Representation theory, for the impatient

$\def\C{\mathbf{C}}$

Let $G$ be a finite group. The group algebra $\C G$ is the complex algebra with basis $G$, with multiplication defined by linearly extending the group law of $G$. The description of $\C G$ as an algebra is called representation theory. Relatedly, we want to understand the structure of $\C G$-modules. We often call $\C G$-modules (complex) representations of $G$, and simple $\C G$-modules irreducible representations.

Lemma: (Schur's lemma) If $V$ and $V'$ are simple $\C G$-modules, then every nonzero homomorphism $V\to V'$ is an isomorphism. Moreover every homomorphism $V\to V$ is a scalar multiple of the identity.

Proof: Suppose that $f:V\to V'$ is a nonzero homomorphism. Then $\ker f$ is a submodule, so $\ker f = 0$, and similarly $f(V) = V'$, so $f$ is an isomorphism. Now suppose $V=V'$. Since every eigenspace of $f$ is a submodule, by simplicity every eigenspace is either $0$ or $V$. From the spectral theorem we deduce that $f$ is a scalar multiple of the identity.

It turns out that we can ask for a unitary structure in $\C G$-modules for free.

Lemma: (Weyl's averaging trick) Every $\C G$-module $V$ has a $G$-invariant inner product. Moreover if $V$ is simple then this inner product is unique up to scaling.

Proof: Let $V$ be a $\C G$-module and let $(,)$ be any inner product on $V$. Define a new inner product $\langle,\rangle$ by

$$\langle u, v\rangle = \frac{1}{|G|} \sum_{g\in G} (g\cdot u, g\cdot v).$$

Clearly $\langle,\rangle$ has the desired properties. Now suppose that $V$ is simple, and that $\langle,\rangle'$ is another $G$-invariant inner product. Let $f:V\to V$ be the adjoint of the formal identity

$$(V,\langle,\rangle)\to(V,\langle,\rangle').$$

In other words let $f$ be the unique function $V\to V$ satisfying

$$\langle u,v\rangle' = \langle u, f(v)\rangle$$

for all $u,v\in V$. Then $f$ is a homomorphism, so by Schur's lemma it must be a multiple of the identity, so $\langle,\rangle'$ must be a multiple of $\langle,\rangle$.

Lemma: Let $U$ be a finite-dimensional $\C G$-module with $G$-invariant inner product $\langle,\rangle$, and for each simple $\C G$-module $V$ let $U_V$ be the sum of all submodules of $U$ isomorphic to $V$ (the isotypic component of $U$ corresponding to $V$). Then $U_V\cong V^{\oplus m}$ for some $m$, and $U$ is the orthogonal direct sum of the submodules $U_V$.

Proof: Since the orthogonal complement of a submodule is a submodule, it follows by induction on dimension that $U$ is an orthogonal direct sum of simple submodules, so $U = \bigoplus U'_V$, where $V$ runs over simple $\C G$-modules and each $U'_V \cong V^{\oplus m}$ for some $m\geq 0$. Moreover by Schur's lemma any two nonisomorphic simple submodules must be orthogonal, as the orthogonal projection from one to the other is a homomorphism, so every submodule of $U$ isomorphic to $V$ must be contained in $U'_V$, so $U'_V=U_V$.

The above lemma is particularly interesting when applied to the $\C G$-module $\C G$ itself, the regular representation. We give $\C G$ a $G$-invariant inner product by declaring the basis $G$ to be orthonormal. From the above lemma we know then that $\C G = \bigoplus \C G_V$, where $V$ runs over simple $\C G$-modules, and $\C G_V\cong V^{\oplus m}$ for some $m\geq 0$. Note then by Schur's lemma that $\dim\text{Hom}(\C G,V)=m$. On the other hand every homomorphism $\C G\to V$ is determined uniquely by the destination of the unit $1$ in $V$, so $\dim\text{Hom}(\C G,V)=\dim V$. We deduce that as $\C G$-modules

$$\C G \cong \bigoplus V^{\oplus \dim V}.$$

In particular there are only finitely many simple $\C G$-modules, and their dimensions obey

$$|G| = \sum (\dim V)^2.$$

One can also prove $\C G_V\cong V^{\oplus\dim V}$ in a more informative way, as follows. Fix an invariant inner product $\langle,\rangle_V$ on $V$, and consider any homomorphism $f:V\to\C G$. The adjoint $f^*:\C G\to V$ is also a homomorphism, so

$$ f(v) = \sum_{g\in G}\langle f(v),g\rangle_{\C G} g = \sum_{g\in G}\langle v, f^*(g)\rangle_V g = \sum_{g\in G}\langle v,g f^*(1)\rangle_V g. $$

Conversely for any $u\in V$ we may define a homomorphism $V\to\C G$ by

$$ f_{V,u}(v) = \sum_{g\in G} \langle v,g\cdot u\rangle_V g.$$

We deduce therefore that $\C G_V$ is the subspace of $\C G$ spanned by the elements $f_{V,u}(v)$, where $u,v\in V$. Moreover using Schur's lemma one can show that the images of $f_{V,u}$ and $f_{V,u'}$ are orthogonal whenever $u$ and $u'$ are orthogonal in $V$, so by letting $u$ range over a basis of $V$ we thus see that $\C G_V$ is the orthogonal direct sum of $\dim V$ copies of $V$.

In any case we now understand the structure of $\C G$ as a $\C G$-module, and we are only a short step away from understanding its structure as an algebra. Consider the obvious map

$$\C G \longrightarrow \bigoplus \text{End}(V),$$

where as always the sum runs over a complete set of irreducible representations up to isomorphism. We claim this map is an isomorphism. Since we already know the dimensions agree, it suffices to prove injectivity. Thus suppose $x\in\C G$ maps to zero, i.e., that $x$ acts trivially on each simple $\C G$-module $V$. Then $x$ acts trivially on $\C G$, so $x = x1 = 0$. Thus we have proved the following theorem.

Theorem: As complex algebras, $\C G \cong \bigoplus \text{End}(V)$.

$\def\tr{\text{tr}}$

Finally, it is useful to understand how to project onto isotypic components. Given $g\in G$, we can compute the trace of $g$ as an operator on $\C G$ in two different ways. On the one hand, by looking at the basis $G$,

$$ \tr_{\C G} (g) = \begin{cases} |G| & \text{if }g=1,\\ 0 &\text{if }g\neq 1.\end{cases}$$

On the other hand from the decomposition $\C G\cong\bigoplus V^{\oplus\dim V}$ we have

$$ \tr_{\C G} (g) = \sum_V (\dim V) \tr_V(g).$$

As a consequence, for every $x\in \C G$ we have

$$ x = \sum_V P_V x, $$

where $P_V : \C G \to \C G$ is the operator defined by

$$ P_V x = \frac{\dim V}{|G|} \sum_{g\in G} \tr_V(xg^{-1}) g = \frac{\dim V}{|G|} \sum_{g\in G} \tr_V(g^{-1}) gx.$$

These identities are most easily verified first for $x\in G$, then extending to all of $\C G$ by linearity. Now if $x \in \C G_U$ for $U\neq V$ then $x$ acts as zero on $V$, so $\tr_V(x g^{-1}) = 0$, so $P_V x = 0$. On the other hand one can verify directly that $P_V$ is a homomorphism, so by Schur's lemma the image of $P_V$ must be contained in $\C G_V$. We deduce therefore from $x = \sum_V P_V x$ that $P_V$ is the orthogonal projection onto $\C G$.

