}(S)$. Hence, Assertion (iii) must hold. $\square$

We now make a key definition. We shall say that a split system $\mathcal S$ on X

• 4-dices X if $|X| < 4$ or for all $Y \in {X \atopwithdelims ()4}$, $\mathcal S|_Y$ contains at least two 2-splits,

• 5-dices X if $|X| < 5$ or for all $Y \in {X \atopwithdelims ()5}$, $\mathcal S|_Y$ contains at least five 2-splits, and

• 6-dices X if $|X| < 6$ or for all $Y \in {X \atopwithdelims ()6}$, $\mathcal S|_Y$ contains at least one 3-split or a triangle of 2-splits, that is, three 2-splits of the form $xy|Y-\{x,y\}$, $xz|Y-\{x,z\}$ and $yz|Y-\{y,z\}$ where x, y, and z are distinct elements in Y.

Note that, in general, if a split system on X k-dices X it need not $k'$-dice X, for $k,k'\in \{4,5,6\}$ distinct. Nevertheless, some interesting relationship between these concepts hold as the next lemma illustrates.

Lemma 3

Suppose $\mathcal S$ is a split system on X.

• (i) If $\mathcal S$ 4-dices X and $|X| \ge 5$ then, for all $Y \in {X \atopwithdelims ()5}$, $\mathcal S|_Y$ contains at least four 2-splits.

• (ii) If $\mathcal S$ 5-dices X and $|X| \ge 6$ then, for all $Y \in {X \atopwithdelims ()6}$, $\mathcal S|_Y$ contains a 3-split or (at least) eight 2-splits.

Proof

(i) Suppose that $\mathcal S$ 4-dices X and that $|X|\ge 5$. Let $Y=\{x,y,z,t,u\}\in {X \atopwithdelims ()5}$ and $Y'=\{x,y,z,t\}\in {X \atopwithdelims ()4}$. Since $\mathcal S$ 4-dices X, $\mathcal S|_{Y'}$ contains at least two 2-splits $S'_1$ and $S'_2$. Hence, $\mathcal S|_Y$ contains two splits $S_1$ and $S_2$ such that $S_1|_{Y'}=S'_1$ and $S_2|_{Y'}=S'_2$. Moreover, since $S'_1$ and $S'_2$ are both 2-splits on $Y'$, the part $A_1$ of $S_1$ and $A_2$ of $S_2$ of size 2 does not contain u. Note that $A_1$ and $A_2$ must be parts of $S'_1$ and $S'_2$, respectively. In particular, since $S'_1$ and $S'_2$ are splits on $Y'$ and $S'_1\ne S'_2$ it follows that $|A_1 \cap A_2|=1$. Without loss of generality, we may assume that $A_1 \cap A_2=\{x\}$. Replacing $Y'$ by $Y''=\{y,z,t,u\}$ and using an analogous argument implies that $\mathcal S|_Y$ also contains two distinct 2-splits on Y, call them $S_3$ and $S_4$, whose parts of size 2 do not contain x. In particular, $S_3$ and $S_4$ are distinct from $S_1$ and $S_2$. In summary, $\mathcal S|_Y$ contains at least four distinct 2-splits.

(ii) Suppose that $\mathcal S$ 5-dices X and that $|X|\ge 6$. Let $Y \in {X \atopwithdelims ()6}$. If $\mathcal S|_Y$ contains a 3-split we are done. Hence, assume $\mathcal S|_Y$ does not contain a 3-split. Since $|Y|=6$ it follows that a split in $\mathcal S|_Y$ must be trivial or a 2-split. We continue with showing that $\mathcal S|_Y$ contains at least eight 2-splits. Let $x\in Y$. Since a split in $\mathcal S|_Y$ is either trivial or a 2-splits, all 2-splits of $\mathcal S|_{Y-\{x\}}$ correspond to the 2-splits of $\mathcal S|_Y$ whose small part does not contain x. We claim that there exists an element $x_0$ of $Y-\{x\}$ that belongs to the small part of at least three 2-splits of Y. To see this, we consider the following two cases: (a) $\mathcal S|_Y$ does not contain a split whose small part contains x and (b) $\mathcal S|_Y$ contains a split whose small part contains x.

In case of (a), let $Y' = \{x,y,a,b,c\}$ be a subset of Y of size 5. Since $\mathcal S$ 5-dices X, it follows that $\mathcal S|_{Y'}$ contains at least five of the ${4\atopwithdelims ()2}=6$ possible 2-splits in $\{ya|{\overline{ya}}, yb|{\overline{yb}}, yc|{\overline{yc}}, ab|{\overline{ab}}, ac|{\overline{ac}},$ $bc|{\overline{bc}}\}$

that might be contained in $\mathcal S|_Y$ and do not have x in their small part. It is now straight-forward to verify that there is some $x_0 \in Y'-\{x\}$ such that $\mathcal S|_{Y'}$ contains three 2-splits whose small part contains $x_0$.

Consider now Case (b). Since $\mathcal S$ 5-dices X and $|X|\ge 6$, $\mathcal S|_{Y-\{x\}}$ contains again at least five 2-splits. Then if there exists an element $x_0 \in Y-\{x\}$ such that $\mathcal S|_{Y-\{x\}}$ contains three 2-splits whose small part contains $x_0$ then the claim follows. If this is not the case, then consideration of all ${5\atopwithdelims ()2}=10$ possible 2-splits in $\mathcal S|_{Y-\{x\}}$ shows that $\mathcal S|_{Y-\{x\}}$ must contain exactly five 2-splits and that all elements of $Y-\{x\}$ must belong to the small part of exactly two 2-splits of $\mathcal S|_{Y-\{x\}}$. In addition, by assumption on x, there exists an element $x_0$ of $Y-\{x\}$ such that $\{x,x_0\}$ is the small part of a split of $\mathcal S|_Y$. Since $x_0$ also belongs to the small part of exactly two 2-splits in $\mathcal S|_{Y-\{x\}}$, it follows that $x_0$ belongs to the small part of exactly three 2-splits of $\mathcal S|_{Y}$. Hence, there is some $x_0 \in Y-\{x\}$ such that $\mathcal S|_{Y}$ contains three 2-splits whose small part contains $x_0$.

