Let $M$ be a compact oriented Riemannian manifold of dimension $n$ with boundary $\partial M$. The aim of this note is to define the divergence and Laplacian operators on $M$ and to clarify the validy and meaning of various formulas such as integration by parts

$\int_M \langle \nabla f, X \rangle = - \int_M f \text{div}X + \int_{\partial M} f \langle X, \nu \rangle$

or Green’s formula

$\int_M (f \Delta g - g \Delta f) = - \int_{\partial M}\left(f\dfrac{\partial g}{\partial \nu} - g\dfrac{\partial f}{\partial \nu}\right)$

which are well-known to hold for domains in $\mathbb{R}^n$.

The Laplace operator acting on functions defined over $\mathbb{R}^n$ is usually defined by the simple formula

$\Delta f = \dfrac{\partial^2 f}{\partial (x^1)^2} + \cdots \dfrac{\partial^2 f}{\partial (x^n)^2}$.

Obviously, this definition is no good to define anything on a manifold, so we formulate it in a more geometric way as

$\Delta f = \text{div}(\text{grad}(f))$.

In a beginning calculus course, this expression is usually understood as some kind of matrix multiplication, and we formally see that it holds. But this equation tells us what the Laplacian really is. Indeed, the divergence of a vector field is a real-valued function that at a point $p$ measures the amount of infinitesimal dilation an infinitesimal object placed at $p$ would experience if it was to flow infinitesimally along the vector field. This is the geometric interpretation of the Laplacian. On a Riemannian manifold, we can define the gradient of a function by duality via its exterior derivative. Thus we only need a notion of divergence. From our informal discussion, it makes sense to define the divergence of a vector field $X$ on an oriented Riemannian manifold as the infinitesimal change of measurement of volumes when the volume form is flowing along $X$, that is we want

$(\text{div}X)d\mu = \left.\dfrac{d}{dt} \left(\Phi_t^{-1}\right)^*d\mu \right|_{t=0}$

where $d\mu$ denotes the volume form associated to the metric and $\Phi_t$ the one-parameter group of diffeomorphisms associated to $X$. The right side of this equation is called the Lie derivative of $d\mu$ along $X$ and is noted $\mathcal{L}_Xd\mu$. Since the bundle of top forms of an oriented Riemannian manifold is trivialised by $d\mu$, i.e. every top form is of the form $\alpha d\mu$ for a unique $C^{\infty}$ function $\alpha$, we actually get a well-defined function $(\text{div}X)$ by the relation

$(\text{div}X)d\mu = \mathcal{L}_Xd\mu$,

given of course that there actually is a volume form, i.e. that our manifold is orientable. We thus define the Laplacian acting on smooth functions as $\Delta = \text{div}(\text{grad}f)$ where $\text{grad}f$ is the gradient of $f$, caracterized by the relation

$g(\text{grad}f, X) = df(X)$.

The only thing that remains undefined is the normal vector field $\nu$. It is not hard to check that if all the transition functions of some atlas on $M$ have positive Jacobian determinants, then their restriction to the boundary also have positive Jacobian determinants, enough so that the boundary is also orientable. Supposing without loss of generality that all charts containing a point of the boundary have their image lying in only one half-space of $\mathbb{R}^n$, say $\{e_1 \geq 0\}$, we can define a notion of “inward-pointing vector” tangent to the boundary. Indeed, at a point $p$ of the boundary, a tangent vector in $T_pM$ will be called “inward-pointing” if its image in any chart is contained in the half-space $\{e_1 > 0\}$. At each point of the boundary, we decompose $T_pM = T_p\partial M \oplus \mathbb{R}$ orthogonally with the metric, and we define the normal unit inward-pointing vector field $\nu$ as the unit vector field lying in the second component of this decomposition that is everywhere inward-pointing. Note that this gives a canonical volume form on the boundary, given by $d\tilde{\mu} = \iota_{\nu}d\mu$ where $\iota$ is the interior product.

Before we start to state and prove theorems, let’s recall Cartan’s formula which says that for every vector field $X$ and differential form $\omega$,

$\mathcal{L}_X\omega = d(\iota_X\omega) + \iota_Xd\omega$.

This identity tells us that $(\text{div}X)d\mu = \mathcal{L}_Xd\mu = d(\iota_Xd\mu)$ because the volume form $d\mu$ is of course closed. This will be useful when paired with Stoke’s theorem, which says that for all forms $\omega$ of degree $n-1$ on $M$,

$\int_M d\omega = \int_{\partial M} i^*\omega$

where $i : \partial M \to M$ is the inclusion. We can now prove the generalization of Gauss’s divergence theorem:

Theorem 1: With the notation introduced above,

$\int_M \text{div}Xd\mu = - \int_{\partial M} g(X,\nu)d\tilde{\mu}$

for all vector fields $X$.

