00footnotetext: Mathematics Classification Number: 14C25, 14D05, 14D20,
14D2100footnotetext: Keywords: Pushforward homomorphism, Theta divisor,
Jacobian varieties, Chow groups, higher Chow groups.

# Zero cycles on Prym varieties

Kalyan Banerjee  VIT Chennai, India<EMAIL_ADDRESS>
###### Abstract.

In this text we prove that if an abelian variety $A$ admits an embedding into
the Jacobian of a smooth projective curve $C$, and if we consider $\Theta_{A}$
to be the divisor $\Theta_{C}\cap A$, where $\Theta_{C}$ denotes the theta
divisor of $J(C)$, then the embedding of $\Theta_{A}$ into $A$ induces an
injective push-forward homomorphism (under certain conditions) at the level of
Chow groups. We show that this is the case for every Prym varietiy arising
from an unramified double cover of smooth projective curves. As a consequence
that there does not exist a universal codimension two cycle on the product of
a very general cubic threefold and the Prym variety associated to it. Hence we
conclude that a very general cubic threefold is stably irrational.

##  1\. Introduction

In the paper [BI] the authors were investigating the following question. Let
$C$ be a smooth projective curve of genus $g$ and let $\Theta$ denote the
theta divisor embedded into the Jacobian $J(C)$ of the curve $C$. Let $j$
denote this embedding. Then the push-forward homomorphism $j_{*}$ at the level
of Chow groups of zero cycles is injective.

In this paper we investigate the following question for an arbitrary
principally polarized abelian variety $A$. Let $A$ be a principally polarized
abelian variety and let $H$ denote a divisor embedded inside $A$. Let $j$
denote this embedding. Then what can we say about the kernel of the push-
forward homomorphism $j_{*}$ at the level of Chow groups of $0$-dimensional
cycles? This question is affirmatively answered in certain cases when the
abelian variety $A$ is embedded in the Jacobian variety $J(C)$ of some smooth
projective curve $C$ and we consider the divisor $\Theta\cap A$ inside $A$,
where $\Theta$ is the theta divisor of $J(C)$, with the assumption of some
extra conditions on the inverse image of the abelian variety under the natural
map ${\rm{Sym}}^{g}C\to J(C)$. It is well-known that any principally polarized
abelian variety is embedded in the Jacobian of a smooth projective curve (see
[BL] corollary $12.2.4$).So for the principally polarized abelian varieties
with the condition on the inverse image of the abelian variety mentioned
above, the above question about the kernel of the push-forward homomorphism at
the level of Chow groups of $0$-dimensional cycle is answered, when the
divisor $H$ is the intersection of the abelian variety $A$ with the theta
divisor of the ambient Jacobian variety, where $A$ is embedded.

######  Theorem 1.1.

(= Theorem 2.8)Let $A$ be a principally polarized abelian variety embedded
into the Jacobian $J(C)$ of a smooth projective curve $C$. Suppose that $A$ is
a proper subvariety of $J(C)$. Let $\Theta_{C}$ denote the theta divisor of
$J(C)$. Pull it back to $A$ and denote the pull-back by $\Theta_{A}$. Suppose
that ${\rm{Sym}}^{g-i}C\cap A^{\prime}$ is singular for some $i\geq 0$ and
suppose also that $C\cap A^{\prime}$ is irreducible, where $A^{\prime}$ is the
inverse image of $A$, under the map ${\rm{Sym}}^{g}C\to J(C)$. Then for the
closed embedding $j:\Theta_{A}\to A$, $j_{*}$ has torsion kernel at the level
of Chow groups of zero cycles.

So in answering our question for an arbitrary principally polarized abelian
variety $A$ and the divisor $\Theta_{A}=\Theta_{C}\cap A$, it is enough to
assume that $C^{\prime}\cap A$ is irreducible.

As an application we get that the embedding of the theta divisor inside a Prym
variety induces injection at the level of Chow groups of zero cycles with
${\mathbb{Q}}$-coefficients. We also show that if we start with a principally
polarized abelian surface $A$ embedded in $J(C)$ (with the condition that
$A^{\prime}\cap C$ is irreducible) and consider the corresponding
$K3$-surface, then the push-forward at the level of Chow groups of zero cycles
(with ${\mathbb{Q}}$-coefficients) induced by the closed embedding of the
divisor associated to $\Theta_{C}\cap A$ inside the $K3$-surface is injective.

Another application of the main theorem is the non-existence of a universal
codimension two cycle on the product of a very general smooth, cubic threefold
and the Prym variety associated to it. As a consequence the integral Chow
theoretic decomposition of the diagonal does not hold on the two fold product
of a very smooth, general cubic threefold and hence a very general smooth,
cubic threefold is stably irrational.

Acknowledgements: The author is indebted to A.Beauville for communicating the
main application of the paper. The author expresses his sincere gratitude to
Jaya Iyer for telling this problem about injectivity of push-forward induced
by the closed embedding of a divisor into an abelian variety, to the author
and for discussions relevant to the theme of the paper. The author thanks
C.Voisin for pointing out the fact that this injectivity of the push-forward
is not true for zero cycles on a very general divisor on an abelian variety
and also for relevant discussion regarding the theme of the paper. Author is
grateful to the anonymous referee for mentioning the non-injectivity phenomena
in the main theorem of the paper, when we consider the embedding of an
elliptic curve into a Jacobian and also for several other improvements.
Finally the author wishes to thank Olivier de Gaay Fortman concerning a
crucial point in the final argument about the singularity of the theta
divisor. The author wishes to thank the ISF-UGC and DAE grant for funding this
project and hospitality of Indian Statistical Institute, Bangalore Center, HRI
Allahabad, VIT Chennai for hosting this project.

##  2\. Abelian varieties embedded in Jacobians

We work over an algebraically closed ground field $k$ of characteristic zero.

First let us fix some notation. Let $P_{0}$ be a $k$-rational point on the
smooth projective curve $C$. Then define the closed embedding of
${\rm{Sym}}^{m}C$ into ${\rm{Sym}}^{n}C$ for $m\leq n$ by the rule:

$[x_{1},\cdots,x_{m}]\mapsto[x_{1},\cdots,x_{m},P_{0},\cdots,P_{0}]$

here $[x_{1},\cdots,x_{m}]$ denote the unordered $m$-tuple of $k$-points on
$C$. Let us defined the map from ${\rm{Sym}}^{n}C$ to $J(C)$ by the rule:

$[x_{1},\cdots,x_{n}]\mapsto\sum_{i}x_{i}-nP_{0}\;.$

Let $A$ be an abelian variety embedded inside the Jacobian of a smooth
projective curve $C$. Let $\Theta_{C}$ denote the theta divisor of the
Jacobian $J(C)$. Consider $\Theta_{A}$ to be $\Theta_{C}\cap A$, and the
closed embedding of $\Theta_{A}$ into $A$, denote it by $j_{A}$, then we prove
that $j_{A*}$ from ${\it{CH}}_{0}(\Theta_{A})$ to ${\it{CH}}_{0}(A)$ has
torsion kernel.

To prove that, first we show that the embedding of $\Theta_{C}$ into $J(C)$
gives rise to an injection at the level of Chow groups of zero cycles.
Although this has been proved in [BI][theorem 3.1], here we present an
alternative proof following [Co] which gives a better understanding of the
picture when we blow up $J(C)$ along some subvariety (in our case finitely
many points).

######  Theorem 2.1.

The embedding of the theta divisor $\Theta_{C}$ into $J(C)$ for a smooth
projective curve $C$, gives rise to an injection at the level of Chow groups
of zero cycles.

###### Proof.

We prove this by induction. Let $\sum_{i=1}^{m}C$ be the image of
${\rm{Sym}}^{m}C$, under the natural map from ${\rm{Sym}}^{m}C$ to $J(C)$. We
prove : for any $m\leq g-1$ the closed embedding of $\sum_{i=1}^{m}C$ into
$J(C)$ gives rise to an injective homomorphism at the level of Chow group of
zero cycles. We use the fact that ${\rm{Sym}}^{g}C$ maps surjectively and
birationally onto $J(C)$ and ${\rm{Sym}}^{m}C$ maps surjectively and
birationally onto $\sum_{i=1}^{m}C$. Let ${\rm{pr}}$ be the natural projection
from $C^{g}$ to $C^{m}$ given by

$(x_{1},\cdots,x_{g})\mapsto(x_{1},\cdots,x_{m})\;.$

We have a natural correspondence $\Gamma$ given by
$\pi_{g}\times\pi_{m}(Graph({\rm{pr}}))$, where $\pi_{i}$ is the natural
quotient morphism from $C^{i}$ to ${\rm{Sym}}^{i}C$. Consider the
correspondence $\Gamma_{1}$ on $J(C)\times\sum_{i=1}^{m}C$ given by
$(f_{2}\times f_{1})(\Gamma)$, where $f_{1},f_{2}$ are natural morphisms from
${\rm{Sym}}^{m}C,{\rm{Sym}}^{g}C$ to $\sum_{i=1}^{m}C,J(C)$ respectively. Now
we define the homomorphism $\Gamma_{1*}$ from ${\it{CH}}_{0}(J(C))$ to
${\it{CH}}_{0}(\sum_{i=1}^{m}C)$ as follows:

$\Gamma_{1*}(z)={\rm{pr}}_{2*}(z\times\sum_{i=1}^{m}C.\Gamma_{1})$

where

$(z\times\sum_{i=1}^{m}C).\Gamma_{1}:=(f_{2}\times f_{1})_{*}[(f_{2}\times
f_{1})^{*}(z\times\sum_{i=1}^{m}C).(f_{2}\times f_{1})^{*}\Gamma_{1}]$

this is to make $\Gamma_{1*}$ well defined, as $J(C)\times\sum_{i=1}^{m}C$ is
singular. So we have to pull the cycles back to
${\rm{Sym}}^{g}C\times{\rm{Sym}}^{m}C$ to have a well defined intersection
product. Since $f_{2},f_{1}$ are not flat we need to define what we mean my
$(f_{2}\times f_{1})^{*}(z\times\sum_{i=1}^{m}C)$, for any zero cycle $z$. So
for $z$ supported in the complement of the exceptional locus of the birational
map $f_{2}$, we have

$(f_{2}\times f_{1})^{*}(z)=z^{\prime}\times{\rm{Sym}}^{m}C$

here $z^{\prime}$ is $f_{2}^{*}(z)$, this is a zero cycle as $f_{2}^{-1}$
restricted to the complement of the exceptional locus is an isomorphism, hence
it takes zero cycles to zero cycles. Let $V_{1}$ be the Zariski open subset of
$J(C)\times\sum_{i=1}^{m}C$ mapping isomorphically onto the Zariski open set
$V_{1}^{\prime}$ of ${\rm{Sym}}^{g}C\times{\rm{Sym}}^{m}C$.

