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Distance hessian and laplacian in space forms

Example

Assume the inherited Axiom of Countable Choice ${A C}_{ω}$ . Let $n \geq 2$ and $k \in R$ , and let $(M, g)$ be a complete, connected, boundaryless $n$ -dimensional Riemannian manifold of constant sectional curvature $k$ , that is, a space form of curvature $k$ in the sense of Constant sectional curvature and space form. Let $p \in M$ , let $r : = r_{p} = d_{g} (p, \cdot)$ be the distance from $p$ , and let $q \in M ∖ ({p} \cup Cut (p))$ be a point off $p$ and off the cut locus of $p$ , with $t_{0} : = r (q) > 0$ and $t_{0} < \frac{π}{\sqrt{k}} when k > 0.$ Then:

the Hessian of the distance is the projection onto the normal hyperplane, $Hess r (W, Z) = {ct}_{k} (t_{0}) (g_{q} (W, Z) - d r (W) d r (Z)) for all W, Z \in T_{q} M;$
the Laplace–Beltrami operator of $g$ satisfies $Δ_{g} r (q) = (n - 1) {ct}_{k} (t_{0});$
the space form attains equality in both comparison theorems: the two curvature hypotheses of the Hessian comparison hold with equality on every normal plane, and the Ricci lower bound of the Laplacian comparison holds with equality, so the bounds of Hessian comparison for distance under sectional curvature bounds and Laplacian comparison for distance under a ricci lower bound are attained with equality in the model geometry.

The explicit comparison cotangent values are ${ct}_{0} (t) = \frac{1}{t}, {ct}_{k} (t) = \sqrt{k} \cot (\sqrt{k} t) (k > 0), {ct}_{k} (t) = \sqrt{- k} \coth (\sqrt{- k} t) (k < 0),$ so in particular Euclidean space gives $Hess r = r^{- 1} (g - d r \otimes d r)$ and $Δ_{g} r = (n - 1) / r$ . No choice beyond the inherited ${A C}_{ω}$ is used; the point $q$ and the geodesic to it are supplied by the cited interfaces and no family of directions is selected.

Facts & Assumptions

Given: The inherited ${A C}_{ω}$ of [A1]; a complete, connected, boundaryless $n$ -dimensional Riemannian manifold $(M, g)$ of constant sectional curvature $k$ with $n \geq 2$ ; a point $p \in M$ ; a point $q \in M ∖ ({p} \cup Cut (p))$ with $t_{0} : = d_{g} (p, q) > 0$ and $t_{0} < π / \sqrt{k}$ when $k > 0$ ; the distance function $r = r_{p}$ ; and the comparison functions ${sn}_{k}$ , ${cs}_{k}$ , ${ct}_{k}$ .

[A1]

The countable-choice premise is the inherited ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), carried by the exponential, cut-locus, curvature and Riccati interfaces cited below; the Jacobi initial-value construction itself requires no choice; the argument makes no selection of directions, bases or curves from a family.

[F1]

Space forms (Constant sectional curvature and space form): a connected, boundaryless, geodesically complete Riemannian manifold of constant sectional curvature $k$ is a space form of curvature $k$ . By Hopf–Rinow theorem, geodesic completeness of the connected manifold $(M, g)$ is equivalent to metric completeness, and any two points of $M$ are joined by a minimizing geodesic.

[F2]

Constant curvature (Curvature tensor of constant sectional curvature, Sectional curvature, Riemann curvature four-tensor): $(M, g)$ has constant sectional curvature $k$ exactly when $R (X, Y) Z = k (g (Y, Z) X - g (X, Z) Y), Rm (X, Y, Z, W) = k (g (Y, Z) g (X, W) - g (X, Z) g (Y, W)),$ and the sectional curvature of the plane spanned by an orthonormal pair $(u, w)$ is $K (u \land w) = Rm (u, w, w, u)$ ; the four-linear form $Rm$ is linear in each argument.

