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.