Recovering the Integral of a Function
The theory of integration on manifolds was built around forms of top degree, not functions, for a reason that
was unavoidable. On a bare smooth manifold a real-valued function has no coordinate-independent integral.
Changing coordinates multiplies the expression for a function by a Jacobian factor that the function alone
cannot absorb, so the number one would compute depends on the chart, which is no number at all. A top-degree
form is exactly the object engineered to absorb that factor, and that is why
integration was defined
for forms rather than functions.
A Riemannian metric removes the obstruction. On an oriented manifold the metric singles out a canonical
top-degree form, the
Riemannian volume
form \(\omega_g\). Its coordinate expression carries precisely the factor \(\sqrt{\det g_{ij}}\)
needed to make the resulting integral coordinate-independent. Given such a form, a function can be turned into a
form simply by multiplication, and then integrated. This section makes that precise and records the one
positivity property that every later estimate rests on.
The Definition
Let \((M, g)\) be an oriented Riemannian manifold with or without boundary, and let \(\omega_g\) denote its
Riemannian volume form. For a compactly supported continuous function \(f\), the product \(f\omega_g\) is a
compactly supported \(n\)-form, so we define the integral of \(f\) over \(M\) to be
\[
\int_M f\, dV_g := \int_M f\,\omega_g,
\]
where the notation \(dV_g\) is the traditional name for the volume form in this role. On a surface it is written
\(dA_g\), and on a curve \(ds_g\). If \(M\) is itself compact, taking \(f \equiv 1\) gives the volume of \(M\),
\[
\operatorname{Vol}(M) := \int_M \omega_g.
\]
One warning about the notation is worth stating once. The symbol \(dV_g\) does not denote the exterior
derivative of anything. It is a single indivisible name for the volume form. On a compact manifold without
boundary the volume form is never exact, because its integral is the positive number \(\operatorname{Vol}(M)\),
whereas an exact top form
integrates to zero.
The \(d\) is purely a typographical inheritance from classical calculus.
Positivity of the Integral
The single analytic fact that distinguishes the integral of a function from that of a form is positivity. A
nonnegative function has a nonnegative integral, and the integral vanishes only when the function does. A form
has no sign, so nothing like this holds at the level of forms. It is the metric, through the positivity of
\(\sqrt{\det g_{ij}}\), that makes it true.
Proposition: Positivity of the Integral
Let \((M, g)\) be a nonempty oriented Riemannian manifold with or without boundary, and let \(f\) be a
compactly supported continuous real-valued function on \(M\) satisfying \(f \geq 0\). Then
\[
\int_M f\, dV_g \geq 0,
\]
with equality if and only if \(f \equiv 0\).
Proof.
Suppose first that \(f\) is supported in the domain of a single oriented smooth chart \((U, \varphi)\). In
coordinates the
volume form is
\(\sqrt{\det g_{ij}}\,dx^1 \wedge \cdots \wedge dx^n\) in a positively oriented chart, so the
integral becomes the ordinary Lebesgue integral
\[
\int_M f\, dV_g = \int_{\varphi(U)} f(x)\,\sqrt{\det g_{ij}}\, dx^1 \cdots dx^n.
\]
The factor \(\sqrt{\det g_{ij}}\) is strictly positive because \(g\) is positive-definite, and \(f \geq 0\)
by hypothesis, so the integrand is nonnegative and the integral is \(\geq 0\). In a negatively oriented
chart the volume form acquires a minus sign, but the definition of integration over a negatively oriented
chart contributes a compensating minus sign, so the same inequality holds.
For general \(f\), choose a partition of unity \(\{\psi_k\}\) subordinate to a cover of the support by
oriented charts. Each \(\psi_k f\) is nonnegative and supported in one chart, so
\(\int_M \psi_k f\, dV_g \geq 0\) by the case just handled, and summing gives \(\int_M f\, dV_g \geq 0\).
For the equality case, suppose \(f\) is positive at some point. By continuity \(f\) is positive on a
nonempty open set, on which some \(\psi_k\) is also positive. The corresponding term
\(\int_M \psi_k f\, dV_g\) is then strictly positive, forcing the total to be strictly positive. Hence a
vanishing integral requires \(f \equiv 0\), and the converse is immediate.
