Hessian and Laplacian Comparison

Throughout, manifolds and maps are smooth unless stated otherwise. On a Riemannian manifold, \(\nabla\) denotes the Levi–Civita connection. The needed local constructions start with coordinate charts.

Away from the base point and the cut locus, the distance function is smooth and its gradient follows radial geodesics. Its Hessian measures how this radial field changes in transverse directions. Jacobi fields relate that derivative to the index form, so curvature comparison gives Hessian comparison.

Taking a trace leads to the Laplacian. We first establish the Hessian identity, then explain why comparison with a constant-curvature model can use a Ricci bound.

Remark 1 (Hessian Operator on Manifolds). The Hessian operator associated with the smooth function \(u : M \to \mathbb R\) is a symmetric, bilinear \(2\)-tensor that associates a pair of vector fields with a smooth function. By definition, \(\operatorname{grad} u\) is a vector field on \(M\) defined through

\[\langle \operatorname{grad} u, v \rangle = \dif u_p (v) ~,\]

where \(p \in M\) and \(v \in T_p M\). Then, the Hessian is defined as the covariant derivative of \(\operatorname{grad} u\), i.e.

\[ \begin{aligned} \operatorname{Hess} u &:& T_p M &\to& & T_p M \\ &:& X &\mapsto& & \nabla_X \operatorname{grad} u ~. \end{aligned}\tag{1}\]

Metric compatibility and torsion-freeness imply symmetry. Indeed,

\[\begin{aligned} \langle \nabla_X \operatorname{grad} u, Y \rangle &= X \langle \operatorname{grad} u, Y \rangle - \langle \nabla_X Y, \operatorname{grad} u \rangle \\ &= X(Yu) - (\nabla_X Y) u \\ &= Y(Xu) + [X,Y]u - (\nabla_X Y) u = \langle \nabla_Y \operatorname{grad} u, X \rangle ~, \end{aligned}\tag{2}\]

where the second equality follows from the fact that the Levi-Civita connection is compatible with the metric, the third equality follows from definition, the fourth through sixth equalities follow from the torsion-free property of the metric connection \([X,Y] = \nabla_X Y - \nabla_Y X\). It follows that the Hessian operator is self-adjoint, and determines a symmetric, bilinear form

\[ \operatorname{Hess}_u (X,Y) = \langle (\operatorname{Hess} u) X, Y \rangle ~~,~~ X,Y \in T_p M ~.\tag{3}\]

Furthermore, we note that the bracket \([V,W]\) applied to \(u\) is

\[\begin{aligned} [V,W] u &= V(W(u)) - W(V(u)) = V \langle \operatorname{grad} u, W \rangle - W \langle \operatorname{grad} u, V \rangle \\ &= \operatorname{Hess}_u (V,W) + \langle \nabla_V W, \operatorname{grad} u \rangle - \operatorname{Hess}_u (W,V) - \langle \nabla_W V, \operatorname{grad} u \rangle \\ &= \operatorname{Hess}_u (V,W) - \operatorname{Hess}_u (W,V) + [V,W]u ~, \end{aligned}\]

which yields another demonstration that the Hessian operator induces a symmetric, bilinear form.

The notation distinguishes the self-adjoint endomorphism \(X\mapsto\nabla_X\operatorname{grad}u\) from its associated bilinear tensor \(\nabla^2u(X,Y)\). The formula \(X(Yu)-(\nabla_XY)u\) is \(C^\infty\)-linear in each argument: the derivative of a coefficient in \(Y\) cancels between the two terms. Thus the value depends only on \(X_p,Y_p\), and in coordinates

\[(\nabla^2u)_{ij}=\partial_i\partial_j u-\Gamma^k_{ij}\partial_k u.\]

This also makes clear why ordinary second partial derivatives alone are not an invariant Hessian away from a critical point.

