# Lecture 7: The Margulis Lemma

*General idea: A subgroup of Lie group generated by elements close to the identity gives an uncomplicated (almost abelian) algebra. Groups generated by small elements are almost abelian.*

First, we state the main result of this lecture:

**Margulis Theorem.** (Margulis, Thurston, Choi, Crampon-Marquis, Cooper-Long-Tillmann)
For each \(n \ge 2\) there exist \(m_n\), \(\varepsilon_n\) with:
Let \(\Omega \subset \mathbb R P^n\) be an open properly convex domain. Let \(\Gamma \subset \mathsf{PGL}(\Omega)\) a discrete group. Let \(x \in \Omega\).
Let \(\varepsilon < \varepsilon_n\) and \(\Gamma_\varepsilon(x) =\langle \gamma \in \Gamma: d_\Omega(x, \gamma(x)) < \varepsilon \rangle\). Then \(\Gamma_\varepsilon(x)\) contains a nilpotent subgroup of index at most \(m_n\).

Note: The most general version, shown by Cooper-Long-Tillman, is what’s stated here.

**Zassenhaus Neighborhood Theorem.**
Let \(G\) be a Lie group. Then there exists an open neighborhood \(U\) of the identity such that if \(\Gamma \subset G\) is a discrete group then \(\langle \Gamma \cap U \rangle\) is nilpotent.

Rough idea: We can locally embed \(G \rightarrow \mathsf{GL}_N(\mathbb R)\) for a sufficiently large \(N\), so we only need to show this for \(\mathsf{GL}_N(\mathbb R)\). Near the identity, we have a map \(\mathsf{GL}_N(\mathbb R) \times \mathsf{GL}_N(\mathbb R) \rightarrow \mathsf{GL}_N(\mathbb R)\) which maps a pair to its group theoretic commutator \((A,B) \mapsto ABA^{-1} B^{-1}\). If \(A\) is near the identity, \(A = I +a\) where \(a\) is small, and \(A^{-1} = I -a + [\text{something else small}]\). So, we have

Thus, in a discrete group, taking commutators must terminate (eventually hit the identity) because it’s always getting smaller.

So far, we have two notions of small: things that don’t move elements in the domain far (Margulis Lemma) and things not far from the identity (Zassenhaus Neighborhood Theorem). We need a way to relate these two notions of small.

**Theorem.** (Cooper-Long-Tillmann)
Let \(d>0\). Then \(K = K_d \subset \mathsf{SL}_{n+1}(\mathbb R)\) compact such that if \((\Omega, 0)\) is in Benzecri position and \(A \in \mathsf{PGL}(\Omega)\) such that \(d_{\Omega}(0,A0) < d\) then \(A \in K\).

**Definition (Benzecri position).** \(\Omega\) is sandwiched between the ball of radius 1 centered at 0, and a ball of large radius.

**Coset Lemma.** Let \(G\) be a group with finitely generated set \(S\) closed under inverses \((S = S^{-1})\) and let \(H \subset G\) such that \([G:H] = k \le m+1\). Then there exists a set of coset representatives \(\{\gamma_1, \gamma_2, \dots, \gamma_k \}\) such that the word length in the generating set is less than or equal to \(m\), i.e. \(\mid\gamma_i\mid_S \le m\) for \(1 \le i \le k\).

Note: This is purely algebraic, even though some readings may not make it seem that way.

Proof: Given a coset \(gH = s_N \dots s_3s_2s_1 H\), where each \(s_i \in S\). Since \(S\) generates \(G\) this always works. We can to show we can choose our \(s_i\) such that \(N \le m\). We’ll split up our word into 3 parts:

We also observe that \(H, s_1H, s_2s_1H, \dots, s_N \dots s_1H\) are cosets, and there an \(N+1\) many of them. By the time we get to \(s_k \dots s_1H\), we’ve listed \(k+1\) cosets, but there are only \(k\) distinct cosets, so by the pigeonhole principle, we’ve listed at least one twice. So, there is some \(a,b\) with \(0 \le a<b\le k \le m_1\) such that \(s_b \dots s_1H = s_a\dots s_1H\). This \(\varepsilon \beta H = \beta H\). So, \(\beta^{-1} \varepsilon B H =H\), i.e. \(\beta^{-1} \varepsilon \beta \in H\). So, we have

which has a shorter word length.