The function $\chi_V(g) = \tr_V(g)$ is usually called the character of $V$. From the relations $P_V^2 = P_V$ and $P_U P_V = 0$ for $U\neq V$ one can deduce the well known orthogonality relations for characters. In fact the distinction between $P_V$ and $\chi_V$ is hardly more than notational. Often we identify functions $f:G\to \C$ with elements $\sum_{g\in G} f(g) g \in \C G$, in which case the operation of convolution corresponds to multiplication in the group algebra. The operator $P_V$ then is just convolution with $(\dim V)\chi_V$. So, in brief, to project onto the $V$-isotypic component you convolve with the character of $V$ and multiply by $\dim V$.

We have kept to almost the bare minimum in the above discussion: the complex numbers $\C$ and finite groups $G$. There are a number of directions we could try to move in. We could replace $\C$ with a different field, say one which is not algebraically closed, or one which has positive characteristic. Alternatively we could replace $G$ with an infinite group, say with a locally compact topology. We mention two such generalisations.

Theorem: (Artin--Wedderburn) Every semisimple ring is isomorphic to a product $\prod_{i=1}^k M_{n_i}(D_i)$ of matrix rings, where the $n_i$ are integers and the $D_i$ are division rings. In particular every semisimple $\C$-algebra is isomorphic to a product $\prod_{i=1}^k M_{n_i}(\C)$. 

When defining unitary representations for compact groups $G$ we demand that the map $G\to U(V)$ be continuous, where $U(V)$ is given the strong operator topology.$\def\HS{\text{HS}}$

Theorem: (Peter--Weyl) Let $G$ be a compact group and $\mu$ its normalised Haar measure. Let $\widehat{G}$ be the set of all irreducible unitary representations of $G$ up to isomorphism. Then $\widehat{G}$ is countable, every $V\in\widehat{G}$ is finite-dimensional, and the algebra $L^2(G)$ of square-integrable functions with the operation of convolution decomposes as a Hilbert algebra as $$ L^2(G) \cong \bigoplus_{V\in\widehat{G}} (\dim V) \cdot \HS(V), $$ where $\HS(V)$ is the space $\text{End}(V)$ together with the Hilbert-Schmidt inner product.

Group limits

A few months ago on his blog Terry Tao explained how one could, by analogy with the theory of graph limits, replace the explicit use of arithmetic regularity with a soft device which he calls an additive limit. Roughly speaking, the additive limit, or Kronecker factor, of a sequence of finite abelian groups $G_i$ is a compact quotient of the ultraproduct $\prod G_i$ which controls the convolutions. The result is that many theorems from additive combinatorics, such as Roth's theorem, which are usually proved using quantitative tools like Fourier analysis can instead be proved using soft tools more like the Lebesgue differentiation theorem.

The purpose of this post is to extend Tao's construction to the nonabelian setting. Tao already stated in his post that this should be possible, so one could say that this is just an exercise in nonabelian Fourier analysis. On the other hand the proof in the nonabelian setting more or less forces a more categorical point of view, so certain points of this exercise are instructive.

1. Measurable Bohr compactification

$\def\B{\text{Bohr}}\def\Ba{\text{Baire}}$Given any topological group $G$ there is a compact group $\B(G)$, called the Bohr compactification of $G$, such every continuous homomorphism from $G$ to a compact group $K$ factors uniquely through $\B(G)$. One can think of $\B(G)$ as the 'largest' compact group in which $G$ has dense image. We need a variant of this definition for groups $G$ endowed only with a $\sigma$-algebra instead of a topology.

By a measurable group we mean a group $G$ together with a $\sigma$-algebra $\Sigma$ of subsets of $G$. Note that we do not make any measurability assumptions about the group operation or even the left- or right-shifts, though certainly it would be sensible to do so in other contexts. For us the only role of $\Sigma$ is to distinguish among all homomorphisms the measurable homomorphisms. The analogue of Bohr compactification for measurable groups is given by the following theorem.

Theorem 1 (Existence of measurable Bohr compactification): For every measurable group $G$ there is a compact group $\B_m(G)$ together with a $(\Sigma,\Ba)$-measurable homomorphism $\pi:G\to \B_m(G)$ such that every $(\Sigma,\Ba)$-measurable homomorphism from $G$ to a compact group $K$ factors uniquely as the composition of $\pi:G\to \B_m(G)$ and a continuous homomorphism $\B_m(G)\to K$.

In category-theoretic terms $\B_m$ is a left adjoint to the functor $\Ba$ from compact groups to measurable groups which replaces a group's topology with its Baire $\sigma$-algebra. By the adjoint functor theorem it suffices to check that the functor $\Ba$ is continuous, which boils down to the following lemma. Incidentally the analogue of this lemma fails for the Borel $\sigma$-algebra, which is why we must consider the Baire $\sigma$-algebra instead.

Lemma 2: For compact Hausdorff spaces $X_i$ we have $\Ba(\prod X_i) = \prod \Ba(X_i)$. In words, the Baire $\sigma$-algebra of the product is the product of the Baire $\sigma$-algebras.

Proof: The containment $\prod \Ba(X_i) \subset \Ba(\prod X_i)$ is immediate from the defintions. To prove the opposite containment it suffices to check that every continuous function $f:\prod X_i \to \mathbf{R}$ is measurable with respect to $\prod\Ba(X_i)$. This is certainly true of functions $f$ which depend on only finitely many coordinates, and thus for all continuous functions $f$ by the Stone-Weierstrass theorem.$\square$

Those who, like me, are not used to thinking in terms of the adjoint functor theorem will appreciate a more pedestrian proof of Theorem 1. To this end, let $\mathcal{K}$ be the set of all pairs $(f,K)$, where $K$ is a compact group and $f:G\to K$ is a $(\Sigma,\Ba)$-measurable homomorphism with $f(G)$ dense in $K$. Then if$$\pi:G\to\prod_{(f,K)\in\mathcal{K}} K$$ is the diagonal map then $\pi:G\to\overline{\pi(G)}$ is the Bohr compactification of $G$. Indeed, the measurability of $\pi$ follows from Lemma 2, and the universal property is essentially obvious: given a $(\Sigma,\Ba)$-measurable homomorphism $f:G\to K$ with $K$ compact, the pair $(f,\overline{f(G)})$ appears in $\mathcal{K}$, so we have continuous maps$$\overline{\pi(G)}\to\prod_{(g,L)\in\mathcal{K}} L \to \overline{f(G)}\to K$$whose composition $h$ satisfies $h\pi = f$; moreover $h$ is unique because $\pi(G)$ is dense in $\overline{\pi(G)}$.

Alternatively, by the Peter-Weyl theorem, $\B_m(G)$ can be defined as the inverse limit of all measurable finite-dimensional unitary representations of $G$.

It is natural to ask what relation the measurable Bohr compactification $\B_m$ bears to the usual Bohr compactification $\B$. In particular, is $\B$ just the composition $\B_m\circ\Ba$? Clearly $\B(G) = \B_m(\Ba(G))$ if and only if every measurable homomorphism $f:G\to K$ from $G$ to a compact group $K$ is continuous. This follows from Steinhaus's theorem if $G$ is locally compact, but it certainly fails in general, for instance for $G=\mathbf{Q}$ with the topology inherited from $\mathbf{R}$.

