Upper half-space model geometry

The metric $g_a$ is Riemannian and $U^n$ is boundaryless. [F1, given] In the global coordinates $g_{ij}=(a^2{y}^2{)}^{-1}{\delta }_{ij}$ is smooth and symmetric, and for $v\mathrm{\ne }0$ $g_a(v,v)=\frac{1}{a^2{y}^2}{\sum }_i(v^i{)}^2>0$ because $y>0$. The coordinate criterion [F1] therefore makes $g_a$ a Riemannian metric on the open set $U^n$, which is an $n$-dimensional boundaryless smooth manifold as an open subset of ${\mathbb{R}}^n$.

The Christoffel symbols. [F1, F2, step 1.1] Write $f=(ay{)}^{-1}$, so that $g_{ij}=f^2{\delta }_{ij}$, and put $f_l:={\mathrm{\partial }}_{l}f$; since $f$ depends only on $y=x^n$ one has $f_l=f^{\mathrm{\prime }}{\delta }_{ln}$ with $f^{\mathrm{\prime }}=-1\mathrm{/}(a{y}^2)=-f\mathrm{/}y$. Substituting ${\mathrm{\partial }}_l{g}_{ij}=2f{f}_l{\delta }_{ij}$ and $g^{kl}=f^{-2}{\delta }^{kl}$ in the formula of [F2] gives ${\mathrm{\Gamma }}^k{}_{ij}=f^{-1}{\delta }^{kl}(f_i{\delta }_{jl}+f_j{\delta }_{il}-f_l{\delta }_{ij})=\frac{f^{\mathrm{\prime }}}{f}({\delta }_{in}{\delta }_j^k+{\delta }_{jn}{\delta }_i^k-{\delta }_n^k{\delta }_{ij})=-\frac{1}{y}({\delta }_{in}{\delta }_j^k+{\delta }_{jn}{\delta }_i^k-{\delta }_{kn}{\delta }_{ij}),$ where the second equality contracts ${\delta }^{kl}{f}^{\mathrm{\prime }}{\delta }_{ln}=f^{\mathrm{\prime }}{\delta }_n^k$ in each of the three terms and the third uses $f^{\mathrm{\prime }}\mathrm{/}f=-y^{-1}$ and ${\delta }_n^k={\delta }_{kn}$. The symbols depend on the point only through the factor $y^{-1}$.