Proof: By Cartan’s formula and Stoke’s theorem, we get

$\int_M \text{div}Xd\mu = - \int_M d\left(\iota_X d\mu\right) = - \int_{\partial M} \iota_X d\mu$.

But if $(\nu, e_2, \ldots, e_n)$ is an orthonormal basis of $T_pM$ for $p \in \partial M$, then we see that

$(\iota_Xd\mu)(e_2, \ldots, e_n) = d\mu(X,e_2, \ldots, e_n)$

is equal to

$g(X,\nu)d\mu(\nu, e_2, \ldots, e_n) = g(X, \nu)d\tilde{\mu}(e_2, \ldots, e_n)$

because $X = g(X,\nu)\nu + \sum_{j=2}^ng(X,e_j)e_j$ and $d\mu(e_j, e_2, \ldots, e_n) = 0$ for all $2 \leq j \leq n$. Thus at all points of the boundary we have the equality

$\iota_Xd\mu = g(X,\nu)d\tilde{\mu}$.

We can now conclude from the first equation in the proof.

QED

Recall that the integration by parts formula for functions $f : \mathbb{R} \to \mathbb{R}$ follows from Leibniz rule $(uv)' = u'v + v'u$ and from the fundamental theorem of calculus:

$u(1) - u(0) = \int_{[0,1]} (uv)' = \int_{[0,1]}u'v + v'u$.

To get the generalized version for oriented Riemannian manifolds, we need to establish the following Leibniz rule for the divergence operator:

Lemma: For all functions $f \in C^{\infty}(M)$ and vector fields $X$,

$\text{div}(fX) = f(\text{div}X) + g(\text{grad}f,X)$.

Proof: This again follows from Cartan’ identity. Indeed, by the very definition of the interior product, we get that $\iota_{fX}\omega = f\iota_X\omega$ for any differential form $\omega$ so we get

$\mathcal{L}_{fX}\omega = d(\iota_{fX}\omega) + \iota_{fX}d\omega = d(f\iota_X\omega) + f\iota_Xd\omega$.

But by the usual Leibniz rule for the exterior derivative, this is equal to

$df \wedge \iota_X \omega + fd (\iota_X \omega) + f\iota_X d\omega$

hence

$\mathcal{L}_{fX}\omega = f\mathcal{L}_X\omega + df\wedge \iota_X\omega$

so it suffices to see that $df \wedge \iota_Xd\mu = g(\text{grad}f,X)$. We proceed as in theorem 1. Let $e_1, \ldots, e_n$ be an orthonormal basis of $T_pM$ at some point $p \in M$. Then

$df \wedge \iota_Xd\mu (e_1, \ldots, e_n) = \sum_i (-1)^{i-1}df(e_i) \iota_Xd\mu(e_1, \ldots, \hat{e_i}, \ldots, e_n)$

which is equal to

$\sum_i (-1)^{i-1}df(e_i)d\mu(X,e_1,\ldots,\hat{e_i}, \ldots, e_n) = \sum_i df(e_i)X^id\mu (e_1, \ldots, e_n)$.

This shows that

$df \wedge \iota_X d\mu = \sum_i df(e_i)X^i d\mu$

but $\sum_i df(e_i)X^i$ is precisely $df(X) = g(\text{grad}f,X)$.

QED

Coupled with the divergence theorem, this immediately yields the

Theorem: For all vector fields $X$ and smooth functions $f$, there is an integration by parts formula

$\int_M g(\text{grad}f,X)d\mu = -\int_M f \text{div}X d\mu + \int_{\partial M} f \cdot g(X, \nu) d\tilde{\mu}$.

Note that for a closed Riemannian manifold, with $\langle X,Y \rangle = \int_M g(X,Y) d\mu$, this shows that

$\langle \text{grad} f, X \rangle = \langle f, -\text{div}X \rangle$

i.e. that minus the divergence operator is kind of a formal adjoint to the gradient operator.  The so-called Green formulas are a simple application of integration by parts. Recall that the Laplacian of a smooth function $f$ is defined as $\Delta f = \text{div}(\text{grad}f)$ and that $\nu$ is the inward-pointing vector field on the boundary. We will denote $g(\text{grad}f, \nu)$ by $\frac{\partial f}{\partial \nu}$.

