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Let $\mathbf{F}_q$ be a finite field of odd characteristic. Let $X_t$ be the hyperelliptic curve over $\mathbf{F}_{q^2}(t)$ with affine equation $$y^2 = \left((x^{(q+1)/2}-(x-1)^{(q+1)/2})^2 - t\right) \left((x^{(q+1)/2}+(x-1)^{(q+1)/2})^2 - t\right).$$ This family of hyperelliptic curves is smooth of genus $q-1$ away from $t=0,1,\infty$. For theoretical reasons I expect the following claim to be true:

The Jacobians of $X_t$ and $X_{1-t}$ are isogenous. Equivalently, there is a nontrivial correspondence between $X_t$ and $X_{1-t}$.

It is easy to verify (with MAGMA for instance) that the zeta functions of $X_t$ and $X_{1-t}$ agree for small values of $q$ and specific values of $t$, to the point that I'm quite convinced of the claim. To prove it, one could try to construct the correspondence explicitly. There is some literature about pairs of hyperelliptic curves with isogenous Jacobians (for instance this paper of Mestre), but it seems to involve finding a congruence between bivariate polynomials, which comes out of nowhere.

Short of constructing the actual correspondence, is there an algorithm for deciding whether two hyperelliptic curves admit an isogeny of given degree between their Jacobians?

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  • $\begingroup$ Did you have luck with this? $\endgroup$ Jan 13, 2021 at 10:45
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    $\begingroup$ Hi Dror, sorry, I didn't check this for awhile. The curves $X_t$ and $X_{1-t}$ have a meaning: they are a Drinfeld modular curve and a Drinfeld-Shimura curve, respectively. I thought it was cute that they had such a simple formula. The Jacquet-Langlands correspondence combined with the Tate conjecture for divisors on abelian varieties over function fields produces the desired isogeny. I wanted to know if there was an explicit correspondence, but maybe this is too much to hope for. $\endgroup$ Apr 23, 2021 at 13:30

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In the most easy case $q=3$, the curve $X_t$ is bielliptic (the bielliptic involution given by $x\mapsto 1-x$), and the Jacobian of $X_t$ is then $(2,2)$-isogenous to the product $E_{1,t}\times E_{2,t}$, where $E_{1,t}$ and $E_{2,t}$ are the elliptic curves $$E_{1,t}: y^2=x(x^2 - (t+1)x + (t- 1)^2)$$ and $$E_{2,t}: y^2=x(x^2 + x - (t^3 - 1))$$ Now, $E_{1,t}$ is 2-isogenous to $$E_{1,t}': y^2= x(x+1)(x-(t-1))$$ and one has that $$E_{1,t}'\cong E_{1,1-t}$$ by sending $x\mapsto 1-x$ and then twisting by $-1$, which is an isomorphic curve over $\mathbb{F}_9$ (this is the only place where we use we are over $\mathbb{F}_9$ and not over $\mathbb{F}_3$). while $E_{2,t}$ is 2-isogenous to $$E_{2,t}': y^2 = x(x^2 + x + t^3)$$ which in turn is clearly equal to $E_{2,1-t}$.

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    $\begingroup$ Thank you Xarles! I am attempting to see whether your construction generalizes. The Jacobian of $X_t$ is isogenous to a product of Jacobians of hyperelliptic curves $C_{1,t}$ and $C_{2,t}$. Evidence suggests that your pattern persists: the Jacobians of $C_{1,t}$ and $C_{1,1-t}$ are isogenous over $\mathbf{F}_{q^2}(t)$, and the Jacobians of $C_{2,t}$ and $C_{2,1-t}$ are isogenous already over $\mathbf{F}_q(t)$. $\endgroup$ Feb 4, 2020 at 3:06
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The Tate conjecture is known for function fields (Zarhin). To check whether the Tate modules are isomorphic you need to check that the image of Frobenius match for a finite set of places that can be bounded a priori (the argument is in Faltings's Mordell paper) although the bound is probably not very small. Checking that the image of Frobenius match at a place is essentially your computation of specializing $t$ and compute zeta functions. This gives, in principle, an algorithm to check that the Jacobians are isogenous for a given $q$. It's probably not practical except for very small $q$. It doesn't give any information of the degree of the isogeny, though.

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