The curvature tensor and the sectional curvature $-a^2$. [F2, F3, step 2.1] Put ${\epsilon }_i:={\delta }_{in}$ and $A^{\mathrm{\ell }}{}_{jk}:={\epsilon }_j{\delta }_k^{\mathrm{\ell }}+{\epsilon }_k{\delta }_j^{\mathrm{\ell }}-{\delta }_{jk}{\epsilon }^{\mathrm{\ell }},$ so that step 2.1 reads ${\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{jk}=-y^{-1}{A}^{\mathrm{\ell }}{}_{jk}$ and ${\mathrm{\partial }}_i{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{jk}=y^{-2}{\epsilon }_i{A}^{\mathrm{\ell }}{}_{jk}$. Substitution in the coordinate formula of [F3] gives ${\mathrm{\partial }}_i{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{jk}-{\mathrm{\partial }}_j{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{ik}=y^{-2}({\epsilon }_i{A}^{\mathrm{\ell }}{}_{jk}-{\epsilon }_j{A}^{\mathrm{\ell }}{}_{ik})=y^{-2}({\epsilon }_i{\epsilon }_k{\delta }_j^{\mathrm{\ell }}-{\epsilon }_j{\epsilon }_k{\delta }_i^{\mathrm{\ell }}-{\epsilon }_i{\epsilon }^{\mathrm{\ell }}{\delta }_{jk}+{\epsilon }_j{\epsilon }^{\mathrm{\ell }}{\delta }_{ik}),$ the second equality expanding the two $A$'s and cancelling the two terms ${\epsilon }_i{\epsilon }_j{\delta }_k^{\mathrm{\ell }}$. Expanding the two quadratic terms likewise gives ${\mathrm{\Gamma }}^m{}_{jk}{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{im}-{\mathrm{\Gamma }}^m{}_{ik}{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{jm}=y^{-2}(A^m{}_{jk}{A}^{\mathrm{\ell }}{}_{im}-A^m{}_{ik}{A}^{\mathrm{\ell }}{}_{jm})=y^{-2}({\delta }_{ik}{\delta }_j^{\mathrm{\ell }}-{\delta }_{jk}{\delta }_i^{\mathrm{\ell }}-{\epsilon }_i{\epsilon }_k{\delta }_j^{\mathrm{\ell }}+{\epsilon }_j{\epsilon }_k{\delta }_i^{\mathrm{\ell }}+{\epsilon }_i{\epsilon }^{\mathrm{\ell }}{\delta }_{jk}-{\epsilon }_j{\epsilon }^{\mathrm{\ell }}{\delta }_{ik}),$ so that the four $\epsilon$-terms of the quadratic expansion are exactly the negatives of the four $\epsilon$-terms of the derivative expansion and cancel them, leaving $R^{\mathrm{\ell }}{}_{kij}=y^{-2}({\delta }_{ik}{\delta }_j^{\mathrm{\ell }}-{\delta }_{jk}{\delta }_i^{\mathrm{\ell }})\mathrm{.}$ Since $g_{ik}=f^2{\delta }_{ik}$ with $f^{-2}=a^2{y}^2$, this is $R({\mathrm{\partial }}_i,{\mathrm{\partial }}_j){\mathrm{\partial }}_k=y^{-2}({\delta }_{ik}{\mathrm{\partial }}_j-{\delta }_{jk}{\mathrm{\partial }}_i)=-a^2(g_{jk}{\mathrm{\partial }}_i-g_{ik}{\mathrm{\partial }}_j)\mathrm{.}$ Tensoriality of $R$ in [F3] upgrades this identity on coordinate frames to $R(X,Y)Z=-a^2(g(Y,Z)X-g(X,Z)Y)$ for arbitrary tangent vectors. Pairing with $X=u$ after putting $Y=Z=v$ and using [F3] gives $\mathrm{Rm}(u,v,v,u)=-a^2(g(u,u)g(v,v)-g(u,v{)}^2),$ so for a basis $(u,v)$ of any tangent two-plane the positive Gram determinant in the denominator cancels and the sectional curvature is $K=-a^2$ at every point of $U^n$.

The vertical lines are complete geodesics. [F2, F4, F6, step 2.1] Let $x_0\in {\mathbb{R}}^{n-1}$ and $y_0>0$, and put $\gamma (t)=(x_0,y_0{e}^{at})$. Then $\gamma$ is smooth, takes values in $U^n$ and is defined on all of $\mathbb{R}$. Its coordinates are ${\gamma }^i=x_0^i$ for $i<n$ and ${\gamma }^n=y_0{e}^{at}=:y(t)$, so ${\dot{\gamma }}^i=0$ for $i<n$ and ${\dot{\gamma }}^n=ay$, and ${\ddot{\gamma }}^i=0$ for $i<n$, ${\ddot{\gamma }}^n=a^{2}y$. Substituting the symbols of step 2.1, for $k<n$ every term contains the factor ${\delta }_j^k$ or ${\delta }_i^k$ or ${\delta }_{kn}$ with $k<n$, so ${\mathrm{\Gamma }}^k{}_{ij}{\dot{\gamma }}^i{\dot{\gamma }}^j=-\frac{1}{y}({\delta }_{in}{\delta }_j^k+{\delta }_{jn}{\delta }_i^k-{\delta }_{kn}{\delta }_{ij}){\dot{\gamma }}^i{\dot{\gamma }}^j=0,$ while for $k=n$ the bracket is ${\dot{\gamma }}^n{\dot{\gamma }}^n+{\dot{\gamma }}^n{\dot{\gamma }}^n-\mathrm{\mid }\dot{\gamma }{\mathrm{\mid }}_{\mathrm{E}}^2={\dot{y}}^2$, so ${\ddot{\gamma }}^n+{\mathrm{\Gamma }}^n{}_{ij}{\dot{\gamma }}^i{\dot{\gamma }}^j=a^{2}y-\frac{1}{y}{\dot{y}}^2=a^{2}y-\frac{1}{y}(ay{)}^2=0.$ Hence the geodesic equation of [F2] holds, and $\gamma$ is a geodesic defined on all of $\mathbb{R}$, of constant speed $g_a(\dot{\gamma },\dot{\gamma })=(a^2{y}^2{)}^{-1}(ay{)}^2=1$.

