The gonality sequence of complete graphs

The gonality sequence $(\gamma_r)_{r\geq1}$ of a finite graph / metric graph / algebraic curve comprises the minimal degrees $\gamma_r$ of linear systems of rank $r$. For the complete graph $K_d$, we show that $\gamma_r = kd - h$ if $r<g=\frac{(d-1)(d-2)}{2}$, where $k$ and $h$ are the uniquely determined integers such that $r = \frac{k(k+3)}{2} - h$ with $1\leq k\leq d-3$ and $0 \leq h \leq k $. This shows that the graph $K_d$ has the gonality sequence of a smooth plane curve of degree $d$. The same result holds for the corresponding metric graphs.


Introduction
Baker and Norine in [BN07] introduced a theory of linear systems on graphs, later generalized by several authors to metric graphs and other combinatorial objects [MZ08,GK08]. It presents strong analogies with the one on algebraic curves. Many remarkable theorems have been proven to have a combinatorial counterpart, for example the Riemann-Roch Theorem and Clifford's Theorem, see [BN07,MZ08,Cop16].
With the notation g r s we indicate a linear system of degree s and rank r. We refer to Section 2 for the definitions. The gonality sequence (γ r ) r≥1 of a finite graph is defined as γ r := min{s such that there exists a g r s }. The gonality sequence was first introduced in [LM12] in the context of algebraic curves and the terminology comes from the integer γ 1 , which is called the gonality. By Riemann-Roch, it follows that γ r = g + r if r ≥ g.
The gonality sequence is known for general chains of loops, as it follows from the results in [CDPR11]. Except for graphs with low gonality, the gonality sequence is still undetermined for other graphs. The main result proven in this paper is the following: Theorem 1. The gonality sequence of K d is given by 2 is the genus of K d , and k and h are the uniquely determined integers with 1 ≤ k ≤ d − 3 and 0 ≤ h ≤ k such that The proof consists of two parts, respectively presented in Section 3 and in Section 4. In Section 3 we verify the existence of divisors of degree s = kd−h and rank (at least) r = k(k+3) 2 −h. In other words, we prove that kd−h is an upper bound for γ r . Herefore, we use an algorithm, presented by Cori and Le Borgne in [CB16], for computing the rank of divisors on complete graphs. The more involving part is showing that the upper bound kd − h for γ r is in fact sharp, which is done in Section 4. We translate this problem into a property of sequences of integers (α i ) i=1,...,d−1 satisfying certain hypotheses. In both parts, reduced divisors play an essential role. We provide an easy characterization of reduced divisors on complete graphs.
In Section 5, we extend the result to the complete metric graph K d with edge lengths equal to one. To be precise, we show that the gonality sequence of this metric graph is the same as the one of the corresponding ordinary graph; hereby answering [Bak08, Conjecture 3.14] in the case of complete graphs. The existence part of the theorem follows immediately from [HKN13, Theorem 1.3], which relates the rank of divisors on graphs and their corresponding metric graphs. The second part is proven in an analogous way as in Section 4, again utilizing the property of integer sequences.
We conclude by remarking that our arguments do not directly extend to complete metric graphs with arbitrary edge lengths. This is done in Section 6, where we give a quick account on the problems we encountered.
While the proof is purely combinatorial, the motivation for considering the question comes from plane curves and the specialization of divisors from curves to graphs.
Baker in [Bak08] showed that there is a close connection between linear systems on curves and on graphs. Indeed, let R be a discrete valuation ring with field of fractions K. A model for a curve X over K is a surface over R whose generic fiber is X. Given a smooth curve X over K and a strongly semistable regular model X over R, it is possible to specialize a divisor on the curve to a divisor on the dual graph of the special fiber of X. The complete graph K d pops up if one takes a model of a smooth plane curve of degree d degenerating to a union of d lines.
The gonality sequence of smooth plane curves has been computed by Ciliberto in [Cil83], and Hartshorne in [Har86]. Therefore a natural question in this setting is whether the complete graph and the smooth plane curve have the same gonality sequence. Theorem 1 provides a positive answer. As we point out in Remark 9, the linear systems with degree s = dk − h and rank r = k(k+3) 2 − h on plane curves specialize to linear systems of the same degree and rank on the graphs. So the first part of Theorem 1 can in fact be deduced from the result on plane curves.

Linear systems and reduced divisors
Let G be a finite graph without loop edges. We denote its vertex set by V (G) and its edge set by E(G).
