Why Differential Forms Are the Right Objects to Integrate
Our model is the integral of a smooth
covector field
along a curve. Smooth covector fields are also called differential \(1\)-forms. Let \(\omega\) be a \(1\)-form on a
smooth manifold \(M\), and let \(\gamma : [a, b] \to M\) be a smooth curve. The line integral
\(\int_a^b \omega_{\gamma(t)}\bigl(\gamma'(t)\bigr)\, dt\) evaluates \(\omega\) on the velocity vector at each
point of the curve and accumulates the readings. Its value depends only on the curve and the form, not on any
choice of coordinates.
This page 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 \(k\)-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. That is the route taken when
integration on Riemannian manifolds is developed in the pages ahead. 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. The value is signed because reversing the vector reverses the sign, and the covector is 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. That is 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 that number as the
signed \(k\)-dimensional volume of the parallelepiped they span, and we write \(\omega(v_1, \dots, v_k)\) for it. 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 value is the signed area of a parallelogram, and the pattern continues 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. 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. In other words, \(\omega\) 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, since it encodes orientation, the higher-dimensional analogue of
the sign distinguishing forward from backward traversal of a curve.
A signed volume meter must be, at each point, an alternating covariant tensor. Fields of alternating \(k\)-tensors
that vary smoothly over a manifold are exactly the differential \(k\)-forms. The rest of this page makes this good.
We define the integral of an \(n\)-form over a domain in Euclidean space and transport it to oriented manifolds.
Stokes's theorem then connects that integral to the classical integral theorems
of vector calculus.
Integrating an \(n\)-Form over a Domain in \(\mathbb{R}^n\)
The construction proceeds in stages. 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, meaning \(n\)-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 key fact, the next
proposition, is that pulling a form back through a coordinate change reproduces the change-of-variables factor
automatically. That 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
\(\omega = f\, dy^1 \wedge \cdots \wedge dy^n\) is an \(n\)-form on \(E\), written in the standard coordinates
\((y^1, \dots, y^n)\) of \(\mathbb{R}^n\), with \(f\) continuous 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
and measure theory, rather than proving it. It is the foundation on which integration of forms is built, not a theorem of
this development, and we record once the precise form in which it will be used. If \(G : A \to C\) is a diffeomorphism
between open subsets of \(\mathbb{R}^n\) and \(f\) is Lebesgue integrable over \(C\), then
\((f \circ G)\,\lvert \det DG \rvert\) is Lebesgue integrable over \(A\) and
\[
\int_C f\, dV = \int_A (f \circ G)\,\lvert \det DG \rvert\, dV.
\]
The proof above used the special case of open domains of integration and coefficients continuous on their closures, where
the Riemann and Lebesgue integrals of a bounded continuous function coincide. 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, we define the integral of a compactly supported \(n\)-form \(\omega\) on an open set
\(U \subseteq \mathbb{R}^n\). Choose any domain of integration \(D\) containing the support and set
\(\int_U \omega := \int_D \omega\), where \(\omega\) is taken to be zero outside its support. The coefficient extended by
zero is continuous on all of \(\mathbb{R}^n\), because the support is a compact subset of the open set \(U\). Two different
enclosing domains both contain the support and differ only where \(\omega\) vanishes, so the value does not depend on the
choice of \(D\). The same convention serves an open subset \(U \subseteq \mathbb{H}^n\), which is what a boundary chart
produces. The coefficient function of a compactly supported \(n\)-form on such a \(U\), extended by zero, is a bounded
function with compact support that is continuous off the measure-zero set \(\partial\mathbb{H}^n\), so its integral against
Lebesgue volume over any domain of integration containing the support is defined, and it equals the integral over the open
subset \(U \cap \operatorname{Int}\mathbb{H}^n\) of \(\mathbb{R}^n\).
Both conventions are instances of a single one. If \(C \subseteq \mathbb{R}^n\) is open and
\(\omega = f\, dy^1 \wedge \cdots \wedge dy^n\) is an \(n\)-form on \(C\) whose coefficient \(f\) is Lebesgue integrable
over \(C\), we write \(\int_C \omega := \int_C f\, dV\). For a compactly supported form on an open subset of
\(\mathbb{R}^n\) this agrees with the definition just given, since the coefficient vanishes off the support, and the
integral attached to an open subset of \(\mathbb{H}^n\) is by construction of this kind. Allowing supports that are not
compact costs only the hypothesis of integrability, and in return diffeomorphism invariance holds between arbitrary open
sets.
