Cartan hadamard for hyperbolic space

$H^n$ is an embedded $n$-submanifold of ${\mathbb{R}}^{n+1}$ and $T_p{H}^n=p^{\perp }$ for $p\in H^n$. [F1, given] Put $f(x)=⟨x,x⟩$, a smooth polynomial function whose differential at $x$ is $Df(x)v=2⟨x,v⟩$. At a point of $f^{-1}(-1)$ one has $x\mathrm{\ne }0$, so $Df(x)$ is not the zero functional and $x$ is a regular point; by [F1] the level set $f^{-1}(-1)$ is an embedded submanifold of dimension $n$ with tangent space $\ker Df(x)=x^{\perp }$. The upper sheet is $H^n=f^{-1}(-1)\cap \{x_{n+1}>0\}$, the intersection of that submanifold with an open subset of ${\mathbb{R}}^{n+1}$, so [F1] gives it the structure of an embedded $n$-submanifold with the same tangent spaces.

The induced form $g^H$ is a Riemannian metric on $H^n$. [F2, step 1.1] Let $\iota :H^n\hookrightarrow {\mathbb{R}}^{n+1}$ be the inclusion and let $⟨\cdot ,\cdot ⟩$ also denote the ambient covariant two-tensor $h(X,Y)=⟨X,Y⟩$, which is smooth, symmetric and $C^{\mathrm{\infty }}$-bilinear. Then $g^H:={\iota }^{\ast }h$ is a smooth symmetric $(0,2)$-tensor field by [F2], and it remains to see that it is positive definite on each tangent space. Write $p=(u,q)$ with $u\in {\mathbb{R}}^n$ and $q=\sqrt{1+\mathrm{\mid }u{\mathrm{\mid }}^2}>0$, and let $v=(v_0,v_{n+1})\in T_p{H}^n=p^{\perp }$ by step 1.1; the orthogonality $⟨v,p⟩=0$ says $v_0\cdot u=v_{n+1}q$, that is $v_{n+1}=(v_0\cdot u)\mathrm{/}q$. Hence $g_p^H(v,v)=\mathrm{\mid }{v}_0{\mathrm{\mid }}^2-v_{n+1}^2=\mathrm{\mid }{v}_0{\mathrm{\mid }}^2-\frac{(v_0\cdot u{)}^2}{1+\mathrm{\mid }u{\mathrm{\mid }}^2}\ge \mathrm{\mid }{v}_0{\mathrm{\mid }}^2(1-\frac{\mathrm{\mid }u{\mathrm{\mid }}^2}{1+\mathrm{\mid }u{\mathrm{\mid }}^2})=\frac{\mathrm{\mid }{v}_0{\mathrm{\mid }}^2}{1+\mathrm{\mid }u{\mathrm{\mid }}^2}\ge 0,$ where [F2] (Cauchy–Schwarz) was used for $(v_0\cdot u{)}^2\le \mathrm{\mid }{v}_0{\mathrm{\mid }}^2\mathrm{\mid }u{\mathrm{\mid }}^2$. Equality forces $v_0=0$ and then $v_{n+1}=0$, so $v=0$; therefore $g^H$ is positive definite and, by [F2], a Riemannian metric on $H^n$.

The graph map is a homeomorphism onto $H^n$. [F2, F5, step 1.1] Define $\mathrm{\Phi }:{\mathbb{R}}^n\to H^n$ by $\mathrm{\Phi }(u)=(u,\sqrt{1+\mathrm{\mid }u{\mathrm{\mid }}^2})$, continuous because the square root is continuous; its image lies in $H^n$, since $\mathrm{\mid }u{\mathrm{\mid }}^2-(1+\mathrm{\mid }u{\mathrm{\mid }}^2)=-1$ and the last coordinate is positive. Conversely every $x\in H^n$ equals $\mathrm{\Phi }(x_1,\dots ,x_n)$, because $x_{n+1}^2=1+{\sum }_{i\le n}{x}_i^2$ and $x_{n+1}>0$ force $x_{n+1}=\sqrt{1+{\sum }_{i\le n}{x}_i^2}$. The inverse ${\mathrm{\Phi }}^{-1}$ is the restriction to $H^n$ of the continuous projection $x\mapsto (x_1,\dots ,x_n)$, so $\mathrm{\Phi }$ is a homeomorphism. Therefore $H^n$ is path connected (the image of the path-connected ${\mathbb{R}}^n$) and hence connected by [F5]; being homeomorphic to the contractible space ${\mathbb{R}}^n$, which is contractible by Every nonempty convex subset of ${\mathbb{R}}^n$ is contractible applied to the nonempty convex set ${\mathbb{R}}^n$, it is contractible and so simply connected by [F5].

