Why Differential Forms Are the Right Objects to Integrate
We already know how to integrate a covector field along a curve: the line integral evaluates a \(1\)-form on the velocity
vector at each point of an oriented curve and accumulates the readings, depending only on the curve and the form, not on
any choice of coordinates. This chapter generalizes that construction to oriented \(k\)-dimensional submanifolds. The
question is which kind of field can play the role covector fields played for curves; the answer, motivated here, is the
differential form.
Functions Cannot Be Integrated Intrinsically
One might expect the object to be integrated to be a function, as in elementary calculus. But a function is not the right
object once coordinates are allowed to change, and that failure is what forces differential forms upon us. Consider the
simplest case. Let \(C \subseteq \mathbb{R}^n\) be a closed ball and let \(f : C \to \mathbb{R}\)
be the constant function \(f(x) \equiv 1\). Integrating it against the ordinary volume element gives the volume of the
ball,
\[
\int_C f\, dV = \operatorname{Vol}(C).
\]
Now apply a linear change of coordinates — say, a map that stretches one axis by a factor of \(2\). The set \(C\) and the
function \(f \equiv 1\) are unchanged as geometric data, yet the volume of the image is twice that of \(C\): the integral
has changed even though we did nothing to the function. The culprit is the determinant of the coordinate change, which
rescales \(dV\) but which the bare symbol \(f\) carries no record of. So \(\int_C f\, dV\) is not an invariant of the
function and the set alone; it depends on the coordinates in which the volume element was written, and for nonlinear
changes the rescaling factor even varies from point to point. A function carries no instruction for how to respond when
the measuring scale is deformed, so there is no coordinate-independent way to integrate it over a region of a manifold.
Why a Riemannian Metric Rescues Functions
The obstruction is the absence of a built-in volume scale, not anything pathological about functions. On an oriented
manifold a Riemannian metric supplies a canonical positively oriented \(n\)-form, the Riemannian volume form, against which
a function can be integrated unambiguously — the route taken in the Riemannian integration developed later in this chapter.
On a bare smooth manifold there is no preferred scale, so one must look instead for fields that carry their own.
A Covector Field Is a Signed Length Meter
The line integral succeeds where the function integral fails, and seeing why points at the general construction. A
covector field assigns to each tangent vector a number, linearly. At each point a covector thus reads off a tangent
vector and returns a signed length — signed because reversing the vector reverses the sign, a "meter" because it measures
every vector in a given direction on a common scale. The word "length" here is not metric length, which would require a
Riemannian metric; it is the reading on the scale the covector itself carries, a coordinate-independent proportionality
factor along each direction. A covector field is a field of such meters, one per tangent space, and the line integral
measures a curve by applying the meter at each point to the velocity and summing. The meter is intrinsic to the form and
consults no coordinate system, so its reading is coordinate-independent — exactly the property the function integral lacked.
Seeking a Signed Volume Meter in Higher Dimensions
To integrate over a \(k\)-dimensional submanifold with \(k \gt 1\), we need the analogous device one dimension higher: a
machine that at each point reads off a \(k\)-tuple of tangent vectors and returns a number we interpret as the signed
\(k\)-dimensional volume of the parallelepiped they span. Write \(\omega(v_1, \dots, v_k)\) for that signed volume. What
algebraic properties must \(\omega\) have to behave like a volume?
The prototype already exists: on \(\mathbb{R}^n\), the determinant returns, up to sign, the volume of the parallelepiped
spanned by \(n\) vectors — for \(n = 2\), the signed area of a parallelogram, and likewise in every dimension. It is no
accident that the determinant is the standard example of an
alternating tensor:
the properties we are about to demand of a general volume meter are exactly the ones the determinant already enjoys.
Volume Forces Multilinearity, and Then Alternation
Two demands pin down the algebraic form of \(\omega\). Scaling one edge by a constant should scale the volume by that
constant; and a parallelepiped whose \(i\)th edge is a sum \(v_i + v_i'\) should have volume equal to the sum of the
volumes built from \(v_i\) and from \(v_i'\) separately, the other edges fixed. In symbols, for each index \(i\),
\[
\begin{align*}
\omega(v_1, \dots, c\,v_i, \dots, v_k) &= c\,\omega(v_1, \dots, v_i, \dots, v_k),\\\\
\omega(v_1, \dots, v_i + v_i', \dots, v_k) &= \omega(v_1, \dots, v_i, \dots, v_k)\\\\
&\quad{} + \omega(v_1, \dots, v_i', \dots, v_k).
\end{align*}
\]
In every slot, these say that \(\omega\) is linear in each argument separately: it is a covariant \(k\)-tensor.
One further property is forced. A linearly dependent \(k\)-tuple spans a degenerate parallelepiped, flattened into fewer
than \(k\) dimensions and so of zero \(k\)-volume; a faithful volume meter must return zero on it. For a covariant
\(k\)-tensor this is no mild side condition: by the
characterization of alternating tensors,
a \(k\)-tensor vanishing on every linearly dependent \(k\)-tuple is necessarily alternating — its value reverses sign
when two arguments are interchanged. That sign reversal is welcome: it encodes orientation, the higher-dimensional
analogue of the sign distinguishing forward from backward traversal of a curve.