The semicircles are complete geodesics. [F2, F4, F6, step 2.1] By a horizontal translation and rotation it suffices to take $\bar{c}=0$ and $u=e_1$. Put $\gamma (t)=(-\rho \tanh (at),0,\dots ,0,\rho \mathrm{sech}(at)),y(t)=\rho \mathrm{sech}(at)>0.$ Then $\gamma$ takes values in $U^n$; by [F6] it is smooth and defined on all of $\mathbb{R}$, with ${\dot{\gamma }}^1(t)=-\rho a{\mathrm{sech}}^2(at)$, ${\dot{\gamma }}^n(t)=-\rho a\mathrm{sech}(at)\tanh (at)$ and all other components zero. Its acceleration is ${\ddot{\gamma }}^1=2\rho a^2{\mathrm{sech}}^2(at)\tanh (at)$ and ${\ddot{\gamma }}^n=\rho a^2(\mathrm{sech}(at){\tanh }^2(at)-{\mathrm{sech}}^3(at))$. By step 2.1 the only nonvanishing symbols with the present velocity are ${\mathrm{\Gamma }}^1{}_{1n}={\mathrm{\Gamma }}^1{}_{n1}=-1\mathrm{/}y$, ${\mathrm{\Gamma }}^n{}_{11}=1\mathrm{/}y$ and ${\mathrm{\Gamma }}^n{}_{nn}=-1\mathrm{/}y$; the remaining coordinates are constant with vanishing Christoffel contributions, since $A^k{}_{ij}=0$ whenever $k\mathrm{\ne }n$ and both $i,j\mathrm{\ne }n$. The $x^1$-component of the geodesic equation is therefore ${\ddot{\gamma }}^1-\frac{2}{y}{\dot{\gamma }}^1{\dot{\gamma }}^n=2\rho a^2{\mathrm{sech}}^2\tanh -\frac{2{\rho }^2{a}^2{\mathrm{sech}}^3\tanh }{\rho \mathrm{sech}}=0,$ and the $y$-component is ${\ddot{\gamma }}^n-\frac{1}{y}(({\dot{\gamma }}^n{)}^2-({\dot{\gamma }}^1{)}^2)=\rho a^2\mathrm{sech}({\tanh }^2-{\mathrm{sech}}^2)-\frac{{\rho }^2{a}^2{\mathrm{sech}}^2({\tanh }^2-{\mathrm{sech}}^2)}{\rho \mathrm{sech}}=0,$ where the two displays use ${\dot{\gamma }}^1=-\rho a{\mathrm{sech}}^2$, ${\dot{\gamma }}^n=-\rho a\mathrm{sech}\tanh$ and $y=\rho \mathrm{sech}$. Hence $\gamma$ satisfies the geodesic equation and is a geodesic defined on all of $\mathbb{R}$, of constant $g_a$-speed $1$, because $g_a(\dot{\gamma },\dot{\gamma })=\frac{{\rho }^2{a}^2{\mathrm{sech}}^2({\mathrm{sech}}^2+{\tanh }^2)}{a^2{\rho }^2{\mathrm{sech}}^2}=1$ by the identity ${\mathrm{sech}}^2+{\tanh }^2=1$ of [F6].

