Hyperbolic space as so zero n one mod so n

The group $O(n,1)$ is closed in ${\mathrm{GL}}_{n+1}(\mathbb{R})$, so [L1] makes it an embedded Lie subgroup. Differentiating $A^{\mathsf{T}}JA=J$ gives $X^{\mathsf{T}}J+JX=0$; conversely this condition implies $\exp (tX{)}^{\mathsf{T}}J\exp (tX)=J$ by differentiation in $t$, so it characterizes the Lie algebra. It consists of $\left(\begin{array}{cc}A& b\\ b^{\mathsf{T}}& 0\end{array}\right)$ with $A^{\mathsf{T}}=-A$. Determinant has values $\pm 1$ on $O(n,1)$, hence equals $1$ on its identity component $G$. This component has the same Lie algebra. Write $X_b=\left(\begin{array}{cc}0& b\\ b^{\mathsf{T}}& 0\end{array}\right)$ and $K_{ij}=E_{ij}-E_{ji}$ for $i<j\le n$.

The level function $F(v)=⟨v,v⟩$ has differential $d{F}_v(w)=2⟨v,w⟩$, nonzero at every $F(v)=-1$. Thus [L2] gives tangent space $v^{\perp }$ and dimension $n$. The upper sheet is the graph $v=(u,\sqrt{1+\mathrm{\mid }u{\mathrm{\mid }}^2})$, hence connected. Its Lorentz tangent metric is positive: if $w=(a,b)\in v^{\perp }$, then $b=u\cdot a\mathrm{/}\sqrt{1+\mathrm{\mid }u{\mathrm{\mid }}^2}$ and $⟨w,w⟩\ge \mathrm{\mid }a{\mathrm{\mid }}^2\mathrm{/}(1+\mathrm{\mid }u{\mathrm{\mid }}^2)>0$ for $w\mathrm{\ne }0$. Every $g\in G$ preserves this sheet, since the sign of the last coordinate of $gv$ cannot change continuously on connected $G$.

The group $\mathrm{SO}(n)$ is compact, being closed and bounded in matrix space and hence compact by Heine--Borel in [L1], and is path connected: plane rotations can carry any unit first column to the first coordinate vector; after doing so the remaining block is in $\mathrm{SO}(n-1)$, and induction ends with $\mathrm{SO}(1)=\{1\}$. Each plane rotation has a path to the identity through its angle. Therefore $K=\{\mathrm{diag}(R,1):R\in \mathrm{SO}(n)\}$ lies in $G$.

The basis $K_{ij},X_i:=X_{e_i}$ satisfies $[K_{ij},X_l]={\delta }_{jl}{X}_i-{\delta }_{il}{X}_j$ and $[X_i,X_j]=K_{ij}$ for $i<j$. For fixed $K_{ij}$, its adjoint square is $-I$ on each two-dimensional span of the rotations joining $i,j$ to a third spatial index, and on $\mathrm{span}(X_i,X_j)$; it vanishes on the remaining basis vectors. Its trace is $-2(n-2)-2=-2(n-1)$. For fixed $X_i$, its adjoint square is $+I$ on each span of $X_j$ and the rotation joining $i,j$ ($j\mathrm{\ne }i$), and zero on the rest, giving trace $2(n-1)$. Mixed Killing pairings of different basis vectors vanish: conjugation by $\mathrm{diag}({ϵ}_1,\dots ,{ϵ}_n,1)$, ${ϵ}_i=\pm 1$, is a Lie-algebra automorphism, preserves the adjoint trace, and acts with distinct sign characters ${ϵ}_i{ϵ}_j$ on $K_{ij}$ and ${ϵ}_i$ on $X_i$. A sign choice therefore negates any mixed pairing while preserving it. This proves on the whole basis, hence bilinearly, $B(X,Y)=(n-1)\mathrm{tr}(XY)$. It is nondegenerate for every $n\ge 2$, including $n=3$, so [L3] proves semisimplicity. The involution $\theta X=-X^{\mathsf{T}}$ has eigenspaces $\mathfrak{k}=\mathrm{span}(K_{ij})$ and $\mathfrak{p}=\{X_b\}$, with $B_{\theta }(X,X)=(n-1)\mathrm{tr}(X{X}^{\mathsf{T}})>0$ for $X\mathrm{\ne }0$. Since $[\mathfrak{p},\mathfrak{p}]=\mathfrak{k}$, no proper ideal contains $\mathfrak{p}$, so the noncompact-type criterion of [L4] is satisfied.

