# A virtually free RACG

(Thanks to Harry Richman for bringing this example up.)

The free group $\langle a, b \rangle$ and the right-angled Coxeter group $\rangle a, b, c, d | a^2, b^2, c^2, d^2 \rangle$ have the same Cayley graph—the infinite 4-valent tree—; hence they are quasi-isometric. We note here that quasi-isometries are not required to group homomorphisms, and indeed the one we have here—an “identity map” of sorts (in an admittedly very loose and imprecise sense) between Cayley graphs—is not.

By results of Stallings and Dunwoody involving ends of groups ad accessibility (although Chapter 20 of Drutu-Kapovich’s draft manuscript seems to be the only reference I can find for this online), free groups are quasi-isometrically rigid, and hence, since finite-index subgroups of free groups are free, the RACG is virtually free (!)

With a little more thought we can explicitly exhibit a finite-index free subgroup: the subgroup of the RACG generated by ab, ac and ad is a nonabelian free group on three generators; we can verify it has finite index via a covering space argument.

Now I wonder if there is a finite-index $F_2$-subgroup of the RACG as well, so that the commensurability need not involve passing to finite-index subgroups on both sides. It seems like there shouldn’t be, although I can’t quite prove this yet.

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# Brouwer’s invariance of domain

I am recording here an elementary* result used in the proof of the Teichmueller Existence Theorem.

Theorem (Brouwer). Any injective continuous map from $\mathbb{R}^n$ to itself is open.

*the statement itself sounds elementary enough, but the proof (all known proofs) seem to require some amount of machinery from algebraic topology. Note that one corollary of the theorem is that Euclidean spaces of different dimensions are not homeomorphic, another statement which appears elementary (and intuitive), but is difficult to prove without additional machinery.

It seems that the category of continuous maps contains enough curious things—Peano and other space-filling curves and the like—to quite thoroughly upset our intuitions, although going to the rather more restrictive category of smooth things does restore much of our intuition.

Proof of Theorem: It suffices to show that any such map sends open balls to open balls, or, more narrowly, that for any injective continuous map $f: B^n \to \mathbb{R}^n$$f(0)$ lies in the interior of $f(B^n)$ (where $B^n$ denotes the closed unit ball.)

Let f be as in the hypothesis of the Theorem. $f: B^n \to f(B^n)$ is a continuous bijection between compact Hausdorff spaces and hence a homeomorphism. $f^{-1}: f(B^n) \to B^n$ is continuous; by the Tietze extension theorem, $f^{-1}$ may be extended to a continuous map $G: \mathbb{R}^n \to \mathbb{R}^n$.

G has a zero at $f(0)$, and moreover we have the following

Lemma: If $\tilde{G}: f(B^n) \to \mathbb{R}^n$ is a continuous map with $\|G - \tilde{G}\|_\infty \leq 1$ (i.e. is a small perturbation of G), then $\tilde{G}$ has a zero in $f(B^n)$.

Proof of Lemma: Applying Brouwer’s fixed-point theorem to the function $x \mapsto x - \tilde{G}(f(x)) = (G - \tilde{G})(f(x))$ (from the closed unit ball to itself) yields a $x \in B^n$ s.t. $x = x - \tilde{G}(f(x))$, i.e. $\tilde{G}(f(x)) = 0$.

If $f(0)$ were not an interior point of $f(B^n)$, we may construct a small perturbation of G that no longer has a zero on $f(B^n)$, contradicting the above lemma.

At this point I refer the interested reader to Terry Tao’s blogpost (where the gist of this proof was also taken from; he was interested in Hilbert’s fifth problem, whose solution also makes use of invariance of domain) for the details of the construction.

Corollary Any proper injective continuous map $f: \mathbb{R}^n \to \mathbb{R}^n$ is a homeomorphism.

Proof of Corollary: By invariance of domain, f is open. Now it suffices to show that f is surjective (and hence bijective) for it to be a homeomorphism; but surjectivity follows from f being a proper map into a metrizable space, and hence closed—given $E \subset X$ closed and a sequence of points $f(x_n) \in f(E)$ accumulating to $y \in Y$ and $\epsilon > 0$, $f(x_n) \in B(y,\epsilon)$ (the closed ball) for all $n \geq N = N(\epsilon) \gg 0$. By properness, we have, for all $n \geq N$, $x_n \in K \subset E$ with K compact. Then, up to subsequence, $x_n \to x \in K$. By continuity, $f(x_n) \to f(x) = y$, so $y = f(x) \in f(E)$.

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# Isomorphic groups via hyperbolic geometry

$\mathrm{PSO}(1, 2)$ is the isometry group of hyperbolic 2-space under the hyperboloid model; $\mathrm{PSL}(2,\mathbb{R})$ is the isometry group of the upper half-plane; $\mathrm{PSU}(1,1)$ is the isometry group of the Poincaré unit disk. Since these are in fact models for the same space, all of these Lie groups are isomorphic.

