Kazhdan’s Property (T) for the group \(SL_n(\mathbb{Z})\)
Juan Giordani
2025-03-20
This week I presented Yeshuda Shalom’s proof that the group \(SL_n(\mathbb{Z})\) has Kazhdan’s Property
(T) for \(n\ge 3\) as a final exam for
an elective class in university, so I decided to write a short summary
on the topic and upload it here. I will mostly be following Bekka, de la
Harpe and Valette’s book,which I’ll call BHV from now on.
Kazhdan’s Property (T) is a rigidity property for topological groups.
In informal terms,this means that if a topological group \(G\) is acting unitarily on a hilbert space
\(\mathcal{H}\), and it has vectors
that are sufficiently close to being invariant under the action of \(G\), then it has invariant vectors.
Definition and basic
properties
Let \(\mathcal{H}\) be a complex
Hilbert space, and \(\mathcal{U}(\mathcal{H})\) its unitary
group. If \(G\) is a topological group,
then a unitary representation is a group morphism \(\pi:G\to\mathcal{U}(\mathcal{H})\)
such that \(G\to\mathcal{H}, g\mapsto\pi(g)\xi\) is
continuous for all \(\xi \in
\mathcal{H}\).
Definition 1. Let \((\pi,\mathcal{H})\) be a representation of
\(G\), a topological group.
\((\pi,\mathcal{H})\) almost
has invariant vectors if it has \((Q,\varepsilon)\)-invariant vectors for
each compact \(Q\subset G\) and \(\varepsilon > 0\).
\((\pi,\mathcal{H})\) has
invariant vectors if there exists \(\xi\neq
0\) in \(\mathcal{H}\) such that
\(\pi(g)\xi-\xi\) for all \(g\in G\).
Definition 2. A topological group \(G\) has Kazhdan’s Property (T) if there is
a compact subset \(Q\) and \(\varepsilon > 0\) such that each unitary
representation that has a \((Q,\varepsilon)\)-invariant vector, has an
invariant vector.
We also call \(Q\) a Kazhdan set and
\((Q,\varepsilon)\) a Kazhdan pair. the
next proposition shows that every \((G,\sqrt{2} )\) is a Kazhdan pair. Thus, if
\(G\) is compact, then \(G\) has Property (T). Compact groups are
the simplest example of groups with this property, finding other
examples is highly non-trivial, as we’ll soon see.
Proposition 1. Let \(G\) be a topological group. \((G,\sqrt{2} )\) is a Kazhdan pair, i.e. if
\((\pi,\mathcal{H})\) has a unitary
\(\xi \in\mathcal{U}\) such that \[\sup\limits_{x\in G}\|\pi(x)\xi-\xi\|<\sqrt{2}
.\] then \(\pi\) has a
non-zero invariant vector.
Proof. Take \(\mathcal{C}\)
the closed convex hull of \(\pi(G)\xi\). Let \(\eta_0\) be the unique element in \(\mathcal{C}\) with minimal norm. \(\mathcal{C}\) is \(G\)-invariant, so \(\eta_0\) is \(G\)-invariant. Now a short calculation
shows that \(\eta_0\neq 0\) : \[\varepsilon = \sqrt{2} - \sup\limits_{x\in
G}\|\pi(x)\xi-\xi\|>0
.\] For each \(x\in G\) we
have: \[2-2\Re\left<\pi(x)\xi,\xi
\right> = \|\pi(x)\xi-\xi\|^{2}\le (\sqrt{2}-\varepsilon)^{2}
.\] Thus, \[\Re\left<\pi(x)\xi,\xi \right>\ge
\frac{2-(\sqrt{2} -\varepsilon)^{2}}{2}=\frac{\varepsilon(2\sqrt{2}
-\varepsilon)}{2}>0
.\] This implies that \[\Re\left<\eta,\xi \right>\ge
\frac{\varepsilon(2\sqrt{2} -\varepsilon)}{2}\quad\forall \eta\in
\mathcal{C}
.\] In particular, taking \(\eta=\eta_0\) we get that \(\eta_0\neq 0\). ◻
We now introduce Property (T) for pairs:
Definition 3. Let \(G\) be a topological group, \(H\subset G\) a closed subgroup. Then \((G,H)\) has Property (T) if for each
representation \((\pi,\mathcal{H})\) of
\(G\) that almost has invariant
vectors, \(\mathcal{H}\) has a non-zero
invariant vector.
