THE DISTRIBUTION OF SELMER RANKS OF QUADRATIC TWISTS OF JACOBIANS OF HYPERELLIPTIC CURVES

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1 THE DISTRIBUTION OF SELMER RANKS OF QUADRATIC TWISTS OF JACOBIANS OF HYPERELLIPTIC CURVES MYUNGJUN YU Abstract. Let C be an odd degree hyperelliptic curve over a number field K and J be its Jacobian. Let J χ be the quadratic character of J by a quadratic character χ Hom(G K, {±1}). For every non-negative integer r, we show the probability that dim F2 (Sel 2 (J χ /K)) = r for a certain family of quadratic twists can be given explicitly conditional on some heuristic hypothesis. 1. Introduction Let E be an elliptic curve over a number field K. Let Sel 2 (E/K) denote the 2-Selmer group of E over K. We write E χ for the quadratic twist of E by a quadratic character χ Hom(G K, {±1}). Define d 2 (E/K) := dim F2 (Sel 2 (E/K)) dim F2 (E(K)[2]), where E(K)[2] is the group of K-rational 2-torsion elements of E. Let Prob( ) denote the probability of an event. Let E 1 denote the elliptic curve y 2 = x 3 x. Heath-Brown [5] proved that (1) Prob(d 2 (E χ 1 /Q)) = d) = f d 2 := 1 2 d (1 + 2 j ) 1 j 1 j=1 2 2 j 1 Swinnerton-Dyer [13] and Kane [6] generalized this by obtaining the same distribution for the family of quadratic twists of any E over Q with E[2] E(Q) with no rational cyclic 4-isogeny. Poonen and Rains [12] noticed that the 2-Selmer groups (in fact the p-selmer groups) can be viewed as an intersection of two maximal isotropic subspaces (or Lagrangian subspaces) A and B in a (infinite dimensional) metabolic space and showed that the probability of two maximal isotropic subspaces having the intersection of dimension r is equal to f d /2. In other words, if one believes that A and B appear randomly in the set of maximal isotropic subspaces, one will get the same distribution as in (1) for the family of all elliptic curves over a general number field K. In a similar fashion to this, Bhargava, Kane, Lenstra, Poonen, and Rains [3] found plausible models for Mordell-Weil groups, Selmer groups, and Shafarevich-Tate groups of elliptic curves in terms of random maximal isotropic subspaces. When we restrict the family of all elliptic curves over K to the family of quadratic twists however, it requires a little bit more care. For an elliptic curve E over K, there could be a bias in the parity of Selmer ranks in the family of quadratic twists. For example, Dokchitser-Dokchitser [4] showed that d 2 (E χ /K) has a constant parity for any quadratic twists E χ if and only if K has no real embedding and E acquires everywhere good reduction over an abelian extension of K. Klagsbrun, Mazur 1

2 2 MYUNGJUN YU and Rubin [7] were able to quantify this bias and called it the disparity. Yu [14] generalized this to the case of Jacobians of odd degree hyperelliptic curves over K. Theorem 1.1 (Klagsbrun-Mazur-Rubin, Yu). Let C be an odd degree hyperelliptic curve and J be its Jacobian. Then we have Prob (dim F2 (Sel 2 (J χ /K)) is even) = δ(j/k), where δ(j/k) 1 is defined in Definition 8.6. We remark here that this theorem has been recently generalized further to the case of principally polarised abelian varieties by Morgan [11]. It seems that 2-Selmer groups of Jacobians of (odd degree) hyperelliptic curves of fixed genus g 2 behave (statistically) similarly to those of elliptic curves. For example, Bhargave and Gross [2] showed that when odd degree hyperelliptic curves of fixed genus g 1 (elliptic curves if g = 1) over Q are ordered by height, the average size of the 2-Selmer groups of their Jacobians is 3. As a corollary, it follows that the average of the 2-Selmer ranks of Jacobians of odd degree hyperelliptic curves of genus g is bounded above by 3/2. For the Jacobians of odd degree hyperelliptic curves, there is a conjecture ([12, Conjecture 1.7]) that predicts the same distribution as in (1). For the Jacobians of even degree hyperelliptic curves, due to the presence of a nontrivial torsor, which contributes systematically on the 2-Selmer groups, we have a conjecture with different numbers. See [12, Conjecture 1.8] and [12, Example 4.20]. For the distribution of 2-Selmer ranks of elliptic curves in the family of quadratic twists, Klagsbrun-Mazur-Rubin showed the following [8]: Theorem 1.2 (Klagsbrun-Mazur-Rubin). Let E be elliptic curve over a number field K with Gal(K(E[2])/K) = S 3 For every m 0 and X > 0 let m B m (X) = k B m,k,x be the fan-structure of collections of quadratic characters of K. Then for every r 0, lim lim m X {χ B m (X) : dim F2 Sel 2 (E χ /K) = r} B m (X) where δ(e/k) is given by Definition 8.6. = { ( δ(e/k))f r if r is odd, ( 1 2 δ(e/k))f r if r is even, They proved Theorem 1.2 by finding a certain Markov model for 2-Selmer ranks of elliptic curves in the family of quadratic twists and showing the density of 2-Selmer ranks is given by the equilibrium distribution. The main goal of this paper is to give an evidence that the distribution of 2-Selmer ranks of Jacobians of (odd degree) hyperelliptic curves in the family of quadratic twists should be the same as that of elliptic curves in the family of quadratic twists. In fact, assuming an equidistribution condition (see Definition 5.7) on certain families of Lagrangian subspaces, we prove Theorem 1.2 holds for Jacobians of odd degree hyperelliptic curves. Namely, we prove in [8]. 1 Note that in [7] and [14], the disparity constant is defined by 2δ(J/K); we follow the definition