In summary, in both Case (a) and (b), there is some $x_0 \in Y-\{x\}$ such that $\mathcal S|_{Y}$ contains three 2-splits whose small part contains $x_0$. Moreover, $\mathcal S|_{Y-\{x_0\}}$ contains at least five 2-splits because $\mathcal S$ 5-dices X and $|Y|=6$. Since the small part of a split in $\mathcal S|_{Y-\{x_0\}}$ is also the small part of a split in $\mathcal S|_Y$ whose small part does not contain $x_0$, it follows that there also exists at least five 2-splits in $\mathcal S|_Y$ whose small part does not contain $x_0$. Hence, $\mathcal S|_Y$ contains at least eight 2-splits. $\square$

To prove the main theorem of this section, we require a further result concerning dicing.

Proposition 2

Suppose $\mathcal S$ is a split system on X with $|X|\ge 4$. Then the following holds.

• (i) $\mathcal S$ 4-dices X if and only if for all $A,B \in {X \atopwithdelims ()3}$ with $|A \cap B|=2$, we have $\phi _A \ne \phi _B$.

• (ii) If $|X|\ge 5$ then $\mathcal S$ 4- and 5-dices X if and only if for all distinct $A,B \in {X \atopwithdelims ()3}$ with $A \cap B \ne \emptyset$, we have $\phi _A \ne \phi _B$.

Proof

(i) Let $A=\{x,y,z\}$ and $B=\{x,y,t\}$ be subsets of X and let $Y=A \cup B$. Assume first that $\mathcal S$ 4-dices X. Then $\mathcal S|_Y$ contains at least two 2-splits because $|Y|=4$. In particular, $\mathcal S|_Y$ contains at least one 2-split S distinct from xy|tz. By Lemma 2(i), it follows that $\phi _A(S) \ne \phi _B(S)$. Consequently, $\phi _A \ne \phi _B$.

Conversely, if $\phi _A \ne \phi _B$, then there exists a split S in $\mathcal S$ such that $\phi _A(S) \ne \phi _B(S)$. By Lemma 2(i), $S|_Y \in \{xz|yt,yz|xt\}$. If $S|_Y = xz|yt$, then consider the set $C=\{x,z,t\}$. Since, by assumption, $\phi _A \ne \phi _C$ there must exist a split $S'$ in $\mathcal S$ such $\phi _A(S') \ne \phi _C(S')$. By Lemma 2(i) it follows that $S'|_Y \ne S|_Y$. If $S|_Y = yz|xt$ then an analogous argument with C replaced by $D=\{y,z,t\}$ implies that there exists a split $S''$ with $S''|_Y \in \{yx|zt,yt|zx\}$. Lemma 2(i) implies again that $S|_Y \ne S''|_Y$. Hence, $\mathcal S|_Y$ contains at least two 2-splits one of which is $S|_Y$ and the other is $S'|_Y$ or $S''|_Y$.

(ii) Assume first that $\mathcal S$ 4-dices and 5-dices X. Let $A,B\in \left( {\begin{array}{c}X\\ 3\end{array}}\right)$ distinct such that $A\cap B\not =\emptyset$. If $|A\cap B|=2$ then, by Proposition 2(i), $\phi _A \ne \phi _B$ must hold. So assume that $|A\cap B|\not =2$. Then $|A\cap B|=1$. Let $A=\{x,y,z\}$ and $B=\{x,p,q\}$. Since $\mathcal S$ 5-dices X and $|X|\ge 5$ it follows that $\mathcal S|_Y$ contains at least five 2-splits where $Y=A\cup B$. Since there are exactly ${5\atopwithdelims ()2}=10$ 2-splits on Y, it follows that $\mathcal S|_Y$ contains at least one of the six 2-splits in $\{yz|xpq, pq|xyz, xy|zpq, xz|ypq, xp|yzq,xq|yzp\}$. By Lemma 2(ii), it follows that $\phi _A \ne \phi _B$.

Conversely, assume that for all distinct $A,B \in {X \atopwithdelims ()3}$ with $A \cap B \ne \emptyset$ we have that $\phi _A \ne \phi _B$. If $|A\cap B|=2$ then $\mathcal S$ 4-dices X in view of Proposition 2(i). To see that $\mathcal S$ also 5-dices X, we need to show in view of $|X|\ge 5$ that for all $Y\in \left( {\begin{array}{c}X\\ 5\end{array}}\right)$ the split system $\mathcal S|_Y$ contains at least five 2-splits.

Let $Y\in \left( {\begin{array}{c}X\\ 5\end{array}}\right)$. Since $\mathcal S$ 4-dices X, it follows by Lemma 3(i) that $\mathcal S|_Y$ contains at least four 2-splits. Assume for contradiction that $\mathcal S|_Y$ contains precisely four 2-splits $S_1, \ldots , S_4$. For all $1\le i\le 4$, let $A_i$ denote the small part of $S_i$. Then the multiset $\mathcal A=A_1 \cup A_2 \cup A_3 \cup A_4$ contains eight elements. We claim that there exists no element $x\in Y$ with multiplicity three or more in $\mathcal A$. To see the claim, assume for contradiction that there exists some $x \in X$ that is contained in three of $A_i$, $1\le i\le 4$. Since, for all $1 \le i \le 4$, the split $S_i|_{Y-\{x\}}$ is a 2-split of $Y-\{x\}$ if and only if $x\not \in A_i$ it follows that $\mathcal S|_{Y-\{x\}}$ contains at most one 2-split. But this is not possible because $\mathcal S$ 4-dices X and $|X|\ge 5$ thereby concluding the proof of the claim. Hence, every element of Y has multiplicity at most two in $\mathcal A$. Since Y contains five elements and $\mathcal A$ has size eight, one of the following two cases must hold: (a) three elements of Y have multiplicity two in $\mathcal A$ and the other two have multiplicity one and (b) four elements of Y have multiplicity two in $\mathcal A$ and one does not appear in $\mathcal A$.

Suppose first that Case (a) holds. Let x and y be the two elements in $\mathcal A$ that appear only once. Then there exists an element $q\in Y- \{x,y\}$ such that neither $\{x,q\}$ nor $\{y,q\}$ is contained in $\{A_1,A_2,A_3,A_4\}$. Since q has multiplicity two in $\mathcal A$ while x and y have multiplicity one each, this implies that there exist $i,j\in \{1,\ldots , 4\}$ distinct such that the two sets $A_i$ and $A_j$ not containing q satisfy $A_i \cup A_j=\{x,y,z,p\}$. It follows that $\mathcal S|_{A_i\cup A_j}$ only contains the split $A_i|A_j$, contradicting the fact that $\mathcal S$ 4-dices X.