Definition 2. A divisor on G is an element of the free abelian group Div(G) on the set V (G). Any divisor can be represented in a unique way as a finite formal combination of vertices of G with integer coefficients: The degree deg(D) of a divisor D is the sum v∈V (G) a v of its coefficients. If a v ≥ 0 for every v ∈ G, the divisor is said to be effective, and this is indicated with D ≥ 0.
Let f : V (G) → Z be an integer-valued function on the vertices of G. We define the principal divisor corresponding to f as The linear system corresponding to D, indicated with |D|, is the set of the effective divisors linearly equivalent to D. In symbols, The rank rk G (D) of a divisor D is defined as −1 if D is not equivalent to any effective divisor, otherwise We will often omit the subscript G in rk G (D) when it is clear from the context on which graph G we are working.
Reduced divisors will play an important role in proving the results of this paper. We briefly repeat the definition presented in [BN07].
Proof. For every A ⊆ V (K d ) \ {v} and for every w ∈ A, we have that outdeg A (w) = d − |A|. Therefore D is v-reduced if and only if each nonempty subset A contains an element w such that D(w) < d − |A|.
Suppose that there exists an ordering such that For the other implication, if we take

An upper bound for the gonality sequence
We provide a completely combinatorial proof of the inequality γ r ≤ kd−h. . We will explicitly construct such a divisor: if V (K d ) = {v 1 , . . . , v d }, then the divisor will do the job.
In [CB16] the authors provide an algorithm to compute the rank of a divisor on the complete graph K d . Although it is stated in a different terminology, it is possible to present it in terms of divisors.
The algorithm takes a divisor as input. The first step consists of computing the v d -reduced divisor D equivalent to it. If D(v d ) < 0, then the divisor has rank −1. Otherwise we take a vertex v i such that D(v i ) = 0 and consider the divisor D 1 = D − (v i ). By the previous lemma rk(D) = rk(D 1 ) + 1. We iterate until we find a divisor of rank −1. The algorithm terminates after at most deg(D) steps.
Proof. We use Lemma 7 and the algorithm described above. For convenience, we write the subsequent divisors appearing in the algorithm as D s,t where s ≥ 1 and 0 ≤ t ≤ s. In this way, we obtain a sequence of divisors where the indexes are in lexicographic order: The first two divisors are: At every step we subtract the divisor (v i ) corresponding to the vertex with zero coefficient and smallest index i. So, The divisor obtained at the next step is Since it is v k+1 -reduced, it is not equivalent to any effective divisor, so it has rank −1. We can conclude that the rank of D is (k+1)(k+2) Remark 9. For the reader familiar with the theory of specialization of divisors from curves to graphs, we remark that it is possible to prove the first part of Theorem 1 using the gonality sequence of smooth plane curves [Cil83] and Baker's Specialization Lemma [Bak08]. In fact, the complete graph K d is the dual graph of the special fiber of a regular strongly semistable model of a smooth plane curve of degree d. In [Cil83], Ciliberto showed that, given an effective divisor E of degree h ≤ k on X and H the generic divisor cut out by a line, the complete linear system The divisors kH − E described above specialize to divisors of the form on K d , with a 1 + a 2 + · · · + a d = h. By the Specialization Lemma, the rank of these divisors is at least k(k+3) It is not difficult to prove that it cannot be strictly bigger. By a relabeling of the vertices, we can assume that a i = 0 for i > h and a i ≥ a j if i < j. We consider the effective divisor This divisor is v 1 -reduced, therefore |D − E| = ∅ and the rank of D is

Sharpness of the upper bound
This section is concerned with proving the inequality In fact, we show that each divisor D of degree dk − h − 1 has rank strictly smaller than k(k+3) 2 − h. It suffices to do this for h = k. This follows from rk(D + (v)) ≤ rk(D) + 1 for each divisor D and vertex v.
Let D be any divisor of degree k(d − 1) − 1 on K d . By the above argumentation, we need to show that its rank is strictly smaller than We may assume that D is reduced with respect to the vertex v d . By Lemma 5, we can label the vertices so that the coefficients satisfy the following inequalities: , we can already conclude that the rank of the divisor cannot be k(k+1) Note that D has a negative coefficient at v d . If we are able to construct an effective divisor E of degree at most k(k+1) 2 which is supported on the vertices v 1 , . . . , v d−1 and such that D − E is v d -reduced, then we can conclude that the rank of D cannot be k(k+1) 2 . By Lemma 5, this happens if To simplify the notation, we define We have that From the above arguments, it follows that the following inequalities are satisfied for every index i ≤ d − 1: In particular, we can deduce that α 1 = a + 1 and α a+2 ≥ 0.
is not always satisfied, as the following example illustrates: In this case, a = k − 1, b = 0 and It is still possible to show that the divisor does not have rank k(k+1) 2 , by considering instead the divisor We remark that in this case the coefficient at v d is not negative. Let E be the following divisor of degree k(k+1) The divisor D − E has a negative coefficient at the vertex v d and is v dreduced. Therefore, also for this example we can conclude that D has not rank k(k+1) 2 .