3 THE DISTRIBUTION OF SELMER RANKS 3 Theorem 1.3. Let C : y 2 = f(x) be an hyperelliptic curve of degree 2g + 1 over a number field K with Gal(K(J[2])/K) = S 2g+1. Suppose C has UDRL (see Definition 5.7). For every m 0 and X > 0 let m B m (X) = k B m,k,x be the fan-structure of collections of quadratic characters of K as in Definition 8.4. Then for every r 0, lim lim m X {χ B m (X) : dim F2 (Sel 2 (J χ /K)) = r} B m (X) where δ(j/k) is given by Definition 8.6. = { ( δ(e/k))f r if r is odd, ( 1 2 δ(e/k))f r if r is even, Since an elliptic curve satisfying Gal(K(E[2])/K) = S 3 has UDRL (see Remark 5.9), this theorem can be regarded as a generalization of Theorem 1.2. As in [8] for elliptic curves, a direct application of Theorem 1.3 is Corollary 1.4. Suppose a hyperelliptic curve C : y 2 = f(x) of degree 2g +1 defined over a number field K has UDRL. Suppose that Gal(K(J[2])/K) = S 2g+1. With notation as in Theorem 1.3, the average Mordell-Weil rank of the quadratic twists of J satisfies lim lim χ B rk(j χ m(x) (K)) < δ(j/k) < , m X B m (X) where rk(j χ (K)) denotes the Mordell-Weil rank of J χ over K. We rely heavily on the theory built in [8], so we keep notation consistent with [8] for the convenience of the reader. 2. Metabolic spaces and Lagrangian subspaces For this section, fix a finite dimensional F p -vector space V. Definition 2.1. A quadratic form on V is a function Q : V F p such that Q(av) = a 2 Q(v) for every a F p and v V, the map (v, w) Q := Q(v + w) Q(v) Q(w) is a bilinear form. We say that X is a Lagrangian subspace or maximal isotropic subspace of V if Q(X) = 0 and X = X, where X denotes the orthogonal complement of X in the bilinear form (, ) Q. Note that it follows that dim Fp (V ) = 2 dim Fp (X). A metabolic space (V, Q) is a vector space such that (, ) Q is nondegenerate and V contains a Lagrangian subspace. When Q is understood or not important, we just call V a metabolic space. Lemma 2.2. (i) Suppose (V, Q) is a metabolic space, X is a Lagrangian space, and W is contained in a Lagrangian space. Then W + W X is a Lagrangian subspace of V. In particular, (W + W X)/W is a Lagrangian subspace of the metabolic space W /W with the quadratic form induced by Q. (ii) Suppose X, Y, and Z are Lagrangian subspaces of V. Then dim Fp ((X + Y ) Z) dim Fp (X Z + Y Z) (mod 2).

4 4 MYUNGJUN YU Proof. (i) already appeared in [12, Remark 2.4(b)] and is not difficult to prove. See [7, Lemma 2.3] for (ii). Fix a finite dimensional F p -vector space T with a continuous action of G K such that there exists bilinear, alternating, nondegenerate and G K -equivariant pairing T T µ p. For every place v of K, the cup product induces a pairing H 1 (K v, T ) H 1 (K v, T ) H 2 (K v, T T ) H 2 (K v, µ p ). For every v, there is a canonical inclusion H 2 (K v, µ p ) F p that is an isomorphism if v is nonarchimedean. The local Tate pairing is the composition (2), v : H 1 (K v, T ) H 1 (K v, T ) F p. Definition 2.3. Suppose v is a place of K. We say that Q is a Tate quadratic form on H 1 (K v, T ) if the bilinear form induced by Q (see Definition 2.1) is, v in (2). Our main interest is when T = J[2], the group of 2-torsion elements of J. In such a case, there is a canonical way to construct a Tate quadratic form q v on H 1 (K v, J[2]) for every place v of K. This quadratic form q v is induced by the Heisenberg group (see [14, Definition 5.5]). We recall for every abelian variety A/K v and ψ Hom(G Kv, {±1}), there is a canonical isomorphism A ψ [2] = A[2] and the Kummer map A(K v )/2A(K v ) H 1 (K v, A[2]). Lemma 2.4. For every place v and ψ C(K v ), the space Im(J ψ (K v )/2J ψ (K v ) H 1 (K v, J ψ [2]) = H 1 (K v, J[2])) is a Lagrangian subspace of (H 1 (K v, J[2]), q v ). Proof. The lemma follows from [12, Proposition 4.11] and [14, Theorem 5.10]. 3. Counting Lagrangian subspaces In this section, we count the number of Lagrangian subspaces of a metabolic space under various conditions, which will turn out to be crucial in our heuristic model. For this section, we fix a metabolic space V of dimension 2n over F p. Definition 3.1. Define L V := {Lagrangian subspaces of V }. Definition 3.2. Let A and B be Lagrangian subspaces of V such that dim Fp (A B) = i. Let b n,i (j) := #{C L V : dim Fp (C B) = 0 and dim Fp (C A) = j}. Remark 3.3. By Definition 3.2, if i + j > n, then b n,i (j) = 0. Remark 3.4. Suppose Y is contained in a Lagrangian subspace of V. Then Y /Y is a metabolic space (with quadratic form induced by that attached to V ). Note that there is a map (Lemma 2.2(i)) Φ Y : L V L Y /Y by sending W to (W Y + Y )/Y. The following lemma compute b n,i (j) by investigating the map Φ Y.

