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.



Newform of a cuspidal Automorphic Representation
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.


