Lattice of Integer Flows and Poset of Strongly Connected Orientations
Abstract
We show that the Voronoi cells of the lattice of integer flows of a finite connected graph in the quadratic vector space of real valued flows have the following very precise combinatorics: the face poset of a Voronoi cell is isomorphic to the poset of strongly connected orientations of subgraphs of . This confirms a recent conjecture of Caporaso and Viviani Torelli Theorem For Graphs and Tropical Curves, Duke Math. J. 153(1) (2010), 129171. We also prove an analogue theorem for the lattice of integer cuts.
This work was done while the author was visiting the California Institute of Technology.
1 Introduction
1.1 The Lattice of Integer Flows
Consider a finite graph with possible parallel edges and loops. All through the paper, we suppose that is connected. Following the classical works (e.g.,[3], [1]), we replace the set of edges of with the set of oriented edges , where each edge is replaces with two oriented edges in opposite directions in . By the abuse of notation, we denote by an element of and by the inverse of .
For an abelian group , let be the free module generated by the set of vertices and let be the quotient of the free module generated by the set of oriented edges by all the relations for any oriented edge . Note that the basis elements of are denoted by for and the ”basis” elements of are denoted by for . The chain complex is the usual simplicial chain complex of : the boundary map is given by for and being respectively the head and the tail of the oriented edge .
Similarly, one can define the cochain complex by letting to be the module of all the functions and to be the module of all the functions , the latter one having the property that . The differential is defined as follows: where and are respectively the head and the tail of the oriented edge . Note that the spaces and are canonically dual for .
The homology of and the cohomology of have a rather simple description:

, i.e., is a free module of rank where is the genus of the (unoriented) graph . .

Similarly, and .
All the other (co)homology groups are trivial.
The first homology group is nothing but the module of all the valued flows in . A flow with values in is the data of for any with the following properties:

For any oriented edge , .