The conclusion is decisive: a signed volume meter must be, at each point, an alternating covariant tensor — and a smoothly
varying field of alternating \(k\)-tensors on a manifold is exactly a differential \(k\)-form. The rest of the chapter
makes this good: we define the integral of an \(n\)-form over a domain in Euclidean space, transport it to oriented
manifolds, and connect it through Stokes's theorem to the classical integral theorems of vector calculus.
Integrating an \(n\)-Form over a Domain in \(\mathbb{R}^n\)
The construction proceeds in stages, exactly as the line integral did: we first integrate over the most concrete domains
in Euclidean space, then extend to open sets, and only afterward transport everything to a manifold. The reason for this
order is that integration of a function over a region of \(\mathbb{R}^n\) is already understood from multivariable
calculus; our task here is only to attach that machinery to forms in the right way and check that it respects
coordinate changes. Throughout, we work with top-degree forms — \(n\)-forms on an open subset of \(\mathbb{R}^n\) — since
those are the forms that have a chance of being integrated over an \(n\)-dimensional region.
The regions we integrate over must be tame enough for multiple integrals to make sense. We need them bounded, so that
the integral is finite, and we need their boundaries to be negligible, so that it does not matter whether boundary
points are included.
Definition: Domain of Integration
A domain of integration in \(\mathbb{R}^n\) is a bounded subset \(D \subseteq \mathbb{R}^n\) whose
boundary \(\partial D\) is a
set of measure zero.
The measure-zero condition is what makes the integral insensitive to the exact specification of the region. A bounded
continuous function is Riemann integrable over such a set, and changing the function on \(\partial D\) — for instance, by
deciding whether the boundary belongs to \(D\) — leaves the integral unchanged, because a set of measure zero contributes
nothing. Rectangles, balls, and the bounded regions of elementary calculus are all domains of integration; their
boundaries are lower-dimensional and hence negligible.
Integration over a Domain: Erasing the Wedges
On a domain of integration the definition of the integral of a top form is as simple as possible. Any \(n\)-form on an open
set containing \(\overline{D}\) can be written in standard coordinates as \(\omega = f\, dx^1 \wedge \cdots \wedge dx^n\)
for a single continuous coefficient function \(f\), since the space of alternating \(n\)-tensors on \(\mathbb{R}^n\) is
one-dimensional. We integrate the form by integrating that coefficient.
Definition: Integral of an \(n\)-Form over a Domain
Let \(D \subseteq \mathbb{R}^n\) be a domain of integration and let \(\omega = f\, dx^1 \wedge \cdots \wedge dx^n\) be an
\(n\)-form with \(f\) continuous on \(\overline{D}\). The integral of \(\omega\) over \(D\) is
\[
\int_D \omega := \int_D f\, dV,
\]
the ordinary multiple integral of \(f\) over \(D\) against Lebesgue volume \(dV\).
In words: to integrate \(f\, dx^1 \wedge \cdots \wedge dx^n\), one erases the wedges and computes the multiple integral of
\(f\). The wedge symbols are not decorative — they record the ordering of the coordinates, which is what carries the sign
information once coordinates are changed.
Diffeomorphism Invariance: the Pullback Carries the Jacobian
The definition above is stated in one fixed coordinate system, so its naturality is not yet visible. The decisive fact,
the next proposition, is that pulling a form back through a coordinate change reproduces the change-of-variables factor
automatically — which is what makes the integral of a form coordinate-independent.
Proposition (Diffeomorphism Invariance on Domains)
Let \(D\) and \(E\) be open domains of integration in \(\mathbb{R}^n\), and let \(G : \overline{D} \to \overline{E}\) be a
smooth map that restricts to an orientation-preserving or orientation-reversing diffeomorphism from \(D\) to \(E\). If
\(\omega\) is an \(n\)-form on \(\overline{E}\), then
\[
\int_D G^*\omega =
\begin{cases}
\phantom{-}\displaystyle\int_E \omega & \text{if } G \text{ is orientation-preserving,}\\\\
-\displaystyle\int_E \omega & \text{if } G \text{ is orientation-reversing.}
\end{cases}
\]
Proof.
Write \((y^1, \dots, y^n)\) for the standard coordinates on \(E\) and \((x^1, \dots, x^n)\) for those on \(D\), and write
\(\omega = f\, dy^1 \wedge \cdots \wedge dy^n\). Suppose first that \(G\) is orientation-preserving, so that its Jacobian
determinant \(\det DG\) is positive throughout \(D\). The classical change-of-variables formula for multiple integrals
relates the integral over \(E\) to an integral over \(D\) weighted by the absolute value of the Jacobian:
\[
\begin{align*}
\int_E \omega
&= \int_E f\, dV\\\\
&= \int_D (f \circ G)\,\lvert \det DG \rvert\, dV.