######  Lemma 2.2.

The pullback cycle $(f_{2}\times f_{1})^{*}(\Gamma_{1})$ is well defined and
it is rationally equivalent to $\Gamma+S$, where $S$ is supported on the
complement $V_{1}^{\prime}$ and it is of same dimension as $(f_{2}\times
f_{1})^{*}(\Gamma_{1})\;.$

We define

$(f_{2}\times f_{1})^{*}(\Gamma_{1}):=[\overline{((f_{2}\times
f_{1})|_{V_{1}}^{*}({\Gamma_{1}}|_{V_{1}}))}]$

Let us explain this. Since $V_{1}^{\prime}\to V_{1}$ is an isomorphism, the
pullback $(f_{2}\times f_{1})|_{V_{1}}^{*}$ is well defined from
${\it{CH}}^{m}(V_{1})$ to ${\it{CH}}^{m}(V_{1}^{\prime})$. Therefore we
restrict the correspondence $\Gamma_{1}$ to $V_{1}$ and consider its pull-back
under ${(f_{2}\times f_{1})}^{*}|_{V_{1}}$ from ${\it{CH}}^{m}(V_{1})$ to
${\it{CH}}^{m}(V_{1}^{\prime})$. Then consider the Zariski closure of each of
the components of $(f_{2}\times f_{1})^{*}|_{V_{1}}(\Gamma_{1}|_{V_{1}})$ in
${\rm{Sym}}^{g}C\times{\rm{Sym}}^{m}C$. Consider the rational equivalence
class of the cycle defined by taking Zariski closures of each of the
components with multiplicity, that is consider:

$[\overline{((f_{2}\times f_{1})|_{V_{1}}^{*}({\Gamma_{1}}|_{V_{1}}))}]\;.$

Now consider the following commutative diagram, where the rows are exact:

|
---|---
$\textstyle{{\it{CH}}_{g}(E^{\prime})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j^{\prime}_{*}}$$\textstyle{{\it{CH}}_{g}({\rm{Sym}}^{g}C\times{\rm{Sym}}^{m}C)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\it{CH}}_{g}(V_{1}^{\prime})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\
}$$\textstyle{{\it{CH}}_{g}(E^{\prime\prime})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\it{CH}}_{g}(J(C)\times\sum_{i=1}^{m}C)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\it{CH}}_{0}(V_{1})}$

Here the rightmost arrows in the rows are onto, $E^{\prime},E^{\prime\prime}$
are the complement of $V_{1}^{\prime},V_{1}$ in
${\rm{Sym}}^{g}C\times{\rm{Sym}}^{m}C$ and $J(C)\times\sum_{i=1}^{m}C$
respectively. The right vertical map is an isomorphism. Therefore if we
consider the cycle

$(f_{2}\times f_{1})^{*}(\Gamma_{1})-\Gamma$

restricted on $V_{1}^{\prime}$, then it is zero because:

${(f_{2}\times f_{1})^{*}(f_{2}\times f_{1})}_{*}(\Gamma)|_{V_{1}}$

is equal to

$(f_{2}\times f_{1})^{*}|_{V_{1}}(f_{2}\times
f_{1})|_{V^{\prime}_{1}*}(\Gamma|_{V_{1}^{\prime}})=\Gamma|_{V_{1}^{\prime}}\;.$

Therefore by the localization exact sequence of the first row of the above
diagram we have that there exists $S$ supported on
${\it{CH}}_{g}(E^{\prime})$, such that

$(f_{2}\times f_{1})^{*}(\Gamma_{1})=\Gamma+S\;.$

So we have

$(f_{2}\times f_{1})^{*}(z\times\sum_{i=1}^{m}C).(f_{2}\times
f_{1})^{*}\Gamma_{1}=z^{\prime}\times{\rm{Sym}}^{m}C.(\Gamma+S)$

where $z^{\prime}\times{\rm{Sym}}^{m}C.S$ is zero because $z^{\prime}.E=0$,
for $E$ being the exceptional locus of the map $f_{2}$. Therefore the above
intersection is

$z^{\prime}\times{\rm{Sym}}^{m}C.\Gamma\;.$

Now we have to check the following:

######  Lemma 2.3.

The above intersection product is well defined.

###### Proof.

So suppose that we have $z_{1}$ rationally equivalent to $z_{2}$ on $J(C)$ and
supported on the Zariski open set $V$, which maps isomorphically to the
Zariski open set $V^{\prime}$ of ${\rm{Sym}}^{g}C$. Then we have to prove that
the lifts $z_{1}^{\prime},z_{2}^{\prime}$ of $z_{1},z_{2}$ supported on
$V^{\prime}$ are rationally equivalent on ${\rm{Sym}}^{g}C$. So let
$C_{1},\cdots,C_{n}$ be irreducible normal curves such that their union
contains the support of $z_{1}-z_{2}$ and there exist rational functions
$f_{i}$ on $C_{i}$ such that

$z_{1}-z_{2}=\sum_{i=1}^{n}j_{i*}[div(f_{i})]$

where $[div(f_{i})]$ is the class of the divisor associated to the rational
function $f_{i}$ on $C_{i}$ and $j_{i}$ is the inclusion of $C_{i}$ into
$J(C)$. Now here we have to prove the following:

######  Sublemma 2.4.

Given $z_{1}-z_{2}$ as above, we always have a finite collection of smooth
projective curves $C_{i}$ and rational functions $f_{i}$ on $C_{i}$, such that
each $C_{i}$ intersect the exceptional locus $E$ of the above bi-rational map
at finitely many points and we have

$z_{1}-z_{2}=\sum_{i=1}^{n}j_{i*}[div(f_{i})]\;.$

###### Proof.

Consider the collection of curves and rational functions on them say
$\\{(C_{i},f_{i})\\}_{i}$ such that the above relation

$z_{1}-z_{2}=\sum_{i}j_{i*}[div(f_{i})]$

holds. By [Fu][example 1.6.3] the above is equivalent to the following: there
exists a morphism from ${\mathbb{P}}^{1}_{k}$ to ${\rm{Sym}}^{n}J(C)$ for some
$n$ and a positive zero cycle $z$ on $J(C)$ such that

$f(0)=z_{1}+z,\quad f(\infty)=z_{2}+z\;.$

That is given a collection of $\\{(C_{i},f_{i})\\}_{i}$ such that

$z_{1}-z_{2}=\sum_{i}j_{i*}[div(f_{i})]$

there exists a morphism $f$ from ${\mathbb{P}}^{1}_{k}$ to a symmetric power
${\rm{Sym}}^{n}J(C)$ and $z$ a positive zero cycle such that the above is
satisfied and vice versa. Now consider the collection of all $(f,z)$ such that
$f\in{\rm Hom}^{v}({\mathbb{P}}^{1}_{k},{\rm{Sym}}^{n+d}J(C))$, where the
superscript $v$ stands for the degree of the morphism from
${\mathbb{P}}^{1}_{k}$ to ${\rm{Sym}}^{n+d}J(C)$ and $z$ in
${\rm{Sym}}^{d}J(C)$ such that

$f(0)=z_{1}+z,\quad f(\infty)=z_{2}+z\;.$

We prove that this collection is a countable union of Zariski closed subsets
in the product ${\rm
Hom}^{v}({\mathbb{P}}^{1}_{k},{\rm{Sym}}^{n+d}J(C))\times{\rm{Sym}}^{d}J(C)$.
It is enough to consider $z_{1},z_{2}$ to be effective cycles of the same
degree $n$. This is because the effective cycles $z_{1}+z,z_{2}+z$ are
connected by a rational curve on the same symmetric power
${\rm{Sym}}^{n+d}J(C)$. Therefore

$\deg(z_{1}+z)=\deg(z_{2}+z)$

which gives

$\deg(z_{1})+\deg(z)=\deg(z_{2})+\deg(z)$

implying that

$\deg(z_{1})=\deg(z_{2})\;.$

Consider the following morphisms:

$\phi:{\rm{Sym}}^{n}J(C)\times{\rm{Sym}}^{n}J(C)\times{\rm{Sym}}^{d}J(C)\to{\rm{Sym}}^{n+d}J(C)\times{\rm{Sym}}^{n+d}J(C)$

given by

$(z_{1},z_{2},z)\mapsto(z_{1}+z,z_{2}+z)$

and

${\it{ev}}:{\rm
Hom}^{v}({\mathbb{P}}^{1}_{k},{\rm{Sym}}^{n+d}J(C))\to{\rm{Sym}}^{n+d}J(C)\times{\rm{Sym}}^{n+d}J(C)$

given by

$f\mapsto(f(0),f(\infty))$

Then consider the fiber product of ${\rm
Hom}^{v}({\mathbb{P}}^{1}_{k},{\rm{Sym}}^{n+d}J(C))$ and
${\rm{Sym}}^{n}J(C)\times{\rm{Sym}}^{n}J(C)\times{\rm{Sym}}^{d}J(C)$ over
${\rm{Sym}}^{n+d}J(C)\times{\rm{Sym}}^{n+d}J(C)$ given by the above two maps.
That is we consider all

$(f,z_{1},z_{2},z)$

such that

$f(0)=z_{1}+z,f(\infty)=z_{2}+z\;.$

Call this fiber product $Z^{n}_{d,v}$, this admits a projection morphism to
${\rm{Sym}}^{n}J(C)\times{\rm{Sym}}^{n}J(C)$. Then for fixed $(z_{1},z_{2})$,
the fiber over it in $Z^{n}_{d,v}$ is the collection of all $(f,z)$ such that

$f(0)=z_{1}+z,\quad f(\infty)=z_{2}+z\;.$

Now consider the scheme

${\mathscr{V}}^{n}_{d,v}:=\\{(x,f,z)|x\in{\rm
im}(f),f(0)=z_{1}+z,f({\infty})=z_{2}+z\\}$

this is a ${\mathbb{P}}^{1}$-bundle over ${Z^{n}_{d,v}}_{z_{1},z_{2}}$ which
is the fiber of the projection of
$Z^{n}_{d,v}\to{\rm{Sym}}^{n}J(C)\times{\rm{Sym}}^{n}J(C)$. Consider the
subscheme of ${\mathscr{V}}^{n}_{d,v}$ given by ${\mathscr{V}}^{n}_{d,v,E}$ to
be the collection of $(x,f)$ in ${\mathscr{V}}^{n}_{d,v}$ such that image of
$f$ intersects ${\rm{Sym}}^{n+d}E$ at finite number of points. Suppose that
the projection map from ${\mathscr{V}}^{n}_{d,v,E}$ to
${Z^{n}_{d,v}}_{z_{1},z_{2}}$ is surjective (observe that this map has finite
fibers) and the dimension of ${\mathscr{V}}^{n}_{d,v}$ is
$\dim({Z^{n}_{d,v}}_{z_{1},z_{2}})+1$, which is
$\dim({\mathscr{V}}^{n}{d,v,E})+1$. So we have

$\dim({\mathscr{V}}^{n}_{d,v})-1=\dim({\mathscr{V}}^{n}_{d,v,E})\;.$

Consider the projection
$\pi_{n+d}:{\mathscr{V}}^{n}_{d,v}\to{\rm{Sym}}^{n+d}J(C)$. Now
${\mathscr{V}}^{n}_{d,v,E}$ is nothing but the inverse image

$\pi_{n+d}^{-1}({\rm{Sym}}^{n+d}E)\;.$

The Zariski closed set, ${\rm{Sym}}^{n+d}E$ is of codimension bigger than two,
say $i$ in ${\rm{Sym}}^{n+d}J(C)$. Let $g$ be the genus of the curve $C$.
Observe that ${{\rm{Sym}}^{n+d}}{\rm{Sym}}^{g}C$ maps surjectively and
birationally onto ${\rm{Sym}}^{n+d}J(C)$. So the inverse image of
${\rm{Sym}}^{n+d}E$ under this map is generically a product of projective
bundles over ${\rm{Sym}}^{n+d}E$ (because the fibers are product of linear
systems of positive dimension). Consider the pullback of this prodcut of
projective bundles to ${\mathscr{V}}^{n}_{d,v,E}$. Denote it by
$\widetilde{{\mathscr{V}}^{n}_{d,v,E}}$. Then the dimension of this pullback
$\widetilde{{\mathscr{V}}^{n}_{d,v,E}}$ is strictly greater than that of
${\mathscr{V}}^{n}_{d,v,E}$. On the other hand since
${{\rm{Sym}}^{n+d}}{\rm{Sym}}^{g}C\to{\rm{Sym}}^{n+d}J(C)$ is birational, the
pull-back of ${\mathscr{V}}^{n}_{d,v}$ along the above map, say
$\widetilde{{\mathscr{V}}^{n}_{d,v}}$ is birational to
${\mathscr{V}}^{n}_{d,v}$ and it contains the pullback
$\widetilde{{\mathscr{V}}^{n}_{d,v,E}}$. Also we have

$\dim(\widetilde{{\mathscr{V}}^{n}_{d,v}})-1=\dim({\mathscr{V}}^{n}_{d,v})-1=\dim({\mathscr{V}}^{n}_{d,v,E})\;.$

This tells us that $\widetilde{{\mathscr{V}}^{n}_{d,v}}$ is a
${\mathbb{P}}^{1}$-bundle over ${\mathscr{V}}^{n}_{d,v,E}$, contradicting the
fact that ${\mathscr{V}}^{n}_{d,v}$ is birational to
$\widetilde{{\mathscr{V}}^{n}_{d,v}}$. Therefore there exists $f$ in ${\rm
Hom}^{v}({\mathbb{P}}^{1}_{k},{\rm{Sym}}^{n+d}J(C))$ such that image of $f$
has empty intersection with ${\rm{Sym}}^{n+d}E$. This finishes the proof
saying that there exist curves $C_{i}$ such that $C_{i}$ is not contained
inside $E$ and we have

$z_{1}-z_{2}=\sum_{i=1}^{n}j_{i*}[div(f_{i})]\;.$

Indeed suppose that there exist curves $C_{i}$ such that some of them are
contained in $E$. Then by the equivalent definition of rational
equivalence[Fu][chapter 1 on rational equivalence] it follows that the curves
$C_{i}$, gives rise to algebraic cycles (of dimension $1$) $V_{i}$ on
$J(C)\times{\mathbb{P}}^{1}$ mapping dominantly onto ${\mathbb{P}}^{1}$, such
that

$\sum_{i}V_{i}[0]=z_{1}+z\;,$ $\sum_{i}V_{i}[\infty]=z_{2}+z\;,$

such that $z$ is a positive zero cycle.

