Volume and rank of lattices

 The guiding philosophy of geometric group theory is that large scale properties of groups are intimately related to the geometry (and topology) of spaces that they act on nicely. We will see another lovely example of this principle in action today, this time in the context of lattices $\Gamma$ in a connected semisimple Lie group $G$. We follow Gelander's lectures.

Let $X$ be the associated symmetric space of $G$. When $\Gamma$ has torsion, the quotient $M \colon= \Gamma \backslash X$ has ramified points, i.e. it is not a manifold, and dealing with the geometry of orbifolds is much more delicate, but nonetheless tractable. For a group $\Gamma$ we denote by $d(\Gamma)$ the minimal cardinality of a generating set. When considering the class of finite index subgroups $\Gamma$ in some given finitely generated group $\Delta$, it is easy to show that $d(\Gamma)$ is at most $(d(\Delta)-1)[\Delta:\Gamma]+1$ and in particular bounded linearly by the index. Using some Morse theory, non-positive curvature, and the Margulis lemma, one can prove the analogous statement when $\Delta$ is replaced by a connected semisimple Lie group $G$.

Theorem 1: Let $G$ be a connected semisimple Lie group without compact factors. Then there is an effectively computable constant $C=C(G)$ such that \[ d(\Gamma)\le C\mathrm{vol}(G/\Gamma)\] for every irreducible lattice $\Gamma\le G$.

The basic idea is to deform the symmetric space $X=G/K$ to nice connected $\Gamma$-invariant subset $Y\subset X$ where the displacement of every $\gamma\in\Gamma\setminus\{1\}$ is bounded below by a uniform constant, and deduce that $\Gamma$ is a quotient of $\pi_1(Y/\Gamma)$ while the topology of $Y/\Gamma$ is bounded by its volume since it is uniformly thick. We will explain the ingredients below before giving a sketch of the proof.

Some notation and background

Let $G$ be a connected centreless semisimple Lie group, $K\le G$ a maximal compact subgroup of $G$ and $X=G/K$ the associated Riemannian symmetric space.
$X$, when equipped with the analytic Riemannian metric coming from the Killing form of $G$, is nonpositively curved, i.e. the distance function $d:X\times X\to \mathbb{R}$ is convex. 

For $g\in G,~x\in X$ we denote by 
$$
 d_g(x)=d(x,g\cdot x)
$$ 
the displacement of $g$ at $x$, and by 
$$
 \mathrm{Min}(g):=\{x\in X:d_g(x)=\inf(d_g)\}
$$ 
the set where $d_g$ attains its infimum. Recall that $g$ is semisimple iff $\mathrm{Min}(g)\ne\emptyset$ (equivalently, iff $g$ belongs to a torus in $G$), and otherwise $g$ is called parabolic. 

Just as in the case of hyperbolic spaces, there is a classification of semisimple isometries. A semisimple element $g$ is elliptic if $\inf d_g=0$ in which case $\mathrm{Min}(g)$ is also denoted by 
$$
 \text{Fix}(g)=\{x\in X:g\cdot x=x\},
$$ 
and otherwise it is hyperbolic and admits an axis on which it translates by a positive amount and the set $\mathrm{Min}(g)$ consists of the union of all its axis. In either case $\mathrm{Min}(g)$ is a complete totally geodesic submanifold. 

$G$ admits a structure of a linear real algebraic group, and an element $g\in G$ is semisimple iff it is diagonalizable, and is unipotent iff all its eigenvalues are $1$. These properties are independent of the chosen representation of $G$. Moreover, a unipotent element $g\in G$ is parabolic with $\inf d_g=0$ (the converse is not true in general).

For $t\ge 0$ we denote by $\{d_g\le t\}$ the sublevel set $\{x:d_g(x)\le t\}$. Since $X$ has nonpositive sectional curvature, $\{ d_g\le t\}$ is convex.

A co-dimension 2 condition

Write $c~\|~c'$ to express that two geodesic lines $c,c'$ are parallel, and denote 
$$
 \mathfrak{I}(c):=\bigcup \{ c'(\mathbb{R}):c'~\|~c\}
$$
the union of all traces of geodesic lines in $X$ parallel to $c$. 