2. The Bohr compactification of an ultrafinite group

Let $G_1,G_2,\dots$ be a sequence of finite groups and let $p\in \beta\mathbf{N} \setminus\mathbf{N}$ be a nonprincipal ultrafilter. We form the ultraproduct $G=\prod_{n\to p} G_n$ and make it into a measurable group by giving it the Loeb $\sigma$-algebra $\mathcal{L}_G$, the $\sigma$-algebra generated by internal sets $\prod_{n\to p} A_n$, where $A_n\subset G_n$. We define the Loeb measure $\mu_G$ on internal sets $\prod_{n\to p}A_n$ by putting$$\mu_G\left(\prod_{n\to p}A_n\right)=\text{st}\lim_{n\to p} |A_n|/|G_n|,$$and we define $\mu_G$ on $\mathcal{L}_G$ by extension.

While the group operation $G\times G\to G$ is not generally measurable with respect to the product $\sigma$-algebra $\mathcal{L}_G\times\mathcal{L}_G$, it is measurable with respect to the larger $\sigma$-algebra $\mathcal{L}_{G\times G}$. Moreover this latter $\sigma$-algebra is still 'product-like' in the sense that all $\mathcal{L}_{G\times G}$-measurable $f:G\times G\to\mathbf{R}_{\geq0}$ obey Fubini's theorem, so we have a sensibly defined convolution operation $L^1(G)\times L^1(G)\to L^1(G)$ given $$f*g(x) = \int f(y)g(y^{-1}x)\,d\mu_G(y).$$

Now consider the Bohr compactification $\B_m(G)$ of $G$. The first thing to notice is that $\pi_*\mu_G$ is a $\pi(G)$-invariant Baire probability measure on $\B_m(G)$. Since $\pi(G)$ is dense in $\B_m(G)$ we conclude that $\pi_*\mu_G$ is in fact $\B_m(G)$-invariant, so by the uniqueness of Haar measure we must have$$\pi_*\mu_G=\mu_{\B_m(G)}.$$

In the remainder of this section we relate the two convolution algebras $L^2(G)$ and $L^2(\B_m(G))$. Given $f\in L^2(\B_m(G))$ we can form the pullback $\pi^*f = f\circ\pi$. Since$$ \|f\circ\pi\|_{L^2(G)}^2 = \int |f\circ\pi|^2\,d\mu_G = \int |f|^2 \,d\mu_{\B_m(G)} = \|f\|_{L^2(\B_m(G))}^2,$$we see that $\pi^*f$ is a well defined element of $L^2(G)$, and in fact $\pi^*$ defines an isometric embedding$$\pi^*:L^2(\B_m(G))\to L^2(G).$$In the other direction we have the pushforward$$\pi_*:L^2(G)\to L^2(\B_m(G)),$$ defined as the adjoint of $\pi_*$. The identities $$\pi_*\pi^*f=f\quad\text{for }f\in L^2(\B_m(G)),$$$$\pi^*(f*g)=\pi^*f*\pi^*f\quad\text{for }f,g\in L^2(\B_m(G)),$$$$\pi_*(f*g)=\pi_*f*\pi_*g\quad\text{for }f,g\in L^2(G)$$are readily verified. For instance, the last of these is verified by the following computation, valid for $h\in L^2(\B_m(G))$ by $\mathcal{L}_{G\times G}$-Fubini:$$\langle h, \pi_*(f*g)\rangle = \langle \pi^* h, f*g\rangle = \int h(\pi(xy)) f(x)g(y)\,d\mu_G(x)\,d\mu_G(y)$$$$=\int h(x'y') \pi_*f(x')\pi_*g(y')\,d\mu_{\B_m(G)}(x')\,d\mu_{\B_m(G)}(y') = \langle h,\pi_*f*\pi_*g\rangle.$$We can summarise the situation as an isometric Banach algebra isomorphism$$L^2(G) \cong L^2(\B_m(G))\oplus \ker\pi_*.$$

The following theorem asserts that in fact $\B_m(G)$ alone determines convolutions in $G$, and thus $\B_m(G)$ will more generally control all 'first-order configurations' in $G$.

Theorem 3: We have $(\ker\pi_*)*L^2(G)=0$. Thus all convolutions in $L^2(G)$ can be computed in $L^2(\B_m(G))$, in the sense that$$f*g=\pi^*(\pi_*f*\pi_*g)$$for all $f,g\in L^2(G)$.

The theorem follows from the following lemma.

Lemma 4: For all $f,g\in L^2(G)$ and $d\in\mathbf{R}$ we have$$\|f*g\|^2_{L^2(G)} \leq \left(\frac{1}{d}\|f\|_{L^2(G)}^2 + d\|\pi_*f\|_{L^2(\B_m(G))}^2\right)\|g\|_{L^2(G)}^2.$$In particular by optimising $d$ we have$$\|f*g\|^2_{L^2(G)} \leq 2\|\pi_*f\|_{L^2(\B_m(G))}\|f\|_{L^2(G)}\|g\|_{L^2(G)}^2.$$

$\def\st{\text{st}}$Proof: We borrow nonabelian harmonic analysis notation from Tao. Certainly we may assume that $f$ and $g$ are internal, say $f=\st\lim_{n\to p}f_n$ and $g=\st\lim_{n\to p}g_n$. Then by nonabelian Plancherel,$$\|f*g\|_{L^2(G)}^2 = \st\lim_{n\to p} \|f_n*g_n\|_{L^2(G_n)}^2= \st\lim_{n\to p} \sum_{\xi\in\widehat{G_n}} \dim V_\xi \|\hat{f_n}(\xi)\hat{g_n}(\xi)\|^2_{\text{HS}(V_\xi)}$$$$\leq\st\lim_{n\to p} \sup_\xi\|\hat{f_n}(\xi)\|^2_{\text{HS}(V_\xi)} \|g_n\|_{L^2(G_n)}^2 = \sup_\xi\st\lim_{n\to p}\|\hat{f_n}(\xi_n)\|^2_{\text{HS}(V_{\xi_n})} \|g\|_{L^2(G)}^2,$$where in the last line the supremum is taken over all $\xi=(\xi_n)$. Fixing some such $\xi$, by Plancherel again$$\|\hat{f_n}(\xi_n)\|^2_{\text{HS}(V_{\xi_n})} \leq \frac{1}{\dim V_{\xi_n}} \|f_n\|_{L^2(G_n)}^2,$$so we may assume that $\dim V_{\xi_n}\leq d$ for $p$-most $n$. But then the representations $\rho_{\xi_n}:G_n\to U(d)$ induce a measurable representation $\rho_\xi:G\to U(d)$, which in turn by the universal property of $\B_m(G)$ factors through a continuous representation $\rho'_\xi:\B_m(G)\to U(d)$. Thus$$\st\lim_{n\to p}\|\hat{f_n}(\xi_n)\|^2_{\text{HS}(V_{\xi_n})} = \st\lim_{n\to p} \int f_n(x)f_n(y) \text{tr}\rho_{\xi_n}(xy^{-1})\,d\mu_{G_n}(x)\,d\mu_{G_n}(y)$$$$= \int f(x) f(y) \text{tr}\rho_\xi(xy^{-1})\,d\mu_G(x)\,d\mu_G(y) = \int f(x) f(y) \text{tr}\rho'_\xi(\pi(xy^{-1}))\,d\mu_G(x)\,d\mu_G(y)$$$$= \int \pi_*f(x') \pi_*f(y') \text{tr}\rho'_\xi(x'y'^{-1})\,d\mu_{\B_m(G)}(x)\,d\mu_{\B_m(G)}(y) \leq d \|\pi_*f\|_{L^1(\B_m(G))}^2\leq d\|\pi_*f\|_{L^2(\B_m(G))}^2.\square$$

