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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$\begingroup$ What is $V_{p}$? $\endgroup$– OlivierCommented Jan 12, 2014 at 6:58
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$\begingroup$ $V_p$ is the map $S_k(\Gamma_0(N))\to S_k(\Gamma_0(Np)) $ given by $f(z)\mapsto f(pz)$. $\endgroup$– Abdullah.YCommented Jan 12, 2014 at 11:38
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1 Answer
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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.
answered Jan 12, 2014 at 8:58
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$\begingroup$ Sorry!Yes, $V_p$ is $f(z)\to f(pz)$. $\endgroup$ Commented Jan 12, 2014 at 12:06
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$\begingroup$ Is there a similar identity if p|N? $\endgroup$ Commented Feb 13, 2014 at 15:32
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$\begingroup$ Yes, the same works if $p \mid N$. $\endgroup$ Commented Feb 13, 2014 at 20:11
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$\begingroup$ 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. $\endgroup$ Commented Feb 13, 2014 at 21:04