Lemma 2: Let $X$ be an irreducible symmetric space of noncompact type. If for some geodesic line $c$, $\mathfrak{I}(c)$ has codimension one, then $X$ is isometric to the hyperbolic plane $\mathbb{H}^2$.

Proof: Let $p^+,p^-\in \partial X$ be the endpoints of $c$ in the ideal boundary of $X$, and let $P^+,P^-\le G=\text{Isom}(X)^\circ$ be the corresponding parabolic subgroups. 
Observe that for $g\in P^+$, $g\cdot c(0)\in\mathfrak{I}(c)$ iff $g\in P^-$. Indeed,  if $g\in P^-\cap P^+$ then $g\cdot c$ is parallel to $c$, and vice versa: if $g\cdot c(0)\in\mathfrak{I}(c)$, the geodesic line through $g\cdot c(0)$ parallel to $c$ has one end $p^+$, and since this geodesic is determined by $g\cdot c(0)$ and $p^+=g\cdot p^+$ it coincides with $g\cdot c$, and its other end point $p^-$ has to be $g\cdot p^-$.
Since $P^+$ acts transitively on $X$, we deduce that 
$$
 \dim P^+/P^+\cap P^-=\dim X-\dim\mathfrak{I}(c)=1.
$$ 
Furthermore, since $P^+P^-$ is open in $G$, being the big cell in a Bruhat decomposition, the orbit $P^+\cdot p^-$ is open in $G\cdot p^-\sim G/P^-$. It follows that $G/P^-$ is one dimensional, hence, being compact, is homeomorphic to the circle $S^1$. However the only centreless simple Lie group that acts nontrivially on $S^1$ is $\mathrm{PSL}_2(\mathbb{R})$. Hence $G\cong\mathrm{PSL}_2(\mathbb{R})$ and $X\cong\mathbb{H}^2$. $\blacksquare$

Corollary 3: Let $G$ be a connected centreless semisimple Lie group without compact factors and not isometric to $\mathrm{PSL}_2(\mathbb{R})$, and let $g\in G$ be an element whose projection to every factor of $G$ is nontrivial. Then $\text{codim}_X\mathrm{Min}(g)\ge 2$.

Proof: If $G$ has more than one factor, $\mathrm{Min}(g)\cong\prod\mathrm{Min}(g_i)$ where $g_i$ are the components of $g$ so the conclusion is trivial since the co-dimension is at least the number of factors. Assume therefore that $G$ is simple. Since $G$ is connected, every $g\in G$ preserves the orientation of $X$, hence $\text{Fix}(g)$ is of codimension $\ge 2$ for every elliptic element $g\in G$. For $g$ hyperbolic, since $\mathrm{Min}(g)\subset\mathfrak{I}(c)$ where $c$ is an axis of $g$,
the result follows from Lemma 2.  $\blacksquare$

The Margulis lemma

I learned this classical fact for real hyperbolic space from Martelli's book. However, we will need these facts in a more general setting, so for my own sake I record the statements. For good measure we also recall a couple of theorems about Lie groups, since they are nice and sort of relevant.

Theorem 4 (Zassenhaus): Let $G$ be a Lie group. There is an open identity neighborhood $\Omega \subset G$ such that every discrete subgroup $\Delta\le G$ which is generated by $\Delta \cap \Omega$ is contained in a connected nilpotent Lie subgroup of $N\le G$. Moreover $\Delta $ is a uniform lattice in $N$.

A set $\Omega$ as in the theorem above is called a Zassenhaus neighborhood.

Jordan-Schur theorem: For a compact Lie group $K$ there is a constant $m\in \mathbb{N}$ such that every finite subgroup $\Delta\le K$ admits an abelian subgroup of index $\le m$. 

The result as stated is due to Jordan but has later been generalised to periodic subgroups by Schur, as well as over general linear groups of an arbitrary field with approaches that both use and avoid the classification of finite simple groups.