\end{align*}
\]
Because \(G\) is orientation-preserving, \(\lvert \det DG \rvert = \det DG\), so the absolute value may be dropped. On the
other hand, the
pullback of a top-degree form
is computed by exactly the same Jacobian factor,
\[
G^*\omega = (f \circ G)\,(\det DG)\; dx^1 \wedge \cdots \wedge dx^n,
\]
and integrating this over \(D\) — again by erasing the wedges — gives \(\int_D (f \circ G)\,(\det DG)\, dV\). Comparing
the two displays, the right-hand sides agree, so
\[
\begin{align*}
\int_D G^*\omega
&= \int_D (f \circ G)\,(\det DG)\, dV\\\\
&= \int_E \omega.
\end{align*}
\]
If \(G\) is orientation-reversing, then \(\det DG \lt 0\), so removing the absolute value in the change-of-variables
formula introduces a sign: \(\lvert \det DG \rvert = -\det DG\). Every step above is unchanged except for this one factor,
and the result is \(\int_D G^*\omega = -\int_E \omega\).
Why the Jacobian Appears Twice — and Cancels
The proof turns on a coincidence that is no coincidence. The change-of-variables formula produces a factor
\(\lvert \det DG \rvert\) from the geometry of how volume distorts under \(G\); the pullback of a top form produces a
factor \(\det DG\) from the algebra of how the wedge \(dy^1 \wedge \cdots \wedge dy^n\) transforms. These two factors are
the same up to sign, and that is exactly why integrating a form is coordinate-independent while integrating a function is
not. The bare function notation drops the Jacobian; the wedge notation carries it. The only discrepancy is the sign,
which is precisely the data of orientation — the form integral remembers the direction in which coordinates are traversed,
whereas unsigned volume forgets it.
Here we have taken the change-of-variables formula for multiple integrals as known background from multivariable calculus,
rather than proving it; it is the foundation on which integration of forms is built, not a theorem of this development. What
the proposition adds is the observation that the language of forms encodes that formula automatically, sign included.
Extending to Compactly Supported Forms on Open Sets
Before integrating over a manifold we need one intermediate setting: a form on an open set, not just on a domain of
integration. The natural class is the compactly supported forms. The idea is to enclose the support inside a domain of
integration and integrate there, so we first record that such an enclosing domain always exists.
Lemma (Existence of an Enclosing Domain)
Let \(U \subseteq \mathbb{R}^n\) be open and let \(K \subseteq U\) be compact. Then there is an open domain of integration
\(D\) with \(K \subseteq D \subseteq \overline{D} \subseteq U\).
Proof.
Each point of \(K\) lies in an open ball whose closure is contained in \(U\); since \(K\) is compact, finitely many such
balls \(B_1, \dots, B_m\) cover it. Let \(D = B_1 \cup \cdots \cup B_m\). This set is open, bounded, and satisfies
\(K \subseteq D \subseteq \overline{D} \subseteq U\) by construction. It remains to check that \(\partial D\) has measure
zero. The boundary of \(D\) is contained in the union of the boundaries of the balls, and the boundary of each ball is a
sphere, a
lower-dimensional submanifold
of \(\mathbb{R}^n\) and therefore of measure zero. A finite union of measure-zero sets has measure zero, so \(\partial D\)
has measure zero and \(D\) is a domain of integration.
With this in hand, the integral of a compactly supported \(n\)-form \(\omega\) on an open set \(U \subseteq \mathbb{R}^n\) is
defined by choosing any domain of integration \(D\) containing the support and setting \(\int_U \omega := \int_D \omega\),
where \(\omega\) is taken to be zero outside its support. Two different enclosing domains agree on the support and differ only
where \(\omega\) vanishes, so the value does not depend on the choice of \(D\). This is the form in which the definition is
ready to be transported to a manifold, one chart at a time — the construction taken up in the next section.
Integration on Manifolds
We can now define the integral of a top-degree form over an oriented manifold. The work of the previous section pays off
here: integration in a single chart is just integration over an open subset of \(\mathbb{R}^n\), pulled back through the
coordinate map, and diffeomorphism invariance is exactly what guarantees that the answer does not depend on which chart we
used. We treat one chart first, then patch charts together with a partition of unity.
Integration in a Single Chart
Let \(M\) be an oriented smooth \(n\)-manifold, with or without boundary, and let \(\omega\) be an \(n\)-form on \(M\).
Suppose for the moment that \(\omega\) is compactly supported inside the domain of a single smooth chart \((U, \varphi)\).
The coordinate map \(\varphi\) carries \(U\) to an open subset \(\varphi(U) \subseteq \mathbb{R}^n\) (or \(\mathbb{H}^n\) if
\(U\) meets the boundary), and pushes the form there: the inverse map \(\varphi^{-1}\) pulls \(\omega\) back to a compactly
supported \(n\)-form \((\varphi^{-1})^*\omega\) on \(\varphi(U)\), which we already know how to integrate. We integrate
\(\omega\) by integrating its coordinate representative, attaching a sign according to whether the chart respects the
orientation.