Here

$V_{i}[P]={\rm{pr}}_{J(C)*}({{\rm{pr}}_{{\mathbb{P}}^{1}}}|_{V_{i}}^{-1}(P))\;.$

We have to define $V_{i}[P]$ this way because the Chow moving lemma may not be
true, when we replace $J(C)$ by a singular variety. Now if some of the $C_{i}$
is contained inside $E$, then we have $V_{i}$ contained in
$E\times{\mathbb{P}}^{1}$. This is because, $V_{i}$ is obtained as the closure
of the graph of the rational function $f_{i}$ on $C_{i}$ added to some other
specific cycle,[Fu][chapter 1]. So if $C_{i}$ is contained in $E$, then the
closure of the graph of corresponding $f_{i}$ is contained in
$E\times{\mathbb{P}}^{1}$. Since $V_{i}$ maps dominantly onto
${\mathbb{P}}^{1}$ the fibers of these maps are finite(possibly after
normalizing $V_{i}$). Therefore the scheme

${{{\rm{pr}}_{\mathbb{P}}^{1}}}|_{V_{i}}^{-1}(P)$

is a finite set of points. Further if $V_{i}$ is in $E\times{\mathbb{P}}^{1}$,
this finite set of points land inside $E$. Following [Fu][chapter 1], this
$V_{i}$’s gives rise to a set theoretic map $f$ from ${\mathbb{P}}^{1}$ To
${\rm{Sym}}^{d}J(C)$ for some $d$, which is a regular morphism of schemes. The
map is obtained as above:

$V_{i}[P]={\rm{pr}}_{J(C)*}({{\rm{pr}}_{{\mathbb{P}}^{1}}}|_{V_{i}}^{-1}(P))\;.$

Since for $C_{i}$ in $E$, we have, $V_{i}$ in $E\times{\mathbb{P}}^{1}$, we
obtaine that image of $f$ actually intersects ${\rm{Sym}}^{d}E$. ∎

Then according to the above theorem there exists a finite collection of smooth
projective curves $C_{i}$ not contained in $E$ and rational functions $f_{i}$
on them such that

$z_{1}-z_{2}=\sum_{i}j_{i*}[div(f_{i})]\;.$

Let $\widetilde{C_{i}}$ be the strict transform of $C_{i}$, under the
birational map $f_{2}$. This means that

$\widetilde{C_{i}}=\overline{f_{2}^{-1}(C_{i}\setminus E)}$

where $E$ is the exceptional locus of the birational map $f_{2}$. Since
$C_{i}$ is birational to $\widetilde{C_{i}}$ and $C_{i}$ is normal, we have
$C_{i}\cong\widetilde{C_{i}}$. So let us consider the map

$\phi:\cup_{i}\widetilde{C_{i}}\to\cup_{i}C_{i}$

this is an isomorphism. So now consider the pull-back of the cycle
$z_{1}-z_{2}$ by $\phi$, that is

$z_{1}^{\prime}-z_{2}^{\prime}=\phi^{*}(z_{1}-z_{2})=\sum_{i=1}^{n}\phi^{*}j_{i*}[div(f_{i})]$

This pullback is well defined as $\phi$ is an isomorphism onto its image and
$z_{1}-z_{2}$ is supported on $V$. Now all we have to prove is that

$\phi^{*}j_{i*}[div(f_{i})]=j_{i*}^{\prime}\phi^{*}[div(f_{i})]\;.$

Now

$\phi^{*}[div(f_{i})]=[div(\phi^{*}(f_{i}))]$

and

$j_{i*}^{\prime}[div(\phi^{*}(f_{i}))]=[div(N(\phi^{*}(f_{i})))]$

Here $N(\phi^{*}(f))$ is the determinant of the linear transformation given by
$\phi^{*}(f)$. This is obtained as follows. Consider the map
$j_{i}^{\prime}:\widetilde{C_{i}}\to\widetilde{D_{i}}$, where
$\widetilde{D_{i}}$ is the image of $\widetilde{C_{i}}$ under $j_{i}^{\prime}$
and it is an isomorphism. So we have $k(\widetilde{C_{i}})\cong
k(\widetilde{D_{i}})$. Hence $N(\phi^{*}(f_{i}))$ is nothing but
multiplication by $\phi^{*}(f_{i})$. Now since we also have $C_{i}\cong
j_{i}(C_{i}):=D_{i}$ and $C_{i}\cong\widetilde{C_{i}}$,

$N(\phi^{*}(f_{i}))=\phi^{*}(N(f_{i}))\;.$

So we get that

$div[N(\phi^{*}(f_{i}))]=div[\phi^{*}(N(f_{i}))]=\phi^{*}div[N(f_{i})]=\phi^{*}j_{i*}[div(f_{i})]\;.$

Therefore we can conclude that $z_{1}^{\prime}-z_{2}^{\prime}$ is rationally
equivalent to zero on ${\rm{Sym}}^{g}C$ and it is supported on $V^{\prime}$. ∎

Now we recall the following lemma from [Voi] to make the above intersection
product well defined on all of $J(C)$.

######  Lemma 2.5.

Let $Y$ be a connected projective variety over an uncountable algebraically
closed field $k$. Let $U\subset Y$ be the complement to a countable union of
proper Zariski closed subsets of $Y$. Then any zero cycle $z$ on $Y$ is
rationally equivalent in $Y$ to some $z^{\prime}$ supported on $U$.

###### Proof.

See [Voi][Fact 3.3]. ∎

Let $j$ denote the embedding of $\sum_{i=1}^{m}C$ to $J(C)$. Similar to above
we can define:

$(z\times\sum_{i=1}^{m}C).(j\times{\rm{id}})^{*}(\Gamma_{1})$ $=(f_{1}\times
f_{1})_{*}((f_{1}\times f_{1})^{*}(z\times\sum_{i=1}^{m}C).(f_{1}\times
f_{1})^{*}(j\times{\rm{id}})^{*}(\Gamma_{1}))$

Now we prove the following:

######  Lemma 2.6.

The homomorphism $\Gamma_{1*}j_{*}$ is induced by
$(j\times{\rm{id}})^{*}(\Gamma_{1})$.

###### Proof.

By the definition of $\Gamma_{1*}j_{*}(z)$ we have

$\Gamma_{1*}j_{*}(z)={\rm{pr}}_{2*}(j_{*}(z)\times\sum_{i=1}^{m}C.\Gamma_{1})$

which by the previous is equal to

${\rm{pr}}_{2*}(f_{2}\times f_{1})_{*}((f_{2}\times
f_{1})^{*}(j_{*}(z)\times\sum_{i=1}^{m}C).(f_{2}\times f_{1})^{*}\Gamma_{1})$

which is nothing but

$f_{1*}({\rm{pr}}_{2*}((f_{2}\times
f_{1})^{*}(j_{*}(z)\times\sum_{i=1}^{m}C).(f_{2}\times
f_{1})^{*}\Gamma_{1}))\;.$

This because of the following commutative diagram:

|
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$\textstyle{{\rm{Sym}}^{g}C\times{\rm{Sym}}^{m}C\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}\times
f_{1}}$$\scriptstyle{{\rm{pr}}_{2}}$$\textstyle{{\rm{Sym}}^{m}C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{1}}$$\textstyle{J(C)\times\sum_{i=1}^{m}C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{\rm{pr}}_{2}}$$\textstyle{\sum_{i=1}^{m}C}$

On the other hand the homomorphism

$(j\times{\rm{id}})^{*}(\Gamma_{1})_{*}(z)$

is equal to

${\rm{pr}}_{2*}(z\times\sum_{i=1}^{m}C.(j\times{\rm{id}})^{*}(\Gamma_{1}))$

which by the previous is equal to

${\rm{pr}}_{2*}(f_{1}\times f_{1})_{*}((f_{1}\times
f_{1})^{*}(z\times\sum_{i=1}^{m}C).(f_{1}\times
f_{1})^{*}(j\times{\rm{id}})^{*}(\Gamma_{1}))\;.$

This is nothing but

$f_{1*}{\rm{pr}}_{2*}((f_{1}\times
f_{1})^{*}(z\times\sum_{i=1}^{m}C).(j^{\prime}\times{\rm{id}})^{*}(f_{2}\times
f_{1})^{*}\Gamma_{1})$

because of the following commutative diagram:

|
---|---
$\textstyle{{\rm{Sym}}^{m}{C}\times{\rm{Sym}}^{m}C\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{1}\times
f_{1}}$$\scriptstyle{j^{\prime}\times{\rm{id}}}$$\textstyle{{\rm{Sym}}^{g}C\times{\rm{Sym}}^{m}C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}\times
f_{1}}$$\textstyle{\sum_{i=1}^{m}C\times\sum_{i=1}^{m}C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j\times{\rm{id}}}$$\textstyle{J(C)\times\sum_{i=1}^{m}C}$

Now we have

$(f_{1}\times
f_{1})^{*}(z\times\sum_{i=1}^{m}C)=f_{1}^{*}(z)\times{\rm{Sym}}^{m}C$

by the commutative diagram

|
---|---
$\textstyle{{\rm{Sym}}^{m}C\times{\rm{Sym}}^{m}C\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\rm{Sym}}^{m}C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\sum_{i=1}^{m}C\times\sum_{i=1}^{m}C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\sum_{i=1}^{m}C}$

So by projection formula we get that

$f_{1*}{\rm{pr}}_{2*}((f_{1}\times
f_{1})^{*}(z\times\sum_{i=1}^{m}C).(j^{\prime}\times{\rm{id}})^{*}(f_{2}\times
f_{1})^{*}\Gamma_{1})$

is equal to

$f_{1*}{\rm{pr}}_{2*}((j^{\prime}\times{\rm{id}})_{*}(f_{1}\times
f_{1})^{*}(z\times\sum_{i=1}^{m}C).(f_{2}\times f_{1})^{*}\Gamma_{1})\;.$

So all we need to prove that is

$(j^{\prime}\times{\rm{id}})_{*}(f_{1}\times
f_{1})^{*}(z\times\sum_{i=1}^{m}C)=(f_{2}\times
f_{1})^{*}(j\times{\rm{id}})_{*}(z\times\sum_{i=1}^{m}C)$

So let $z$ be a zero cycle on $\sum_{i=1}^{m}C$, the above will follow from

$j^{\prime}_{*}f_{1}^{*}(z)=f_{2}^{*}j_{*}(z)\;.$

By Lemma 2.5 we can assume that $z$ is supported on the complement of the
exceptional locus of the map ${\rm{Sym}}^{m}C\to\sum_{i=1}^{m}C$. Now
supposing the above this follows because of Lemma 2.3 and of the following: we
have a Cartesian square

|
---|---
$\textstyle{U^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{V^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{U\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{V}$

where $U^{\prime}$ is Zariski open in ${\rm{Sym}}^{m}C$ and maps
isomorphically onto $U$ in $\sum_{i=1}^{m}C$, and $V^{\prime}$ is Zariski open
in ${\rm{Sym}}^{g}C$ and maps birationally onto $V$ in $J(C)$. ∎

Proof of theorem 2.1:

Now we come back to the proof of the theorem 2.1.

