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## Hot answers tagged abelian-varieties

30

Recall that any smooth projective variety of dimension $g$ embeds into $\mathbf{P}^{2g+1}$. Consider now an abelian variety $A$ of dimension $g$ which embeds into $\mathbf{P}^{2g}$. Van de Ven proves (essentially by applying the self-intersection formula to the normal bundle of $A$ in $\mathbf{P}^{2g}$) that the degree of $A$ in $\mathbf{P}^{2g}$ is given by ...

24

Weil introduced the term "polarization" in connection with his study of abelian varieties with complex multiplication. His definition is slightly different from what one sees today; one might call it a polarization up to isogeny instead of a polarization. One can find a discussion in Weil's article "On the theory of complex multiplication" ([1955d] in volume ...

23

This is a theorem of Lang's from 1956. Here's an online document giving a proof (in the form $H^1(A,k)=0$): Lecture 14: Galois Cohomology of Abelian Varieties over Finite Fields, William Stein. http://wstein.org/edu/2010/582e/lectures/582e-2010-02-12/582e-2010-02-12.pdf Stein notes that there is a "more modern proof" in the first few sections of Chapter ...

22

I don't have any contribution for the intuition beyond the fact that, I can't construct something outside the image of (1) so I hope it's surjective. Here is a sketch of the central idea of Tate's proof. Consider $A = A' \times A''$ and try to find an endomorphism of $A$ from an endomorphism of its Tate module. The endomorphism of the Tate module gives you ...

21

Really this is mostly just consolidating what has been (implicitly) said in the comments and cleaning it up a bit (e.g. using the Chow ring instead of singular cohomology), but might as well make it an answer... Let $A$ be an abelian variety of dimension $d$. We prove that $A$ cannot embed in $\mathbb{P}^{2d-1}$ and can only embed in $\mathbb{P}^{2d}$ if $d=... 18 Let me answer your last question "How can I think geometrically (in the lattice) about fixing a polarization?". I will follow the treatment given in [Birkenhake-Lange, Complex Abelian Varieties, Chapter 3]. Let$X = V / \Lambda$be a complex torus of dimension$g$and$L$a line bundle on$X$with first Chern class$H$. Then$H$is an Hermitian form on$V$, ... 18 I am not aware of anybody seriously considering hyperelliptic curves for actual real-world usage, beyond toys, and I would be rather surprised to hear differently from anyone. As you say, hyperelliptic provide comparatively few (if any!) advantages over elliptic curves but have the huge disadvantage that virtually nobody has well-tested, battle-hardened ... 18 Welcome new contributor. Yes, that is true. Let$k$be any field, let$A$be an Abelian variety over$k$, and let$U\subset A$be a dense open affine. Denote by$D\subset A$the complementary divisor with its induced reduced structure. Denote the invertible sheaf of this divisor by$\mathcal{L}:=\mathcal{O}_A(D).$Denote by$s$the global section of$\...

17

Embed the dual abelian variety into projective space. Take a smooth hyperplane section and interate until it's one-dimensional, obtaining a smooth curve $C$. By Lefschetz $C$ is irreducible, and the natural map $H_1(C, \mathbb Z) \to H_1(A^\vee, \mathbb Z)$ is surjective. Because $H_1(C, \mathbb Z)= H_1(J(C), \mathbb Z)$, the natural map $H_1(J(C), \mathbb Z)... 16 Actually, this is an exercise in Serre's Lectures on Mordell--Weil Theorem:$K(A[n])$always contains$\mu_n$if$char(K)$does not divide$n$and$A$is an abelian variety of positive dimension over$K$. (You don't need to assume that$K$is a number field) Here is a solution. First, it suffices to check the case when$n=\ell^m$is a power of a prime$\...

16

No. It is always difficult to "prove" that something is "not known", but this may do: I claim it is not even known when $K=\mathbb Q$, $A$ is an elliptic curve $E$. In fact in this case, the result you ask for is not even known when $E$ has CM. In fact, even in this very special case, it is not known that the $l$-primary torsion subgroup of the Sha is ...

