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. The metric singles out, on an oriented manifold, a canonical
top-degree form — the
Riemannian volume
form \(\omega_g\) — whose 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 — written \(dA_g\) on a
surface and \(ds_g\) on a curve. 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, and on a compact manifold
without boundary the volume form is never exact — 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 setting in which the divergence theorem, taken up next, lives.
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, the right-hand side 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 \(X \lrcorner\, \omega_g\),
\[
\int_M (\operatorname{div} X)\, dV_g = \int_M d\bigl(X \lrcorner\, \omega_g\bigr)
= \int_{\partial M} \iota_{\partial M}^{*}\bigl(X \lrcorner\, \omega_g\bigr).
\]
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 \(\operatorname{div} X > 0\) on a dense subset — with the analogous
statements for the reverse inequalities.
Proof.
For a compact regular domain \(D\), the flow carries \(D\) diffeomorphically onto \(\theta_t(D)\), and 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,
\[
\frac{d}{dt}\bigg|_{t=t_0} \operatorname{Vol}(\theta_t(D))
= \int_D \theta_{t_0}^{*}\bigl(\mathcal{L}_X \omega_g\bigr)
= \int_{\theta_{t_0}(D)} (\operatorname{div} X)\, dV_g,
\]
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:
if \(\operatorname{div} X > 0\) on a dense set, then by continuity \(\operatorname{div} X \geq 0\)
everywhere, so the derivative is nonnegative, and the integral of \(\operatorname{div} X\) over the open
set \(\theta_{t_0}(\operatorname{Int} D)\) is strictly positive because the integrand is positive
somewhere on it; hence the volume is strictly increasing. The contrapositive arguments for the converses
run identically, using that a divergence which is negative somewhere makes the flow volume-decreasing on
a neighborhood.
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 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 \(n\). 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 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 same boundary computation used for the divergence theorem shows that
\(dA = \iota_S^{*}(N \lrcorner\, \omega_g)\), so this flux integral is itself the integral over \(S\) of the
\(2\)-form \(N \lrcorner\, \omega_g\) weighted by \(\langle X, N \rangle_g\) — the surface-integral analogue
of the boundary integrand from the divergence theorem.
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 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, by the two translations established
above and just below. 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 — gradient, curl, or divergence — through which the exterior
derivative is read.