I am looking at how one maps the space of cusp forms $S_k(N,\psi)$ into the space of automorphic forms on $GL_2$ over the Adeles. And I have a question about the definition of $K$-finiteness. I am following Gelbart's Automorphic forms on adele groups.

My problem is the following. When one takes a classical cusp form and then makes an automorphic form out of this, the resulting function $\phi$ satisfies the following (among other conditions):

1) $\phi(gz)=\phi(zg)=\psi(z)\phi(g), \qquad \text{for all } z \in Z_\mathbb{A}$;

2) $\phi(gk_0)=\phi(g)\psi(k_0)$ and $\phi(gr(s))=e^{-iks}\phi(g)$ for $k_0 \in \prod_{p< \infty} K_p'$ and $r(s)= \left ( \begin{smallmatrix} \cos(s) &-\sin(s)\\ \sin(s)&\cos(ss) \end{smallmatrix} \right )$

with $$K_p'= \{ \left ( \begin{smallmatrix} a&b\\ c&d \end{smallmatrix} \right) \in K_p | c \equiv 0 \mod N\mathbb{Z}_p \},$$ with $N$ any positive integer. ($K_p=GL_2(\mathcal{O}_p)$, with the usual convention when $p$ is infinite)

Now my guess is that these conditions are the right $K_{\infty} \prod_{p < \infty} K_p$-finite, condition in the definition of an automorphic form (See Gelbart p.44). But is seems to me that the above (1),(2) only show its $$S \mathcal{O}_2(\mathbb{R})\prod_{p < \infty} K_p'-finite.$$ And even though $K_p'=K_p$ for almost all $p$ what about the places where this isnt true? Is the space of right translates still finite-dimensional? and why? (Also I think I can do the infinite part of this to go from $S \mathcal{O}_2(\mathbb{R})-finite$ to $\mathcal{O}_2(\mathbb{R})-finite$, but its the finite places that have we worried.)

(I hope this question makes sense)

Thank you


1 Answer 1


The index of $\prod_p K'_p$ in $\prod_p K_p$ is finite, and so if $V$ is the space of translates by the first group (which you agree is finite dim'l), then the space of translates by the second (larger) group can be written as $\sum_{k} k \cdot V,$ where $k$ runs over a set of coset reps for the first group in the second. Since the index is finite, this is a finite sum of finite dim'l spaces, hence is again finite dim'l. (The passage from $SO(2)$ to $O(2)$, which you say you are happy with, is done the same way: $SO(2)$ has finite index in $O(2)$.)

  • $\begingroup$ Ah I see, that makes sense. Thank you very much. $\endgroup$ Jan 10, 2014 at 12:41

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