Theorem: (Green formulas) For any two functions $u,v \in C^{\infty}(M)$,

$\int_M u\Delta v \; d\mu + \int_M g\left( \text{grad}u, \text{grad}v \right) d\mu = \int_{\partial M} u \dfrac{\partial v}{\partial \nu} \; d\tilde{\mu}$

and hence

$\int_M(u\Delta v - v\Delta u) \; d\mu = \int_\partial M \left(u\dfrac{\partial v}{\partial \nu} - v \dfrac{\partial u}{\partial \nu}\right)d\tilde{\mu}$.

Proof: Integrating by parts, we get

$\int_M g\left( \text{grad}u, \text{grad}v \right) d\mu = -\int_M u \text{ div}\left(\text{grad }v\right) d\mu + \int_{\partial M} u\dfrac{\partial v}{\partial \nu} \; d\tilde{\mu}$

hence the first formula. The second evidently follows from the first.

QED

As an immediate application, we can show the

PropositionConsider the equation $\Delta f = 0$ on a compact Riemannian manifold with boundary $(M,g)$ subject to either Dirichlet (i.e. $f(x)|_{\partial M} = H(x)$) or Neumann (i.e. $\frac{\partial f(x)}{\partial \nu} = H(x)$) boundary conditions. Then if a smooth solution exists, it is unique in the case of Dirichlet conditions and unique up to a constant  in the case of Newmann conditions.

Proof: Consider two solutions $u$ and $v$ to the problem and consider their difference $w = u-v$. Then since $\Delta = 0$, it follows that $\int_M w\Delta w = 0$. Integrating by parts, we thus find

$\int_M g\left(\text{grad }w, \text{grad }w \right) d\mu = \int_{\partial M} w \dfrac{\partial w}{\partial \nu} \; d\tilde{\mu}$

so if either the Dirichlet or Neumann boundary conditions are satisfied, the integral on the right vanishes and we deduce that $w$ is a constant, i.e. that $u = v + C$. In the case of Dirichlet boundary conditions, the function $w$ has to vanish on the boundary hence everywhere, so $u=v$.

QED

In the same vein, if $f$ is a harmonic function on a closed Riemannian manifold (i.e. $\Delta f = 0$ and $\partial M = \emptyset$), then $\int_M g(\text{grad }f, \text{grad }f)d\mu = 0$ so the only harmonic functions on a closed Riemannian manifold are the constants.

From now on, let us denote $\text{grad }f$ by $\nabla f$ and $\int_M g\left(\text{grad }f, \text{grad }f\right) d\mu$ by $\int_M |\nabla f|^2 d\mu$. Another immediate application of integration by parts yield results about the spectrum of $\Delta$:

Proposition: Let $(M,g)$ be a closed Riemannian manifold. Then (with our definition), the eigenvalues of the Laplacian are non-positive. Moreover, eigenfunctions corresponding to different eigenvalues are orthogonal for the $L^2$ inner product.

Proof: For the first assertion, suppose $\Delta u = \lambda u$. Then integrating by parts gives

$\lambda \int_M u^2 \; d\mu = \int_M u\Delta u \; d\mu = -\int_M |\nabla u|^2 \; d\mu \leq 0$.

The second assertion follows from Green’s formula since if the eigenfunctions $u$ and $v$ correpond to eigenvalues $\lambda$ and $\lambda'$, then

$\lambda \int_M uv \; d\mu = \int u\Delta v \; d\mu = \int_M v \Delta u \; d\mu = \lambda' \int_M uv \; d\mu$.

QED

This last proposition is not surprising in view of the fact that integrating by parts actually shows that $\Delta$ is formally self-adjoint with respect to the $L^2$ inner product on a closed manifold. Indeed, in this context Green’s formula reads $\langle \Delta u, v \rangle = \langle u, \Delta v \rangle.$ The first assertion of this proposition is the reason why geometers often define the Laplacian as $-\Delta$, in order to get a positive spectrum. Finally, from the first equation of the proof, we see that eigenvalues of (minus) the Laplacian satisfy

$\lambda = \dfrac{\int_M |\nabla u|^2 \; d\mu}{\int_M u^2 \; d\mu}$

with $u$ is their associated eigenfunctions, which one could recognize as a renormalized Dirichlet energy functional. With this point of view, the variational min/max principle which says that the $(k+1)$‘st eigenvalue is given by the infimum of that functional over the functions orthogonal to the eigenfunctions associated to the first $k$ eigenvalues makes plenty of sense. This min/max principle in turn explains the link between the first eigenvalue of the Laplacian and the important Poincaré’s inequality, which says that for some constant $C$, all smooth functions integrating to $0$ (i.e. orthogonal to the constants) satisfy

$||u||_{L^2} \leq C||\nabla u||_{L^2}$,

the optimal constant being exactly $\frac{1}{\sqrt{\lambda_1}}$, attained by the first eigenfunctions of the Laplacian.