Two consequences are routine but constantly used. Applied to \(|f|\), positivity yields the integral triangle
inequality \(\left| \int_M f\, dV_g \right| \leq \int_M |f|\, dV_g\). Applied to a difference, it shows that
integration is monotone. These are the properties that let the Riemannian integral behave like the integral of
elementary calculus, and they are what make the volume form the right measure for analysis on a curved space.
The divergence theorem, taken up next, lives in exactly that setting.
The Divergence Theorem
The classical divergence theorem of vector calculus equates the integral of the divergence of a vector field
over a region with the flux of that field through the boundary. With the Riemannian integral in hand, the same
statement holds on any oriented Riemannian manifold, and its proof is a single application of Stokes's theorem
once the right differential-form translations are in place. The work is entirely in setting up two pieces of
geometry: a coordinate-free definition of divergence, and a clean identification of the boundary integrand.
The Divergence Operator
The volume form provides a coordinate-free way to measure how a vector field distorts volume. Contracting it
against a vector field \(X\) by
interior
multiplication produces an \((n-1)\)-form \(X \lrcorner\, \omega_g\), and its exterior derivative
is again a top-degree form, hence a function multiple of \(\omega_g\). That multiple is the divergence.
Definition: The Divergence of a Vector Field
Let \((M, g)\) be an oriented Riemannian manifold with or without boundary, with volume form \(\omega_g\).
The divergence of a smooth vector field \(X\) is the unique smooth function
\(\operatorname{div} X\) satisfying
\[
d\bigl(X \lrcorner\, \omega_g\bigr) = (\operatorname{div} X)\, \omega_g.
\]
Because \(\omega_g\) is nowhere zero, this equation determines \(\operatorname{div} X\) uniquely at each
point.
This definition makes no reference to coordinates or to a choice of orientation. Reversing the orientation flips
the sign of \(\omega_g\) on both sides, leaving \(\operatorname{div} X\) unchanged, so the divergence is well
defined on any Riemannian manifold, orientable or not. Its meaning is made transparent by Cartan's formula.
Since \(\omega_g\) is a top-degree form, \(d\omega_g = 0\), and the
Cartan formula
\(\mathcal{L}_X \omega_g = X \lrcorner\, d\omega_g + d(X \lrcorner\, \omega_g)\) collapses to
\[
\mathcal{L}_X \omega_g = d(X \lrcorner\, \omega_g) = (\operatorname{div} X)\, \omega_g.
\]
The divergence is therefore exactly the rate at which the flow of \(X\) stretches or compresses the volume form.
It is the proportionality factor in the Lie derivative of \(\omega_g\) along \(X\).
In standard coordinates on Euclidean space, where \(\omega_g = dx^1 \wedge \cdots \wedge dx^n\), expanding
\(d(X \lrcorner\, \omega_g)\) reproduces the familiar formula
\(\operatorname{div} X = \sum_i \partial X^i / \partial x^i\), so the definition extends the classical one.
The Boundary Integrand
To turn Stokes's theorem into a flux statement, the restriction of \(X \lrcorner\, \omega_g\) to the boundary
must be identified. The answer is geometrically exactly what one would hope. Only the component of \(X\) normal
to the boundary contributes, weighted by the boundary's own volume form.
Lemma: Restriction of the Contracted Volume Form
Let \((M, g)\) be an oriented Riemannian manifold with boundary, let \(N\) be the outward unit normal along
\(\partial M\), and let \(\widetilde g\) be the induced metric on \(\partial M\). For any smooth vector
field \(X\) on \(M\),
\[
\iota_{\partial M}^{*}\bigl(X \lrcorner\, \omega_g\bigr) = \langle X, N \rangle_g\, \omega_{\widetilde g},
\]
where \(\omega_{\widetilde g}\) is the volume form of \(\partial M\) with the Stokes orientation.
Proof.
Along \(\partial M\), decompose \(X\) into its normal and tangential parts. Writing
\(X^{\perp} = \langle X, N \rangle_g\, N\) for the normal component and \(X^{\top} = X - X^{\perp}\) for the
tangential one, the contraction splits by linearity:
\[
X \lrcorner\, \omega_g = X^{\perp} \lrcorner\, \omega_g + X^{\top} \lrcorner\, \omega_g.