3. Quasirandomness

From an additive combinatorics point of view, nonabelian groups obey a structure versus randomness principle: the asymptotic behaviour with respect to linear configurations can usually be described as some combination of abelian and random-like behaviour. Following Gowers, we call a sequence of finite groups $(G_n)$ quasirandom if the least dimension of a nontrivial representation of $G_n$ tends to infinity with $n$. For example for $n\geq 7$ every nontrivial representation of the alternating group $\text{Alt}(n)$ has dimension at least $n-1$, so the sequence $(\text{Alt}(n))$ is quasirandom.

Theorem 5: The ultrafinite group $G_p=\prod_{n\to p} G_n$ has trivial Bohr compactification if and only if the least dimension of a nontrivial representation of $G_n$ tends to infinity as $n\to p$. In particular, $(G_n)$ is quasirandom if and only if $\B_m(G_p)$ is trivial for every $p\in\beta\mathbf{N}\setminus\mathbf{N}$.

First we need a simple lemma.

Lemma 6: Let $G$ and $H$ be groups with $G$ finite and $f:G\to H$ a map which satisfies $f(xy)=f(x)f(y)$ for $1-o(1)$ of the pairs $(x,y)\in G^2$. Then there is a homomorphism $h:G\to H$ such that $f(x)=h(x)$ for $1-o(1)$ of the points $x\in G$.

Proof: For every $x\in G$ and for $1-o(1)$ of the pairs $(y,z)\in G^2$ we have$$f(xyz)f(yz)^{-1} = f(xy)f(z)(f(y)f(z))^{-1} = f(xy)f(y)^{-1},$$so for each $x\in G$ there is a unique $h(x)\in H$ such that$$h(x) = f(xy)f(y)^{-1}$$ for $1-o(1)$ of the points $y\in G$. Clearly $h(x)=f(x)$ for $1-o(1)$ of the points $x\in G$, and for $x_1,x_2\in G$ we have$$h(x_1)h(x_2)^{-1} = f(x_1y)f(y)^{-1}f(y)f(x_2 y)^{-1} = f(x_1y)f(x_2y)^{-1}=h(x_1x_2^{-1})$$for $1-o(1)$ of the points $y\in G$, in particular for at least one $y\in G$, so $h$ is a homomorphism.$\square$

Proof of Theorem 5: Suppose we have a sequence of nontrivial homomorphisms $f_n:G_n\to U(d)$ for all $n$ on some neighbourhood of $p$. Then $(f_n)$ induces a measurable homomorphism $f:G_p\to U(d)$, and since $U(d)$ has no small subgroups the induced homomorphism $f:G_p\to U(d)$ will also be nontrivial, so $\B_m(G_p)$ must be nontrivial.

Conversely suppose the least dimension of a nontrivial representation of $(G_n)$ tends to infinity as $n$ tends to $p$, and let $f:G_p\to U(d)$ be a measurable homomorphism. By a countable saturation argument there is an internal function $g:G_p\to U(d)$, say $g=\st\lim_{n\to p}g_n$, such that $f=g$ almost everywhere. Then $g_n$ satisfies $g_n(xy)=g_n(x)g_n(y)$ for $1-o(1)$ of the pairs $(x,y)\in G_n$, so by the lemma there is a homomorphism $h_n:G_n\to U(d)$ such that $g_n(x) = h_n(x)$ for $1-o(1)$ of the points $x\in G_n$. But by assumption any such homomorphism $h_n$ must be trivial for $n$ near enough to $p$, so we must have $g=1$ almost everywhere, so $f=1$ almost everywhere. Moreover since $f(x) = f(xy)f(y)^{-1}$ we must in fact have $f=1$ identically. Since the Peter-Weyl theorem implies that $\B_m(G_p)$ is an inverse limit of matrix groups this implies that $\B_m(G_p)$ is trivial.$\square$

Equations are generally easy to solve in quasirandom groups. We illustrate this point with the following theorem.

Theorem 7: Let $(G_n)$ be quasirandom and let $\epsilon>0$. Then there exists $n_0$ such that if $n\geq n_0$ and $A_n\subset G_n$ has density $|A_n|/|G_n|\geq\epsilon>0$ then we can find $x,y,z\in A_n$ with $xy=z$.

Proof: Let $p\in\beta\mathbf{N}\setminus\mathbf{N}$, let $G=\prod_{n\to p}G_n$, and let $f$ be the internal function $\st\lim_{n\to p} 1_{A_n}$. Then $\int f\,d\mu_G \geq \epsilon$, so if $\pi:G\to\B_m(G)$ is the Bohr compactification then $\int\pi_* f\,d\mu_{\B_m(G)}\geq\epsilon$. But by the previous theorem $\B_m(G)$ is trivial, so $\pi_* f$ is a constant $\geq\epsilon$, so by Theorem 3 we have$$\langle f*f,f\rangle_{L^2(G)} = \langle \pi^*(\pi_*f*\pi_*f),f\rangle_{L^2(G)} = \langle \pi_*f*\pi_*f,\pi_*f\rangle_{L^2(\B_m(G))} \geq \epsilon^3.$$In other words the number of pairs $(x,y)\in G_n^2$ such that $x,y,xy\in A_n$ is at least $(\epsilon^3-o(1))|G_n|$ as $n\to p$, but since $p$ was arbitrary this must hold as $n\to\infty$.$\square$

Here is another nice criterion for quasirandomness, which can be found in Gowers's original paper: $(G_n)$ is not quasirandom if and only if the groups $G_n$ have nontrivial abelian quotients or nontrivial small quotients. In our setup we can write this the following way.

Theorem 8: Let $G$ be an ultrafinite group. Then one of the following three alternatives hold:
1. $\B_m(G)$ is trivial.
2. $\B_m(G)$ has a nontrivial abelian quotient.
3. $\B_m(G)$ has a nontrivial finite quotient.

Proof: Suppose $G=\prod_{n\to p} G_n$. By the previous theorem if $\B_m(G)$ is nontrivial then the groups $G_n$ have bounded-dimensional nontrivial representations $\pi_n:G_n\to U(d)$ as $n\to p$. By Jordan's theorem, $\pi_n(G_n)$ has a normal abelian subgroup $A_n$ of bounded index. If $A_n=\pi_n(G_n)$ as $n\to p$ then 2 holds, while if $A_n<\pi_n(G_n)$ as $n\to p$ then 3 holds.$\square$

Using this theorem one can prove a sort of converse to Theorem 7. If $(G_n)$ is not quasirandom then there are arbitrarily large $n$ and product-free subsets $A_n\subset G_n$ of density bounded away from $0$. We leave the details to the reader.

4. Roth's theorem

As an application of group limits we can prove the following version of Roth's theorem.