Remark 2. We denote the Hessian operator not as in (2), but instead use the following notation: for some function \(u\) acting on vector fields \(X\) and \(Y\) on \(M\), we write \(\nabla^2 u(X,Y)\) for the Hessian of \(u\).

Theorem 3 (Hessian comparison). Let \(M,\widetilde M\) be complete Riemannian manifolds of the same dimension \(n\geq2\), and let \(\gamma,\widetilde\gamma:[0,b]\to M,\widetilde M\) be unit-speed minimizing geodesics. Put \(p=\gamma(0),\ \widetilde p=\widetilde\gamma(0)\). Suppose every radial sectional curvature along \(\widetilde\gamma\) is no greater than every radial sectional curvature along \(\gamma\) at the same time:

\[\widetilde K^+(t):=\max_{\widetilde\Pi\ni\widetilde\gamma'(t)}\widetilde K(\widetilde\Pi) \leq K^-(t):=\min_{\Pi\ni\gamma'(t)}K(\Pi).\]

For \(0<a<b\), put \(q=\gamma(a),\ \widetilde q=\widetilde\gamma(a)\). If

\[|X|=|\widetilde X|,\qquad \langle X,\gamma'(a)\rangle=\langle\widetilde X,\widetilde\gamma'(a)\rangle,\]

then

\[\nabla^2d_p(X,X)\leq\widetilde\nabla^2\widetilde d_{\widetilde p}(\widetilde X,\widetilde X).\]

The strict interior condition \(0<a<b\) places both endpoints before their cut times, where these distance functions are smooth.

Proof. Write \(T=\gamma'\), \(\widetilde T=\widetilde\gamma'\). On a normal neighborhood along the interior segment, \(\operatorname{grad}d_p=T\), \(|\operatorname{grad}d_p|=1\), and \(\nabla_TT=0\). Thus \(\nabla^2d_p(T,Y)=0\) for every \(Y\); the same is true in the model. Subtract the radial parts of \(X,\widetilde X\), which have equal lengths. The remaining normal vectors also have equal lengths, so it suffices to consider normal vectors.

There are unique Jacobi fields \(J,\widetilde J\) with initial value zero and terminal values \(X,\widetilde X\), because there is no conjugate point before the cut time. They are normal. A radial geodesic variation with endpoint derivative \(X\) realizes \(J\). Torsion-freeness interchanges the two covariant derivatives, and differentiating the terminal unit radial vector gives

\[\nabla^2d_p(X,X)=\langle J'(a),X\rangle=I_M(J,J).\]

For completeness, one may parameterize the varied geodesics on the fixed interval \([0,a]\). The derivative of their speed at the central curve is \(\langle X,T(a)\rangle/a=0\), so normalizing their velocities introduces no extra term. The last equality follows from integration by parts, \(J(0)=0\), and \(J''+R(T,J)T=0\).

Choose a parallel isometry \(\Phi\) from the normal bundle along \(\widetilde\gamma\) to that along \(\gamma\), mapping \(\widetilde X\) to \(X\) at \(a\). Set \(V=\Phi\widetilde J\). It has the same endpoints as \(J\). The index lemma and the radial curvature bound give the two distinct inequalities

\[I_M(J,J)\leq I_M(V,V) =\int_0^a\bigl(|\widetilde J'|^2-K_M(T,V)|V|^2\bigr)\,dt \leq I_{\widetilde M}(\widetilde J,\widetilde J).\]

When \(V=0\), its curvature integrand is interpreted as zero. Applying the Hessian identity on \(\widetilde M\) proves the conclusion. ◻

As we move onto other comparison theorems, we need the relation between the Laplacian and the trace of the Hessian on a manifold \(M\).

Definition 4. Let \(M\) be a differentiable manifold. The Laplacian of \(M\) is an operator, denoted \(\Delta\), defined on the space of real-valued, smooth functions on \(M\):

\[\begin{aligned} \Delta &:& \mathscr F (M) &\to& & \mathscr F (M) \\ &:& u &\mapsto& & \operatorname{div} \operatorname{grad} u ~. \end{aligned}\]

A Riemannian metric is part of this definition: it determines both the gradient and the divergence. Our sign is \(\Delta=\sum_i\partial_i^2\) on Euclidean space.