For any vertex , let be the set of all the oriented edges emanating from , i.e., with tail equal to . Then . (In other words: the amount of the flow coming out of is equal to the amount of the flow entering .)
Let be the vector space of realvalued flows in , i.e., . And let be the lattice of integervalued flows in , i.e., , . Note that is a subvector space of dimension in and has rank .
1.2 The Positive Quadratic Form
There is a natural innerproduct on (and a natural inner product on ) given by the simplicial structure of . This is given on the basis as follows
(Similarly on , for vertices .)
It is straightforward to check that the above pairings naturally identify the dual spaces and and the adjoint of gets identified with .
The above innerproduct defines by restriction an innerproduct on , denoted again by ; let be the corresponding positive quadratic form on given by , for .
1.3 Voronoi Diagram of in
Consider a discrete subset in the real vector space . For a point in , we define the Voronoi cell of with respect to as
The Voronoi diagram is the decomposition of induced by the cells , for It is straightforward to see that each cell of the Voronoi diagram is a polytope or a polyhedron (in the case there are cells of infinite volume).
Consider the discrete subset of , which is of full rank. By translationinvariance of the distances defined by , the cells of the Voronoi diagram are all simply translations of each other, i.e., for a point in , , where denotes the origin. Thus, the Voronoi cell decomposition of is completely understood by the Voronoi cell , which is easily seen to be a polytope.
Consider now the set of all faces of the polytope ordered by inclusion. They form a finite poset that we denote by . Our aim in this paper is to give a precise description of this poset in terms of the original graph , that we now explain.
1.4 Poset of Strongly Connected Orientations of Subgraphs
Let be an orientation of a connected graph . is called strongly connected if any pair of vertices and in are connected by an oriented path from to and an oriented path from to . An orientation of a (non necessarily connected) graph is called strongly connected if the orientation induced on each of the connected components is strongly connected.
Let be a given graph (with possibly multiple edges and loops). Define the following poset , that we call the poset of strongly connected orientations of subgraphs of : the elements of are all the pairs where is a subgraph of and is a strongly connected orientation of the edges of . (Note that the vertices and the edges of are labeled, so parallel edges are distinguished in dealing with subgraphs.) A partial order is defined on as follows: given two elements and in , we have if and only if and the orientation is the orientation induced by on . It is quite straightforward to see that is a graded poset, and the grading is given by minus the genus of the underlying subgraph. Note that is the maximum element of .
The main result of this paper is the following theorem.
Theorem 1.
The two posets and are isomorphic.
Remark 2.
This is precisely Conjecture 5.2.8 of [2] which answers a question asked by Bacher et al. in [1]. The two posets and are the posets and in the notations of [2] where is the metric graph with all lengths equal to one. The proof of our theorem can be in a very straightforward way extended to general tropical curves, thus, Conjecture 5.2.8 of [2] holds. We have presented the proofs in the unweighted case to simplify the presentation. Note that Conjecture 5.2.8 of [2] has a second part (explained in Section 3): we will establish a precise bijection between the elements of the two posets, from this bijection the second part of the conjecture easily follows.
2 Proof of Theorem 1
A (nonnecessarily connected) graph is Eulerian if every vertex has an even degree. An orientation of an Eulerian graph is called Eulerian if each vertex has the same in and outdegrees. It is clear that every Eulerian orientation of a graph is necessarily strongly connected (in the sense we describes above). Also note that an Eulerian orientation of a graph has a canonical flow denoted by . This is defined by setting if (thus, if ) and if neither nor is in . For simplicity, we say a flow in a graph is Eulerian if it is of the form for some an Eulerian orientation of a subgraph of . The genus of an Eulerian flow , for an Eulerian orientation of an Eulerian graph , is by definition the genus of . In the case is disconnected, this latter is the sum of the genus of the connected components of .
Given a flow , by the support of , denoted by , we mean the set of all the oriented edges with . Clearly the oriented graph defined by is strongly connected. (Different flows can however have the same support.) Note also that for an Eulerian flow , is the Eulerian orientation with .
Lemma 3 (Intersecting Voronoi Cells).
A Voronoi cell , for , intersects the Voronoi cell of the origin if and only if is Eulerian. For two points and in , if and only if is Eulerian.
Proof.
We first show that if , then . This will then easily implies the lemma.
The claim is a particular property of the Voronoi diagrams defined by the lattices. Suppose that . This means there exists a point such that , in other words, or equivalently, . Consider now the parallelogram defined by , , and (which is in ). We show that a point is certainly closer in distance to one of or , and this will be a contradiction. By assumption , which gives . Since , we have . Combining the two (in)equalities, we have
This shows that , which is a contradiction to the assumption . And the claim follows.
Suppose now a lattice point such that . This latter condition is equivalent to the following:
For all ,
Let . We observed that is a strongly connected orientation. For any oriented cycle in , consider the flow in . We must have By combining these two inequalities, we infer that the equality holds, and so, for all . Since every oriented edge in is contained in an oriented cycle , we have for all . This shows that is Eulerian and the first claim of the lemma follows. The second claim follows by translation invariance. ∎
2.1 Codimension One Faces of : Description of the Delaunay Edges
We provide now a description of the faces of codimension one in ; this clearly provides a description of the dual Delaunay edges. (Recall that the onesequelleton of the Delaunay dual is the graph obtained on by joining two points and if is a face of codimension one in both and .)
Let be a face of codimension one and let denote a generic point of (an arbitrary point in the relative interior of ). Since is of codimension one, there is a Voronoi cell such that . By Lemma 3, is Eulerian.
Lemma 4.
The intersection lies in an affine plane of codimension equal to the genus of . In particular, if and intersect in a face of codimension one, then is a circuit (an orientation of a cycle) and .
Proof.
A point in the intersection has the following properties:

or equivalently , and

for all , .
By Lemma 3, is Eulerian; is thus an Eulerian oriented graph and can be decomposed into a disjoint union of oriented cycles. In other words, there exist Eulerian , for some , such that

for , is a circuit;