Definition 4. Let \(G\) be a topological group, \(H\subset G\) a closed subgroup. Suppose
that there exists \(Q\subset G\) and
\(\varepsilon>0\) such that each
representation \((\pi,\mathcal{H})\) of
\(G\) that has a \((Q,\varepsilon)\)-invariant vector also has
a non-zero invariant vector. In that case, we call \((Q,\varepsilon)\) a Kazhdan pair for \((G,H)\).
Note that \(G\) has Property (T) if
and only if \((G,G)\) has Property (T).
Property (T) for pairs is a weaker version of Property (T), since asking
for a \(H\)-invariant vector is less
than asking for a \(G\)-invariant
vector.
Property (T)
for the group \(SL_n(\mathbb{Z}),\quad n\ge
3\).
The main result of the talk will be to establish Property (T) for the
group \(SL_n(\mathbb{Z})\) of \(n\times n\) matrices with entries in \(\mathbb{Z}\) of determinant \(1\). We will follow a paper of Y. Shalom.
First, we’ll need to relate Property (T) to the following property of
groups:
Definition 5. Let \(G\) be a group. We say that \(G\) is boundedly generated if there exist a
finite \(S\subset G\) and \(\nu\in \mathbb{N}_0\) such that each \(g\in G\) can be written as \[g=g_1^{k_1}\cdots g_{\nu}^{k_{\nu}} \quad g_i\in
S,k_i\in \mathbb{Z}\quad \forall i=1,\ldots,\nu
.\]
We will also need the following result:
Theorem 1 (Carter-Keller, ’84). For \(n\ge 3\), \(SL_n(\mathbb{Z})\) is boundedly generated
by the elementary matrices and \[\nu_n\le
\frac{1}{2}(3n^{2}-n)+36
.\]
The proof for this can be found in Shalom’s paper "Bounded generation
and Kazhdan’s Property (T)". Note that having an explicit upper bound
for \(\nu_n\) is what will allow us to
get an explicit Kazhdan constant for \(SL_n(\mathbb{Z})\) later.
Property (T)
for \((SL_2(\mathbb{Z})\rtimes\mathbb{Z}^{2},\mathbb{Z}^{2})\)
We define the following matrices in \(SL_2(\mathbb{Z})\)\[U^{\pm}=\begin{pmatrix} 1 & \pm 1 \\ 0 &
1 \end{pmatrix} \quad L^{\pm}=\begin{pmatrix} 1 & 0 \\ \pm 1 & 0
\end{pmatrix}
.\]
Lemma 1 (Burger, ’91). Let \(\nu\) be a probability measure on \(\mathcal{B}(\mathbb{R}^{2}\backslash
\{0\}\). There exist a borel subset \(M\subset \mathbb{R}^{2}\backslash \{0\}\)
and \(\gamma \in \{U^{\pm}L^{\pm}\}\)
such that: \[|\nu(\gamma .M)-\nu(M)|\ge
\frac{1}{4}
.\]
Proof. Begin by dividing the plane into sections \(A,B,C\) and \(D\) contained within the x-axis, the y-axis
and the graphics of the functions \(x=y\) and \(x=-y\) as shown in the picture below.
Assume toward contradiction that \(|\nu(\gamma M)-\nu(M)|<\frac{1}{4}\) for
any Borel subset \(M\) of \(\mathbb{R}\backslash \{0\}\) and for all
\(\gamma\in \{U^{\pm},L^{\pm}\}\). For
\(M=a\cup B\) we have \[\nu(A)=\nu(A\cup B)-\nu(B)=\nu(A\cup
B)-\nu(L^{+}(A\cup B))<\frac{1}{4}
.\] And similarly, \(\nu(B),\nu(C),\nu(D)<\frac{1}{4}\). But
this contradicts that \(\nu\) is a
probability measure on \(\mathbb{R}^{2}\backslash \{0\}\). ◻
Recall that \(\hat{\mathbb{Z}^{2}}\)
can be identified with the \(2\)
dimensional torus \(\mathbb{T}^{2}\) by
identifying each \((e^{2\pi ix},e^{2\pi
iy})\in \mathbb{T}^{2}\) with the character: \[(m,n)\mapsto e^{2\pi i(mx+ny)} \quad \forall
(m,n)\in \mathbb{Z}^{2}
.\] With this identification we get an action \(SL_2(\mathbb{Z})\curvearrowright
\hat{\mathbb{Z}^{2}}, \quad \gamma . \chi(x) =
\chi(\gamma^{-1}x\gamma)\). We define \[e^{\pm}=\begin{pmatrix} \pm 1 \\ 0 \end{pmatrix}
\quad f^{\pm}=\begin{pmatrix} 0 \\ \pm 1 \end{pmatrix}
.\] and let \(Q\) be the
generating subset of \(SL_2(\mathbb{Z})\rtimes\mathbb{Z}^{2}\)
given by: \[Q=\{U^{\pm},L^{\pm},e^{\pm},f^{\pm}\}
.\] This next theorem is the central piece of the talk, and we’ll
be able to derive Property (T) for \(SL_n(\mathbb{Z})\) as a corollary of
it.