Proof: Let $\Omega$ be a Zassenhaus neighborhood in $K$, let $U$ be a symmetric identity neighborhood satisfying $U^2\subset\Omega$, and set $m:=\frac{\mu(K)}{\mu(U)}$. Given a finite subgroup $F\le K$ set $A=\langle F\cap\Omega\rangle$. Since compact connected nilpotent groups are abelian, $A$ is abelian. Now if $f_1,\ldots,f_{m+1}$ are $m+1$ elements in $F$ then for some $1\le i\ne j\le m+1$ we have $f_iU\cap f_jU\ne\emptyset$ implying that $f_i^{-1}f_j\in F\cap\Omega\subset A$. Thus $[F:A]\le m$. $\blacksquare$

Margulis Lemma: Let $G$ be a Lie group acting properly by isometries on a Riemannian manifold $X$. Given $x\in X$ there are $\epsilon=\epsilon(x)>0$ and $m=m(x)\in\mathbb{N}$ such that if $\Delta\le G$ is a discrete subgroup which is generated by the set 
$$
 \Sigma_{\Delta,x,\epsilon}:=\{\gamma\in\Delta:d(\gamma\cdot x,x)\le\epsilon\}
$$
then $\Delta$ admits a subgroup of index $\le m$ which is contained in a (closed) connected nilpotent Lie group. Furthermore, if $G$ acts transitively on $X$ then $\epsilon$ and $m$ are independent of $x$. 
\end{thm}

Proof: The properness of the action implies that the set 
$$
 C=\{ g\in G: d(g\cdot x,x)\le 1\}
$$ 
is compact. Let $V\subset G$ be a relatively compact open symmetric set such that $V^2$ is a Zassenhaus neighborhood. Setting 
$$
 m=[\frac{\mathrm{vol}(C\cdot V)}{\mathrm{vol}(V)}]~\text{and}~\epsilon=1/m
$$ 
one can prove the theorem arguing as in the proof of Jordan's theorem. An extra complication arises from the fact that $\Delta$ and $F=\langle\Delta\cap V^2\rangle$ are infinite, but this can be taken care of by observing that whether a connected graph has more than $m$ vertices or not, can be seen by looking at a ball of radius $m$ in the graph. Thus, assuming in contrary that the Schreier graph of $\Delta/F$ has more than $m$ vertices, we could find $m+1$ elements $\gamma_1,\ldots,\gamma_{m+1}$ in the $m$-ball $(\Sigma_{\Delta,x,\epsilon})^m$ which belong to mutually different cosets of $F$. However, by the choice of $\epsilon$ we have that $(\Sigma_{\Delta,x,\epsilon})^m\subset C$, hence for some $1\le i\ne j\le m+1$ we have $\gamma_iV\cap\gamma_jV\ne\emptyset$, i.e. $\gamma_i^{-1}\gamma_j\in V^2\cap\Delta\subset F$, a contradiction. $\blacksquare$

Defining the Morse function

We now explain how to define a useful Morse function, which is the crucial part of the argument. Set
$$
 \epsilon=\frac{\epsilon_G}{2}~~~~~\text{and}~~~~~m=m_G!
$$
We will also use the following:

Lemma 5: There is an integer $\nu=\nu(G)$ such that for every ascending sequence of length $\nu$ of centralizers of elements $g_1,\ldots,g_\nu\in G$ 
$$
 C_G(g_1)\subset C_G(g_2)\subset\cdots\subset C_G(g_\nu),
$$
for some $i<\nu$ we have $C_G(g_i)=C_G(g_{i+1})$.

Proof: Considering $G$ as an algebraic subgroup of $\mathrm{GL}_n(\mathbb{C})$, the centralizer 
$$
 C_G(g)=\{h:hgh^{-1}=g\}\cap G
$$ 
is defined by $n^2$ quadratic polynomials and the polynomials defining $G$. Since $C_G(g)$ is a group, its irreducible components are its connected components.
Hence, by Bezout's theorem, the number of connected components of $C_G(g)$ is bounded uniformly (independently of $g$). Now if $C_G(g_i)$ and $C_G(g_{i+1})$ have the same dimension and the same number of components, they must coincide, and the lemma is proved since $G$ is finite dimensional. $\blacksquare$