The tangential projection of the flat connection is the Levi-Civita connection of $g^H$. [F3, step 1.1, step 2.1] Write $D$ for the coordinate directional derivative on ${\mathbb{R}}^{n+1}$ and $p$ for the position field, so that $D_{X}p=X$ for every smooth ambient field $X$. Extend tangent fields on $H^n$ smoothly to ${\mathbb{R}}^{n+1}$ (locally, by extending their coordinate expressions). As in step 1.1 the tangent space at $x\in H^n$ is $x^{\perp }$, so the metric orthogonal projection of an ambient vector $w$ onto $T_x{H}^n=x^{\perp }$ along the line $\mathbb{R}x$ is $w-\frac{⟨w,x⟩}{⟨x,x⟩}x=w+⟨w,x⟩x,$ because $⟨x,x⟩=-1$. For tangent fields $X,Y$ differentiate the identically vanishing function $⟨Y,p⟩$ in the direction $X$: $0=X⟨Y,p⟩=⟨D_{X}Y,p⟩+⟨Y,D_{X}p⟩=⟨D_{X}Y,p⟩+⟨X,Y⟩,$ using $D_{X}p=X$. Hence the tangential projection of $D_{X}Y$ is ${\mathrm{\nabla }}_X^{H}Y:=D_{X}Y+⟨D_{X}Y,p⟩p=D_{X}Y-⟨X,Y⟩p,$ which is tangent because $D_{X}Y$ and $p$ are ambient and the correction removes the normal component. The assignment ${\mathrm{\nabla }}^H$ is a connection on $H^n$: it is $C^{\mathrm{\infty }}$-linear in $X$, additive in $Y$, and ${\mathrm{\nabla }}_X^H(fY)=f{\mathrm{\nabla }}_X^{H}Y+(Xf)Y$ for smooth $f$, all read off from the same properties of $D$. It is torsion-free, since for tangent fields ${\mathrm{\nabla }}_X^{H}Y-{\mathrm{\nabla }}_Y^{H}X=D_{X}Y-D_{Y}X=[X,Y]$ by [F3] (the correction terms cancel because $⟨X,Y⟩$ is symmetric), and $[X,Y]$ is tangent to $H^n$ because it is a difference of tangential derivatives. It is metric compatible, since $X⟨Y,Z⟩=⟨D_{X}Y,Z⟩+⟨Y,D_{X}Z⟩$ and $⟨{\mathrm{\nabla }}_X^{H}Y,Z⟩+⟨Y,{\mathrm{\nabla }}_X^{H}Z⟩=⟨D_{X}Y,Z⟩-⟨X,Y⟩⟨p,Z⟩+⟨Y,D_{X}Z⟩-⟨X,Z⟩⟨Y,p⟩=⟨D_{X}Y,Z⟩+⟨Y,D_{X}Z⟩,$ both correction terms vanishing because $Y,Z\perp p$. By the uniqueness clause of [F3], ${\mathrm{\nabla }}^H$ is the Levi-Civita connection of the Riemannian metric $g^H$ of step 2.1.