Every maximal geodesic is one of these and is defined for all time. Let $p=({\bar{x}}_0,y_0)\in U^n$ and let $v\in T_p{U}^n$ be a unit vector with respect to $g_a$, decomposed as $v=(v_H,v_n)$ with $v_H\in {\mathbb{R}}^{n-1}$ and $v_n\in \mathbb{R}$, so that $\mathrm{\mid }{v}_H{\mathrm{\mid }}^2+v_n^2=a^2{y}_0^2$. If $v_H=0$, then $v=\pm a{y}_0{\mathrm{\partial }}_n$; the vertical line of step 3.2 passes through $p$ with velocity $a{y}_0{\mathrm{\partial }}_n$ at $t=0$, and its time reversal $t\mapsto \gamma (-t)$ (an affine reparametrization, [F4]) passes through $p$ with velocity $-a{y}_0{\mathrm{\partial }}_n$. So in this case the maximal geodesic with initial datum $(p,v)$ is that line, and its domain is $\mathbb{R}$. Otherwise $v_H\mathrm{\ne }0$. Put $u:=-v_H\mathrm{/}\mathrm{\mid }{v}_H\mathrm{\mid }$, let $s\in \mathbb{R}$ be the unique solution of $\sinh s=-v_n\mathrm{/}\mathrm{\mid }{v}_H\mathrm{\mid }$ (unique because $\sinh$ is strictly increasing and onto, [F6]), and set $\rho :=y_0\cosh s>0,\bar{c}:={\bar{x}}_0+\rho \tanh (s)u\mathrm{.}$ Consider the semicircle of step 3.3 with data $(\bar{c},u,\rho )$: $\gamma (t)=(\bar{c}-\rho \tanh (at)u,\rho \mathrm{sech}(at))=:({\gamma }_H(t),{\gamma }^n(t))\mathrm{.}$ At $t_1:=s\mathrm{/}a$ its height is ${\gamma }^n(t_1)=\rho \mathrm{sech}s=y_0\cosh s\mathrm{sech}s=y_0$, so its horizontal part is $\bar{c}-\rho \tanh (s)u={\bar{x}}_0$, that is, $\gamma (t_1)=p$. Its velocity there is $\dot{\gamma }(t_1)=a{y}_0(-\mathrm{sech}(s)u,-\tanh (s)),$ because $\rho a{\mathrm{sech}}^{2}s=a{y}_0\mathrm{sech}s$ and $\rho a\mathrm{sech}s\tanh s=a{y}_0\tanh s$. The identities ${\cosh }^{2}s=1+{\sinh }^{2}s$ and $\sinh s=-v_n\mathrm{/}\mathrm{\mid }{v}_H\mathrm{\mid }$ give $\mathrm{\mid }{v}_H\mathrm{\mid }=a{y}_0\mathrm{sech}s,v_n=-\mathrm{\mid }{v}_H\mathrm{\mid }\sinh s=-a{y}_0\tanh s,$ using $\tanh s\cosh s=\sinh s$; together with $u=-v_H\mathrm{/}\mathrm{\mid }{v}_H\mathrm{\mid }$ this shows $\dot{\gamma }(t_1)=(v_H,v_n)=v$. Hence the parameter-translated curve $t\mapsto \gamma (t+t_1)$ is a geodesic defined on all of $\mathbb{R}$ with initial datum $(p,v)$, so by uniqueness in [F4] it is the maximal geodesic of that initial datum, and its domain is $\mathbb{R}$. For an arbitrary nonzero initial velocity $w$, apply the preceding construction to $v=w\mathrm{/}\mathrm{\mid }w{\mathrm{\mid }}_{g_a}$ and reparametrize by $t\mapsto \mathrm{\mid }w{\mathrm{\mid }}_{g_a}t$; this gives a geodesic on $\mathbb{R}$ with velocity $w$, which is maximal by uniqueness. For zero initial velocity the constant curve solves [F2] on $\mathbb{R}$ and is maximal by [F4]. Thus every maximal geodesic of $(U^n,g_a)$ has domain $\mathbb{R}$: the manifold is geodesically complete, and by Hopf–Rinow [F5] it is a complete metric space.

Simple connectedness and the boundary cases. The half-space $U^n=\{y>0\}$ is convex: for $x,z\in U^n$ and $t\in [0,1]$ the point $(1-t)x+tz$ has last coordinate $(1-t)x^n+t{z}^n>0$. Hence $U^n$ is contractible by [F7], so its fundamental group is trivial at every basepoint and it is path connected; by [F7] it is simply connected and connected. Together with completeness from step 4.1 and constant curvature $-a^2$ from step 3.1, $(U^n,g_a)$ is a complete, simply connected space form of curvature $k=-a^2$. Boundary cases: the degenerate scale $a=0$ is excluded by hypothesis, since $g_0$ is not defined; the limiting boundary hyperplane $y=0$ is not part of $U^n$, and every geodesic of steps 3.2 and 3.3 is finite at every finite time but approaches the boundary only as $t\to \pm \mathrm{\infty }$; the case $n=2$ exhibits the semicircles as the only nonvertical geodesics, and for $n>2$ each nonvertical geodesic lies in a two-plane spanned by its initial horizontal direction and ${\mathrm{\partial }}_n$ while the remaining horizontal coordinates stay constant. Every geodesic in steps 3.2 and 3.3 is defined for all real times, so no endpoint of a maximal geodesic is finite. The only choice used is the inherited ${\mathrm{A}\mathrm{C}}_{\omega }$ of [A1], inherited through sectional curvature in step 3.1, geodesic existence and uniqueness in step 4.1, and Hopf–Rinow in step 4.1; the charts, the geodesics and the reparametrizations are explicit.