Thus, for any semisimple $g\in G$ there is $0\le j< \nu$ such that $C_G(g^{m^j})=C_G(g^{m^{j+1}})$. Denote by $j(g)$ the minimal such $j$. Set 
$$
\mu=m^\nu.
$$
Let $\Gamma$ be an irreducible lattice in $G$. Denote by $M=\Gamma\backslash X$ the corresponding orbifold, and by $\pi: X\to M$ the associated (ramified) covering map. For a subset $Y\subset M$ let $\widetilde Y=\pi^{-1}(Y)$ be its preimage in $X=\widetilde M$. 
For $\gamma\in\Gamma\setminus\{ 1\}$ let 
$$
 \overline{\gamma}=\gamma^{m^{j(\gamma)}}~\text{if}~\gamma~\text{is hyperbolic, and}~\overline{\gamma}=\gamma~\text{otherwise}.
$$ 
Obviously $d_{\overline{\gamma}}(x)\le\mu d_\gamma(x)$.
Since for a hyperbolic element $\mathrm{Min}(g)$ is determined by $C_G(g)$ we have 
\begin{equation}\label{eq:min=min}
\mathrm{Min}(\overline{\gamma})=\mathrm{Min}(\overline{\gamma}^m),~\text{for every hyperbolic element}~\gamma\in\Gamma\setminus\{1\}.
\end{equation}
Additionally, for $\gamma_1,\gamma_2$ hyperbolic $[\overline{\gamma_1}^m,\overline{\gamma_2}^m]=1\implies [\overline{\gamma_1},\overline{\gamma_2}]=1$.

Note that for every $\gamma_0\in\Gamma$, the set $\{\gamma\in\Gamma:\overline{\gamma}=\gamma_0\}$ is finite, being compact and discrete.  Let
$$
 \widetilde N:=\cup\{ \min(\overline{\gamma}):\gamma\in\Gamma\setminus\{1\},~\inf d_{\overline{\gamma}}<\epsilon\}.
$$
Clearly $\widetilde N$ is a $\Gamma$-invariant union of totally geodesic submanifolds, and since $\Gamma$ is discrete, this union is locally finite. 
By Corollary 3, unless $G\cong\mathrm{PSL}_2(\mathbb{R})$, $\text{codim}_X\widetilde N\ge 2$, implying that $X\setminus\widetilde N$ is connected. 
$N=\pi(\widetilde N)\subset M$ is a singular submanifold consisting of all ramified points and closed geodesics shorter than $\epsilon$. 

Let $f:\mathbb{R}^{>0}\to\mathbb{R}^{\ge 0}$ be a smooth function which tends to $\infty$ at $0$, strictly decreases on $(0,\epsilon]$ and is identically $0$ on $[\epsilon,\infty)$. For $\gamma\in\Gamma\setminus\{ 1\}$ and $x\in X\setminus\widetilde N$ define 
\[
\begin{array}{lll}
\phi_\gamma(x)=&f(d_\gamma(x))&~\text{if}~\gamma~\text{is parabolic}\\
              &f(\mu d_\gamma(x))&~\text{if}~\gamma~\text{is elliptic}\\
              &f(d_\gamma(x)-\inf d_\gamma)&~\text{if}~\gamma~\text{is hyperbolic and}~\inf d_\gamma<\epsilon\\
              &0&~\text{if}~\gamma~\text{is hyperbolic and}~\inf d_\gamma\ge\epsilon,
\end{array}
\]
and set
$$
 \widetilde\psi(x):=\sum_{\gamma\in\Gamma\setminus\{ 1\}}\phi_{\overline{\gamma}}(x).
$$
There are only finitely many nonzero summends for each $x$, since $\phi_{\overline{\gamma}}(x)\ne 0\implies d_{\overline{\gamma}}(x)\le 2\epsilon$, and $\Gamma$ is discrete.
Thus $\widetilde\psi$ is a well defined smooth function on $X\setminus \widetilde N$. Note that $\widetilde\psi(x)$ tends to $\infty$ as $x$ approaches $\widetilde N$. Since $\widetilde\psi$ is $\Gamma$-invariant it induces a function on $M\setminus N$ 
$$
 \psi(x)=\widetilde\psi(\pi^{-1}(x)).
$$

