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Let $\text{S}_k^{new}(\Gamma_0(N),\chi)$ be the space of newforms. We call $f\in\text{S}_k^{new}(\Gamma_0(N))$ a newform if $f$ is a Hecke eigenform i.e $\text{T}_nf=\lambda_nf$ ($\text{T}_n$ hecke operator) for all $n$ and $f$ is normalized by setting the first coefficient $a(1)=1$ in the Fourier expansion of $f$. It is well known, that two different newforms in $\text{S}_k^{new}(\Gamma_0(N),\chi)$ are orthogonal relative to the Petersson inner product.

Let $f\in\text{S}_k^{new}(\Gamma_0(N),\chi)$, $g\in \text{S}_k^{new}(\Gamma_0(M),\chi)$ be newforms and $n$, $m$ positive integers, is then $<V_nf,V_mg>=0$? Where $V_d$ is the map $V_d:\text{S}_k^{new}(\Gamma_0(N),\chi)\to \text{S}_k^{new}(\Gamma_0(dN),\chi)$ given by $f(z)\mapsto f(dz)$.

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  • $\begingroup$ Do you want some hypothesis about $n$ and $m$ being different, and/or some hypothesis about $f$ being different from $g$? $\endgroup$ Apr 24, 2014 at 16:28
  • $\begingroup$ The case $M=N$ and $f=g$ is clear for me. If $M\not=N$ then of course $f$ and $g$ are different. There are no conditions for $n$ and $m$. $\endgroup$
    – Abdullah.Y
    Apr 24, 2014 at 16:49
  • $\begingroup$ What do you mean by the Petersson inner product? By definition it should be of the form $\langle f,g\rangle = \frac{1}{[\Gamma_0(1) : \Gamma_0(N)]}\int_{\Gamma_0(N) \backslash \mathbb{H}}f(z)\overline{g(z)} y^k \, \frac{dx\, dy}{y^2}$, and in particular depends on $N$ (and $k$). So it only makes sense to take the inner product of two modular forms of the same weight and level. $\endgroup$ Apr 24, 2014 at 17:54
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    $\begingroup$ If $f$ have level $N$ and $g$ have level $M$, but of the same weight $k$, than f and g can be viewed as modular forms of level weight $k$ and level $MN$. $\endgroup$
    – Abdullah.Y
    Apr 24, 2014 at 18:01

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