On Weierstrass semigroups arising from finite graphs

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1 On Weierstrass semigroups arising from finite graphs Justin D. Peachey Department of Mathematics Davidson College October 3, 2013

2 Finite graphs Definition Let {P 1, P 2,..., P n } be the set of vertices of a graph G. A divisor is an element of the free abelian group on {P 1, P 2,..., P n }, Div(G), i.e., D = n a i P i, i=1 where a i Z. Furthermore, Div k +(G) = {D Div(G) : D 0 and deg(d) = n i=1 a i = k}.

3 Finite graphs Definition Let {P 1, P 2,..., P n } be the set of vertices of a graph G. A divisor is an element of the free abelian group on {P 1, P 2,..., P n }, Div(G), i.e., D = n a i P i, i=1 where a i Z. Furthermore, Div k +(G) = {D Div(G) : D 0 and deg(d) = n i=1 a i = k}. Definition Let {P 1, P 2,..., P n } be the vertices of a graph G. Then, M(G) = {f : f (P i ) Z for all P i }.

4 Finite graphs Definition Let f M(G). Then, (f ) = v (f )v v V (G) where v (f ) = (f (v) f (w)). e=wv E(G) Note that if Q is the Laplacian of G, [ (f )] = Q [f ].

5 Example Consider the graph G below and f such that f (P 1 ) = 1, f (P 3 ) = 1, f (P 2 ) = 0, and f (P 4 ) = 0. P 1 P 2 P 3 P 4

6 Example Consider the graph G below and f such that f (P 1 ) = 1, f (P 3 ) = 1, f (P 2 ) = 0, and f (P 4 ) = 0. P 1 P 2 P 3 Then, P1 (f ) = 1 0 P2 (f ) = (0 ( 1)) + (0 ( 1)) P3 (f ) = ( 1 0) + ( 1 0) P4 (f ) = 0 ( 1) Thus, (f ) = 2P 2 + P 4 P 1 2P 3. P 4

7 Divisors on function fields Let F/K be a function field meaning that F is a finite algebraic extension of K (x) for some element x F which is transcendental over K.

8 Divisors on function fields Let F/K be a function field meaning that F is a finite algebraic extension of K (x) for some element x F which is transcendental over K. Recall that a divisor D is an element of the free abelian group on the set of places of F/K, meaning D = P n P P with n P Z, almost all n P = 0.

9 Divisors on function fields Let F/K be a function field meaning that F is a finite algebraic extension of K (x) for some element x F which is transcendental over K. Recall that a divisor D is an element of the free abelian group on the set of places of F/K, meaning D = P n P P with n P Z, almost all n P = 0. Definition Given f F \ {0}, the divisor of f is (f ) = v P (f )P. P P F

10 Riemann-Roch on function fields Definition For a divisor D, set L(D) := {f F : (f ) D} {0}. Let l(d) = dim L(D).

11 Riemann-Roch on function fields Definition For a divisor D, set L(D) := {f F : (f ) D} {0}. Let l(d) = dim L(D). Riemann-Roch Theorem For a divisor D, where W is a canonical divisor. Remark l(d) = deg D + 1 g + l(w D), If D is a divisor of F /K of degree 2g 1 where g denotes the genus of F /K, then l(d) = deg D + 1 g.

12 Riemann-Roch on finite graphs Definition The dimension of a divisor D Div(G) is r(d) = max { k : f M(G) such that (f ) E D E Div k +(G) }.

13 Riemann-Roch on finite graphs Definition The dimension of a divisor D Div(G) is r(d) = max { k : f M(G) such that (f ) E D E Div k +(G) }. Riemann-Roch Theorem for a finite graph (Baker, Norine) Let G be a graph, D Div(G), K = (deg(v) 2)v, and g = E(G) V (G) + 1. Then, v V (G) r(d) = deg(d) + 1 g + r(k D).

14 Weierstrass semigroups on function fields Definition Let P be a rational place of F /K. Then, the Weierstrass semigroup of P is { } f F with (f ) = A αp H(P) := α N :. where A 0 and P / suppa Also, the Weierstrass gap set of P is G(P) := N \ H(P).

15 Weierstrass semigroups on function fields Definition Let P be a rational place of F /K. Then, the Weierstrass semigroup of P is { } f F with (f ) = A αp H(P) := α N :. where A 0 and P / suppa Also, the Weierstrass gap set of P is G(P) := N \ H(P). Fact Given a rational place P, H(P) = {α N : l(αp) = l((α 1)P) + 1}. Remark Thus, by the Riemann-Roch Theorem if α 2g, α H(P). Hence, G(P) is finite; in fact, G(P) = g.

16 Analogues of Weierstrass semigroups for finite graphs Definition Let G be a graph and P V (G). Then, and H f (P) = H r (P) = {α N : r(αp) = r((α 1)P) + 1} { α N : f M(G) such that (f ) = A αp where A 0 and P / suppa }.

