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.
Integrating its coordinate expression in a different chart introduces, through the change-of-variables formula,
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 proves the positivity of
the resulting integral.
The Definition
Let \((M, g)\) be an oriented Riemannian \(n\)-manifold with or without boundary, and let \(\omega_g\) denote
its Riemannian volume form. When \(n = 0\), we take \(\omega_g\) to be the orientation function, equal to \(+1\)
at positively oriented points and \(-1\) at negatively oriented points. For a compactly supported continuous
function \(f\), the product \(f\omega_g\) is a compactly supported continuous \(n\)-form, which the
integral of forms already covers, since that construction uses
only the continuity of the coefficients. 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 nonempty compact manifold
without boundary, of dimension \(n \geq 1\), the volume form is never exact. Its integral \(\operatorname{Vol}(M)\)
is positive by part (c) of the
properties of the integral,
since \(\omega_g\) is a positively oriented orientation form, 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
top-degree form, by contrast, can be called nonnegative only relative to a choice of orientation. It is the
metric, through the positivity of \(\sqrt{\det g_{ij}}\), that makes this positivity hold for functions.
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.
If \(f \equiv 0\), then \(f\omega_g = 0\) and the integral is \(0\). Otherwise \(f(p) \gt 0\) at some point
\(p\), and we show that the integral is strictly positive, which proves both assertions at once. The form
\(f\omega_g\) is everywhere non-negative with respect to the orientation, in the sense of part (c) of the
properties of the integral.
Indeed, for \(n \geq 1\), at each point \(q\) the
volume form
takes the value \(1\) on an oriented orthonormal basis of \(T_qM\), and any positively oriented basis differs
from that one by a matrix of positive determinant, so \(\omega_g\) is positive on every positively oriented
basis, while \(f(q) \geq 0\). For \(n = 0\) the volume form is the orientation sign, and the same conclusion
holds in the zero-dimensional reading of part (c). At \(p\) the form \(f\omega_g\) is strictly positive, so
part (c) gives \(\int_M f\, dV_g \gt 0\).
For \(n \geq 1\), in coordinates this is the positivity of \(\sqrt{\det g_{ij}}\). In a positively oriented
chart the
volume form is
\(\sqrt{\det g_{ij}}\,dx^1 \wedge \cdots \wedge dx^n\), so a function supported in the domain of
such a chart \((U, \varphi)\) integrates to \(\int_{\varphi(U)} f\,\sqrt{\det g_{ij}}\, dx^1 \cdots dx^n\), the integral
of a nonnegative function, the factor \(\sqrt{\det g_{ij}}\) being strictly positive because \(g\) is positive definite.
In a negatively oriented chart the coordinate expression acquires a minus sign, which the definition of the integral
over such a chart cancels.
Two consequences are routine but constantly used. Applied to the nonnegative functions \(|f| - f\) and
\(|f| + f\), positivity and linearity give \(-\int_M |f|\, dV_g \leq \int_M f\, dV_g \leq \int_M |f|\, dV_g\), the
integral triangle inequality \(\left| \int_M f\, dV_g \right| \leq \int_M |f|\, dV_g\). Applied to a difference
\(h - f\) with \(f \leq h\), positivity and linearity show 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, is a
statement about this integral.
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 for compactly supported vector fields on any positive-dimensional oriented Riemannian manifold
with boundary, 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 \(n\)-manifold, \(n \geq 1\), 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.
The resulting function depends neither on coordinates nor on the choice of orientation. Reversing the
orientation flips the sign of \(\omega_g\) on both sides, leaving \(\operatorname{div} X\) unchanged. On a
nonorientable manifold one applies the definition on each orientable open subset, such as a chart domain, with
any orientation of it. Since \(d\) and interior multiplication are local, and the sign argument applies on each
component of an overlap, the resulting functions agree on overlaps. So the divergence is well defined on any
Riemannian manifold, orientable or not. When \(M\) has no boundary, so that the flow of \(X\)
and the Lie derivative along it are available, 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 vectors \(V_1, \dots, V_{n-1}\) tangent to \(\partial M\) at a point, 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 a positive-dimensional 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 divergence theorem
of vector calculus, often called the Gauss-Ostrogradsky theorem.
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. We first say what the
volume behavior of a flow means.