Suppose now that Case (b) holds. Let x be the element of Y not present in $\mathcal A$. Since each element of $\{y,z,p,q\}$ appears twice in $\mathcal A$, it follows that, up to potentially having to relabel the elements of $Y-\{x\}$, $S|_Y=\{yp|xzq, yq|xzp, zp|xyq, zq|xyp\}$. We can now use Lemma 2(ii) to conclude that $\phi _A=\phi _B$, which contradicts our assumption that $\phi _A\not =\phi _B$. $\square$

Note that the assumption that $\mathcal S$ 4-dices X is necessary for the characterization in Proposition 2 (ii) to hold. For example, the split system on $X=\{1, \ldots 5\}$ whose set of non-trivial splits equals

$$\begin{aligned} \{12|345, 23|451, 34|512, 45|123\} \end{aligned}$$

does not 5-dice X but $\phi _A \ne \phi _B$ holds for all $A,B \in {X \atopwithdelims ()3}$ with $|A \cap B|=1$.

We now show that injectivity of a split system can be characterized by considering at most 6-points.

Theorem 2

Suppose $\mathcal S$ is a split system on X, $|X| \ge 3$. Then $\mathcal S$ is injective if and only if $\mathcal S$ 4-, 5- and 6-dices X.

Proof

If $|X|=3$, then the equivalence trivially holds. Hence, we may assume for the following that $|X| \ge 4$.

Assume first that $\mathcal S$ 4-, 5- and 6- dices X, and let $A,B \in {X \atopwithdelims ()3}$ distinct. If $|X|=4$, then $|A \cap B|=2$. In that case, Proposition 2(i) implies that $\phi _A \ne \phi _B$. If $|X|=5$, then $A \cap B \ne \emptyset$. In that case, Proposition 2(ii) implies that $\phi _A \ne \phi _B$. Finally, suppose that $|X| \ge 6$. In view of Proposition 2(ii), we have that $\phi _A \ne \phi _B$ holds in case $A \cap B \ne \emptyset$. It remains to show that $\phi _A \ne \phi _B$ also holds when $A \cap B=\emptyset$. To see this, let $A=\{x,y,z\}$ and $B=\{t,u,v\}$ be subsets of X such that $A\cap B=\emptyset$. Let $Y=A \cup B$. Since $|X|\ge 6$ and $\mathcal S$ 6-dices X, the split system $\mathcal S|_Y$ contains either a 3-split or a triangle of 2-splits. In both cases, we can use Lemma 2(iii) to conclude that $\phi _A \ne \phi _B$.

Conversely, assume that $\mathcal S$ is injective. Then $\phi _A \ne \phi _B$ for all distinct $A, B \in {X \atopwithdelims ()3}$ with $A \cap B \ne \emptyset$. By Proposition 2(ii), it follows that $\mathcal S$ 4-dices and 5-dices X. To see that $\mathcal S$ also 6-dices X, suppose that $|X| \ge 6$ and let $Y=\{x,y,z,t,u,v\}$ be a subset of X of size 6. Since $\mathcal S$ 5-dices X Lemma 3 implies that $\mathcal S|_Y$ contains either a 3-split or at least eight 2-splits. We claim that if $\mathcal S|_Y$ does not contain a 3-split then $\mathcal S|_Y$ must contain a triangle of 2-splits. To see the claim, we remark first that if $\mathcal S|_Y$ contains ten 2-splits or more, then it must contain a triangle of 2-splits. Employing a case analysis, we obtain that, up to potentially having to relabel the elements of Y, a split system on Y containing eight 2-splits or more without containing a triangle of 2-splits is either (a) the split system $\mathcal S_1$ whose set of non-trivial splits is $\{xy|{\overline{xy}}, xz|{\overline{xz}}, xt|{\overline{xt}}, xu|{\overline{xu}}, yv|{\overline{yv}}, zv|{\overline{zv}}, tv|{\overline{tv}}, uv|{\overline{uv}}\}$ or (b) a subset of the split system $\mathcal S_2$ whose set of non-trivial splits is $\{xy|{\overline{xy}}, yz|{\overline{yz}}, zt|{\overline{zt}}, tu|{\overline{tu}}, uv|{\overline{uv}}, vx|{\overline{vx}}, xt|{\overline{xt}}, yu|{\overline{yu}}, zv|{\overline{zv}}\}$. Since $\mathcal S_1$ does not 5-dice X because $\mathcal S_1|_{Y-\{x\}}$ contains only four 2-splits it follows that $\mathcal S|_Y\not =\mathcal S_1$. Hence, Case (a) cannot hold. But Case (b) cannot hold either since if $\mathcal S|_Y$ is a subset of $\mathcal S_2$ then Lemma 2(iii) implies $\phi _{uxz}= \phi _{tvy}$. But this is impossible because $\mathcal S|_Y$ is injective. Hence, $\mathcal S|_Y$ must contain a triangle of 2-splits, as claimed. Thus, $\mathcal S$ also 6-dices X. $\square$

As an important consequence of the last result, we see that injectivity of a split system is well behaved with respect to restriction:

Corollary 1

Suppose $\mathcal S$ is a split system on X with $|X|\ge 3$. If $\mathcal S$ is injective, then $\mathcal S|_Y$ is injective, for all $Y \subseteq X$ with $|Y|\ge 3$.

Proof

Suppose that $Y \subseteq X$ with $|Y|\ge 3$ and that $\mathcal S$ is injective. Then, by Theorem 2, $\mathcal S$ 4-, 5- and 6-dices X. So $\mathcal S|_Y$ 4-, 5- and 6-dices Y. By Theorem 2, it follows that $\mathcal S|_Y$ is injective. $\square$

5 The Injective Dimension

Recall that the dimension $\dim (\mathcal S)$ of a split system $\mathcal S$ is defined as the dimension of the largest hypercube in $\mathcal B(\mathcal S)$ or, equivalently, the size of the largest incompatible subset of $\mathcal S$. For $n\ge 3$, we define the injective dimension ${{\,\textrm{ID}\,}}(n)$ of n to be

$$\begin{aligned} {{\,\textrm{ID}\,}}(n) = \min \{ \dim (\mathcal S) :\mathcal S \text{ is } \text{ an } \text{ injective } \text{ split } \text{ system } \text{ on } \{1, \ldots , n\} \}. \end{aligned}$$

(2)

Note that since Theorem 1 implies that for all X with $n=|X|\ge 4$ there exists an injective split system on X, the quantity ${{\,\textrm{ID}\,}}(n)$ is well-defined. We are interested in ${{\,\textrm{ID}\,}}(n)$ since its value gives a lower bound for the number of vertices in the Buneman graph of any injective split system on X. In particular, if ${{\,\textrm{ID}\,}}(n)=m$ then the Buneman graph $\mathcal B(\mathcal S)$ of any injective split system $\mathcal S$ on X must contain an m-cube as a subgraph. Hence, $\mathcal B(\mathcal S)$ must contain at least $2^m$ vertices.