17 Analogues of Weierstrass semigroups for finite graphs Definition Let G be a graph and P V (G). Then, and H f (P) = H r (P) = {α N : r(αp) = r((α 1)P) + 1} { α N : f M(G) such that (f ) = A αp where A 0 and P / suppa }. Remark If α 2g, α H r (P). Hence, G r (P) is finite; in fact, G r (P) = g.

18 Analogues of Weierstrass semigroups for finite graphs Theorem Let G be a graph and P be a vertex of G. Then, H r (P) H f (P).

19 Analogues of Weierstrass semigroups for finite graphs Theorem Let G be a graph and P be a vertex of G. Then, H r (P) H f (P). Proof. Let α H r (P). Then, r((α 1)P) = k and r(αp) = k + 1. Thus, since r((α 1)P) = k, there exists E 0 Div+ k+1 (G) so that for all f M(G), (α 1)P E 0 + (f ) 0. Now, for all E Div k+1 + (G), there exists f M(G) so that αp E + (f ) 0. Thus, there exists h M(G) so that αp E 0 + (h) 0. Thus, it must be that P (h) = α and v (h) 0 for all v V (G) \ {P}. Hence, α H f (P).

20 Example Consider G as below. P 7 P 1 P 4 P 2 P 6 P 3 P 5

21 Example Consider G as below. P 7 P 1 P 4 P 2 P 6 P 3 P 5 Note that g = 2. Then, 1 < G r (P) for all P V (G).

22 Example Consider G as below. P 7 P 1 P 4 P 2 P 6 P 3 P 5 Note that g = 2. Then, 1 < G r (P) for all P V (G). Now, consider f M(G) where f (P i ) = 0 for i = 1, 2, 3, 7 and f (P i ) = 1 for i = 4, 5, 6. Then, (f ) = P 5 P 7. Thus, 1 H f (P 7 ).

23 Analogues of Weierstrass semigroups for finite graphs Proposition Let G be a graph and P be a vertex of G such that G \ {P} is connected. Then, deg(p) is the smallest nonzero element of H f (P).

24 Analogues of Weierstrass semigroups for finite graphs Proposition Let G be a graph and P be a vertex of G such that G \ {P} is connected. Then, deg(p) is the smallest nonzero element of H f (P). Proposition The set H f (P) is a numerical semigroup; that is, if α, β H f (P), then α + β H f (P).

25 Example Consider the given graph G. P 1 P 2 P 3 P 4

26 Example Consider the given graph G. P 1 P 2 P 3 Since deg(p 1 ) = 1, 1 H f (P 1 ). Thus, H f (P 1 ) = N. P 4

27 Example Consider the given graph G. P 1 P 2 P 3 Since deg(p 1 ) = 1, 1 H f (P 1 ). Thus, H f (P 1 ) = N. However, g = E(G) V (G) + 1 = = 1. P 4

28 Example Consider the given graph G. P 1 P 2 P 3 Since deg(p 1 ) = 1, 1 H f (P 1 ). Thus, H f (P 1 ) = N. However, g = E(G) V (G) + 1 = = 1. Thus, 1 / H r (P 1 ). Hence, P 4 H r (P 1 ) H f (P 1 ).

29 Example Consider the given graph G. P 1 P 2 P 3 Since deg(p 1 ) = 1, 1 H f (P 1 ). Thus, H f (P 1 ) = N. However, g = E(G) V (G) + 1 = = 1. Thus, 1 / H r (P 1 ). Hence, P 4 H r (P 1 ) H f (P 1 ). Remark We will use the following notation: { n } a 1, a 2,..., a n = c i a i : c i N. i=1

30 Cycles Proposition Consider C n. Then, H r (P) = H f (P) = 2, 3 for all P V (G).

31 Cycles Proposition Consider C n. Then, H r (P) = H f (P) = 2, 3 for all P V (G). Proof. Consider P V (C n ). By the Riemann-Roch theorem for finite graphs and the fact that g = 1, we know that if α 2, α H r (P). Thus, H r (P) = 2, 3. Now, since H r (P) H f (P), 2, 3 H f (P). It remains to show that 1 / H f (P). We know that deg(p) = 2 is the minimum element of H f (P) \ {0} since C n \ {P} is connected. Thus, 1 / H f (P).

32 Complete graphs Proposition Consider K n. Then, H r (P) = H f (P) = n 1, n for all P V (G).

33 Complete graphs Proposition Consider K n. Then, H r (P) = H f (P) = n 1, n for all P V (G). Sketch of the Proof Consider P V (K n ). Then, show that H f (P) = n 1, n and N \ n 1, n = g. A simple counting argument shows N \ n 1, n = g and a class of indicator functions show that n 1, n H f (P). Specifically, we define the class of indicator functions to be {f Pj } n i=1 where f Pj (P i ) = δ ij. Then, (f Pj ) = i j P i (n 1)P j.