Theorem 9: Let $G$ be a finite group on which the squaring map $s:y\mapsto y^2$ is $O(1)$-to-$1$. Let $A\subset G$ be a subset of density $|A|/|G|\geq\epsilon>0$. Then there are $\gtrsim_\epsilon |G|^2$ solutions to $y^2=xz$ in $A$. Equivalently, there are $\gtrsim_\epsilon|G|^2$ pairs $(a,b)\in G^2$ such that $a,ab,bab\in A$.

Proof: If the theorem fails then we have finite groups $G_n$, some $\epsilon>0$, and subsets $A_n\subset G_n$ of density $|A_n|/|G_n|\geq\epsilon$ for which there are fewer than $|G_n|^2/n$ pairs $(a,b)\in G_n^2$ such that $a,ab,bab\in A_n$. Let $G=\prod_{n\to p} G_n$ and $f = \st\lim_{n\to p} 1_{A_n}$. Then $\int_G f \,d\mu_G \geq \epsilon$ but$$\int_{G^2} f(a)f(ab)f(bab)\,d\mu_G^2 = 0.(*)$$But note$$\int_{G^2} f(a)f(ab)f(bab)\,d\mu_G^2 = \int_{G^2} f(x) f(y) f(x^{-1}y^2)\,d\mu_G^2= \langle f,(f*f)\circ s\rangle_{L^2(G)},$$and by Theorem 3 this becomes$$\langle f,\pi^*(\pi_*f*\pi_*f)\circ s\rangle_{L^2(G)} = \langle f, \pi^*((\pi_*f*\pi_*f)\circ s)\rangle_{L^2(G)}$$$$= \langle \pi_*f,(\pi_*f*\pi_*f)\circ s\rangle_{L^2(B)} = \int_{B^2} \pi_*f(a)\,\pi_*f(ab)\,\pi_*f(bab)\,d\mu_B^2,(**)$$where $B=\B_m(G)$.

Fix a small $\delta>0$, and let $g:B\to[0,1]$ be a continuous function such that $\|\pi_*f-g\|_{L^1(G)} \leq \delta$. Since a continuous function on a compact space is uniformly continuous we can find a neighbourhood $U$ of $1$ such that$$|g(x)-g(ux)|\leq \delta\quad\text{and}\quad|g(x)-g(xu)|\leq\delta$$ for all $x\in B$ and $u\in U$. Now note$$\int_B\int_{xU\times U} |\pi_*f(a)-g(a)| \,d\mu_B(a)d\mu_B(b)d\mu_B(x) = \mu_B(U)^2\|\pi_*f-g\|_{L^1(B)} \leq \mu_B(U)^2\delta,$$$$\int_B\int_{xU\times U} |\pi_*f(ab)-g(ab)| \,d\mu_B(a)d\mu_B(b)d\mu_B(x) = \mu_B(U)^2\|\pi_*f-g\|_{L^1(B)} \leq \mu_B(U)^2\delta,$$$$\int_B\int_{xU\times U} |\pi_*f(bab)-g(bab)| \,d\mu_B(a)d\mu_B(b)d\mu_B(x) = \mu_B(U)^2\|\pi_*f-g\|_{L^1(B)} \leq \mu_B(U)^2\delta,$$so if $R$ is the set of $x\in B$ such that$$\int_{xU\times U}|\pi_*f(a)-g(a)|\,d\mu_B(a)d\mu_B(b) \leq \delta^{1/2}\mu_B(U)^2,$$$$\int_{xU\times U}|\pi_*f(ab)-g(ab)|\,d\mu_B(a)d\mu_B(b) \leq \delta^{1/2}\mu_B(U)^2,$$$$\int_{xU\times U}|\pi_*f(bab)-g(bab)|\,d\mu_B(a)d\mu_B(b) \leq \delta^{1/2}\mu_B(U)^2,$$then $\mu_B(R^c)\leq 3\delta^{1/2}$. Thus$$\int_R g(x)\,d\mu_B(x) \geq \int_B g(x)\,d\mu_B(x) - 3\delta^{1/2}\geq \epsilon-4\delta^{1/2},$$
so there is some $x\in R$ such that$$g(x)\geq\epsilon-4\delta^{1/2}.$$Now using all our information about $x$ we can bound $(**)$ below:$$\int \pi_*f(a)\pi_*f(ab)\pi_*f(bab) \,d\mu_B(a)d\mu_B(b)\geq \int_{xU\times U} \pi_*f(a)\pi_*f(ab)\pi_*f(bab)\,d\mu_B(a)d\mu_B(b)$$$$\geq \int_{xU\times U} g(a)g(ab)g(bab)\,d\mu_B(a)d\mu_B(b) - 3\delta^{1/2}\mu_B(U)^2$$$$\geq \int_{xU\times U} g(x)^3\,d\mu_B(a)d\mu_B(b) - 6\delta\mu_B(U)^2 - 3\delta^{1/2}\mu_B(U)^2$$$$\geq\left( (\epsilon-4\delta^{1/2})^3 - 6\delta - 3\delta^{1/2}\right)\mu_B(U)^2.$$Now if $\delta$ is sufficiently small depending on $\epsilon$ this figure is positive, contradicting $(*)$.$\square$

We chose to count configurations of the form $(a,ab,bab)$ precisely because they are alternatively described by the rather simple equation $y^2=xz$. If instead we chose to count the more "obvious" nonabelian analgues of three-term arithmetic progressions, namely configurations of the form $(a,ab,ab^2)$, then we would be counting solutions to the more complicated equation $z = yx^{-1}y$. The problem is that the count of these configurations is not obviously controlled by convolutions, so we can't easily transport the problem to the Bohr compactification. In fact the situation is delicate and not completely understood: see for example this paper of Tao for the case of $\text{SL}_d(F)$.

Leon Green's theorem

The fundamental objects of study in higher-order Fourier analysis are nilmanifolds, or in other words spaces given as a quotient $G/\Gamma$ of a connected nilpotent Lie group $G$ by a discrete cocompact subgroup $\Gamma$. Starting with Furstenberg's work on Szemeredi's theorem and the multiple recurrence theorem, work by Host and Kra, Green and Tao, and several others has gradually established that nilmanifolds control higher-order linear configurations in the same way that the circle, as in the Hardy-Littlewood circle method, controls first-order linear configurations.

Of basic importantance in the study of nilmanifolds is equidistribution: one needs to know when the sequence $g^n x$ equidistributes and when it is trapped inside a subnilmanifold. It turns out that this problem was already studied by Leon Green in the 60s. To describe the theorem note first that the abelianisation map $G\to G/G_2$ induces a map from $G/\Gamma$ to a torus $G/(G_2\Gamma)$ which respects the action of $G$, and recall that equidistribution on tori is well understood by Weyl's criterion. Leon Green's beautiful theorem then states that $g^n x$ equidistributes in the nilmanifold if and only if its image in the torus $G/(G_2\Gamma)$ equidistributes.

Today at our miniseminar Aled Walker showed us Parry's nice proof of this theorem, which is more elementary than Green's original proof. During the talk there was some discussion about the importantance of various hypotheses such as 'simply connected' and 'Lie'. It turns out that the proof works rather generally for connected locally compact nilpotent groups, so I thought I would record the proof here with minimal hypotheses. The meat of the argument is exactly as in Aled's talk and, presumably, Parry's paper.