To be able to present some upper and lower bounds for ${{\,\textrm{ID}\,}}(n)$ (Theorem 3), we first show that ${{\,\textrm{ID}\,}}: {\mathbb {N}}_{\ge 3} \rightarrow {\mathbb {N}}$ is a monotone increasing function.

Lemma 4

For any two integers n and m with $n \ge m \ge 3$, we have ${{\,\textrm{ID}\,}}(n) \ge {{\,\textrm{ID}\,}}(m)$.

Proof

Let $\mathcal S$ be an injective split system on some set X with $|X|=n$ such that $\dim (\mathcal S)={{\,\textrm{ID}\,}}(n)$. Let Y be a subset of X of size m. By Corollary 1, the split system $\mathcal S|_Y$ is injective, so ${{\,\textrm{ID}\,}}(m) \le \dim (\mathcal S|_Y)$. To see that $\dim (\mathcal S|_Y) \le dim(\mathcal S)$ also holds it suffices to remark that if two splits S and $S'$ in $\mathcal S$ are such that $S|_Y$ and $S'|_Y$ are incompatible then S and $S'$ are also incompatible. Hence, an incompatible subset of $\mathcal S|_Y$ naturally induces an incompatible subset of $\mathcal S$ of the same size. It follows that ${{\,\textrm{ID}\,}}(m) \le \dim (\mathcal S|_Y) \le \dim (\mathcal S)={{\,\textrm{ID}\,}}(n)$, as desired. $\square$

We now give upper and lower bounds for ${{\,\textrm{ID}\,}}(n)$ where $n=|X|\ge 4$. As we shall see in the proof, the upper bound comes from the fact that a maximal circular split system on X is injective by Theorem 1(ii) and that in [8] it was shown that the maximum dimension of a hypercube in $\mathcal B(\mathcal S)$ is $\lfloor \frac{n}{2} \rfloor$. Note that the split system $\mathcal S$ formed by all splits of X of size two or less is injective by Theorem 1(i) and, by [8], has dimension $n-1$. Indeed, two splits S and $S'$ in $\mathcal S$ are incompatible if there exists an element $x \in X$ such that x belongs to the small part of both S and $S'$. Hence, the largest incompatible subsets of $\mathcal S$ are the subsets of the form $xy|{\overline{xy}}\,:\, y \in X-\{x\}\}$, some $x \in X$, and these subsets have size $n-1$.

Theorem 3

For all integers $n \ge 4$, we have ${{\,\textrm{ID}\,}}(n) \le \lfloor \frac{n}{2} \rfloor$. Moreover, ${{\,\textrm{ID}\,}}(3)=1$, ${{\,\textrm{ID}\,}}(4)={{\,\textrm{ID}\,}}(5)=2$, ${{\,\textrm{ID}\,}}(6)={{\,\textrm{ID}\,}}(7)={{\,\textrm{ID}\,}}(8)=3$, and for all $n \ge 9$, ${{\,\textrm{ID}\,}}(n) \ge 3$.

Proof

Let $X=\{1, \ldots , n\}$. To see that the first statement holds, let $\mathcal S$ be a maximal circular split system on X. If $n\ge 4$ then Theorem 1(ii) implies that $\mathcal S$ is injective. Hence, ${{\,\textrm{ID}\,}}(n) \le \dim (\mathcal S)$. By [8], the Buneman graph $\mathcal B(\mathcal S')$ of a maximal circular split system $\mathcal S'$ on X contains an $\lfloor \frac{n}{2} \rfloor$-cube, and all other subcubes in $\mathcal B(\mathcal S')$ have no larger dimension. Hence, $\dim (\mathcal S')=\lfloor \frac{n}{2} \rfloor$. Thus, ${{\,\textrm{ID}\,}}(n) \le \lfloor \frac{n}{2} \rfloor$.

To see the remainder of the theorem, note first that ${{\,\textrm{ID}\,}}(3)=1$ since, as was mentioned in Sect. 3 already, the unique split system on X is injective and $\mathcal B(\mathcal S)$ is a phylogenetic tree on X

To see that ${{\,\textrm{ID}\,}}(4)={{\,\textrm{ID}\,}}(5)=2$ holds, we first remark that in view of the first statement of the theorem, we have ${{\,\textrm{ID}\,}}(4) \le 2$ and ${{\,\textrm{ID}\,}}(5) \le 2$. Now, let X be such that $n \in \{4,5\}$ and assume for contradiction that there exists an injective split system $\mathcal S$ on X with $\dim (\mathcal S)=1$. In particular, $\mathcal S$ is compatible. Then $\mathcal B(\mathcal S)$ is a phylogenetic tree on X and has $|\mathcal S|+1$ vertices. Moreover, since a compatible split system on X has at most $2n-3$ elements (see e.g. [12, Theorem 3.3]), it follows that $\mathcal B(\mathcal S)$ has at most 2 internal vertices if $n=4$, and at most 3 internal vertices if $n=5$. But $\mathcal S$ is injective, so $\mathcal B(\mathcal S)$ must have at least ${4 \atopwithdelims ()3}=4$ internal vertices if $n=4$, and at least ${5 \atopwithdelims ()3}=10$ internal vertices if $n=5$, a contradiction. Hence, ${{\,\textrm{ID}\,}}(4)={{\,\textrm{ID}\,}}(5)=2$.