Theorem 2. The pair \((Q,\frac{1}{10})\) is a Kazhdan pair for
\((SL_2(\mathbb{Z})\rtimes\mathbb{Z}^{2},\mathbb{Z}^{2})\).
In particular, \((SL_2(\mathbb{Z})\rtimes\mathbb{Z}^{2},\mathbb{Z}^{2})\)
has Property (T).
Proof. Let \((\pi,\mathcal{H})\) be a representation of
\(SL_2(\mathbb{Z})\rtimes\mathbb{Z}^{2}\)
with a \((Q,\frac{1}{10})\)-invariant
vector.
Assume, by way of contradiction, that there is no \(Z^{2}\)-invariant vector in \(\mathcal{H}\).
By the SNAG theorem, applied to the restriction \(\pi\vert_{\mathbb{Z}^{2}}\) and the locally
compact abelian group \(\hat{\mathbb{Z}^{2}}=\mathbb{T}^{2}\), we
have that there exists a unique projection-valued measure \(E:\mathcal{B}(\mathbb{T}^{2})\to
Proj(\mathcal{H})\) such that \[\pi(x)=\int_{\mathbb{T}}\hat{x}(x)dE(\hat{x})
\quad \forall x\in \mathbb{Z}^{2}
.\]
For each \(\gamma\in SL_2(\mathbb{Z}),B\in
\mathcal{B}(\mathbb{T}^{2})\), we have that
\[E(\gamma
.B)=\pi(\gamma)E(B)\pi(\gamma^{-1})\]
Indeed, for each \(x\in
\mathbb{Z}^{2},\gamma\in SL_2(\mathbb{Z})\) we have
\[\begin{aligned}
\int_{\mathbb{T}^{2}}\hat{x}(x)d(\pi(\gamma)E(\cdot)\pi(\gamma^{-1}))(\hat{x})
&= \pi(\gamma)\pi(x)\pi(\gamma^{-1}) \\
&=
\pi(\gamma x \gamma^{-1}) \\
&=
\int_{\mathbb{T}^{2}}(\gamma^{-1}.\hat{x})(x)dE(\hat{x}) \\
&=
\int_{\mathbb{T}^{2}}\hat{x}(x)dE(g.\hat{x})
\end{aligned}\] By the uniqueness of the measure given by the
SNAG theorem, we get that \(B\mapsto\pi(\gamma)E(B)\pi(\gamma^{-1})\)
and \(B\mapsto E(\gamma .B)\) must be
equal. Let \(\varepsilon=\frac{1}{10}\), \(\xi \in H\) be a \((Q,\varepsilon)\)-invariant vector and
\(\mu_{\xi}\) be the probability
measure in \(\mathbb{T}^{2}\) given by:
\[\mu_{\xi}(B)=\left<E(B)\xi,\xi
\right> \quad B\in \mathcal{B}(\mathbb{T}^{2})
.\] Note that since \(\pi\)
has no \(\mathbb{Z}^{2}\)-invariant
vectors, then \(E(\{0\})=0\). Indeed,
we argue by contradiction. Suppose that \(E(\{0\})\neq 0\), and let \(\eta\in Ran(E(\{0\}))\) with \(\|\xi\|=1\). \[E_{\eta,\eta}(\hat{Z}^{2})=\|\xi\|^{2}=1
.\]\[1 =
\|\eta\|^{2}=\left<E(\{0\})\xi,\xi \right> = E_{\xi,\xi}(\{0\})
.\] From this, we conclude that \(E_{\xi,\xi}(\hat{\mathbb{Z}^{2}}\backslash \{0\}
)\). In section D of the appendix to BHV, we can find the
following identity: \[\|\pi(\xi)-\xi\|^{2}=\int_{\hat{G}}|\overline{\hat{x}(x)}-1|^{2}dE_{\xi,\xi}(\hat{x})
.\] Valid for every locally compact abelian group \(G\), \((\pi,\mathcal{H})\) representation of \(G\) and \(\xi \in
\mathcal{H}\).