\]
For the normal part, \(X^{\perp} = \langle X, N \rangle_g\, N\), so pulling back to the boundary and using
that the boundary
volume form is \(\iota_{\partial M}^{*}(N \lrcorner\, \omega_g)\) gives
\[
\iota_{\partial M}^{*}\bigl(X^{\perp} \lrcorner\, \omega_g\bigr)
= \langle X, N \rangle_g\, \iota_{\partial M}^{*}\bigl(N \lrcorner\, \omega_g\bigr)
= \langle X, N \rangle_g\, \omega_{\widetilde g}.
\]
For the tangential part, it suffices to show the pullback of \(X^{\top} \lrcorner\, \omega_g\) vanishes.
Evaluated on any \(n-1\) vectors tangent to \(\partial M\), the form \((X^{\top} \lrcorner\, \omega_g)\)
feeds \(\omega_g\) the \(n\) vectors \(X^{\top}, V_1, \dots, V_{n-1}\), all of which are tangent to
\(\partial M\). But \(\partial M\) is \((n-1)\)-dimensional, so these \(n\) tangent vectors are linearly
dependent, and \(\omega_g\), being alternating, vanishes on any linearly dependent set. Hence the tangential
term contributes nothing, and the displayed identity follows.
The Theorem
With both pieces in place, the divergence theorem is immediate.
Theorem: The Divergence Theorem
Let \((M, g)\) be an oriented Riemannian manifold with boundary. For any compactly supported smooth vector
field \(X\) on \(M\),
\[
\int_M (\operatorname{div} X)\, dV_g = \int_{\partial M} \langle X, N \rangle_g\, dV_{\widetilde g},
\]
where \(N\) is the outward unit normal along \(\partial M\) and \(\widetilde g\) is the induced metric on
the boundary. In dimension three over a compact region of space, this is the classical theorem of Gauss.
Proof.
By the definition of divergence, the interior integrand is
\((\operatorname{div} X)\, \omega_g = d(X \lrcorner\, \omega_g)\). Applying
Stokes's theorem to the
\((n-1)\)-form \(X \lrcorner\, \omega_g\), we obtain
\[
\int_M (\operatorname{div} X)\, dV_g = \int_M d\bigl(X \lrcorner\, \omega_g\bigr)
= \int_{\partial M} \iota_{\partial M}^{*}\bigl(X \lrcorner\, \omega_g\bigr).
\]
The preceding lemma identifies the boundary integrand as \(\langle X, N \rangle_g\, \omega_{\widetilde g}\),
which is exactly \(\langle X, N \rangle_g\, dV_{\widetilde g}\). Substituting completes the proof.
The Volume Interpretation Made Precise
The reading of \(\operatorname{div} X\) as the rate of volume change under the flow can be sharpened into an
exact dictionary between the sign of the divergence and the volume behavior of the flow.
Proposition: Geometric Interpretation of the Divergence
Let \((M, g)\) be an oriented Riemannian manifold, let \(X\) be a smooth vector field, and let \(\theta\) be
its flow. Then the flow is volume-preserving if and only if \(\operatorname{div} X = 0\) everywhere,
volume-nondecreasing if and only if \(\operatorname{div} X \geq 0\) everywhere, and strictly
volume-increasing if and only if \(\operatorname{div} X \gt 0\) on a dense subset. The analogous statements
hold for the reverse inequalities.
Proof.
Let \(D \subseteq M\) be a compact
regular domain.
The flow domain is open, so compactness of \(D\) gives an interval
around \(0\) on which \(\theta_t\) is defined on all of \(D\), and there the
flow carries \(D\)
diffeomorphically onto \(\theta_t(D)\). The change-of-variables formula for the volume form
lets the volume of the image be written as an integral over \(D\) of the pulled-back form
\(\theta_t^{*}\omega_g\). Differentiating under the integral sign at \(t = t_0\) and using that the
\(t\)-derivative of a pullback along a flow is the pullback of the Lie derivative, we obtain
\[
\frac{d}{dt}\bigg|_{t=t_0} \operatorname{Vol}(\theta_t(D))
= \int_D \theta_{t_0}^{*}\bigl(\mathcal{L}_X \omega_g\bigr)
= \int_{\theta_{t_0}(D)} (\operatorname{div} X)\, dV_g,
\]
the last step substituting \(\mathcal{L}_X \omega_g = (\operatorname{div} X)\, \omega_g\) and changing
variables back.
