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Let $\mathrm{G}$ be a reductive group over a number field $F$, but for simplicity we can think about $\mathrm{G}=\mathrm{GL_n}$ for $n>2$ and $F =\mathbb{Q}$.

Then for an automorphic form, $\varphi : \mathrm{G}(F)\backslash\mathrm{G}(\mathbb{A}) \to \mathbb{C}$ is said to be cuspidal if for each parabolic subgroup $\mathrm{P}=\mathrm{MU}$,

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\int_{\mathrm{U}(F)\backslash\mathrm{U}(\mathbb{A})} \varphi(ug)du = 0$,

where $\mathrm{U}$ is the unipotent radical of $\mathrm{P}$.

In the case that $\mathrm{G}=\mathrm{GL_2}$, then we can unravel the adelic language to the more classical setting, where this becomes

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\int_0^1f(x+iy)dx = 0$, (ie: the constant term vanishes)

where $f$ is a Maass form which for simplicity I will assume has level 1. In this setting, we view $f$ as a function on the upper half plane,

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad H \cong Z\backslash\mathrm{GL_2}(\mathbb{R})/\mathrm{SO}(2)$,

and $H$ comes equipped with a "canonical" hyperbolic metric $ds^2 = \frac{dxdy}{y^2}$. Passing to the quotient, $\mathrm{SL_2}(\mathbb{Z})\backslash H$, we get a punctured sphere with one geometric cusp at infinity, and the requirement that $f$ be a cusp form reduces to $f$ "vanishing at the cusp".

For more general $\mathrm{G}$ (eg: $\mathrm{G}=\mathrm{GL_3}$), I am curious about analogous geometric understanding of unipotent radicals of various parabolic subgroups corresponding to the "cusp at infinity". I don't know if this is better to think about this on the adelic quotient, $\mathrm{G}(F)\backslash\mathrm{G}(\mathbb{A})$, or perhaps on a locally symmetric space like $ Z\backslash\mathrm{G}/\mathrm{K}$.

Specifically,

Is there a geometric way of understanding how the unipotent radicals correspond to "cusps" beyond simply as failure of a fundamental domain to be compact, perhaps with respect to some nice metric, for groups beyond $\mathrm{GL_2}$?

I am interested in answers in either the classical or adelic language.

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    $\begingroup$ For groups of rank $\geq 2$, the boundary of the locally symmetric space has more structure. The exact relation between the structure of "the cusp at infinity" and the unipotent radicals of parabolic subgroups is captured exactly by the Borel-Serre compactification. This is a manifold with corners, where the corners are built exactly from unipotent radicals of parabolic subgroups. $\endgroup$ – Matthias Wendt Feb 25 '15 at 21:18
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    $\begingroup$ @MatthiasWendt, could you please expand on this? I googled the Borel-Serre compactification, but it is unclear to me what you mean, and this sounds very interesting $\endgroup$ – WSL Feb 25 '15 at 22:55
  • $\begingroup$ See dept.math.lsa.umich.edu/~lji/head.pdf from page 391. In page 392, there is something similar to Matthias comment. The boundary face of a parabolic subgroup $e (P)$ collapses the unipotent radical. $\endgroup$ – user40276 Feb 27 '15 at 2:12
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It's actually a lot easier to write a short comment than to say something more precise. In any case, I can only say something in the classical language of locally symmetric spaces.

The shortest answer is that the geometric understanding of the cusps is revealed in the various compactifications of locally symmetric spaces that can be considered. The book of Borel and Ji (the link was already given in user40276's answer) gives a good overview of the compactifications that have been considered and their relations.

For the question, there seem to be at least two compactifications which are relevant.

On the one hand, the metric setting you described generalizes to the geodesic compactification. Every locally symmetric space $\Gamma\backslash G/K$ inherits a metric, and you can compactify it by adding geodesics going off to infinity (but probably this works better if you consider torsion-free arithmetic groups $\Gamma$). The structure at infinity that you add in the compactification is the Tits building for the algebraic group $G(\mathbb{Q})$ modulo the action of $\Gamma$. In the case of $SL_2\mathbb{Z}$ you end up adding one point, because $\mathbb{Z}$ having class number one implies that the action of $GL_2\mathbb{Z}$ is transitive on the parabolic subgroups of $GL_2\mathbb{Q}$. However, the unipotent radicals are not really visible in this compactification.

To see some relation to unipotent subgroups, you can consider the Borel-Serre compactification. One of the main features of the Borel-Serre compactification is that the inclusion of the locally symmetric space into its compactification is a homotopy equivalence. The compactification adds part of the Langlands decomposition of the $\mathbb{R}$-points of parabolic subgroups modulo the action of the arithmetic subgroup $\Gamma$. In the case of $SL_2\mathbb{Z}$, the locally symmetric space is a punctured sphere, and the Borel-Serre compactification adds an $S^1$ at infinity. This $S^1$ is given as the $\mathbb{R}$-points of the unipotent radical of the (unique up to $SL_2\mathbb{Z}$-conjugacy) Borel subgroup of $SL_2\mathbb{Q}$ modulo the action of the $\mathbb{Z}$-points. Some of this generalizes to the higher rank cases, but it is better to look up the exact definitions in the book of Borel and Ji.

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  • $\begingroup$ Thank you! Is there a way of understanding a relationship between the two compactifications you mention? $\endgroup$ – WSL Mar 1 '15 at 13:23
  • $\begingroup$ A detailed discussion of the relation between the compactifications can be found in the book of Borel and Ji (Section III.15). In the case $SL_2\mathbb{Z}$, the boundary $S^1$ of the Borel-Serre compactification is mapped to the cusp point of the geodesic compactification. In the higher rank situations, the Borel-Serre compactification still has the geodesic compactification as a quotient, but I am not sure about the fibers. $\endgroup$ – Matthias Wendt Mar 4 '15 at 13:30

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