Question: is there a proof of this isomorphism that doesn’t proceed through this identification as isometry groups? (Probably, via the Lie algebras for instance, but I’m a little too lazy to try and work it out at the moment.)

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# A quick and dirty definition of an arithmetic subgroup?

Or, at least, of an arithmetic subgroup of $G \leq \mathrm{SL}(n,\mathbb{R})$ a semisimple Lie group:

1. The “subgroup of integer points” $G_{\mathbb{Z}} = G \cap \mathrm{SL}(n,\mathbb{Z})$ is an arithmetic group …
2. … if embedded in a reasonable way which doesn’t distort the arithmetic structure—slightly more precisely, if it is [essentially] an algebraic group over the rationals.
3. Anything isomorphic to a finite extension is arithmetic too.

For a more precise definition—Witte Morris’ book is a good place to start; he fudges over a few things too, but actually has references to fill those in …

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# A (small) regularity zoo

The image is a imagemap (which doesn’t scale because WordPress.com doesn’t allow Javascript widgets and so everything probably looks off on your screen): move your cursor around it and explore!

Some general (imprecise) notes:

• Function spaces in the left two columns are defined by regularity conditions
• Those in the right three columns are defined by integrability conditions (BMO, the space of functions with bounded mean oscillation, is the dual of the Hardy space $H^1$)
• Sobolev spaces $W^{k,p}$ are defined by some mix of integrability and regularity conditions.
• Most of the spaces above are Banach spaces; some (e.g. $L^2$) are Hilbert spaces.
• Functions in any of the above spaces are (a fortiori) measurable.
• Sobolev spaces embed into $C^k$ / Hoelder and $L^p$ spaces; the precise statements can get tricky.
• This map is in no way comprehensive, it only sketches some links between some common families.
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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 ${}^C\Lambda$.

[[ some pictures to be inserted here ]]

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.

[[ some more pictures here ]]

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.

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# The many faces of the hyperbolic plane

$\mathbb{H}^2$ is the unique (up to isometry) complete simply-connected 2-dimensional Riemann manifold of constant sectional curvature -1.

1. It is diffeomorphic to $\mathrm{SL}(2,\mathbb{R}) / \mathrm{SO}(2)$ as a topological space (or, indeed, isometric as a Riemannian manifold, if we give $\mathrm{SL}(2,\mathbb{R}) / \mathrm{SO}(2)$ a left-invariant metric): to show this, we note that $\mathbb{H}^2$ has isometry group $\mathrm{SL}(2,\mathbb{R}) / \pm\mathrm{id}$, and the subgroup of isometries which stabilize any given $p \in \mathbb{H}^2$ is isomorphic to $\mathrm{SO}(2)$.
2. Since any positive-definite binary quadratic form is given by a symmetric 2-by-2 matrix with positive eigenvalues, and since the group of linear transformations on $\mathbb{R}^2$ preserving any such given form is isomorphic to $\mathrm{O}(2)$, $\mathrm{SL}(2,\mathbb{R}) / \mathrm{SO}(2)$ is also the space of positive-definite binary quadratic forms of determinant 1, via the map from $\mathrm{SL}(2,\mathbb{R})$ to the symmetric positive-definite 2-by-2 matrices given by $g \mapsto g^T g$.
3. $\mathrm{SL}(2,\mathbb{R}) / \mathrm{SO}(2) \cong \mathbb{H}^2$ is also the moduli space of marked Riemann surfaces of genus 1, i.e. the Teichmüller space Teich(S) of the torus S. One way to prove this is to note that any such marked surface is the quotient of $\mathbb{R}^2$ by a $\mathbb{Z}^2$ action; after a suitable conformal transformation, we may assume that the generators of this $\mathbb{Z}^2$ act as $z \mapsto z + 1$ and $z \mapsto z + \tau$ for some $z \in \mathbb{H}^2$ (in the upper half-plane model.) But now $\tau$ is the unique invariant specifying this point in our moduli space.
4. Since any marked Riemann surface of genus 1 has a unique flat metric (inherited as a quotient manifold of the Euclidean plane), $\mathrm{SL}(2,\mathbb{R}) / \mathrm{SO}(2)$ is also the moduli space of marked flat 2-tori of unit area.
5. Since there is a unique unit-covolume marked lattice associated to each marked complex torus in the above, $\mathrm{SL}(2,\mathbb{R}) / \mathrm{SO}(2)$ is also the space of marked lattices in $\mathbb{R}^2$ with unit covolume.
6. Note we may go directly between marked Riemann surfaces and quadratic forms by considering intersection forms.
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