Definition: Volume-Preserving and Volume-Increasing Flows
Let \((M, g)\) be an oriented Riemannian \(n\)-manifold without boundary, \(n \geq 1\), and let \(\theta\) be the
flow of a smooth vector field on \(M\), with \(M_t\) the domain of \(\theta_t\). For a compact
regular domain
\(D \subseteq M\), let \(J_D\) be the set of \(t\) with \(D \subseteq M_t\). The flow is
volume-preserving if \(\operatorname{Vol}(\theta_t(D)) = \operatorname{Vol}(D)\) for every
compact regular domain \(D\) and every \(t \in J_D\). It is volume-nondecreasing, respectively
volume-nonincreasing, if for every compact regular domain \(D\) the function
\(t \mapsto \operatorname{Vol}(\theta_t(D))\) is nondecreasing, respectively nonincreasing, on \(J_D\). It is
strictly volume-increasing, respectively strictly volume-decreasing, if this
function is strictly increasing, respectively strictly decreasing, on \(J_D\) for every nonempty compact
regular domain \(D\).
Two remarks make the definition meaningful. First, \(J_D\) is an open interval containing \(0\). It is an
interval because the maximal integral curve through each point is defined on an open interval containing \(0\),
so if \(D \subseteq M_t\) then \(D \subseteq M_s\) for every \(s\) between \(0\) and \(t\). It is open because for
\(t_0 \in J_D\) the compact set \(\{t_0\} \times D\) lies in the open flow domain, so some product
\((t_0 - \varepsilon, t_0 + \varepsilon) \times D\) does too. Second, for \(t \in J_D\) the map \(\theta_t\) is a
diffeomorphism from the open set \(M_t\) onto the open set \(M_{-t}\), so it carries \(D\) onto a compact
embedded codimension-\(0\) submanifold with boundary \(\theta_t(D)\). Being compact, this is a closed subset of
\(M\), hence properly embedded, that is, a compact regular domain. Its volume \(\int_{\theta_t(D)} \omega_g\), taken
with the orientation and volume form restricted from \(M\), is defined. The empty set is excluded from the strict notions
only because its volume is identically \(0\).
Proposition: Geometric Interpretation of the Divergence
Let \((M, g)\) be an oriented Riemannian \(n\)-manifold without boundary, \(n \geq 1\), let \(X\) be a
smooth vector field, and let \(\theta\) be its flow. Then, in the sense of the
definition above, 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.
For \(t\) in the open interval \(J_D\) of the definition, \(\theta_t\) is defined on all of \(D\), and the
flow carries \(D\)
diffeomorphically onto \(\theta_t(D)\). The change-of-variables formula for the volume form, part (d) of the
properties of the integral,
lets the volume of the image be written as an integral over \(D\) of the pulled-back form
\(\theta_t^{*}\omega_g\) (the flow maps \(\theta_t\) are orientation-preserving, because on \(D\) we have
\(\theta_t^{*}\omega_g = f_t\,\omega_g\) for a nowhere-vanishing function \(f_t\) that depends
continuously on \(t\) and equals \(1\) at \(t = 0\), so \(f_t \gt 0\) on the interval \(J_D\)). We now differentiate under the
integral sign at \(t = t_0\). This is legitimate: computing \(\int_D \theta_t^{*}\omega_g\) with a fixed
finite family of charts of \(D\) and a subordinate partition of unity, each term is the integral over a
fixed compact set of a coefficient that, together with its \(t\)-derivative, is continuous in \((t, x)\) for
\(t\) near \(t_0\), because the flow is smooth in \((t, p)\). For such integrands the derivative of the
integral is the integral of the derivative, a standard fact of multivariable calculus that we take as
background. The \(t\)-derivative of a pullback along a flow is the pullback of the Lie derivative,
\(\frac{d}{dt}\big|_{t = t_0} \theta_t^{*}\omega_g = \theta_{t_0}^{*}\bigl(\mathcal{L}_X \omega_g\bigr)\), the
identity displayed in the proof of the
characterization of invariance by a vanishing Lie derivative.
