# Truncation and weighted orbital integrals in hyperbolic term of trace formula for $\mathrm{GL}(2)$

$$\DeclareMathOperator\GL{GL}\DeclareMathOperator\PGL{PGL}$$I am looking at Gelbart--Jacquet's article in the first Corvallis volume (the article entitled Forms of $$\GL(2)$$ from an analytic point of view).

To wrap my head around exactly what is going on with truncation, I am working out the case where the test function is cuspidal (also because this is the situation we are in when we try to derive the Eichler–Selberg trace formula for traces of Hecke operators from the general trace formula for $$\GL(2)$$).

Basically, my question is about why truncation is necessary even when the test function is nice enough to make all the terms depending on the truncation parameter vanish (it seems to me like this is true: no matter what, you end up with a weighted orbital integral which doesn't depend on the truncation parameter but does involve the height function — the only way I know of how to think about this is as arising from truncation).

Let $$G = \GL_2$$, $$N$$ be the upper-triangular unipotent radical, $$M$$ the diagonal matrices, and $$P = MN$$ the upper-triangular Borel subgroup. Let $$\varphi$$ be the test function on $$G$$, which is assumed to be cuspidal, smooth, and compact modulo center.

In Gelbart--Jacquet, the truncated hyperbolic term (which is ultimately integrated over $$g$$ in $$\PGL_2(\mathbf{Q})\backslash \PGL_2(\mathbf{A}_\mathbf{Q})$$) is $$\sum_{\gamma \neq 1} \varphi(g^{-1}\gamma g) - \sum_{\xi \in P(\mathbf{Q})\backslash G(\mathbf{Q})} \int_{N(\mathbf{A}_\mathbf{Q})} \sum_{\alpha \in \mathbf{Q}^\times - \{1\}} \varphi\left(g^{-1}\xi^{-1}\left(\begin{matrix}\alpha & 0 \\ 0 & 1\end{matrix}\right)n\xi g\right)\chi_{[T, \infty)}(H(\xi g))\, dn$$ where $$\chi_{[T, \infty)}$$ is the characteristic function of being $$\geq T$$, $$H$$ is the usual logarithmic height function using the Iwasawa decomposition, and the sum is over non-identity hyperbolic elements $$\gamma$$.

Anyway, I know/thought that the point is that this thing is supposed to be absolutely integrable (thanks to the construction, which is essentially subtracting off the constant term in the Fourier expansion in a neighborhood except averaged so that it maintains automorphicity), and then one can use Fubini when integrating to end up with $$(\log T)\mathrm{vol}(\mathbf{Q}^\times \backslash \mathbf{A}_\mathbf{Q}^{\times, 1})\int_K \int_{N(\mathbf{A}_\mathbf{Q})} \sum_{\alpha \in \mathbf{Q}^\times - \{1\}} \varphi\left(k^{-1}\left(\begin{matrix}\alpha & 0 \\ 0 & 1\end{matrix}\right)nk\right)\, dn dk$$ minus $$\frac{1}{2}\mathrm{vol}(\mathbf{Q}^\times \backslash\mathbf{A}_\mathbf{Q}^{\times, 1})\int_K \int_{N(\mathbf{A}_\mathbf{Q})} \sum_{\alpha \in \mathbf{Q}^\times - \{1\}} \varphi \left(k^{-1}n^{-1}\left(\begin{matrix}\alpha & 0 \\ 0 & 1\end{matrix}\right)nk\right)\log H(wnk)\, dndk,$$ where $$w$$ is the matrix you conjugate a diagonal matrix by to switch the two entries. If $$\varphi$$ was nice enough to begin with, then the first term in this final result vanishes (not surprising -- the truncation doesn't matter, so any term depending on $$T$$ has to vanish). On the other hand, for the same reason, the truncation in the thing we started out with also has no effect. In particular, the correction term in the integrand (i.e. the thing being subtracted in the first display equation above) vanishes, and $$\sum_{\gamma \neq 1}\varphi(g^{-1}\gamma g)$$ is supposed to already be absolutely integral modulo center. But this can't be right -- then if you carry out the computation, the integral you end up with won't have anything to do with the height function $$H$$. Indeed, you end up with (by the Iwasawa decomposition)

$$\int_K \int_{N(\mathbf{A}_\mathbf{Q})} \int_{\mathbf{Q}^\times \backslash \mathbf{A}_\mathbf{Q}^\times} \sum \cdots \, da dn dk,$$ the inside integral of which diverges badly if the sum on the inside ever doesn't vanish.

So am I mistaken that truncation has no effect on the integrand for nice test functions? I suppose that I haven't checked that you can commute the sum and the integral over $$N$$. However, it definitely has to be true that the final orbital integral doesn't depend on $$T$$ when the test function is cuspidal, which would suggest that I am right about the vanishing of both of the correction terms (that is, the terms that depend on $$T$$ in both the original integrand and final value of the orbital integral). All of this is made more confusing by the fact that Knightly-Li's book Traces of Hecke operators says multiple times that truncation has no effect on the thing being integrated, and that the use of truncation is purely pedagogical.

Apologies for the long question. I think there is something very simple that I do not understand, so hopefully an experienced person will easily be able to tell me where I am going wrong.