We continue with showing that ${{\,\textrm{ID}\,}}(6)\ge 3$ from which it then follows by Lemma 4 and the first statement of the theorem that ${{\,\textrm{ID}\,}}(6)={{\,\textrm{ID}\,}}(7)=3$ and that ${{\,\textrm{ID}\,}}(n) \ge 3$, for all $n \ge 8$. Suppose that $\mathcal S$ is an injective split system on $X=\{1,\ldots , 6\}$. Bearing in mind that, by Theorem 2, $\mathcal S$ 4-, 5- and 6-dices X we next perform a case analysis on the number of 3-splits in $\mathcal S$. If $\mathcal S$ contains three 3-splits or more then $\dim (\mathcal S) \ge 3$ since all 3-splits of X are pairwise incompatible.

If $\mathcal S$ contains two 3-splits, say 123|456 and 234|561, then since $\mathcal S$ 4-dices X it follows that there must exist a split $S\in \mathcal S$ such that $S(2) \ne S(3)$ and $S(5) \ne S(6)$. Since the splits S, 123|456, and 234|561 are pairwise incompatible, we obtain $\dim (\mathcal S) \ge 3$.

If $\mathcal S$ contains one 3-split, say 123|456, then one of the following two cases must hold. If there exists an element $x \in X$ and three splits $S_1$, $S_2$, and $S_3$ in $\mathcal S$ containing x in their small part then $\dim (\mathcal S) \ge 3$ because $\{S_1,S_2,S_3\}$ is incompatible. If no such element x exists then $\mathcal S$ contains at most six 2-split. An exhaustive search shows that, up to potentially having to relabel the elements in $\{1,2,3\}$, there exists only one such split system that is injective i. e. $\mathcal S$ is the split system whose subset of non-trivial splits is the set

$$\begin{aligned} \{123|456, 15|{\overline{15}}, 16|{\overline{16}}, 24|{\overline{24}}, 26|{\overline{26}}, 34|{\overline{34}}, 35|{\overline{35}}\}. \end{aligned}$$

One can then easily verify that $\{123|456,15|{\overline{15}}, 16|{\overline{16}}\}$ is incompatible. Hence, $\dim (\mathcal S)\ge 3$ in this case.

Finally, if $\mathcal S$ does not contain a 3-split, then it must contain a triangle of 2-splits because $\mathcal S$ 6-dices X. Since the three splits in such a triangle are pairwise incompatible it follows that $\dim (\mathcal S) \ge 3$. This concludes the proof that ${{\,\textrm{ID}\,}}(6)\ge 3$.

To show that ${{\,\textrm{ID}\,}}(8) =3$, we employed Theorem 2 and used a computer program to verify that $\mathcal S$ is the split system whose subset of non-trivial splits is

$$\begin{aligned}{} & {} \{1234|5678,1357|2468,123|{\overline{123}},246|{\overline{246}},478|{\overline{478}},\\{} & {} 156|{\overline{156}},12|{\overline{12}}, 34|{\overline{34}},56|{\overline{56}},78|{\overline{78}},26|{\overline{26}},\\{} & {} 35|{\overline{35}},17|{\overline{17}},48|{\overline{48}},68|{\overline{68}},57|{\overline{57}},23|{\overline{23}}\} \end{aligned}$$

is injective. Since $\dim (\mathcal S)=3$, it follows that ${{\,\textrm{ID}\,}}(8)=3$. $\square$

Note that as ${{\,\textrm{ID}\,}}(8)=3$, the upper bound for ${{\,\textrm{ID}\,}}(n)$ given in Theorem 3 is not tight even for $n=8$. In general, it appears to be difficult to find a better upper or lower bounds for ${{\,\textrm{ID}\,}}(n)$, however in the next two sections we shall give improved bounds for two variants of the injective dimension.

6 The Injective 2-Split-Dimension

To help better understand the injective dimension of a split system, in this section we shall consider a restricted version of this quantity that is defined as follows. For $n\ge 3$, let ${\mathbb {S}}_2(n)$ be the set of all injective split systems on $X=\{1,\dots , n\}$ whose non-trivial splits all have size 2. As mentioned in the introduction, we define ${{\,\textrm{ID}\,}}_2(n)$ for $n\ge 3$ as

$$\begin{aligned} {{\,\textrm{ID}\,}}_2(n) = \min \{ dim(\mathcal S) :\mathcal S\in {\mathbb {S}}_2(n) \}. \end{aligned}$$

(3)

By Theorem 1 (i), ${{\,\textrm{ID}\,}}_2(n)$ is well-defined. Clearly ${{\,\textrm{ID}\,}}_2(n) \ge {{\,\textrm{ID}\,}}(n)$ and equality holds for $n =3,4,5$ since every non-trivial split of a set X of size 3, 4, or 5 is a 2-split. In the main result of this section (Theorem 4), we provide upper and lower bounds for ${{\,\textrm{ID}\,}}_2(n)$. To prove it, we shall use two lemmas.

For $\mathcal S$ a split system on X, we denote by $P(\mathcal S)$ the graph with vertex set X and with edge set all the pairs $\{x,y\}$ such that $xy|{\overline{xy}} \in \mathcal S$. We also denote the degree of a vertex $x \in X$ in $P(\mathcal S)$ by $\deg _{P(\mathcal S)}(x)$. If $\mathcal S$ contains only trivial splits and 2-splits then $P(\mathcal S)$ and dicing are related as stated as in Lemma 5. We omit its straight-forward proof but remark in passing that Lemma 5 is a strengthening of Theorem 2 for split systems in ${\mathbb {S}}_2(n)$, for all $n\ge 3$.

Lemma 5

Let $\mathcal S \in {\mathbb {S}}_2(|X|)$ be a split system on X with $|X| \ge 3$. Then,

• $\mathcal S$ 4-dices X if and only if $|X| \le 4$ or for all $Y \in {X \atopwithdelims ()4}$, the restriction $P(\mathcal S|_Y)$ contains two edges that share a vertex.

• $\mathcal S$ 5-dices X if and only if $|X| \le 5$ or for all $Y \in {X \atopwithdelims ()5}$, the restriction $P(\mathcal S|_Y)$ contains five edges or more.

• $\mathcal S$ 6-dices X if and only if $|X| \le 6$ or for all $Y \in {X \atopwithdelims ()6}$, the restriction $P(\mathcal S|_Y)$ contains a 3-clique.

In terms of the dimension of a split system in ${\mathbb {S}}_2(n)$, $n\ge 3$, we also have the following result.