If \(\operatorname{div} X = 0\), this derivative vanishes for every \(D\) and every \(t_0\), so the volume
is constant and the flow is volume-preserving. The inequality versions follow the same way. For the strict
and converse directions, positivity of the integral of a function does the work. Suppose
\(\operatorname{div} X \gt 0\) on a dense set. By continuity \(\operatorname{div} X \geq 0\) everywhere, so
the derivative is nonnegative. The
manifold interior of a regular domain is its topological interior,
so \(\operatorname{Int} D\) is open in \(M\) and its image under the diffeomorphism
\(\theta_{t_0}\) is open as well. The integral of \(\operatorname{div} X\) over
\(\theta_{t_0}(\operatorname{Int} D)\) is therefore strictly positive, because the integrand is
nonnegative and positive somewhere on an open set. Hence the volume is strictly increasing.
The converses are contrapositives. If \(\operatorname{div} X\) is nonzero at a point, continuity makes it of
one sign on a neighborhood. That neighborhood
contains a compact regular domain
with nonempty interior, and the volume of that domain then changes, so the
flow is not volume-preserving, and in the negative case not volume-nondecreasing either. If instead
\(\operatorname{div} X\) fails to be positive on a dense set, some nonempty open set carries
\(\operatorname{div} X \leq 0\), and a compact regular domain inside it has nonincreasing volume, so the
flow is not strictly volume-increasing.
Surface Integrals and the Classical Stokes Theorem
The theorem that originally bore the name of Stokes concerned the circulation of a vector field around the
boundary of a surface in space, equated with the flux of its curl through the surface. That classical result is
the three-dimensional shadow of the general theorem, recovered once the curl is given an intrinsic definition.
Unlike divergence, which makes sense in every dimension, the curl is special to dimension three, and seeing why
is half the content of this section.
Gradient and Curl from the Metric
On a Riemannian manifold the metric converts between vector fields and \(1\)-forms through the musical
isomorphisms. Lowering an
index sends a vector field \(X\) to the \(1\)-form \(X^\flat = \langle X, \cdot \rangle_g\), and
raising an
index sends a \(1\)-form back to a vector field.
The gradient of a
function is defined by raising the index on its differential,
\[
\operatorname{grad} f = (df)^\sharp,
\]
so that \(\langle \operatorname{grad} f, Y \rangle_g = df(Y) = Yf\) for every \(Y\). This is the coordinate-free
version of the classical gradient, and it works in every dimension.
The curl is more delicate. Given a vector field \(X\), the \(1\)-form \(X^\flat\) has an exterior derivative
\(d(X^\flat)\), which is a \(2\)-form. To turn a \(2\)-form back into a vector field we would need a
correspondence between \(2\)-forms and vector fields, and on an \(n\)-manifold the space of \(2\)-forms has
dimension \(\binom{n}{2}\) while vector fields correspond to \(1\)-forms, of dimension \(n\). For \(n \geq 1\)
these match, \(\binom{n}{2} = n\), only when \(n = 3\). In that dimension alone, contraction with the volume
form gives a bijection between vector fields and \(2\)-forms, and the curl is defined by inverting it.
Definition: The Curl of a Vector Field
Let \((M, g)\) be an oriented Riemannian \(3\)-manifold. The curl of a smooth vector field
\(X\) is the unique smooth vector field \(\operatorname{curl} X\) satisfying
\[
(\operatorname{curl} X) \lrcorner\, \omega_g = d\bigl(X^\flat\bigr).
\]
The contraction \(Y \mapsto Y \lrcorner\, \omega_g\) is a bijection from vector fields to \(2\)-forms in
dimension three, so \(\operatorname{curl} X\) is well defined.
