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I am trying to understand the following situation for $G=GL(2)$, when going from the compact trace formula to the non-compact case.

The integral over $G(\mathbb{A})^1_\gamma \backslash G(\mathbb{A})^1$ may diverge for a given conjugacy class $\gamma$. Say $\gamma$ is the matrix of the translation by $1$. Then $G(\mathbb{A})_\gamma$ is the set of upper triangular matrices with same diagonal entries. I wan to compute the integral $$\int_{G(\mathbb{A})^1_\gamma \backslash G(\mathbb{A})^1} f(x^{-1}\gamma x)dx$$ to see why it diverges. In Arthur's notes, it reduces to $$\int_{G_\gamma(\mathbb{A}) \backslash P(\mathbb{A}) } \int_{P(\mathbb{A}) \backslash G(\mathbb{A})} f((pk)^{-1} \gamma pk) dpdk$$ and this reduces apparently to a constant times $$\prod_p (1-p^{-1})^{-1}$$ which is infinite (except when the constant is zero). I do not understand this computation (that I would also like to see for more general groups).

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  • $\begingroup$ Have you read Gelbart's book/notes or Knapp's article? $\endgroup$
    – Kimball
    Commented Mar 22, 2023 at 19:14

1 Answer 1

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I take it that the question is "Why does this iterated integral diverge?," but correct me if that is not the question.

The issue here is the outer integral: when $\gamma$ is as you describe, then if $P$ is the group of upper triangular matrices, the quotient $G_\gamma\backslash P$ is a rank $1$ split torus, the adelic integral over which is infinite. Things get more complicated for regular unipotent elements in higher rank, but the problem is always that the orbit is not closed in $G$. In this case, the matrix $I_2$ lies in the closure for example.

This non-closed orbit issue is morally why the adelic integral diverges. This is visible in the calculation, so I'll quickly sketch it. The idea is that if this converges, it converges to the product of all the local integrals (which we will see do converge). That is, we can assume that $f= \prod_p f_p$ is a pure tensor in the space of test functions, and consider the local integrals $$ \int_{G_\gamma(\mathbb{\mathbb{Q}_p}) \backslash P(\mathbb{Q}_p) } \int_{P(\mathbb{Q}_p) \backslash G(\mathbb{Q}_p)} f_p((bk)^{-1} \gamma bk) dkdb $$ for each $p\leq \infty$. For this to make sense, we need $f_p = \mathbb{1}_{GL_2(\mathbb{Z}_p)}$ to be the characteristic function of the integral points for almost all $p$.

The inner integral is compact since $P\backslash GL_2$ is just a projective line, so won't impact whether the iterated integral diverges or not. In particular, there is a smooth function $f'_p$ (just shoving the compact integral into the notation) such that $$ \int_{G_\gamma(\mathbb{Q}_p) \backslash P(\mathbb{Q}_p) } \int_{P(\mathbb{Q}_p) \backslash G(\mathbb{Q}_p)} f_p((bk)^{-1} \gamma bk) dkdb =\int_{G_\gamma(\mathbb{Q}_p) \backslash P(\mathbb{Q}_p) } f'_p(b^{-1} \gamma b) db. $$ Note that we still have $f'_p = \mathbf{1}_{GL_2(\mathbb{Z}_p)}$ for almost every prime $p$. At such primes, the integral is $$ \int_{\mathbb{Q}_p^\times } \mathbb{1}_{GL_2(\mathbb{Z}_p)}\left(\begin{pmatrix}t& \\ &1\end{pmatrix} \begin{pmatrix}1&1 \\ &1\end{pmatrix}\begin{pmatrix}t^{-1}& \\ &1\end{pmatrix}\right) dt =\int_{\mathbb{Q}_p^\times } \mathbb{1}_{GL_2(\mathbb{Z}_p)}\left(\begin{pmatrix}1&t \\ &1\end{pmatrix}\right) dt . $$ If we normalize the Haar measure on $\mathbb{Q}_p^\times$ so that $vol(\mathbb{Z}_p^\times)=1$, this last integral is $\sum_{i=0}^\infty p^{-i}=(1-p^{-1})^{-1}$, converging to the local zeta value $\zeta_p(1)$. Notice that this integral ``sees'' the limit point of $G_\gamma\backslash G$, since as $t\to 0$, $$\begin{pmatrix}1&t \\ &1\end{pmatrix} \longmapsto \begin{pmatrix}1& \\ &1\end{pmatrix}. $$

Formally taking the product over all $p$, you see the divergent product $\prod_p(1-p^{-1})^{-1}$ times some factor coming from finitely many places where $f'_p\neq\mathbb{1}_{GL_2(\mathbb{Z}_p)}$. This factor could vanish for a given $f$, but it doesn't always so that the integral generally diverges.

In general, orbital integrals for non-semisimple elements will be integrals over non-closed orbits, so even though the local integrals make sense, the global need some notion of regularization to give a well-defined distribution.

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