[F3]

The exponential map off the cut locus (The exponential map is a diffeomorphism on the open tangent cut domain): $D_{p} = {t v : v \in S_{p} M, 0 < t < c_{p} (v)}$ is open in $T_{p} M$ and $\exp_{p} ∣_{D_{p}}$ is a diffeomorphism onto $M ∖ ({p} \cup Cut (p))$ .

[F4]

Gradient of the distance (Gradient of the distance is the outward unit radial field off the base point and the cut locus): if $v \in S_{p} M$ and $0 < t < c_{p} (v)$ , then with $γ (t) = \exp_{p} (t v) = q$ one has $grad r (q) = \dot{γ} (t)$ , the segment $γ ∣_{[0, t]}$ is the minimizing radial geodesic from $p$ to $q$ , and in particular $d_{g} (p, q) = t$ .

[F5]

The gradient represents $d r$ (Riemannian gradient): $g_{x} (grad f (x), W) = d f_{x} (W)$ for every smooth $f$ , every $x \in M$ and every $W \in T_{x} M$ .

[F6]

Hessian of the distance off the cut locus (Hessian of distance in terms of radial jacobi fields, The Hessian of a $C^{2}$ scalar field is symmetric, Distance from p is smooth off p and the cut locus): the distance function $r$ is smooth at $q$ , its Levi-Civita Hessian $\nabla^{2} r$ is a symmetric bilinear form there, and for $T : = \dot{γ} (t_{0}) = grad r (q)$ one has $Hess r (T, Y) = 0$ for every $Y \in T_{q} M$ .

[F7]

Hessian comparison (Hessian comparison for distance under sectional curvature bounds): let $γ$ be the minimizing unit-speed geodesic from $p$ to $q$ with $\dot{γ} (t_{0}) = grad r (q)$ , and put $N : = {\dot{γ} (t_{0})}^{⊥}$ . If $Rm (X, \dot{γ} (s), \dot{γ} (s), X) \geq k ∣ X ∣^{2}$ for all $s \in (0, t_{0}]$ and all $X \in {\dot{γ} (s)}^{⊥}$ , then $Hess r (X, X) \leq {ct}_{k} (t_{0}) g_{q} (X, X)$ for all $X \in N$ ; if the reverse curvature inequality holds, then $Hess r (X, X) \geq {ct}_{k} (t_{0}) g_{q} (X, X)$ there; and $Hess r (grad r, \cdot) = 0$ .

[F8]

Laplacian comparison (Laplacian comparison for distance under a ricci lower bound): if $Ric \geq (n - 1) k g$ , then $Δ_{g} r (q) \leq (n - 1) {ct}_{k} (t_{0})$ .

[F9]

Laplace–Beltrami operator (Laplace–Beltrami operator as the trace of the Hessian): $Δ_{g} f = {tr}_{g} Hess f = \sum_{i = 1}^{n} Hess f (e_{i}, e_{i})$ in a $g$ -orthonormal basis $(e_{1}, \dots, e_{n})$ of the tangent space.

[F10]

Ricci curvature (Ricci curvature, Ricci curvature is symmetric and basis independent): in an orthonormal basis $(e_{1}, \dots, e_{n})$ of $T_{x} M$ , with $Rm (A, B, C, D) = g (R (A, B) C, D)$ , $Ric (X, Y) = \sum_{i = 1}^{n} Rm (e_{i}, X, Y, e_{i})$ .

[F11]

Comparison functions (Comparison sine, cosine and cotangent functions): ${sn}_{k} = \sin (\sqrt{k} \cdot) / \sqrt{k}$ , ${cs}_{k} = \cos (\sqrt{k} \cdot)$ for $k > 0$ , ${sn}_{0} (t) = t$ , ${cs}_{0} = 1$ , and ${sn}_{k} = \sinh (\sqrt{- k} \cdot) / \sqrt{- k}$ , ${cs}_{k} = \cosh (\sqrt{- k} \cdot)$ for $k < 0$ , with ${ct}_{k} = {cs}_{k} / {sn}_{k}$ on the positive domain.

