# Normalisation of $L$-function for classical modular forms and automorphic representations

I found that the normalisation of $$L$$-functions of classical modular forms and corresponding automorphic representations is somewhat confusing for me. Recall that if $$f\in S_k(\Gamma_0(N))$$ is a classical cusp form with respect to $$\Gamma_0(N)$$, then $$f$$ has a $$q$$-expansion $$f=\sum a_n q^n$$, and the (classical) partial $$L$$-function (non-archimedean part) of $$f$$ is defined as $$L(s,f)=\sum \frac{a_n}{n^{s}}.$$ After adding an archimedean $$L$$-factor, we obtain the completed $$L$$-function $$\Lambda(s,f)$$. If $$f$$ is a newform (Hecke eigenform and orthogonal to oldforms), then we know that $$\Lambda(s,f)$$satisfies a functional equation, with central line $$Re(s)=k/2$$, c.f. GTM228, Diamond-Shurman section 5.10.

On the other hand, we can lift $$f$$ as an automorphic form on $$G(\mathbb{Q})\backslash G(\mathbb{A})$$, where $$\mathbb{A}=\mathbb{A}_\mathbb{Q}$$, $$G:=GL(2)$$. This process is rather standard: we have to start from the automorphy factor $$j(g, \tau)=\operatorname{det}(g)^{-\frac{1}{2}}(c \tau+d), \quad g=\left(\begin{array}{ll} a & b \\ c & d \end{array}\right) ,$$ and lift it to $$GL(2,\mathbb{R})^+$$ by the following formula $$\phi\left(g_{\infty}\right)=\phi_f\left(g_{\infty}\right)=f\left(g_{\infty}(i)\right) j\left(g_{\infty}, i\right)^{-k},$$ and finally we use the strong approximation to obtain $$G(\mathbb{Q}) \backslash G(\mathbb{A}) / K^{\prime} \simeq \Gamma^{\prime} \backslash G(\mathbb{R})^{+},$$ where $$K^{\prime}=K_0(N)=\left\{k=\left(\begin{array}{ll} a & b \\ c & d \end{array}\right) \mid c \equiv 0 \bmod (N)\right\}$$ and $$\Gamma^{\prime}=G(\mathbb{Q}) \cap G(\mathbb{R})^{+} K^{\prime}=\Gamma_0(N)$$. Then we obtain an adelic automorphic form $$\phi$$ and it generates an automorphic representation. Finally one defines its **automorphic $$L$$-function **according to the Godement-Jacquet theory of zeta integrals. Here the zeta interal is defined as $$Z(\phi, h, s)=\int_{\mathrm{GL}_2(\mathbb{A})} \phi(g) h(g)|\operatorname{det}(g)|^{s+1 / 2} \mathrm{~d} g,$$ where $$h(g)$$ is a matrix coefficient of this automorphic representation. In some sense the $$L$$-function $$L(s,\pi_f)$$ is the "greatest common denominator" of these integrals (but here there's a huge bunch of tough technical problems, which I didn't mention here).

What makes me puzzled is: finally we have $$L(s,\pi_f)=\Lambda(s+(k-1)/2,f).$$ Here the left hand side is independent of the **weight ** of the modular form and the right hand side does depend on the weight of $$f$$. Intuitively, the weight is "packaged" in the lifting formula given above, and finally the weight $$k$$ appeared in the action of $$K_\infty = O(2)\subset G(\mathbb{R})$$, namely the type of the archimedean $$(\mathfrak{g}_\infty,K_\infty)$$-module. If the modern adelic theory generalises the classical theory, then I feel confused that left hand side is independent of the weight $$k$$. Because the normalisation of the automorphic $$L$$-function only requires a translation $$1/2$$ in the variable $$s$$ which has nothing to do with $$k$$, but the classical $$L$$-function is not the case!

Can any veteran explain this to me in some detail? I found the notes by Buzzard before: https://www.ma.imperial.ac.uk/~buzzard/maths/research/notes/automorphic_forms_for_gl2_over_Q.pdf but it seems not to explain my question here: the central character comes from the "nebentype" character of the modular form, not the weight?

It is the $$j(g_\infty,i)^{-k}$$ factor which is doing the weight normalization for the automorphic side. The obtained function $$\phi:GL_2(\Bbb{A_Q})\to \Bbb{C}$$ is left $$GL_2(\Bbb{Q})$$-invariant (its diagonal embedding into $$GL_2(\Bbb{A_Q})$$)
Easier to check first that the function $$F:SL_2(\Bbb{R})\to \Bbb{C},\qquad F(g)= g'(i)^{k/2} f(gi)$$ is left $$\Gamma_0(N)$$ invariant.
$$\gamma'(z)=(cz+d)^{-2}$$ so $$f(\gamma z)= \gamma'(z)^{-k/2} f(z)$$ and
$$F(\gamma g) = (\gamma g)'(i)^{-k/2} f(\gamma gi) = (\gamma'(gi) g'(i))^{k/2} \gamma'(gi)^{-k/2} f(gi)=F(g)$$