**Proof of the Margulis Lemma:** Assume that \((\Omega, x)\) is in Benzecri posotion. Let \(K\) be as in the Theorem by Cooper-Long-Tillman for \(d=1\). Let \(U\) be the Zassenhaus neighborhood for \(\mathsf{SL}_{n+1}(\mathbb R)\) and let \(U' \subset \mathsf{SL}_{n+1}(\mathbb R)\) be a symmetric (meaning \(U' = (U')^{-1}\)) neighborhood of the identity such that \((U')^2 \subset U\). (\(U'\) is slightly smaller than \(U\).)
Since \(K\) is compact, we can cover \(K\) with \(m\) translates of \(U'\) for some finite number \(m\).

Let \(m_n = m\) and \(\varepsilon_n =\frac1m\). Let

Notice: \(w_\Omega = w_\Omega^{-1}\) and \(w_\Omega^m \subset K\). So, by our hypothesis, \(\Gamma_{\varepsilon _n}(x) = \langle \Gamma_\varepsilon(x) \cap w_\Omega\rangle\). Let \(\Gamma_U = \langle \Gamma_\varepsilon(x) \cap U \rangle\), and notice that \(\Gamma_u\) is nilpotent and \(\Gamma_u \subset \Gamma_\varepsilon(x)\).

**Claim:** \([\Gamma_\varepsilon(x): \Gamma_U ] \le m\)
Suppose otherwise. Then without loss of generality, assume that \([\Gamma_\varepsilon(x): \Gamma_U] = m+1\) (we can induct down to this step).
By the coset lemma, we can find coset representatives \(C = \{ \gamma_1 \dots, \gamma_{m+1}\}\) such that \(\mid \gamma_i \mid_{\Gamma \cap w} \le m\), so \(\gamma_i \in w_\Omega^m \subset K\). Since \(K\) is covered by \(m\) translates of \(U'\), we can find distinct \(g, g'\) in \(C\) and \(h \in \mathsf{SL}_{n+1}(\mathbb R)\) and \(u,u' \in U'\) such that \(g = hu\) and \(g' = hu'\). i.e. \(g^{-1} g' = u^{-1} u' \in \Gamma \cap U\), so \(g \Gamma u = g' \Gamma{u'}\), a contradiction, thus concluding our proof.

Now, we recall that \(S^n = (\mathbb R^{n+1} - \{0\}) /(x \sim \lambda m, \lambda>0)\) covers \(\mathbb R P^n\) by a 2 to 1 map \(\pi: S^n \rightarrow \mathbb R P^n\). Then, the spaces \(\mathsf{SL}_{n+1}^{\pm}(\mathbb R)\) and \(\mathsf{PGL}_{n+1}(\mathbb R)\) act on \(S^n\) and \(\mathbb R P^n\), respectively. There is a map \(f: \mathsf{SL}_{n+1}^{\pm}(\mathbb R) \rightarrow \mathsf{PGL}_{n+1}(\mathbb R)\) such that \(f(A) = [A]\).

We’d like to explore what \(f\) is doing.

Case 1: \(n\) is even. Then \(x \mapsto -x\) on \(\mathbb R^{n+1}\) is orientation reversing. So, \(\mathbb RP^n\) is non-orientable. We also know that \(\mathsf{PGL}_{n+1}(\mathbb R)\) is connected, and \(\mathsf{PGL}_{n+1}(\mathbb R) \cong \mathsf{SL}_{n+1}(\mathbb R)\). Intuitively, you can take odd roots of negative numbers, so it doesn’t mess up anything.

Case 2: \(n\) is odd. Then \(x \mapsto -x\) is orientation preserving, and \(\mathbb R P^n\) is orientable. We know \(\mathsf{PGL}_{n+1}(\mathbb R)\) has 2 components, so there exists a 2 to 1 cover \(\mathsf{SL}_{n+1}(\mathbb R) \rightarrow \mathsf{PGL}^\circ_{n+1}(\mathbb R)\), and so \(\mathsf{PGL}^\circ_{n+1} (\mathbb R) \cong \mathsf{PSL}_{n+1}(\mathbb R)\).

If \(\Omega \subset \mathbb R P^n\) is properly convex, then \(\pi^{-1}(\Omega)\) has two components: \(C_\Omega\) and \(-C_\Omega\). If \(A \in \mathsf{PGL}(\Omega)\) then there are two lifts, \(\widetilde A\) and \(- \widetilde A\), to \(\mathsf{SL}_{n+1}^\pm(\mathbb R)\). Thus, \(\widetilde A\) preserves the components if and only if \(-\widetilde A\) reverses them. So, \(\mathsf{PGL}(\Omega) \cong \mathsf{SL}(\Omega)\subset \mathsf{SL}_{n+1}^\pm (\mathbb R)\).

**Previous Post:** Lecture 6: Benzécri's compactness theorem