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

$$AB A^{-1} B^{-1} \approx I - a^2 -b^2$$.

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:

$$\alpha = s_n \dots, s_{b+1}, \varepsilon = s_b \dots, s_{a+1},\text{ and }\beta = s_a \dots s_2s_1$$.

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

$$\alpha \varepsilon \beta H = \alpha \beta \beta^{-1} \varepsilon \beta H = \alpha \beta H$$

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

$$w=w_\Omega = \{A \in \mathsf{PGL}(\Omega): d_\Omega(x, Ax) < \varepsilon_n\}$$.

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, $$\mathsf{PGL}(\Omega) \cong \mathsf{SL}(\Omega)\subset \mathsf{SL}_{n+1}^\pm (\mathbb R)$$.

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