5 THE DISTRIBUTION OF SELMER RANKS 5 Lemma 3.5. We have (i) If i 1, b n,i (j) = p n 1 b n 1,i 1 (j). i(i+1) ni (ii) If i 1, b n,i (j) = p 2 b n i,0 (j). (iii) If j 1, b n,0 (j) = b n j,0 (0) j k=1 p n k+1 1 p k 1 (iv) n k=0 b n,0(k) = p n(n 1)/2. (v) If n i + j (mod 2), b n,i (j) = 0. (vi) b 2n,0 (0) = p (2n)(2n 1)/2 n k=1 (b 2n 2k,0(0) 2k p 2n+1 i 1 i=1 (vii) b 2n,0 (0) = n k=1 (p2k 1 1)p 2k 2. = b n j,0 (0) n j p n k+1 1 k=1. p k 1 p 2k+1 i 1 ). Proof. Choose two Lagrangian subspaces A and B of V so that dim Fp (A B) = i. We let A is spanned by a 1, a 2,, a i, a i+1,, a n B is spanned by a 1, a 2,, a i, a n+1,, a 2n i. Let (a 1 ) denote the subspace generated by a 1. Consider the map (Remark 3.4) Φ (a1) : L V L (a1) /(a 1). Put A = A/(a 1 ) and B = B/(a 1 ). Note that dim Fp (A B ) = i 1. If D {C L V dim Fp (C B) = 0 and dim Fp (C A) = j}, then Φ (a1)(d) is a Lagrangian subspace of (a 1 ) /(a 1 ), dim Fp (Φ (a1)(d) B ) = 0, and dim Fp (Φ (a1)(d) A ) = j. Conversely, for C L (a1) /(a 1) such that dim Fp (C B ) = 0 and dim Fp (C A ) = j, there are p n elements in the fiber of C in the map Φ (a1) : L V L (a1) /(a 1) by [12, Proposition 2.6(a)]. Every Lagrangian subspace F in the fiber satisfies dim Fp (F B) = 0 and dim Fp (F A) = j except only one that contains a 1. In other words, there is a p n 1 -to-one correspondence between the following two sets: {C L V : dim Fp (C B) = 0 and dim Fp (C A) = j}, and {C L (a1) /(a 1) : dim Fp (C B ) = 0 and dim Fp (C A ) = j}. Therefore, (i) follows. (ii) follows easily from (i). Now we show (iii). We suppose A and B are Lagrangian subspaces of V such that A B = {0}. Note that {M F n p : dim Fp (M) = j} = j k=1 p n k+1 1 p k. 1 Fix a dimension j subspace D of A. Put A = A/D and B = (B D + D)/D. The map Φ D : L V L D /D takes a Lagrangian subspace C (of V ) such that C B = {0} and C A = D to a Lagrangian subspace Φ D (C) of D /D such that Φ D (C) B = {0} and Φ D (C) A = {0}. Conversely, it is easy to see that a Lagrangian space W of D /D such that W A = {0} and W B = {0} has a unique element C W in

6 6 MYUNGJUN YU the fiber of W satisfying C W B = {0} and C W A = D. In other words, there is an one-to-one correspondence between the following two sets: {C L V : C A = D and C B = {0}}, and {C L D /D : C A = {0} and C B = {0}}. Therefore (iii) follows. To see (iv), combine Proposition 2.6 (b) and (e) in [12]. (v) follows from Lemma 2.2(ii). We obtain (vi) by combining (iii), (iv), and (v). Finally Lemma 3.8 implies (vii). To prove Lemma 3.8, we recall some equalities from q-binomial coefficients. Definition 3.6. Define (a) n := (a; q) n := (1 a)(1 aq) (1 aq n 1 ), Lemma 3.7. We have (i) [ ] { n (q) n (q) 1 m (q) 1 n m if 0 m n := m 0 otherwise. m [ ] m ( 1) j = j j=0 (ii) [ ][ 2n j 2n ] [ k j = j+k ][ 2n j j+k]. (iii) (1) N = 0 if N > 0 and (1) 0 = 1. (iv) (z) N = N ] j=0 ( 1) j z j q j(j 1)/2. [ N j { (q; q 2 ) n if m = 2n 0 if m is odd. Proof. (ii) and (iii) are easy to prove by the definition. (i) follows from [1, Theorem 2.3.4] and (iv) follows from [1, Theorem 2.3.3]. Lemma 3.8. Let d 0 = 0. Then d 2n = n k=1 (p2k 1 1)p 2k 2 for all n 1 if and only if ( ) n 2k d 2n = p 2n(2n 1)/2 p 2n+1 i 1 d 2n 2k p 2k+1 i for all n 1. 1 k=1 Proof. This proof was suggested by Dennis Eichhorn. In the proof, we take q = 1/p. Assuming d 2n = n k=1 (p2k 1 1)p 2k 2, we show that d 2n = p 2n(2n 1) 2 n k=1 i=1 (p 2n 1)(p 2n 1 1) (p 2n 2k+1 1) d 2n 2k (p 2k 1)(p 2k 1 1) (p 1, 1) and the converse follows by induction. If we put e 2n = d 2n /p 2n(2n 1)/2, it is equivalent to proving n [ ] 2n e 2n = 1 e 2n 2k (1/p) 2k(2k 1)/2. 2k k=1 Note that e 2k = (1/p; 1/p 2 ) k. Letting e 2k 1 = 0, it is equivalent to proving

7 THE DISTRIBUTION OF SELMER RANKS 7 1 = 2n j=0 e 2n j [ 2n j ] (1/p) j(j 1)/2 = = 2n 2n j j=0 k=0 2n 2n j j=0 k=0 ( 1) k [ 2n j k ][ ] 2n (1/p) j(j 1)/2 j [ ][ ] j + k 2n ( 1) k (1/p) j(j 1)/2. j j + k Let the last summation be [D]. The second equality comes from (i) of Lemma 3.7. The third equlity follows from (ii) of Lemma 3.7. Putting m = j + k, we re-index the double sum so that 2n [ ] 2n m [ ] m [D] = ( 1) m ( 1) j (1/p) j(j 1)/2. m j m=0 j=0 Thus, by (iii) and (iv) of Lemma 3.7, the inner sum of [D] vanishes unless m = 0, in which case the inner sum = 1. Therefore [D] = Local conditions induced by local characters and Selmer groups Let C be a hyperelliptic curve over a number field K of degree 2g + 1 and J be its Jacobian. From now on, we restrict our attention to the case when p = 2. We fix a finite set Σ of places of K containing all primes where J has bad reduction, all primes above 2, and all archimedean places. For the rest of the paper, we assume that Gal(K(J[2])/K) = S 2g+1, where S 2g+1 denotes the symmetric group with 2g + 1 letters. Definition 4.1. Let L be either a local field or a global field. Define C(L) := Hom(G L, {±1}). If L is a local field, we often identify C(L) with Hom(L, {±1}) via the local reciprocity map. We let C ram (L) denote the set of ramified quadratic characters. Definition 4.2. Define N(q) as the norm of a prime q of K. Define P n := {q : q / Σ and dim F2 (J(K q )[2]) = n} for 0 n 2g, P := P 0 P1 P2 P2g = {q : q / Σ}, P n (X) := {q : q P n and N(q) < X}. Define the width function w : P {0, 1, 2,, 2g} by w(q) := n if q P n. Remark 4.3. In the definition of P n, the condition dim F2 (J(K q )[2]) = n is equivalent to dim F2 (J(K q )/2J(K q )) = n by Lemma [14, Lemma 2.11(i)]. Recall that we attach a Tate canonical quadratic form q v to H 1 (K v, J[2]). See the paragraph after Definition 2.3. Definition 4.4. Let v be a place of K. Define and if v / Σ, define H(q v ) := {Lagrangian subspaces of (H 1 (K v, J[2]), q v )}, H ram (q v ) := {X H(q v ) : X H 1 ur(k v, J[2]) = {0}}, where H 1 ur(k v, J[2]) := ker(h 1 (K v, J[2]) H 1 (K ur v, J[2])).