For $\delta\ge 0$ denote 
$$
 \widetilde\psi_{\le\delta}=\{ x\in X:\widetilde\psi(x)\le\delta\}~~\text{and}~~\psi_{\le\delta}=\{ x\in m:\psi(x)\le\delta\}=\pi(\widetilde\psi_{\le\delta}).
$$
 Then $d_\gamma(x)\ge\frac{\epsilon}{\mu}$ for every $x\in \widetilde\psi_{\le 0}$ and $\gamma\in \Gamma\setminus\{ 1\}$. Thus, the injectivity radius in $M$ at any point of $\psi_{\le 0}$ is at least $\frac{\epsilon}{2\mu}$.
 

The main proposition

The proof of Theorem 1 relies on the existence of a deformation retract from $M\setminus N$ into an arbitrarily small neighborhood of $\psi_{\le 0}$. The main part consists in showing that the gradient of $\psi$ does not vanish outside $\psi_{\le 0}$ and hence defines a smooth vector field.

Proposition 6: For $x\in M\setminus N$, $\nabla \psi(x)=0$ iff $\psi(x)=0$.

It is more convenient to work in the universal cover where the proposition translates to: for $x\in X\setminus\widetilde N$, $\nabla \widetilde\psi(x)=0$ iff $\widetilde\psi(x)=0$. For the sake of simplicity we assume that $X$ is the hyperbolic space $\mathbb{H}^n$, but the same ideas extend to the general case. 

Proof:  At any point $x\in \widetilde Y$ with $\widetilde\psi (x)\ne 0$ we will find a tangent vector $\hat{n}_x$ at which the directional derivative of $\widetilde\psi$ is nonzero. Let 
$$
 \Sigma_x=\{\gamma\in\Gamma_\circ: f(d_\gamma(x)-|\gamma|)\ne 0\},~\text{let}~\Delta_x=\langle \Sigma_x\rangle
$$
and let $N_x$ be a normal subgroup of finite index in $\Delta_x$ which is contained in a connected nilpotent Lie subgroup of $G$. In view of Selberg's lemma (that every finitely generated linear group is virtually t.f.) we may also suppose that $N_x$ is torsion free. Let $Z_x$ denote the centre of $N_x$. We distinguish between 3 cases. 

Case 1: Suppose first that $\Delta_x$ is finite. Let $y\in X$ be a fixed point for $\Delta_x$ and let $\hat{n}_x$ be the unit tangent at $x$ to the geodesic ray $c:[0,\infty)\to X$ emanating from $y$ through $x$. Thus $\hat{n}_x=\dot{c}(d(x,y))$. Since $x\in \widetilde Y$ it follows that $d_\gamma(x)>0,~\forall \gamma\in\Gamma\setminus\{1\}$, and since $d_\gamma\circ c$ is a convex function (as $X$ is non-positively curved) we deduce that 
$$
 \frac{d}{dt}|_{t=d(x,y)}d_\gamma(c(t))>0,
$$
for every $\gamma\in\Sigma_x$ and hence
$$
 \nabla\widetilde\psi(x)\cdot\hat{n}_x=\frac{d}{dt}|_{t=d(x,y)}\widetilde\psi(c(t))=\sum_{\gamma\in\Sigma_x}f'(d_\gamma(x))\frac{d}{dt}_{t=d(x,y)}d_\gamma(c(t))<0
$$ 
since $\gamma\in\Sigma_x$ implies $d_\gamma(x)<\epsilon$ and $f$ has negative derivative on $(0,\epsilon)$.

In the next two cases $N_x$ and $Z_x$ are nontrivial hence infinite, being torsion free.