The curvature tensor of $g^H$. [F3, step 3.1] For tangent fields $X,Y,Z$ on $H^n$ extended as above, use ${\mathrm{\nabla }}_X^{H}Y=D_{X}Y-⟨X,Y⟩p$ and $D_{X}p=X$ to expand ${\mathrm{\nabla }}_X^H{\mathrm{\nabla }}_Y^{H}Z=D_X{D}_{Y}Z-⟨Y,Z⟩X-(X⟨Y,Z⟩+⟨X,D_{Y}Z⟩)p,$ because the derivative of the function $⟨Y,Z⟩$ is the function $X⟨Y,Z⟩$ and the derivative of $p$ in the direction $X$ is $X$. Substituting this and the same expression with $X,Y$ interchanged into the definition of the curvature tensor, and adding the term $-{\mathrm{\nabla }}_{[X,Y]}^{H}Z=-D_{[X,Y]}Z+⟨[X,Y],Z⟩p$, the $D$-difference is zero by the flatness of $D$ in [F3], and the $p$-coefficient is $-X⟨Y,Z⟩-⟨X,D_{Y}Z⟩+Y⟨X,Z⟩+⟨Y,D_{X}Z⟩+⟨[X,Y],Z⟩=0,$ the four scalar terms expanding, by metric compatibility of $⟨\cdot ,\cdot ⟩$ with $D$, into $-⟨[X,Y],Z⟩$ and therefore cancelling the bracket term $+⟨[X,Y],Z⟩$ contributed by $-{\mathrm{\nabla }}_{[X,Y]}^{H}Z$. What remains is the purely tangential identity $R^H(X,Y)Z=⟨X,Z⟩Y-⟨Y,Z⟩X\mathrm{.}$

Every sectional curvature of $g^H$ equals $-1$. [F3, step 4.1] For $p\in H^n$ and an orthonormal pair $(X,Y)$ in $T_p{H}^n=p^{\perp }$ with respect to $g^H$, step 4.1 gives $\mathrm{Rm}(X,Y,Y,X)=⟨R^H(X,Y)Y,X⟩=⟨X,Y⟩⟨Y,X⟩-⟨Y,Y⟩⟨X,X⟩=0-1=-1,$ where the last two equalities use the orthonormality $(g_p^H(X,X)=1$, $g_p^H(Y,Y)=1$, $g_p^H(X,Y)=0)$. Since the Gram determinant in the denominator of the sectional curvature is $1\cdot 1-0^2=1$, the sectional curvature of every tangent two-plane is $K=-1$, so in particular $K\le 0$; in dimension $n\ge 2$ every tangent space carries such a pair.

The explicit curves are the geodesics, defined for all time. [F3, step 3.1, step 4.1] Fix $p\in H^n$ and $v\in T_p{H}^n$ with $g_p^H(v,v)=1$, and define $\gamma :\mathbb{R}\to {\mathbb{R}}^{n+1}$ by $\gamma (t)=\cosh tp+\sinh tv$. Then $⟨\gamma ,\gamma ⟩={\cosh }^{2}t⟨p,p⟩+2\sinh t\cosh t⟨p,v⟩+{\sinh }^{2}t⟨v,v⟩=-{\cosh }^{2}t+{\sinh }^{2}t=-1$ and $⟨{\gamma }^{\mathrm{\prime }},{\gamma }^{\mathrm{\prime }}⟩={\sinh }^{2}t(-1)+{\cosh }^{2}t(1)=1$, so $\gamma$ is a unit-speed curve in the level set $f^{-1}(-1)$. Its last coordinate is ${\gamma }_{n+1}(t)=\cosh t{p}_{n+1}+\sinh t{v}_{n+1}$; writing $p=(u,q)$ as in step 2.1 one has $\mathrm{\mid }{v}_{n+1}\mathrm{\mid }\le \mathrm{\mid }u\mathrm{\mid }<q=p_{n+1}$: indeed $v_{n+1}=(v_0\cdot u)\mathrm{/}q$ and $g_p^H(v,v)=1$ give $v_{n+1}^2\le \mathrm{\mid }{v}_0{\mathrm{\mid }}^2\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}{q}^2=(1+v_{n+1}^2)\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}{q}^2$, hence $v_{n+1}^2(q^2-\mathrm{\mid }u{\mathrm{\mid }}^2)\le \mathrm{\mid }u{\mathrm{\mid }}^2$, that is $v_{n+1}^2\le \mathrm{\mid }u{\mathrm{\mid }}^2$; consequently $p_{n+1}\pm v_{n+1}\ge q-\mathrm{\mid }u\mathrm{\mid }>0$ and, since $\cosh t\mp \sinh t>0$ for every $t$, the last coordinate ${\gamma }_{n+1}(t)=\frac{1}{2}(e^t(p_{n+1}+v_{n+1})+e^{-t}(p_{n+1}-v_{n+1}))$ is positive. Hence $\gamma$ takes values in the upper sheet $H^n$. Its componentwise second derivative is ${\gamma }^{\mathrm{\prime }\mathrm{\prime }}=\cosh tp+\sinh tv=\gamma (t)$, which is $⟨\cdot ,\cdot ⟩$-orthogonal to $T_{\gamma (t)}{H}^n=\gamma (t{)}^{\perp }$; therefore the tangential component of ${\gamma }^{\mathrm{\prime }\mathrm{\prime }}$ vanishes: ${\mathrm{\nabla }}_{{\gamma }^{\mathrm{\prime }}}^H{\gamma }^{\mathrm{\prime }}=({\gamma }^{\mathrm{\prime }\mathrm{\prime }}{)}^{\mathrm{\top }}=0$ by the projection formula of step 3.1. Thus $\gamma$ is a geodesic of $(H^n,g^H)$, defined on all of $\mathbb{R}$, with $\gamma (0)=p$ and ${\gamma }^{\mathrm{\prime }}(0)=v$. For a general initial vector $w\in T_p{H}^n$, $w=0$ gives the constant geodesic and $w\mathrm{\ne }0$ is handled by affine reparametrization [F4] of the unit-speed case with $v=w\mathrm{/}\mathrm{\mid }w\mathrm{\mid }$.

