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I was looking for some information related to the values of the characteristic polynomial $\chi(t)$ of the Frobenius of a Jacobian of a hyperelliptic curve $C$ of genus 2 over $\mathbb{F}_q$ and in some sources it says that it provides information of #$\mathcal{J}_C(\mathbb{F}_{q^k})$ for every $k$ , I just proved that $\chi(-1)=$ #$\mathcal{J}_{C^{TW}}(\mathbb{F}_q)$ which is the jacobian of the quadratic twist of the hyperelliptic curve, so , the value $\chi(-1)$ is not giving me information about the jacobian over $\mathbb{F}_{q^k}$, even that I know that the twist of $C$ and $C^{TW}$ are isomorphic over $\mathbb{F}_{q^2}$.

One question is, where I can find more information about the values of $\chi$.

I am interested in the value of $\chi(2)$, what I am doing is considering the degree of the map $\Phi+[\mathcal{n}]$ (where $[1],\Phi\in End_{\mathbb{F}_q}(\mathcal{J}_{C})$ are the identity and Frobenius, respectively), it is known that $\Phi-[1]$ is a separable map, so # $ker(\Phi+[-1])=$ # $\mathcal{J}_C(\mathbb{F}_q)$ , in fact $\chi(n)=\partial (\Phi+[n])$,

what can be said about $\chi(2)$?, or about $\chi(n)$?

I am working with an specific example to gain some intuition, in fact is $C_{\mathbb{F}_{101}}$ given by $y^2 = x^5 + x + 1$ where the characteristic polynomial of the Frobenius is given by $\chi(t) = t^{2g}L^C_{\mathbb{F}_q}(1/t)$ , which in this case is: $t^4 + 2t^3 + 26t^2 + 202t + 10201$, and $\chi(2)=10741$.

I also saw in the slide 22 of this presentation that #$\mathcal{J}_C(\mathbb{F}_p)=\frac{1}{2}$#$C(\mathbb{F}_{p^2})+\frac{1}{2}$#$C(\mathbb{F}_p)-p$ which is false. (see magma example below)

http://www.skidmore.edu/fq12/uploads/Eisentraeger.pdf

Example of magma, showing that the last equality in slide 22 is false.


> p := 101;
> Fp := GF(p);
> _ := PolynomialRing(Fp);
> f := x^5 + x + 1;
> CFp := HyperellipticCurve(f);
> CFp2 := BaseExtend(CFp,GF(p^2));
> J := Jacobian(CFp);
> #J;   
10432
> ((#CFp+#CFp2)/2)-p;
5076
> #CFp;#CFp2;
104
10250
> CFp;CFp2;J;
Hyperelliptic Curve defined by y^2 = x^5 + x + 1 over GF(101)
Hyperelliptic Curve defined by y^2 = x^5 + x + 1 over GF(101^2)
Jacobian of Hyperelliptic Curve defined by y^2 = x^5 + x + 1 over GF(101)
> 

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The equation for $\# J_C(\mathbb F_p)$ that you quote contains a typo: they must have meant that $\# J_C(\mathbb F_p) = \frac 1 2 \# C(\mathbb F_{p^2}) + \frac 1 2 \# C(\mathbb F_p)^2 - p$, which is consistent with your example. That there is an equation like this is a direct consequence of the Grothendieck-Lefschetz trace formula: use that $H^i$ of an abelian variety is $\wedge^i$ of $H^1$, and that $H^1$ of a jacobian is isomorphic to $H^1$ of the curve.

If you think about it in terms of symmetric functions it's not hard to see that $\chi(n)$ is always going to be some universal expression in $\#C(\mathbb F_p)$ and $\# C(\mathbb F_{p^2})$. The coefficients in the characteristic polynomials are symmetric functions of the Frobenius eigenvalues, and are therefore expressible in terms of the power sums of the Frobenius eigenvalues. But knowing the power sums of the Frobenius eigenvalues is exactly the same as knowing $\#C(\mathbb F_{p^k})$ for all $k$. Since we are in genus two we only need $k=1,2$ to determine all Frobenius eigenvalues (using Poincaré duality/the functional equation).

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  • $\begingroup$ Very nice answer, Do you have a good source to read in this context in more detail, for example using Étale Cohomology groups 1,2,3 to get the Coefficients of the characteristic polynomial. And also it would be nice if there's some type of paper explaining the values of $\chi(n)$ for $n\in \mathbb{Z}$. $\endgroup$ – Eduardo R. Duarte Feb 14 '16 at 20:11
  • $\begingroup$ Hi, I don't know a great reference off the top of my head that doesn't go into much more advanced things than what you are probably interested in. Maybe if you google "Weil conjectures for curves" you'll find some worked out examples to get you started. $\endgroup$ – Dan Petersen Feb 15 '16 at 13:56
  • $\begingroup$ Anyway, expressing the coefficients of the characteristic polynomial in terms of the Frobenius eigenvalues has not so much to do with étale cohomology - it's an exercise about symmetric functions. $\endgroup$ – Dan Petersen Feb 15 '16 at 13:57

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