34 Complete bipartite graphs Proposition Consider K m,n. Then, H f (P) = H r (P) = n, (m 1)n+1, (m 1)n+2,..., (m 1)n+(n 1), where (U, V ) is the natural partition of the vertices of K m,n, U = m, and P U. Proof. The proof is similar to the proof for K n.

35 Regular graphs

36 Regular graphs Example Consider the cube graph Q 3.

37 Regular graphs Example Consider the cube graph Q 3. Note 3 H f (P), but 3 / H r (P).

38 A generalization to multiple vertices Lemma Let α N m and let P 1,..., P m be distinct rational places of F/K. Then, the following are equivalent: 1 α H(P 1,..., P m ), i.e., there exists f F with (f ) = A m α i P i for some A 0 and P i / suppa i=1 2 l ( m i=1 α ip i ) l ( (α j 1) P j + m i=1,i j α ip i ), 1 j m.

39 A generalization to multiple vertices Definition Let G be a graph and P, Q V (G). Then, { } H r (P, Q) = (α, β) N 2 r(αp + βq) = r((α 1)P + βq) + 1 : r(αp + βq) = r(αp + (β 1)Q) + 1

40 A generalization to multiple vertices Definition Let G be a graph and P, Q V (G). Then, { H r (P, Q) = (α, β) N 2 r(αp + βq) = r((α 1)P + βq) + 1 : r(αp + βq) = r(αp + (β 1)Q) + 1 } and H f (P, Q) = (α, β) N2 : f M(G) such that (f ) = A αp βq where A 0 and P, Q / suppa.

41 A generalization to multiple vertices Theorem Let n 5 and P 1, P 2 V (C n ). Then, H r (P 1, P 2 ) = H f (P 1, P 2 ).

42 A generalization to multiple vertices Theorem Let n 5 and P 1, P 2 V (C n ). Then, H r (P 1, P 2 ) = H f (P 1, P 2 ). Sketch of the Proof If α + β 2, (α, β) H r (P 1, P 2 ).

43 A generalization to multiple vertices Theorem Let n 5 and P 1, P 2 V (C n ). Then, H r (P 1, P 2 ) = H f (P 1, P 2 ). Sketch of the Proof If α + β 2, (α, β) H r (P 1, P 2 ). It suffices to show (2, 0), (3, 0), (0, 2), (0, 3), (1, 1), (1, 2), (2, 1) H f (P 1, P 2 ). Question Must n 5? If so, why?

44 A generalization to multiple vertices Theorem Let n 5 and P 1, P 2 V (C n ). Then, H r (P 1, P 2 ) = H f (P 1, P 2 ). Sketch of the Proof If α + β 2, (α, β) H r (P 1, P 2 ). It suffices to show (2, 0), (3, 0), (0, 2), (0, 3), (1, 1), (1, 2), (2, 1) H f (P 1, P 2 ). Question Must n 5? If so, why? Does this hold for other families of graphs?

45 Example Consider K 5 and P 1, P 2 V (K 5 ). Thus, H r (P 1, P 2 ) H f (P 1, P 2 ) H f (P 1, P 2 )

46 Conclusions In the classical setting, H f (P) = H r (P); however, the structures of H f (P) and H r (P) differ in the case of a finite graph.

47 Conclusions In the classical setting, H f (P) = H r (P); however, the structures of H f (P) and H r (P) differ in the case of a finite graph. In certain cases it can be shown that H f (P) = H r (P) for a finite graph.

48 Conclusions In the classical setting, H f (P) = H r (P); however, the structures of H f (P) and H r (P) differ in the case of a finite graph. In certain cases it can be shown that H f (P) = H r (P) for a finite graph. Even when H f (P) = H r (P) for a finite graph, this does not imply that H f (P, Q) = H r (P, Q). Furthermore, while H r (P) H f (P), it is not even true that H r (P, Q) H f (P, Q).

49 References E. Arbarello, M. Cornalba, P. Griffiths, and J. Harris. Geometry of Algebraic Curves. Springer-Verlag, M. Baker. Specialization of linear systems from curves to graphs. Algebra & Number Theory, 2(6): , M. Baker and S. Norine. Riemann-Roch and Abel-Jacobi theory on a finite graph. Advances in Mathematics, 215: , R. Cori and D. Rossin. On the sandpile group of dual graphs. European Journal of Combinatorics, 21(4): , Dino Lorenzini. Smith normal form and Laplacians. Journal of Combinatorial Theory, Series B, 98(6): , 2008.

The Structure of the Jacobian Group of a Graph. A Thesis Presented to The Division of Mathematics and Natural Sciences Reed College

The Structure of the Jacobian Group of a Graph A Thesis Presented to The Division of Mathematics and Natural Sciences Reed College In Partial Fulfillment of the Requirements for the Degree Bachelor of