Definition: Integral over a Single Chart
With \(M\), \(\omega\), and \((U, \varphi)\) as above, where the chart is either positively or negatively
oriented, the
integral of \(\omega\) over \(M\) is
\[
\int_M \omega := \pm \int_{\varphi(U)} \bigl(\varphi^{-1}\bigr)^*\omega,
\]
with the positive sign when \((U, \varphi)\) is positively oriented and the negative sign when it is negatively
oriented.
The sign is essential. A positively oriented chart presents \(M\) as its orientation prescribes, and the Euclidean integral
then reads off the correctly signed volume; a negatively oriented chart reverses that presentation, and the compensating
minus sign undoes the reversal. Without it the value would flip according to the handedness of the chart, and the integral
would not be well defined.
Independence of the Chart
For this definition to be legitimate, two charts containing the support of \(\omega\) must give the same value. This is
where the diffeomorphism invariance established earlier does its work.
Proposition (Independence of Chart)
With \(\omega\) compactly supported in the domain of a single chart, the value of \(\int_M \omega\) given by the
definition above does not depend on which such chart is used.
Proof.
Let \((U, \varphi)\) and \((\tilde U, \tilde\varphi)\) be two smooth charts whose domains both contain the support of
\(\omega\). Consider first the case where the two charts have the same orientation — both positively oriented or both
negatively oriented. Then the transition map \(\tilde\varphi \circ \varphi^{-1}\) is an orientation-preserving
diffeomorphism between open subsets of \(\mathbb{R}^n\) (or \(\mathbb{H}^n\)), since same-handedness charts have
positive transition Jacobian.
Both definitions carry the same sign, so it suffices to show that the two Euclidean integrals agree:
\[
\begin{align*}
\int_{\tilde\varphi(\tilde U)} \bigl(\tilde\varphi^{-1}\bigr)^*\omega
= \int_{\varphi(U)} \bigl(\varphi^{-1}\bigr)^*\omega.
\end{align*}
\]
Set \(G = \tilde\varphi \circ \varphi^{-1}\). A short pullback computation rewrites the left-hand integrand in terms of
the right one. Since \(\tilde\varphi^{-1} = \varphi^{-1} \circ G^{-1}\), functoriality of the pullback gives
\(\bigl(\tilde\varphi^{-1}\bigr)^* = (G^{-1})^* \circ \bigl(\varphi^{-1}\bigr)^*\) — equivalently, applying \(G^*\) to the
right-hand integrand returns the left-hand one. By the
diffeomorphism invariance on domains,
and because \(G\) is orientation-preserving,
\[
\begin{align*}
\int_{\varphi(U)} \bigl(\varphi^{-1}\bigr)^*\omega
&= \int_{\varphi(U)} G^*\Bigl(\bigl(\tilde\varphi^{-1}\bigr)^*\omega\Bigr)\\\\
&= \int_{\tilde\varphi(\tilde U)} \bigl(\tilde\varphi^{-1}\bigr)^*\omega,
\end{align*}
\]
which is the desired equality.
If instead the two charts have opposite orientations, the two definitions carry opposite signs. But now
\(G = \tilde\varphi \circ \varphi^{-1}\) is orientation-reversing, so the same invariance introduces a compensating
minus sign into the computation above. The two sign reversals cancel, and the two definitions again agree.
Integration over the Whole Manifold
A general compactly supported form need not live inside one chart, but it can always be cut into pieces that do. The tool
for this is a partition of unity: a family of functions, each supported in a chart, that sum to \(1\). Multiplying
\(\omega\) by each function localizes it to a chart, where the single-chart integral applies, and summing the pieces
reassembles the whole.
Definition: Integral over an Oriented Manifold
Let \(M\) be an oriented smooth \(n\)-manifold, with or without boundary, and let \(\omega\) be a compactly supported
\(n\)-form on \(M\). Choose a finite open cover \(\{U_i\}\) of \(\operatorname{supp}\omega\) by domains of positively or
negatively oriented smooth charts, and let \(\{\psi_i\}\) be a smooth
partition of unity
subordinate to this cover. The integral of \(\omega\) over \(M\) is
\[
\int_M \omega := \sum_i \int_M \psi_i\, \omega.
\]
Each summand makes sense: the form \(\psi_i\,\omega\) is supported inside the single chart \(U_i\), so its integral is the
single-chart integral already defined, and the finiteness of the cover keeps the sum finite. Two features of the setup
deserve comment. A partition of unity subordinate to the cover always
exists, so the
construction is never vacuous. And negatively oriented charts are deliberately allowed, because on some manifolds with
boundary — a closed interval, for instance — it is not always possible to cover the boundary with positively oriented
charts; permitting either handedness, with the sign convention built into the single-chart integral, sidesteps that
obstruction.
