Let $G$ be an adjoint semisimple group over $\mathbb Q$ with parabolic subgroup $P = MN$ in good position relative to a compact subgroup $U= \prod\limits_v K_v$ of $G(\mathbb A)$. Let $L$ be the space of square integrable functions on $G(\mathbb Q) \backslash G(\mathbb A)$ which are invariant on the right by $U$. In Euler Products, Langlands defines a cusp form in $L$ to be an element $\phi$ satisfying

$$\int\limits_{N(\mathbb Q) \backslash N(\mathbb A)} \phi(ng)dn = 0\tag{1}$$

for almost all $g \in G(\mathbb A)$. However, it is not clear to me why the left hand side converges at all. All we know is that

$$\int\limits_{G(\mathbb Q) \backslash G(\mathbb A)} |\phi(g)|^2 dg < \infty$$ Using the Iwasawa decomposition, we can write $G(\mathbb A) = N(\mathbb A)M(\mathbb A)K$, so that, at least formally,

$$\int\limits_{G(\mathbb A)} \phi(g)dg = \int\limits_{M(\mathbb A)} \int\limits_{N(\mathbb A)} \int\limits_K \phi(nmk) \delta_P(m) dk dm dn = \operatorname{vol}(K)\int\limits_{M(\mathbb A)} \int\limits_{N(\mathbb A)}\phi(nm) \delta_P(m)dndm$$

One can probably finagle from here something like:

$$\int\limits_{G(\mathbb Q) \backslash G(\mathbb A)}|\phi(g)|^2 dg = \operatorname{vol}(K)\int\limits_{M(\mathbb Q)\backslash M(\mathbb A)} \int\limits_{N(\mathbb Q) \backslash N(\mathbb A)} |\phi(nm)|^2 \delta_P(m)dn dm$$

from which we should have

$$\int\limits_{N(\mathbb Q) \backslash N(\mathbb A)} |\phi(ng)|^2 dn < \infty$$ for almost all $g \in G$. However, this says nothing about the convergence of $n \mapsto \phi(ng)$, only of $n \mapsto |\phi(ng)|^2$.

  • $\begingroup$ Isn't $\mathrm{N}(\mathbb{Q}) \backslash \mathrm{N}(\mathbb{A})$ compact? $\endgroup$ Aug 14, 2019 at 19:59
  • 1
    $\begingroup$ Yes, but $\phi$ is not necessarily continuous there $\endgroup$
    – D_S
    Aug 14, 2019 at 21:17
  • $\begingroup$ Hmmm, I guess usually one only considers smooth functions. $\endgroup$ Aug 14, 2019 at 21:18
  • 1
    $\begingroup$ Maybe Langlands should say the cusp forms are the closure of the space generated by such smooth functions $\endgroup$
    – D_S
    Aug 14, 2019 at 21:19
  • 2
    $\begingroup$ It's alright to use arbitrary $L^2$ functions here. Like you wrote, the $L^2$ norm over $N({\mathbb Q})\backslash N({\mathbb A})$ is finite for almost all $g$, then, as the quotient is compact, the constant function is $L^2$, and therefore the inner product of $|\phi|$ and the constant function exists for almost all $g$. $\endgroup$
    – user130903
    Aug 16, 2019 at 5:06

2 Answers 2


I think its convergence is proven by the following method;

claim 1 Let $G$ be a 2nd countable locally compact topological group and $\Gamma$ be a discrete subgroup.
Also, let $A$ be a measureble set of $G$ and $B$ be a measurable set of $\Gamma \backslash G$ such that projection of $A$ contains $B$.
Then there exists a Borel measurable subset $A'$ of $A$ such that $A'$ projects $B$ one-to-one onto.

The proof is trivial.

claim 2 Let $f$ be the element of $L^1_{loc}$(G($\mathbb{Q}$)\G($\mathbb{A}$)), then its constant term $f_P$ along parabolic P is in $L^1_{loc}$(U($\mathbb{A}$)M($\mathbb{Q}$)\G($\mathbb{A}$)).