16

This follows from the following well-known lemma. Lemma. Let $A$ be an abelian variety over $k$. Then any map $f \colon \mathbb P^1 \to A$ is constant. Proof 1. The map $f$ induces a map on the Albanese $f_* \colon \operatorname{Alb}_{\mathbb P^1} \to \operatorname{Alb}_A$ sitting in a commutative diagram $$\begin{array}{ccc}\mathbb P^1 & \stackrel{f}\... 15 Let me put it this way, Tate's conjecture for abelian varieties is known to imply the Hodge conjecture for abelian varieties, and the last is very much open for this class. For the implication, see the article by Deligne and Milne on "Hodge cycles on abelian varieties" (you can get a copy off of Milne's website). There are lot's of interesting cases known ... 15 Abelian varieties over the rationals are modular if and only if they are of "GL_2"-type, which is a notion introduced by Ribet who proved that this statement is a consequence of Serre's conjecture which, as you know, has since been proved. Here is a link to Ribet's paper: http://math.berkeley.edu/~ribet/Articles/korea.pdf Generalizing the statement of ... 14 A bit of overkill, but it follows from the Weil conjectures. The structure of cohomology (H^i = \wedge^i H^1) is computed over the algebraic closure and it follows that the number of points is \prod(\alpha_i-1) where the \alpha_i are the eigenvalues of Frobenius on H^1 so |\alpha_i| = q^{1/2} and the product is therefore not zero. 14 First, when defining the stack you will have the issue that there are formal deformations of abelian varieties which do not extend to families of abelian varieties over any reduced scheme. These are the deformations that do not respect any polarization. (In the complex analytic world these correspond to deformations of complex tori) So unless you have some ... 14 No. The picture over a general base is this: let A_0\to S_0 be an ordinary abelian variety over a characteristic p scheme S_0, and let S_n (n\geq 0) be compatible flat liftings of S_0 over \mathbf{Z}/p^{n+1}. Let F\colon S_0\to S_0 be the Frobenius. Then the pull-back (F^n)^* A_0 has a canonical lifting A_{n,\rm can} to S_n. Thus if ... 13 Another proof that L = \,\overline{\bf \!Q\!}\,: Clearly L is contained in \,\overline{\bf \!Q\!}\,, so we need only show L contains every algebraic number x \notin \bf Q. Let P(X) be the minimal polynomial of x. If \deg P is odd, then the class of ((x,0)) - (\infty) is a 2-torsion point on the Jacobian of the elliptic or hyperelliptic ... 13 The answer is affirmative (in Serre's formulation via principal homogeneous spaces) for proper, geometrically reduced, and geometrically connected schemes X over any field k, giving a strong mapping property relative to working over all k-schemes. That is: Theorem There exists a map f:X \rightarrow E to a torsor for an abelian variety A over k ... 13 For a prime p and an elliptic curve E/\mathbb{Q}, we have the exact sequence$$\displaystyle 0 \rightarrow E(\mathbb{Q})/p E(\mathbb{Q}) \rightarrow S_p(E) \rightarrow \text{Sha}_E[p]\rightarrow 0,$$where S_p(E) is the p-Selmer group of E and \text{Sha}_E is the Tate-Shafarevich group of E and \text{Sha}_E[p] is the p-part of it. This ... 13 Let me give an answer for k = \mathbb{C}. By a theorem of Matsusaka, every abelian variety A over an algebraic closed field k is a quotient of a Jacobian. Now just apply Matsusaka's theorem to A^{\vee}, and dualize. Since we are over \mathbb{C}, dualization sends surjective morphisms of Abelian varieties into injective ones, so we are done. I ... 13 We give a uniform approach to p \leq 61 by applying analytic discriminant bounds to the Hilbert class field. To be sure this is not entirely "conceptual", but then some computation is needed even to deal with p < 36 using Minkowski. If p = 4k+1 is prime then K = {\bf Q}(\sqrt{p}) has odd class number h, so either h=1 or h \geq 3. If h \... 12 I think the following should give a counterexample. Let \mathcal{O} be an order in an imaginary quadratic field K and \mathcal{O}_K, the ring of integers. Then it's not too hard to find a (non-split) short exact sequence of \mathcal{O}-modules:$$0 \to \mathcal{O}_K \to \mathcal{O} \oplus \mathcal{O} \to \mathcal{O}_K \to 0,$$e.g. if 1, \omega... 12 If \lambda\in\overline{\mathbb{Q}}, the elliptic curve$$ E_\lambda\colon y^2=x(x-1)(x-\lambda) $$has (\lambda,0) as 2-torsion point and is defined over (a subfield of) L=\mathbb{Q}(\lambda). Its Weil restriction A_\lambda:=\operatorname{Res}_{L/\mathbb{Q}}(E_\lambda) is an abelian variety defined over \mathbb{Q} and shares the same points of ... 12 There are non-isotrivial families of supersingular abelian varieties of dimension g over \mathbb P^1_{\overline {\mathbb F_p}} if g\geq 2; see Goren, E. Z.(3-MGL); Oort, F. Stratifications of Hilbert modular varieties. (English summary) J. Algebraic Geom. 9 (2000), no. 1, 111–154. There are many other papers of Goren and Oort with explicit ... 12 The arguments of Serre can be in fact made to work over any separably closed field. The result in the general case can then be deduced using Galois descent. Details can be found in Section 2 and the appendix of: Olivier Wittenberg - On Albanese torsors and the elementary obstruction. This is in particular shows the existence of the Albanese torsor and ... 12 A conjecture of Coleman asserts that only finitely many rings arise as the endomorphism ring of an abelian variety of given dimension defined over a number field of given degree. See [1] for an account of this conjecture. In your case, the relevant conjecture is denoted there by C(1,2). To my knowledege, the only results on Coleman's conjecture in ... 11 The answer to question (*) is yes. It is Theorem 1.2.2 in the following preprint. 11 Not in general. For instance, if A is simple then this is not possible. In fact, any isogeny$$f \colon E_1 \times \cdots \times E_n \longrightarrow A$$would give a dual isogeny$$f^{\ast} \colon A^{\ast} \longrightarrow E_1^{\ast} \times \cdots \times E_n^{\ast} and the pullback via $f^*$ of any elliptic curve in $E_1^{\ast} \times \cdots \times E_n^{... 11 Let$B$be a smooth projective curve over an algebraically closed field of characteristic zero. Let$K$be the function field of$B$. Let$S$be a finite set of closed points of$B$. You might find the following reformulation of Faltings's theorem less confusing. Theorem 1. (Faltings, geometric Shafarevich conjecture) Let$g\$ be an integer. Then the set ...

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