We can generalize Example 10 as follows If we are able to construct an effective divisor E of degree at most k(k+1) then we can conclude that the rank of D cannot be k(k+1) 2 . We claim that (at least) one of the two terms t 1 and t 2 is at most k(k+1) 2 .
Remark 12. To summarize our strategy, we add the principal divisor to D until it either has a negative value at v d , or one time before that, and (at least) one of the two resulting divisors is v d -reduced after subtracting an effective divisor of degree k(k+1)
The idea of proof of Claim 11 is as follows: first we introduce a specific integer sequence, which we show to be the "worst-case scenario". Afterwards, we prove the claim for this particular sequence.
Remark 13. Let α = (α i ) i=1,...,d−1 be a sequence of integers satisfying ( * ). For every i we have the following inequalities: Moreover for q > p + 2 it holds that It follows that for each sequence α satisfying ( * ), there exists a unique sequence it is possible to compute p and q. Indeed, since Since the difference between two subsequent terms (corresponding to p + 1 and p) is there is a unique p for which q ∈ {p + 2, . . . , d}. Note that p ≥ k − a, since q ≥ p + 2.
Lemma 15. Let k, d, a, b be integers such that Let α = (α i ) i=1,...,d−1 be a sequence of integer numbers that satisfies the conditions α i ≤ a + 1, α i+1 ≥ α i − 1 and If p, q are integers such that the sequence β p,q satisfies d−1 Proof. Consider sequences α and β (p,q) that satisfy the conditions in the statement. By construction, the sequence β (p,q) = (β i ) i=1,...,d−1 is taken in such a way that it minimizes the number of places i with positive values for β i , fixing the value for d−1 i=1 β + i . Suppose that q < a + p + 2, hence β q > 0. Let i 1 , . . . , i c be the indexes i, ordered from small to big, for which α i has negative values, and j 1 , . . . , j d−a−p−3 the same for β i (so j h = h+a+p+2). Then d−a−p−3 ≥ c since the β i are negative in as much places as possible. Moreover, we have that α i h ≥ β j h for each h ∈ {1, . . . , c}. Indeed, if α i h < β j h = −h, then α i 1 < −1 by repeatedly using the rule α i+1 ≥ α i − 1, hence α i 1 −1 < 0, a contradiction. Therefore, the sum of the negative α i is at least the sum of the negative β i .
, then the number of places i where α i ≥ 0 is more than a + p. From this, it follows that the sum of the negative α i is at least the sum of the negative β i . We obtain a contradiction We conclude this section by proving Claim 11.
Case (1). Remark that β + i = β i for every i. We compute t 2 : This condition is satisfied since k ≤ d − 3.
Case (2). Remark that β q ≥ 1. We compute t 2 also in this case: This is satisfied since p ≥ k − a.
Case (3). First we consider the special situation p = k − a. We compute the term t 1 : If p > k − a, we compute t 2 : This condition is satisfied since p > k − a.

Metric graphs
If G is a graph, we define the metric graph Γ corresponding to G as the metric graph with V (Γ) = V (G), E(Γ) = E(G) and edge lengths l(e) = 1, see [HKN13]. In this section we explain how to extend the previous results to the complete metric graphs corresponding to the graphs K d , which we will also denote by K d . We start by briefly recalling the main definitions regarding linear systems on metric graphs. We refer to [MZ08,GK08,HKN13] for further reading.
Definition 16. Let Γ be a metric graph. A divisor on Γ is an element of the free abelian group Div(Γ) on the points of the graph, so The degree of D, denoted by deg(D), is the sum of its coefficients. As before, if a p ≥ 0 for every p ∈ Γ, the divisor is said to be effective. The support of a divisor D is the set of points p of Γ such that a p = 0 and it is indicated with supp(D).
