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Samir Canning
Samir Canning
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Let $A$ be a $g$-dimensional, complex abelian variety, let $H$ be an ample divisor, let $D\in Pic^0(A)$, and let $0\leq k\leq g$.

Question 1: Does $D^k\neq 0\in CH^k(A,\mathbb{Q})$ imply that $H^{g-k}\cdot D^k\neq 0\in CH^g(A,\mathbb{Q})$?

(As Jason points out, it is necessary to work with $\mathbb{Q}$-coefficients.)

Question 2: Assuming the answer to Question 1 is "yes", is there a simple proof of this fact?

Primarily, we are interested in abelian varieties over $\mathbb{C}$. (As ulrich points out, KunnermanKunnemann answered Question 1 affirmatively for abelian varieties over finite fields.)

Let $A$ be a $g$-dimensional, complex abelian variety, let $H$ be an ample divisor, let $D\in Pic^0(A)$, and let $0\leq k\leq g$.

Question 1: Does $D^k\neq 0\in CH^k(A,\mathbb{Q})$ imply that $H^{g-k}\cdot D^k\neq 0\in CH^g(A,\mathbb{Q})$?

(As Jason points out, it is necessary to work with $\mathbb{Q}$-coefficients.)

Question 2: Assuming the answer to Question 1 is "yes", is there a simple proof of this fact?

Primarily, we are interested in abelian varieties over $\mathbb{C}$. (As ulrich points out, Kunnerman answered Question 1 affirmatively for abelian varieties over finite fields.)

Let $A$ be a $g$-dimensional, complex abelian variety, let $H$ be an ample divisor, let $D\in Pic^0(A)$, and let $0\leq k\leq g$.

Question 1: Does $D^k\neq 0\in CH^k(A,\mathbb{Q})$ imply that $H^{g-k}\cdot D^k\neq 0\in CH^g(A,\mathbb{Q})$?

(As Jason points out, it is necessary to work with $\mathbb{Q}$-coefficients.)

Question 2: Assuming the answer to Question 1 is "yes", is there a simple proof of this fact?

Primarily, we are interested in abelian varieties over $\mathbb{C}$. (As ulrich points out, Kunnemann answered Question 1 affirmatively for abelian varieties over finite fields.)

Made the question more precise.
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edit approved May 30, 2018 at 20:34
Yosemite Stan
edit approved May 30, 2018 at 20:34
Yosemite Stan
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Let $A$ be a $g$-dimensional, complex abelian variety, let $H$ be an ample divisor, let $D\in Pic^0(A)$, and let $0\leq k\leq g$. Then if

Question 1: Does $D^k\neq 0$$D^k\neq 0\in CH^k(A,\mathbb{Q})$ imply that $H^{g-k}\cdot D^k\neq 0\in CH^g(A,\mathbb{Q})$?

(As Jason points out, is it true thatis necessary to work with $H^{g-k}\cdot D^k\neq 0$$\mathbb{Q}$-coefficients.)

Question 2: Assuming the answer to Question 1 is "yes", is there a simple proof of this fact? We

Primarily, we are workinginterested in the Chow group ofabelian varieties over $A$$\mathbb{C}$. (cycles modulo rational equivalence) with rational coefficientsAs ulrich points out, but I'm interested in the case with integer coefficients as wellKunnerman answered Question 1 affirmatively for abelian varieties over finite fields.)

Let $A$ be a $g$-dimensional abelian variety, $H$ an ample divisor, $D\in Pic^0(A)$, and $0\leq k\leq g$. Then if $D^k\neq 0$, is it true that $H^{g-k}\cdot D^k\neq 0$? We are working in the Chow group of $A$ (cycles modulo rational equivalence) with rational coefficients, but I'm interested in the case with integer coefficients as well.

Let $A$ be a $g$-dimensional, complex abelian variety, let $H$ be an ample divisor, let $D\in Pic^0(A)$, and let $0\leq k\leq g$.

Question 1: Does $D^k\neq 0\in CH^k(A,\mathbb{Q})$ imply that $H^{g-k}\cdot D^k\neq 0\in CH^g(A,\mathbb{Q})$?

(As Jason points out, it is necessary to work with $\mathbb{Q}$-coefficients.)

Question 2: Assuming the answer to Question 1 is "yes", is there a simple proof of this fact?

Primarily, we are interested in abelian varieties over $\mathbb{C}$. (As ulrich points out, Kunnerman answered Question 1 affirmatively for abelian varieties over finite fields.)

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Samir Canning
Samir Canning
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Intersections with a Power of an Ample Divisor on an Abelian Variety

Let $A$ be a $g$-dimensional abelian variety, $H$ an ample divisor, $D\in Pic^0(A)$, and $0\leq k\leq g$. Then if $D^k\neq 0$, is it true that $H^{g-k}\cdot D^k\neq 0$? We are working in the Chow group of $A$ (cycles modulo rational equivalence) with rational coefficients, but I'm interested in the case with integer coefficients as well.