Let $\mathbb{G}$ be a reductive group defined over a number field $K$, let $Z$ be its center, and let $\mathbb{A}:=\mathbb{A}_K$ be the ring of adeles of $K$. Reasonably, we care about the $\mathbb{G}(\mathbb{A})$-representation: $L^2(\mathbb{G}(K)\backslash \mathbb{G}(\mathbb{A}))$. It naturally contains the sub-representations $$L^2(Z(\mathbb{A})\mathbb{G}(K)\backslash\mathbb{G}(\mathbb{A}),\omega):=\{f\in L^2(Z(\mathbb{A})\mathbb{G}(K)\backslash\mathbb{G}(\mathbb{A}))|\,\,\,|f|\in L^2(Z(\mathbb{A})\mathbb{G}(K)\backslash \mathbb{G}(\mathbb{A})), \forall z\in Z(\mathbb{A}), g \in \mathbb{G}(\mathbb{A})\,\,\, f(zg)=\omega(z)f(g)\} $$
for every $\omega$ a unitary character of $Z(\mathbb{A})$. In fact $L^2(\mathbb{G}(K)\backslash \mathbb{G}(\mathbb{A}))$ is the direct integral of these subrepresentations.
I understand that it is generally desirable to deal with $L^2(\mathbb{G}(K)\backslash \mathbb{G}(\mathbb{A}))$ by decomposing it into the cuspidal part, which is going to be discrete, and the Eisenstein part, which is (I think!) continuous. In order to define this cuspidal part, people define $L^2_0(Z(\mathbb{A})\mathbb{G}(K)\backslash \mathbb{G} (\mathbb{A}),\omega)$ to be the subrepresentation of $L^2(Z(\mathbb{A})\mathbb{G}(K)\backslash \mathbb{G} (\mathbb{A}),\omega)$ of all of the functions $f$ such that for every $K$-parabolic subgroup $\mathbb{P}$ of $\mathbb{G}$, whose unipotent radical we will call $N$, satisfies that for almost all $g\in\mathbb{G}(\mathbb{A})$ the integral $\int_{N(K)\backslash N(\mathbb{A})} f(gn)dn$ is $0$.
The definition of a cuspidal representation is then an irreducible unitary subrepresentation of $L^2_0(Z(\mathbb{A})\mathbb{G}(K)\backslash \mathbb{G} (\mathbb{A}), \omega)$ for some central character $\omega$.
I feel that I really do not understand the intuition behind the condition with the parabolic subgroups. Parabolic subgroups and their unipotent radicals seem like very formal constructions to me, but I bet there is some geometric intuition that I'm missing. Is there some geometry that should be in the back of my mind that explains the condition $\int_{N(K)\backslash N(\mathbb{A})} f(gn)dn=0$? How does this condition relate to being zeros at the cusps via the classic definition of cusp-forms?
Best Answer
In addition to Paul Garrett's answer, I address your last paragraph in a special example:
Strong approximation gives a homeomorphism $SL_2(Z) \backslash H \cong Z(A) GL_2(Q) \backslash GL_2(A) / \prod_p GL_2(Z_p) \times O(2)$.
Lets $f$ corresponds to $\tilde{f}$. This translates
$$ \int_{0}^1 f( y + t)\; d t = \int\limits_{N(F)\backslash N(A)} \tilde{f}(ng_y)\; dn.$$
This implies that the zero-th Fourier coefficient vanishes, but that does not imply that the functions vanish at the cusps for Maass forms. It only does for modular forms.