# Tag Info

## Hot answers tagged jacobians

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)... 15 To get a model with good reduction at$2$, take$y = 2Y + x^3 + x^2 + 1$, subtract$(x^3+x^2+1)^2$from both sides, and divide by$4$to get $$Y^2 + (x^3+x^2+1) \, Y = -x^5-x^3+x^2-x.$$ (A similar tactic of un-completing the square is well-known for elliptic curves.) 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 ... 12 Over$\mathbb C$, there's also Mumford's beautiful monograph Curves and their Jacobians, The University of Michigan Press, Ann Arbor, Mich., 1975. But a monograph covering all of the topics in your wish list is going to be rather superficial. It's simply too big a subject. And I appreciate your wanting to "understand it as quickly as [you] can," but as ... 12 Any fully faithful functor from$D^b(\mathcal{A})$has adjoints (because$D^b(\mathcal{A})$is a smooth and proper category), so its image is an admissible subcategory. A recent result from Dmitrii Pirozhkov shows that any admissible subcategory in$D^b(\mathbb{P}^2)$is generated by one or two exceptional objects obtained from the standard exceptional ... 9 In fact we know the list of all$N$, whether even or odd, for which$J_0(N)$is isogenous to a product of elliptic curves. See Takuya Yamauchi, On$\mathbb Q$-simple factors of Jacobian varieties of modular curves, Yokohama Math. J. 53 (2007), no. 2, 149-160. An alternative proof (which also corrects a minor error in Yamauchi's list) is given in section ... 9 The answer, due to Jean-Pierre Serre, can be found in an unpublished note of Henri Cohen where he characterizes the odd integers$N$such that$J_0(N)$is isogenous to a product of elliptic curves. For your question,$N$is prime, and only$p=11$,$13$,$17$,$19$et$37$satisfy this condition. (For$p=13$,$J_0(p)$has dimension$0$.) In the general ... 9 I elaborate a bit on Roy Smith's comment. One can show that the general curve of genus$g>1$does not map onto a curve of genus$h>0$by a dimension count, at least for complex curves. Let$f\colon C\to D$be a map of degree$d$from a curve of genus$g$to a curve of genus$h>0$and let$B$the branch divisor of$f$(a point$P\in D$appears in$...

9

Such maps are conformal. A theorem of Liouville says that if $n\geq 3$, the only conformal maps (defined in some region in $R^n$) are Mobius. A Mobius map is a composition of inversions in spheres. For example $x\mapsto x/|x^2|$ is the inversion in the unit sphere. Inversions in all spheres generate the Mobius group. Derivative of a conformal map is a ...

9

NEW (Oct 27, 2017) I can now show that the recurrence given in the Question is correct. In fact, I can deduce a simpler and shorter recurrence relation. For this, we consider the Kummer Surface $K$ associated to $J$. This is a quartic surface in ${\mathbb P}^3$, and there is a map $\kappa \colon J \to K$ that identifies a point $P$ and its negative $-P$ (...

9

No. Choose a field $k$ and $D=k[u^2,u^3,v^2,v^3,uv]\subset k[u,v]$, so $D$ is a noetherian domain. In $D[x,y]$, choose $f=(ux+vy)^2$ and $g=(ux+vy)^3$; they are clearly algebraically dependent (but $ux+vy\notin D[x,y]$). Write $K=k(u,v)=\mathrm{Frac}(D)$. Claim: there is no $P\in D[x,y]$ such that $f,g\in D[P]$. By contradiction, let $P\in D[x,y]$ such ...

9

This is (exactly as stated in your question) for example in Proposition 10.1 of Oda, Tadao; Seshadri, C. S. Compactifications of the generalized Jacobian variety. Trans. Amer. Math. Soc. 253 (1979), 1–90. Although as the idea (by Mumford) of compactifying the Jacobian by torsion free sheaves is older, this can surely be found explicitly in older works, ...

9

I think $y^2=x^9-x$ over $\mathbb{F}_3$ has $J_C(\mathbb{F}_3)$ isomorphic to $(\mathbb{Z}/2)^6$ but please check. The $2$-torsion in $J_C$ over the algebraic closure is $(\mathbb{Z}/2)^{2g}$ (or smaller in characteristic two). On the other hand, $\#J_C(\mathbb{F}_q) \ge (\sqrt{q} -1)^{2g}$, so for $q > 9$, the latter is bigger than the former (and ...

9

As mentioned by the previous answers, this cannot be true for $n \ge 3$ by size considerations. When you identify the points $(x_i, 0)$ of $C$ inside its jacobian $J$, you are implicitly using some base-point. I will assume that $n$ and $\deg f$ are coprime, so that there is exactly one rational point at infinity (which I will denote by $\infty$). So you are ...

9

The Torelli morphism being a double cover is purely a stacky phenomenon. It is not visible on coarse moduli spaces. The involution you ask about is supposed to act on the fibers of the Torelli morphism. Here is how this works out. Pick a geometric point of $A_3$, corresponding to a ppav $(A,\Theta)$. The fiber over this point consists of pairs $(C,\phi)$ ...

