# Identifying elements in the kernel of an explicit endomorphism of a Jacobian variety

I hope this question fits here.

Let $$H/k$$ be a genus $$2$$ curve and $$J$$ its Jacobian variety. Since $$J(k)\cong \text{Pic}^0(H)(k)$$ we have that its generic point looks like $$[(x_1,y_1)+(x_2,y_2)-2\infty]\in J$$. In Mumford coordinates we can see it as $$g:=\langle x^2 -Ax + B,Cx+D\rangle:=\langle u(x),v(x)\rangle\in J$$ with $$u(x_i)=0$$ and $$v(x_i)=y_i$$. This means that $$A:=x_1+x_2, B:=x_1x_2, C:=\tfrac{y_1-y_2}{x_1-x_2}, D:=\tfrac{x_2y_1-x_1y_2}{x_1-x_2}$$.

I calculated explicitly an element $$\gamma\in\text{End}_k(J)$$ using Mumford coordinates for the generic point, that is $$\gamma(g)=$$ where $$A_1,A_2,B_1,B_2,C_1,C_2,D_1,D_2\in k[J]$$.

To formulate my question, note that since the endomorphism $$\gamma$$ works fine for the generic point, and $$J$$ has dimension $$2$$, we have that if for some $$D\in J(k)$$, its image under $$\gamma$$, namely $$\gamma(D)$$ is of the form $$[(x_1,y_1)-\infty]$$, then $$\gamma$$ won't be defined for $$D$$, since some of the denominators will be zero.

How can I distinguish when $$\gamma(D)$$ is 0 or when it is non-zero non-generic of the form $$[(\tilde x,\tilde y)-\infty]$$?

I know I could use a constant divisor $$D_0$$ and calculate $$\gamma(D+D_0)$$ when $$\gamma(D)$$ is not defined and then subtract with $$\gamma(D_0)$$ using Cantor addition. However, in my situation this is not possible since I am testing if $$k$$ is a field using some arithmetic geometry. I want to be able to use the functions defining $$\gamma$$ over $$\mathbb{Q}$$ to distinguish the situation of $$\gamma(D)$$ being $$$$ or being of the form $$[(\tilde x,\tilde y)-\infty]$$

I have noted that when the ALL the denominators are 0, it looks like the image 0 in fact, but when it lies in the "theta divisor", (the image is a point of the form $$[(x_1,y_1)-\infty]$$), some of the denominators are non-zero. However I do not know how to distinguish this formally or maybe my examples are just "lucky" examples.

Is there a way with this information to distinguish when $$\gamma(D)$$ is exactly  ?

What I did

I tried to calculate the formula of $$\gamma$$ using MAGMA via the function field of $$J$$, using the usual relations for the Jacobian of $$H$$ plus the denominators of $$\gamma$$ as relations, but the computation does not finish and eats all my memory eventually.

I just need to know if a point in the image is 0 or non-generic when $$\gamma(D)$$ has 0's in the denominators using the information that I have, or maybe using an element of the function field of $$J$$ that can distinguish if a divisor is 0 or if it has the point at infinity with multiplicity 1.

I would suggest that you work with the Kummer surface $$K$$ of $$J$$ instead of using Mumford coordinates. The advantage is that $$K$$ is a quartic surface in $$\mathbb P^3$$; in the case you are considering when the curve has a unique point at infinity, the vanishing of the first coordinate means that the point is in the theta divisor, whereas it is the origin when the first three coordinates vanish (using the standard Kummer coordinates as in the book by Cassels and Flynn). Since your endomorphism commutes with multiplication by $$-1$$, it induces an endomorphism of $$K$$. This will be given by a quadruple of homogeneous polynomials of some degree $$d$$ in the four coordinates; it should not be too hard to figure out what they are from the generic representation in terms of the Mumford representation. Then your problem comes down to checking whether the first of these polynomials vanishes, and if so, whether the next two also vanish. (This assumes that all four polynomials do not vanish simultaneously at some point on $$K$$.)