By the projection formula it follows that

${\Gamma_{1}}_{*}j_{*}$

is induced by $(j\times id)^{*}(\Gamma_{1})$ where $j$ is the closed embedding
of $\sum_{i=1}^{m}C$ into $J(C)$. Now we compute the cycle

$(j\times id)^{*}(\Gamma_{1})$

that is nothing but the collection of rational equivalence classes of divisors

$([D_{1}],[D_{2}])$

such that $D_{1}$ is linearly equivalent to $\sum_{i=1}^{m}x_{i}-mp$ and
$D_{2}$ is linearly equivalent to $\sum_{i=1}^{m}y_{i}-mp$, where

$([x_{1},\cdots,x_{m},p,p\cdots,p],[y_{1},\cdots,y_{m}])\in\Gamma\;.$

Therefore without loss of generality we can assume that elements in $(j\times
id)^{*}(\Gamma_{1})$ are classes of effective divisors on $C$ of the form

$([x_{1}+\cdots+x_{m}+(g-m)p-gp],[y_{1}+\cdots+y_{m}+(g-m)p-gp])$

such that

$x_{1}+\cdots+x_{m}+(g-m)p=x_{1}^{\prime}+\cdots+x_{m}^{\prime}+(g-m)p+div(f)$

and

$y_{1}+\cdots+y_{m}+(g-m)p=y_{1}^{\prime}+\cdots+y_{m}^{\prime}+(g-m)p+div(g)$

such that

$([x^{\prime}_{1},\cdots,x^{\prime}_{m},p,\cdots,p],[y^{\prime}_{1},\cdots,y^{\prime}_{m}])\in\Gamma\;.$

Now observe that the $g$-tuple
$(x_{1}^{\prime},\cdots,x_{m}^{\prime},\cdots,p)$ projects down under
${\rm{pr}}$ to $(x^{\prime}_{1},\cdots,x_{m}^{\prime})$ and the $g$-tuples
$(x^{\prime}_{1},x^{\prime}_{2},\cdots,x^{\prime}_{i-1},p,x^{\prime}_{i},\cdots,x^{\prime}_{m},p,\cdots,p)$
projects down to
$(x^{\prime}_{1},\cdots,x^{\prime}_{i-1},p,x^{\prime}_{i},\cdots,x^{\prime}_{m})$.
Since

$([x^{\prime}_{1},\cdots,x^{\prime}_{m},p,\cdots,p],[y^{\prime}_{1},\cdots,y^{\prime}_{m}])\in\Gamma$

we have

$[y^{\prime}_{1},\cdots,y^{\prime}_{m}]=[x_{1}^{\prime},\cdots,x_{m}^{\prime}]$

or

$[y_{1}^{\prime},\cdots,y_{m}^{\prime}]=[p,\cdots,p,x^{\prime}_{i+1},\cdots,x_{m}^{\prime}]$

Then we get that either

$x^{\prime}_{i}=y^{\prime}_{i}$

for all $i$ or

$y^{\prime}_{i}=p$

for some $i$. Since we are only considering the divisor classes and divisor of
the above form represents a divisor class in
$(j\times{\rm{id}})^{*}\Gamma_{1}$, we get that $(j\times id)^{*}\Gamma_{1}$
is equal to

$\Delta_{\sum_{i=1}^{m}C}+Y$

where $\Delta_{\sum_{i=1}^{m}C}$ is the diagonal in
$\sum_{i=1}^{m}C\times\sum_{i=1}^{m}C$ and $Y$ is supported on
$\sum_{i=1}^{m}C\times\sum_{i=1}^{m-1}C$. The fact that the multiplicity of
$\Delta_{\sum_{i=1}^{m}C}$ in $(j\times id)^{*}(\Gamma_{1})$ is $1$ follows
from the fact that ${\rm{Sym}}^{g}C$ maps surjectively and birationally onto
$J(C)$, and the computations following [Co]. Also here the intersection
product of a cycle of the form $z\times\sum_{i=1}^{m}C$ and
$(j\times{\rm{id}})^{*}(\Gamma_{1})$ is well defined by Lemma 2.2, 2.3(because
the sublemma 2.4 holds for $\sum_{i=1}^{m}C$).

Let’s give details about the computation of the cycle $\Gamma_{1*}j_{*}(z)$.
Let $U^{\prime}$ be the open subset of ${\rm{Sym}}^{m}C$ mapping birationally
onto $U$ in $\sum_{i=1}^{m}C$. Consider the inverse image of two cycles
$\Delta_{\sum_{i=1}^{m}C}+Y$ on $\sum_{i=1}^{m}C$ and $z\times\sum_{i=1}^{m}C$
(where $z$ is a zero cycle on $\sum_{i=1}^{m}C$) by $f_{1}\times f_{1}$. Then
by lemma 2.2,2.3,2.6, 2.5, we know that, on
${\rm{Sym}}^{m}C\times{\rm{Sym}}^{m}C$ the intersection of $(f_{1}\times
f_{1})^{*}(\Delta_{\sum_{i=1}^{m}C}+Y)$ and $(f_{1}\times
f_{1})^{*}(z\times\sum_{i=1}^{m}C)$ is well defined. So

$(\Delta_{\sum_{i=1}^{m}C}+Y).(z\times\sum_{i=1}^{m}C)=(f_{1}\times
f_{1})_{*}((f_{1}\times f_{1})^{*}(\Delta_{\sum_{i=1}^{m}C}+Y).(f_{1}\times
f_{1})^{*}(z\times\sum_{i=1}^{m}C))$

We know that the natural map from ${\rm{Sym}}^{m}C$ to $\sum_{i=1}^{m}C$ is
birational for all $m\leq g$, [M][Theorem 5.1]. Let $E$ be the exceptional
locus of this map. Then the inverse image scheme $(f_{1}\times
f_{1})^{-1}(\Delta_{\sum_{i=1}^{m}C})$ consists of points $(x,y)$ in
${\rm{Sym}}^{m}C\times{\rm{Sym}}^{m}C$ such that

$f(x)=f(y)$

Since $f$ is a regular birational map, $f(x)=f(y)$ only when $x=y$ or $(x,y)$
is in the same linear system of divisors. In details, we have the fibers of
the natural map ${\rm{Sym}}^{m}C\to\sum_{i=1}^{m}C$ are the linear systems of
effective divisors linearly equivalent to an effective divisor $D$ of degree
$m$. Write

$x=\sum_{i=1}^{m}P_{i},\quad y=\sum_{i=1}^{m}Q_{i}$

since $f(x)=f(y)$ we have

$\sum_{i=1}^{m}P_{i}=\sum_{i=1}^{m}Q_{i}+div(f)$

for a non-zero rational function $f$ on $C$. Therefore we have that
$\sum_{i=1}^{m}P_{i}$ belongs to the linear system of the effective divisor
$\sum_{i=1}^{m}Q_{i}$ and $\sum_{i=1}^{m}Q_{i}$ is contained in the linear
system of the effective divisor $\sum_{i=1}^{m}P_{i}$. The product of this two
linear systems in contained in the product $E\times E$ because $E$ consists of
all effective divisors of degree $m$ such that the corresponding linear system
is positive dimensional,[M][Theorem 5.1]. Therefore we have

$(f_{1}\times
f_{1})^{*}(\Delta_{\sum_{i=1}^{m}C}+Y)=\Delta_{{\rm{Sym}}^{m}C}+Z+Y^{\prime}$

where $Z$ is supported on the exceptional locus $E\times E$ of the birational
map
${\rm{Sym}}^{m}C\times{\rm{Sym}}^{m}C\to\sum_{i=1}^{m}C\times\sum_{i=1}^{m}C$
(where it is not birational) of the map $f_{1}\times f_{1}$ and $Y^{\prime}$
is the inverse image of $Y$ under $f_{1}\times f_{1}$. Now for $(f_{1}\times
f_{1})^{*}(z\times\sum_{i=1}^{m}C)$, we have

$(f_{1}\times
f_{1})^{*}(z\times\sum_{i=1}^{m}C)=z^{\prime}\times{\rm{Sym}}^{m}C$

where $z^{\prime}$ is a zero cycle.

So we have

$(\Delta_{{\rm{Sym}}^{m}C}+Z+Y^{\prime}).(z^{\prime}\times{\rm{Sym}}^{m}C)=z^{\prime}\times
z^{\prime}+z^{\prime}\times{\rm{Sym}}^{m}C.Y^{\prime}$

This is because of the following. First observe that ${\rm{Sym}}^{m}C$ is non-
singular and hence the Chow moving lemma holds there. Therefore we can move
the support of $z^{\prime}$ away from $E$. Also this is possible by Lemma 2.5
to first move $z$ to general position and then obtain $z^{\prime}$ as the
pull-back of $z$ and the pull-back is well defined by Lemma 2.3. Now the
codimension of $z^{\prime}\times{\rm{Sym}}^{m}C.Y^{\prime}$ is $2m$, after
removing the part coming from $E\times E$ from it. Let’s give details on this.
Following the argument as in Lemma 2.2 we have that

$(j\times{\rm{id}})^{*}(\Gamma_{1})=(j^{\prime}\times{\rm{id}})^{*}(\Gamma)+Z$

where $Z$ is supported on $E\times E$. Now following the computation of [Co],
we have that

$(j^{\prime}\times{\rm{id}})^{*}(\Gamma)=\Delta_{{\rm{Sym}}^{m}C}+Y^{\prime}$

where $Y^{\prime}$ is of codimension $m$, this is because
${\rm{Sym}}^{g}C\times{\rm{Sym}}^{m}C$ is smooth hence
$j^{\prime}\times{\rm{id}}$ is a locally complete intersection morphism.
Therefore the intersection of $Y^{\prime}$ and
$z^{\prime}\times{\rm{Sym}}^{m}C$ is of codimension $2m$ by the Chow moving
lemma for the non-singular variety ${\rm{Sym}}^{m}C\times{\rm{Sym}}^{m}C$.
Therefore we have

$(\Delta_{\sum_{i=1}^{m}C}+Y).(z\times\sum_{i=1}^{m}C)=z\times
z+Y.(z\times\sum_{i=1}^{m}C)$

and it gives rise to a cycle of correct codimension.

So we have

$\Gamma_{1*}j_{*}(z)=z+z^{\prime}$

if $z$ is in $U$ and $z^{\prime}$ is supported on $\sum_{i=1}^{m-1}C$ and
hence the above equality is true for all $z$ in $\sum_{i=1}^{m}C$.

So let

$\rho:U^{\prime\prime}\to\sum_{i=1}^{m}C$

be the open embedding of the complement of $\sum_{i=1}^{m-1}C$ in
$\sum_{i=1}^{m}C$. Then we have

$\rho^{*}{\Gamma_{1}}_{*}j_{*}(z)=\rho^{*}(z+z^{\prime})=\rho^{*}(z)$

where $z^{\prime}$ is supported on $\sum_{i=1}^{m-1}C$. This follows since
$(j\times id)^{*}(\Gamma_{1})=\Delta_{\sum_{i=1}^{m}C}+Y$, where $Y$ is
supported on $\sum_{i=1}^{m}C\times\sum_{i=1}^{m-1}C$ and from the previous
computation. Now consider the following commutative diagram.

|
---|---
$\textstyle{{\it{CH}}_{0}(\sum_{i=1}^{m-1}C)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j^{\prime}_{*}}$$\textstyle{{\it{CH}}_{0}(\sum_{i=1}^{m}C)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho^{*}}$$\scriptstyle{j_{*}}$$\textstyle{{\it{CH}}_{0}(U^{\prime\prime})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\
}$$\textstyle{{\it{CH}}_{0}(\sum_{i=1}^{m-1}C)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j^{\prime\prime}_{*}}$$\textstyle{{\it{CH}}_{0}(J(C))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\it{CH}}_{0}(V^{\prime\prime})}$

Here $U^{\prime\prime},V^{\prime\prime}$ are complements of
$\sum_{i=1}^{m-1}C$ in $\sum_{i=1}^{m}C,J(C)$ respectively. Now suppose that
$j_{*}(z)=0$, then from the previous it follows that

$\rho^{*}{\Gamma_{1}}_{*}j_{*}(z)=\rho^{*}(z)=0$

by the localisation exact sequence it follows that there exists
$z^{\prime\prime}$ in $\sum_{i=1}^{m-1}C$ such that
$j^{\prime}_{*}(z^{\prime\prime})=z$. Now consider the induction hypothesis
for all $n\leq m-1$, the embedding $\sum_{i=1}^{n}C$ into $J(C)$ gives rise to
an injective push-forward at the level of Chow groups. The induction starts as
the embedding of a closed point in $J(C)$ gives rise to an injective push-
forward homomorphism by the following reason: the point must be contained in a
smooth projective curve in $J(C)$ and then the equality

$np=div(f)$

for $f$ a rational function on the curve and $p$ being the closed point would
imply that $n=0$. This is because the divisor of rational functions on a
smooth projective curve are of degree zero.

Then by the commutativity of the above diagram it follows that

$j^{\prime\prime}_{*}(z^{\prime})=0$

would imply $z^{\prime}=0$, since by induction hypothesis the embedding of
$\sum_{i=1}^{m-1}C$ into $J(C)$ induce an injection at the level of Chow group
of zero cycles . So we get that $z^{\prime}=0$ hence $z=0$. So $j_{*}$ is
injective.

Now putting $m=g-1$, we get the desired result. ∎

Now we prove the following:

######  Theorem 2.7.

Let $A$ be a principally polarized abelian variety embedded into the Jacobian
$J(C)$ of a smooth projective curve $C$. Suppose that $A$ is a proper
subvariety of $J(C)$. Let $\Theta_{C}$ denote the theta divisor of $J(C)$.
Pull it back to $A$ and denote the pull-back by $\Theta_{A}$. Suppose that
${\rm{Sym}}^{g-i}C\cap A^{\prime}$ is smooth for all $i\geq 0$ and suppose
also that $C\cap A^{\prime}$ is irreducible, where $A^{\prime}$ is the inverse
image of $A$, under the map ${\rm{Sym}}^{g}C\to J(C)$. Then for the closed
embedding $j:\Theta_{A}\to A$, $j_{*}$ has torsion kernel at the level of Chow
groups of zero cycles.