8 8 MYUNGJUN YU Definition 4.5. For every place v of K, define α v : C(K v ) H(q v ) sending ψ C(K v ) to Im(J ψ (K v )/2J ψ (K v ) H 1 (K v, J ψ [2]) = H 1 (K v, J[2])), where the last isomorphism is induced by the canonical isomorphism J ψ [2] = J[2]. The map α v is well-defined by Lemma 2.4. Lemma 4.6. Suppose v Σ and ψ C(K v ). If ψ C(K v ) is unramified, then α v (ψ) = H 1 ur(k v, J[2]). If ψ C ram (K v ), then α v (ψ) H ram (q v ). Proof. It is well known that if J has good reduction at v, then we have α v (1 v ) = Hur(K 1 v, J[2]). The first assertion follows from [14, Lemma 2.15]. The second assertion follows from [14, Lemma 2.16]. Remark 4.7. If v P 1, then H ram (q v ) = 1 by Lemma 3.5(iv). Therefore if v P 1, the map α v sends every element in C ram (K v ) to the single element in H ram (q v ). If C is an elliptic curve and v P 2, then the map α v gives a bijection C ram (K v ) H ram (q v ) by [7, Proposition 5.8]. Definition 4.8. Let D := {squarefree products of primes q P 1 P 2 P 2g }, and if d D let d i be the product of all primes dividing d that lie in P i, so d = d 1 d 2 d 2g. For every d D, define w(d) := q d w(q) = 2g n=1 n {q : q d n}, the width of d, Σ(d) := Σ {q : q d} Σ P 1 P 2g, Ω d := v Σ C(K v) q d C ram(k q ), Ω S d := S q d C ram(k q ) for every subset S Ω 1 = v Σ C(K v), η d,q : Ω S dq Ω S d the projection map, if dq D. Definition 4.9. For every d D and ω = (ω v ) v Ω d, we define a Selmer structure S(ω) as follows. Let Σ S(ω) := Σ(d). If v Σ then let H 1 S(ω) (K v, J[2]) := α v (ω v ) H(q v ), and If v d, let H 1 S(ω) (K v, J[2]) := α v (ω v ) H ram (q v ), where α v is as in Definition 4.5. If ω Ω d we will also write Sel 2 (J, ω) := H 1 S(ω) (K, J[2]) := {x H1 (K, J[2]) res v (x) α v (ω v ) if v Σ(d) and res v (x) H 1 ur(k v, J[2]) otherwise}, where res v : H 1 (K, J[2]) H 1 (K v, J[2]) is the restriction map. We also define Definition Define rk(ω) := dim F2 (Sel 2 (J, ω)). Sel 2 (J, ω) (q) := ker H 1 (K, J[2]) res v H 1 (K v, J[2])/HS(ω) 1 (K v, J[2]), v q ( Sel 2 (J, ω) (q) := ker Sel 2 (J, ω) ) res q H 1 (K q, J[2]).

9 THE DISTRIBUTION OF SELMER RANKS 9 Lemma Let q P n. Then we have dim F2 (Sel 2 (J, ω) (q) ) dim F2 (Sel 2 (J, ω) (q) ) = n. Proof. The assertion follows from the Poitou-Tate duality. For example, see [10, Theorem 2.3.4]. It follows from Lemma 4.11 that if ω η 1 d,q (ω) for q P n and ω Ω d, then rk(ω) rk(ω ) n. If we let η 1 : Ω d Ω 1 be the projection, it follows by induction. Therefore we have rk(ω) rk(η 1 (ω)) + w(d) Proposition Let d D and ω Ω d. We have rk(ω) w(d) + max{rk( ω) : ω Ω 1 }. 5. The heuristic assumption We continue to assume that Gal(K(J[2])/K) = S 2g+1. The goal of this section is to explain our heuristic assumption. Definition 5.1. Fix ω Ω d. For every prime q Σ(d), define V q (ω) := Im(Sel 2 (J, ω) (q) H 1 (K q, J[2])), V q,ur := H 1 ur(k q, J[2]), P n,i := {q P n : dim F2 (V q (ω) V q,ur ) = i}, P n,i (X) := P n,i P n (X), CP n,i := C ram (K q ). q P n,i Then it follows that V q,ur and V q (ω) are Lagrangian subspaces of a metabolic space H 1 (K q, J[2]) from Lemma 4.6 and the following lemma, respectively. Lemma 5.2. For every prime q and ω Ω d, the space V q (ω) is a Lagrangian subspace of H 1 (K q, J[2]). Proof. Let x Sel 2 (J, ω) (q). Recall that for every place v, q v is a Tate quadratic form on H 1 (K v, J[2]). We have q v (x) = 0 for v q by Lemma 2.4 and the definition of Lagrangian spaces. Together with this fact, [14, Lemma 5.8] implies res q (x) = 0. Lemma 4.11 shows V q (ω) is of dimension dim F2 (H 1 (K q, J[2]))/2, so the lemma follows from Definition 2.1. Remark 5.3. Let V be a metabolic space of dimension 2n over F 2. Let A and B be Lagrangian subspaces of V (so dim F2 (A) = dim F2 (B) = n) such that dim(a B) = i. Now suppose a Lagrangian subspace X is randomly chosen in the set {X L V : X B = {0}}. Then the probability that dim F2 (X A) = j is given by b n,i (j) d n,i (j) := n i m=0 b (for p = 2). n,i(m) Here we explain our heuristic model. Suppose ω Ω d is fixed. If q P n,i, we have dim F2 (V q (ω) V q,ur ) = i. For ψ C ram (K q ), we have α q (ψ) V q,ur = {0} by

