Square integrable functions on $\Gamma \backslash G$

I am trying to understand proposition 2.1.6 in Bump's book Automorphic forms and Representations.

Let $G=GL(2,\mathbb{R})^+$ and define $G_1=G/Z^+$, where $Z^+$ denotes the center, and define $\Gamma=SL(2,\mathbb{Z})$. Let $\chi: \Gamma \to S^1$ be a group homomorphism. He writes, "Let $L^2(\Gamma\backslash G, \chi)$ be the space of measurable functions satisfying $$f(\gamma g u)=\chi(\gamma)f(g)\qquad \gamma\in\Gamma, u\in Z^+,g\in G$$ that are square integrable with respect to Haar measure on $G_1$."

Clearly no function which is periodic with respect to $\Gamma$ can be square integrable on $G_1$.

Can someone please explain to me what the right definition should be.

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I couldn't check in the book. But it should be easy: instead of square integrable on $G_1$'', require square integrable on $\Gamma\backslash G_1$ (observe that the modulus of $f$ is left-invariant under $\Gamma$). – Alain Valette Jan 21 '12 at 18:06
$\chi$ is supposed to be the central character of $f$, so $\chi(\gamma)$ should be $\chi(u)$. – B R Jan 21 '12 at 19:11
@ BR, no Bump wrote it like this. – Marc Palm Jan 25 '12 at 17:10
pm, huh! You are correct. I had never seen $L^2(\Gamma\backslash G,\chi)$ used in that way before. I want to add to your second comment that since $G$ is unimodular, any quotient of $G$ by a discrete subgroup will have a unique invariant measure (since discrete groups are unimodular). And we typically ask for $\Gamma$ to be co-finite, instead of co-compact (to allow $SL_2(\mathbb Z)$ and congruence subgroups). – B R Jan 25 '12 at 19:19
Oh yes, clearly $SL(2, \mathbb{Z})$ is not cofinite! I am not sure why I was talking about cocompact lattices, but unimodularity of the big group comes for free in this case. So for the OP, the invariant measure exists. Sorry for the confusion. @BR: You propbably have seen $L^2( \Gamma \backslash G, \chi)$: 1. this is an induced representation, 2. sometimes one considers Maass/modular forms $\Gamma_0(N)$ with a character of $\Gamma_0(N)$, since $L^2( \Gamma_1(N) \backslash G) \cong \bigoplus_{\chi : \Gamma_1(N) \backslash \Gamma_0(N)} L^2( \Gamma_0(N) \backslash G, \chi)$. – Marc Palm Jan 26 '12 at 6:33

The right definition goes as follows: An element $f$ in $L^2( \Gamma \backslash G , \chi)$ is a function $f : G \rightarrow \mathbb{C}$ with measurable with respect to the Haarmeasure

$$f(\gamma z g) = \chi(\gamma) f(g),$$

and

$$\int\limits_{\Gamma Z \backslash G} |f(g)|^2 d \mu(g) < \infty.$$

Here $\mu(g)$ is the unique (up-to-scaling) right invariant Radon measure on $\Gamma Z \backslash G$.

For the existsence and uniqueness of such measure, you can consider Theorem 1.5.2 on page 22 in Deitmar-Echterhoff "Principles on Harmonic Analysis", but this theorem should be found in any book about the analysis of locally compact groups. Actually, I think it should be in Bump as well.

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