A rational function f : Γ → R on Γ is continuous, piecewise linear with integer slopes and only finitely many pieces. The principal divisor div(f ) associated to f is the divisor whose coefficient at p is given by the sum of the incoming slopes of f at p. Only at a finite number of points, the coefficients are not zero.
The linear system of D, indicated with |D|, is the set of the effective divisors linearly equivalent to D, The rank rk Γ (D) of a divisor is defined as −1 if D is not equivalent to any effective divisor, otherwise Again, reduced divisor on metric graphs will play an important role. We recall the definition from [HKN13] and [Luo11].
Definition 17. Let Γ be a metric graph and X be a closed connected subset of Γ. Given p ∈ ∂X, the outgoing degree outdeg X (p) of X at p is defined as the maximum number of internally disjoint segments in Γ\X with an open end in p. Let D be a divisor on Γ. A boundary point p ∈ ∂X is saturated with respect to X and D if D(p) ≥ outdeg X (p), and non-saturated otherwise. A divisor D is p-reduced if it is effective in Γ\{p} and each closed connected subset X ⊆ Γ \ {p} contains a non-saturated boundary point.
Theorem 18 (Proposition 7 in [MZ08]). Let D be a divisor on a metric graph Γ. For every point p ∈ Γ there exists a unique p-reduced divisor linearly equivalent to D.
Let K d be the complete metric graphs on d vertices. We present a similar characterization of reduced divisors on K d as in Lemma 5. Given a vertex v of K d and an ordering v 1 , . . . , v d−1 of the vertices in V (K d ) \ {v}, for every 1 ≤ i ≤ d − 1 we define Figure 1 for an example.
v (1) D is effective in K d \ {v}; (2) for every edge e ∈ E(K d ), (3) there exists an ordering v 1 , . . . , v d−1 of the vertices in Proof. We start by proving the 'if' part, so assume that K d satisfies the three conditions. We need to show that every closed connected subset X of K d \{v} contains a non-saturated boundary point. If X contains a boundary point p that is not in the support of D, than p is non-saturated. So we focus on the case that the boundary points of X are all in the support of D. In case X = {p}, with p a point in the interior of an edge, then p is non-saturated because of condition (2). Suppose that X satisfies the hypothesis that if two vertices are contained in X, then also the edge connecting them is contained in X. We define A vertex w of X is not-saturated if and only if D(w) + q ∈ ∂X : q ∈ (w, w ) with w ∈ A < d − |A|.
Let j = min{i | v i ∈ A}, so j ≤ d − |A|. By condition (3), we know that D(v j ) ≤ j − 1. We show that v j is non-saturated: If the hypothesis on the closed subset X is not satisfied, consider a closed connected subset X ⊂ K d \ {v} containing X and that satisfies the hypothesis. For each vertex v ∈ X it holds that outdeg X (v) ≤ outdeg X (v). Therefore if v is non-saturated for X , it is also non-saturated for X.
For the 'only if' part, suppose that D is v-reduced. It is clear that the conditions (1) and (2) are satisfied. We now show condition (3). As a first step, we claim that there exists a vertex v 1 = v such that Indeed, if not, there is a point p ∈ (v, w] ∩ Supp(D) for each vertex w = v, and we can consider a connected subset X ⊂ K d \{v} for which the boundary consists of all these points p. Note that X contains the complete subgraph on the vertices in V (K d ) \ {v}. Since each boundary point p ∈ ∂X is saturated, this is in contradiction with the fact that D is v-reduced.
In a similar way, we can find a vertex v 2 = v, v 1 such that the sum of the coefficients D(p) with p ∈ (v, v 2 ] ∪ (v 1 , v 2 ] is at most 1. In order to show this, we work with a connected component X for which the boundary consists of points p ∈ Supp(D) lying on either (v, w] or (v 1 , w] for some vertex w = v, v 1 . This component contains the complete subgraph on the vertices in V (K d ) \ {v, v 1 }.
Iterating this argument, we find back condition (3).
Example 20. Consider the graph and the divisor D drawn in Figure 2. The divisor is v-reduced, since we have The definitions of linear systems g r s and gonality sequences translate to the setting of metric graphs. We prove that the gonality sequence of complete metric graphs K d is the same as the one of the ordinary complete graphs.
Theorem 21. The gonality sequence of the metric graph K d is where k and h are the uniquely determined integers with 1 ≤ k ≤ d − 3 and 0 ≤ h ≤ k such that In the proof of the theorem we will use the following result from [HKN13], which relates the rank on graphs with the rank on metric graphs: Theorem 22. Let D be a divisor on a loopless graph G and let Γ be the metric graph corresponding to G. Then, rk G (D) = rk Γ (D).