###### Proof.

As previous we prove the injectivity by induction, that is we prove that for
any $m\leq g-1$, the closed embedding $\sum_{i=1}^{m}C\cap A\to A$ induces an
injection at the level of Chow groups of zero cycles. We have a natural
correspondence as previous, $\Gamma$ given by
$\pi_{g}\times\pi_{m}(Graph({\rm{pr}}))$ on
${\rm{Sym}}^{g}C\times{\rm{Sym}}^{m}C$, where ${\rm{pr}}$ is the projection
from $C^{g}$ to $C^{m}$, $\pi_{i}$ is the natural morphism from $C^{i}$ to
${\rm{Sym}}^{i}C$. Consider the correspondence $\Gamma_{1}$ on
$J(C)\times\sum_{i=1}^{m}C$ given by $(f_{2}\times f_{1})(\Gamma)$, where
$f_{1},f_{2}$ are natural morphisms from ${\rm{Sym}}^{m}C,{\rm{Sym}}^{g}C$ to
$\sum_{i=1}^{m}C,J(C)$ respectively. Consider the restriction of $\Gamma_{1}$
to $A\times(\sum_{i=1}^{m}C\cap A)$. Call it $\Gamma_{1}^{\prime}$. Let $j$
denote the embedding of $\sum_{i=1}^{m}C\cap A$ into $A$. Then by projection
formula it follows that

${\Gamma_{1}^{\prime}}_{*}j_{*}$

is induced by $(j\times id)^{*}(\Gamma_{1}^{\prime})$, (this is similar to the
arguments as in Lemma 2.6 and by the assumption that $A^{\prime}$ is smooth).

Now we compute the cycle

$(j\times id)^{*}(\Gamma_{1}^{\prime})\;.$

This is nothing but the restriction of the cycle
$(j\times{\rm{id}})^{*}(\Gamma_{1})$ by the commutative diagram (here $j$
denote the embedding of $\sum_{i=1}^{m}C$ into $J(C)$).

|
---|---
$\textstyle{\sum_{i=1}^{m}C\cap A\times\sum_{i=1}^{m}C\cap
A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\times\sum_{i=1}^{m}C\cap
A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\sum_{i=1}^{m}C\times\sum_{i=1}^{m}C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{J(C)\times\sum_{i=1}^{m}C}$

Therefore this cycle is the restriction of $\Delta_{\sum_{i=1}^{m}C}+Y$ (by
Theorem 2.1) to $(\sum_{i=1}^{m}C\cap A)\times(\sum_{i=1}^{m}C\cap A)$. Hence
it is

$d\Delta_{\sum_{i=1}^{m}C\cap A}+Y^{\prime}$

where $Y^{\prime}$ is supported on $(\sum_{i=1}^{m}C\cap
A)\times(\sum_{i=1}^{m-1}C\cap A)$.

Here $d$ is a positive natural number. Also here the intersection product for
the cycles of form $z\times(\sum_{i=1}^{m}C\cap A)$ and
$(j\times{\rm{id}})^{*}(\Gamma_{1}^{\prime})$ holds for $\sum_{i=1}^{m}C\cap
A\times\sum_{i=1}^{m}C\cap A$ by the Lemmas’ 2.5,2.3,2.6. Let us give the
details. Let us consider a zero cycle $z$ on
${\it{CH}}_{0}(\sum_{i=1}^{m}C\cap A)$. Then consider the cycles
$z\times\sum_{i=1}^{m}C\cap A$ and $d\Delta_{\sum_{i=1}^{m}C\cap A}+Y$. Since
the inverse image ${\rm{Sym}}^{m}C\cap A^{\prime}$ is smooth, Chow moving
lemma holds on the self product ${\rm{Sym}}^{m}C\cap
A^{\prime}\times{\rm{Sym}}^{m}C\cap A^{\prime}$. Let $f_{1}$ be the map from
$A^{\prime}\to A$. Then we define the product (as in the beginning of Theorem
2.1)

$(z\times\sum_{i=1}^{m}C\cap A).(d\Delta_{\sum_{i=1}^{m}C\cap A}+Y)$

to be

$:=(f_{1}\times f_{1})_{*}((f_{1}\times f_{1})^{*}(z\times\sum_{i=1}^{m}C\cap
A).(f_{1}\times f_{1})^{*}(d\Delta_{\sum_{i=1}^{m}C\cap A}+Y^{\prime}))$

Now

$(f_{1}\times f_{1})^{*}(d\Delta_{\sum_{i=1}^{m}C\cap
A})=d\Delta_{{\rm{Sym}}^{m}C\cap A^{\prime}}+Z^{\prime}$

where $Z^{\prime}$ is supported on the product ${\rm{Sym}}^{m}C\cap
A^{\prime}\cap E\times{\rm{Sym}}^{m}C\cap A^{\prime}\cap E$, where $E$ is the
exceptional locus of the birational map ${\rm{Sym}}^{g}C\to J(C)$. Let
$z^{\prime}$ be a zero cycle on ${\rm{Sym}}^{m}C\cap A^{\prime}$. Then
following the computation as in Theorem 2.1, we have that

$z^{\prime}.E=0$

because we can move the support of the zero cycle $z^{\prime}$ away from
$Z^{\prime}$ in ${\rm{Sym}}^{m}C\cap A^{\prime}$. Therefore we get that

$(z^{\prime}\times{\rm{Sym}}^{m}C\cap
A^{\prime}).(d\Delta_{{\rm{Sym}}^{m}C\cap A^{\prime}}+Z^{\prime}+Y^{\prime})$

is equal to

$z^{\prime}\times z^{\prime}+z^{\prime}\times Y^{\prime\prime}$

where $Y^{\prime\prime}$ is supported on ${\rm{Sym}}^{m}C\cap
A^{\prime}\times{\rm{Sym}}^{m-1}C\cap A^{\prime}$. It is the cycle
$(f_{1}\times f_{1})^{*}Y^{\prime}$. Then arguing as in Theorem 2.1 we have

$(z\times\sum_{i=1}^{m}C\cap A).(d\Delta_{\sum_{i=1}^{m}\cap
A}+Y^{\prime})=z\times z+(z\times\sum_{i=1}^{m}C\cap A).Y^{\prime}$

Moreover the cycle $z\times\sum_{i=1}^{m}C\cap A.Y^{\prime}$ is of correct
codimension.

Therefore we have

$\Gamma_{1*}^{\prime}j_{*}(z)=z+z^{\prime}$

for all $z$ in $\sum_{i=1}^{m}C\cap A$, and $z^{\prime}$ supported on
$\sum_{i=1}^{m-1}C\cap A$.

So let

$\rho:U\to\sum_{i=1}^{m}C\cap A$

be the open embedding of the complement of $\sum_{i=1}^{m-1}C\cap A$ in
$\sum_{i=1}^{m}C\cap A$. Then we have

$\rho^{*}{\Gamma_{1}^{\prime}}_{*}j_{*}(z)=\rho^{*}(dz+z_{1})=d\rho^{*}(z)$

where $z_{1}$ is supported on $\sum_{i=1}^{m-1}C\cap A$. This follows since
$(j\times id)^{*}(\Gamma_{1}^{\prime})=d\Delta_{\sum_{i=1}^{m}C\cap
A}+Y^{\prime}$, where $Y^{\prime}$ is supported on $\sum_{i=1}^{m}C\cap
A\times\sum_{i=1}^{m-1}C\cap A$. Now consider the following commutative
diagram.

|
---|---
$\textstyle{{\it{CH}}_{0}(\sum_{i=1}^{m-1}C\cap
A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j^{\prime}_{*}}$$\textstyle{{\it{CH}}_{0}(\sum_{i=1}^{m}\cap
A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho^{*}}$$\scriptstyle{j_{*}}$$\textstyle{{\it{CH}}_{0}(U)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\
}$$\textstyle{{\it{CH}}_{0}(\sum_{i=1}^{m-1}C\cap
A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j^{\prime\prime}_{*}}$$\textstyle{{\it{CH}}_{0}(A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\it{CH}}_{0}(V)}$

Here $U,V$ are complements of $\sum_{i=1}^{m-1}C\cap A$ in
$\sum_{i=1}^{m}C\cap A,A$ respectively. Now suppose that $j_{*}(z)=0$, then
from the previous it follows that

$\rho^{*}{\Gamma_{1}^{\prime}}_{*}j_{*}(z)=\rho^{*}(dz)=0$

by the localisation exact sequence it follows that there exists $z^{\prime}$
supported on $\sum_{i=1}^{m-1}C\cap A$ such that
$j^{\prime}_{*}(z^{\prime})=dz$. Now assume the induction hypothesis: that is
for all $n\leq m-1$ we have the closed embedding $\sum_{i=1}^{n}C\cap A\to A$
induces a push-forward at the level of Chow groups of zero cycles whose kernel
is torsion. The induction starts with the assumption that $C\cap A$ is
irreducible and hence a closed point (because $A$ is a proper abelian
subvariety of $J(C)$), therefore arguments as in Theorem 2.1 would prove that
the embedding of this point into $A$, induces an injection at the level of
Chow groups of zero cycles.

Now by the commutativity of the previous diagram it follows that

$j^{\prime\prime}_{*}(z^{\prime})=0$

By the induction hypothesis we have that the closed embedding
$\sum_{i=1}^{m-1}C\cap A\to A$, induces a push-forward at the level of Chow
groups of zero cycles whose kernel is torsion. So we get that
$d^{\prime}z^{\prime}=0$ hence $dd^{\prime}z=0$. So $j_{*}$ has torsion
kernel.

Now putting $m=g-1$ we get the desired result.

∎

######  Theorem 2.8.

Let $A$ be a principally polarized abelian variety embedded into the Jacobian
$J(C)$ of a smooth projective curve $C$. Suppose that $A$ is a proper
subvariety of $J(C)$. Let $\Theta_{C}$ denote the theta divisor of $J(C)$.
Pull it back to $A$ and denote the pull-back by $\Theta_{A}$. Suppose that
${\rm{Sym}}^{g-i}C\cap A^{\prime}$ is singular for some $i\geq 0$ and suppose
also that $C\cap A^{\prime}$ is irreducible, where $A^{\prime}$ is the inverse
image of $A$, under the map ${\rm{Sym}}^{g}C\to J(C)$. Then for the closed
embedding $j:\Theta_{A}\to A$, $j_{*}$ has torsion kernel at the level of Chow
groups of zero cycles.