Lemma 5. Let \(M\) be a differentiable manifold, and \(f\) be a smooth function on \(M\). In a geodesic frame at \(p \in M\), \(\Delta f(p) = \sum_{1 \leq i \leq n}E_i(E_i (f)) (p)\).

All identities using \(\nabla_{E_i}E_j=0\) in the proof are evaluated at the chosen point \(p\). For an arbitrary local orthonormal frame the formula is

\[\Delta f=\sum_i\bigl(E_i(E_if)-(\nabla_{E_i}E_i)f\bigr).\]

A geodesic orthonormal frame removes the correction only at its center, not throughout a curved neighborhood.

Proof. Pick any \(p \in M\). Let \(\{E_i\}_{1 \leq i \leq n}\) denote a geodesic frame at \(p\). First, the divergence along a vector field \(X\) is the trace of the linear map \(Y \mapsto \nabla_Y X\). Since the vector field \(X\) admits an orthonormal expansion \(X = \sum_{k=1}^n f_kE_k\), we have

\[\begin{aligned} \nabla_{E_i}X = \nabla_{E_i}\left(\sum_{k=1}^n f_kE_k\right) = \sum_{k=1}^n E_i(f_k)E_k + \sum_{k=1}^n f_k \nabla_{E_i}E_k ~. \end{aligned}\]

By the fact that \(\{E_i\}_{1 \leq i \leq n}\) is a geodesic frame, we have \((\nabla_{E_i}E_k)|_p = 0\). Therefore,

\[\begin{aligned} \operatorname{div} X &=\sum_{1 \leq i, j \leq n}\langle\nabla_{E_j}(f_iE_i),E_j\rangle = \sum_{1 \leq i, j \leq n}\bigl(E_j(f_i)\langle E_i,E_j\rangle + f_i\langle\nabla_{E_j}E_i,E_j\rangle\bigr)=\sum_{1 \leq i \leq n}E_i(f_i) ~. \end{aligned}\]

Then, by definition,

\[\begin{aligned} {\operatorname{grad}} f = \sum_{1 \leq i \leq n} \langle {\operatorname{grad}} f,E_i\rangle E_i = \sum_{1 \leq i \leq n} {\rm d}f(E_i)E_i = \sum_{1 \leq i \leq n} E_i(f)E_i ~. \end{aligned}\]

Since the Laplacian is defined as \(\Delta f = \operatorname{div} \operatorname{grad} f\) for some smooth function \(f\) defined on \(M\), we have the formula

\[\begin{aligned} \Delta f(p) = \operatorname{div} \left( \sum_{1 \leq i \leq n} E_i(f)E_i \right)(p) = \sum_{1 \leq i \leq n}E_i(E_i (f)) (p) ~, \end{aligned}\]

as desired. ◻

Lemma 6. Let \(M\) be a differentiable manifold, and \(f : M \to \mathbb R\). The Laplacian operator \(\Delta\) applied to \(f\) is given by the formula \(\Delta f = \operatorname{Tr} \nabla^2 f\).

Here \(f\) is smooth and the trace is taken with respect to the Riemannian metric, so it is independent of the chosen orthonormal basis.

Proof. Pick any \(p \in M\), and let \(\{E_i\}_{1 \leq i \leq n}\) be an orthonormal basis on \(T_p M\). Then,

\[\begin{aligned} \Delta f &= \operatorname{div}_M \operatorname{grad} f = \sum_{1 \leq i \leq n} \langle \nabla_{E_i} \operatorname{grad} f, E_i \rangle = \sum_{1 \leq i \leq n} \langle (\nabla^2 f)E_i, E_i \rangle = \operatorname{Tr} \nabla^2 f ~, \end{aligned}\]

as desired. ◻

Corollary 7. Under the same assumptions as in Theorem 3, \(\Delta d_p (q) \leq \widetilde \Delta \widetilde d_{\widetilde p} ({\widetilde q})\).