Let $G$ be an arbitrary locally compact connected nilpotent group, say with lower central series$$G=G_1\geq G_2 \geq\cdots\geq G_s\geq G_{s+1}=1,$$and let $\Gamma\leq G$ be a closed cocompact subgroup. Under these conditions the Haar measure $\mu_G$ of $G$ induces a $G$-invariant probability measure $\mu_{G/\Gamma}$ on $G/\Gamma$. We say that $x_n\in G/\Gamma$ is equidistributed if for every $f\in C(G/\Gamma)$ we have$$\frac{1}{N}\sum_{n=0}^{N-1} f(x_n) \to \int f \,d\mu_{G/\Gamma}.$$We fix our attention on the sequence$$x_n = g^n x$$for some $g\in G$ and $x\in G/\Gamma$. As before we have an abelianisation map$$\pi: G/\Gamma\to G/G_2\Gamma$$from the $G$-space $G/\Gamma$ to the compact abelian group $G/G_2\Gamma$. We define equidistribution on $G/G_2\Gamma$ similarly. The theorem is then the following.

Theorem: (Leon Green) For $g\in G$ and $x\in G/\Gamma$ the following are equivalent.
1. The sequence $g^n x$ is equidistributed in $G/\Gamma$.
2. The sequence $\pi(g^n x)$ is equidistributed in $G/G_2\Gamma$.
3. The orbit of $\pi(g)$ is dense in $G/G_2\Gamma$.
4. $\chi(\pi(g))\neq 0$ for every nontrivial character $\chi:G/G_2\Gamma\to\mathbf{R}/\mathbf{Z}$.

Item 1 above trivially implies every other item. The implication 4$\implies$3 (a generalised Kronecker theorem) follows by pulling back any nontrivial character of $(G/G_2\Gamma)/\overline{\langle\pi(g)\rangle}$. The implication 3$\implies$2 (a generalised Weyl theorem) follows from the observation that every weak* limit point of the sequence of measures$$ \frac{1}{N}\sum_{n=0}^{N-1} \delta_{\pi(g^n x)}$$must be shift-invariant and thus equal to the Haar measure. So the interesting content of the theorem is 2$\implies$1.

A word about the relation to ergodicity: By the ergodic theorem the left shift $\tau_g:x\mapsto gx$ is ergodic if and only if for almost every $x$ the sequence $g^n x$ equidistributes; on the other hand $\tau_g$ is uniquely ergodic, i.e., the only $\tau_g$-invariant measure is the given one, if and only if for every $x$ the sequence $g^n x$ equidistributes. Thus to prove the theorem above we must not only prove that $\tau_g$ is ergodic but that it is uniquely ergodic. Fortunately one can prove these two properties are equivalent in this case.

Lemma: If $\tau_g:G/\Gamma\to G/\Gamma$ is ergodic then it's uniquely ergodic.

Proof: (Furstenberg) By the ergodic theorem the set $A$ of $\mu_{G/\Gamma}$-generic points, in other words points $x$ for which$$\frac{1}{N}\sum_{n=0}^{N-1} f(g^n x) \to \int f \,d\mu_{G/\Gamma}$$for every $f\in C(G/\Gamma)$, has $\mu_{G/\Gamma}$-measure $1$, and clearly if $x\in A$ and $c\in G_s$ then $xc\in A$, so $A = p^{-1}(p(A))$, where $p$ is the projection of $G/\Gamma$ onto $G/G_s\Gamma$.

Now let $\mu'$ be any $\tau_g$-invariant ergodic measure. By induction we may assume that $\tau_g:G/G_s\Gamma\to G/G_s\Gamma$ is uniquely ergodic, so we must have $p_*\mu' = p_*\mu_{G/\Gamma}$, so$$\mu'(A) = p_*\mu'(p(A)) = p_*\mu_{G/\Gamma}(p(A)) = \mu_{G/\Gamma}(A) = 1.$$But by the ergodic theorem the set of $\mu'$-generic points must also have $\mu'$-measure $1$, so there must be some point which is both $\mu_{G/\Gamma}$- and $\mu'$-generic, and this implies that $\mu'=\mu_{G/\Gamma}$.$\square$

We need one more preliminary lemma about topological groups before we really get started on the proof.

Lemma: If $H$ and $K$ are connected subgroups of some ambient topological group then $[H,K]$ is also connected.

Proof: Since $(h,k)\mapsto [h,k]=h^{-1}k^{-1}hk$ is continuous certainly $C = \{[h,k]:h\in H,k\in K\}$ is connected, so $C^n = CC\cdots C$ is also connected, so because $1\in C^n$ for all $n$ we see that $[H,K]=\bigcup_{n=1}^\infty C^n$ is connected.$\square$

Thus if $G$ is connected then every term $G_1,G_2,G_3,\dots$ in the lower central series of $G$ is connected.

Proof of Leon Green's theorem: As noted it suffices to prove that $\tau_g$ acts ergodically on $G/\Gamma$ whenever it acts ergodically on $G/G_2\Gamma$. By induction we may assume that $\tau_g$ acts ergodically on $G/G_s\Gamma$. So suppose that $f\in L^2(G/\Gamma)$ is $\tau_g$-invariant. By decomposing $L^2(G/\Gamma)$ as a $\overline{G_s\Gamma}/\Gamma$-space we may assume that $f$ obeys$$f(cx)=\gamma(c)f(x)\quad(c\in G_s, x\in G/\Gamma)$$for some character $\gamma:G_s\Gamma/\Gamma\to S^1$. In particular $|f|$ is both $G_s$-invariant and $\tau_g$-invariant, so it factors through a $\tau_g$-invariant function $G/G_s\Gamma\to\mathbf{R}$, so it must be constant, say $1$. Moreover for every $b\in G_{s-1}$ the function$$\Delta_bf(x) = f(bx)\overline{f(x)}$$is $G_s$-invariant, and also a $\tau_g$ eigenvector:$$\Delta_bf(gx) = \gamma([b,g])\Delta_bf(x).$$By integrating this equation we find that either $\gamma([b,g])=1$, so $\Delta_bf$ is constant, or $\int\Delta_bf \,d\mu_{G/\Gamma}= 0$, so either way we have$$\int \Delta_bf\,d\mu_{G/\Gamma}\in \{0\}\cup S^1.$$But since $\int\Delta_bf\,d\mu_{G/\Gamma}$ is a continuous function of $b$ and equal to $1$ when $b=1$ we must have $\gamma([b,g])=1$ for all sufficiently small $b$, and thus for all $b$ by connectedness of $G_{s-1}$ and the identity$$[b_1b_2,g]=[b_1,g][b_2,g].$$Thus setting $\gamma(b)=\Delta_bf$ extends $\gamma$ to a homomorphism $G_{s-1}\to S^1$. In fact we can extend $\gamma$ still further to a function $G\to D_1$, where $D_1$ is the unit disc in $\mathbf{C}$, by setting$$\gamma(a) = \int \Delta_af\,d\mu_{G/\Gamma}.$$Now if $a\in G$ and $b\in G_{s-1}$ then$$\gamma(ba) = \int f(bax) \overline{f(x)}\,d\mu_{G/\Gamma} = \int \gamma(b)f(ax)\overline{f(x)}\,d\mu_{G/\Gamma}=\gamma(b)\gamma(a),$$and$$\gamma(ab) = \int f(abx)\overline{f(x)}\,d\mu_{G/\Gamma} = \int f(ax) \overline{f(b^{-1}x)}\,d\mu_{G/\Gamma} = \int f(ax) \overline{\gamma(b^{-1})}\overline{f(x)}\,d\mu_{G/\Gamma} = \gamma(b)\gamma(a),$$so$$\gamma(b)\gamma(a)=\gamma(ab)=\gamma(ba[a,b]) = \gamma(ba)\gamma([a,b])=\gamma(b)\gamma(a)\gamma([a,b]).$$Since $|\gamma(b)|=1$ we can cancel $\gamma(b)$, so$$\gamma(a)(\gamma([a,b])-1) = 0.$$Finally observe that $\gamma(a)$ is a continuous function of $a$, and $\gamma(1)=1$, so we must have $\gamma([a,b])=1$ for all sufficiently small $a$, and thus by connectedness of $G$ and the identity$$[a_1a_2,b]=[a_1,b][a_2,b]$$we must have $\gamma([a,b])=1$ identically. But this implies that $\gamma$ vanishes on all $s$-term commutators and thus on all of $G_s$, so in fact $f$ factors through $G/G_s\Gamma$, so it must be constant.$\square$