Everything is complete and simply connected. [F3, F4, F5, step 2.2, step 5.1, step 5.2] Every initial datum $(p,w)\in T{H}^n$ is the initial datum of the geodesic constructed in step 5.2, which is defined on all of $\mathbb{R}$; by the uniqueness clause of [F4] the unique maximal geodesic with those initial data has domain $\mathbb{R}$. Hence $(H^n,g^H)$ is geodesically complete, and it is metrically complete by Hopf–Rinow [F5], the manifold being nonempty, connected (step 2.2) and boundaryless. By step 2.2 it is simply connected, and by step 5.1 it has $K=-1\le 0$.

Cartan–Hadamard gives the global diffeomorphism. [F6, step 6.1] The manifold $(H^n,g^H)$ is complete, connected, boundaryless, nonpositively curved and simply connected by step 6.1, so [F6] applies and ${\exp }_p:(T_p{H}^n,{\exp }_p^{\ast }{g}^H)\to (H^n,g^H)$ is a diffeomorphism for every $p\in H^n$ — in particular bijective, in accordance with the Cartan–Hadamard prediction.

The model fields and the boundary cases. [F7, step 7.1] For a unit-speed geodesic, step 4.1 gives $R(J,T)T=-J$ for normal $J$. Thus its normal Jacobi equation is $D_t^{2}J-J=0$ (Jacobi field). Given $J(0)=0$, let $E$ be the unique parallel field with $E(0)=D_{t}J(0)$ (Existence and uniqueness of parallel sections); its initial value is normal by differentiating $g(J,T)=0$, so $E$ stays normal. By [F7], $\sinh (t)E(t)$ solves the same equation and has the same initial data. Jacobi uniqueness (Existence and uniqueness of jacobi fields from initial data) gives $J(t)=\sinh (t)E(t)$. Such a field has no positive zero when it is nonzero, consistent with $K\le 0$. For a constant speed $c>0$ the corresponding formula is $J(t)=\sinh (ct)P_t{D}_{t}J(0)\mathrm{/}c$; the factor $c^2$ in the curvature term explains why the unit-speed qualification is essential. In dimension $n=1$ the hyperboloid has no tangent two-plane and no sectional curvature to compute, and it is not a Cartan–Hadamard surface; the case $n\ge 2$ is the one stated. The case $v=0$ of step 5.2 is the constant geodesic; the case $w\mathrm{\ne }0$ is reduced to unit speed by reparametrization; and the two connected components of $\{x:⟨x,x⟩=-1\}$ are separated by the sign of $x_{n+1}$, which is why the upper sheet rather than the whole level set is the model. No choice beyond the inherited [A1] is used: the charts, the projection formula and the explicit geodesics are all canonical.