Let N be a prime integer. We know that the element c=(0)-(\infty) $c=(0)-(\infty)$ generates the torsion subgroup of J_0(N) $J_0(N)$ and it has order Num( (N-1)/12). Now, there is a natural map $\pi^*:J_0(N) \rightarrow J_1(N)$, coming from the covering map $\pi:X_1(N) \rightarrow X_0(N)$. My question is what is the image of c under this map? Specifically, is it possible for $\pi^*(c)=0$?
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Let N be a prime integer. We know that the element c=(0)-(\infty) generates the torsion subgroup of J_0(N) and it has order Num( (N-1)/12). Now, there is a natural map $\pi^*:J_0(N) \rightarrow J_1(N)$, coming from the covering map $\pi:X_1(N) \rightarrow X_0(N)$. When $N=11$, we can verify explicitly that $\pi^* (c)=0$. We also know that degree of $\pi$ is equal to the order of $c$. My question is thenwhat is the image of c under this map? Specifically, is it true that $\pi^*(c)=0$ possible for all $N$? How would one prove that?\pi^*(c)=0$? |
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Image of the cuspidal subgroup of J_0(N) in J_1(N)Let N be a prime integer. We know that the element c=(0)-(\infty) generates the torsion subgroup of J_0(N) and it has order Num( (N-1)/12). Now, there is a natural map $\pi^*:J_0(N) \rightarrow J_1(N)$, coming from the covering map $\pi:X_1(N) \rightarrow X_0(N)$. When $N=11$, we can verify explicitly that $\pi^* (c)=0$. We also know that degree of $\pi$ is equal to the order of $c$. My question is then, is it true that $\pi^*(c)=0$ for all $N$? How would one prove that?
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