[F12]

Orthonormal bases (Every finite-dimensional real or complex inner product space has an orthonormal basis): every finite-dimensional inner product space has an orthonormal basis.

[F13]

Standard realizations (Round sphere model geometry, The round sphere has positive constant sectional curvature, Euclidean space has zero curvature, $R$ and $R^{n}$ for $n \geq 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $R$ , Upper half-space model geometry, Cartan hadamard for hyperbolic space): the round sphere $S_{R}^{n}$ has constant sectional curvature $1 / R^{2}$ , is compact hence geodesically complete, and its cut locus at a point is the antipodal singleton, so $S_{1 / \sqrt{k}}^{n}$ has constant sectional curvature $k$ for $k > 0$ ; Euclidean $R^{n}$ has vanishing curvature tensor and is metrically complete; the upper-half-space metric of the half-space model proposition has constant sectional curvature $- a^{2}$ for every $a > 0$ , that is, $- 1 / r^{2}$ on the scale $r = 1 / a$ , for $n \geq 2$ ; and the hyperboloid model of hyperbolic $n$ -space is an $n$ -dimensional, geodesically and metrically complete manifold of constant sectional curvature $- 1$ .

Proof

technique · direct: since the sectional curvature is identically $k$, both the one-sided curvature hypotheses of the Hessian comparison hold, so the two inequalities collapse to the single normal-Hessian identity; adding the known vanishing on the radial direction gives the full tensor $\operatorname{ct}_k(r)(g-dr\otimes dr)$; tracing it in the orthogonal decomposition $T_qM=N\oplus\mathbb R\operatorname{grad}r$ gives the Laplacian, and tracing the constant-curvature tensor gives the Ricci equality that identifies the space form as an equality case of both comparison theorems

1.1F1F2F3F4F5given

The minimizing geodesic, the normal hyperplane and the curvature data. [F1, F2, F3, F4, F5, given] By [F3] there are a unit vector $v \in S_{p} M$ and a time $t_{1}$ with $0 < t_{1} < c_{p} (v)$ and $\exp_{p} (t_{1} v) = q$ ; let $γ (t) : = \exp_{p} (t v)$ for $t \geq 0$ . By [F4] the segment $γ ∣_{[0, t_{1}]}$ is minimizing from $p$ to $q$ , so $t_{1} = d_{g} (p, q) = t_{0}$ ; hence $q = γ (t_{0})$ with $0 < t_{0} < c_{p} (v)$ and $T : = \dot{γ} (t_{0}) = grad r (q)$ . Put $N : = T^{⊥} \subseteq T_{q} M$ , the orthogonal complement of $T$ . Then $T_{q} M = N \oplus R T$ , and by [F5], $d r (W) = g_{q} (grad r (q), W) = g_{q} (T, W) (W \in T_{q} M),$ so $d r (T) = ∣ T ∣_{g}^{2} = 1$ and $d r$ vanishes on $N$ ; moreover $r (q) = t_{0} > 0$ and $t_{0} < π / \sqrt{k}$ when $k > 0$ by hypothesis, so the domain restriction of [F7] and [F8] is satisfied. Finally, for every $s \in (0, t_{0}]$ and every nonzero $X \in {\dot{γ} (s)}^{⊥}$ the vector $u : = X / ∣ X ∣_{g}$ is a unit vector with $g (u, \dot{γ} (s)) = 0$ , so $(u, \dot{γ} (s))$ is an orthonormal pair and [F2] gives $Rm (X, \dot{γ} (s), \dot{γ} (s), X) = ∣ X ∣_{g}^{2} Rm (u, \dot{γ} (s), \dot{γ} (s), u) = ∣ X ∣_{g}^{2} K (u \land \dot{γ} (s)) = k ∣ X ∣_{g}^{2},$ the last equality because the sectional curvature is constantly $k$ ; the equality holds also for $X = 0$ by four-linearity. Hence both $Rm (X, \dot{γ} (s), \dot{γ} (s), X) \geq k ∣ X ∣_{g}^{2}$ and the reverse inequality hold for every such $X$ , and the geodesic $γ$ is the minimizing unit-speed geodesic from $p$ to $q$ with $\dot{γ} (t_{0}) = grad r (q)$ required by [F7].