10 10 MYUNGJUN YU Lemma 4.6. Now suppose that q P n,i and ψ C ram (K q ) are chosen at random. Our heuristic model for α q (ψ), since there is no other systematic restriction on α q (ψ), is then a random Lagrangian subspace in the set An easy consequence of this is that {X L H1 (K q,j[2]) : X V q,ur = {0}}. Prob(dim F2 (α q (ψ) V q (ω)) = j) = d n,i (j). Although this is just a heuristic assumption, we give a name for C satisfying this assumption (Definition 5.7) to ease the statement of our results. We also remark here that the randomness of Lagrangian subspaces was used to model Selmer groups in [12] and [3]. Remark 5.4. Let ω Ω d and q P n,i. For ω ηd,q 1 (ω), we have (3) rk(ω ) = rk(ω) + dim F2 (α q (ω q) V q (ω)) dim F2 (V q (ω) V q,ur ), where ω q C ram (K q ) is the q-component of ω. Note that dim F2 (V q (ω) V q,ur ) = i. Therefore, knowing the distribution of dim F2 (α q (ω q) V q (ω)) is equivalent to knowing that of rk(ω ). We record here some properties of d n,i (j), which will be used in the proof of Theorem 6.9. Lemma 5.5. We have (i) If n i + j (mod 2), d n,i (j) = 0. (ii) d n,i (j) = d n i,0 (j). 2i (iii) d k+i,0 (k i) = c2i 2 k+i h+1 1 h=1 2 h 1, where c 2 (k+i)(k+i 1)/2 2i = b 2i,0 (0) for p = 2 (iv) d k+i 1,0 (k i 1) = c2i 2i h=1 2 k+i h 1 2 h 1 2 k+i 1. 2 (k+i 1)(k+i 2)/2 = d k+i,0 (k i) (2k i 1)2 k+i 1 (v) d k+i+1,0 (k i 1) = d k+i,0 (k i) (2k+i+1 1)(2 k i 1) 2 k i (2 2i+2 1). Proof. (i) follows from Lemma 3.5(v). (ii) follows from Lemma 3.5(ii). Lemma 3.5(iii) and Lemma 3.5(iv) imply (iii) and the first equality of (iv), so the second equality of (iv) follows. (v) is a consequence of (iii) and Lemma 3.5(vii). Definition 5.6. We fix ω = (ω v ) v Ω d. For every q Σ(d), let ( )) res q t(q) := dim F2 (Im Sel 2 (J, ω) H 1 ur (K q, J[2]). In particular, q P n,i if and only if q P n and t(q) = i. Definition 5.7. We say that a hyperelliptic curve C has UDRL 2 if there exists a function L : [1, ) [1, ) such that if X > L(Y ), then {ψ C ram (K q ) dim F2 (α q (ψ) V q (ω)) = j} q P n(x),q d,t(q)=i d n,i (j) < 1 C ram (K q ) 9gY q P n,q d,t(q)=i for any ω Ω d, 0 n 2g and 0 j n i. 2 UDRL stands for uniformly distributed ramified Lagrangians.

11 Remark 5.8. The constant 1 9g THE DISTRIBUTION OF SELMER RANKS 11 have been chosen by any other constant less than 1 9g and Lemma 6.8). in Definition 5.7 is not very crucial, which could (see the proof of Theorem 6.9 Remark 5.9. One can check that elliptic curves E with Gal(K(E[2])/K) = S 3 have UDRL. This follows from Remark 4.7, Lemma 5.2, and Lemma 5.5(i). However, in the hyperelliptic curve of g 2 case, if v P n for n 3, it follows from Lemma 3.5(iv) that there are 2 n(n 1)/2 (> 2) Lagrangian subspaces in H ram (q v ), whereas C ram (K v ) has only two elements. Because of this, it seems not easy to check whether or not C has UDRL. We nevertheless expect the following statement is true. Conjecture Let C be a hyperellipitc curve over a number field K of genus g 2. Suppose that Gal(K(J[2])/K) = S 2g+1. Then C has UDRL. 6. Selmer ranks controlled by primes in P 1 The main goal of this section is to prove Theorem 6.9. Roughly speaking, Theorem 6.9 means the 2-Selmer ranks in the family of local quadratic twits are controlled by the primes in P 1. We start this section by recalling the effective version of Chebotarev density theorem. Theorem 6.1. There is a nondecreasing function L : [1, ) [1, ) such that for we have every Y 1, every d D with N(d) < Y, every Galois extension F of K that is abelian of exponent 2 over K(J[2]), and unramified outside of Σ(d), every pair of subsets S, S Gal(F/K) stable under conjugation, with S nonempty, and every X > L(Y ), {q / Σ(d) : N(q) X, Frob q (F/K) S } {q / Σ(d) : N(q) X, Frob q (F/K) S} S S 1 9gY (and in particular {q / Σ(d) : N(q) X, Frob q (F/K) S} is nonempty). Proof. See [8, Theorem 8.1]. One can define L as for example, the composition of L and 9gY, for L as in [8, Theorem 8.1]. Definition 6.2. Supppse that d D and ω Ω d. restriction map H 1 (K, J[2]) Hom(G K(J[2]), J[2]) Gal(K(J[2])/K). Let Res K(J[2]) denote the Let F d,ω be the fixed field of c Sel2(J,ω) ker(res K(J[2]) (c)). Then F d,ω is a Galois extension over K. The following proposition is proved for elliptic curves E with Gal(K(E[2])/K) = S 3 in [8, Proposition 9.3]. Recall that we assume Gal(K(J[2])/K) = S 2g+1. Proposition 6.3. For every d D and ω Ω d, we have