Lemma 6

Let $\mathcal S \in {\mathbb {S}}_2(|X|)$ be a split system on X with $|X| \ge 3$. Then,

• (i) If $P(\mathcal S)$ does not contain a 3-clique then $\dim (\mathcal S)=\max _{x \in X} \{\deg _{P(\mathcal S)}(x)\}$.

• (ii) If $P(\mathcal S)$ contains a 3-clique then $\dim (\mathcal S)=\max \{\max _{x \in X} \{\deg _{P(\mathcal S)}(x)\}, 3\}$.

Proof

We prove (i) and (ii) together. For this, put $n=|X|$. If $n=3$ then $P(\mathcal S)$ consists of three isolated vertices. So Assertion (i) holds. Since $\mathcal S$ only contains trivial splits, it follows that Assertion (ii) holds vacuously. So assume that $n\ge 4$. Let $\mathcal S \in {\mathbb {S}}_2(n)$. Then a maximal incompatible subset $\mathcal S'$ of $\mathcal S$ must be of one of the following two types:

• (a) A triangle of 2-splits.

• (b) The set of all 2-splits in $\mathcal S$ containing some $x\in X$ in their small part.

To see that these are the only two possible types, it suffices to remark that any subset $\mathcal S'\subseteq \mathcal S$ with $|\mathcal S'|\ge 4$ is incompatible if and only if there exists some $x\in X$ such that all splits of $\mathcal S'$ contain x in their small part.

If $\mathcal S'$ is of Type (a) then $\mathcal S'$ corresponds to a 3-clique in $P(\mathcal S)$ and $|\mathcal S'|=3$. If $\mathcal S'$ is of Type (b) then $\mathcal S'$ corresponds to the set of edges of $P(\mathcal S)$ that are incident with x. Hence, $|\mathcal S'|=\deg _{P(\mathcal S)}(x)\ge 3$. Thus, if $P(\mathcal S)$ has a vertex x with $\deg _{P(\mathcal S)}(x)\ge 3$ or if $P(\mathcal S)$ does not contain a 3-clique then $\dim (\mathcal S)=\max _{x \in X} \{\deg _{P(\mathcal S)}(x)\}$. Otherwise, $\dim (\mathcal S)=3$. $\square$

We now prove the main result of this section.

Theorem 4

For all $n \ge 5$,

$$\begin{aligned} \left\lfloor \frac{n}{2}\right\rfloor \le {{\,\textrm{ID}\,}}_2(n) \le n-3. \end{aligned}$$

Proof

We first show that ${{\,\textrm{ID}\,}}_2(n) \le n-3$ by constructing an injective split system $\mathcal S_n$ on $X_n=\{1, \ldots , n\}$ with $\dim (\mathcal S)=n-3$. For this, let $\sigma _n$ denote some circular ordering of the elements of $X_n$. Let $\mathcal S_n$ denote the set of all splits $xy|X_n-\{x,y\}$ such that $x,y \in X_n$ are not consecutive under $\sigma _n$. By definition of $\mathcal S_n$, all vertices of $P(\mathcal S_n)$ have degree $n-3$. If $n \ge 6$, it follows by Lemma 6 that $\dim (\mathcal S_n)=n-3$. If $n=5$, it is straight-forward to check that $P(\mathcal S_n)$ does not contain a 3-clique. So, by Lemma 6, $\dim (\mathcal S_n)=n-3$ holds in this case too. Thus, it remains to show that $\mathcal S_n$ is injective. In view of Theorem 2, we do this by showing that $\mathcal S_n$ 4-, 5- and 6-dices $X_n$.

To see that $\mathcal S_n$ 4-dices $X_n$, let $Y \in {X_n \atopwithdelims ()4}$ which exists as $n\ge 5$. By Lemma 5, it suffices to show that there exists an element of Y that has degree 2 or more in $P(\mathcal S_n|_Y)$. Let $x \in Y$. If $\deg _{P(\mathcal S_n|_Y)}(x)\ge 2$, we are done by the definition of $\mathcal S_n$. Otherwise, Y contains two elements y and z such that y and z precede and follow x under $\sigma _n$, respectively. Let t be the fourth element of Y. Then $\{x,t\}$ is an edge in $P(\mathcal S_n)$. Moreover, since $n \ge 5$ and $t \ne x$, there must be at least one of y, z that is adjacent with t in $P(\mathcal S_n)$. Thus, $\deg _{P(\mathcal S_n|_Y)}(t)\ge 2$, as required.

To see that $\mathcal S_n$ 5-dices $X_n$, let $Y \in {X_n \atopwithdelims ()5}$ which again exists because $n\ge 5$. By Lemma 5, it suffices to show that $P(\mathcal S_n|_Y)$ contains at least five edges. To see this, note first that, for all $x \in X$, there are at most two elements in $Y-\{x\}$ that do not form an edge with x in $P(\mathcal S_n|_Y)$ because $\mathcal S_n$ is circular. For all $x \in Y$, it follows that $\deg _{P(\mathcal S_n|_Y)}(x)\ge 2$. Since Y contains five elements, this implies that $P(\mathcal S_n|_Y)$ contains at least five edges, as required.

Finally, to see that $\mathcal S_n$ 6-dices $X_n$, note first that we may assume that $|X|\ge 6$ as otherwise $\mathcal S_n$ 6-dices $X_n$ by definition. Let $Y \in {X_n \atopwithdelims ()6}$. By Lemma 5, it suffices to show that $P(\mathcal S_n|_Y)$ contains a 3-clique. To see this, let $x \in Y$. Then, by the definition of $P(\mathcal S_n)$, there exist at least three elements in Y, say y, z and t, that form an edge with x in $P(\mathcal S_n)$. Moreover, at least two of y, z and t, say y and z, must form an edge $\{y,z\}$ in $P(\mathcal S_n)$ since y, z and t cannot all be consecutive with each other under $\sigma _n$. It follows that $\{x,y,z\}$ is the vertex set of a 3-clique in $P(\mathcal S_n|_Y)$, as required. This concludes the proof that ${{\,\textrm{ID}\,}}_2(n) \le n-3$.