2.1F6F7step 1.1

The Hessian on the normal hyperplane. [F6, F7, step 1.1] By step 1.1 both curvature hypotheses of [F7] hold along $γ$ , with $X$ ranging over $N$ and $t_{0} < π / \sqrt{k}$ when $k > 0$ . Applying the first assertion of [F7] (lower curvature bound) and then the second (upper curvature bound) to the same $X \in N$ gives $Hess r (X, X) \leq {ct}_{k} (t_{0}) g_{q} (X, X) \leq Hess r (X, X) .$ The symmetric bilinear forms therefore agree on the diagonal, and polarization gives equality for every $X, Y \in N$ . In addition $Hess r (grad r, \cdot) = 0$ by [F6] (equivalently by the last assertion of [F7]); in particular $Hess r (T, T) = 0$ and, by the symmetry of the Hessian in [F6], $Hess r (T, Y) = Hess r (Y, T) = 0$ for every $Y \in T_{q} M$ .

3.1F5step 1.1step 2.1

The full Hessian tensor. [F5, step 1.1, step 2.1] Let $W, Z \in T_{q} M$ and decompose $W = d r (W) T + X, Z = d r (Z) T + Y,$ where $X : = W - d r (W) T$ and $Y : = Z - d r (Z) T$ lie in $N$ because $d r (X) = d r (W) - d r (W) d r (T) = 0$ and $d r (Y) = 0$ by $d r (T) = 1$ and $d r ∣_{N} = 0$ from step 1.1. Bilinearity of the Hessian, the vanishing of all Hessian entries involving $T$ from step 2.1, and the normal identity of step 2.1 give $Hess r (W, Z) = d r (W) d r (Z) Hess r (T, T) + d r (W) Hess r (T, Y) + d r (Z) Hess r (X, T) + Hess r (X, Y) = {ct}_{k} (t_{0}) g_{q} (X, Y) .$ Since $T ⊥ N$ , orthogonality gives $g_{q} (W, Z) = d r (W) d r (Z) + g_{q} (X, Y)$ , hence $g_{q} (X, Y) = g_{q} (W, Z) - d r (W) d r (Z)$ and $Hess r (W, Z) = {ct}_{k} (t_{0}) (g_{q} (W, Z) - d r (W) d r (Z))$ for all $W, Z \in T_{q} M$ , which is the first displayed conclusion.

4.1F9F12step 3.1

The trace: the Laplacian. [F9, F12, step 3.1] By [F12] the finite-dimensional space $N$ has an orthonormal basis $(e_{1}, \dots, e_{n - 1})$ , and with $e_{n} : = T$ this is an orthonormal basis of $T_{q} M$ because $T_{q} M = N \oplus R T$ , $T ⊥ N$ and $∣ T ∣_{g} = 1$ (step 1.1). By the trace definition [F9] and the Hessian identity of step 3.1, $Δ_{g} r (q) = \sum_{i = 1}^{n - 1} {ct}_{k} (t_{0}) (g_{q} (e_{i}, e_{i}) - d r (e_{i})^{2}) + {ct}_{k} (t_{0}) (g_{q} (T, T) - d r (T)^{2}),$ where $g_{q} (e_{i}, e_{i}) = 1$ and $d r (e_{i}) = 0$ for $i \leq n - 1$ and $g_{q} (T, T) = 1$ , $d r (T) = 1$ . Hence $Δ_{g} r (q) = {ct}_{k} (t_{0}) ((n - 1) - 0) + {ct}_{k} (t_{0}) (1 - 1) = (n - 1) {ct}_{k} (t_{0}),$ which is the second displayed conclusion.