12 12 MYUNGJUN YU (i) there is a Gal(K(J[2])/K)-module isomorphism Gal(F d,ω /K(J[2])) = J[2] rk(ω). (ii) the map Res K(J[2]) : Sel 2 (J, ω) Hom(G K(J[2]), J[2]) induces isomorphisms Sel 2 (J, ω) = Hom(Gal(F d,ω /K(J[2])), J[2]) Gal(K(J[2])/K) Gal(F d,ω /K(J[2])) = Hom(Sel 2 (J, ω), J[2]) (iii) F d,ω /K is unramified outside of Σ(d). Proof. Note that the following statements are true: (1) J[2] is a simple Gal(K(J[2])/K)-module. (2) dim F2 (Hom Gal(K(J[2])/K) (J[2], J[2])) = 1. (3) H 1 (Gal(K(J[2])/K), J[2]) = 0. (1) and (2) are easy to check and left to the reader. For (3), see [14, Lemma 3.2]. Now the proposition follows exactly as in the proof of [8, Proposition 9.3]. Proposition 6.4. Fix d D and ω Ω d. Let rk(ω) = r and define i 1 E n,i,r := (2 r ) n i (1 2 r+h ) h=0 n i m=1 2 i+m 1 2 m 1. If L is a function as in Theorem 6.1, then for every Y > N(d) and every X > L(Y ), we have {q P n (X) : q d, t(q) = i} E n,i,r < 1 {q P n (X) : q d} 9gY. Proof. Define S := {τ Gal(F d,ω /K) dim F2 (J[2]/(τ 1)J[2]) = n}, S := {τ S dim F2 (Im(loc τ : Sel 2 (J, ω) J[2]/(τ 1)J[2]) = i}, S := {σ Gal(K(J[2])/K) dim F2 (J[2]/(σ 1)J[2]) = n}, where loc τ is evaluation at τ. By Theorem 6.1, it is enough to show that S S = E n,i,r. Clearly S = Gal(F d,ω /K(J[2])) S = 2 2gr S by Proposition 6.3(i). For τ Gal(F d,ω /K), if ξ Gal(F d,ω /K(J[2])), then We define J[2]/(τ 1)J[2] = J[2]/(τξ 1)J[2]. λ τ : Sel 2 (J, ω) J[2]/(τ 1)J[2], λ τξ : Sel 2 (J, ω) J[2]/(τξ 1)J[2] = J[2]/(τ 1)J[2], given by evaluations at τ and τξ, respectively. For c Sel 2 (J, ω), we have By Proposition 6.3(ii), We also have λ τξ (c) = c(τξ) = c(τ) + τc(ξ) = λ τ (c) + c(ξ). Gal(F d,ω /K(J[2])) = Hom(Sel 2 (J, ω), J[2]) Hom(Sel 2 (J, ω), J[2]/(τ 1)J[2]). ξ (c c(ξ))

13 THE DISTRIBUTION OF SELMER RANKS 13 Therefore, for any τ K(J[2]) S, there are 2 (2g n)r {φ Hom F2 (F r 2 F n 2 ) dim F2 (Im(φ)) = i} embeddings ξ Gal(F d,ω /K(J[2])) such that dim F2 (Im(λ τξ )) = i. Therefore S = 2 (2g n)r S {φ Hom F2 (F r 2 F n 2 ) dim F2 (Im(φ)) = i}. It is easy to prove that i 1 {φ Hom F2 (F r 2 F n 2 ) dim F2 (Im(φ)) = i} = (2 r 2 h ) which completes the proof. h=0 n i m=1 2 i+m 1 2 m 1, Remark 6.5. Let rk(ω) = r and ω Ω d. Suppose q P 1, ω η 1 δ,q (ω). Let rk(ω ) = s. Recall that (i) b 1,0 (0) + b 1,0 (1) = 1, (ii) α q (ω q) V q,ur = {0}, (iii) dim F2 (α q (ω q) V q (ω)) dim F2 (V q (ω) V q,ur ) (mod 2), which follow from Lemma 3.5(iv), Lemma 4.6, and Lemma 3.5(v), respectively. Then the following can be checked by (3) in Remark 5.4. { r 1 if t(q) = 1, s = r + 1 if t(q) = 0. In virtue of Proposition 6.4, this means Prob(s = r 1) = 1 2 r and Prob(s = r + 1) = 2 r. Definition 6.6. For r, s 0, we define a matrix M L = [m r,s ] by 1 2 r if s = r 1, m r,s = 2 r if s = r + 1, 0 otherwise. Define m (n) r,s 0 so that M n L = [m(n) r,s ]. Definition 6.7. Suppose C has UDRL. Define the function L from [1, ) to itself L(Y ) := max(l(y ), L(Y )). Lemma 6.8. For 0 n 2g, 0 k n, and ɛ < 1, let 0 a k, b k, c k, d k 1 be such that a k c k < ɛ and b k d k < ɛ. Then n k=0 a kb k c k d k < 9ɛg. Proof. It is easy to prove and left to the reader. Recall that for ω ηd,q 1 (ω), we have rk(ω ) = rk(ω) + dim F2 (V q (ω) α q (ω q)) dim F2 (V q (ω) V q,ur ). Note that if we let q P n, e = dim F2 (V q (ω) V q,ur ), r = rk(ω), and s = rk(ω ), we have 0 e n, 0 s + e r n, 0 s + 2e r n.