Sketch of the proof
We may assume that $f$ is positive. Take a compact set $C$ in U($\mathbb{A}$)M($\mathbb{Q}$)\G($\mathbb{A}$).
By Fubini theorem for quotient measures,

\begin{align} \int_{U(\mathbb{A})M(\mathbb{Q})\backslash G(\mathbb{A})} \chi_{C}(g) \int _{U(\mathbb{Q}) \backslash U(\mathbb{A})}f(ug) du dg & = \int_{P(\mathbb{Q}) \backslash G(\mathbb{A})}f(g)\chi_{C}(g)dg \\ & =\int_{G(\mathbb{Q})\backslash G(\mathbb{A})} f(g) \int _{P(\mathbb{Q}) \backslash G(\mathbb{Q})} \chi_{C}(\gamma g)d\gamma dg \\ \end{align} Let $C_1$ be $G(\mathbb{Q})\backslash G(\mathbb{Q})C$, then it is compact in G($\mathbb{Q}$)\G($\mathbb{A}$) and if the above integrand is not zero, then $g$ in $C_1$.
Moreover, we can take the measurable set $C_2$ in G($\mathbb{A}$), such that $C_2$ projects $C_1$ and $C_2$ is relatively compact by claim 1.

Hence, the above formula equals \begin{align} \int_{G(\mathbb{Q})\backslash G(\mathbb{A})} \chi_{C_{1}}(g)f(g) \int _{P(\mathbb{Q}) \backslash G(\mathbb{Q})} \chi_{C}(\gamma g)d\gamma dg & = \int_{C_{2}} f(g) \int _{P(\mathbb{Q}) \backslash G(\mathbb{Q})} \chi_{C}(\gamma g)d\gamma dg \\ & = \sharp \{\gamma \in P(\mathbb{Q}) \backslash G(\mathbb{Q}) : \gamma \in CC_{2}^{-1} \} \|f\|_{C_1} \end{align} The term $\sharp \{\gamma \in P(\mathbb{Q}) \backslash G(\mathbb{Q}) : \gamma \in CC_{2}^{-1} \}$ essentially depends only on $C$ and is finite because discrete compact set is finite. Hence the claim follows.
(I have not seen the complete proof of this fact and think this proof by myself. So it may contain some mistakes...)


You are entirely correct that there are several analytical issues with a naive presentation of this... although the naive presentation does present the intention, which is the most important thing.

So, yes, if we're trying to make "constant term (map)" a reasonable thing, we surely want to specify what space it maps from, and what to, and surely these (vector) spaces of functions should have topologies (or, depending on taste, bornologies...) to make the constant term map(s) continuous.

I hesitate to inject here my own "personal" precisification of this, but/so anyone who wants to see one way to make it all precise (without tooooo much silliness, I hope) can see my CUP book, with a legal version at http://www.math.umn.edu/~garrett/m/v/current_version.pdf

  • 1
    $\begingroup$ Yes, for example ... surely not the only way... the way that I myself finally feel secure/safe with "the constant term map" is as a map from $L^2$ (or even a bigger space) to distributions on $U_k\backslash G_{\mathbb A}$ (or whatever). After that, yes, we can (as in my book) prove continuity in a finer topology on the target space, but it's not obvious, etc. Mercifully, everything works fine, in the usual quasi-miraculous idiot-proof-ness of most of mathematics, so that, in particular, obliviousness of these issues seems not to easily lead to disaster. :) $\endgroup$ Oct 27, 2020 at 23:36
  • 1
    $\begingroup$ And, yes, $L^1_{loc}$'s can work, too, but I think these are just an approximation to spaces of distributions... I feel that spaces of distributions are more robust. Tastes vary. :) $\endgroup$ Oct 27, 2020 at 23:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.