Clifford`s Theorem

10. Clifford’s Theorem
In this section we consider natural relations between the degree and rank of a divisor on a metric graph. Our primary reference is Yoav Len’s “Hyperelliptic graphs on
metrized complexes” (2016). See also Melody Chan’s “Tropical hyperelliptic curves”
(2013), Marc Coppens’s “Clifford’s theorem for graphs” (2016), and Kawaguchi and
Yamaki’s “Rank of divisors on hyperelliptic curves and graphs under specialization”
By Riemann-Roch, the rank of a divisor D is bounded below by min{−1, deg(D) −
g}, with equality if deg(D) < 0 or deg(D) > 2g − 2. On the other hand, the rank of a
divisor is bounded above by its degree. The first part of Clifford’s Theorem provides
a stronger upper bound on the rank of a divisor in terms of its degree.
Proposition 10.1. Let Γ be a metric graph of genus g, and let D be a divisor on Γ
satisfying 0 < deg(D) < 2g − 2. Then deg(D) ≥ 2r(D).
Proof. When D has negative rank, the result is obvious, so we assume throughout
that r(D) ≥ 0. First, consider the case where r(K − D) = −1. Since deg(D) < 2g,
by Riemann-Roch we have
r(D) = deg(D) − g < deg(D).
Now, consider the case where r(K − D) ≥ 0. Let E be an effective divisor of degree
r(D) + r(K − D) and let E 0 and E 00 be effective divisors of degree r(D) and r(K − D),
respectively, such that E = E 0 +E 00 . Note that D −E 0 and K −D −E 00 are equivalent
to effective divisors, so K − E is equivalent to an effective divisor. It follows that
r(D) + r(K − D) ≤ r(K) = g − 1.
By Riemann-Roch, we have
r(D) − r(K − D) = deg(D) − g + 1.
Adding these, we obtain
2r(D) ≤ deg(D).
Classically, Clifford’s Theorem for algebraic curves consists of two parts. The first
part is the inequality above, while the second part states that equality is obtained if
and only if the curve is hyperelliptic. This second statement is less straightforward
in the case of metric graphs, and will be our primary focus for the remainder of this
Definition 10.2. A metric graph Γ is hyperelliptic if it has a divisor of degree 2 and
rank 1.
Date: March 6, 2016,
Speaker : Nicholas Wawrykow,
Scribe: Netanel Friedenberg.
Figure 1. A divisor of degree 2 and rank 1 on a metric graph
Example 10.3. Consider the graph pictured in Figure 1 with three edges of arbitrary
length. Let D be the divisor of degree 2 with one chip on each of the trivalent vertices.
It is easy to see that after a chip is subtracted at any point on the graph, the resulting
divisor is still equivalent to an effective divisor. Thus, the divisor D has rank 1, and
the metric graph is hyperelliptic.
Before we proceed to Clifford’s theorem for metric graphs, we note the following
simple fact about hyperelliptic graphs.
Proposition 10.4. Let Γ be a hyperelliptic metric graph of genus g > 1. If D and
D0 are divisors of degree 2 and rank 1 on Γ, then D and D0 are equivalent.
Proof. Since rank is a superadditive function for divisors of nonnegative rank,
r((g − 1)D) ≥ g − 1
r((g − 2)D + D0 ) ≥ g − 1.
By Riemann-Roch, it follows that
(g − 1)D ∼ (g − 2)D + D0 ∼ K,
hence D ∼ D0 .
Our goal is to prove the following theorem.
Theorem 10.5. Let Γ be a metric graph of genus g. If there exists a divisor D of
rank r on Γ such that 0 < r < g − 1 and deg(D) = 2r, then Γ is hyperelliptic.
By the classification of special divisors on a hyperelliptic graph, the class of any such
D must be equal to r times the class of the unique g21 . See the paper of Kawaguchi
and Yamaki (2015).
The proof will use geometric properties of the Jacobian Jac(Γ) and the theta divisor
Wg−1 (Γ). Interestingly, however, in the case where g − 1 is prime, we can provide a
purely combinatorial proof.
Proof of Theorem 10.5 in the case that g − 1 is prime. We prove this by induction
on r, the case r = 1 being obvious. If r 6= 1, then since g − 1 is prime, there is
an integer n such that nr < g − 1 < (n + 1)r. Note that r(nD) ≥ nr, and since
deg(nD) = 2nr, by Proposition 10.1 we see that r(nD) = nr. Let r0 = g − 1 − nr. By
Riemann-Roch, we then have r(K − nD) = r0 and deg(K − nD) = 2r0 . Since r0 < r,
the result follows by induction.