14 14 MYUNGJUN YU Theorem 6.9. Let ω Ω d. If C has UDRL and Gal(K(J[2])/K) = S 2g+1, then for X > L(Y ) we have q P n(x),q d {ω ηd,q 1 (ω) : dim F 2 (Sel 2 (J, ω ) = s)} q P n(x),q d η 1 d,q (ω) m (n) rk(ω),s < 1 Y. Proof. Let rk(ω) = r. Let A n,r,s := {e N 0 : 0 e n, 0 s + e r n, and 0 s + 2e r n}. By (3), Theorem 6.1, Proposition 6.4 and Lemma 6.8, we only need to show that (4) F n,r (s) := e A n,r,s d n,e (s + e r)e n,e,r = (M n L) r,s. If r + s + n is odd then (M n L ) r,s = 0, and F n,r (s) is also equal to zero by Lemma 5.5(ii). Moreover, if r s > n, both (M n L ) r,s = 0, and F n,r (s) = 0. Hence it remains to show (4) when n + r + s is even and r s n. We will show (4) by induction on n. When n = 1, it is clear (see Remark 6.5). For 0 k n, it follows from Lemma 5.5(ii) (and the fact that E n,n k i,r = 0 if n k i < 0) that We want to show that F n,r (r n + 2k) = (M n L) r,r n+2k = k d k+i,0 (k i)e n,n k i,r. i=0 k d k+i,0 (k i)e n,n k i,r. i=0 By induction hypothesis for n 1, we have (ML) n r,r n+2k =(1 2 r )(M n 1 L ) r 1,(r 1) (n 1)+2k + 2 r (M n 1 L k =(1 2 r ) d k+i,0 (k i)e n 1,n k i 1,r 1 Let i=0 k + 2 r d k+i 2,0 (k i)e n 1,n k i+1,r+1. i=1 S 1 = k i=0 d k+i,0(k i)e n,n k i,r, S 2 = (1 2 r ) k i=0 d k+i,0(k i)e n 1,n k i 1,r 1, S 3 = 2 r k i=1 d k+i 2,0(k i)e n 1,n k i+1,r+1. We want to show that S 1 = S 2 + S 3. We have S 1 S 2 = = k d k+i,0 (k i) ( E n,n k i,r (1 2 r ) )E n 1,n k i 1,r 1 i=0 k n k i 1 d k+i,0 (k i)(2 r ) k+i i=0 h=0 k+i 1 (1 2 r+h ) m=1 ) r+1,(r+1) (n 1)+2(k 1) 2 n k i+m 1 2 m, 1

15 S 3 = 2 r Let = THE DISTRIBUTION OF SELMER RANKS 15 k d k+i 2,0 (k i)e n 1,n k i+1,r+1 i=1 k n k i 2 r d k+i 2,0 (k i)(2 (r+1) ) k+i 2 i=1 h=0 d k+i,0 (k i)u i := i-th summand of S 1 S 2, V i := i-th summand of S 3, g i := d k+i,0 (k i) 2k i 1 2 k+i 1, and h i := d k+i,0 (k i) g i = d k+i,0 (k i) 2k+i 2 k i 2 k+i 1. Now it is enough to show that (5) g i U i + h i+1 U i+1 = V i+1. k+i 2 (1 2 r 1+h ) m=1 2 n k i+1+m 1 2 m. 1 One can show h i+1 = d k+i,0 (k i) 2k i 1 2 by Lemma 5.5(v). Using this and Lemma 5.5(iv) for V i+1, it is straightforward to check (5) and left to the reader. 7. Local and global twists All the results in this section and the next section are already done in the elliptic curve case in Section 10 and Section 11 of [8]. We will explain how their results can be applied to the hyperelliptic curve case as well when it is necessary. We continue to assume Gal(K(J[2])/K) = S 2g+1. For the rest of this paper we assume that the set Σ is large enough to satisfy (6) Pic(O K,Σ ) = 0, where O K,Σ is the ring of Σ-integers of K. By [14, Lemma 4.9], we have (7) O K,Σ /(O K,Σ )2 K v /(K v ) 2 is injective. v P 0 Recall that C is given by an affine model y 2 = f(x) of degree 2g + 1. Let := f := the discriminant of f. Lemma 7.1. Define the subgroup A K /(K ) 2 by A := ker(k /(K ) 2 K(J[2]) /(K(J[2]) ) 2 ). Then A = Z/2Z, generated by O K,Σ. Proof. If b A, then there exists x K(J[2]) such that x 2 = b. Note that K( ) is the only nontrivial quadratic subextension of K(J[2]) over K since S 2g+1 has only one normal subgroup of index 2, so the lemma follows. Lemma 7.2. Let q P n. (i) If n is even and χ C(K q ), then χ( ) = 1. (ii) If n is odd and χ C(K q ), then χ( ) = 1 if and only if χ is unramified. Proof. This is [14, Lemma 6.1].

16 16 MYUNGJUN YU Definition 7.3. If χ C(K) and v is a place of K, we let χ v C(K v ) denote the restriction of χ to G Kv. For d D, define C(d) := {χ C(K) : χ is ramified at all q dividing d For X > 0 define and unramified outside of Σ(d) P 0 }. C(X) = {χ C(K) : χ is unramified outside of Σ {q : N(q) < X}}, C(d, X) := C(d) C(X). Let η d : C(d) Ω d be the natural map χ (..., χ v,...) v Σ(d), where χ v C(K v ) is the restriction of χ to G Kv. Lemma 7.4. Suppose d D, α O K,Σ(d) /(O K,Σ(d) )2, and α 1. If α, then there exists a prime q P 0 with N(q) L(N(d)) such that α / (O q ) 2. Proof. Note that K( α) K(J[2]) = K. Choose τ Gal(K(J[2], α)/k) such that τ K(J[2]) Gal(K(J[2])/K) = S 2g+1 is a single orbit of length 2g + 1, and τ K( α) 1. By Theorem 6.1 applied with F = K(J[2], α) and S equal to the conjugacy class of τ, we see that there exists q / Σ(d) with N(q) L(N(d)) whose Frobenius in Gal(K(J[2], α)/k) is in the conjugacy class of τ. We have α / (O q ) 2 and it follows from [14, Lemma 2.12] that q P 0. Definition 7.5. Define sign : Ω 1 {±1} by sign (..., ω v,...) := v Σ ω v( ). Define S + := {ω Ω 1 : sign (ω) = 1}, S := {ω Ω 1 : sign (ω) = 1}. We will abbreviate Ω + d = ΩS+ d and Ω d = ΩS d. Proposition 7.6. Suppose that d D and X > L(N(d)). { Ω + d if w(d) is even (i) η d (C(d, X)) = Ω d if w(d) is odd. (ii) For every ω η d (C(d, X)) we have Proof. Let {χ C(d, X) : η d (χ) = ω} C(d, X) Q 1 := {q : q P 0, N(q) X}, = 2/ Ω d. Q 2 := {q : q P 1 P 2 P 2g, q d} {q : q P 0, N(q) X}. Note that by Lemma 7.2, for q P n, we have (K q) 2 n is even, generates O q/(o q) 2 = Z/2Z n is odd. Then the proposition follows as in the proof of [8, Proposition 10.7].