Proof of Theorem 21. Again for every k ≥ 1 and for every 0 ≤ h ≤ k there are two statements that need to be shown: • there exists a divisor of degree kd − h and rank k(k+3) 2 − h; • there does not exist a divisor of degree strictly smaller than kd − h and rank k(k+3) 2 − h. The first statement follows directly from Theorem 1 and Theorem 22. We consider the second statement. As before, it is enough to set h = k and to show that the rank of every divisor D of degree dk − k − 1 is strictly smaller than k(k+3) Let D be v d -reduced. By Lemma 19, we can label the vertices so that the following inequalities are satisfied: We write

Consider the divisors
We define a i := D(v i ) + a − (i − 2) and a + i := max{0, a i }. The sequence a = (a i ) i=1,...,d−1 will play the role of the sequence α in Section 4. In particular, we have that and the inequalities in ( * ) are satisfied. Because of Claim 11, either . This allows us to construct an effective divisor E of degree 6. What about arbitrary edge lengths?
In this section, we will focus on complete metric graphs with arbitrary edge lengths, which we will denote by K d ( ) for a vector ∈ (R >0 ) ( d 2 ) . The arguments presented in the previous sections do not directly extend to the graphs K d ( ). Therefore, computing the gonality sequence of K d ( ) is still an open problem in general. Below, we briefly explain the obstructions we encountered.
In Section 3, we proved that the divisor Theorem 22 allowed us to conclude that the divisor D viewed on the corresponding metric graph Γ = K d has the same rank. This result does not hold if the lengths of the edges are arbitrary. Moreover, our computation of the rank on G = K d relies on [CB16, Lemma 3], which uses the symmetries of complete graphs.
Even though we can not use the arguments from Section 3, we still expect that the divisor D = d i=1 k(v i ) on Γ = K d ( ) has rank k(k+3)

2
, if 1 ≤ k ≤ d−3. This statement holds for k = d−3: in this case, the divisor D coincides with the canonical divisor K Γ , which has rank g − 1 = d(d−3) 2 . It is an easy exercise to check this statement for k = 1 and (by using Riemann-Roch) k = d − 4. The following results provide extra motivation. Proof. Consider the effective divisor which has degree k(k+3) 2 + 1. The divisor D − E is v 1 -reduced (Lemma 19 also holds for arbitrary edge lengths) and has a negative coefficient at v 1 , hence |D − E| = ∅ and rk Γ (D) ≤ k(k+3) 2 .
Proof. We need to show that |D − E| = ∅ for every effective divisor E of degree 5. We may assume that E is supported on the vertices, since the vertex set is rank-determining (see [Luo11]). If E(v i ) ≤ 2 for all i, then D − E ≥ 0 and the statement follows directly. Hence, we may assume that E(v i ) ≥ 3 for some index i, so E is of the form 3(v i )+(v j )+(v k ), 3(v i )+2(v j ), 4(v i ) + (v j ) or 5(v i ) for some i = j = k = i. We will only handle the latter case; the others are proven in an analogous manner.
Suppose that E = 5(v i ). Denote by δ the minimal length of the edges (v i , v j ) with j = i and by p j ∈ (v i , v j ) the point on distance δ from v j . Then If p j = v i for more than one index j, then we are done since F ≥ 0. Otherwise, this means that the divisor F has coefficient −1 at v i . Say that k is the unique index for which p k = v i . Now we consider the linearly equivalent divisor with s j , t j ∈ (v j , v i ) the points on equal and maximal distance from p j , where s j is in between p j and v j , and t j is in between p j and v i . Because of the maximality assumption, either t j = v i or s j = v j for all j = i. If t j = v i for at least one j, we can conclude since F ≥ 0. Otherwise, we have that s j = v j for all j. Then we can consider a linearly equivalent divisor with t j ∈ (v j , v i ), s j on the path (v j , v k , v i ) and t j = v i or s j = v i for a certain j, hence F ≥ 0.
Also the proof of the sharpness of the upper bound cannot directly be extended to the graphs Γ = K d ( ). Indeed, the divisor D and D introduced in (1) and (2) are no longer linearly equivalent to D. It is natural to consider instead the following divisors with p i ∈ (v d , v i ) and p j = v j for at least one index j. But then we run into problems while constructing the effective divisor E of degree k(k+1) 2 , since in general D − E and D − E are not v d -reduced.