TITLE: Where do the real analytic Eisenstein series live?
QUESTION [17 upvotes]: In obtaining the spectral decomposition of $L^2(\Gamma \backslash G)$ where $G=SL_2(\mathbb{R})$, and $\Gamma$ is an arithmetic subgroup (I am satisfied with $\Gamma = SL (2,\mathbb{Z})$) we have a basis of eigenfunctions of the hyperbolic Laplacian, and orthogonal to that we have the space spanned by the incomplete Eisenstein series, 
$$
E(z,\psi) = \sum_{\Gamma_\infty \backslash \Gamma} \psi (\Im(\gamma z)) = \frac{1}{2\pi i}\int_{(\sigma)} E(z,s)\tilde{\psi}(s)\mathrm{d}s 
$$
where $\psi \in C_c^\infty(\mathbb{R}^+)$, $\tilde{\psi}$ is its Mellin transform, and $E(z,s) = \sum_{\gamma \in \Gamma_\infty \backslash \Gamma} \Im(\gamma(z))^s$ is the usual Eisenstein series.
My question is, where does $E(z,s)$ itself live with respect to the vector space $V = L^2(\Gamma \backslash G)$ which can be considered as the vector space of the right regular representation of $G$, and what is this parameter $s$?
A similar question of course goes for $\mathbb{R}$, where does $e^{2\pi i x}$ live with respect to $(L^2(\mathbb{R}), \rho)$?
I would appreciate a representation theoretic flavored answer, that is why I mentioned representations, but any other answer would also be an addition to my understanding of this.
In general, is there an associated space to $(V,\pi)$, an automorphic representation, such that the elements of the vector space are of moderate or rapid growth, instead of decay.

REPLY [2 votes]: I can give an additional point of view, which is coming from the theory of parabolic induction. Parabolic induction plays a prominent role in representation theory and gives you a better intuition for the higher rank situation. This point of view is often better stressed in the adelic theory than in the classical picture.
Let $G =PSL_2(\mathbb{R})$ with standard parabolic $B$ and $\Gamma$ a cofinite lattice.
My intuition is that te analytic Eisenstein series on  $\Gamma \backslash G$ are vectors of the induced representation:
$$ Ind_{\Gamma N}^G 1,$$
but there is one major issue with this, namely that $\Gamma N$ is not a group.
Rigorously seen they are given as $P$ series, i.e.
$$ E: C_c^\infty(N \backslash G) \rightarrow L^2(\Gamma \backslash G),$$
by defining the $B$ series 
$$E(f) (g)= \sum\limits_{B \cap \Gamma \backslash \Gamma} f(\gamma g).$$
The image of $E$ generates a dense subspace of the orthocomponent to the cuspidal forms.
Now, we one notices that $C_c^\infty(N \backslash G)$ is a dense subspace of $Ind_N^G 1$. Induction by steps gives a decomposition
$$ Ind_N^G 1 \cong Ind_B^G Ind_N^B 1.$$
Now for $Ind_N^B 1 \cong L^2(B/N) = L^2(M)$, where $M$ are the diagonal matrices. Pontryagin duality gives you a direct integral decomposition of $L^2(M)$, and we have as a result
$$ Ind_N^G 1 \cong \int\limits_{\Re s = 0}^{\oplus} Ind_B^G | \cdotp |^s. $$
Certainly one hopes that $E$ extends to $Ind_B^G | \cdotp |^s$, but convergence only happens $\Re s >1/2$, and the operators has to be defined by analytic continuation to make sense on $\Re s = 0$.
Perhaps it useful to give at least one definition here: functions $f \in Ind_B^G | \cdotp |^s$ are defined as $f(bg) = |b_{1,1} / b_{2,2}|^{s+1/2} f(g)$ for $b \in B$ with $f|_K\in L^2(K)$ for $K= PSO(2)$.