SLn(R) has property (T) for n at least three

Fix distinct indices $i,j$ and $k\mathrm{\notin }\{i,j\}$ in $\{1,\dots ,n\}$, using [F4]'s coordinate labels. In the coordinate order $(e_i,e_k,e_j)$, define ${\theta }_{ij,k}(A,v)$ to have block $\left(\begin{array}{cc}A& v\\ 0& 1\end{array}\right)$ and to fix all other basis vectors. Its determinant is $\det A=1$, and block multiplication gives $\theta (A,v)\theta (A^{\mathrm{\prime }},v^{\mathrm{\prime }})=\theta (A{A}^{\mathrm{\prime }},v+A{v}^{\mathrm{\prime }})$. The map and its inverse (coordinate restriction) are continuous by [F1], so its image $G_{ij,k}$ is a topological subgroup isomorphic to ${\mathrm{S}\mathrm{L}}_2(\mathbb{R})⋉{\mathbb{R}}^2$. Its translation subgroup $N_{ij,k}={\theta }_{ij,k}(\{I\}\times {\mathbb{R}}^2)$ is closed and normal, and contains $E_{ij}(t)$ as $v=(t,0)$.

Let $(Q_{\mathrm{r}\mathrm{e}\mathrm{l}},\delta )$ be the relative Kazhdan pair supplied by [F2], and let $Q_{ij,k}={\theta }_{ij,k}(Q_{\mathrm{r}\mathrm{e}\mathrm{l}})$. Each $(Q_{ij,k},\delta )$ is a relative Kazhdan pair for $(G_{ij,k},N_{ij,k})$. Let $Q_{\ast }:={\bigcup }_{i\mathrm{\ne }j,k\mathrm{\notin }\{i,j\}}{Q}_{ij,k}$; this is a finite set and hence compact. Every transvection $E_{ij}(t)$ belongs to at least one $N_{ij,k}$, since for each $i\mathrm{\ne }j$ there is a remaining index $k$.

Put $M=M(n)$ from [F4] and $\epsilon =\delta \mathrm{/}(4M)$. Let $\pi$ be a strongly continuous unitary representation of ${\mathrm{S}\mathrm{L}}_n(\mathbb{R})$ with a $(Q_{\ast },\epsilon )$-invariant unit vector $\xi$. For each triple, $Q_{ij,k}\subseteq Q_{\ast }$ gives ${\mathrm{\Delta }}_{Q_{ij,k}}^{\pi }(\xi )\le \epsilon$. Applying the second inequality [F3] to the restriction of $\pi$ to $G_{ij,k}$ yields ${\sup }_{v\in N_{ij,k}}\parallel \pi (v)\xi -\xi \parallel \le 2\epsilon \mathrm{/}\delta =1\mathrm{/}(2M)$. Hence every elementary transvection moves $\xi$ by at most $1\mathrm{/}(2M)$.

Every $g\in {\mathrm{S}\mathrm{L}}_n(\mathbb{R})$ is a product of at most $M$ transvections by [F4]. Telescoping along such a product and using [F5] gives $\parallel \pi (g)\xi -\xi \parallel \le M\mathrm{/}(2M)=1\mathrm{/}2$. Therefore the orbit $\pi (G)\xi$ lies in the closed ball of radius $1\mathrm{/}2$ centered at $\xi$. Let $C$ be the intersection of all closed convex subsets of $H$ containing this orbit. By [F6], $C$ is nonempty, closed and convex, and it is contained in that closed ball.

For each $g\in G$, the set $\pi (g)C$ is closed and convex and contains the orbit, so minimality of the intersection gives $\pi (g)C\supseteq C$; applying the same argument to $g^{-1}$ gives equality. By [F7], $C$ has a unique point $\eta$ of least norm. Since $C$ is $G$-invariant and $\pi (g)$ preserves norms, uniqueness implies $\pi (g)\eta =\eta$ for every $g$. Moreover $\parallel \eta -\xi \parallel \le 1\mathrm{/}2$, so the reverse triangle inequality and $\parallel \xi \parallel =1$ give $\parallel \eta \parallel \ge 1\mathrm{/}2>0$. Thus every representation with a $(Q_{\ast },\epsilon )$-invariant unit vector has a nonzero invariant vector; almost invariance provides such a vector because $Q_{\ast }$ is compact, so $G$ has property (T) by [F8]. AC is used through the relative-(T), normal-distance, and least-norm suppliers as stated in the axiom audit.