# Why are characters orthogonal to cusp forms?

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.

• You might want "linear characters" or something in the title. (Although I guess these are the only sensible characters in the adèlic case, in the local case we have characters also for non-1-dimensional representations.) Mar 18, 2021 at 16:12

We may assume that $$\omega$$ is unitary by tensoring a central character.
If $$\pi$$ is one of the irreducibles, then orthogonal projection from $$\chi$$ to $$\pi$$ is G-equivariant and not isomorphism, hence 0, by Schur’s lemma, so $$\chi$$ is orthogonal to L^2-cusp forms.