# On Gromov’s theorem on groups of polynomial growth

This article documents my presentation of Gromov’s theorem on groups of polynomial growth at the MIT combinatorics reading group. The presentation is based on Gromov’s 1981 paper, Groups of polynomial growth and expanding maps, Kleiner’s 2007 paper, A new proof of Gromov’s theorem on groups of polynomial growth, and Tao’s 2009 blog post, A finitary version of Gromov’s polynomial growth theorem.

#### Introduction

Throughout, we fix a finitely generated group G and a finite symmetric generating set S (that is \forall x \in S. x^{-1}\in S). For every group x \in G, the word length \lVert x \rVert of x is the shortest length n of a word s_1s_2 \dots s_n in S that expresses x.

Gromov’s theorem connects a group property of G with the growth of the cardinality of the ball B(r) := \left\{x \in G : \lVert x \rVert \le r\right\} of radius r. The ball B(r) can be seen as the set of vertices within distance r from the identity element of G in the Cayley graph of (G, S).

Definition (nilpotent and virtually nilpotent). A group H is nilpotent of class n if there is a lower central series H = H_0 \rhd H_1 \rhd \dots \rhd H_n = \{e\}, where H_{i+1} = [H_i, H]. A group G is virtually nilpotent if there is a finite index subgroup H of G that is nilpotent.

Example 1. When G is abelian, G is nilpotent and \lvert B(r) \rvert = r^{\mathrm{rank}(G)}.

Example 2. When G is the discrete Heisenberg group, that is G = \left\{ \begin{pmatrix} 1 & a & b \\ & 1 & c \\ & & 1 \end{pmatrix} : a, b, c \in \mathbb{Z} \right\}, we have the lower central series G > \langle \begin{pmatrix} 1 & & 1 \\ & 1 & \\ & & 1\end{pmatrix} \rangle > \{e\}, and the growth of \lvert B(r) \rvert is bounded by r^4.

Theorem (Gromov 1981). For every group G generated by a finite symmetric set S, \lvert B(r)\rvert is at most polynomial in r if and only if G is virtually nilpotent.

Gromov’s proof uses the deep Montgomery–Zippin–Yamabe structure theory of locally compact groups on Hilbert’s fifth problem. Colding and Minicozzi, solving a conjecture of Yau, showed that the space of harmonic functions with polynomial growth on an open manifold with non-negative Ricci curvature. To state (a weak version of) the discrete analog of their result, we need the notion of Lipschitz harmonic on the group G with the finite symmetric generating set S.

Definition (Lipschitz and harmonic). A function f\colon G\to\mathbb{R} is Lipschitz if \sup_{g \in G, s \in S}\lvert f(gs) - f(g) \rvert is finite, and is harmonic if f(g) = \frac{1}{\lvert S \rvert} \sum_{s \in S} f(gs) for all g \in G.

Theorem (Colding and Minicozzi 1997, Kleiner 2010). If \vert B(r)\lvert is at most polynomial in r, then the linear space of Lipschitz harmonic functions on G is finite dimensional.

Kleiner provided a new proof of Gromov’s theorem by establishing the Colding–Minicozzi theorem from scratch. Later Shalom and Tao pushed Kleiner’s methods to obtain the following quantitative version of Gromov’s theorem.

Theorem (Shalom and Tao 2010). There exists an absolute constant c such that if \lvert B(r) \rvert \le r^d for some r > \exp(\exp(cd^c)) then G contains a finite index subgraph H which is nilpotent of class \le c^d.

In the rest of the article, we will prove the Colding–Minicozzi theorem through the Poincaré inequality and the reverse Poincaré inequality.

#### Poincaré inequality

Lemma (Poincaré inequality). For every function f \colon G \to \mathbb{R}, if f has mean 0 on B(r), the l^2-norm of f on B(r) is bounded by the fluctuation of f on B(2r): \sum_{x\in B(r)}f(x)^2 \le \frac{\lvert B(2r) \rvert}{\lvert B(r) \rvert}\cdot 2r^2 \sum_{x,y\in B(2r), x\sim y}(f(x) - f(y))^2.

Proof. One can check that the left hand side is equal to \frac{1}{2\lvert B(r)\rvert}\sum_{x,y\in B(r)}(f(x)-f(y))^2.

For each z \in B(2r), we fix a shortest path e = z_0, z_1, \dots, z_{\lVert z \rVert} = z from e to z in the Cayley graph of (G, S). Given x,y\in B(r), let z = x^{-1}y \in B(2r) and get f(x) - f(y) = \sum_{i=1}^{\lVert z \rVert}f(xz_{i-1})-f(xz_i) \\ \implies (f(x)-f(y))^2 \le \lVert z \rVert \sum_{i=1}^{\lVert z \rVert}(f(xz_{i-1})-f(xz_i))^2.

When summing over the last inequality over all x,y\in B(r), we can regroup the summands by z and i as follows: \sum_{z \in B(2r)}\lVert z\rVert\sum_{i=1}^{\lVert z \rVert} \left(\sum_{x\in B(r) : xz \in B(r)}(f(xz_{i-1})-f(xz_i))^2\right).