When $$\gamma$$ is multiplication by 2, for example, the polynomials are of degree 4 and can be obtained via

KummerSurface(J)Delta;
`

in Magma.

For the curve $$C \colon y^2 = x^5 + 10\,,$$ one choice of polynomials giving multiplication by $$\sqrt{5}$$ on the Kummer surface is $$\begin{array}{r@{\,}c{\,}l} P_1(x_1,x_2,x_3,x_4) &=& 8000 x_1^3 x_2^2 + 400 x_1^2 x_2 x_4^2 + 200 x_1^2 x_3^2 x_4 + 400 x_1 x_2^2 x_3 x_4 - 600 x_1 x_2 x_3^3 + 5 x_1 x_4^4 + 200 x_2^3 x_3^2 + 10 x_2 x_3 x_4^3 + 10 x_3^3 x_4^2 \\ P_2(x_1,x_2,x_3,x_4) &=& 8000 x_1^4 x_4 + 8000 x_1^3 x_2 x_3 + 400 x_1 x_2^2 x_4^2 + 200 x_1 x_2 x_3^2 x_4 + 400 x_1 x_3^4 - 400 x_2^3 x_3 x_4 + 200 x_2^2 x_3^3 - 5 x_2 x_4^4 + 10 x_3^2 x_4^3 \\ P_3(x_1,x_2,x_3,x_4) &=& 40 x_1^3 x_2 x_4 + 8000 x_1^3 x_3^2 - 8040 x_1^2 x_2^2 x_3 - 200 x_1^2 x_4^3 - 7960 x_1 x_2^4 - 996 x_1 x_2 x_3 x_4^2 + 200 x_1 x_3^3 x_4 + 399 x_2^3 x_4^2 - 598 x_2^2 x_3^2 x_4 - x_2 x_3^4 + 5 x_3 x_4^4 \\ P_4(x_1,x_2,x_3,x_4) &=& 64000 x_1^5 + 2720 x_1^3 x_3 x_4 + 8000 x_1^2 x_2^2 x_4 + 5280 x_1^2 x_2 x_3^2 + 2720 x_1 x_2^3 x_3 - 200 x_1 x_2 x_4^3 - 1528 x_1 x_3^2 x_4^2 + 1600 x_2^5 - 68 x_2^2 x_3 x_4^2 - 64 x_2 x_3^3 x_4 + 52 x_3^5 + x_4^5 \end{array}$$ (they are not unique, since we can add multiples of the defining equation of the Kummer surface).
• Prof. Stoll. Thanks for the answer. In fact I retook a problem, using also your advice from some time ago, which was to extend the function field of the jacobian to the denominators). My profile in mathoverflow changed since univ. mail does not exists anymore. Related to your answer, I would need then to calculate the endomorphism mapped to the Kummer Surface, in my case I have to map concretely $\sqrt{5}\in\text{End}(J)$ where $J$ is the Jacobian of $H:y^2=x^5 + 10$ over $\mathbb{Q}$. Any suggestions? (technically talking) do you know if someone has done this for explicit endomorphisms? – Eduardo R. Duarte Mar 28 '19 at 16:16
• Prof, this is really nice, I just accepted the answer. I assume that $x_1,x_2, x_3,x_4$ are standard coordinates found in Prolegomena Flynn's book. I am really interested in the MAGMA computation in fact. Why the formulas are so small ? When I calculated $\sqrt{5}\in\text{End}(J)$ I used the generic point of $g\in J$ and then the action of $\zeta_1,\zeta_4\in\text{End}(J)$ in $g$. The final formulas are obtained through $2(\zeta_1(g)+\zeta_4(g))+g=\sqrt{5}(g)$. These formulas are huge. Yours are very compact, is there something I am missing? – Eduardo R. Duarte Mar 31 '19 at 19:55