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Abelian varieties are projective algebraic varieties endowed with an Abelian group structure. Over the complex numbers, they can be described as quotients of a vector space by a lattice of full rank. They are analogs in higher dimensions of elliptic curves, and play an important role in algebraic geometry and number theory.
15
votes
Accepted
Torsion points of abelian variety as zeros of a section of a vector bundle?
The crucial case is $m=1$: if you have a vector bundle $E$ on $A$ of rank $\dim(A)$ and a section $s$ of $E$ whose zero locus is $\{0\} $, pulling back $(E,s)$ by multiplication by $m$ gives the gener …
7
votes
Accepted
Difference between stabilizer and automorphism group of subvariety of an abelian variety
They have absolutely no reason to be equal. Consider the case where $A$ is the Jacobian of a genus 2 curve $C$, and $X=C$ embedded in $A$ by $x\mapsto [x]-[p]$ for some fixed point $p\in C$. Then $X$ …
9
votes
Accepted
Surjectivity of the Abel-Prym map
First of all, note that your definition is not correct: when $d$ is odd, the image of your map does not land in the Prym variety -- you have to add a constant term. When this is done, the answer is ye …
3
votes
Existence of divisor in the Jacobian of smooth curve of genus two whose intersection with th...
I think I understand your question now: you want your $D$ to be effective, hence irreducible since $\theta $ is ample. Now the index theorem $(D\cdot \theta )^2\geq D^2\cdot\theta ^2\ $ implies $D^2= …
9
votes
Accepted
Pull-back of an irreducible ample divisor via an isogeny of abelian varieties
I think the Proposition is not true if $D$ is singular. Take a smooth curve $C$ of genus 2, and $X=JC$; embed $C$ in $X$ (say, by choosing a point of $C$). Let $\alpha$ be a point of order 2 in $X$; t …
9
votes
Accepted
Identifying the canonical principal polarization of a Jacobian
The answer depends very much on how you define the homomorphism $p$ associated to the polarization. My personal choice is $p(a)=\mathcal{O}(\Theta _a-\Theta )$, where $\Theta $ is a theta divisor and …
3
votes
Accepted
Derived equivalence of families of dual abelian varieties
Yes if $X$ is an abelian scheme over $B$: the Fourier-Mukai functor provides an equivalence of the derived categories. This is Theorem 1.1 in Mukai's Fourier functor and its application to the moduli …
1
vote
Accepted
Is there a unique line bundle in the Kummer surface which pulls back to a totally symmetric ...
1) Yes. If there is another one, it differs from $L'$ by a line bundle $M$ with $M^{2}\cong \mathcal{O}_Y$. Consider the resolution $\pi :\hat{Y}\rightarrow Y$ obtained by blowing up the double points …
3
votes
Accepted
A curve in an abelian surface and its image in the Kummer surface
Let $b:\hat{X}\rightarrow X$ the blowing up of the 16 2-torsion points, and $E_1,\ldots ,E_{16}$ the exceptional $(-1)$ curves on $\hat{X}$. The involution $x\mapsto -x$ lifts to an involution $\sigm …
1
vote
Action of $(\mathbb{Z}/2g\mathbb{Z})$ on quadratic forms on $\mathbb{Z}/2\mathbb{Z}$-vector ...
Let me call $m$ the symmetric biIinear form associated to $M$. In your first question you mean "all quadratic forms whose associated bilinear form is $m$"; then the answer is yes. There is indeed a ca …
1
vote
Interpretation of $H_1(A_\mathbb{C}^{top},\mathbb{Q})$
I am not sure I fully understand your question, but : $A_{\mathbb{C}}$ is the quotient of a complex vector space $V$ by a lattice $\Gamma $, which is canonically isomorphic to $H_1(A_{\mathbb{C}},\mat …
3
votes
Accepted
Divisors on an abelian surface
The fiber of the projection to $E'_i$, say $F_i$, is isomorphic to $E_j$ ($j\neq i$), not to $E'_j$. We have indeed $(F_1.F_2)=2$, and $F_1,F_2$ generate the group of divisors on $A$ up to numerical e …
1
vote
Accepted
Linear system on an abelian surface
The situation is different for abelian surfaces: if $A$ contains no elliptic curves, a linear system $|C|$ on $A$ is very ample as soon as the genus of $C$ is $\geq 8$, see this paper of Ramanan, or …
6
votes
Accepted
Algebraic Hodge decomposition of CM abelian varieties
Suppose, to simplify, that $A$ is defined over $\mathbb{Q}$. Then $H^1(A_{\mathbb{C}},\mathbb{C})$ has two $\mathbb{Q}$-structures, one coming from singular cohomology, the other one from the algebrai …
1
vote
can all CM types be realized by Jacobians?
It is certainly not what is expected : a conjecture of Coleman predicts that for $g\geq N$ (see below) there are only finitely many Jacobians of genus $g$ which are CM. Coleman's original conjecture …