###### Proof.

The proof here goes similarly as in Theorem 2.7 but the only matter of concern
is the injectivity of ${\it{CH}}_{0}(\sum_{l=1}^{m}C\cap
A)\to{\it{CH}}_{0}(A)$ for $m\leq g-1$, where ${\rm{Sym}}^{m}C\cap A^{\prime}$
is singular. So suppose that ${\rm{Sym}}^{m}C\cap A^{\prime}$ is singular. Let
$\widetilde{{\rm{Sym}}^{m}C\cap A^{\prime}}$ is the resolution of
singularities of it. Then considering embedding
$j^{\prime}:{\rm{Sym}}^{m}C\cap A^{\prime}\to A^{\prime}$ and the induced
embedding $j:\sum_{l=1}^{m}C\cap A\to A$ we have (following the arguments of
Lemma 2.6) $\Gamma^{\prime}_{1*}j_{*}$ is induced by
$(j\times{\rm{id}})^{*}(\Gamma^{\prime}_{1})$. Following Theorem 2.7 we have

$(j\times{\rm{id}})^{*}(\Gamma^{\prime}_{1})=d\Delta_{\sum_{l=1}^{m}C\cap
A}+Y$

where $\Delta_{V}$ is the diagonal of $V$ and $Y$ is supported on
$\sum_{l=1}^{m}C\cap A\times\sum_{l=1}^{m-1}C\cap A$. Let’s put
$V=\sum_{l=1}^{m}C\cap A$. Now as before consider the cycles $z\times V$ and
$d\Delta_{V}+Y$. Since the variety $\widetilde{{\rm{Sym}}^{m}C\cap
A^{\prime}}$ is smooth, Chow moving lemma holds on the self product
$\widetilde{{\rm{Sym}}^{m}C\cap
A^{\prime}}\times\widetilde{{\rm{Sym}}^{m}C\cap A^{\prime}}$ and we have the
inverse images (as defined in the proof of theorem 2.1,under the map from
$\widetilde{{\rm{Sym}}^{m}C\cap
A^{\prime}}\times\widetilde{{\rm{Sym}}^{m}C\cap A^{\prime}}$ to
$\sum_{l=1}^{m}C\cap A\times\sum_{l=1}^{m}C\cap A$) of $z\times V$ and
$d\Delta_{V}+Y$ intersect properly on the product. Let $f_{1}$ be the map from
$\widetilde{{\rm{Sym}}^{m}C\cap A^{\prime}}\to\sum_{l=1}^{m}C\cap A$. Then we
define the product

$(z\times V).(d\Delta_{V}+Y)$

to be

$:=(f_{1}\times f_{1})_{*}((f_{1}\times f_{1})^{*}(z\times V).(f_{1}\times
f_{1})^{*}(d\Delta_{V}+Y))\;.$

Note that

$(f_{1}\times f_{1})^{*}(d\Delta_{V}+Y)=\Delta_{\widetilde{{\rm{Sym}}^{m}C\cap
A^{\prime}}}+{Z^{\prime}}+Z^{\prime\prime}+Y^{\prime}$

where $Z^{\prime}$ is supported on the inverse image of $E^{\prime}\times
E^{\prime}$ where, $E^{\prime}=E\cap{\rm{Sym}}^{m}C\cap A^{\prime}$ under the
map $\widetilde{{\rm{Sym}}^{m}C\cap
A^{\prime}}\times\widetilde{{\rm{Sym}}^{m}C\cap
A^{\prime}}\to{\rm{Sym}}^{m}C\cap A^{\prime}\times{\rm{Sym}}^{m}C\cap
A^{\prime}$. The cycle $Z^{\prime\prime}$ is supported on
$E^{\prime\prime}\times E^{\prime\prime}$, $E^{\prime\prime}$ being the
exceptional divisor associated with the resolution of singularities of
${\rm{Sym}}^{m}C\cap A^{\prime}$. Suppose that $(f_{1}\times
f_{1})^{*}(z\times V)=z^{\prime}\times\widetilde{{\rm{Sym}}^{m}C\cap
A^{\prime}}$, where $z^{\prime}$ is a zero cycle. Then the intersection:

$(f_{1}\times f_{1})^{*}(z\times V).(f_{1}\times f_{1})^{*}(d\Delta_{V}+Y)$
$=z^{\prime}\times z^{\prime}+(z^{\prime}\times\widetilde{{\rm{Sym}}^{m}C\cap
A^{\prime}}).({Z^{\prime}}+Z^{\prime\prime})+Y^{\prime}.(z^{\prime}\times{\rm{Sym}}^{m-1}C\cap
A^{\prime})\;.$

Here $Y^{\prime}$ is supported on $({\rm{Sym}}^{m}C\cap
A^{\prime})\times({\rm{Sym}}^{m-1}C\cap A^{\prime})$. Then moving the support
away from $E^{\prime\prime}$ we have
$z.E^{\prime\prime}=0,z.\hat{E^{\prime}}=0$ ($\hat{E^{\prime}}$ is the inverse
image of $E^{\prime}$ under the map $\widetilde{{\rm{Sym}}^{m}C\cap
A^{\prime}}\to{\rm{Sym}}^{m}C\cap A^{\prime}$). Hence the above intersection
comes to

$z^{\prime}\times
z^{\prime}+Y^{\prime}.(z^{\prime}\times\widetilde{{\rm{Sym}}^{m}C\cap
A^{\prime}})$

similar reasons as in Theorem 2.1 we have the second part
$Y^{\prime}.(z^{\prime}\times\widetilde{{\rm{Sym}}^{m}C\cap A^{\prime}})$ is
of correct codimension. Therefore

$(f_{1}\times f_{1})_{*}(z^{\prime}\times
z^{\prime}+Y^{\prime}.(z^{\prime}\times{\rm{Sym}}^{m-1}C\cap A^{\prime}))$

is well defined.

Therefore we have

$\Gamma_{1*}^{\prime}j_{*}(z)=z+z^{\prime}$

for all $z$ in $\sum_{l=1}^{m}C\cap A$ and $z^{\prime}$ supported on
$\sum_{l=1}^{m-1}C\cap A$.

So let

$\rho:U\to\sum_{l=1}^{m}C\cap A$

be the open embedding of the complement of $\sum_{l=1}^{m-1}C\cap A$ in
$\sum_{l=1}^{m}C\cap A$. Then we have

$\rho^{*}{\Gamma_{1}^{\prime}}_{*}j_{*}(z)=\rho^{*}(dz+z_{1})=d\rho^{*}(z)$

where $z_{1}$ is supported on $\sum_{l=1}^{m-1}C\cap A$. This follows since
$(j\times id)^{*}(\Gamma_{1}^{\prime})=d\Delta_{\sum_{i=1}^{m}C\cap A}+Y$,
where $Y$ is supported on $\sum_{l=1}^{m}C\cap A\times\sum_{l=1}^{m-1}C\cap
A$.

Now consider the following commutative diagram.

|
---|---
$\textstyle{{\it{CH}}_{0}(\sum_{l=1}^{m-1}C\cap
A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j^{\prime}_{*}}$$\textstyle{{\it{CH}}_{0}(\sum_{l=1}^{m}C\cap
A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho^{*}}$$\scriptstyle{j_{*}}$$\textstyle{{\it{CH}}_{0}(U)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\
}$$\textstyle{{\it{CH}}_{0}(\sum_{l=1}^{m-1}C\cap
A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j^{\prime\prime}_{*}}$$\textstyle{{\it{CH}}_{0}(A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\it{CH}}_{0}(V)}$

Here $U,V$ are complements of $\sum_{l=1}^{m-1}C\cap A$ in
$\sum_{l=1}^{m}C\cap A,A$ respectively. Now suppose that $j_{*}(z)=0$, then
from the previous it follows that

$\rho^{*}{\Gamma_{1}^{\prime}}_{*}j_{*}(z)=\rho^{*}(dz)=0$

by the localisation exact sequence it follows that there exists $z^{\prime}$
supported on $\sum_{l=1}^{m-1}C\cap A$ such that
$j^{\prime}_{*}(z^{\prime})=dz$. Now assume by induction hypothesis that for
all $n\leq m-1$ we have the closed embedding $\sum_{l=1}^{n}C\cap A\to A$
induces a push-forward at the level of Chow groups of zero cycles whose kernel
is torsion. The induction starts with the assumption that $C\cap A$ is
irreducible and hence a closed point (because $A$ is a proper abelian
subvariety of $J(C)$), therefore arguments as in Theorem 2.1 would prove that
the embedding of this point into $A$, induces an injection at the level of
Chow groups of zero cycles.

Now by the commutativity of the previous diagram it follows that

$j^{\prime\prime}_{*}(z^{\prime})=0$

By the induction hypothesis we have that the closed embedding
$\sum_{l=1}^{n}C\cap A\to A$, for $n\leq m-1$ induces a push-forward at the
level of Chow groups of zero cycles whose kernel is torsion. So we get that
$d^{\prime}z^{\prime}=0$ hence $dd^{\prime}z=0$. So $j_{*}$ has torsion
kernel.

∎

The previous theorem gives rise to the following corollary:

######  Corollary 2.9.

Let $\widetilde{C}\to C$ be an unramified double cover of smooth projective
curves such that the genus of $C$ is atleast $2$. Consider the Prym variety
associated to this double cover, denote by $P(\widetilde{C}/C)$. Consider the
embedding of $P(\widetilde{C}/C)$ into $J(\widetilde{C})$. Let
$\Theta^{\prime}$ be the pullback of the theta divisor on $J(\widetilde{C})$
to $P(\widetilde{C}/C)$. Then the closed embedding $\Theta^{\prime}\to
P(\widetilde{C}/C)$ induces a pushforward homomorphism at the level of Chow
groups which has torsion kernel.

###### Proof.

We need to check that the Prym variety and the restriction of the theta
divisor on $J(\widetilde{C})$ satisfies the assumption of Theorem 2.8.

Let $g$ be the genus of $C$. So by Riemann-Hurwitz formula the genus of
$\widetilde{C}$ is $2g-1$. We have the following commutative square.

|
---|---
$\textstyle{{\rm{Sym}}^{2g-1}\widetilde{C}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\widetilde{q}}$$\textstyle{{\rm{Sym}}^{2g-1}C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{q_{C}}$$\textstyle{J(\widetilde{C})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{J(C)}$

Then first of all the Prym variety is the image under $\widetilde{q}$ of the
double cover $P^{\prime}$ of a projective space ${\mathbb{P}}^{g-1}$. This
follows from the Riemann-Roch and the very definition of the Prym variety. Now
consider the intersection of ${\rm{Sym}}^{2g-i}\widetilde{C}$ with
$P^{\prime}$, where $i\geq 2$. This intersection may be singular for a general
copy of ${\rm{Sym}}^{2g-i}\widetilde{C}$ in ${\rm{Sym}}^{2g-1}\widetilde{C}$
but they fits into a family of algebraic varities in the following way.
Consider the family

${\mathscr{U}}:=\\{([x_{1},\cdots,x_{2g-2},x_{2g-1}],p)\in
P^{\prime}\times\widetilde{C}|p\in[x_{1},\cdots,x_{2g-1}]\\}$

and the projection from ${\mathscr{U}}$ to $\widetilde{C}$. Then
${\mathscr{U}}$ is a family over $\widetilde{C}$ with fibers
${\rm{Sym}}^{2g-2}\widetilde{C_{p}}\cap P^{\prime}$ over $\widetilde{C}$, for
$p\in\widetilde{C}$ and $\widetilde{C_{p}}\cong\widetilde{C}$. Now
$P^{\prime}$ is an unramified double cover of the projective space
${\mathbb{P}}^{g-1}$. Therefore $P^{\prime}$ cannot be non-singular because if
it is non-singular then $P^{\prime}$ is an étale cover of ${\mathbb{P}}^{g-1}$
making it isomorphic to ${\mathbb{P}}^{g-1}$. Then ${\mathbb{P}}^{g-1}$ maps
to an abelian variety so the map should be constant making the Prym variety
trivial, which is not the case.

Therefore ${\mathscr{U}}$ is a singular variety and its tangent space at a
point is a hyperplane in the product of tangent spaces

$T_{[x_{1},\cdots,x_{2g-1}]}P^{\prime}\times T_{p}(\widetilde{C})\;.$

So we resolve the singularity of ${\mathscr{U}}$ and construct a non-singular
variety ${\mathscr{U}}^{\prime}$. Consider the map from
${\mathscr{U}}^{\prime}\to\widetilde{C}$.

Following the theorem 2.27 in [Sha], over a Zariski open, non-empty subset $V$
of $\widetilde{C}$, the map
${\mathscr{U}}^{\prime}_{V}:={\mathscr{U}}^{\prime}\times_{\widetilde{C}}V\to\widetilde{V}$
is a submersion, hence the fibers ${\mathscr{U}}^{\prime}_{p}$ over $p\in V$
are non-singular subvarities of ${\mathscr{U}}^{\prime}_{V}\to V$.

Hence for a general $p$ in $\widetilde{C}$, ${\mathscr{U}}^{\prime}_{p}$ is
smooth which is nothing but a resolution of singularities of
${\rm{Sym}}^{2g-2}\widetilde{C}_{p}\cap P^{\prime}$.

Similarly we can prove that for a general
${\rm{Sym}}^{2g-i}\widetilde{C_{p}}\cap P^{\prime}$ its resolution of
singularity also fits into a family of algebraic varieties.

Also considering the family

${\mathscr{U}}^{\prime}:=\\{([x_{1},x_{2},\cdots,x_{2g-1}],p)\in
P^{\prime}\times\widetilde{C}|x_{2}=x_{3}=\cdots=x_{2g-1}=p\\}$

the fiber of ${\mathscr{U}}^{\prime}\to\widetilde{C}$ is isomorphic to
$\widetilde{C_{p}}\cap P^{\prime}$ for any point $p\in\widetilde{C}$, here
$\widetilde{C_{p}}\cong\widetilde{C}$. The collection of points $x$ in
$\widetilde{C_{p}}\cap P^{\prime}$ satisfies that

$(i(x)-i(p))=(-x+p)$

or $(i(x)+x)=(i(p)+p)$ on $J(\widetilde{C})$ for a fixed point
$p\in\widetilde{C}$. Therefore on $J(C)$ we have

$2x^{\prime}=2p^{\prime}\;.$

Here $x^{\prime},p^{\prime}$ are the images of $x,p$ under the quotient map
from $\widetilde{C}$ to $C$.