; and

for all , .
In addition, for one may choose any oriented circuit of and define .
(To see this, note that one way to obtain the decomposition is to proceed greedily, and in each step, choose an arbitrary circuit in the remaining oriented graph, delete it and proceed.)
Applying Property II above to each and using and above, one sees that
Since there exists a decomposition such that for an arbitrary circuit of , we infer that
Let be a basis for the cycle space of and let be the affine plane defined by the equations for , for . Here is the genus of . Since the elements are independent, and the inner product is nondegenerate, has codimension and it contains . The lemma follows. ∎
Proposition 5.
Let be the set of all Eulerian elements of of genus . forms a system of generates of . The onesequeletton of the Delaunay dual of is isomorphic to .
Proof.
Let . We show the existence of such that each is of the form for a circuit and in addition (and in addition ). This clearly proves the proposition.
Let be the Eulerian oriented graph obtained by replacing each arc in with different parallel arcs. Since is Eulerian, it admits a decomposition into circuits, for some . Looking at the circuits in the original oriented graph , one obtains circuits , and it is clear that for . ∎
Hyperplane Arrangement Defined by Circuits
Let be the set of all the Eulerian elements of of genus one, and let be an element of . Let be the hyperplane in defined by the equation
More generally, for an arbitrary lattice point and an Eulerian of genus one, let be the affine hyperplane . By the results of the previous section, the Voronoi diagram is the hyperplane arrangement of all the hyperplanes for and . In addition, the Voronoi cell (and more generally ) is the cell containing (resp. ) in the hyperplane arrangement of the hyperplanes (resp. ) for .
2.2 Definition Of The Map From To
Each face in is obtained as an intersection of some of the hyperplanes for .
Lemma 6.
Let be a face of , and let and be two different elements of such that . Then and are consistent in their orientations: i.e., there is no oriented edge such that and .
Proof.
Let and , and for the sake of a contradiction, suppose there exists an oriented edge with and . We show that for every point in , there exists an element which is strictly closer to . This will show that the intersection is empty, which will be a contradiction, and thus, the lemma follows.
Let . We have
Thus,
(1) 
Let . Observe that neither nor belong to and so the sum is strictly less than . By (the proof of) Proposition 5, can be written as a sum such that and . This shows that
(2) 
If for all , , then by summing up all these inequalities and applying Inequality 2, we should have
(3) 
which is impossible by Equation 1. Thus, there exists an such that . And clearly, in this case , contradicting the assumption that . ∎
Definition of the injective map from to .
Applying Lemma 6, we now construct a map as follows. Let , and consider the family . Define . By Lemma 6, two different and are consistent in their orientations, so the union in the definition of defines a welldefined orientation of a subgraph of . In addition, it is quite straightforward to check that is a strongly connected orientation of that subgraph, i.e., lies in .
Lemma 7.
Let be a face in , and and be defined as above. For any with , we have , i.e., . More precisely, if and only if .
Proof.
Define the flow . By applying Lemma 6 and the definition of , we have
(4) 
and since , we have
(5) 
As the prove of Proposition 5 and Lemma 4 shows, since , there exists another decomposition of with and such that . This shows that . Equation 4 and Inequalities 5 for then imply that for all , , i.e., for all , , and in particular . The second part of the lemma follows from the first part and the definition of . ∎
Lemma 8.
The map is injective and orderpreserving.
Proof.
Injectivity follows easily from the previous lemma: if for two faces and we have , then . Since is the cell containing the origin in the arrangement of the hyperplanes for , this shows that .
To show that the map is order preserving, note that if for two faces in , then , thus, and so by the definition of the partial order in , we have . ∎
2.3 Explicit Description of the Faces in
We now show that the map defined in the previous section is surjective.
Let be a strongly connected orientation of a subgraph of . Let be the affine plane defined as the intersection of all the affine hyperplanes for and . From the analysis we did before, we know that , the genus of . We show that there is a face of of codimension which is included in and . This proves the surjectivity of and Theorem 1 follows.
In what follows, by the abuse of the notation, by a strongly connected orientation of the whole graph we simply mean a strongly connected orientation of the graph minus the set of bridges. (Thus, if does not have any bridge, i.e., is 2edge connected, the orientation is indeed defined on the whole graph.)
Lemma 9.
If is a strongly connected orientation of the whole graph , then consists of precisely a vertex of , denoted by . All the vertices of are of this form.
Proof.
We already know that is of codimension . In other words, for a point . We now show that is a vertex of . To this end, we have to show that for all , , or equivalently . Let . Since is a strongly connected orientation of , the circuits of form a basis for . There exists thus a set of coefficients for with such that
We have
In addition, the last inequality is strict if and only if . We infer that and it is a vertex of .
To show that a vertex of is of this form, consider the image of the face by . This is a strongly connected orientation of a subgraph of , and since the intersection is of codimenion , the genus of should be equal to that of , i.e., and thus, and the lemma follows. ∎
Consider now the general case, being a strongly connected orientation of a subgraph of . Let be the family of all the extensions of the orientation of to a strongly connected orientation of the whole graph , i.e., the family of all the strongly connected orientations of such that . For each element , let the vertex of given by the previous lemma. By the convexity of and , we have
Lemma 10.
The convexhull of for defines a face of and we have . Thus, the map is surjective.
Proof.
Let . We first show that is a face of , the second statement easily follows. By the previous lemma, any vertex of which lies in is of the form for a strongly connected orientation of . Since , and and are strongly connected orientation, we infer that , i.e., , and so all the vertices of which lie in are of the form for . It follows that is a face of . To show that , note that for any , by the proof of the previous lemma, we have if , then for every , , i.e., . And clearly if , then for every , , and since , we have for every , i.e., . We conclude that , and the lemma follows. ∎
The proof of Theorem 1 is now complete. Note that, as the proof shows, the codimension of a face is equal to the genus of .
3 Quotient Posets and
The Voronoi diagram is naturally equipped with the action of obtained by translation. (More precisely this is defined as follows: for an element , and a face of a Voronoi cell , is the corresponding face of the Voronoi cell under the natural isomorphism obtained by translation). The poset is defined as the quotient of under this action. Another way to define is to start from and identify those faces of which differ by a translation by an element of .
There is also a quotient poset associated to obtained roughly by identifying those strongly connected orientations of subgraphs of which differ by reversing the orientation of an Eulerian orientation of an Eulerian subgraph of . More precisely, is defined as follows.
Let be a strongly connected orientation of a subgraph of . The outdegree function is define by sending a vertex of to the number of oriented edges emanating from in the orientated graph , i.e., the number of arcs in whose tail is . The equivalence relation is defined on as follows. For two strongly connected orientations of a subgraph and of a subgraph of , we set if an only if