We now show that $\lfloor \frac{n}{2} \rfloor \le {{\,\textrm{ID}\,}}_2(n)$. We begin by showing that ${{\,\textrm{ID}\,}}_2(n+2)>{{\,\textrm{ID}\,}}_2(n)$, for all $n \ge 3$. Assume that $n\ge 3$. Also, assume that $\sigma _{n+2}$ is the natural ordering of $X_{n+2}=\{1,2,\ldots , n,n+1, n+2\}$. Let $\mathcal S\in {\mathbb {S}}_2(n+2)$ denote a split system on $X_{n+2}$ that attains ${{\,\textrm{ID}\,}}_2(n+2)$. Let $\mathcal S'$ denote a maximal incompatible subset of $\mathcal S$. We claim that $\mathcal S'$ must contain a non-trivial split that separates the elements $n+1$ and $n+2$. Clearly, $\mathcal S$ must contain such a split as otherwise Lemma 1 implies that $\phi _Y(S)=\phi _{Y'}(S)$ holds for all $S \in \mathcal S$ and all $Y,Y' \in {X_{n+2} \atopwithdelims ()3}$ with $Y \cap Y'=\{n+1,n+2\}$. Hence, $\mathcal S$ is not injective which is impossible. Choose a split $S_0\in \mathcal S$ such that $S_0(n+1)\not =S_0(n+2)$. Assume for contradiction that all splits $S \in \mathcal S'$ satisfy $S(n+1)=S(n+2)$. Then $\mathcal S_0$ is incompatible with every split in $\mathcal S'$ because $S_0$ and every split in $\mathcal S'$ have size two. Hence, $\mathcal S'\cup \{S_0\}$ is an incompatible subset of $\mathcal S$ that contains $\mathcal S'$ as a proper subset which contradicts the choice of $S'$.

Consider now the restriction $\mathcal S_n$ of $\mathcal S$ to $X_n$. By Corollary 1, $\mathcal S_n$ is injective because $\mathcal S$ is injective. Moreover, since all maximal incompatible subsets of $\mathcal S$ contain a split separating $n+1$ and $n+2$ by the previous claim, it follows that no maximal incompatible subset of $\mathcal S_n$ has size equal to $\dim (\mathcal S)$. Hence, $\dim (\mathcal S_n)< \dim (\mathcal S)$. Since $\dim (\mathcal S)={{\,\textrm{ID}\,}}_2(n+2)$ by the choice of $\mathcal S$, and $\dim (\mathcal S_n) \ge {{\,\textrm{ID}\,}}_2(n)$ by the injectivity of $\mathcal S_n$, it follows that ${{\,\textrm{ID}\,}}_2(n+2)>{{\,\textrm{ID}\,}}_2(n)$, as required.

We conclude with showing that ${{\,\textrm{ID}\,}}_2(n) \ge \lfloor \frac{n}{2} \rfloor$ holds by performing induction on n. If $n =5$ then ${{\,\textrm{ID}\,}}_2(n)={{\,\textrm{ID}\,}}(n)$ since all non-trivial splits on $X_n$ are 2-splits and ${{\,\textrm{ID}\,}}(n)= \lfloor \frac{n}{2} \rfloor$ holds by Corollary 3. This implies the stated inequality in this case. Now, let $n>5$ and assume that the stated inequality holds for all $5\le n'<n$. Since ${{\,\textrm{ID}\,}}_2(n)>{{\,\textrm{ID}\,}}_2(n-2)$ it follows by induction hypothesis that ${{\,\textrm{ID}\,}}_2(n)> {{\,\textrm{ID}\,}}_2(n-2)\ge \lfloor \frac{n-2}{2} \rfloor$. Hence, ${{\,\textrm{ID}\,}}_2(n) \ge \lfloor \frac{n-2}{2} \rfloor +1=\lfloor \frac{n}{2} \rfloor$, as desired. $\square$

7 Rooted Injective Dimension

In this section, we consider another variant of the injective dimension which behaves quite differently from ${{\,\textrm{ID}\,}}(n)$. Let X denote a set with $|X|=n$. Choose some element $r \in X$. For $Z \in {X-\{r\} \atopwithdelims ()2}$, put $Z_r=Z \cup \{r\}$. We say that a split system is rooted-injective (relative to r) if

$$\begin{aligned} \phi _{Z_r} \ne \phi _{Z'_r} \end{aligned}$$

for all $Z,Z' \in {X \atopwithdelims ()2}$ distinct. This concept is closely related to the rooted median graphs considered in [7]. Note that if $X=3$ then the (unique) split system on X is r-rooted injective for any choice of $r\in X$. Also, note that if $\mathcal S$ is injective, then $\mathcal S$ is rooted-injective relative to r, for all $r \in X$. The converse, however, does not hold. For example, the split system $\mathcal S$ on $X=\{1, \ldots , 6\}$ whose set of non-trivial splits is:

$$\begin{aligned} \{14|{\overline{14}}, 15|{\overline{15}}, 16|{\overline{16}}, 24|{\overline{24}}, 25|{\overline{25}}, 26|{\overline{26}}, 34|{\overline{34}}, 35|{\overline{35}}, 36|{\overline{36}}\} \end{aligned}$$

is not injective because $\mathcal S$ does not 6-dice X and so Theorem 2 does not hold. But $\mathcal S$ is rooted-injective relative to r, for all $r \in X$.

For $n \ge 3$, X a set with $|X|=n$ and some $r\in X$, we define the rooted-injective dimension ${{\,\textrm{ID}\,}}^r(n)$ to be

$$\begin{aligned} {{\,\textrm{ID}\,}}^r(n) = \min \{ dim(\mathcal S): \mathcal S \text{ is } \text{ a } \text{ rooted-injective } \text{ split } \text{ system } \text{ on } \text{ X } \text{ relative } \text{ to } \text{ r } \}. \end{aligned}$$

Our next result (Theorem 5) shows that ${{\,\textrm{ID}\,}}^r(n)$ is well-defined for all $n\ge 3$, and that, in contrast to ${{\,\textrm{ID}\,}}_2(n)$, ${{\,\textrm{ID}\,}}^r(n)$ is always equal to 2 when $n \ge 4$.

Theorem 5

Suppose that X is such that $n=|X|\ge 4$ and that $r \in X$. Then there exists a rooted-injective split system $\mathcal S$ on X relative to r with $\dim (\mathcal S)=2$. Moreover ${{\,\textrm{ID}\,}}^r(n) =2$.