5.1F2F7F8F10step 2.1step 4.1

Equality in both comparison theorems. [F2, F7, F8, F10, step 2.1, step 4.1] The Ricci tensor of a manifold of constant sectional curvature $k$ is $Ric = (n - 1) k g$ : indeed, in an orthonormal basis $(e_{1}, \dots, e_{n})$ of $T_{x} M$ at an arbitrary point $x$ , [F10] and the second display of [F2] give $Ric (X, Y) = \sum_{i = 1}^{n} Rm (e_{i}, X, Y, e_{i}) = \sum_{i = 1}^{n} k (g (X, Y) g (e_{i}, e_{i}) - g (e_{i}, Y) g (X, e_{i})) = k (n g (X, Y) - g (X, Y)),$ that is, $Ric (X, Y) = (n - 1) k g (X, Y)$ . Thus the hypothesis of [F8] holds with equality, and [F8] gives $Δ_{g} r (q) \leq (n - 1) {ct}_{k} (t_{0})$ , which step 4.1 upgrades to equality; so the Laplacian comparison bound is attained. Likewise step 2.1 shows that $Hess r = {ct}_{k} (t_{0}) g$ on $N$ , which is simultaneously the upper bound of the first assertion and the lower bound of the second assertion of [F7]: the model attains equality in both one-sided Hessian comparisons.

6.1F11F13step 3.1step 4.1step 5.1∎

Conclusion, explicit values and the standard realizations. [F11, F13, step 3.1, step 4.1, step 5.1] This proves the three displayed conclusions for every point $q$ off $p$ and off the cut locus with $t_{0} < π / \sqrt{k}$ when $k > 0$ . The explicit values of the comparison cotangent follow from the piecewise formulas of [F11]: ${ct}_{0} (t) = \frac{1}{t}, {ct}_{k} (t) = \frac{\sqrt{k} \cos (\sqrt{k} t)}{\sin (\sqrt{k} t)} = \sqrt{k} \cot (\sqrt{k} t) (k > 0), {ct}_{k} (t) = \sqrt{- k} \coth (\sqrt{- k} t) (k < 0)$ for $t$ in the positive domain, so in Euclidean space the first conclusion reads $Hess r = r^{- 1} (g - d r \otimes d r)$ and the second reads $Δ_{g} r = (n - 1) / r$ , while on the sphere and in hyperbolic space it reads $Δ_{g} r = (n - 1) \sqrt{k} \cot (\sqrt{k} r)$ and $Δ_{g} r = (n - 1) \sqrt{- k} \coth (\sqrt{- k} r)$ . By [F13] the round sphere $S_{1 / \sqrt{k}}^{n}$ for $k > 0$ , Euclidean $R^{n}$ for $k = 0$ , and the constant-curvature models of curvature $k < 0$ built from the hyperbolic half-space metric and the hyperboloid model are realizations of the hypotheses, so the formulas above are their distance Hessians and Laplacians off the pole and the cut locus. The case $n = 2$ is included; the restriction $t_{0} < π / \sqrt{k}$ is vacuous for $k \leq 0$ and excludes the spherical pole for $k > 0$ ; the value $t_{0} = 0$ is excluded because $q \neq p$ . Every object used is supplied by the cited interfaces and no family of directions, bases or curves is selected from a set, so no choice beyond [A1] is used.

Source locator

Datar §§26.1–26.2 and 28.1, pp.191–197 and 205–209, computes the Hessian and Laplacian of the distance function in the constant-curvature model, with the shape operator ${ct}_{k} (r)$ and the saturated comparison at the model pole. Eschenburg §§2–4, pp.6–16, derives the Riccati and Hessian comparison operators whose equality case is the constant-curvature model used here. The two comparison items applied in both directions are the in-run Hessian and Laplacian comparison theorems; the model values are the piecewise comparison functions of Comparison sine, cosine and cotangent functions, and the standard realizations cited in [F13] record the curvature and completeness of the round sphere, Euclidean space and hyperbolic space.

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Sources

Ved Datar, Lectures on Riemannian Geometry (2025) (standard reference, not scraped)
J.-H. Eschenburg, Comparison Theorems in Riemannian Geometry (standard reference, not scraped)