Applying it to \(G=\mathbb{Z}^{2}\)
we find that: \[0=\|\pi(x)\xi-\xi\|^{2}=\int_{\hat{\mathbb{Z}^{2}}}|\overline{\hat{x}(x)}-1|^{2}dE_{\xi,\xi}(\hat{x})
\quad \forall x\in \mathbb{Z}^{2}
.\] Because each representation \(\hat{x}\in \hat{Z}^{2}\) is an exponential
function, and \(\overline{\hat{x}(x)}-1=e^{-2\pi i\omega\cdot
x}-1=0\) if \(\omega =0\). thus
\(\pi\vert_{\mathbb{Z}^{2}}\) has a
non-zero invariant vector \(\eta\), a
contradiction. We have the following two claims:
\(\mu_{\xi}(X)\ge
1-\varepsilon^{2}\)
If \(\nu(B)=\frac{\mu_{\xi}(B\cap
X)}{\mu_{\xi}(X)} \quad B\in \mathcal{B}(\mathbb{T}^{2})\). Then
\[|\nu(\gamma .B)-\nu(B)|<\frac{1}{4}
\quad \forall B\in \mathcal{B}(\mathbb{T}^{2},\gamma \in
\{U^{\pm},L^{\pm}\}
.\]
Suppose that they both hold. Since \(E(\{0\})=0\), then \(\mu_{\xi}(\{0\})=0\) and \(\nu\) is a measure in \(\mathbb{R}^{2}\backslash \{0\}\). Now \(\gamma .X\subset
(-\frac{1}{2},\frac{1}{2}]^{2}\). Since \(\nu(X)=1\), then claim 2 contradicts
Burger’s lemma and this ends the proof.
Let’s prove claim 1: Since \(\xi\)
is \((Q,\varepsilon)\)-invariant and
\(e^{\pm}\in Q\)\[\|\pi(e^{\pm})\xi-\xi\|^{2}=\int_{(-\frac{1}{2},\frac{1}{2}]^{2}}|e^{\pm
2\pi ix}-1|^{2}d\mu(x,y)\le\varepsilon^{2}
.\] If \(\frac{1}{4}\le|t|\le\frac{1}{2}\) we get
\[|e^{\pm 2\pi it}-1|^{2}=2-2\cos 2\pi t =
4\sin ^{2}\pi t \ge 2
.\]\[\varepsilon^{2}\ge
\int_{\substack{(x,y)\in (-\frac{1}{2},\frac{1}{2}]^{2} \\ |x| \ge
\frac{1}{4}}}4\sin ^{2}(\pi x) d\mu_{\xi}(x,y) \ge 2\mu_{\xi}(\{|x|\ge
\frac{1}{4}\}).\] So \(\mu_{\xi}(\{|x|\ge \frac{1}{4}\}) \le
\frac{\varepsilon^{2}}{2}\).
