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.

  • $\begingroup$ 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.) $\endgroup$
    – LSpice
    Mar 18, 2021 at 16:12

1 Answer 1


For the second measure theoretic question, I don’t know the answer, but I think, for the first question about the orthogonality is resolved as follows:

We may assume that $\omega$ is unitary by tensoring a central character.

By the Gelfand-Piatetski-Shapiro theorem, the space of L^2 cusp forms is decomposed to a direct sum of irreducible unitary representations with finite multiplicities (e.g. use Corvallis, Borel-Jacquet, 2.2 and 4.3, Wallach real reductive groups I 1.6.6).

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.


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