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Let $f\in S_k(\Gamma_0(N))$, $(p,N)=1$ and $T_r:S_k(\Gamma_0(Np))\longrightarrow S_k(\Gamma_0(N))$ the trace map. Does then $T_r(f\mid V_p) = f\mid T_p$ hold?

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What is $V_{p}$? –  Olivier Jan 12 at 6:58
    
$V_p$ is the map $S_k(\Gamma_0(N))\to S_k(\Gamma_0(Np)) $ given by $f(z)\mapsto f(pz)$. –  Abdullah.Y Jan 12 at 11:38

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You don't define your notations, as Olivier points out, but I'm going to assume that $V_p$ is the map $S_k(\Gamma_0(N)) \to S_k(\Gamma_0(Np))$ given by $f(z) \mapsto f(pz)$.

Then the answer is "yes" (at least up to a constant factor, something like $p^{k-1}$, that depends on your choice of conventions); in fact, this is essentially the definition of the $T_p$ operator -- pullback along one degeneracy map followed by pushforward along the other.

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Sorry!Yes, $V_p$ is $f(z)\to f(pz)$. –  Abdullah.Y Jan 12 at 12:06
    
Is there a similar identity if p|N? –  Abdullah.Y Feb 13 at 15:32
    
Yes, the same works if $p \mid N$. –  David Loeffler Feb 13 at 20:11
    
Where can I read this. I tried to show this but its not easy to give a full set of the cosets for $\Gamma_0(pN)$ in $\Gamma_0(N)$. I need this to compute the trace map. –  Abdullah.Y Feb 13 at 21:04

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