Fix z and i for a moment. One can show that both xz_{i-1} and xz_i are in B(2r), and moreover the directed edges (xz_{i-1}, xz_{i}) are distinct when x varies in B(r). Thus \sum_{x\in B(r) : xz \in B(r)}(f(xz_{i-1})-f(xz_i))^2 \le \sum_{x,y\in B(2r), x\sim y}(f(x)-f(y))^2.

We obtain the Poincaré inequality by putting everything together and noticing that \lVert z \rVert \le 2r. [qed]

#### Reverse Poincaré inequality

Lemma (Reverse Poincaré inequality). For every harmonic function f\colon G \to \mathbb{R}, the fluctuation of f on B(r) is bounded by the l^2-norm of f on B(2r): \sum_{x,y\in B(r), x\sim y}(f(x) - f(y))^2 \le \lvert S \vert \cdot 4r^{-2}\sum_{x\in B(2r)}f(x)^2.

To facilitate the proof, we introduce the following notations. Given a function f\colon G\to \mathbb{R} and s\in S, write f_s(x) := f(xs) and \partial_s f := f_s - f. It is easy to see that

1. \sum_{s\in S}\partial_{s^{-1}}\partial_s f = 0 when f is harmonic, and
2. \sum_{x\in G}f(x)\partial_s g(x) = \sum_{x\in G}\partial_{s^{-1}}f(x) g(x) when f or g is finitely supported.

Proof. Fix the harmonic function f\colon G\to \mathbb{R} and let \phi\colon G \to [0,1] be defined by \phi(x) = \begin{cases} 1 & \text{if }\lVert x\rVert \le r,\\ 2 - \lVert x\rVert/r & \text{if }r < \lVert x\rVert < 2r, \\ 0 & \text{if }\lVert x\rVert \ge 2r.\end{cases}

For every s \in S, note that \partial_s (f\phi^2) = (\partial_s f)\phi^2 + f_s(\partial_s \phi^2) and \partial_s \phi^2 = (\partial_s \phi)\phi + \phi_s (\partial_s\phi) = (\partial_s \phi)(2\phi + \partial_s \phi). We obtain \begin{aligned}\partial_s f \partial_s (f\phi^2) & = (\partial_s f)^2 \phi^2 + (\partial_s f)f_s(\partial_s \phi)(2\phi + \partial_s\phi) \\ & \ge \tfrac{1}{2}(\partial_s f)^2\phi^2 - 2(f_s)^2(\partial_s \phi)^2 + (\partial_s f)f_s(\partial_s\phi)^2 \\ & = \tfrac{1}{2}(\partial_s f)^2\phi^2 - f_s(f_s + f)(\partial_s\phi)^2 \\ & \ge \tfrac{1}{2}(\partial_s f)^2\phi^2 - \tfrac{1}{2}(3f_s^2 + f^2)(\partial_s\phi)^2. \end{aligned}

When summing the above inequality over all s\in S and x \in G, by noticing that \sum_{s\in S}\sum_{x\in G}\partial_s f(x) \partial_s (f(x)\phi(x)^2) = \sum_{s\in S}\sum_{x\in G}(\partial_{s^{-1}}\partial_s f(x)) f(x)\phi(x)^2 \\ = \sum_{x\in G}\left(\sum_{s\in S}\partial_{s^{-1}}\partial_s f(x)\right) f(x)\phi^2(x) = 0, we get \sum_{x\in G}\sum_{s \in S}(\partial_s f(x))^2\phi(x)^2 \le \sum_{x\in G}\sum_{s \in S}(3f_s(xs)^2+f(x)^2)(\partial_s \phi(x))^2 .

The left hand side of the last inequality is at least \sum_{x,y\in B(r), x\sim y}(f(x)-f(y))^2, whereas the right hand side is at most 4\lvert S\rvert \tfrac{1}{r^2}\sum_{r \le \lVert x\rVert \le 2r}f(x)^2 because (\partial_s\phi)^2 \le 1/r^2 and \partial_s\phi(x)^2 > 0 only if both x and xs are in \{x \in G : r\le \lVert x\rVert \le 2r\}. [qed]

#### Colding–Minicozzi theorem

To simplify the presentation, we will assume the doubling constant \lvert B(2r) \rvert / \lvert B(r) \rvert is uniformly bounded at all scales r, which, for example, is indeed the case when \lvert B(r)\rvert = \Theta(r^d). In general, one needs the pigeonhole principle to select the correct radii for the argument below to work.

Proof assuming the doubling constant is uniformly bounded. Suppose for the sake of contradiction, the dimension of the linear space consisting of Lipschitz harmonic functions on G is at least n, where the parameter n will be determined later. Denote by V the n-dimensional linear subspace.