Case 2: Suppose now that $Z_x$ contains an hyperbolic element $\gamma_0$ and let $A$ be the axis of $\gamma_0$. It follows that all elements of $N_x$ preserve $A$ and hence attain their minimal displacement on $A$. Thus $A=\cap\min_{\gamma\in N_x}(\gamma)$ and since $N_x$ is normal it follows that $A$ is also $\Delta_x$ invariant, and hence all elements in $\Delta_x$ attain their minimal displacement on $A$. Let $y=\pi_A(x)$ be the nearest point to $x$ in $A$, let $c:[0,\infty)\to X$ be the ray from $y$ through $x$ and let $\hat{n}_x$ be the tangent to $c$ at $x$. Since $A$ is convex and $\gamma$-invariant $d_\gamma(x)\ge d_\gamma(y),~\forall\gamma\in\Sigma_x$. Moreover it also follow from convexity and the fact that $\mathbb{H}^n$ admits no constant-distance geodesics that for every $\gamma\in\Sigma_x$ we have $d_\gamma(x)>d_\gamma(y)$. Thus one can proceed arguing as in case 1.

Case 3:We are left with the case that $Z_x$ contains a parabolic element $\gamma_0$. Since $Z_x$ is characteristic in $N_x$ and $N_x$ is normal, also $Z_x$ is normal in $\Delta_x$. Moreover, since $N_x$ is of finite index in $\Delta_x$ the element $\gamma_0$ has only finitely many conjugates in $\Delta_x$ and they are all in $Z_x$. Denote these elements by $\gamma_0,\gamma_1,\ldots,\gamma_k$. All of them are parabolics and since they commute with each other, for every $t$ the corresponding sublevel sets intersect nontrivially:
$$
 B_t=\cap_{i=0}^k\{p\in X:d_{\gamma_i}(p)\le t\}\ne\emptyset.
$$
Taking $t<\min\{d_{\gamma_i}(x):i=0,\ldots,k\}$ we get a nonempty $\Delta_x$-invariant closed convex set $B_t$ not containing $x$. Taking $y=\pi_{B_t}(x)$ and proceeding as in the previous cases allows us to complete the proof. $\blacksquare$

In higher rank, the difficulty lies in finding these invariant, closed, convex sets. One proves successively the next few results and applies them in the appropriate places. For details see Gelander's paper.

Proposition 7: Let $g$ be an isometry of $X$, let $\mathcal{C}\subset X$ be a closed convex $g$-invariant set and let $x\in X$. Then $d_g(x)\ge d_g(P_\mathcal{C}(x))$.

Lemma 8: Let $\mathcal{C}\subset X$ be a closed convex set and let $g$ be an isometry of $X$ which preserves $\mathcal{C}$. Let $c:[0,\infty)\to X$ be a 
geodesic ray satisfying 
$$
 c(0)=c([0,\infty))\cap \mathcal{C}=P_\mathcal{C}(c(1)),
$$ 
where $P_\mathcal{C}:X\to\mathcal{C}$ denotes the projection to the nearest point.
Then $h(t):=d_g(c(t))$ is a nondecreasing smooth convex function on $(0,\infty)$. In particular, if $h(t_0)>h(0)$ then $h'(t_0)>0$.

Lemma 9: 

1.  The intersection of a finite collection of nonempty sub-level sets corresponding to the displacement functions of commuting isometrics of $X$ is nonempty.
2.  If $A\leq G$ is an abelian subgroup consisting of semisimple elements. Then $$ \cap_{g\in A}\mathrm{Min} (g)\ne\emptyset.$$

Retraction

Theorem 10: For every $\delta>0$ there is a deformation retract $r_\delta:M\setminus N\to \psi_{\le\delta}$.

Proof: By Proposition 6 the smooth function $\psi(x)$ has no positive critical values. We claim furthermore that $\psi:M\setminus N\to \mathbb{R}^{\ge 0}$ is a proper map. Indeed, for $x\in X\setminus\widetilde N$, $\widetilde\psi(x)\le a$ implies that $\phi_{\overline{\gamma}}(x)\le a$ for every $\gamma\in\Gamma\setminus\{1\}$ which implies that $d_\gamma(x)\ge f^{-1}(a)$ if $\gamma$ is parabolic, and that $d_\gamma(x)\ge \frac{f^{-1}(a)}{\mu}$ if $\gamma$ is elliptic or hyperbolic. Thus the injectivity radius at $\pi (x)$ is at least $\frac{f^{-1}(a)}{2\mu}$. It follows that $\psi^{-1}([0,a])$ is contained in the $\big(\frac{f^{-1}(a)}{2\mu}\big)$-thick part of $M$, which is compact since $M$ has finite volume. The result follows by standard Morse theory. $\blacksquare$

An immediate consequence of Theorem 10 is that $\psi_{\le\delta}$ is nonempty for every $\delta>0$. 