A remark is in order about the possibility that some of the groups $G_i$ and $G_i\Gamma$ are not closed. This should not matter. One could either read the above proof as it is written, noting carefully that I never said groups should be Hausdorff, or, what's similar, instead modify it so that whenever you quotient by a group $H$ you instead quotient by the group $\overline{H}$.

Embarrassingly, it's difficult to come up with a non-Lie group to which this generalised Leon Green's theorem applies. It seems that many natural candidates have the property that $G$ is not connected but $G/\Gamma$ is: for example consider$$\left(\begin{array}{ccc}1&\mathbf{R}\times\mathbf{Q}_2&\mathbf{R}\times\mathbf{Q}_2\\0&1&\mathbf{R}\times\mathbf{Q}_2\\0&0&1\end{array}\right)/\left(\begin{array}{ccc}1&\mathbf{Z}[1/2]&\mathbf{Z}[1/2]\\0&1&\mathbf{Z}[1/2]\\0&0&1\end{array}\right).$$So it would be interesting to know whether the theorem extends to such a case. Or perhaps there are no interesting non-Lie groups for this theorem, which would be a bit of a let down.

Erdős-Turán statistical group theory

What is Erdős-Turán "statistical group theory"? Erdős and Turán published a series of seven papers with this title, from 1965 to 1972, in which they proved many beautiful statistical or counting results about permutations. For example, if $|\sigma|$ denotes the order of a permutation $\sigma\in S_n$, they showed that when $\sigma$ is chosen uniformly at random $\log|\sigma|$ is approximately normally distributed, with mean $(\log n)^2/2$ and variance $(\log n)^3/3$. Another typical result, though much simpler, is that if $A$ is any subset of $\{1,\dots,n\}$, then the probability that a random permutation contains no cycle with length in $A$ is at most $2\left(\sum_{a\in A} 1/a\right)^{-1}$.

Although most of their results are of the above approximate nature, they prove at least one beautiful exact counting result, and I thought I might relate it here.

Theorem: If $q$ is a prime power then the proportion of $\sigma\in S_n$ with order not divisible by $q$ is exactly

\[\left(1-\frac{1}{q}\right)\left(1-\frac{1}{2q}\right)\cdots\left(1-\frac{1}{\lfloor n/q\rfloor q}\right).\]

Proof: The number of $\sigma\in S_n$ with $m_1$ cycles of length $v_1$, $m_2$ cycles of length $v_2$, $\dots$, and $m_k$ cycles of length $v_k$, where $m_1 v_1 + \cdots m_k v_k = n$, is
\[\frac{n!}{m_1!\cdots m_k! v_1^{m_1}\cdots v_k^{m_k}}:\]
indeed, one can partition $\{1,\dots,n\}$ into $m_i$ sets of size $v_i$ for each $i$ in
\[\frac{n!}{m_1!\cdots m_k! v_1!^{m_1}\cdots v_k!^{m_k}}\]
ways, and then one can arrange each of the $m_i$ sets of size $v_i$ for each $i$ into cycles in
\[(v_1 - 1)!^{m_1} \cdots (v_k - 1)!^{m_k}\]
ways. Moreover, the order of every such $\sigma$ is $\text{lcm}(v_1,\dots,v_k)$, so the order of $\sigma$ is divisible by $q$ if and only if some $v_i$ is divisible by $q$. (This is the where we use the hypothesis that $q$ is a prime power.) Thus the proportion of $\sigma\in S_n$ with order not divisible by $q$ is
\[\sum\frac{1}{m_1!\cdots m_k! v_1^{m_1} \cdots v_k^{m_k}},\]
where the sum runs over all $k\geq 0$ and $2k$-tuples $(m_1,\dots,m_k,v_1,\dots,v_k)$ of positive integers such that $m_1 v_1 + \cdots m_k v_k = n$ and such that no $v_i$ is divisible by $q$. But this is just the coefficient of $X^n$ in
\[\prod_{v:q\nmid v} \sum_{m\geq 0} \frac{X^{mv}}{m! v^m}\\=\prod_{v:q\nmid v} \exp\left(\frac{X^v}{v}\right)\\= \exp\left(\sum_{v:q\nmid v} \frac{X^v}{v}\right)\\= \exp\left(\sum_v \frac{X^v}{v} - \sum_v \frac{X^{qv}}{qv}\right)\\= \exp\left(-\log(1-X) + \log(1-X^q)/q\right)\\ = \frac{(1-X^q)^{1/q}}{1-X}\\= (1 + X + \cdots + X^{q-1}) (1-X^q)^{-\frac{q-1}{q}}\\= (1+X+\cdots+X^{q-1}) \left(1 + \left(1-\frac{1}{q}\right) X^q + \left(1-\frac{1}{q}\right)\left(1-\frac{1}{2q}\right)X^{2q} + \cdots\right),\]
where the last equality follows from the binomial formula. This completes the proof.

Such an explicit formula should make us feel foolish for having used generating functions. Is there a direct combinatorial proof?

By the way, for those who are not already aware of this invaluable resource, almost all of Erdős's papers have been made freely available by the Rényi institute here: http://www.renyi.hu/~p_erdos/Erdos.html.

How many lattices are there?

The basic question is the one in the title: How many lattices are there? I intend to answer two quite different interpretations of this question.

1. How many sublattices of the standard lattice $\mathbf{Z}^n$ are there? Here sublattice just means subgroup. There are infinitely many of course. Less trivially, how many sublattices of $\mathbf{Z}^n$ are there of index at most $m$?
2. How many lattices are there in $\mathbf{R}^n$ of covolume $1$? Here a lattice means a discrete subgroup of rank $n$, and covolume refers to the volume of a fundamental domain. Again, the answer is infinitely many. Less trivially, what (in an appropriate sense) is the volume of the set of covolume-$1$ lattices?

Question number 2 obviously needs some explanation. Covolume, as can easily be proved, is equal to the determinant of any complete set of spanning vectors, from which it follows that covolume-$1$ lattices are in some way parameterised by $\text{SL}_n(\mathbf{R})$. Actually, we are counting each lattice several times here, because each lattice has a stabliser isomorphic to $\text{SL}_n(\mathbf{Z})$. Thus, the space of covolume-$1$ lattices can be identified with the quotient space $\text{SL}_n(\mathbf{R})/\text{SL}_n(\mathbf{Z})$.