Let k be a natural number to be determined soon, and fix r for a moment. Let \mathcal{A}_r be a maximal collection of disjoint balls of radius r/2 with centers in B(kr), and let \mathcal{B}_r be the collection of balls with the same centers of the balls in \mathcal{A}_r, but of radius r. Let V_r be the linear subspace of V consisting of harmonic functions in V that average to 0 on each ball in \mathcal{B}_r. Note that the co-dimension of V_r as a subspace of V is at most \lvert \mathcal{B}_r \rvert = \lvert \mathcal{A}_r \rvert, which is at most \lvert B(kr+r/2) \rvert / \lvert B(r/2) \rvert = O(1) =: C.

For every harmonic function f in V_r, using the fact that B(kr) \subseteq \cup\mathcal{B}_r, the fact that each point in G is covered by 2B for at most \lvert B(2r+r/2) \rvert / \lvert B(r/2) \rvert = O(1) many B \in \mathcal{B}_r, the Poincaré inequality and the reverse Poincaré inequality, we get \begin{aligned}\sum_{x \in B(kr)}f(x)^2 & \le \sum_{B \in \mathcal{B}(r)}\sum_{x\in B}f(x)^2 \\ & \lesssim r^2\sum_{B \in \mathcal{B}(r)}\sum_{x,y \in 2B, x\sim y}(f(x)-f(y))^2 \\ & \lesssim r^2\sum_{x, y \in B(kr+2r), x\sim y}(f(x)-f(y))^2 \\ & \lesssim \frac{1}{(k+2)^2}\sum_{x \in B(2(k+2)r)}f(x)^2.\end{aligned} Now take k large enough (depending only on the group G) so that for all f\in V_r, 3^d\sum_{x\in B(kr)}f(x)^2 \le \sum_{x\in B(3kr)}f(x)^2.

Consider the quadratic form Q_r on V defined by Q_r(f) := \sum_{x\in B(r)}f(x)^2. Since the kernels of Q_r‘s form a descending chain of vector spaces, there exists r_0 such that Q_r is positive-definite for all r \ge r_0.

For every r \ge r_0, let q(r) be the volume of the ellipsoid E_r induced by Q_r. To be more precise, after fixing a basis \{f_1, \dots, f_n\} of V, the ellipsoid is defined by E_r := \{(c_1, \dots, c_n) \in \mathbb{R}^n : Q_r(c_1f_1 + \dots c_nf_n) \le 1\}. By scaling and translation, we may assume that f_i is 1-Lipschitz and f_i(e)=0 (whenever f_i is non-constant). Since \lvert B(r) \rvert is at most polynomial in r, there exists a natural number d (depending only on the group G) such that \sum_{x \in B(r)} f_i(x)^2 \le r^d for all i \in [n]. By Cauchy–Schwarz inequality, we have \begin{aligned}Q_r(c_1f_1 + \dots + c_nf_n) & = \sum_{x\in B(r)}(c_1f_1(x)+\dots+c_nf_n(x))^2 \\ & \le n\sum_{x\in B(r)}c_1^2f_1(x)^2 + \dots c_n^2f_n(x)^2 \\ & \le (c_1^2+\dots+c_n^2)n^2r^d.\end{aligned} Therefore E_r contains the ball of radius (nr^{d/2})^{-1}, hence q_r \ge v_n(nr^{d/2})^{-n}, where v_n is the volume of the n-dimensional Euclidean unit ball.

Although the volume q(r) of the ellipsoid is not intrinsic to Q_r, the ratio between q(r) and q(r') does not depend on the choice of the basis.

After a linear transformation, we may assume the symmetric matrices associated to Q_{kr} and Q_{3kr} are of the form \begin{pmatrix}A_1 & B_1 \\ B_1^T & C_1 \end{pmatrix}, \begin{pmatrix}A_3 & \\ & C_3\end{pmatrix}, where A_1, A_3 act on V_r. Using the Schur complement, we obtain that the volume ratio q_{3kr}/q_{kr} is \frac{\det Q_{kr}}{\det Q_{3kr}} = \frac{\det A_1 \det (C_1 - B_1^TA_1^{-1}B_1)}{\det A_3 \det C_3}. As B_1^TA_1^{-1}B_1 is positive semi-definite, we obtain \det(C_1 - B_1^TA_1^{-1}B_1) \le \det C_1 and so q_{3kr}/q_{kr} is at most \frac{\det A_1\det C_1}{\det A_3\det C_3}. Recall that 3^dQ_{kr}(f) \le Q_{3kr}(f) for all f \in V_r and clearly Q_{kr}(f) \le Q_{3kr}(f) for all f \in V. In other words, 3^dA_1 \preceq A_3 and C_1 \preceq C_3 and so q_{kr}/q_{3kr} \ge (3^d)^{\dim V_r} \ge (3^d)^{n-C}.

Choose n so that (3^d)^{n-C} \ge 2^{dn} hence 2^{dn}q(3kr) \le q(kr). Repeatedly apply the last inequality to obtain q(kr) \ge 2^{mdn}q(3^mkr) \ge 2^{mdn}v_n(n(3^mkr)^{d/2})^{-n} = \Omega_{d,n}((2/\sqrt{3})^{mdn}), which is absurd for m sufficiently large. [qed]