Graphs admitting only constant splines
Abstract.
We study the generalized graph splines introduced by Gilbert, Tymoczko, and Viel [3]. We show that if the graph has a particular type of cutset (the simplest example being a bridge) then the space of splines has a natural decomposition as a direct sum. We also consider graphs that only admit constant splines, and provide a characterization for splines on such graphs over the ring .
1. Introduction
In 1946, mathematicians appropriated the term spline from shipbuilding, where engineers modeled the hull of a ship by anchoring thin strips of wood, known as splines, to select points on the hull and allowed the strips to naturally flex to approximate the rest of the shape of the hull. Mathematical splines arise in many areas: topology, analysis, algebraic geometry, and algebraic combinatorics. Regardless of the context, we can view the notion of a spline as a sort of interpolation that fits some given data.
We will work in the setting of generalized graph splines, as introduced by Gilbert, Tymoczko, and Viel [3]:
Definition 1.1.
Given a graph and a commutative ring with unity, an edge labeling of is a function , where is the set of ideals of . A spline on is a vertexlabeling such that for each edge , the difference .
For a graph , let denote the set of all splines on with labels from and edgelabeling . Then is a ring and an module [3]. When and are clear from the context, we will use the notation . A fundamental problem in the theory of graph splines is to describe a “basis” for , by which we mean a minimal spanning set, i.e., a spanning set having the fewest elements. It is not difficult to do this when the graph is a tree, but it becomes a significantly more difficult problem for graphs containing cycles.
We will show how the ring of splines for a graph can sometimes be determined by looking at particular subgraphs. In Section 2, we begin by observing that, in certain cases, deleting or contracting an edge preserves the ring of splines. We then show how some decompositions of the graph (such as into two subgraphs connected by a bridge) induce a decomposition of the ring of splines. In Section 3, we consider changing the ring rather than changing the graph. We first show that a spline on over a ring induces a spline on over the ring , where is an ideal of . We then show that, with additional conditions, we can lift a spline over to a spline over . In Section 4, we combine these perspectives to characterize graphs which admit only trivial splines over in terms of the existence of certain spanning trees. Finally, in Section 5, we apply our results to graphs arising from particular triangulations of a plane.
2. Decomposing Graphs
2.1. Contraction and Deletion
In this section we relate splines on a graph to splines on subgraphs. In Lemmas 2.1 and 2.3 we observe that we can contract any edge labeled by the trivial ideal and delete any edge labeled by the ideal without changing the ring of splines.
Lemma 2.1.
[3] If is the trivial ideal (generated by in ), then , where is the graph obtained by contracting edge .
Since contractions sometimes produce loops or multiple edges, it is useful to also define splines for multigraphs, where there can be multiple edges between two vertices. In this case, the spline condition must be satisfied for each edge individually. The next lemma shows that the splines of a multigraph can also be viewed as the splines of a simple graph. (Since loops start and end at the same vertex, the spline condition is trivially satisfied, so loops can be removed without changing the space of splines.)
Lemma 2.2.
Suppose is a multigraph, with two edges and between vertices and . Suppose is the graph obtained by replacing and with a single edge , and setting (for all other edges, ). Then .
Proof.
First, suppose . Then and . Thus , so . Conversely, if , then , so is in both ideals . Thus, and hence . ∎
Lemma 2.3.
If contains every edge label in , then , provided is connected. In particular, this is true if (and, in this case, we can drop the requirement that is connected).
Proof.
Clearly, every spline on is also a spline on , so . Now suppose that . Since is connected, there is a path from to in . Then
Since , this means is a sum of elements from the ideals (here we let and ). Then , since contains every other edge label. Hence is also a spline for , so . We conclude that . ∎
2.2. Bridges.
We now turn to the case of graphs that contain a bridge; i.e., an edge whose deletion disconnects the graph. We will show that the ring of splines on the graph is (almost) the direct sum of the rings for the two components after the bridge is deleted. To do this, it is convenient to consider splines based at a vertex:
Definition 2.4.
Given a vertex of a graph , we say that a spline is based at if . We denote the subring of splines based at by (or just if and are clear) and call the basepoint of this subring.
Note that there may be distinct vertices and with , if for every spline . Our next lemma shows that is almost the same as . Here is the constant spline that takes the value 1 on every vertex.
Lemma 2.5.
For any vertex of , .
Proof.