Definition 10.6. Let Γ be a metric graph, and let D1 and D2 be divisors on Γ. We
say that D1 contains D2 if D1 − D2 is an effective divisor.
Definition 10.7. For a metric graph Γ we define Wdr (Γ) to be the polyhedral subset
of Picd (Γ) of equivalence classes of divisors of degree d and rank at least r. For the
sake of notational convenience we write Wd (Γ) for Wd0 (Γ).
Definition 10.8. A divisor D on a metric graph Γ is said to be rigid if it is the
unique effective divisor in its class. We say that a divisor class [D] is rigid if it
contains a unique effective representative.
Remark 10.9. Every rigid divisor has rank 0. On an algebraic curve, a divisor is
rigid if and only if it has rank 0, but on a metric graph this is not the case.
The following lemma tells us that for a metric graph Γ of genus g, the set of rigid
divisors of degree g − 1 is open and dense in Wg−1 (Γ).
Lemma 10.10. Let Γ be a metric graph of genus g > 0. Then there exists a divisor
P of degree g − 1 such that both [P ] and [K − P ] are rigid. Moreover, the set of such
divisors is open and dense in Wg−1 (Γ).
Proof. Note that Wg−1 (Γ) is a connected polyhedral subset of pure dimension g − 1.
Let B ⊂ Γ be the set of points of valence different from 2. We show that, if D ∈
Wg−1 (Γ) is not rigid, then Supp|D| ∩ B 6= ∅. To see this, recall that (Supp|D|)c is a
disjoint union of YL sets. Let U be a YL set and X a connected component of Γ r U .
If X ∩B = ∅, then by definition there exists v ∈ X such that outdegX (v) = val(v) = 2,
hence X = {v} is an isolated point. It follows that, if Supp|D| ∩ B = ∅, then Supp|D|
is a finite union of isolated points, so D is rigid.
In other words, the set of non-rigid divisors is contained in the image of B ×
Symg−2 Γ under the Abel-Jacobi map. Since B is finite, it follows that this set has
dimension at most g − 2, hence its complement is dense in Wg−1 (Γ). The same
argument shows that the set of divisors P ∈ Wg−1 (Γ) such that [K − P ] is rigid is
open and dense. Since any two open dense sets must intersect, the claim follows. We will use σ to denote the open subset of Wg−1 (Γ) consisting of divisor classes
[P ] such that both [P ] and [K − P ] are rigid. Let µ : σ → σ denote the map that
sends [P ] in σ to [K − P ] in σ.
From now on [P ] will denote an element of σ, [Q] will denote [K − P ], P will
denote the rigid divisor in [P ], and Q will denote the rigid divisor in [Q].
We now define the intersection and union of two divisors.
Definition 10.11. For divisors D and D0 let
(D ∩ D0 )(v) = min(D(v), D0 (v))
(D ∪ D0 )(v) = max(D(v), D0 (v)).
For a divisor D, define its P -part as DP = D ∩ P and its Q-part as DQ = D ∩ Q.
For the rest of the paper let Γ be a metric graph of genus g. We assume that there
exists a divisor class δ of degree 2r and rank r where 0 < r < g − 1. We aim to show
that Γ is hyperelliptic.
Lemma 10.12. Let D ∈ δ such that deg(DP ) = r. Then D is supported on P + Q.
In particular, D = DP + DQ .
Proof. By Riemann-Roch we have r(K − D) = g − 1 − r. Hence there exists a divisor
E ∼ K − D such that E − (P − DP ) is effective.
We wish to show that D + E − P = Q. We may write
D + E − P = (D − DP ) + (E − P + DP ).
By the definition of the P -part, D contains DP , and by the definition of E we know
that E contains P − DP . Thus, D + E − P is effective. Because D + E − P ∼ Q is
rigid, we see that in fact D + E − P = Q.
Note that, in Lemma 10.12, the Q-part DQ is determined uniquely by the P -part
D . Indeed, if DP + DQ ∼ DP + D0Q , then DQ ∼ D0Q implies DQ = D0Q by the
rigidity of Q. Lemma 10.12 therefore provides us with a correspondence between
divisors of degree r supported on P and divisors of degree r supported on Q. We
want to know that this correspondence is respected by the intersection of two or more
divisors. To see this, we will consider how the correspondence changes as we deform
the divisor P .
Proposition 10.13. Let D1 , . . . , Dn be effective representatives of δ such that deg(DiP ) =
r for all i. Then
deg( DiQ ) = deg( DiP ).