Proof.
Write \(\omega = f\, dy^1 \wedge \cdots \wedge dy^n\). By the
pullback of a top-degree form,
\(G^*\omega = (f \circ G)\,(\det DG)\, dx^1 \wedge \cdots \wedge dx^n\). The hypothesis on \(G\) says that
\(\det DG\) is positive at every point of \(A\) or negative at every point of \(A\), so
\((f \circ G)\det DG = \pm (f \circ G)\lvert \det DG \rvert\) with a single sign throughout \(A\), even when \(A\) is
disconnected. By the change-of-variables formula recorded above, this function is integrable over \(A\), and its
integral is \(\pm \int_C f\, dV = \pm \int_C \omega\).
This is the form in which the definition is ready to be transported to a manifold, one chart at a time. That
construction is 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 \(n \ge 1\), with or without boundary, and let \(\omega\) be an
\(n\)-form on \(M\). (For a \(0\)-manifold, whose orientation is a choice of sign at each point, the integral is fixed by
a separate convention below.) 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 \((U, \varphi)\) is a boundary chart), 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 and attach a sign according to whether the chart respects the orientation.
Precisely, the chart is positively oriented if its coordinate frame \((\partial_1, \dots, \partial_n)\)
is a positively oriented frame
on \(U\), and negatively oriented if that frame is negatively oriented.
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
\(\operatorname{supp}\omega\), and set \(U_0 = U \cap \tilde U\). Write \(f\) and \(\tilde f\) for the coefficients
of \(\bigl(\varphi^{-1}\bigr)^*\omega\) on \(\varphi(U)\) and of \(\bigl(\tilde\varphi^{-1}\bigr)^*\omega\) on
\(\tilde\varphi(\tilde U)\). Each is continuous and vanishes outside the compact set \(\varphi(\operatorname{supp}\omega)\),
respectively \(\tilde\varphi(\operatorname{supp}\omega)\), so each is bounded and Lebesgue integrable.
We first shrink both Euclidean integrals to a common region. Set
\[
\begin{align*}
V_0 &= \varphi(U_0 \cap \operatorname{Int} M),\\\\
\tilde V_0 &= \tilde\varphi(U_0 \cap \operatorname{Int} M).
\end{align*}
\]
If \((U, \varphi)\) is an interior chart, every point of \(U\) is an interior point of \(M\), so \(V_0 = \varphi(U_0)\).
If it is a boundary chart, the points of \(U\) carried into \(\partial\mathbb{H}^n\) are boundary points, which by
invariance of the boundary
are not interior points, while those carried into \(\operatorname{Int}\mathbb{H}^n\) are interior points. Hence
\(V_0 = \varphi(U_0) \cap \operatorname{Int}\mathbb{H}^n\). In either case \(V_0\) is open in \(\mathbb{R}^n\), and it
differs from \(\varphi(U_0)\) at most by a subset of the measure-zero hyperplane \(\partial\mathbb{H}^n\). Since \(f\)
vanishes on \(\varphi(U) \setminus \varphi(U_0)\), the convention of the previous section gives
\[
\int_{\varphi(U)} \bigl(\varphi^{-1}\bigr)^*\omega = \int_{V_0} \bigl(\varphi^{-1}\bigr)^*\omega,
\]
and likewise with tildes. The transition map \(G = \tilde\varphi \circ \varphi^{-1}\) is a diffeomorphism from
\(\varphi(U_0)\) onto \(\tilde\varphi(U_0)\), and it carries \(V_0\) onto \(\tilde V_0\), because both are images of the
same set \(U_0 \cap \operatorname{Int} M\). Since \(\tilde\varphi^{-1} = \varphi^{-1} \circ G^{-1}\), functoriality of the
pullback gives \(G^*\bigl(\bigl(\tilde\varphi^{-1}\bigr)^*\omega\bigr) = \bigl(\varphi^{-1}\bigr)^*\omega\) on
\(\varphi(U_0)\).