This subgroup $\text{SL}_n(\mathbf{Z})$ itself is also considered a lattice, in the group $\text{SL}_n(\mathbf{R})$. Generally, let $G$ be a locally compact group, and recall that $G$ possesses a unique (up to scale) left-invariant regular Borel measure, its Haar measure $\mu$. Suppose moreover that $G$ is unimodular, i.e., that $\mu$ is also right-invariant. If $H\leq G$ is a unimodular subgroup, and we fix a Haar measure on $H$ as well, then $\mu$ descends to a unique $G$-invariant regular Borel measure on the homogeneous space $G/H$ which we also denote by $\mu$. If $\mu(G/H)<\infty$, we say that $H$ has finite covolume. For example, if $H$ is a discrete subgroup then we can fix the counting measure as a bi-invariant Haar measure. If in this case $H$ has finite covolume then we say that $H$ is a lattice.

It is a classical theorem due to Minkowski, which we prove in a moment, that $\text{SL}_n(\mathbf{Z})$ is a lattice in the unimodular group $\text{SL}_n(\mathbf{R})$. We thus consider the volume of $\text{SL}_n(\mathbf{R})/\text{SL}_n(\mathbf{Z})$, for a natural choice of Haar measure, to be the answer to question 2 above, at least in some sense. The actual computation of this volume, I understand, is originally due to Siegel, but I learnt it from this MO answer.

Let $\lambda$ be the usual Lebesgue measure on $\mathbf{R}^{n^2}$, normalized (as usual) so that $\lambda(\mathbf{R}^{n^2}/\mathbf{Z}^{n^2}) = 1$, and note that $\lambda$ is invariant under the left multiplication action of $\text{SL}_n(\mathbf{R})$. Given closed $E\subset\text{SL}_n(\mathbf{R})$, we denote $\text{cone}(E)$ the union of all the line segments which start at $0\in\mathbf{R}^{n^2}$ and end in $E$, and we define
$$\mu(E) = \lambda(\text{cone}(E)).$$
This function $\mu$ extends to a nontrivial Borel measure in the usual way, and inherits regularity and both left- and right-invariance from $\lambda$. Thus $\text{SL}_n(\mathbf{R})$ is unimodular, and $\mu$ is a Haar measure. We consider this $\mu$ to be the natural choice of Haar measure.

Now let $E\subset\text{SL}_n(\mathbf{R})$ be a measurable fundamental domain for $\text{SL}_n(\mathbf{Z})$. Then, for $R>0$,
\[
\mu(\text{SL}_n(\mathbf{R})/\text{SL}_n(\mathbf{Z})) = \mu(E) = \lambda(\text{cone}(E)) = \frac{\lambda(R\,\text{cone}(E))}{R^{n^2}}.
\]
But $\lambda(R\,\text{cone}(E))$ is asymptotic, as $m\rightarrow\infty$, to $|R\,\text{cone}(E)\cap M_n(\mathbf{Z})|$, which is just the number of sublattices of $\mathbf{Z}^n$ of index at most $R^n$. Thus we have reduced question 2 to question 1.

To finish the proof of Minkowski's theorem, it suffices to show that there are $O(R^{n^2})$ sublattices of $\mathbf{Z}^n$ of index at most $R^n$, and this much can be accomplished by a simple inductive argument. If $\Gamma\leq\mathbf{Z}^n$ has index at most $R^n$, then by Minkowski's convex bodies theorem there exists some nonzero $\gamma\in\Gamma$ of $\|\gamma\|_\infty\leq R$. Without loss of generality, $\Gamma\cap\mathbf{R}\gamma = \mathbf{Z}\gamma$. But then $\Gamma^\prime = \Gamma/(\mathbf{Z}\gamma)$ is a lattice in $\mathbf{Z}^n/(\mathbf{Z}\gamma)\cong\mathbf{Z}^{n-1}$ of index at most $R^n$. Since there are no more than $(2R+1)^n$ choices for $\gamma$ and, by induction, at most $O(R^{n(n-1)})$ choices for $\Gamma^\prime$, there are at most $O(R^{n^2})$ choices for $\Gamma$.

But with a more careful analysis, this calculation can actually be carried out explicitly. An elementary(-ish) calculation shows that the number $a_m(\mathbf{Z}^n)$ of subgroups of $\mathbf{Z}^n$ of index exactly $m$ is the coefficient of $m^{-s}$ in the Dirichlet series of
\[
\zeta_{\mathbf{Z}^n}(s) = \zeta(s)\zeta(s-1)\cdots\zeta(s-n+1),
\]
where $\zeta(s) = \sum_{n\geq1} n^{-s}$ is the Riemann zeta function. Thus, by Perron's formula, if $x\notin\mathbf{Z}$,
\[
\frac{1}{x^n}\sum_{1\leq m< x} a_m(\mathbf{Z}^n) = \frac{1}{i2\pi} \int_{n+1/2-i\infty}^{n+1/2+i\infty} \zeta_{\mathbf{Z}^n}(z) \frac{x^{z-n}}{z}\,dz.
\]
But, by Cauchy's residue theorem, this integral differs from
\[
\text{Res}_{z=n}\left(\frac{\zeta_{\mathbf{Z}^n}(z) x^{z-n}}{z}\right) = \frac{1}{n}\zeta(2)\zeta(3)\cdots\zeta(n)
\]
by
\[
\frac{1}{i2\pi} \int_{n-1/2-i\infty}^{n-1/2+i\infty} \zeta_{\mathbf{Z}^n}(z) \frac{x^{z-n}}{z}\,dz,
\]
which tends to $0$ as $x\to\infty$. Thus there are about
\[
\frac{1}{n}\zeta(2)\zeta(3)\cdots\zeta(n) m^n
\]
sublattices of $\mathbf{Z}^n$ of index at most $m$, and
\[
\mu(\text{SL}_n(\mathbf{R})/\text{SL}_n(\mathbf{Z})) = \frac{1}{n}\zeta(2)\zeta(3)\cdots\zeta(n).
\]

Generating direct powers

Before continuing, prove or disprove: If $G$ is a group, the direct power $G^{\times n}$ is never generated by fewer than $n$ elements.

Certainly this is the case if $G$ is abelian, as in this case $G^{\times n}$ has a quotient of the form $\mathbf{F}_p^n$, i.e., an $n$-dimensional vector space over $\mathbf{F}_p$. This is also the case if $n\leq 2$. Examples of  small size are therefore a little hard to find. 

Here is a nice example: Denote the $m$th prime by $p_m$, and let $G = A_{p_m}$. Then I claim that $G^{\times(m-1)}$ can be generated with $3$ elements. Indeed, take $\alpha,\beta\in G^{\times(m-1)}$ to be in each factor equal to some two generators $\tilde{\alpha},\tilde{\beta}$ of $G$, and take $\gamma$ to be an element of the form  (3-cycle, 5-cycle, 7-cycle, ... , $p_m$-cycle). Then the $(p_2\cdots p_j p_{j+2}\cdots p_m)$th power of $\gamma$ is a $p_{j+1}$-cycle in the $j$th factor. With $\alpha$ and $\beta$ we can generate all the conjugates of this cycle, and these together generate the whole $j$th factor by simplicity.