# 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
• The link to the notes is unavailable now. Aug 28 at 6:46

## 1 Answer

Yes, the classical version of this adelic newform is the newform in the sense of Atkin-Lehner, and vice versa. See Casselman: On some results of Atkin and Lehner (Math. Ann. 201 (1973), 301-314), especially Theorem 4 there. Another important reference is Miyake: On automorphic forms on GL_2, and Hecke operators (Ann. of Math. 94 (1971), 174-189.). See also Sections 5.3-5.4 in Goldfeld-Hundley: Automorphic Representations and L-Functions for the General Linear Group, Volume 1.

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
• @ShubhodipMondal: Yes, this is a consequence of the multiplicity one theorem (which is ultimately an analytic result), see my added section. The multiplicity one result is very important: it is present in Casselman's paper (Theorem 2) and in Miyake's paper (Corollary to Theorem A) as well. It says that Hecke eigenvalues (or Hecke eigenvalues outside any finite set of places) identifies the cuspidal representation (and hence its 1-dimensional newspace as well). Apr 16, 2017 at 16:12
• 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