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# Two (co)compactness criteria for lattices

Here and H are Lie groups, $\Gamma$ and $\Lambda$ are lattices in one of those Lie groups, and C is a subset of the Lie group.

### Detecting (non)compactness via accumulation at the identity

Conjugates by compact sets do not accumulate: The image of $C \subset H$ in $H / \Lambda$ is precompact iff the identity element $\mathrm{id}_H$ is not an accumulation point of the conjugated lattice ${}^C\Lambda$.

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Proof: ${}^C\Lambda \cap U^{-1}U = \{e\}$ iff we cannot find distinct $u_1, u_2 \in U$ s.t. $u_1c \Gamma = u_2c \Gamma$ for all $c \in C$, iff the translated orbit maps (covering maps!) $H \to H/\Lambda$ given by $h \mapsto hc\Lambda$ are injective on U for every $c \in C$.

Now if our image is precompact, then we may cover it by finitely many pancakes (i.e. neighborhoods in $H / \Lambda$ diffeomorphic to each connected component of their preimage in the cover H); by this finiteness, and H-equivariance, we can find some small open neighborhood U so that the desired injectivity condition on our translated orbit maps holds.

If our image is not precompact, then given any small open neighborhood U we can find a sequence of elements in C escaping any compact translate $U^{-1}K$. By an inductive / diagonal argument (involving larger and larger compact sets, each of which contains a previous $Uc_n\Lambda$) we have a sequence $(c_n) \subset C$ with $Uc_n\Lambda$ pairwise disjoint. Since $\Lambda$ has finite covolume, these cannot all have the same (nonzero) volume, and so (by fundamental domain considerations) not all of the translated orbit maps are injective on U.

Mahler’s compactness criterion is the special case where $H = \mathrm{SL}(\ell,\mathbb{R})$ and $\Lambda = \mathrm{SL}(\ell,\mathbb{Z})$.

This can be proven without using that $\Lambda$ is a lattice, by considering $\mathrm{SL}(\ell,\mathbb{R}) / \mathrm{SL}(\ell,\mathbb{Z})$ as the moduli space of unit-covolume lattices in $\mathbb{R}^\ell$ (or: “wow, the lattices are back at a whole new level!”): then the result says that a set of unit-covolume lattices is compact iff it does not contain arbitrarily short vectors.

The forward implication is fairly immediate after a moment’s thought; the reverse implication follows by considering the shortest vectors which appear in the lattices of any given sequence from our set, observing that those vectors form bounded orbits, taking (sub)sequential limits in those orbits using compactness in $\mathbb{R}^\ell$, and finally passing back to (sub)sequential limits in the sequence of lattices.

### Detecting (non)compactness with unipotents

Cocompact lattices have no unipotents: The homogeneous space $G / \Gamma$ is compact only if $\Gamma$ has no nontrivial unipotents.

One proof [of the contrapositive] follows from (the general version of) Mahler’s criterion above, via the Jacobson-Morosov Lemma which gives a map $\phi: \mathrm{SL}(2,\mathbb{R}) \to G$ sending our favorite unipotent $\left( \begin{array}{cc} 1 & 1 \\ & 1 \end{array} \right)$ to a nontrivial unipotent $u \in \Gamma$. Call a the image under $phi$ of our favorite hyperbolic element $\left( \begin{array}{cc} 2 \\ & 1/2 \end{array}\right)$. Then $a^nua^{-n}$ accumulates to the identity, and so $G / \Gamma$ is not precompact by the first criterion above.

More is true if we have more structure on our lattices:

Godement: Suppose G is defined over a real algebraic number field F (e.g. $\mathbb{Q}$), and let $\mathcal{O}$ be the ring of integers in F. The homogeneous space $G / G_{\mathcal{O}}$ (in the case of $F = \mathbb{Q}$, $G / G_{\mathbb{Z}}$) is compact iff $G_{\mathbb{Z}}$ has no nontrivial unipotents.

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Proof of the new (reverse) direction uses the arithmetic structure on $G_{\mathbb{Z}}$. Again we prove the contrapositive. Suppose $G / G_{\mathbb{Z}}$ is not compact. By (Mumford’s generalisation of) the Mahler criterion above, this implies ${}^gG_{\mathbb{Z}}$ accumulates at the identity for some $g \in G$. By continuity, the characteristic polynomials of the elements in an accumulating sequence $g\gamma_n g^{-1}$ converge (in the sense of coefficients; this makes sense if we are working within some fixed ambient $\mathrm{SL}(\ell, \mathbb{R})$, for example) to $(x-1)^\ell$.

But this implies, by matrix similarity, that the characteristic polynomials of the $\gamma_n$ converge to the same; and now we use the fact that $\gamma$ is an integer matrix to say that this implies the $\gamma_n$, for n sufficiently large, has characteristic polynomial $(x-1)^\ell$, and hence is unipotent.