I am trying to understand the Borel-Harish Chandra theorem about arithmetic subgroups being lattices.

Suppose $G$ is an algebraic group inside $GL_n(\mathbb{C})$ such that it is definable as a zero set of polynomials with coefficients in $\mathbb{Q}$. Let $G_\mathbb{R} = G \cap G_\mathbb{R}$ and $G_\mathbb{Z} = {G} \cap GL_n(\mathbb{Z})$.

The Borel-Harish Chandra theorem says that $X_\mathbb{Q}(G^0) = \{ e \} \Leftrightarrow G_\mathbb{R} / G_\mathbb{Z}$ has a finite natural measure. Here $X_\mathbb{Q}(G^0)$ is the multiplicative group of $\mathbb{Q}$-morphisms of the algebraic group $G^0$ to $GL_1(\mathbb{C})$.

I am wondering if the "$\Leftarrow$" part can be proven shown easily under the assumption that $G$ is closed. Here is my attempt:

Since $G$ is closed, any $\mathbb{Q}$-morphism $\chi:G^0 \rightarrow GL_1(\mathbb{C})$ can be written as a polynomal, instead of a regular function. However, on the points of $G_\mathbb{Z}^0$, this polynomial will actually take values in $N^{-1} \mathbb{Z}$ for some $N \in \mathbb{N}$ (the lcm of the denominators of all the rational coefficients) but since the image is a multiplicatively closed set of $\mathbb{Q}^*$, $\chi$ can only be $\{ \pm 1\}$ on $G_\mathbb{Z}^0$. But then, $\chi:G^0_\mathbb{R}/G^0_\mathbb{Z} \rightarrow \mathbb{R}^*$ is a continuous map that might induce a finite Haar measure on $\mathbb{R}^*$.

This approach might work, but saying that all $\mathbb{Q}$-characters on $G_\mathbb{Z}^0$ are trivial seems like a big deal and I could not find any such statements in Borel's "Intro. to Arithmetic groups". Do you have any counterexamples that show that the integer points of closed connected $\mathbb{Q}$ algebraic groups may admit a non-trivial $\mathbb{Q}$-character?

  • 2
    $\begingroup$ While your definition of $G_{\mathbb Z}$ is legal, I have a feeling that your hypotheses are too weak to guarantee that it is a sensible construction. (For example, I think that you can arrange that $G_{\mathbb Z}$ is trivial even for very large groups $G$.) $\endgroup$
    – LSpice
    Aug 9, 2020 at 0:07
  • $\begingroup$ This is not the question I expected from the title. $\endgroup$
    – Kimball
    Aug 9, 2020 at 1:53
  • $\begingroup$ @Kimball I am sorry. I have fixed the title. $\endgroup$ Aug 9, 2020 at 9:41
  • $\begingroup$ @LSpice Can you please give me an example of what you are saying? $\endgroup$ Aug 9, 2020 at 9:44
  • 1
    $\begingroup$ I think the example to keep in mind is the trivial algebraic group $GL_1(\mathbb{C})$. The integer points are $\pm 1$, so it is not a lattice. But I don't know the proof, except in some special cases (certain lattices can be proved to be cocompact using the Mahler compactness theorem, which is more elementary than the proof for non-uniform lattices). A reductive Lie group is semisimple iff it has no characters. The construction of arithmetic lattices that I'm familiar with just proves it for semisimple algebraic groups. See Theorem 1.3.9 of Witte-Morris' book. deductivepress.ca $\endgroup$
    – Ian Agol
    Aug 11, 2020 at 2:19


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