Proof. Choose orthonormal bases \(T(a),E_2,\ldots,E_n\) and \(\widetilde T(a),\widetilde E_2,\ldots,\widetilde E_n\). Apply Theorem 3 to each pair \(E_i,\widetilde E_i\). The radial entries on both sides are zero, and summing the remaining \(n-1\) inequalities gives the traces by Lemma 6. Equal dimensions ensure that the sums have the same number of terms. ◻

1. Laplacian Comparison Theorem

There are situations where we want to obtain a Laplacian comparison as in Corollary 7, but would like to relax the sectional curvatures requirements in Theorem 3—namely, to imposing requirements on the Ricci curvature instead. Comparison with a constant-curvature model admits such a result. The reason is that summing the curvature terms in the index forms gives Ricci curvature.

We will suppress dependence on the ambient dimension below. Explicitly, when the dimension of \(M\) is known (e.g. \(\dim M = n\)), then \(M_k = M^n_k\) denotes the space of \(n\)-dimensional space forms with constant sectional curvature \(k\).

Theorem 8 (Laplacian comparison). Let \(M^n\) be a complete connected Riemannian manifold, \(n\geq2\), with

\[\operatorname{Ric}(V,V)\geq(n-1)k|V|^2\quad\text{for all }V.\]

Let \(\gamma\) be unit speed from \(p\). At \(t>0\) strictly before the cut time along \(\gamma\), and also \(t<\pi/\sqrt{k}\) if \(k>0\),

\[\Delta d_p(\gamma(t))\leq(n-1)\frac{s_k'(t)}{s_k(t)} =\Delta_{M_k^n}d_{\widetilde p}(\widetilde\gamma(t)),\]

where \(M_k^n\) is the complete simply connected space form of curvature \(k\) and \(\widetilde\gamma\) is unit speed from \(\widetilde p\). This is a classical pointwise assertion on the smooth part of the distance function; it is not a statement about its classical derivatives at the base point or cut locus.

Proof. Fix such a \(t\), and parallel transport an orthonormal frame \(T,E_2,\ldots,E_n\) along \(\gamma|_{[0,t]}\). For \(i\geq2\), let \(J_i(0)=0,\ J_i(t)=E_i(t)\) be the unique normal Jacobi field. The Hessian identity from Theorem 3 gives

\[\Delta d_p(\gamma(t))=\sum_{i=2}^n I(J_i,J_i).\]

Transfer the explicit model Jacobi fields as test fields

\[V_i(u)=\frac{s_k(u)}{s_k(t)}E_i(u),\qquad 0\leq u\leq t.\]

They have the same endpoints as \(J_i\), and the denominator is positive on the stated interval. The index lemma gives \(I(J_i,J_i)\leq I(V_i,V_i)\). Summing before estimating the curvature terms is the key step:

\[\begin{aligned} \sum_{i=2}^n I(V_i,V_i) &=\frac1{s_k(t)^2}\int_0^t \left((n-1)s_k'(u)^2-s_k(u)^2\operatorname{Ric}(T,T)\right)\,du\\ &\leq\frac{n-1}{s_k(t)^2}\int_0^t(s_k'^2-ks_k^2)\,du\\ &=(n-1)\frac{s_k'(t)}{s_k(t)}. \end{aligned}\]

Here our curvature convention gives \(\operatorname{Ric}(T,T)=\sum_{i=2}^n\langle R(T,E_i)T,E_i\rangle\). Finally \((s_ks_k')'=s_k'^2-ks_k^2\) and \(s_k(0)=0\), proving the last equality. In the model all \(V_i\) themselves are the endpoint Jacobi fields and its Ricci curvature equals \((n-1)k\), so every inequality becomes equality there. ◻

2. The Bochner Formula

On Euclidean space \(\mathbb R^n\), partial derivatives commute, hence

\[ \frac12 \Delta |\nabla u|^2 = |\operatorname{Hess}_u|^2 + \langle \nabla u, \nabla \Delta u \rangle ~.\tag{4}\]

On a Riemannian manifold \(M\), derivatives do not commute. The curvature tensor measures the failure of covariant derivatives to commute. After tracing, the curvature contribution becomes the Ricci term in Bochner’s formula.

