Let $N \geq 1$ be an integer and let $p$ be a prime not dividing $N$. For $r \geq 1$, let $M\_2(\Gamma\_0(Np^r))$ denote the space of weight $2$ modular forms of level $\Gamma\_0(Np^r)$. Let $$U\_p: M\_2(\Gamma\_0(Np^r)) \to M\_2(\Gamma\_0(Np^r))$$ denote the $p$-th Hecke operator, which acts on $q$-expansions by sending $\sum a\_n q^n$ to $\sum a\_{np} q^n$.


I did some numerical calculations on Sage, and it looks like $U\_p$ takes values in $M\_2(\Gamma\_0(Np))$, not merely in level $Np^r$. That is, for all $r \geq 1$, it looks like $U\_p$ is actually a map from $M\_2(\Gamma\_0(Np^r))$ to $M\_2(\Gamma\_0(Np))$.


Is this true? And if so, could anyone sketch a proof / point to a reference?



What is the image of the Hecke operator $U_p$?
The statement, as claimed, is false.


Let $p = 2, N = 11$, and let $f\_0$ be the unique normalised eigenform in $S\_2(\Gamma\_0(11))$; and set $f(\tau) = f\_0(8\tau)$. Then $f \in M\_2(\Gamma\_0(Np^3))$, but $U\_p(f) = f\_0(4\tau)$ is not in $M\_2(\Gamma\_0(Np))$.


However, $U\_p(f)$ is in $M\_2(\Gamma\_0(Np^2))$. The correct general statement is that for $r \ge 2$, $U\_p$ maps $M\_2(\Gamma\_0(Np^r))$ to $M\_2(\Gamma\_0(Np^{r-1}))$ (and likewise for forms of any weight $k$, not just weight $2$). But in general it doesn't go all the way from level $\Gamma\_0(Np^r)$ to $\Gamma\_0(Np)$.


PS. You asked for an outline of the proof. The idea is to check that (under the hypotheses of my last paragraph) the double coset $\Gamma\_0(Np^r) \begin{pmatrix} 1 & 0 \\ 0 & p \end{pmatrix} \Gamma\_0(Np^r)$ is actually invariant under right-translation by $\Gamma\_0(Np^{r-1})$, not merely by $\Gamma\_0(Np^r)$. (This kind of argument is very important in Hida theory.)


EDITED TO ADD. One can check that the composite of $U\_p$ with the twisted map $M\_2(\Gamma\_0(Np^{r-1})) \to M\_2(\Gamma\_0(Np^r))$, $f(\tau) \mapsto f(p\tau)$, is multiplication by a power of $p$. So $U\_p$ is *surjective* as a map $M\_2(\Gamma\_0(Np^r)) \to M\_2(\Gamma\_0(Np^{r-1}))$, answering the slightly more refined question in the title of Adithya's post.