With gradient, curl, and divergence all expressed through the exterior derivative, the three classical operators
are revealed as a single operator \(d\) read in three successive degrees. Gradient is \(d\) on functions, curl
is \(d\) on \(1\)-forms, and divergence is \(d\) on \((n-1)\)-forms, all transported across the musical
isomorphisms and the volume-form contraction. This is the intrinsic form of the grad-curl-div correspondence
that appears for Euclidean space. The two classical
identities of vector calculus, that the curl of a gradient and the divergence of a curl both vanish, are once
again the single fact \(d \circ d = 0\).
Surface Integrals
Let \(S\) be a compact oriented \(2\)-dimensional
embedded submanifold with boundary
inside an oriented Riemannian \(3\)-manifold \(M\), and let \(N\) be a smooth unit normal field along \(S\)
determining its orientation. The induced metric gives \(S\) its own area form \(dA\), and the surface integral of a vector field \(X\) over \(S\)
is the integral of the normal component,
\[
\int_S \langle X, N \rangle_g\, dA,
\]
the flux of \(X\) through \(S\).
The volume form of a
hypersurface gives \(dA = \iota_S^{*}(N \lrcorner\, \omega_g)\), so this flux integral is itself
the integral over \(S\) of the \(2\)-form \(N \lrcorner\, \omega_g\) weighted by \(\langle X, N \rangle_g\).
That weighting is the surface-integral analogue of the boundary integrand from the divergence theorem. The proof
of the divergence lemma used only that \(N\) is a unit normal along a hypersurface. The same decomposition
therefore applies with \(\partial M\) replaced by \(S\), giving
\(\iota_S^{*}(Y \lrcorner\, \omega_g) = \langle Y, N \rangle_g\, dA\) for every smooth vector field \(Y\).
The Classical Stokes Theorem
Everything is now in place to recover the surface form of Stokes's theorem. The flux of the curl through a
surface equals the circulation of the field around its boundary.
Theorem: The Classical Stokes Theorem
Let \(M\) be an oriented Riemannian \(3\)-manifold, and let \(S \subseteq M\) be a compact oriented
\(2\)-dimensional embedded submanifold with boundary, with unit normal \(N\) determining its
orientation. For any
smooth vector field \(X\) on \(M\),
\[
\int_S \langle \operatorname{curl} X, N \rangle_g\, dA = \int_{\partial S} \langle X, T \rangle_g\, ds,
\]
where \(T\) is the positively oriented unit tangent field along \(\partial S\) and \(ds\) its arc-length
form.
Proof.
Apply Stokes's theorem
to the \(1\)-form \(X^\flat\) on \(S\):
\[
\int_S d\bigl(X^\flat\bigr) = \int_{\partial S} X^\flat.
\]
The two integrands are exactly the two sides of the claimed identity.
On the left, the surface analogue of the divergence lemma gives
\(\iota_S^{*}\,d(X^\flat) = \langle \operatorname{curl} X, N \rangle_g\, dA\). The curl was defined by
\((\operatorname{curl} X) \lrcorner\, \omega_g = d(X^\flat)\), and restricting that \(2\)-form to \(S\)
extracts the normal component of \(\operatorname{curl} X\) against the area form, exactly as the divergence
lemma extracted \(\langle X, N \rangle_g\). On the right, restricting \(X^\flat\) to the curve
\(\partial S\) gives \(\langle X, T \rangle_g\, ds\). The \(1\)-form \(\iota_{\partial S}^{*}X^\flat\) on a
\(1\)-manifold equals \(f\, ds\) for some function \(f\), and evaluating on the unit tangent \(T\), where
\(ds(T) = 1\), gives \(f = X^\flat(T) = \langle X, T \rangle_g\). Substituting both translations yields the
stated equality.
This is the theorem of classical vector calculus, now seen as one more reading of
\(\int_M d\omega = \int_{\partial M}\omega\). Together with the divergence theorem and Green's theorem, it
completes the recovery of the integral theorems of three-dimensional analysis from a single statement about
forms. Each is that one equation, specialized to a degree and dressed in the language of the metric. What
changes from one to the next is only the dimension and the dictionary entry through which the exterior
derivative is read. That entry is the gradient, the curl, or the divergence.