Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $p$ be a primes that does not divide $N$, then $T_p$ defined an endomorphism $J_0(N)\to J_0(N)$. what is $T_p^\vee$? In other words, we naturally have $J_0(N)^\vee \xrightarrow{T_p^\vee} J_0(N)^\vee$, and therefore $J_0(N) \xrightarrow{T_p^\vee} J_0(N)$. So, what is this dual map?

The reason that I ask this question is the following.

Let $Y_0(N)= \Gamma_0(N)\backslash \mathbb{H}$ be the usual modular curve of level $N$. Given $p\nmid N$, we have two maps from $Y_0(pN)$ to $Y(N)$. Let's call them $B_1$ and $B_p$. By definition, $B_1$ is induced by the identity map of the upper half plane, and $B_p$ is induced by $z\mapsto pz$. Now $B_1$ and $B_p$ uniquely determines maps from $X_0(pN)$ to $X_0(N)$. With an abuse of notation, I am going to denote the map between the Jacobian of the modular curves $J_0(pN)\to J_0(N)$ induced by Albanese functoraility still by $B_1$ or $B_p$, and denote the maps induced by the Picard functoriality by $B_1^\vee$ or $B_p^\vee$.

I was able to show that $B_1\circ B_p^\vee$ coincide with the Hecke operator $T_p$. This can be explained by the following fact. If we let $\alpha =\left(\begin{smallmatrix} 1 & 0 \\ 0 & p \end{smallmatrix} \right)$. Then $$\alpha^{-1}\Gamma_0(N)\alpha \cap \Gamma_0(N)= \alpha^{-1}\Gamma_0(pN) \alpha.$$

However, I am having trouble to figure out what is $B_p\circ B_1^\vee$.

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.