For any spline , let ; then , so . So and together span . Suppose and . Then ; but since , this means . And then , so . This shows the two factors are independent, so . ∎
If a graph has a bridge, its space of splines is (almost) the direct sum of the spaces of splines of the two components created by removing the bridge. We prove this by giving a more general result, one in which removing a subgraph more complicated than an edge disconnects the graph. We still require this generalized bridge to meet each component of the disconnected graph in a single vertex.
More formally, suppose is an edgelabeled graph with a subgraph such that each connected component of the graph contains exactly one vertex of , as in Figure 1. If has vertices , let be the component of containing vertex . Note that any spline on can be extended to a spline on that takes the constant value on . Similarly, any spline on that is zero at can be extended to a spline on by setting it to zero off of . We will let and denote the subrings of formed by extending the splines in and , respectively. The following result expresses as a direct sum of of these subrings.
Theorem 2.6.
Suppose is an edgelabeled graph with a subgraph such that each connected component of the graph contains exactly one vertex of , as in Figure 1. Suppose has vertices , and is the component of containing . Then can be written as an internal direct sum:
Proof.
Suppose is a spline in . Then . Let be the restriction of to , so and . Now let be the restriction of to , so . Then
Now we show that this decomposition is unique. Suppose that , where and for . Note that and, in particular, . Since , this means . Then , since each is 0 on , and is constant on each . But since , then , so . Similarly, for each , . So the decomposition is unique. ∎
We give the result in the case of a graph with a bridge that is labeled by a principal ideal; it generalizes to any graph with a bridge if the generator is replaced by a collection of generators, one for each of the generators of the ideal labeling the bridge.
Corollary 2.7.
Let be a connected graph with bridge and edgelabeling such that the bridge is labeled by the principal ideal . Suppose that has two components; denote the component that contains by and the component that contains by . Let represent the function (not necessarily a spline) that is identically 1 on the vertices of and 0 on the vertices of . Then
Proof.
This follows directly from the previous theorem once we note that the graph consisting simply of the edge is generated as an module by the identity spline together with the spline that is on and on , and that is the extension of this latter spline to all of . ∎
As a consequence of Lemma 2.3 and Corollary 2.7, we can easily compute the space of splines for rings that have the property that all the ideals are in a nested hierarchy (i.e., for any two ideals and , either or ); these are known as uniserial rings (the rings and , where is a field, are such rings). In fact, we need only that the edge labels satisfy this nested relationship (rather than all ideals in the ring). In this case, on any finite graph , there is an edge with a maximal label (i.e., the ideal labeling that edge contains every other edge label), and we can delete this edge by Lemma 2.3 unless it is a bridge. If it is a bridge, then we can delete it using Lemma 2.7. Continuing in this way (and keeping track of the labels on any bridges deleted), we can reduce the graph to a collection of isolated vertices and obtain a basis for the space of splines.
Example 2.8.
Let . Suppose we have the graph with edge labeling as given below. Let and .
Since contains every other edge label, by Lemma 2.3, we have . Because contains the only other remaining edge label, we now wish to delete the edge , which is a bridge, leaving the graph . Let be the component of containing and be the component of containing . Then has a basis and (where the coordinates are listed ). Then by Lemma 2.7, has a basis .
3. Quotients
In the previous section, we looked at connections between splines on a graph over a ring and splines on subgraphs over . In this section, we will keep the graph the same, and instead change the ring. In particular, we will explore connections between splines on a graph over a ring and splines on over a quotient ring . This connection will be a key ingredient in our characterization of rank one graphs over in Section 4.
Our first observation is that any spline on a graph over a ring induces a spline on over , where is an ideal of . More precisely, suppose is an edgelabeled graph over a ring , and let be an ideal in . Let be an edgelabeling of over defined by .
Lemma 3.1.
If is a spline on , then the function defined by is a spline on .
Proof.
For any edge , . Since , this means . Hence is a spline on . ∎
Unfortunately, a spline on over will not always lift to a spline on over , as we see in Example 3.2.
Example 3.2.
However, with additional conditions, we can use a spline over to define a spline over , and even lift the spline from to .
Theorem 3.3.
Suppose is an edgelabeled graph over a ring , and let be an ideal in . Suppose further that there is a second ideal in such that . Then