Corollary 11: For every $\delta\ge 0$, $\psi_{\le\delta}$ is nonempty and compact.

Proof: Since 
$$
 \widetilde\psi_{\le 0}=\bigcap_{\delta>0}\widetilde\psi_{\le\delta}.
$$
we derive that the same hold for $\delta=0$. $\blacksquare$

Moreover, when $X$ is not $\mathbb{H}^2$ we get:

Corollary 12: Assume that $\dim X>2$. Then is $\psi_{\le\delta}$ connected for every $\delta\ge 0$.

Proof: Since $\Gamma\le G$ is irreducible and $G$ is centreless, every $\gamma\in\Gamma\setminus\{1\}$ projects nontrivially to every factor of $G$ 
By Corollary 3 $\text{codim}_MN\ge 2$ and hence $M\setminus N$ is connected. Thus by Theorem 10 $\psi_{\le\delta}$ is connected for $\delta>0$, and since a descending intersection of compact connected sets is connected, the same holds for $\delta=0$. $\blacksquare$

For $G=\mathrm{PSL}_2(\mathbb{R})$ theorem 1 follows from the Gauss--Bonnet formula and the explicate presentation of surface groups, so we assume that $G$ is not locally isomorphic to $\mathrm{PSL}_2(\mathbb{R})$.

Since $\widetilde\psi_{\le\delta}$ is connected nonempty and $\Gamma$ acts freely on it (as all the fixed points of nontrivial elements are in the singular submanifold $\widetilde N$), we conclude, from standard covering theory, that:

Corollary 13: For every $\delta\ge 0$, $\Gamma$ is a quotient of $\pi_1(\psi_{\le\delta})$ --- the fundamental group of $\psi_{\le\delta}=\pi (\widetilde\psi_{\le\delta})=\Gamma\backslash\widetilde\psi_{\le\delta}$.

Proof of theorem 1

Instead of rescaling the Haar measure of $G$, let us suppose that it is fixed and corresponds to the Riemannian measure of $X$, and prove that there is a constant $C(G)$ such that $d(\Gamma)\le C(G)\mathrm{vol}(G/\Gamma)$.
Recall from the proof of Theorem 10 that $\psi_{\le 0}$ is contained in the $\alpha$-thick part of $M$ for $\alpha=\frac{\epsilon}{2\mu}$. Let $\mathcal{S}$ be a maximal $\alpha$-discrete subset of $\psi_{\le 0}$. For $t>0$ denote by $v (t)$ the volume of a $t$-ball in $X$.
Since the $\alpha/2$ balls centreed at points of $\mathcal{S}$ are pairwise disjoint and isometric to a $\alpha/2$ ball in $X$, the size of $\mathcal{S}$ is bounded by $\mathrm{vol} (M)/v(\frac{\alpha}{2})$. Moreover, the union of the $\alpha$ balls centreed at points of $\mathcal{S}$ covers $\psi_{\le 0}$. Denote this union by $U$. Choose $\delta_0>0$ sufficiently small so that $\psi_{\le\delta_0}$ is contained in $U$.

Claim: $\pi_1(\psi_{\le\delta_0})$ is a quotient of $\pi_1(U)$.

Proof of claim: The inclusion $i:\psi_{\le\delta_0}\to U$ induces a map $i_*:\pi_1(\psi_{\le\delta_0})\to\pi_1(U)$, and the deformation retract $r=r_{\delta_0}$ restricted to $U$, $r:U\to \psi_{\le\delta_0}$ induces a map $r_*:\pi_1(U)\to\pi_1(\psi_{\le\delta_0})$. Since $r\circ i$ is the identity on $\psi_{\le\delta_0}$, we see that $r_*\circ i_*$ is the identity on $\pi_1(\psi_{\le\delta_0})$. It follows that $r_*:\pi_1(U)\to\pi_1(\psi_{\le\delta_0})$ is onto. $\blacksquare$