The action is transitive: for $v=(u,\sqrt{1+\mathrm{\mid }u{\mathrm{\mid }}^2})$ with $u\mathrm{\ne }0$, put $a=u\mathrm{/}\mathrm{\mid }u\mathrm{\mid }$ and $t\ge 0$ with $\sinh t=\mathrm{\mid }u\mathrm{\mid }$. The matrix exponential gives $\exp (t{X}_a)e_0=(\sinh ta,\cosh t)=v$ and belongs to $G$; $v=e_0$ uses the identity. The stabilizer of $e_0$ consists exactly of $\mathrm{diag}(R,1)$ with $R\in \mathrm{SO}(n)$, since it preserves $e_0^{\perp }$ and determinant one; these matrices are in $G$ by step 1.3. Thus it is $K$.

The smooth orbit map factors through the quotient submersion to a smooth bijection $\mathrm{\Phi }:G\mathrm{/}K\to H^n$ by [L5] and step 2.2. Its derivative at $eK$, using $\mathfrak{g}\mathrm{/}\mathfrak{k}\cong \mathfrak{p}$, is $X_b\mapsto (b,0)$, an isomorphism. Equivariance makes the derivative an isomorphism everywhere, so the inverse function theorem gives a local diffeomorphism everywhere; a bijective local diffeomorphism has a smooth inverse.

The center of $G$ is trivial. If $z$ is central, $z{e}_0$ is fixed by $K$; the only spatial vector fixed by all spatial rotations for $n\ge 2$ is zero. Since $z{e}_0\in H^n$, it equals $e_0$, so $z=\mathrm{diag}(R,1)\in K$. Commuting with every $\exp (t{X}_b)$ and differentiating forces $Rb=b$ for every $b$, hence $z=I$. The group automorphism $\mathrm{\Theta }(g)=(g^{\mathsf{T}}{)}^{-1}=JgJ$ preserves $G$, is involutive and differentiates to $\theta$. Its fixed elements lie in both $O(n+1)$ and $O(n,1)$, hence commute with $J$ and have block form $\mathrm{diag}(R,c)$; the upper-sheet condition gives $c=1$ and determinant one gives $R\in \mathrm{SO}(n)$. Thus $G^{\mathrm{\Theta }}=K$. Together with step 2.1 this verifies all hypotheses of the symmetric-pair interface [L4].

Step 3.2 proves that $G$ is connected semisimple with finite center, that $\mathrm{\Theta }$ is a global Cartan involution, and that $G^{\mathrm{\Theta }}=K$. Therefore [L7] applies directly and makes $K$ maximal compact.

The metric of [L4] is now applicable by step 3.2. At the origin $B_{\theta }(X_a,X_b)=B(X_a,X_b)=2(n-1)a\cdot b$ by step 2.1. Under $d\mathrm{\Phi }$, the Lorentz metric is $a\cdot b$. Both metrics are $G$-invariant, so the Cartan metric is $2(n-1)$ times the Lorentz metric everywhere. For independent $a,b$ let $D=\mathrm{\mid }a{\mathrm{\mid }}^2\mathrm{\mid }b{\mathrm{\mid }}^2-(a\cdot b{)}^2>0$. The bracket is $[X_a,X_b]=\mathrm{diag}(a{b}^{\mathsf{T}}-b{a}^{\mathsf{T}},0)$, whose squared $B_{\theta }$-norm is $2(n-1)D$, while the Gram determinant of $X_a,X_b$ is $4(n-1{)}^{2}D$. The sectional formula in [L4] therefore gives $-1\mathrm{/}(2(n-1))$ at the origin and, by transitivity, everywhere.

Scaling a metric by a constant $c>0$ preserves its Levi-Civita connection and its curvature operator of type $(1,3)$: the same connection remains torsion free and metric compatible. The sectional numerator scales by $c$ and its Gram denominator by $c^2$. Thus $(2(n-1){)}^{-1}{B}_{\theta }$, the Lorentz metric from step 4.2, has curvature $-1$. At $n=2$ the Cartan curvature is $-1\mathrm{/}2$; at $n=3$ the direct trace proof remains valid. Rank $n=1$ is excluded because the algebra is abelian with zero Killing form and there are no tangent two-planes. AC covers the Lie-group, quotient, maximal-compact and symmetric-space interfaces; the finite matrix computations require no further choice.