Conversely suppose that we have for a point $x^{\prime}$ in ${C}$ such that

$2(x^{\prime}-p^{\prime})=0$

Consider the pre-images $x,i(x)$ of $x^{\prime}$ and $p,i(p)$ of $p$ in
$\widetilde{C}$. Then

$(x-p)+(i(x)-i(p))$

is mapped to

$2(x^{\prime}-p^{\prime})=0\;.$

Therefore we have $x,i(x)\in\widetilde{C_{p}}\cap P^{\prime}$. Therefore
$\widetilde{C_{p}}\cap P^{\prime}/i$ is in bijection with
${\mathscr{L}}((4g-2)p^{\prime})\cap C$, where
${\mathscr{L}}((4g-2)p^{\prime})$ is the linear system associated to the cycle
$(4g-2)p^{\prime}$. By Riemann-Roch this linear system is of dimension $3g-1$.
Therefore we have ${\rm{Sym}}^{4g-2}C$ is a projective bundle over $J(C)$. Now
consider the family

${\mathscr{U}}=\\{((4g-2)p,q)|q\in{\mathscr{L}}((4g-2)p)\\}\subset{\rm{Sym}}^{4g-2}C\times
C$

over ${\rm{Sym}}^{4g-2}C$. Then for a $p$ in $C$ realised as an element of
${\rm{Sym}}^{4g-2}C$, the inverse image under the above projection is
$C\cap{\mathscr{L}}((4g-2)p)$. Now the above family can be realised as the
pull-back to $C$ of the projective bundle on ${\rm{Sym}}^{4g-2}C$ given by

$\\{(\sum_{i=1}^{4g-2}P_{i},\sum_{j=1}^{4g-2}Q_{j})|\sum_{j=1}^{4g-2}Q_{j}\in{\mathscr{L}}(\sum_{i=1}^{4g-2}P_{i})\\}\subset{\rm{Sym}}^{4g-2}C\times{\rm{Sym}}^{4g-2}C\;.$

Therefore ${\mathscr{U}}$ is irreducible, hence for a general $p$,
${\mathscr{L}}((4g-2)p)\cap C$ is irreducible, [Sha][theorem 2.26].

Hence it is a point, otherwise we have
$C\subset{\mathscr{L}}((4g-2)p^{\prime})$, which in turn gives us that all
points of $\widetilde{C}$ are $i$-antiinvariant. Therefore $C$ is rational
contradicting the assumption that genus of $C$ is atleast $2$. Since
$\widetilde{C_{p}}\cap P^{\prime}/i$ is in bijection with
${\mathscr{L}}((4g-2)p^{\prime})\cap C$, we have $\widetilde{C_{p}}\cap
P^{\prime}$ consists of two points say $x,i(x)$. We have

$i(x)-i(p)=-x+p\;.$

We have to prove that the pushforward from
${\it{CH}}_{0}(\\{x-p,i(x-p)\\})\cong{\mathbb{Z}}\oplus{\mathbb{Z}}$ to
${\it{CH}}_{0}(P(\widetilde{C}/C))$ is injective. Consider $z$ in the Chow
group ${\it{CH}}_{0}({x-p,i(x-p)})$, such that $j_{*}(z)$ is rationally
equivalent to zero on $P(\widetilde{C}/C)$. Here $j$ is the closed embedding
of the two points $\\{x-p,i(x-p)\\}$ into $P(\widetilde{C}/C)$. Then by the
definition of rational equivalence there exist finitely many smooth projective
curves $X_{i}$ in $P(\widetilde{C}/C)$ and rational functions $f_{i}$ on
$X_{i}$ such that

$j_{*}(z)=\sum_{i}div(f_{i})$

Since $div(f_{i})$ has degree zero for each $i$, we have $j_{*}(z)$ is degree
zero, hence so is $z$. So $z$ will look like

$n(x-p)-ni(x-p)\;.$

Now from the previous we have that the zero cycle

$n(x-p)+ni(x-p)$

maps to zero under the albanese map. So combining the previous two we have

$2n(x-p)$

maps to zero under the albanese map. So the albanese image of $x-p$ is
$2n$-torsion. Therefore by the Roitman’s torsion theorem the zero cycle $x-p$
is a $2n$-torsion. Therefore the zero cycle of $i(x-p)$ is also $2n$-torsion.
But then by the definition of rational equivalence it follows that the degree
of the cycles

$2n(x-p)$

and

$2ni(x-p)$

are zero. Which means that $n=0$. Therefore $z=0$ in
${\it{CH}}_{0}(\\{x-p,i(x-p)\\})$.

Hence we have the assumption of the Theorem 2.8 is satisfied, whence the
conclusion follows. ∎

Now we prove that if we blow up $J(C)$ at finitely many points and denote the
blow up by $\widetilde{J(C)}$ and let $\widetilde{\Theta_{C}}$ denote the
strict transform of $\Theta_{C}$, then the closed embedding of
$\widetilde{\Theta_{C}}$ into $\widetilde{J(C)}$ induces injective push-
forward homomorphism at the level of Chow groups.

######  Theorem 2.10.

Let $\widetilde{J(C)}$ be the blow up of $J(C)$ at some non-singular
subvariety $Z$ whose inverse image is $E$. Let $Z^{\prime}$ be the inverse
image of $Z$ under the map ${\rm{Sym}}^{g}C\to J(C)$. Consider the blow-up of
${\rm{Sym}}^{g}C$ along $Z^{\prime}$, denote it by
$\widetilde{{\rm{Sym}}^{g}C}$. Suppose that the blow-up of ${\rm{Sym}}^{g-i}C$
along $Z^{\prime}\cap{\rm{Sym}}^{g-i}C$ (that is the strict transform of
${\rm{Sym}}^{g-i}C$ under the above blow-up) is smooth. Let
$\widetilde{\Theta_{C}}$ denote the strict transform of $\Theta_{C}$ in
$\widetilde{J(C)}$. Then the closed embedding of $\widetilde{\Theta_{C}}$ into
$\widetilde{J(C)}$ induces injective push-forward homomorphism at the level of
Chow groups of zero cycles with ${\mathbb{Q}}$ coefficients.

###### Proof.

We prove it by induction. That is we prove that the closed embedding of the
strict transform $\widetilde{\sum_{i=1}^{m}C}$ of $\sum_{i=1}^{m}C$ into
$\widetilde{J(C)}$ induces injective push-forward at the level of Chow group
of zero cycles for all $m\leq g-1$. Let us consider $\pi$ to be the morphism
from $\widetilde{J(C)}$ to $J(C)$. Consider the correspondence
$(\pi\times\pi^{\prime})^{*}(\Gamma_{1})$, where $\pi^{\prime}$ is the
restriction of $\pi$ to $\widetilde{\sum_{i=1}^{m}C}$. Call this
correspondence $\Gamma^{\prime}$. Then $\Gamma^{\prime}_{*}\widetilde{j}_{*}$
is induced by $(\widetilde{j}\times id)^{*}\Gamma^{\prime}$ (following Lemma
2.6), where $\widetilde{j}$ is the closed embedding of
$\widetilde{\sum_{i=1}^{m}C}$ into $\widetilde{J(C)}$. Consider the
commutative square.

|
---|---
$\textstyle{\widetilde{\sum_{i=1}^{m}C}\times\widetilde{\sum_{i=1}^{m}C}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi^{\prime}\times\pi^{\prime}}$$\scriptstyle{\widetilde{j}\times
id}$$\textstyle{\widetilde{J(C)}\times\widetilde{J(C)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi\times\pi}$$\textstyle{\sum_{i=1}^{m}C\times\sum_{i=1}^{m}C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j\times
id}$$\textstyle{J(C)\times J(C)}$

This gives us that

$(\widetilde{j}\times{\rm{id}})^{*}\Gamma^{\prime}=(\pi^{\prime}\times\pi^{\prime})^{*}(j\times
id)^{*}\Gamma_{1}=(\pi^{\prime}\times\pi^{\prime})^{*}(\Delta_{\sum_{i=1}^{m}C}+Y)$

where $Y$ is supported on $\sum_{i=1}^{m}C\times\sum_{i=1}^{m-1}C$. Now

$(\pi^{\prime}\times\pi^{\prime})^{*}(\Delta_{\sum_{i=1}^{m}C})=\Delta_{\widetilde{\sum_{i=1}^{m}C}}+V$

where $E$ is the exceptional locus of $\pi$ and $V$ is supported on
$(E\cap\widetilde{\sum_{i=1}^{m}C})\times(E\cap\widetilde{\sum_{i=1}^{m}C})$.

Let $\widetilde{{\rm{Sym}}^{m}C}$ denote the fiber product of
${\rm{Sym}}^{m}C$ and $\widetilde{\sum_{i=1}^{m}C}$ over $J(C)$. Since
${\rm{Sym}}^{m}C$ is birational to $\sum_{i=1}^{m}C$, we have that
$\widetilde{{\rm{Sym}}^{m}C}$ is birational to $\widetilde{\sum_{i=1}^{m}C}$.
Now $\widetilde{{\rm{Sym}}^{m}C}$ is a blow-up of ${\rm{Sym}}^{m}C$ along
$Z^{\prime}\cap{\rm{Sym}}^{m}C$, which we assume to be smooth. Therefore the
Chow moving lemma holds on
$\widetilde{{\rm{Sym}}^{m}C}\times\widetilde{{\rm{Sym}}^{m}C}$, therefore
arguing as in Theorem 2.1, we can define the intersection of two cycles on
$\widetilde{\sum_{i=1}^{m}C}\times\widetilde{\sum_{i=1}^{m}C}$. So let us
consider the cycles $\Delta_{\widetilde{\sum_{i=1}^{m}C}}+V+Y$ and
$z\times\widetilde{\sum_{i=1}^{m}C}$. We consider the pullback of them by
$f\times f$, where $f$ is the birational map from
$\widetilde{{\rm{Sym}}^{m}C}\to\widetilde{\sum_{i=1}^{m}C}$. Then first note
that

$(f\times
f)^{*}(\Delta_{\widetilde{\sum_{i=1}^{m}C}}+V+Y)=\Delta_{\widetilde{{\rm{Sym}}^{m}C}}+(f\times
f)^{*}(V)+V^{\prime}+Y^{\prime}$

where $V^{\prime}$ is supported on the exceptional divisor $E^{\prime}$ of the
map $\widetilde{{\rm{Sym}}^{m}C}\to\widetilde{\sum_{i=1}^{m}C}$ and
$Y^{\prime}$ is supported on
$\widetilde{{\rm{Sym}}^{m}C}\times\widetilde{{\rm{Sym}}^{m-1}C}$. Then as
before following computation as in Theorem 2.8 for a zero cycle $z$ on
$\widetilde{\sum_{i=1}^{m}C}$ we have

$(f\times f)_{*}((f\times
f)^{*}(\Delta_{\widetilde{\sum_{i=1}^{m}C}}+V+Y).(f\times
f)^{*}(z\times\widetilde{\sum_{i=1}^{m}C}))=z\times
z+Y.(z\times\widetilde{\sum_{i=1}^{m}C})\;.$

Consider the following commutative diagram.

|
---|---
$\textstyle{{\it{CH}}_{0}(A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\widetilde{j^{\prime}}_{*}}$$\textstyle{{\it{CH}}_{0}(\widetilde{\sum_{i=1}^{m}C})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho_{0}^{*}}$$\scriptstyle{\widetilde{j}_{*}}$$\textstyle{{\it{CH}}_{0}(U)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\
}$$\textstyle{{\it{CH}}_{0}(A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\widetilde{j^{\prime\prime}}_{*}}$$\textstyle{{\it{CH}}_{0}(\widetilde{J}(C))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\it{CH}}_{0}(V)}$

Here $A=\widetilde{\sum_{i=1}^{m-1}C}$. Now suppose that
$\widetilde{j}_{*}(z)=0$. By the previous computation we get that
$\rho^{*}(z)=0$, so by the localisation exact sequence we get that there
exists $z^{\prime}$ in ${\it{CH}}_{*}(A)$ such that
$\widetilde{j^{\prime}}_{*}(z^{\prime})=z$.

Assume that for all $n\leq m-1$, we have the map
${\it{CH}}_{0}(\widetilde{\sum_{i=1}^{n}C})\to{\it{CH}}_{0}(J(C))$ is
injective. The induction starts because $\widetilde{p}$ for a closed point on
$J(C)$, is either the point itself or it is empty, so argument similar to
Theorem 2.1 applies here.

