Dual abelian variety

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In mathematics, a dual abelian variety can be defined from an abelian variety A, defined over a field K.

The theory was first put into a good form, when K was the complex number field. In that case there is a general form of duality between the Albanese variety of a complete variety V, and its Picard variety; this was realised, for definitions in terms of complex tori, as soon as André Weil had given a general definition of Albanese variety. For an abelian variety A, the Albanese variety is A itself, so the dual should be Pic⁰(A), the connected component of what in contemporary terminology is the Picard scheme.

For the case of the Jacobian variety J of a compact Riemann surface C, the choice of a principal polarization of J gives rise to an identification of J with its own Picard variety. This in a sense is just a consequence of Abel's theorem. For general abelian varieties, still over the complex numbers, A is in the same isogeny class as its dual. An explicit isogeny can be constructed by use of an invertible sheaf L on A (i.e. in this case a holomorphic line bundle), when the subgroup

K(L)

of translations on L that take L into an isomorphic copy is itself finite. In that case, the quotient

A/K(L)

is isomorphic to the dual abelian variety Â.

This construction of Â extends to any field K of characteristic zero (David Mumford, Abelian Varieties (1970), pp.74-80). In terms of this definition, the Poincaré bundle, a universal line bundle can be defined on

A × Â.

The construction when K has characteristic p uses scheme theory. The definition of K(L) has to be in terms of a group scheme that is a scheme-theoretic stabilizer, and the quotient taken is now a quotient by a subgroup scheme. (Mumford, Abelian Varieties, p.123 onwards).

[edit] Dual isogeny (elliptic curve case)

Given an isogeny

$f : E \rightarrow E'$

of elliptic curves of degree  $n$ , the dual isogeny is an isogeny

$\hat{f} : E' \rightarrow E$

of the same degree such that

$f \circ \hat{f} = [n].$

Here $[n]$ denotes the multiplication-by- $n$ isogeny $e\mapsto ne$ which has degree $n 2 .$

[edit] Construction of the dual isogeny

Often only the existence of a dual isogeny is needed, but the construction is explicit as

$E'\rightarrow \mbox{Div}^0(E')\to\mbox{Div}^0(E)\rightarrow E\,$

where $Div 0$ is the group of divisors of degree 0. To do this, we need maps $E \rightarrow {\mbox{Div}}^0(E)$ given by $P\to P - O$ where $O$ is the neutral point of $E$ and ${\mbox{Div}}^0(E) \rightarrow E\,$ given by $\sum n_P P \to \sum n_P P.$

To see that $f \circ \hat{f} = [n]$ , note that the original isogeny $f$ can be written as a composite

$E \rightarrow {\mbox{Div}}^0(E)\to {\mbox{Div}}^0(E')\to E'\,$

and that since $f$ is finite of degree $n$ , $f * f *$ is multiplication by $n$ on $Div 0 (E').$

Alternatively, we can use the smaller Picard group $Pic 0$ , a quotient of $Div 0 .$ The map $E\rightarrow {\mbox{Div}}^0(E)$ descends to an isomorphism, $E\to{\mbox{Pic}}^0(E).$ The dual isogeny is

$E' \to {\mbox{Pic}}^0(E')\to {\mbox{Pic}}^0(E)\to E\,$

Note that the relation $f \circ \hat{f} = [n]$ also implies the conjugate relation $\hat{f} \circ f = [n].$ Indeed, let $\phi = \hat{f} \circ f.$ Then $\phi \circ \hat{f} = \hat{f} \circ [n] = [n] \circ \hat{f}.$ But $\hat{f}$ is surjective, so we must have $φ = [n].$