Since the sectional curvature is nonpositive on $M$ the $\alpha$-balls centreed at points of $\mathcal{S}$ are convex, and hence any nonempty intersection of such is convex, hence contractible. Thus these balls form a good cover of $U$, in the sense that for any finite set such that the intersection is non-empty, the intersection is homeomorphic to $\mathbb{R}^n$. Furthermore, the nerve $\mathfrak{N}$ of this cover is homotopic to $U$. Now $\pi_1(U)\cong\pi_1(\mathfrak{N})$ has a generating set of size $E(\mathfrak{N})$ --- the number of edges of the $1$-skeleton $\mathfrak{N}^1$. To see this one may choose a spanning tree $\mathcal{T}$ to the graph $\mathfrak{N}^1$ and pick one generator for each edge belonging to $\mathfrak{N}^1\setminus\mathcal{T}$. 
Finally note that the edges of $\mathfrak{N}$ correspond to pairs of points in $\mathcal{S}$ which are of distance at most $2\alpha$. Thus the degree of each vertex of $\mathfrak{N}$ is at most $v(2.5\alpha)/v(\alpha/2)$. Thus
$$
 d(\Gamma)\le |E(\mathfrak{N})|\le\frac{\mathrm{vol}(M)\cdot v(2.5\alpha)}{2v(\alpha/2)^2},
$$
and we can take $C(G)=\frac{v(2.5\alpha)}{2v(\alpha/2)^2}$. $\blacksquare$

Consequences of Theorem 1

Theorem 1 implies in particular the well known but nontrivial fact that every lattice is finitely generated, as well as the classical Kazhdan--Margulis theorem:

Corollary 14 (Kazhdan-Margulis):For any semisimple Lie group without compact factors $G$, there is a positive lower bound on the covolume of lattices.  

Proof:$$
 \mathrm{vol} (G/\Gamma)\ge\frac{d(\Gamma)}{C}\ge\frac{2}{C},
$$
for every lattice $\Gamma\le G$. $\blacksquare$

 
It can be shown that the minimal co-volume $v_0$ is attained, but in general it is very hard to obtain a good estimate of it. 

The nonemptiness of $\psi_{\le 0}$ ($\psi$ being the Morse function used in the proof of Theorem 1) implies the stronger version of the Kazhdan--Margulis theorem that there is an identity neighborhood in $G$ which intersects trivially a conjugate of every lattice. Since the injectivity radius at points of $\psi_{\le 0}$ is at least $\frac{\epsilon}{2\mu}$, the pre-image in $G$ of the $\frac{\epsilon}{2\mu}$-neighborhood of $K$ in $G/K$ is an open set which trivially intersect a conjugate of every lattice. Furthermore, this set contains a maximal compact subgroup of $G$.

In general, the linear estimate in Theorem 1 cannot be improved. For instance, for every $n\ge 2$ the group $G=\mathrm{SO}(n,1)$ admits a lattice $\Gamma$ which projects onto the free group $F_2$, hence taking $\Gamma_k$ in $\Gamma$ to be the pre-image of an index $k$ subgroup of $F_2$, we get $d(\Gamma_k)\ge k+1$ while $\mathrm{vol}(G/\Gamma_k)=k\cdot\mathrm{vol} (G/\Gamma)$. On the other hand, in general one cannot give a lower bound on $d(\Gamma)$ which tends to $\infty$ with $\mathrm{vol}(G/\Gamma)$. Moreover, it is known that when $\mathrm{rank}_\mathbb{R}(G)\ge 2$, every nonuniform lattice in $G$ admits a $3$-generated finite index subgroup. 

Another immediate consequence of Theorem 1 is that the first Betti number grows at most linearly with the covolume:

Corollary 15: The first Betti number $b_1(\Gamma,A)$ of a lattice $\Gamma\le G$, with respect to an arbitrary ring $A$, is at most $C\mathrm{vol} (G/\Gamma)$. 

Proof: $$
 b_1(\Gamma,A)\le d(\Gamma)\le C\mathrm{vol} (G/\Gamma). 
$$$\blacksquare$

We may discuss in a sequel (i.e. when I get round to it) an application of theorem 1 to counting conjugacy classes of lattices of covolume at most a given bound.