En Passant VII

4 Responses to En Passant VII

1. Dick Gross says:

   April 27, 2020 at 5:45 pm

   Frank,

   You are probably right that this is known. However, since the golf courses and beaches are still closed in my neighborhood, I’ll make a comment. The reason that the simple factors A of the Jacobian of the curve X_ns(p)^+ should have positive rank over Q is that they should all have “odd” rank: rank(A) = (2n+1)dim(A). This is because an eigenform form of weight 2 contributes to the cotangent space of the Jacobian if it is new of level p^2 and has local epsilon factor +1 at p. Hence the L-function vanishes to odd order at s = 1, as the only non-trivial epsilon factor is at infinity. By BSD and the conjecture that the order of vanishing at s = 1 is the same for conjugate L-series, you get the prediction for the rank of A. If the L-functions all vanish to order 1, this is known. I don’t know the first prime p where one of these L-functions vanishes to order 3.

   I suspect the same may be true for X_ns(N)^+ for any N — you get newforms of level N^2 whose local epsilon factors are all +1 except at infinity, so give L-functions vanishing to odd order. One could test this by computing the genera of the associated modular curves, but I haven’t done so. Anyhow, the fact that there is no “Eisenstein quotient” of the Jacobian with rank 0 is what makes it so hard to explicitly determine the rational points on these curves. Fortunately, Heegner only had to determine the integral points to solve the class number 1 problem.

   Reply
   • Persiflage says:

     April 27, 2020 at 8:18 pm

     I always found sign computations a little tricky to do; starting with the fact that I always get confused between the relationship between the three signs corresponding to the eigenvalues of \(U_N\), \(w_N\), and the sign of the functional equation. I suspected that in this question it would come down to the fact that \((1)^2 = 1\) and so ranks would still be odd when \(N =pq\), but there was the outside possibility that \((-1)^2=1\ne -1\) would be the relevant identity.

     Maybe the one way I can remember it is to think about the Eisenstein ideal; at prime level \(N\) one has \(U_{N} – 1 \in \mathscr{I}\), and the rank of the Eisenstein quotient is zero. But for \(N = 11\) the curve \(X_0(11)^{+}\) has genus zero, so \(w_{11} = -1\) and \(U_{11} = +1\) and the sign of the functional equation is \(+1\). So probably the sign of the functional equation is the eigenvalue of \(w_N\) and then indeed \((+1)^2=+1\) seems to exactly confirm your remarks. (And just to confirm a theory with a computation; the space of newforms at level \(35^2 = 1225\) with signs \((+,+)\) has dimension \(13\) which is the correct genus according to a paper of Burcu Baran).

     Reply
2. Dick Gross says:

   April 27, 2020 at 9:40 pm

   You have to remember the local epsilon at infinity, to get the global root number. For weight 2k the local epsilon at infinity is (-1)^k, so for weight 2 it is -1. The local epsilon at p is the eigenvalue of W_p. When p exactly divides N, the eigenvalue of W_p is -a(p), so W_p + U_p = 0. When p^2 divides N, you can’t get W_p from the Fourier expansion.

   What I said in the second paragraph isn’t quite correct. For example, when N = pq you also get newforms of level p^2 and q^2 with finite local epsilons = +1. In any case, the global sign is -1 and the L-function vanishes to odd order.

   Reply
3. JSE says:

   April 30, 2020 at 4:20 pm

   Hey Frank — I had to deal with stuff like this in my old paper on A^4 + B^2 = C^p, but not for exactly this case. What did happen there, though, is that you got some mileage out of levels being composite instead of prime, because you have some quotients coming from the old part. We train ourselves to think just about newforms, but for the Mazur-type formal immersion, a rank-0 quotient from oldforms is just as good as one from newforms. Indeed (and please forgive me if this is wrong, it’s from memory) the “reason” split is easier than non-split is that under the isogeny described by Chen and Edixhoven, J_0^split corresponds to a piece of J_0(p^2)/w_p which has some oldforms in it, while J_0^ns exactly gets ride of those oldforms and you’re left with something that’s totally Atkin-Lehner positive and thus odd in rank throughout.

   Reply