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A subvariety $V$ of an abelian variety $X$ is geometrically nondegenerate if it meets any subvariety of $X$ of dimension bigger than or equal $codim(V)$.

My question is about the Prym varieties as subvarieties of the Jacobians of curves. Are they geometrically nondegenerate?

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Pryms are geometrically degenerate. Let $f:Y\to X$ be an étale double cover of a genus $g>1$ curve $X$. Up to taking isogenous quotients, which will not affect "geometric degeneracy", the Prym of $f$ is the kernel of $\text{Alb}_f : \text{Alb}_Y\to \text{Alb}_X$, where the domain is an Abelian variety of dimension $2g-1$, and where the target is an Abelian variety of dimension $g$. The Prym has dimension $g-1$, so codimension $g$ in $\text{Alb}_Y$. So now let $C\subset \text{Alb}_X$ be any curve that does not contain the origin. Then the inverse image $\text{Alb}_f^{-1}(C)$ is a subvariety of $\text{Alb}_Y$ of dimension $g$ that is disjoint from the Prym.

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  • $\begingroup$ How about cover of degree $r > 2$? $\endgroup$
    – Z.A.Z.Z
    Mar 26, 2016 at 17:40
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    $\begingroup$ "How about cover of degree $r>2$?" It will be the same. What the argument is actually showing is that every proper sub-Abelian variety (of positive dimension) of an Abelian variety is algebraically degenerate. For a hypersurface in the quotient Abelian variety that does not contain the origin, the inverse image is a hypersurface that is disjoint from the sub-Abelian variety. $\endgroup$
    – Jason Starr
    Mar 26, 2016 at 17:43

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