Similarly, replacing \(e^{\pm}\)
with \(f^{\pm}\) we get \(\mu_{\xi}(\{|y| \ge \frac{1}{4}\} )\le
\frac{\varepsilon^{2}}{2}\)
We now prove claim 2 using equation (1) \[\begin{aligned}
|\mu_{\xi}(\gamma .B)-\mu_{\xi}(B)| &=
|\left<\pi(\gamma)E(B)\pi(\gamma^{-1})\xi,\xi
\right>-\left<E(B)\xi,\xi \right>| \\
&\le
|\left<\pi(\gamma)E(B)\pi(\gamma^{-1})\xi,\xi
\right>-\left<\pi(\gamma)E(B)\xi,\xi \right>| \\ &+
|\left<\pi(\gamma)E(B)\xi,\xi \right>-\left<E(B)\xi,\xi
\right>| \\
&=
|\left<\pi(\gamma)E(B)(\pi(\gamma^{-1})\xi-\xi),\xi \right>| \\
&+ \mid
\left<E(B)\xi,(\pi(\gamma^{-1})\xi-\xi) \right>| \\
&\le
\|\pi(\gamma)E(B)\|\|\pi(\gamma^{-1})\xi-\xi\|+\|E(B)\|\|\pi(\gamma^{-1})\xi-\xi\|
\\
&\le \varepsilon +
\varepsilon = 2\varepsilon
\end{aligned}\] By claim 1: \[0 \le
\mu_{\xi}(B)-\mu_{\xi}(B\cap X) \le \varepsilon^{2}
.\]\[\begin{aligned}
\mu_{\xi}((\gamma .B)\cap X)-\mu_{\xi}(B\cap X) &=
(\mu((\gamma .B)\cap X)-\mu_{\xi}(\gamma .B)) \\
&+
(\mu_{\xi}(\gamma .B)-\mu_{\xi}(B)) \\
&+
(\mu_{\xi}(B)-\mu_{\xi}(B\cap X)) \\
&\le
\varepsilon^{2}+2\varepsilon+\varepsilon^{2}
\end{aligned}\] Since this works for both \(B\) and \(\gamma
.B\) we get that \[|\mu_{\xi}((\gamma
.B)\cap X)-\mu_{\xi}(B\cap X)| \le 2\varepsilon + 2\varepsilon^{2}
.\] By claim 1 and using that \(\varepsilon=\frac{1}{10}\)\[|\nu(\gamma .B)-\nu(B)| \le
\frac{2\varepsilon+2\varepsilon^{2}}{1-\varepsilon^{2}}=\frac{22}{99}<\frac{1}{4}
.\] ◻
We’ll now use that \((Q,\frac{1}{10})\) is a Kazhdan pair for
\((SL_2(\mathbb{Z})\rtimes
\mathbb{Z}^{2},\mathbb{Z}^{2})\) to prove the following
corollary:
Corollary 1. Let \(\varepsilon >0\), \((\pi,\mathcal{H})\) a representation of
\(SL_2(\mathbb{Z})\rtimes
\mathbb{Z}^{2}\). If \(\pi\) has
a \((Q,\frac{\varepsilon}{20})\)-invariant
vector \(\xi\), then \(\xi\) is \((\mathbb{Z}^{2},\varepsilon)\)-invariant.
Proof. Let \(\mathcal{H}_0 \subset
H\) be the subspace of \(\mathbb{Z}^{2}\)-invariant vectors in \(\mathcal{H}\), let \(H_1=H_0^{\perp}\). Since \(\mathbb{Z}^{2}\) is normal in \(\Gamma=SL_2(\mathbb{Z})\rtimes\mathbb{Z}^{2}\),
both subspaces are \(\Gamma\)-invariant. Certainly, if \(\gamma \in \Gamma,t\in \mathbb{Z}^{2}\)
then \[\begin{aligned}
\pi(\gamma)H_0=\pi(\gamma)\pi(t)H_0=\pi(\gamma
t)H_0=\pi(\tilde{t})\pi(\gamma)H_0
\end{aligned}\] For some \(\widetilde{t}\in \mathbb{Z}^{2}\). As \(\widetilde{t}\) runs through \(\mathbb{Z}^{2}\) when \(t\) runs through \(\mathbb{Z}^{2}\). Then, \(\pi(\gamma)H_0\) is \(Z^{2}\)-invariant, ie \(\pi(\gamma)H_0\subset H_0\). \(\pi(\gamma)H_1\subset H_1\), so we can use
the same argument to show that \(\pi(t)H_1\subset H_1 \quad \forall t\in
\mathbb{Z}^{2}\).