, and

the two corresponding outdegree functions are equal, i.e., .
Lemma 11.
Two strongly connected orientations and of a subgraph of are equivalent if and only there exists an Eulerian subgraph of and an Eulerian orientation of such that and is the orientation obtained by replacing each edge in by , i.e., .
Proof.
The proof is quite easy: consider the set of all oriented edges such that . This defines an Eulerian oriented graph and lemma easily follows. ∎
The poset is the quotient of obtained by identifying equivalent elements of , in other words . Note that this is indeed a poset: if , and , then by the definition of the partial order and the lemma above, there exists an Eulerian orientation of a subgraph of such that and is obtained from by reversing the orientation of the oriented edges of . Now reversing the orientation of the oriented edges of in , one obtained an element such that and . So the quotient partial order is welldefined.
Theorem 12.
The two posets and are isomorphic.
Proof.
The proof follows from the proof of our Theorem 1 and Lemma 3. Namely, consider the bijective map defined in the previous section. Let and be two faces in which are identified in , i.e., there exists an element such that . Since is also a face of , we infer that , and by Lemma 3, is Eulerian. Let , note that is Eulerian. We show that and is obtained by reversing the orientation of in , from this we obtain a welldefined quotient map , which becomes automatically bijective, and the theorem follows.
By the definition of , we have to show that for every with an oriented cycle in , we have . Let be such that . Since is a face of , for all , we have . It follows that
Since , and is a face of , the above inequality is indeed an equality for all , or equivalently for all , i.e., for all , . Thus, . We infer that .
Let , and the element of obtained from by reversing the orientation of every oriented edge in . We aim to show that .
In the proof of Theorem 1, we show that (resp. ) is the convex hull of all the vertices of the Voronoi cell for (resp. ), i.e., for being a strongly connected orientation of which contains (resp. ) as an oriented subgraph. On the other hand, since , we have . We now show that the set coincides with , i.e., with the set of all the strongly connected orientations of which contains as an oriented subgraph. From this, and the proof of Theorem 1, it easily follows that .
Recall that has the property that for every with ,
Since is in , the set of ’s with generate . And since the above equations are all linear, we easily have for all (not only with ),
where (resp. ) is the number of edges of with (resp. with ). From this, we have for all ,
(6) 
We claim that is equal to for the orientation obtained by reversing the orientation of in (note that ). For this, we have to show that for all with , we have . But this easily follows from the Equation 6 above. Indeed, for such a we have
The proof of the theorem is now complete. ∎
4 Lattice of Integer Cuts and Poset of Coherent Acyclic Orientations of Cut Subgraphs: a Dual Theorem
It is possible to drive a result analogue to Theorem 1, using similar ideas, for the lattice of integer cuts in a given graph. For more details on the duality between this lattice and the lattice , we refer to [1]; see also the short discussion in Section 1.
We continue to use the terminology we introduced in the previous sections, , , , and the quadratic form have the same meaning as before, and is connected. Consider the chain complex . Recall that is the module of all the functions and is the module of all the functions , the latter one having the property that . The differential is defined as follows: where and are respectively the head and the tail of the oriented edge .
Let and be the integer lattice . We note that is of dimension where is the number of vertices of and is of full rank in . The lattice is called the lattice of integer cuts in . It is easy to see that is generated by the functions for , is the characteristic function of , i.e., if and otherwise. Let be cut defined by : the set of all oriented edges from to . We have
I.e., is the cut defined by .
An element of is called a cutelement if it is of the form for a subset .
Note also that an element of , with , is in if and only if for any orientated cycle of , we have . These equations guarantee the existence of a function with .
Let