17 THE DISTRIBUTION OF SELMER RANKS 17 For ω Ω d, recall that 8. The distribution of 2-Selmer ranks rk(ω) = dim F2 (Sel 2 (J, ω)). If χ C(K) then χ C(d) for a (unique) d D, and we define Sel 2 (J, χ) := Sel 2 (J, η d (χ)) and rk(χ) := rk(η d (χ)), where η d : C(d) Ω d is as in Definition 7.3. Then Sel 2 (J, χ) is the classical 2-Selmer group of J χ /K, so rk(χ) = dim F2 (Sel 2 (J χ /K)). Definition 8.1. Suppose d D. Let Ω + d and Ω d be the sets given by Definition 7.5. Let E d ± W be the probability distribution corresponding to Ω± d so that E d + (n) := {ω Ω+ d : rk(ω) = n} Ω + d, Ed (n) := {ω Ω d : rk(ω) = n} Ω d. Definition 8.2. Define a sequence of real valued functions {L n (Y )} n 1 by L 1 (Y ) := L(Y ), L n+1 (Y ) := max{l( j n L j(y )), Y L n (Y )}, n 1. If m, k Z 0 and X R >0, define the fan D m,k,x := {δ D : w(δ) = k and δ = {q 1,..., q m } with N(q j ) < L j (X) for all j}. Recall f n = n (1 + 2 j ) j 1. j=1 Definition 8.3. Define E +, E by { E + f n if n is even (n) := 0 if n is odd, j=1 E (n) := { 0 if n is even f n if n is odd. Definition 8.4. For m, k 0, define B m,k,x := C(d, L(L m+1 (X))) C(K). d D m,k,x We call B m,k,x a fan structure on C(K). We collect here several lemmas in [8] due to Klagsbrun, Mazur, and Rubin. Lemma 8.5. (i) If X > L(N(d)), then {χ C(d, X) : rk(χ) = n} C(d, X) {χ B m,k,x : rk(χ) = n} lim = X B m,k,x = { E d + (n) Ed (n) if w(d) is even if w(d) is odd. (ii) Suppose d D. For every X > L(N(d)) we have C(d, X) = C(1, X). (iii) If m, k, n 0 and X D m,k,x is nonempty, then { M k (E 1 + )(n) if k is even, M k (E 1 )(n) if k is odd.

18 18 MYUNGJUN YU Proof. (i) follows from Proposition 7.6(i) just as [8, Proposition 11.2] follows from [8, Proposition 10.7]. (ii) is [8, Lemma 11.3]. For (iii), see [8, Theorem 11.6]. In the proof, the authors used [8, Corollary 7.3] (Proposition 4.12 in this paper) to apply [8, Theorem 4.3]. Definition 8.6. If ψ, ψ C(K v ), let h(ψ, ψ ) := dim F2 (α v (ψ)/(α v (ψ) α v (ψ ))) where α v : C(K v ) H(q v ) is given by Definition 4.5, and define δ v = 1 C(K v ) γ v (ψ) := ( 1) h(1v,ψ) ψ( ) {±1}, ψ C(K v) γ v (ψ), and δ(j/k) := ( 1)rk(1) 2 δ v. Definition 8.7. If Σ n Z 0 f(n) = 1, we define the parity ρ(f) of f by ρ(f) := f(n). n odd Lemma 8.8. Suppose that Gal(K(J[2])/K) = S 2g+1 and that Σ contains a prime q 2 where J has good reduction and dim F2 (J(K q )[2]) is odd. Then ρ(e + 1 ) = 1 2 δ(j/k) and ρ(e 1 ) = δ(j/k). Proof. The lemma follows exactly as in the proof of [8, Lemma 11.10]. In the proof, the authors chose ϕ Ω 1 such that ϕ q ( ) = 1 and ϕ v = 1 v if v q. Note that they cited [9, Proposition 3] to prove v Σ ( 1) h(1q,ωq) = ω q ( ), ( 1) h(1q,ωqϕq) = ω q ϕ q ( ) = ω q ( ), which can be checked by combining Lemma 7.2(ii) and Lemma 4.6. Theorem 8.9. Suppose C over K has UDRL and Gal(K(J[2])/K) = S 2g+1. Suppose that Σ contains a prime q 2 where J has good reduction and dim F2 (J(K q [2])) is odd. Let B m (X) := k B m,k,x. Then for every n 0 we have lim lim m X {χ B m (X) rk(χ) = n} B m (X) = ( δ(j/k)) E + (n)+ ( 1 2 δ(j/k)) E (n). Proof. The theorem follows from Lemma 8.5(iii), Lemma 8.8, and [8, Proposition 2.4] just as [8, Corollary 11.12] follows from [8, Corollary 11.8] (which is deduced by using [8, Theorem 11.6] and [8, Proposition 2.4]) and [8, Lemma 11.10]. If we assume Conjecture 5.10, we obtain the following. Conjecture Theorem 8.9 holds withdout the UDRL condition. Acknowledgements The author is grateful to Karl Rubin and Dennis Eichhorn for discussions.

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