Together these give
\[
\frac{d}{dt}\bigg|_{t=t_0} \operatorname{Vol}(\theta_t(D))
= \int_D \theta_{t_0}^{*}\bigl(\mathcal{L}_X \omega_g\bigr)
= \int_{\theta_{t_0}(D)} (\operatorname{div} X)\, dV_g,
\]
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 for nonempty \(D\) the set \(\operatorname{Int} D\) is an open subset of \(M\), nonempty because a nonempty
manifold with boundary has interior points, and its image
under the diffeomorphism \(\theta_{t_0}\) is open as well. By density, \(\operatorname{div} X\) is positive
at some point of \(\theta_{t_0}(\operatorname{Int} D)\). The
positivity of the integral,
applied on the compact manifold with boundary \(\theta_{t_0}(D)\), therefore makes
\(\int_{\theta_{t_0}(D)} (\operatorname{div} X)\, dV_g\) strictly positive. Hence the derivative is positive
at every \(t_0 \in J_D\), and 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
\(D\) with nonempty interior. The derivative of its volume at \(t = 0\) is
\(\int_D (\operatorname{div} X)\, dV_g\), which is nonzero with the sign of \(\operatorname{div} X\) there, by
the positivity of the integral. So the volume of \(D\) changes, and the flow is not volume-preserving. In the
negative case \(\operatorname{Vol}(\theta_t(D)) \lt \operatorname{Vol}(D)\) for small \(t \gt 0\), so the flow is
not volume-nondecreasing either. If
instead \(\operatorname{div} X\) fails to be positive on a dense set, some nonempty open set \(W\) carries
\(\operatorname{div} X \leq 0\). Choose a compact regular domain \(D \subseteq W\) with nonempty interior.
The set of \((t, p)\) in the flow domain with \(\theta(t, p) \in W\) is open and contains \(\{0\} \times D\),
so by compactness of \(D\) there is \(\delta \gt 0\) with \((-\delta, \delta) \times D\) inside that set, which
places \((-\delta, \delta)\) inside \(J_D\) and gives \(\theta_t(D) \subseteq W\) for \(|t| \lt \delta\).
For such \(t\) the derivative \(\int_{\theta_t(D)} (\operatorname{div} X)\, dV_g\) is \(\leq 0\), so the
volume of \(D\) is nonincreasing on \((-\delta, \delta)\), and the flow is not strictly volume-increasing.
The derivative at \(t = 0\) alone would not suffice here, since a strictly increasing function can have zero
derivative at a point.
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 the
dimension count below shows why.
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. Pointwise, on an \(n\)-manifold the alternating
\(2\)-tensors on \(T_pM\) form a space of dimension \(\binom{n}{2}\), while \(T_pM\) has dimension \(n\). For
\(n \geq 1\) these dimensions agree, \(\binom{n}{2} = n\), 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, with or without boundary. 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. At each point the contraction is a linear map
from \(T_pM\) to the alternating \(2\)-tensors on \(T_pM\). This map is injective. If \(v \in T_pM\) is
nonzero, extend it to a basis \((v, v_2, v_3)\). Then
\((v \lrcorner\, \omega_g)(v_2, v_3) = \omega_g(v, v_2, v_3) \neq 0\). Both spaces have dimension three, so
the contraction is an isomorphism on each tangent space.
The pointwise inverse is smooth. To see this, we take a local oriented orthonormal frame \((E_i)\), obtained
by Gram-Schmidt from a positively oriented local frame, with
dual coframe
\((\varepsilon^i)\). In this frame \(\omega_g = \varepsilon^1 \wedge \varepsilon^2 \wedge \varepsilon^3\),
because both sides are \(3\)-forms that take the value \(1\) on \((E_1, E_2, E_3)\). The contraction therefore
sends \(E_1, E_2, E_3\) to \(\varepsilon^2 \wedge \varepsilon^3\), \(\varepsilon^3 \wedge \varepsilon^1\),
\(\varepsilon^1 \wedge \varepsilon^2\). A smooth \(2\)-form has smooth coefficients in this basis and is
therefore the image of a smooth vector field.
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 \(M\), with or without boundary, 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
restriction lemma used only that
\(N\) is a unit normal along a hypersurface, together with the boundary volume form identity. The same
decomposition therefore applies with \(\partial M\) replaced by \(S\) and that identity replaced by the
hypersurface identity just quoted, 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, with or without boundary, 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 \(\partial S\) carries the Stokes orientation induced from \(S\), \(T\) is the unit tangent field along
\(\partial S\) that is positively oriented for it, and \(ds\) is the Riemannian volume form of \(\partial S\)
for the induced metric and that orientation, its arc-length form.
Proof.
Apply Stokes's theorem to
the \(1\)-form \(X^\flat\) on \(S\):
\[
\int_S d\bigl(X^\flat\bigr) = \int_{\partial S} X^\flat.
\]
Here both sides are integrals of pullbacks, to \(S\) and to \(\partial S\). Stokes's theorem is applied to
\(\iota_S^{*}X^\flat\), and by
naturality of the exterior derivative,
\(d(\iota_S^{*}X^\flat) = \iota_S^{*}\,d(X^\flat)\). The two integrands are exactly the two sides of the
claimed identity.
On the left, the surface version of the restriction 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
restriction 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
fundamental theorem for line integrals and the divergence
theorem, it completes the recovery of the integral theorems of three-dimensional vector calculus 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.