Suppose is a finite, connected, edge-weighted graph, i.e. there is an associated weight function . One can make into a metric space by identifying each edge with a closed line segment of length , setting
and declaring to be the length of the shortest path between and when traveling along the edges of . On the space of finite, edge-weighted graphs, there is the equivalence relation if and are isometric. A metric graph is the unique (up to isometry) metric space associated with an equivalence class in , and any graph in the equivalence class is called a model for .
We begin with , the group (under pointwise addition) of continuous, piecewise linear functions for which every linear piece has integer slope. This is a linearized analogue of the space of meromorphic functions on a Riemann surface, over which divisor theory is classically developed. We denote by the free abelian group on , i.e.
In other words, is the group of integer-valued functions on having finite support. With this interpretation, we introduce the notation for a divisor .
There is a homomorphism given by , and is called the degree of . For any , we denote by the set of divisors with degree . Only when does form a subgroup of , and is the kernel of .
With each , there is an associated divisor , which records information on the “zeros” and “poles” of . Namely, for each , we set to be the sum of the slopes of in all directions of the graph emanating from . Then we define the principal divisor as
The principal divisors form a subgroup , and one observes for every , so that . This begs to form the quotient
which is called the Picard group of . More generally, two divisors , are linearly equivalent, denoted , if , and one can form the sets .
There is a combinatorial/dynamical interpretation of linear equivalence. Given a divisor on , one can interpret as the number of dollars person has (if , then person is in debt). For any star-shaped subset and sufficiently small , there is a function taking value 0 on , taking value outside the -neighborhood of , and with slope 1 along each outgoing direction from . Then , and this equivalence has the effect of each person on the boundary of giving one dollar to the individual situated exactly distance away. Every element of may be written as a finite -linear combination of ‘s, so two divisors are linearly equivalent precisely when connected by a sequence of these “chip-firing” moves.
In terms of the chip-firing game, is the space of inequivalent degree 0 wealth distributions on . We would like to understand the structure of this space, both algebraically and geometrically. As it turns out, one can learn quite a bit about through the integer homology group .
Recall that is constructed as follows. Fix a model for , fix an orientation of , fix a spanning tree of , and fix a base vertex . Then, we denote by the -vector space with basis , and by the -vector space with basis . The boundary map given by extends linearly to a map of vector spaces. One defines the real homology to be the kernel of . Given the choices we have made, there is a canonical basis for , consisting of simple cycles. For every , there is a unique cycle beginning and ending at that traverses no edges other than and those in . Recording each of these cycles as a sum of directed edges gives the aforementioned basis. Then is defined to be the free abelian group on this canonical basis, i.e. the associated integer lattice inside .
The Abel-Jacobi theorem states there is an isomorphism
We will not prove this fact, as it is rather difficult, but we will show how this isomorphism gives rise to interesting geometric features. One defines the Abel-Jacobi map via
By the isomorphism above, descends to a map from into the -dimensional torus
where is the genus of . This can be described in very geometric terms: it essentially draws a skeletonized version of on the torus .
To specify the map , we will need to equip the ambient edge space with an inner product given by
This allows us to talk about orthogonality of vectors in , and in particular, define an orthogonal projection . A line segment in can be specified by providing a vector and a base point onto which the tail of the vector should be translated. Accordingly, let denote the line segment in obtained by translating vector to . We describe the image of by constructing the map inductively. Assume in the basis the edges are ordered such that for every , the edge shares a vertex with some having . Furthermore, assume the tail of is . Begin by identifying with the segment (so that is defined on ). Now, if for shares a vertex with , where , we identify with either the segment or the segment , depending on if is the tail of or the head of , respectively. Once this process is complete, the entire image of is contained in a unique fundamental domain of the lattice , hence actually gives a map .
How much information does preserve? The only edges mapped into non-injectively are those which orthogonally project to , i.e. those edges inhabiting the orthogonal complement of . There is a nice characterization of this space when for all . Recall that a cut of is any disjoint partition , and any cut determines a cut set, consisting of the edges connecting a vertex in to a vertex in . For each cut set , there is an associated cut vector in encoding that set, namely . where if is oriented to flow from to and if is oriented to flow from to . We claim the cut space
is precisely the orthogonal complement of the cycle space .
To prove this, we first note that any cut vector is orthogonal to any cycle vector. Suppose represents a simple cycle (a cycle that traverses no edge twice) and is a cut vector. Let if traverses along its orientation, let if traverses against its orientation, and let if does not traverse . Then
But simple cycles form a basis for , so . Since has dimension , to complete the argument it will suffice to exhibit linearly independent vectors in . Fix an ordering of the vertices, and set . Then for each , the partition is a cut, with a corresponding cut set . Since always contains a vertex that , , all do not, the cut vectors corresponding to these cuts are linearly independent.
Hence, if and only if is a bridge, an edge whose removal would disconnect the graph. This means the only information lost by is the existence of bridges, which are collapsed into a point when the graph is drawn on ; elsewhere on , the map is injective. Because of the isomorphism between and , this geometric observation can be translated to an equivalent algebraic fact about : if , then either or there is some bridge such that , .
(Acknowledgment: the nice diagrams above are due to Samir Khan, UChicago ’19)