Posted by Jason Polak on 25. August 2015 · Write a comment · Categories: algebraic-geometry, number-theory · Tags: ,

In the Arthur-Selberg trace formula and other formulas, one encounters so-called ‘orbital integrals’. These integrals might appear forbidding and abstract at first, but actually they are quite concrete objects. In this post we’ll look at an example that should make orbital integrals seem more friendly and approachable. Let $k = \mathbb{F}_q$ be a finite field and let $F = k( (t))$ be the Laurent series field over $k$. We will denote the ring of integers of $F$ by $\mathfrak{o} := k[ [t]]$ and the valuation $v:F^\times\to \mathbb{Z}$ is normalised so that $v(t) = 1$.

Let $G$ be a reductive algebraic group over $\mathfrak{o}$. Orbital integrals are defined with respect to some $\gamma\in G(F)$. Often, $\gamma$ is semisimple, and regular in the sense that the orbit $G\cdot\gamma$ has maximal dimension. One then defines for a compactly supported smooth function $f:G(F)\to \mathbb{C}$ the orbital integral
\Ocl_\gamma(f) = \int_{I_\gamma(F)\backslash G(F)} f(g^{-1}\gamma g) \frac{dg}{dg_\gamma}.

Here, $dg$ is a Haar measure on $G(F)$, which exists since $G$ is reductive, $I_\gamma$ is the centraliser of $\gamma$, and $dg_\gamma$ is a Haar measure on $I_\gamma(F)$. For simplicity, let’s assume that $I_\gamma$ is a torus and $\gamma$ is regular semisimple so that this is all defined and the integral converges.

That sounds a bit complicated, but is less so once we glimpse a computation of such a beast. In general of course, computation of these orbital integrals is quite tricky, which is why any progress is manipulating orbital integrls in the Arthur-Selberg trace formula is quite a feat. Nonetheless, simple examples can easily be computed.

Let’s do my favourite easy-to-compute example, which is $G = \mathrm{GL}_2$ and $\gamma$ is the diagonal element $\gamma = \mathrm{diag}(x,y)$, the diagonal matrix with nonzero entries $x$ and $y$. The orbit of $\gamma$ has maximal dimension when the centraliser of $\gamma$ is as small as possible, which in this case happens exactly when $x\not=y$. In our computation, let’s just do the case when $f$ is the characteristic function of $G(\mathfrak{o})$, which we will denote by $\mathbf{1}$.

One then has to choose Haar measures. The standard choice is giving $G(\mathfrak{o})$ unit volume. The reason for this is then the corresponding integral just counts points on a certain projective variety called the affine Springer fiber, the geometry of which is closely related to the theory of orbital integrals such as the theory of endoscopy and the fundamental lemma.

To compute the orbital integral, one then uses the Iwasawa decomposition. The centraliser of $\gamma$ in this case is the diagonal torus, which we will denote by $T$. The torus $T$ sits inside the Borel subgroup of upper triangular matrices of $\mathrm{GL}_2$ and if $U$ denotes the unipotent radical of the Borel subgroup, then we have the decomposition $\mathrm{GL}_2(F) = T(F)U(F)G(\mathfrak{o})$. The important point about this decomposition is that with the compatible Haar measures assigning unit volume to $T(\mathfrak{o})$ and $U(\mathfrak{o})$, we can rewrite the orbital integral now as
\Ocl_\gamma(\mathbf{1}) = \int_{U(F)} \mathbf{1}(u^{-1}\gamma u)du.
Now $G(\mathfrak{o})$ disappeared since conjugating by an element of $G(\mathfrak{o})$ does not affect the integrality of anything. Also, $U$ is isomorphic to the additive group $\mathbb{G}_a$. Explicitly, the isomorphism is given on points:
\mathbb{G}_a(R) &\longrightarrow U(R)\\
a&\longmapsto \begin{bmatrix} 1 & a\\ 0 & 1\end{bmatrix}\end{align*}
for an $\ofr$-algebra $R$. Hence, the integral becomes
\int_{F} \mathbf{1}\left(
x & a(x-y)\\
0 & y
We see first that $\Ocl_\gamma(\mathbf{1})$ is nonzero only when $x,y\in \mathfrak{o}$. So let’s assume this. Then we also need $a(x-y)\in \mathfrak{o}$. In other words, we need
v(a) \geq -v(x-y).
We thus need to determine the Haar measure of the set $\{ a\in F : v(a) \geq -v(x-y)\}$. Remember that the Haar measure was chosen so that $\mathfrak{o}$ has unit volume. Since the Haar measure is translation invariant, we can use this to take the measure of all sorts of sets. For example, the Haar measure of the set $\{ a\in F : v(a) \geq 1\}$ is $q^{-1}$ since it takes $q$ cosets of this set to cover $\mathfrak{o}$. Similarly, the measure of $\{a \in F : v(a) \geq -v(x-y)\}$ is $q^{v(x-y)}$. Hence
\Ocl_\gamma(\mathbf{1}) = q^{v(x-y)}.$$
And of course $v(x-y) \geq 0$, since we assumed that $x,y\in\mathfrak{o}$, so $x-y\in \mathfrak{o}$ as well.

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