Independence of the Cover and Partition
The definition refers to a choice of cover and partition of unity, so once more we must check that the value is independent
of those choices.
Proposition (Independence of Cover and Partition)
The value of \(\int_M \omega\) defined above does not depend on the choice of open cover or subordinate partition of
unity.
Proof.
Let \(\{U_i\}\) with partition \(\{\psi_i\}\) and \(\{\tilde U_j\}\) with partition \(\{\tilde\psi_j\}\) be two such
choices. Because each \(\psi_i\,\omega\) is supported in the single chart \(U_i\), and \(\sum_j \tilde\psi_j = 1\) on
\(\operatorname{supp}\omega\), we may insert the second partition inside the integral of \(\psi_i\,\omega\) and split the
sum, each term \(\tilde\psi_j\,\psi_i\,\omega\) being supported in a single chart and hence unambiguously integrable:
\[
\begin{align*}
\int_M \psi_i\, \omega
&= \int_M \Bigl(\sum_j \tilde\psi_j\Bigr) \psi_i\, \omega\\\\
&= \sum_j \int_M \tilde\psi_j\, \psi_i\, \omega.
\end{align*}
\]
Summing over \(i\) gives
\[
\sum_i \int_M \psi_i\, \omega = \sum_{i,j} \int_M \tilde\psi_j\, \psi_i\, \omega.
\]
The right-hand side is symmetric in the two partitions: running the identical argument starting from
\(\sum_j \int_M \tilde\psi_j\, \omega\), and using \(\sum_i \psi_i = 1\), produces the same double sum. Therefore the two
choices yield the same value for \(\int_M \omega\).
Boundaries, Submanifolds, and the Zero-Dimensional Case
Three conventions extend the reach of the definition and will be used repeatedly. First, when \(S \subseteq M\) is an
oriented immersed \(k\)-dimensional submanifold and \(\omega\) is a \(k\)-form on \(M\) whose restriction to \(S\) is
compactly supported, we write \(\int_S \omega\) to mean \(\int_S \iota_S^*\omega\), where \(\iota_S : S \hookrightarrow M\)
is the inclusion and \(\iota_S^*\omega\) is the pullback of \(\omega\) to \(S\). Integration over a submanifold is thus
integration of the restricted form on the submanifold in its own right.
Second, in the most important instance of this, \(M\) is a compact oriented \(n\)-manifold with boundary and \(\omega\) is an
\((n-1)\)-form on \(M\). The boundary \(\partial M\) is an \((n-1)\)-dimensional manifold, and we always give it the induced
orientation. Then \(\int_{\partial M}\omega\) is unambiguous: it is the integral over \(\partial M\), with that orientation,
of the pullback of \(\omega\) under the boundary inclusion. This is the integral that will appear on the right-hand side of
Stokes's theorem.
Third, the zero-dimensional case is handled by a special convention, since a \(0\)-manifold has no coordinate charts in the
usual sense. An oriented \(0\)-manifold is a discrete set of points, each tagged with a sign \(+\) or \(-\) as its
orientation, and a compactly supported \(0\)-form is just a function \(f\) that is nonzero at finitely many of them. Its
integral is the signed sum
\[
\int_M f = \sum_{p \in M} \pm f(p),
\]
taking \(+f(p)\) at positively oriented points and \(-f(p)\) at negatively oriented ones. This is the degenerate case in
which integration reduces to addition, and it is exactly what makes the one-dimensional Stokes's theorem reproduce the
fundamental theorem of calculus, with the two endpoints of an interval serving as an oriented \(0\)-manifold.
Properties of the Integral
With the integral now defined on every oriented manifold, we record the basic properties it satisfies. Each follows
directly from the definition, but each will be used silently in later computations, so it is worth stating them precisely
and seeing why they hold. Throughout, \(M\) and \(N\) are nonempty oriented smooth \(n\)-manifolds, with or without boundary,
and \(\omega, \eta\) are compactly supported \(n\)-forms on \(M\).
Proposition (Properties of the Integral)
The integral of compactly supported top-degree forms has the following properties.
(a) Linearity.
For all \(a, b \in \mathbb{R}\),
\[
\int_M (a\,\omega + b\,\eta) = a \int_M \omega + b \int_M \eta.
\]
(b) Orientation reversal.
If \(-M\) denotes \(M\) with the opposite orientation, then
\[
\int_{-M} \omega = -\int_M \omega.
\]
(c) Positivity.
If \(\omega\) is everywhere non-negative with respect to the orientation — meaning
that in every positively oriented chart its coefficient function is \(\ge 0\) — and is strictly positive at some point,
then \(\int_M \omega \gt 0\).
(d) Diffeomorphism invariance.
If \(F : N \to M\) is an
orientation-preserving or orientation-reversing
diffeomorphism, then
\[
\int_N F^*\omega =
\begin{cases}
\phantom{-}\displaystyle\int_M \omega & \text{if } F \text{ is orientation-preserving,}\\\\
-\displaystyle\int_M \omega & \text{if } F \text{ is orientation-reversing.}
\end{cases}
\]
Proof.