Let \(\xi=\xi_0 + \xi_1\) be the
orthogonal decomposition. \[\|\pi(\gamma)\xi-\xi\|^{2}=\|\pi(\gamma)\xi_0-\xi_0\|^{2}+\|\pi(\gamma)\xi_1-\xi_1\|^{2}
.\] For \(\gamma\in
\Gamma\), we have: \[\|\pi(\gamma)\xi_1-\xi_1\|^{2}\le
\|\pi(\gamma)\xi-\xi\|^{2}<\bigg(\frac{\varepsilon}{20}\bigg)^{2}
\quad \forall \gamma \in Q
.\] As there are no non-zero \(\mathbb{Z}^{2}\)-invariant vectors in \(H_1\), from the previous theorem it follows
that \[\|\pi(\gamma)\xi_1-\xi_1\|\ge\bigg(\frac{\|\xi\|}{10}\bigg)^{2}
.\] Combining both inequalities we get \[\bigg(\frac{\|\xi_1\|}{10}\bigg)^{2}<\bigg(\frac{\varepsilon}{20}\bigg)^{2}
.\] ie \(\|\xi_1\|<\frac{\varepsilon}{2}\). Thus,
as \(\xi_0\) is \(\mathbb{Z}^{2}\)-invariant \[\|\pi(t)\xi-\xi\|=\|\pi(t)\xi_1-\xi_1\|\le
2\|\xi_1\|<\varepsilon \quad \forall t\in \mathbb{Z}^{2}
.\] ◻
Property (T) for \(SL_n(\mathbb{Z}), n\ge 3\)
Let us fix integer \(n\ge 3\), the
following lemma, whose proof can be found in section 4.2 of the BHV
tells us that each elemntary matrix of \(SL_n(\mathbb{Z})\) is included in a copy of
\(SL_2(\mathbb{Z})\rtimes\mathbb{Z}^{2}\).
Lemma 2. Let \(R\) be a commutative ring with unity. Let
\(i\neq j\) be integers \(1\le i,j \le n\). Then, there exists an
injective homomorphism \(\alpha:SL_2(R)\rtimes
R^{2}\to SL_n(R)\) such that \(E_{i,j}(t)\in SL_n(R)\) is contained in
\(\alpha(R^{2})\) and \[\alpha(SL_2(R)) = \begin{pmatrix} I_k & 0
& 0 \\ 0 & SL_2(R) & 0 \\ 0 & 0 & I_{n-k-2}
\end{pmatrix} \quad k\in [n-2]
.\]
Remember \(SL_n(\mathbb{Z})\) is
boundedly generated by the elementary matrices, with \(\nu_n \le \frac{1}{2}(3n^{2}-n)+36\)
Theorem 3. The group \(SL_n(\mathbb{Z})\) has Property (T) for
\(n\ge 3\). More precisely, let \(Q_n\subset SL_n(\mathbb{Z})\) the subset of
the \(n^{2}-n\) matrices \(E_{i,j}(1), \quad 1\le i,j\le n, i\neq j\).
Then \((Q_n,\frac{1}{20\nu_n})\) is a
Kazhdan pair for \(SL_n(\mathbb{Z})\).
Proof. Let \(T_n=Q_n\cup
Q_n^{-1}\). It suffices to prove that \((T_n,\frac{1}{20\nu_n})\) is a Kazhdan
pair. Let \((\pi,\mathcal{H})\) be a
representation of \(SL_n(\mathbb{Z})\)
with a unitary, \((T_n,\frac{1}{20\nu_n})\)-invariant vector
\(\xi\). Let \(\gamma\in SL_n(\mathbb{Z})\) be an
elementary matrix.
By the Lemma, there exists an embedding \(\alpha\) of \(SL_2(\mathbb{Z})\rtimes\mathbb{Z}^{2}\) in
\(SL_n(\mathbb{Z})\) such that \(\gamma\in \alpha(\mathbb{Z}^{2})\) and
\(\alpha(Q)=T_n\cap im(\alpha)\).
By corollary 1, \(\xi\) is \((\alpha(\mathbb{Z}^{2}),\frac{1}{\nu_n})\)-invariant,
so \[\|\pi(\gamma)\xi-\xi\|<\frac{1}{\nu_n} \quad
\forall \gamma \text{ elementary matrix}
.\] We now take \(\gamma\in
SL_n(\mathbb{Z})\) to be an arbitrary matrix. By bounded
generation, there exist \(N\le \nu_n\)
and \(\gamma_1,\ldots,\gamma_N\)
elementary matrices such that \(\gamma=\gamma_1,\ldots\). Then \[\begin{aligned}
\|\pi(\gamma)\xi-\xi\| &\le
\sum\limits_{i=0}^{N-1}\|\pi(\gamma_1\cdots\gamma_{N-i})\xi-\pi(\gamma_1\cdots\gamma_{N-i-1})\xi\|
\\
&=
\sum\limits_{i=0}^{N}\|\pi(\gamma_i)\xi-\xi\| \\
&\le \frac{N}{\nu_n}\le 1
\end{aligned}\] So \(\xi\) is
\((SL_n(\mathbb{Z}),1)\)-invariant. By
Proposition 1, we get that \(\pi\) has
non-zero invariant vectors. ◻
