# Newform of a cuspidal Automorphic Representation

I was going through these notes https://www.dpmms.cam.ac.uk/~ty245/2008_AGR_Fall/2008_agr_week1.pdf . There, Theorem 9.2 states that: If $\pi ^{\infty}$ is a cuspidal automorphic representation of $\text{GL}_2(\mathbb A^{\infty})$ (on $V$), then there exists $N \in \mathbb Z _{>0}$ with $V^{U_1(N)} \ne 0$ and for minimal such $N$, we have $\dim_{\mathbb C}V^{U_1(N)} = 1$. The cusp form $\varphi \in \mathcal A_k( U_1(N))$ generating (unique upto scalar) $V^{U_1(N)}$ is defined to be a "newform".

Later it is shown that this $\varphi$ gives a classicial cusp form w.r.t. $\Gamma_1(N)$. My question is:

Quetion: Does it follow that $\varphi$ is a newform according to the classical definition (i.e., the one involving Petersson inner product)?

I understand that $\varphi$ is a Hecke eigenform w.r.t. $T_n$ and $<n>$ for $n>0$. So it follows that $\varphi$ is either a newform or an old form. If it is an old form one can associate a newform $f$ to it with some conductor $M \mid N$. I was trying to show that $f \in V$, which, then by minimality of $N$ would answer the question.

Thank you.

• What do you mean by $\varphi$ being a classical newform? Atkin-Lehner defined classical newforms for $\Gamma_0(N)$, not for $\Gamma_1(N)$. Apr 16, 2017 at 23:08
• I'm using the notion from Diamond Shurman. Apr 16, 2017 at 23:22
• Ah, I thought by $\Gamma_1$ you meant unipotent mod $N$, but the notes you refer to work with another definition of $\Gamma_1$. So there's no problem. Apr 17, 2017 at 0:41

Added. To your last paragraph: the adelization of $f$ lies in $V$, because it has the same Hecke eigenvalues as $\varphi$. The multiplicity one theorem says (or implies) that adelic (almost) Hecke cusp forms with the same Hecke eigenvalues (outside any finite set of places) generate the same cuspidal representation.
• Thanks, I will definitely take a detailed look at the sources you cited. Meanwhile, I would like to ask if it is possible to prove it the way I was trying to, i.e., does there exist a $g \in \text{GL}_2(\mathbb A^{\infty})$ such that $g \cdot \varphi = f?$ (In above notations) Apr 16, 2017 at 16:02
• Does one really need to use the multiplicity one theorem here? Can't one find the relevant $g \in GL_2(\mathbb A^{\infty})$ explicitly, as $\begin{pmatrix} p & 0 \\ 0 & 1 \end{pmatrix} \in GL_2(\mathbb Q_p)$ or something like that? Apr 16, 2017 at 16:35
• @ShubhodipMondal: Right. Let $\tilde\varphi$ (resp. $\tilde f_M$) be the adelization of $\varphi$ (resp. $f_M=f$). Then $\tilde\phi(g)=\sum_{\substack{M\mid N\\M<N}}\sum_{n\mid\frac{N}{M}} \tilde f_M\left(g\left(\begin{smallmatrix}n^{-1}&\\&1\end{smallmatrix}\right)\right)$. The inner sum is orthogonal to $\pi$, because it lies in a cuspidal representation of conductor $M\neq N$ (here I use multiplicity one). Hence the whole sum is orthogonal to $\pi$, hence also to $\tilde\phi$ (i.e. to itself), which is a contradiction. Apr 16, 2017 at 17:19
• @ShubhodipMondal: A coefficient $c_{n,M}$ (which is your $c_n$) should be included in my inner sum (in the previous comment). I am lazy to retype. Apr 16, 2017 at 17:27