Proof. Let σi be the subset of σ that consists of rigid divisors classes whose rigid
representatives contain P − DiP . Similarly, let τi be the subset of σ that consists of
rigid divisor classes whose rigid representatives contain Q − DiQ . Note that
σi ) = deg(
DiP ),
and similarly for the τi . It therefore suffices to show that
dim( τi ) = dim( σi ).
In order to see this, we first prove that µ(σi ) ⊆ τi for all i.
Let P 0 be a rigid divisor of degree g − 1 that contains P − DiP . That is, P 0 is an
element of σi . Set Ei = P 0 − P + DiP . Since P 0 contains P − DiP , Ei is effective, and
since deg(Ei ) = deg(P 0 ) − deg(P ) + deg(DiP ), deg(Ei ) = r. Since δ has degree 2r
and rank r, there exists a divisor Di0 ∈ δ such that Di0 contains Ei . By Lemma 10.12,
Di is contained in P + µ(P ), so P + µ(P ) + Di0 − Di is effective and equivalent to
K. By looking at the definition of µ, we note that P + µ(P ) ∈ [K]. Moreover, since
P + µ(P ) + Di0 − Di contains P 0 and both P 0 and µ(P 0 ) are rigid, this divisor must
be equal to P 0 + µ(P 0 ). Thus, µ(P 0 ) contains Q − DiQ , and is therefore contained in
τi . It follows that
µ( σi ) ⊆
τi .
Since µ is a bijection, we may argue similarly that
µ( τi ) ⊆
σi .
τi ) = dim(
σi ).
Now we are in a position to prove Clifford’s theorem for metric graphs.
Proof of Theorem 10.5. Choose [P ] ∈ σ, and let P be its rigid representative. For
each point pi ∈ SuppP , let Si = {D ∈ δ| deg(DP ) = r, pi ∈ D}. By Lemma 10.12,
we can decompose D ∈ Si as D = DP + DQ , where Q is the rigid representative of
[Q] = [K − P ]. Let φ be the map that assigns a rigid subdivisor A of P of degree r to
the unique rigid subdivisor B of Q of degree r such that A + B ∈ δ. By Proposition
10.13, ∩D∈Si DQ consists of a single point qi . Thus, a divisor D ∈ δ with P -part of
degree r contains pi if and only if it contains qi .
We now show that the divisors pi + qi are equivalent for all i. Let pi 6= pj . Choose
Di , Dj ∈ δ such that Di ∈ Si r Sj , Dj ∈ Sj r Si , and deg(DiP ∩ DjP ) = r − 1. By
Proposition 10.13, deg(DiQ ∩ DjQ ) = r − 1, and by Lemma 10.12, Di and Dj are
supported on P + Q. Thus, deg(Di ∩ Dj ) = deg(DiP ∩ DjP ) + deg(DiQ ∩ DjQ ) = 2r − 2.
Since Di and Dj are equivalent, Di does not contain pj + qj , and Dj does not contain
pi + qi , we have the following equivalence:
0 ∼ Di − Dj = qi + pi − qj − pj .
Therefore, pi + qi ∼ pj + qj .
To show that p1 + q1 has rank 1, we need to extend SuppP = {p1 , . . . , pg−1 } to a
rank determining set. Choose a point pg such that Γ r {p1 , . . . , pg } is a tree, and such
that P 0 = P − p1 + pg ∈ σ. Note that the maps sending DP to DQ and DP to DQ
coincide for any divisor D whose P -part D is contained in P ∩ P . It follows that,
for any pi ∈ P ∩ P 0 , the associated point qi does not depend on which of the two sets
P, P 0 we use to define the bijection. Thus p1 + q1 ∼ pi + qi ∼ pg + qg . Since p1 + q1
moves, Supp(p1 + q1 ) contains a point pg+1 ∈
/ {p1 , . . . , pg }. Since {p1 , . . . , pg+1 } is a
rank determining set contained in Supp|p1 + q1 |, we see that p1 + q1 must have rank
at least 1.
[Cha13] M. Chan. Tropical hyperelliptic curves. J. Algebraic Combin., 37(2):331–359, 2013.
[Cop16] M. Coppens. Clifford’s theorem for graphs. Adv. Geom., 16(3):389–400, 2016.
[KY15] S. Kawaguchi and K. Yamaki. Rank of divisors on hyperelliptic curves and graphs under
specialization. Int. Math. Res. Not. IMRN, (12):4121–4176, 2015.
[Len16] Y. Len. Hyperelliptic graphs and metrized complexes. preprint arXiv:1601.01968, 2016.