By the induction hypothesis
${\it{CH}}_{*}(A)\to{\it{CH}}_{*}(\widetilde{J(C)})$ is injective. So we get
that $z^{\prime}=0$, hence $z=0$ giving $\widetilde{j_{*}}$ injective.

Putting $m=g-1$ we get the result.

∎

Now let $A$ be an abelian surface which is embedded in some $J(C)$. Let $i$
denote the involution of $A$. Then $i$ has $16$ fixed points. We blow up $A$
along these fixed points. Then we get $\widetilde{A}$ on which we have an
induced involution, call it $i$. Let $\widetilde{\Theta_{A}}$ denote the
strict transform of $\Theta_{A}$ in $\widetilde{A}$. Suppose that the abelian
surface $A$ intersects the curve $C$ at one point. Then the techniques as in
Theorem 2.10 tells us the following.

######  Theorem 2.11.

Let $A$ be as above. Then the closed embedding of $\widetilde{\Theta_{A}}$
into $\widetilde{A}$ induces injective push-forward homomorphism at the level
of Chow groups of zero cycles.

Now $\widetilde{A}/i$ is the Kummer’s K3 surface associated to $A$. Suppose
that we choose $\Theta_{C}$ such that it is $i$ invariant. Then
$\widetilde{\Theta_{A}}$ will be $i$ invariant. The above theorem gives us:

######  Theorem 2.12.

Let $A$ be as above then the closed embedding of $\widetilde{\Theta_{A}}/i$
into $\widetilde{A}/i$ induces injective push-forward homomorphism at the
level of Chow groups of zero cycles with ${\mathbb{Q}}$-coefficients.

Note that all these techniques can be repeated if we consider the group of
algebraic cycles modulo algebraic equivalence. Then therefore the closed
embedding $\widetilde{\Theta_{A}}/i$ into $\widetilde{A}/i$ induces injection
at the level of zero cycles modulo algebraic equivalence.

##  3\. Application of the above result

Let us consider a smooth, cubic threefold $X$ embedded in ${\mathbb{P}}^{4}$.
Let us consider a line $l$ in $X$ and project from that line $l$ onto
${\mathbb{P}}^{2}$. Let us consider the discriminant curve $C$ embedded in
${\mathbb{P}}^{2}$. Let us consider the étale double covering of this curve
$C$ say $\widetilde{C}$ inside $F(X)$, the Fano variety of lines of $X$. Let
us consider the Prym variety of this double covering and the theta divisor of
the principally polarized Prym variety. Call it $P(\widetilde{C}/C)$ and the
theta divisor as $\Theta$. Let us consider the universal line correspondence
$L$ on $X\times F(X)$ and consider the correspondence induced on $X\times
P(\widetilde{C}/C)$ by this, say L. Then we consider the $L_{*}$ from
$A_{0}(P(\widetilde{C}/C))$ to $A_{1}(X)$. Consider the pull back of $\Theta$
to $X\times P(\widetilde{C}/C)$ and intersect it with the correspondence $L$
and then push down the intersection to $X$. Call the support of the cycle
obtained this way as $L(\Theta)$. Consider the embedding

$j^{\prime}:L(\Theta)\to X$

and consider the homomorphism induced at the level of algebraically trivial
1-cycles

$j^{\prime}_{*}:A_{1}(L(\Theta))\to A_{1}(X)\;.$

From the equality as in 2.8

$(j\times{\rm{id}})^{*}\Gamma_{1}=d\Delta_{\Theta}+Y$

where $Y$ is supported on $\Theta\times\sum_{i=1}^{g-2}\widetilde{C}\cap
P(\widetilde{C}/C)$, $g$ the genus of $\widetilde{C}$. Then same calculation
as in 2.8, proves that the homomorphism $j^{\prime}_{*}$ has torsion kernel.

Now with the help of this we prove that:

######  Theorem 3.1.

There does not exist a universal codimension 2 cycle $Z$ on
$P(\widetilde{C}/C)\times X$ such that $a\mapsto AJ(Z_{*}(a))=Id$

###### Proof.

Suppose that there exists such a cycle $Z$ such that $AJ(Z_{*}(a))=Id$. Then
we have $Z_{*}$ factoring through $A_{0}(P(\widetilde{C}/C))\to A_{1}(X)$
(this is by the very definition of $Z_{*}$) and the universal-ness tells us
that the homomorphism from $P(\widetilde{C}/C)$ to $A_{1}(X)$ is injective.
Consider the square

|
---|---
$\textstyle{\Theta\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P(\widetilde{C}/C)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A_{1}(L(\Theta))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A_{1}(X)}$

Now considering a very general cubic threefold $X$, we have the abelian
variety $A_{1}(X)\cong P(\widetilde{C}/C)$ is simple. Consider the
desingularization of $L(\Theta)$, call it $\widetilde{L(\Theta)}$. Therefore
the image of
$A_{1}(\widetilde{L(\Theta)})\cong{\rm{Pic}}^{0}(\widetilde{L(\Theta)})$ is
either $\\{0\\}$ or all of $A_{1}(X)$. It can’t be zero. Then it will follow
by the existence of the universal codimension two cycle that, $\Theta$ is
embedded into the trivial abelian variety , which implies $\Theta$ is finite
and hence $P(\widetilde{C}/C)$ is one dimensional, which is not true.

Now suppose that the image of $A_{1}(\widetilde{L(\Theta)})$ in $A_{1}(X)$ is
all of $A_{1}(X)$. The singularity of the theta divisor of the Prym variety
(which is identified with $A_{1}(X)$) is of zero dimension and consists of a
unique singular point [H][Corollary 4.6]. The map from
$A_{1}(\widetilde{L(\Theta)})$ to $P(\widetilde{C}/C)$ is a map of principally
polarized abelian varieties. Observe that the theta divisor $\Theta$ is
embedded in $A_{1}(\widetilde{L(\Theta)})$ by the existence of the universal
codimension two cycle. Consider the pull-back of $\Theta$ to
$A_{1}(\widetilde{L(\Theta)})$, say $f^{*}(\Theta)$, where $f$ is the isogeny
from $A_{1}(\widetilde{L(\Theta)})$ to $A_{1}(X)\cong P(\widetilde{C}/C)$.
Since by the above diagram $\Theta$ is mapped to $\Theta$ by the composition
of the inclusion of $\Theta$ and $AJ\circ Z_{*}$, we have

$\Theta\subset f^{*}(\Theta)\;.$

We Prove that $n\Theta=f^{*}(\Theta)$ for some integer $n>0$. Suppose that

$f^{*}(\Theta)=\Theta+\Theta_{1}+\cdots+\Theta_{n}$

Suppose $g$ is the map from $\Theta$ to $A_{1}(\widetilde{L(\Theta)})$. We
have

$f_{*}f^{*}(\Theta)=m\Theta$

where $m$ is the degree of the isogeny $f$. This gives us

$f_{*}(\Theta_{1}+\cdots+\Theta_{n})=(m-1)\Theta$

as divisors on $P(\widetilde{C}/C)$. Also since $f_{*}(\Theta)=\Theta$, we
have

$f_{*}(\Theta_{1}+\cdots+\Theta_{n}-(m-1)\Theta)=0$

and since the kernel of $f_{*}$ is torsion we have

$k(\Theta_{1}+\cdots+\Theta_{n})=k(m-1)\Theta\;,$

modulo rational equivalence.

Therefore $f^{*}(k\Theta)=km\Theta$ for some positive integer $m$. Therefore
$km\Theta$ is singular at a unique point on the Prym variety and hence
$f^{*}(k\Theta)=km\Theta$ is smooth except one singular point on the Picard
variety ${\rm{Pic}}^{0}(\widetilde{L(\Theta)})$. Note that the above equality
is only true if $g$ is an injective map, because otherwise the image of
$\Theta$ under $g$ might be having different singularities. This is a crucial
point of difference with the case when there exists a codimension two cycle
$Z$, such that $Z_{*}\circ AJ=N{\rm{id}}$. This is always the case for a cubic
threefold as in this case we have a decomposition of diagonal of the cubic
(whether integral decomposition exists or not is the question). Therefore we
have

$km\Theta=f^{*}(k\Theta)$

where $\Theta$ is realized as embedded in
$A_{1}(\widetilde{L(\Theta)})\cong{\rm{Pic}}^{0}(\widetilde{L(\Theta)})$.
Since $km\Theta$ is smooth except one singular point we have $f^{*}(k\Theta)$
is smooth except one singular point.

But now observe that $L(\Theta)$ can either be a curve or a surface. It can’t
be all of the cubic threefold because that $\Theta$ is a divisor in
$P(\widetilde{C}/C)$. Because then any 1-cycle on $X$ is supported on
$L(\Theta)$ as a cycle. So any point on the Prym variety is coming from a
point on $\Theta$. Now the composition $\Theta\to A_{1}(L(\Theta))\to
P(\widetilde{C}/C)$ induced by the correspondence $L$ is a regular map of
varieties, which is surjective, by the fact that $L(\Theta)=X$. So a variety
of dimension $g-2$ maps surjectively to a variety of dimension $g-1$. This
cannot happen. It can’t be a curve, because the singularity locus of the theta
divisor of the Picard variety of a curve is of dimension greater than or equal
to $g-4$. Observe that the Jacobian of the curve is in isogeny with
$P(\widetilde{C}/C)$ by the simplicity of the Prym variety (as the $\Theta$
divisor is embedded in the Jacobian, the Jacobian can’t be isogenous to the
trivial abelian variety). So it is a curve of genus equal to five, in such a
case, the singularity locus of the theta divisor is positive dimensional. So
the only possibility that $L(\Theta)$ is a surface. Then by Bertini’s theorem
consider a smooth hyperplane section of $\widetilde{L(\Theta)}$ which is a
smooth projective curve of genus $g$. The Jacobian of the smooth projective
curve is such that the kernel of the homomorphism from the Jacobian to the
Albanese of the surface is simple [for reference please see [Voi](Proposition
1.4)]. Now observe that the Jacobian of the curve is isogenous to the two-fold
product of the kernel and the Albanese variety of the desingularization
$\widetilde{L(\Theta)}$. If both the factors in the product is non-trivial,
then the singularity locus of the theta divisor of the Jacobian must have an
irreducible component of codimension two in the Jacobian. So the singularity
locus of the theta divisor contains an irreducible component of dimension
$g-2$. So the dimension of the singularity of the theta divisor of the
Jacobian is $g-2$. Now the contribution from the singularity locus of the
theta divisor of the Albanese is of codimension $5$ (because the singularity
of the theta divisor of the Albanese is of zero dimensional). Therefore the
codimension two component of the singularity of the theta divisor must come
from singularity locus of the theta divisor of the kernel. Note that the
kernel is a simple abelian variety. Therefore there cannot be a codimension
two component of the singularity locus of the theta divisor of the kernel,
this is by [EL]. Hence the dimension of the singularity locus of the theta
divisor of the curve is less than $g-3$, so it is either of dimension $g-3$ or
$g-4$. In that case again using [EL] the kernel must be trivial. So the
Jacobian is isogenous to the Albanese of the desingularisation of $L(\Theta)$,
which again is isogenous to the Prym variety. So the curve is of genus five
and hence the singularity locus of the theta divisor of the Jacobian is
positive dimensional but that of the Albanese is of zero dimensional. This is
a contradiction. Therefore there does not exist an universal codimensional two
cycle on the product $P(\widetilde{C}/C)\times X$. ∎

######  Corollary 3.2.

For a very general cubic threefold, the integral Chow theoretic decomposition
does not hold. Consequently a very general cubic threefold is not stably
rational.

###### Proof.

If the integral Chow theoretic decomposition holds then we have the existence
of a universal codimension two cycle on $P(\widetilde{C}/C)\times X$ by
[Voi1][Theorem 1.6], which is false by the previous theorem. Since stable
rationality implies integral Chow theoretic decomposition of the diagonal, we
can conclude that a very general cubic threefold is not stably rational. ∎

######  Corollary 3.3.

For a very general cubic threefold the cohomology class of $\Theta^{4}/4!$ is
not algebraic on the intermediate Jacobian of $X$.

###### Proof.

According to Theorem 1.7 of [Voi1], the universally triviality of Chow group
of a smooth cubic threefold is equivalent to the algebraic-ness of the minimal
class of $\Theta^{4}/4!$ is algebraic. As the universal triviality does not
hold a very general smooth cubic threefold, the cohomology class of
$\Theta^{4}/4!$ is not algebraic on the intermediate Jacobian of $X$. ∎

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