If is a spline on , for each vertex , and , then the function defined by is a welldefined spline on (i.e., does not depend on the choice of representative ).

Furthermore, if , there is a unique such that for every spline on (i.e., is a lift of ).
Proof.
We first want to show that is welldefined. Suppose . Then there is some such that . So . But, since and are ideals, , so . Hence , and is welldefined.
To see that is a spline, observe that for any edge , . Since , there is some such that . Then , so . Hence is a spline on .
For the second part, we assume . If , then , and every spline over is constant. In this case, is a constant spline. On the other hand, if , then , so we may simply choose . So we may assume that and , so there are nonzero and such that . Since , and are unique. We will show for any spline on .
For any vertex , for some . Since , , so and . Then
Hence . Conversely, if for every , then for every possible . This happens if and only if , or . But this means that , so is the unique element of that will lift every spline on . ∎
4. Graphs with rank one
In this section, we consider labeled graphs that admit only constant splines, which we call graphs with rank one:
Definition 4.1.
Given a graph with edge labeling , a constant spline on takes the same value on every vertex of . The graph over a ring has rank one if it admits only constant splines.
Graphs with rank one are particularly interesting from the point of view of Theorem 2.6. If the “bridge” graph has rank one, then its contribution to the space of splines is trivial. Our main result is a characterization of which labeled graphs over have rank one.
We begin by giving a sufficient condition for a graph to not have rank one (i.e., to admit nonconstant splines).
Lemma 4.2.
If has a bridge whose label is not , then admits a nonconstant spline.
Proof.
Suppose that is a bridge, and . Then there is some such that . Since is a bridge, is the disjoint union of graphs (containing ) and (containing ). Label all vertices in (including ) with and all vertices in (including ) with 0. This is a nonconstant spline of . ∎
As a corollary, we can characterize when a labeled tree has rank one (over any ring).
Corollary 4.3.
If is a tree, it is rank one if and only if every edge is labeled (0).
Proof.
If every edge is labeled (0), then the graph can be contracted to a single vertex without changing the space of splines, by Lemma 2.1. Hence the space of splines will be generated by the constant splines, and the graph has rank one.
Conversely, since every edge of a tree is a bridge, if there is an edge whose label is not (0), then by Lemma 4.2 there is a nonconstant spline, and the graph does not have rank one. ∎
For general graphs, we do not have such a nice characterization of graphs with rank one. However, we can give a general necessary condition.
Lemma 4.4.
Suppose has rank one. Then, at every vertex , the intersection of the ideals labeling the edges incident to is .
Proof.
Suppose there is a vertex incident to edges labeled , and there is some with . Then define by and for every other vertex . If is adjacent to , then for some , and . Otherwise, for any pair of adjacent vertices and , we have . Then is a nonconstant spline on , and the graph does not have rank one. ∎
Remark 4.5.
The condition in Lemma 4.4 is necessary but not sufficient. For example, in the ring there are two proper ideals and . Note that . In the 4cycle shown below, the intersection of the edge labels at each vertex is trivial; however, there is a nonconstant spline, as shown.
However, if we restrict the ring used to label the edges and vertices, then we can obtain stronger results. We begin with the quotient , where is prime.
Lemma 4.6.
Let , where is a prime. Then a graph has rank 1 if and only if has a spanning tree with all edges labeled .
Proof.
By Lemma 2.1, we can contract all edges labeled without changing the space of splines. This will contract the graph to a single vertex exactly when there is a spanning tree with all edges labeled ; in this case, the graph will have rank one.
Suppose the graph does not contract to a single vertex. Recall that the ideals of are . By Lemma 2.3, we can delete all edges labeled , then all edges labeled , and so forth, until we encounter an edge which is a bridge. But then we have a bridge whose label is not , so by Lemma 4.2 the graph does not have rank one. ∎
We can extend this result to any ring using the machinery of quotient rings developed in Section 3.
Theorem 4.7.
Let , where the prime factorization of is . Then a graph has rank one if and only if contains spanning trees such that all edge labels of are contained in the ideal . The ’s need not be edgedisjoint (or even distinct).
Proof.
We first suppose that has spanning trees as described, and that is a spline on . Since every pair of vertices and is connected by a path in , . But since the ’s are relatively prime, the intersection of all the ideals is in . Hence is a constant spline.
Conversely, suppose only admits constant splines. Take the quotient of the graph by , as described in Theorem 3.3, to get a graph over the ring . By Lemma 4.6, if the quotient graph does not have a spanning tree with all edges labeled , then it has a nonconstant spline . Since , and the two ideals intersect trivially, there is some such that . Since is nonconstant, this means is a nonconstant spline on , which is a contradiction.
Hence the quotient graph has a spanning tree with all edges labeled . This means that in every edge of the spanning tree has a label in . Repeat the process for each to obtain the desired set of spanning trees. ∎
Example 4.8.
Let , where and are distinct primes. Then the ideals in are . Consider the graph shown below:
The thick and thin edges indicate two spanning trees, one with all edges labeled and the other with all edges labeled . Hence, by Theorem 4.7, this graph has rank one. Note that this graph has rank one even though none of the edges are labeled .
5. Applications
As an application we consider splines on certain triangulated rectangular grids studied by Zhou and Lai [6]. They describe a rectangular grid, with each square divided into two triangles by a diagonal, with the triangles then subdivided by a CloughTocher refinement as shown on the left in Figure 3. We will denote this graph as (starting with an grid). The splines discussed in [6] correspond to splines on the graph dual to the triangulated grid (ignoring the exterior region), shown on the right in Figure 3, which we denote by . Notice that the graph will always contain at least two bridges. If the graph has rank one, then all of these bridges must be labeled . In the following statement we use Theorem 4.7 to give a sharper lower bound on the number of edges.
Proposition 5.1.
If has at most 3 distinct prime factors, and is a labeling of the edges of over such that has rank one, then a lower bound on the number of edges labeled is

if ,

if , and

if .
Proof.
Our argument depends on counting the vertices and edges in . Observe that has regions and edges. However, we are not interested in the exterior edges, leaving interior edges. So will have vertices and edges.
If has only one prime factor, then Proposition 5.1 holds by Lemma 4.6. Next suppose . By Theorem 4.7, for any labeling over , has rank one if and only if contains spanning trees and such that all edge labels of are in and all edge labels of are in . Since these ideals only intersect in , and can only overlap on edges labeled .
Since has vertices, each spanning tree will have edges. Since there are only in all, the two spanning trees must overlap, and so there must be edges labeled for to have rank one. In fact, there must be at least such edges.
Now suppose , where are distinct primes. Then has rank one if and only if there are three spanning trees such that all labels of are in . contains at least edges, all with labels contained in . Hence contains at least edges, which must all be labeled . ∎
References
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