8

This argument is mostly contained in t3suji's comments, but with some of the proofs somewhat expanded. As a convention (consistent with that of the theory of Chow motives), all actions of correspondences $\alpha \in \operatorname{CH}^*(X \times Y)$ on cohomology (or Jacobians) are covariant: $\alpha_* \colon H^*(X) \to H^*(Y)$, etc. Lemma. If $X$ and $Y$ ...

8

Any matrix $A$ can be written as $B.U$ where $B=\sqrt{AA^\top}$ is posititive semidefinite and $U$ is orthogonal (polar decomposition). Thus $|\det(A)| =|\det(B)|.|\det(U)|$ is the product of the eigenvalues of $B$ which are $\ge 0$ and which are also called the singular values or s-numbers of $A$. Your max.min formula multiplies the two singular values of ...

8

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 yes, for the following reason. Let $X$ be the image of $\tilde{C}$ in $P:=\operatorname{Prym}(\tilde{C}/C )$. Let me put $h:=g-1=\dim P$. What you want to prove ...

7

Yes, there is such a result for Neron models of abelian varieties with semiabelian (aka semistable) reduction: "the number of components of the special fiber cannot decrease after base change". This is a special case of Prop. 3 in section 7.4 of Bosch, Lutkebohmert, Raynaud "Neron models". Beyond the semiabelian case though, a similar claim is false: ...

7

The equation for $\# J_C(\mathbb F_p)$ that you quote contains a typo: they must have meant that $\# J_C(\mathbb F_p) = \frac 1 2 \# C(\mathbb F_{p^2}) + \frac 1 2 \# C(\mathbb F_p)^2 - p$, which is consistent with your example. That there is an equation like this is a direct consequence of the Grothendieck-Lefschetz trace formula: use that $H^i$ of an ...

7

The reverse implication is true in a considerably more general setting (Burchnall-Chaundy theory). Namely, for any pair $(U,V)$ of commuting meromorphic coefficient differential operators in one variable of order at least one, there is a two-variable polynomial $P(z,w)$ such that $P(U,V)=0$ (the polynomial evaluation is unambiguous because $U$ and $V$ ...

7

You can find a detailed proof here (theorem 1.2) in the case of principally polarized abelian varieties. One reduces to this case using the Zarhin's trick. The assumption of $k$ being infinite should not be necessary (see remark 1.3 in the paper)

7

$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}$No. As explained in this question, in $\mathbb{Q}[x,y]$, the condition $\operatorname{Jac}(f,g)=0$ implies that there exists an $h \in \mathbb{Q}[x,y]$ such that $f$ and $g$ are in $\mathbb{Q}[h]$. We now need some lemmas that are basically variants of Gauss's lemma, with multiplication replaced by composition. Recall ...

6

Given $(K,\Phi)$ , there always exists a $g$ -dimensional abelian variety $A$ such that $End(A)\otimes Q=K$ and that $K$ acts on $H^0(A,\Omega^1)$ through $\Phi$. One easily constructs as a complex torus, starting from the embedding $K\subset C^g$ given by $\Phi$." Actually, this is not always the case. For example, if $K$ is a quartic CM-field ...

6

For torsion subgroups, one can consider the "rational cuspidal subgroup", which is the subgroup of degree zero divisors generated by $\mathbb{Q}$-rational divisors coming from cusps. (The cusps themselves need not be $\mathbb{Q}$-rational, but linear combinations of them can be.) This gives a way of constructing lots of torsion points on $J_{0}(\mathbb{Q})$ ...

6

This is not true, as soon as $g\geq 3$. Taking Chern classes this would imply that $c_{g-1}(\mathcal{O}_{a(C)})$ is an integral multiple of $\ \Theta ^{g-1}$ in $\ H^{2g-2}(JC,\mathbb{Z})$. But $c_{g-1}(\mathcal{O}_{a(C)})=(-1)^{g}(g-2)![a(C)]=(-1)^g\frac{\Theta ^{g-1}}{g-1}\$, a contradiction.

6

Such curves are constructed in my paper "Familles de courbes et de variétés abéliennes sur $\mathbb{P}^1$, II", Astérisque vol. 86 (1981).

5

This is a really big topic. So I'll treat it entirely as reference request. The Jacobian of a smooth projective curve $X$ of genus $g$ is an abelian variety of dimension $g$ whose group of points is the divisor class group of $X$. More information about the algebraic constructions and properties can be found in Milne's article on Jacobian Varieties in the ...

5

I presume I am making a very basic error somewhere but I don't see where. Yes: you are confusing the automorphism group $\operatorname{Aut} A = (\operatorname{End} A)^{\times}$ with the automorphism group of the polarized abelian variety $(A,a)$. The former can be infinite as soon as $g = \operatorname{dim} A \geq 2$ but the latter is always finite. Let ...

5

First of all, the image of your homomorphism is invariant under the Galois group $G:=\mathrm{Gal}(\bar{K}/K)$. So the right question is to ask whether the induced homomorphism $\mathrm{Pic}(X_{K})\rightarrow \mathrm{Pic}(X_{\bar{K}})^G$ is surjective. In our answer to this question of yours, we (Daniel Loughran and me) give elements for the answer: the ...

Only top voted, non community-wiki answers of a minimum length are eligible