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Hi.

What are the simplest examples to have in mind of non-isomorphic smooth (complex) algebraic curves with isomorphic jacobian variety?

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2 Answers 2

up vote 5 down vote accepted

It is not easy to give equations of non-isomorphic Riemann surfaces whose Jacobians are isomorphic as (unpolarized) Abelian varieties.

Some explicit examples can be found in the paper by Everett W. Howe Infinite Families of Pairs of Curves Over $\mathbb{Q}$ with Isomorphic Jacobians, J. London Math. Soc. (2005) 72 (2), 327-350.

A preprint version of the paper is also available in the arXiv website, see http://arxiv.org/abs/math.AG/0304471. In Section 8 you can find the examples you are looking for.

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Thanks Francesco! But this does not exactly answers to the question: what are the SIMPLEST examples... But one can also argue that the question is not rigourously stated, I agree. –  BadaBing Apr 24 '13 at 12:44
    
You are welcome. Right, it is not clear what "simplest" should mean in this case. At any rate, in Howe's paper there are examples given by pairs of genus $2$ curves, which is of course the lowest possible genus. In this sense, such examples might be considered "simple". –  Francesco Polizzi Apr 24 '13 at 12:50

Slightly tangential -- the following paper of Bjorn Poonen offers some interesting computational perspectives on curves of genus at least $2$ which is as Francesco mentions, the lowest possible genus where such examples exist. This seems like it might be relevant.

http://www-math.mit.edu/~poonen/papers/ants2.pdf

I think the following is an interesting question to ask over a fixed number field $K$: Can one make tables of genus $2$ curves over $K$ ordered in some lexicographic fashion by the invariants? Perhaps the curves would be modelled by singular plane curves $y^2 = f(x)$ where $f$ has degree $5$ or $6$. The first pair $(X,Y)$ with isomorphic Jacobian variety to occur in this table might satisfy the request that the pair $(X,Y)$ be the "simplest" such pair to be defined over $K$.

Poonen notes that not every not every hyperelliptic curve has a model of the form $y^2 = f(x)$ over $k$ a field, because the quotient of the curve by its hyperelliptic involution might be a twist of $\mathbb{P}^1$ i.e. birational over $k$ to a conic in $\mathbb{P}^2$ without a $k$-rational point. However when $g$ is even, it seems to turn out (I'm not sure why) that the $k$-form of $\mathbb{P}^1$ has a $k$-point and the hyperelliptic curve has a model $y^2 = f(x)$, so for $g = 2$ we should be okay.

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