Proposition 9 (Bochner formula). On a Riemannian manifold \(M\),

\[ \frac12 \Delta |\nabla u|^2 = |\operatorname{Hess}_u|^2 + \langle \nabla u, \nabla \Delta u \rangle + \operatorname{Ric}_M (\nabla u, \nabla u) ~.\tag{5}\]

Proof of Proposition 9. Let \(\{E_i\}_{1 \leq i \leq n}\) denote a local orthonormal frame of \(M\). Then,

\[\Delta v = \sum_{1 \leq i \leq n} \nabla^2 v(E_i,E_i) = \sum_{1 \leq i \leq n} \langle \nabla_{E_i} \nabla v, E_i \rangle ~.\]

Putting \(v = \frac12 |\nabla u|^2\), we have

\[\begin{aligned} \langle \nabla v, E_j \rangle = \langle \nabla_{E_j} \nabla u, \nabla u \rangle &= \operatorname{Hess}_u (E_j, \nabla u) \\ &= \operatorname{Hess}_u (\nabla u, E_j) = \langle \nabla_{\nabla u} \nabla u, E_j \rangle ~, \end{aligned}\]

where the first equality follows from compatibility with the metric, second equality follows from definition, third equality follows from symmetry, and the final equality follows from definition. Since this is true for any local coordinates \(E_j\), we have

\[\nabla v = \nabla_{\nabla u} \nabla u ~.\]

A direct calculation reveals

\[\begin{aligned} \nonumber \Delta v &= \sum_{1 \leq i \leq n} \langle \nabla_{E_i} \nabla_{\nabla u} \nabla u, E_i \rangle \\ &= \sum_{1 \leq i \leq n} \langle \nabla_{\nabla u} \nabla_{E_i} \nabla u - \nabla_{[\nabla u, E_i]} \nabla u + R^M (\nabla u, E_i) \nabla u, E_i \rangle ~, \end{aligned}\tag{6}\]

where \(R^M\) is the Riemann curvature tensor on \(M\). Expanding (6), we note that the last term \(\sum_{1 \leq i \leq n} \langle R^M (\nabla u, E_i) \nabla u, E_i \rangle = \operatorname{Ric}_M (\nabla u, \nabla u)\), by definition. The second term, by symmetry of the connection, becomes

\[\begin{aligned} \nonumber - \sum_{1 \leq i \leq n} \langle \nabla_{[\nabla u, E_i]} \nabla u, E_i \rangle &= - \sum_{1 \leq i \leq n} \langle \nabla_{E_i} \nabla u,[\nabla u, E_i] \rangle = \sum_{1 \leq i \leq n} \langle \nabla_{E_i} \nabla u, \nabla_{E_i} \nabla u - \nabla_{\nabla u} E_i \rangle \\ &= |\operatorname{Hess}_u|^2 - \sum_{1 \leq i \leq n} \langle \nabla_{E_i} \nabla u, \nabla_{\nabla u} E_i \rangle ~. \end{aligned}\tag{7}\]

The first term can be rewritten by appealing to the compatibility of the connection with the metric, and becomes

\[\begin{aligned} \nonumber \sum_{1 \leq i \leq n} \langle \nabla_{\nabla u} \nabla_{E_i} \nabla u, E_i \rangle &= \sum_{1 \leq i \leq n} \nabla u \langle \nabla_{E_i} \nabla u, E_i \rangle - \sum_{1 \leq i \leq n} \langle \nabla_{E_i} \nabla u, \nabla_{\nabla u} E_i \rangle \\ &= \langle \nabla u, \nabla \Delta u \rangle - \sum_{1 \leq i \leq n} \langle \nabla_{E_i} \nabla u, \nabla_{\nabla u} E_i \rangle ~. \end{aligned}\tag{8}\]