Proof

Put $X = \{1,2,\dots ,n-1,r\}$. First note that ${{\,\textrm{ID}\,}}^r(n)\ge 2$, since if ${{\,\textrm{ID}\,}}^r(n) =1$, then there would be a rooted-injective split system $\mathcal S$ on X relative to r with $\dim (\mathcal S)=1$. But this is not possible since then the Buneman graph $B(\mathcal S)$ associated to $\mathcal S$ would be a phylogenetic tree on X with $|\mathcal S|+1$ edges. Using a similar argument to the one used to show that ${{\,\textrm{ID}\,}}(4)={{\,\textrm{ID}\,}}(5)=2$ in the proof of Theorem 3, it is straight-forward to check that then $\mathcal S$ is not rooted-injective which is impossible.

Now, define the split system $\mathcal S$ on X whose subset of non-trivial splits is equal to $\mathcal S_1 \cup \mathcal S_2$, where:

$$\begin{aligned} \mathcal S_1=\{ \{n-1-i,\ldots ,n-1\} | \overline{\{n-1-i,\ldots ,n-1\}} \cup \{r\} :\, 0 \le i \le n-3\} \end{aligned}$$

and

$$\begin{aligned} \mathcal S_2=\{ \{n-1-i,\ldots ,1\} | \overline{\{n-1-i,\ldots ,1\}} \cup \{r\} :\, 0 \le i \le n-3\}. \end{aligned}$$

Fig. 3

The Buneman graph of a split system on $\{1,2,3,4,5, 6,r\}$ that is rooted injective relative to r that is constructed as described in the proof of Theorem 5

For example, for $n=7$, the Buneman graph $B(\mathcal S)$ of $\mathcal S$ is the half-grid pictured in Fig. 3. More precisely, in that figure, the splits in $\mathcal S_1$ and $\mathcal S_2$ are the splits associated to edges oriented downwards from left to right and from right to left, respectively.

To see that $\dim (\mathcal S)=2$, it suffices to remark that $\mathcal S_1$ and $\mathcal S_2$ are compatible, so a maximal incompatible subset of $\mathcal S$ has size at most 2. Since $\mathcal S$ is not compatible, it follows that $\dim (\mathcal S)=2$.

We next show that $\mathcal S$ is rooted-injective relative to r. To see this, let $Z, Z' \in {X-\{r\} \atopwithdelims ()2}$ distinct. Also, let $x^-=\min (Z \cup Z')$ and $x^+=\max (Z \cup Z')$. Since $Z \cup Z'$ has size at least 3, we have that $x^-$ and $x^+$ are distinct. Furthermore, $x^- \le n-3$ and $x^+ \ge 3$ must hold. In particular, the splits $S^-=\{x^-+1, \ldots , n-1\}|\{1, \ldots , x^-\} \cup \{r\}$ and $S^+=\{1, \ldots , x^+-1\}|\{x^+, \ldots , n-1\} \cup \{r\}$ belong to $\mathcal S_1$ and $\mathcal S_2$ respectively, so both splits belong to $\mathcal S$. Moreover, $Z \cap Z'$ contains at most one element, so at least one of $x^-$ and $x^+$ does not belong to $Z \cap Z'$. If $x^- \notin Z \cap Z'$ then $S^-$ satisfies $\phi _{Z_r}(S^-) \ne \phi _{Z'_r}(S^-)$, and if if $x^+ \notin Z \cap Z'$ then $S^+$ satisfies $\phi _{Z_r}(S^+) \ne \phi _{Z'_r}(S^+)$. So, $\mathcal S$ is rooted-injective relative to r. $\square$

Remark 1

The proof that the split system $\mathcal S$ is rooted-injective relative to r in Theorem 5 gives an alternative proof that the extended half-grid for $(n+1)$ in [7, p.7] can be used to represent a symbolic map, since the Buneman graph $B(\mathcal S)$ with the pendant edge containing r contracted is isomorphic to the extended half-grid on n.

Note that the rooted-injective split system $\mathcal S$ in the proof of Theorem 5 is the union of two split systems $\mathcal S_1$ and $\mathcal S_2$ whose associated Buneman graphs are phylogenetic trees. In general, if $\mathcal S$ is a split system on X with this property then $\dim (\mathcal S)\le 2$ (since every 3-subset of $\mathcal S$ must contain at least one pair of splits that is contained in one of the split systems, and so this pair of splits must be compatible). Hence, by Theorem 3, $\mathcal S$ cannot be injective in case $|X| \ge 6$.

8 Discussion

In this paper we have defined and explored the concept of injective split systems, that is, splits systems $\mathcal S$ on a set X such that two distinct sets of three elements of X have distinct median vertex in the Buneman graph $\mathcal B(\mathcal S)$ associated to $\mathcal S$. Making use of the notion of dicing, we have shown that a given split system is injective if and only if its subsets of size 6 or less are injective, from which we derived a characterization of injective split systems. We also studied the injective dimension of an integer $n \ge 3$, that is, the minimal dimension of an injective split system on some set of n elements. On this topic, it remains an open question whether there is a lower bound for ${{\,\textrm{ID}\,}}(n)$ that is linear in n.

The notion of an injective split system also suggests to consider a matching concept of surjective split systems. We call a split system ${\mathcal S}$ on some set X with $|X|\ge 3$ surjective if the vertex set of $B(\mathcal S)$ is equal to

$$\begin{aligned} \{\phi _x :\, x \in X\} \cup \left\{ \phi _Y :\, Y \in {X \atopwithdelims ()3}\right\} , \end{aligned}$$

(4)

In other words, every non-leaf vertex in $B({\mathcal S})$ is the median of three leaves in $B({\mathcal S})$. Note that every split system whose Buneman graph is a phylogenetic tree is surjective but, for example, the split system corresponding to the Buneman graph in example in Fig. 2(ii) is not surjective because the central vertex in the graph is not the median of any three leaves. The general properties of surjective split systems remain to be investigated.

Naturally, one may want to study bijective split system ${\mathcal S}$ that are both injective and surjective. We conjecture that a split system ${\mathcal S}$ on some set X with $|X|\ge 3$ is bijective if and only if either $|X|=3$ and $|{\mathcal S}|=3$ or $|X|=4$, $|{\mathcal S}|=6$ (i. e. the Buneman graph associated to ${\mathcal S}$ is a three-leaved phylogenetic tree or—up to leaf relabelling—the graph in Fig. 2(i), respectively). A proof or counter-example for this conjecture might use concepts that are related to the so-called median stabilization degree of a median algebra—see e.g. [3, 13].

Finally, another interesting open problem is the following: Can we develop a modular decomposition theory for Buneman graphs along the lines described in [7]?