Each property reduces to the single-chart case, where it follows from the corresponding statement for integrals over
domains in \(\mathbb{R}^n\); the partition of unity then carries it to all of \(M\) by linearity of the finite sum
defining \(\int_M\).
(a) In a single chart the integral is the Euclidean integral of the coefficient function, and
\((a\,\omega + b\,\eta)\) has coefficient \(a\,f + b\,g\) when \(\omega, \eta\) have coefficients \(f, g\); ordinary
integrals are linear, so the single-chart integral is linear in the form. Summing the partition-of-unity pieces, which
is again a linear operation, gives linearity on \(M\).
(b) Reversing the orientation of \(M\) turns every positively oriented chart into a negatively oriented
one and vice versa. In the single-chart definition this flips the sign attached to each Euclidean integral, so each term
in the partition-of-unity sum changes sign, and therefore so does the total.
(c) Suppose \(\omega\) is non-negative with respect to the orientation and strictly positive at some
point \(p\). In a positively oriented chart the coefficient of \((\varphi^{-1})^*\omega\) is \(\ge 0\), and the single-chart
integral takes the positive sign; in a negatively oriented chart the coefficient is \(\le 0\) and the compensating negative
sign again makes the contribution \(\ge 0\). For a subordinate partition of unity \(\{\psi_i\}\) with \(\psi_i \ge 0\), each
term \(\int_M \psi_i\,\omega\) is therefore nonnegative, so \(\int_M \omega \ge 0\). At the point \(p\) the coefficient is
strictly positive, and some \(\psi_i\) is positive near \(p\) since the \(\psi_i\) sum to \(1\); that term contributes a
strictly positive integral. Hence \(\int_M \omega \gt 0\).
(d) It suffices to treat the case where \(\omega\) is supported in a single positively oriented chart
\((U, \varphi)\) of \(M\), since the general case follows by applying this to each piece of a partition of unity and using
linearity. When \(F\) is orientation-preserving, \(\bigl(F^{-1}(U), \varphi \circ F\bigr)\) is a positively oriented chart
of \(N\) whose domain contains the support of \(F^*\omega\). Computing both integrals in these charts reduces the claim to
the
diffeomorphism invariance on domains
established earlier, which gives \(\int_N F^*\omega = \int_M \omega\). If \(F\) is orientation-reversing, the chart
\(\varphi \circ F\) is negatively oriented, and the extra sign in the single-chart definition — equivalently, property
(b) — produces \(\int_N F^*\omega = -\int_M \omega\).
The Integral Sees Only the Form and the Orientation
Taken together, these properties say that the integral is a linear functional on compactly supported top forms that
depends on the manifold through nothing but its smooth structure and its orientation. Property (d) is the precise sense
in which the integral is coordinate-free: any orientation-preserving diffeomorphism — a relabeling of points that respects
handedness — leaves it unchanged, and reversing handedness merely flips the sign. This is exactly the invariance that the
function integral failed to have, now secured for every oriented manifold.
Computing Integrals: Parametrizations and the Sphere
The partition-of-unity definition is ideal for proving theorems but useless for computing a specific integral: writing down
an explicit smooth partition of unity is rarely feasible, and integrating against one is harder still. In practice one
computes an integral by cutting the manifold into a few pieces, each the image of a single parametrization from a domain in
\(\mathbb{R}^n\), and integrating on each piece by pulling the form back to that domain. The next proposition justifies this,
and an example then carries it out.
Proposition (Integration over Parametrizations)
Let \(M\) be an oriented smooth \(n\)-manifold, with or without boundary, and let \(\omega\) be a compactly supported
\(n\)-form on \(M\). Suppose \(D_1, \dots, D_k\) are open domains of integration in \(\mathbb{R}^n\), and for each \(i\) we
are given a smooth map \(F_i : \overline{D_i} \to M\) such that
- \(F_i\) restricts to an orientation-preserving diffeomorphism from \(D_i\) onto an open subset \(W_i \subseteq M\);
- \(W_i \cap W_j = \varnothing\) when \(i \neq j\);
- \(\operatorname{supp}\omega \subseteq \overline{W_1} \cup \cdots \cup \overline{W_k}\).
Then
\[
\int_M \omega = \sum_{i=1}^{k} \int_{D_i} F_i^*\omega.
\]
Proof.
The images \(W_i\) are pairwise disjoint open sets whose closures cover the support of \(\omega\), and the leftover set
\(M \setminus (W_1 \cup \cdots \cup W_k)\) is contained in the union of the boundaries \(\partial W_i\). Each
\(\partial W_i\) is the image under \(F_i\) of \(\partial D_i\), a set of measure zero, and
smooth maps send sets of measure zero to sets of measure zero;
so the part of \(M\) not covered by the open pieces is itself negligible and contributes
nothing to the integral. Restricted to each \(W_i\), the integral of \(\omega\) is the integral over a set parametrized by
the single orientation-preserving diffeomorphism \(F_i : D_i \to W_i\). By the
diffeomorphism invariance on domains,
pulling back through \(F_i\) converts the integral over \(W_i\) into one over \(D_i\),
\[
\int_{W_i} \omega = \int_{D_i} F_i^*\omega,
\]
with no sign because \(F_i\) is orientation-preserving. Summing over the pieces, and using that the overlaps and the
uncovered remainder have measure zero, gives \(\int_M \omega = \sum_i \int_{D_i} F_i^*\omega\).
