Let $G = \operatorname{GL}_2$, and let $V = L^2(Z(\mathbb A)G(\mathbb Q) \backslash G(\mathbb A),\omega)$, for $\omega$ a character of the ideles $\mathbb A^{\ast}$, identified with a central character. For a character $\mu$ of $\mathbb A^{\ast}/\mathbb Q^{\ast}$ such that $\mu^2 = \omega$, let $\chi = \mu \circ \operatorname{det}$. Then $\chi$ is an element of $V$.

How does one see (or intuit) that $\chi$ should be orthogonal to all cusp forms in $V$? Recall a cusp form is an element $f \in V$ such that

$$\int\limits_{N(\mathbb A)/N(\mathbb Q)} f(ng) dn = 0$$

for almost all $g \in G(\mathbb A)$, where $P = TN$ is the usual Borel subgroup of $G$ with its Levi decomposition.

My idea was to take the usual maximal compact subgroup $K$ of $G(\mathbb A)$ and say that, just as we have $\int\limits_{G(\mathbb A)} = \int\limits_{N(\mathbb A)} \int\limits_{T(\mathbb A)} \int\limits_K$, we should also have something like

$$\int\limits_{Z(\mathbb A)G(\mathbb Q) \backslash G(\mathbb A)} f(g) \overline{\chi(g)}dg = \int\limits_{N(\mathbb Q) \backslash N(\mathbb A)} \space \int\limits_{ Z(\mathbb A) T(\mathbb Q) \backslash T(\mathbb A)} \space \int\limits_{[ Z(\mathbb A) G(\mathbb Q) \cap K] \backslash K} f(ntk) \overline{\chi(ntk)} \space dk dt dn$$ which should come out to $0$, using the fact that $\chi(ng) = \chi(g)$ for all $n \in N(\mathbb A)$. The problem is, I don't know if the integration over $Z(\mathbb A)G(\mathbb Q) \backslash G(\mathbb A)$ should decompose like this. I tried to prove this for some time, but I am not sure whether this step can be made sound. Is such a decomposition of measures ''legit?'' I would be grateful for any explanation or reference.