Combining (6), (7), and (8), we have

\[\Delta v = |\operatorname{Hess}_u|^2 + \langle \nabla u, \nabla \Delta u \rangle + \operatorname{Ric}_M (\nabla u, \nabla u) - 2 \sum_{1 \leq i \leq n} \langle \nabla_{E_i} \nabla u, \nabla_{\nabla u} E_i \rangle ~. \tag{9}\]

To complete the proof, we claim that \(\sum_{1 \leq i \leq n} \langle \nabla_{E_i} \nabla u, \nabla_{\nabla u} E_i \rangle = 0\). The claim follows from a series of direct calculations. Expanding both \(\nabla_{E_i} \nabla u\) and \(\nabla_{\nabla u} E_i\) in the basis \(\{E_j\}_{1 \leq j \leq n}\), we have

\[\begin{aligned} \nabla_{E_i} \nabla u &= \sum_{1 \leq j \leq n} a_{ij} E_j \\ \nabla_{\nabla u} E_i &= \sum_{1 \leq j \leq n} b_{ij} E_j \end{aligned}\]

where

\[\begin{aligned} a_{ij} &= \langle \nabla_{E_i} \nabla u, E_j \rangle ~, \\ b_{ij} &= \langle \nabla_{\nabla u} E_i, E_j \rangle ~. \end{aligned}\]

It follows that

\[\begin{aligned} \sum_{1 \leq i \leq n} \langle \nabla_{E_i} \nabla u, \nabla_{\nabla u} E_i \rangle = \sum_{1 \leq i \leq n} \sum_{1 \leq j \leq n} a_{ij} b_{ij} ~. \end{aligned}\]

Note \(a_{ij}\) is symmetric, by the symmetry of the Hessian operator. By compatibility of the connection with the metric, \(b_{ij} = \langle \nabla_{\nabla u} E_i, E_j \rangle = \nabla u \langle E_i, E_j \rangle - \langle E_i, \nabla_{\nabla u} E_j \rangle = - \langle E_i, \nabla_{\nabla u} E_j \rangle = -b_{ji}\). Finally, it follows that

\[\begin{aligned} \sum_{1 \leq i \leq n} \langle \nabla_{E_i} \nabla u, \nabla_{\nabla u} E_i \rangle = \sum_{1 \leq i \leq n} \sum_{1 \leq j \leq n} a_{ij} b_{ij} &= - \sum_{1 \leq i} \sum_{1 \leq j \leq n} a_{ij} b_{ji} ~, \end{aligned}\]

and since \(b_{ij} + b_{ji} = 0\), it follows that the left-most expression evaluates to zero. This concludes the proof of the Bochner formula. ◻

More explicitly, if \(S=\sum_{i,j}a_{ij}b_{ij}\), then relabeling the two indices gives

\[S=\sum_{i,j}a_{ji}b_{ji}=-\sum_{i,j}a_{ij}b_{ij}=-S,\]

using symmetry of \(a\) and skew-symmetry of \(b\). Thus \(S=0\). The skew-symmetry uses a frame orthonormal on its whole domain, so \(\nabla u\langle E_i,E_j\rangle=0\). The norm in the formula is \(|\operatorname{Hess}_u|^2=\sum_{i,j}a_{ij}^2\). In Euclidean coordinates, the preliminary identity follows directly from \(\frac12\sum_{i,j}\partial_i^2(\partial_j u)^2 =\sum_{i,j}(\partial_{ij}u)^2+\sum_j(\partial_ju)\partial_j\Delta u\).

Reading: Carmo (1992), Riemannian geometry.