The proposition says exactly what one would hope: to integrate over a region covered by a parametrization \(F\), pull the
form back to the parameter domain and integrate there. The pullback \(F^*\omega\) is again a top form on a domain in
\(\mathbb{R}^n\), so its integral is computed by erasing the wedges. The only bookkeeping is to check that the
parametrization preserves orientation; if it reverses orientation, a single minus sign is attached.
Worked Example: A Two-Form on the Sphere
We compute the integral over the unit sphere \(S^2\) of a particular \(2\)-form, giving \(S^2\) the orientation it inherits
as the
boundary
of the closed unit ball \(\overline{\mathbb{B}}^3\) — the outward-pointing orientation. Let
\[
\omega = x\, dy \wedge dz + y\, dz \wedge dx + z\, dx \wedge dy
\]
on \(\mathbb{R}^3\), and integrate its restriction to \(S^2\).
Parametrize the sphere, away from a set of measure zero, by spherical coordinates. Let
\(D = (0, \pi) \times (0, 2\pi)\) and define \(F : \overline{D} \to S^2\) by
\[
F(\varphi, \theta) = (\sin\varphi \cos\theta,\ \sin\varphi \sin\theta,\ \cos\varphi).
\]
The map \(F\) restricted to \(D\) is an orientation-preserving diffeomorphism onto the sphere minus a half-meridian (a
measure-zero arc), so it meets the hypotheses of the previous proposition with a single parametrization. To pull \(\omega\)
back we first pull back the coordinate differentials. Differentiating each component,
\[
\begin{align*}
F^*dx &= \cos\varphi \cos\theta\, d\varphi - \sin\varphi \sin\theta\, d\theta,\\\\
F^*dy &= \cos\varphi \sin\theta\, d\varphi + \sin\varphi \cos\theta\, d\theta,\\\\
F^*dz &= -\sin\varphi\, d\varphi.
\end{align*}
\]
Wedging these in pairs and substituting the coordinate functions \(x = \sin\varphi\cos\theta\), and so on, every term
combines into a single multiple of \(d\varphi \wedge d\theta\). Carrying out the algebra, the three contributions add to
\[
F^*\omega = \sin\varphi\, d\varphi \wedge d\theta.
\]
The coefficient is the familiar area element of the sphere in spherical coordinates, which is a reassuring sign that the
form \(\omega\) is measuring surface area correctly.
The integral is now an ordinary double integral, obtained by erasing the wedge:
\[
\begin{align*}
\int_{S^2} \omega
&= \int_D \sin\varphi\, d\varphi \wedge d\theta\\\\
&= \int_0^{2\pi} \int_0^{\pi} \sin\varphi\, d\varphi\, d\theta\\\\
&= \int_0^{2\pi} 2\, d\theta = 4\pi.
\end{align*}
\]
The value \(4\pi\) is the surface area of the unit sphere, which is what this particular \(\omega\) was built to compute: it
is the restriction to \(S^2\) of the form whose integral over any closed surface returns the flux of the radial field
through it, and for the unit sphere that flux equals the area.
Why This Is the Practical Method
Notice what the computation did not require: no partition of unity, no atlas, no abstract machinery — only one
parametrization, three derivatives, a wedge product, and a double integral from elementary calculus. This is the pattern
for essentially every concrete integral over a manifold. The theoretical definition guarantees that the answer is
well defined and orientation-invariant; the parametrization method is how one actually produces the number. The single
check that cannot be skipped is orientation: had we parametrized \(S^2\) with \(\varphi\) and \(\theta\) interchanged, the
map would reverse orientation and the same arithmetic would yield \(-4\pi\), the correct integral with respect to the
opposite orientation.
Integration on Lie Groups: the Haar Volume Form
There is one class of manifolds on which a canonical integral exists, requiring no metric and no arbitrary choices: a Lie
group. The symmetry of the group singles out a preferred top-degree form, unique once we ask that the total volume be one,
and integrating against it is the construction on which harmonic analysis on groups — and, further along the curriculum, the
representation theory behind equivariant neural networks — is built. The key is to look for forms that are unchanged when
the group acts on itself by translation.
Recall that left translation by a group element \(g\) is the diffeomorphism \(L_g : G \to G\) given by \(L_g(h) = gh\). A
covariant tensor field \(A\) on \(G\) — in particular, a differential form — is called left-invariant if it
is unchanged by pulling back along every left translation:
\[
L_g^* A = A \qquad \text{for all } g \in G.
