TROPICAL BRILL-NOETHER THEORY

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1 TROPICAL BRILL-NOETHER THEORY 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 (2015). 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 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 < 1 2 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 and E be effective divisors of degree r(d) and r(k D), respectively, such that E = E +E. Note that D E and K D E are equivalent to effective divisors, so K E is equivalent to an effective divisor. It follows that By Riemann-Roch, we have Adding these, we obtain r(d) + r(k D) r(k) = g 1. r(d) r(k D) = deg(d) g r(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 section. Definition 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. 1

2 2 TROPICAL BRILL-NOETHER THEORY Figure 1. A divisor of degree 2 and rank 1 on a metric graph Example 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 Let Γ be a hyperelliptic metric graph of genus g > 1. If D and D are divisors of degree 2 and rank 1 on Γ, then D and D are equivalent. Proof. Since rank is a superadditive function for divisors of nonnegative rank, r((g 1)D) g 1 and r((g 2)D + D ) g 1. By Riemann-Roch, it follows that (g 1)D (g 2)D + D K, hence D D. Our goal is to prove the following theorem. Theorem 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 g 1 2. See the paper of Kawaguchi and Yamaki (2015). The proof will use geometric properties of the Jacobian Jac(Γ) and the theta divisor W g 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 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 r = g 1 nr. By Riemann-Roch, we then have r(k nd) = r and deg(k nd) = 2r. Since r < r, the result follows by induction.

3 TROPICAL BRILL-NOETHER THEORY 3 Definition Let Γ be a metric graph, and let D 1 and D 2 be divisors on Γ. We say that D 1 contains D 2 if D 1 D 2 is an effective divisor. Definition For a metric graph Γ we define Wd r (Γ) to be the polyhedral subset of Pic d (Γ) of equivalence classes of divisors of degree d and rank at least r. For the sake of notational convenience we write W d (Γ) for Wd 0(Γ). Definition 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 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 W g 1 (Γ). Lemma 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 W g 1 (Γ). Proof. Note that W g 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 W g 1 (Γ) is not rigid, then Supp D B. 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 U. If X B =, then by definition there exists v X such that outdeg X (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 Sym g 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 W g 1 (Γ). The same argument shows that the set of divisors P W g 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 W g 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 For divisors D and D let (D D )(v) = min(d(v), D (v)) and (D D )(v) = max(d(v), D (v)). For a divisor D, define its P -part as D P = D P and its Q-part as D Q = 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.

4 4 TROPICAL BRILL-NOETHER THEORY Lemma Let D δ such that deg(d P ) = r. Then D is supported on P + Q. In particular, D = D P + D Q. Proof. By Riemann-Roch we have r(k D) = g 1 r. Hence there exists a divisor E K D such that E (P D P ) is effective. We wish to show that D + E P = Q. We may write D + E P = (D D P ) + (E P + D P ). By the definition of the P -part, D contains D P, and by the definition of E we know that E contains P D P. 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 D Q is determined uniquely by the P -part D P. Indeed, if D P + D Q D P + D Q, then D Q D Q implies D Q = D Q by the rigidity of Q. Lemma 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 Let D 1,..., D n be effective representatives of δ such that deg(di P ) = r for all i. Then n deg( D Q i ) = deg( Di P ). Proof. Let σ i be the subset of σ that consists of rigid divisors classes whose rigid representatives contain P Di P. Similarly, let τ i be the subset of σ that consists of rigid divisor classes whose rigid representatives contain Q D Q i. Note that dim( σ i ) = deg( Di P ), and similarly for the τ i. It therefore suffices to show that dim( τ i ) = dim( σ i ). In order to see this, we first prove that µ(σ i ) τ i for all i. Let P be a rigid divisor of degree g 1 that contains P Di P. That is, P is an element of σ i. Set E i = P P + Di P. Since P contains P Di P, E i is effective, and since deg(e i ) = deg(p ) deg(p ) + deg(di P ), deg(e i) = r. Since δ has degree 2r and rank r, there exists a divisor D i δ such that D i contains E i. By Lemma 10.12, D i is contained in P + µ(p ), so P + µ(p ) + D i D i 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 ) + D i D i contains P and both P and µ(p ) are rigid, this divisor must be equal to P + µ(p ). Thus, µ(p ) contains Q D Q i, and is therefore contained in τ i. It follows that µ( σ i ) τ i.

5 TROPICAL BRILL-NOETHER THEORY 5 Since µ is a bijection, we may argue similarly that µ( τ i ) σ i. Hence dim( τ i ) = dim( σ i ). Now we are in a position to prove Clifford s theorem for metric graphs. Proof of Theorem Choose [P ] σ, and let P be its rigid representative. For each point p i SuppP, let S i = {D δ deg(d P ) = r, p i D}. By Lemma 10.12, we can decompose D S i as D = D P + D Q, 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 D Q consists of a single point q i. Thus, a divisor D δ with P -part of degree r contains p i if and only if it contains q i. We now show that the divisors p i + q i are equivalent for all i. Let p i p j. Choose D i, D j δ such that D i S i S j, D j S j S i, and deg(di P Dj P ) = r 1. By Proposition 10.13, deg(d Q i D Q j ) = r 1, and by Lemma 10.12, D i and D j are supported on P + Q. Thus, deg(d i D j ) = deg(di P DP j ) + deg(dq i DQ j ) = 2r 2. Since D i and D j are equivalent, D i does not contain p j +q j, and D j does not contain p i + q i, we have the following equivalence: 0 D i D j = q i + p i q j p j. Therefore, p i + q i p j + q j. To show that p 1 + q 1 has rank 1, we need to extend SuppP = {p 1,..., p g 1 } to a rank determining set. Choose a point p g such that Γ {p 1,..., p g } is a tree, and such that P = P p 1 + p g σ. Note that the maps sending D P to D Q and D P to D Q coincide for any divisor D whose P -part D P is contained in P P. It follows that, for any p i P P, the associated point q i does not depend on which of the two sets P, P we use to define the bijection. Thus p 1 + q 1 p i + q i p g + q g. Since p 1 + q 1 moves, Supp(p 1 + q 1 ) contains a point p g+1 / {p 1,..., p g }. Since {p 1,..., p g+1 } is a rank determining set contained in Supp p 1 + q 1, we see that p 1 + q 1 must have rank at least 1. References [Cha13] M. Chan. Tropical hyperelliptic curves. J. Algebraic Combin., 37(2): , [Cop16] M. Coppens. Clifford s theorem for graphs. Adv. Geom., 16(3): , [KY15] S. Kawaguchi and K. Yamaki. Rank of divisors on hyperelliptic curves and graphs under specialization. Int. Math. Res. Not. IMRN, (12): , [Len16] Y. Len. Hyperelliptic graphs and metrized complexes. preprint arxiv: , 2016.