\]
A left-invariant form is completely determined by its value at the identity, since left translation carries that value to
every other point; this is what makes left-invariant forms rigid enough to be unique.
Proof.
If \(\dim G = 0\) the group is a finite set of points, and we take \(\omega_G\) to be the constant function \(1/k\), where
\(k\) is the number of points; its integral, the signed sum over the points, is \(1\). Assume now \(\dim G = n \ge 1\). A
Lie group is
parallelizable by left translation:
a basis of the tangent space at the identity extends to a
left-invariant global frame
\(E_1, \dots, E_n\), the basis of the Lie algebra. Replacing \(E_1\) by \(-E_1\) if necessary, we may take this frame to be
positively oriented.
Let \((\varepsilon^1, \dots, \varepsilon^n)\) be the dual
coframe,
characterized by \(\varepsilon^i(E_j) = \delta^i_j\). Each \(\varepsilon^i\) is itself left-invariant. To see this, recall
that the frame vectors are left-invariant vector fields, \((L_g)_* E_j = E_j\), and evaluate the pulled-back covector on a
frame vector at an arbitrary point \(h\), writing \(d(L_g)_h\) for the differential of \(L_g\) at \(h\):
\[
\begin{align*}
\bigl(L_g^* \varepsilon^i\bigr)_h\bigl(E_j|_h\bigr)
&= \varepsilon^i|_{gh}\bigl(d(L_g)_h\, E_j|_h\bigr)\\\\
&= \varepsilon^i|_{gh}\bigl(E_j|_{gh}\bigr) = \delta^i_j.
\end{align*}
\]
Since \(L_g^*\varepsilon^i\) and \(\varepsilon^i\) take the same values on the frame that spans the cotangent space at every
point, they are equal; hence \(\varepsilon^i\) is left-invariant.
Now set \(\omega_G = \varepsilon^1 \wedge \cdots \wedge \varepsilon^n\). Pullback commutes with the wedge product, so
\[
\begin{align*}
L_g^* \omega_G
&= \bigl(L_g^*\varepsilon^1\bigr) \wedge \cdots \wedge \bigl(L_g^*\varepsilon^n\bigr)\\\\
&= \varepsilon^1 \wedge \cdots \wedge \varepsilon^n = \omega_G,
\end{align*}
\]
so \(\omega_G\) is left-invariant. It is nowhere zero — indeed \(\omega_G(E_1, \dots, E_n) = 1 \gt 0\) — so it is an
orientation form, positively oriented for the given orientation. This much used nothing about compactness: every Lie group
carries a left-invariant orientation form, unique up to a positive constant multiple. (Two left-invariant orientation forms
agree at the identity up to a positive scalar, hence everywhere by left-invariance.)
Compactness enters only now. Because \(G\) is compact and \(\omega_G\) is a positively oriented orientation form, its
integral \(\int_G \omega_G\) is a finite positive number. Rescaling by its reciprocal, the form
\[
\tilde\omega_G = \left( \int_G \omega_G \right)^{-1} \omega_G
\]
is again left-invariant, positively oriented, and now satisfies \(\int_G \tilde\omega_G = 1\). Any left-invariant
positively oriented form is a positive multiple of \(\omega_G\), and the normalization fixes that multiple, so
\(\tilde\omega_G\) is the unique such form of total volume \(1\).
Definition: Haar Volume Form and Haar Integral
The unique normalized form of the preceding proposition is the Haar volume form on the compact Lie group
\(G\), written \(\omega_G\). The associated linear functional on continuous functions,
\[
f \longmapsto \int_G f\, \omega_G,
\]
is the Haar integral. Because \(\omega_G\) is left-invariant, the Haar integral is unchanged under
translating the integrand: \(\int_G (f \circ L_g)\, \omega_G = \int_G f\, \omega_G\) for every \(g \in G\).
The translation-invariance of the Haar integral is the property that makes it indispensable. It allows one to average any
function — or any object built from a representation of the group — over the whole group in a way that respects the group
structure, and a normalized average of total mass \(1\) behaves like a probability measure on \(G\). This is the device that
will let us decompose functions on a compact group into pieces adapted to its symmetries, the group-theoretic analogue of
expanding a periodic function into its Fourier components.
Left-Invariance, Two-Sided Invariance, and What Compactness Buys
Two features of the construction are worth separating. The existence of a left-invariant orientation form, unique up to
scale, holds for every Lie group; compactness is used only to make the total volume finite, and hence to normalize the
scale to \(1\). On a noncompact group the form still exists but cannot be normalized this way, since its integral may be
infinite. Second, the form built here is left-invariant by construction, but not obviously invariant under right
translation. For compact groups it turns out that the Haar volume form is in fact two-sided invariant — a property called
unimodularity — and that two-sided invariance is what the representation theory of compact groups ultimately requires. That
refinement is established where it is needed, in the representation-theoretic development later in the curriculum; here we
record only the left-invariant normalized form and its integral, which is the foundation on which the rest is built.