Consider first the case where the two charts have the same orientation, both positively oriented or both negatively
oriented. Then the two definitions carry the same sign, and \(G\) has
positive transition Jacobian
throughout \(\varphi(U_0)\). Indeed, at each point the two coordinate frames are bases of the same tangent space, related
by the Jacobian matrix of \(G\), and they lie in the same orientation class. The restriction \(G : V_0 \to \tilde V_0\) is
therefore an orientation-preserving diffeomorphism between open subsets of \(\mathbb{R}^n\), and the
diffeomorphism invariance on open sets
gives
\[
\begin{align*}
\int_{V_0} \bigl(\varphi^{-1}\bigr)^*\omega
&= \int_{V_0} G^*\Bigl(\bigl(\tilde\varphi^{-1}\bigr)^*\omega\Bigr)\\\\
&= \int_{\tilde V_0} \bigl(\tilde\varphi^{-1}\bigr)^*\omega,
\end{align*}
\]
which is the desired equality.
If instead the two charts have opposite orientations, the two definitions carry opposite signs. But now the coordinate
frames lie in opposite orientation classes, so \(\det DG \lt 0\) throughout and \(G\) is orientation-reversing, and 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 \(n \ge 1\), with or without boundary, and let \(\omega\) be a
compactly supported \(n\)-form on \(M\). Choose finitely many positively or negatively oriented smooth charts
\((U_1, \varphi_1), \dots, (U_k, \varphi_k)\) whose domains cover \(\operatorname{supp}\omega\), and let
\(\{\psi_0, \psi_1, \dots, \psi_k\}\) be a smooth
partition of unity
subordinate to the open cover \(\{M \setminus \operatorname{supp}\omega, U_1, \dots, U_k\}\) of \(M\). The
integral of \(\omega\) over \(M\) is
\[
\int_M \omega := \sum_{i=1}^{k} \int_M \psi_i\, \omega.
\]
Each summand makes sense. The support of \(\psi_i\,\omega\) lies in
\(\operatorname{supp}\psi_i \cap \operatorname{supp}\omega\), a compact subset of \(U_i\), so \(\psi_i\,\omega\) is
compactly supported inside the single chart \(U_i\), and its integral is the single-chart integral already defined. The
auxiliary function \(\psi_0\) vanishes on \(\operatorname{supp}\omega\), since its support lies in \(M \setminus
\operatorname{supp}\omega\). Hence \(\psi_1 + \cdots + \psi_k = 1\) on \(\operatorname{supp}\omega\), and \(\omega =
\sum_{i=1}^{k} \psi_i\,\omega\), which is why \(\psi_0\) is left out of the sum. Suitable charts always exist: every point
has a chart with connected domain, and such a chart is positively or negatively oriented, because the set where its
coordinate frame is positively oriented is open and closed in the domain by continuity of the orientation. Compactness of
the support then provides finitely many. Two further features of the setup deserve comment. A partition of unity subordinate
to the cover always
exists, so the
construction is never vacuous.
Negatively oriented charts are also deliberately allowed. On some manifolds with boundary, such as a closed interval, 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 \(\{U_i\}_{i=1}^{k}\) with partition \(\{\psi_i\}_{i=0}^{k}\) and \(\{\tilde U_j\}_{j=1}^{l}\) with partition
\(\{\tilde\psi_j\}_{j=0}^{l}\) be two such choices, all sums below running over \(i, j \ge 1\). Because
\(\sum_j \tilde\psi_j = 1\) on \(\operatorname{supp}\omega\), we have
\(\psi_i\,\omega = \sum_j \tilde\psi_j\,\psi_i\,\omega\). Every form in this identity is supported in the single chart
\(U_i\), where the integral is the Euclidean integral of the coefficient in the chart \(\varphi_i\), so the linearity of
Euclidean integrals splits the sum:
\[
\begin{align*}
\int_M \psi_i\, \omega
&= \int_M \Bigl(\sum_j \tilde\psi_j\Bigr) \psi_i\, \omega\\\\
&= \sum_j \int_M \tilde\psi_j\, \psi_i\, \omega.
\end{align*}
\]
Summing over \(i\) gives
\[
\sum_i \int_M \psi_i\, \omega = \sum_{i,j} \int_M \tilde\psi_j\, \psi_i\, \omega.
\]
Each term on the right is the integral of a form supported in both \(U_i\) and \(\tilde U_j\), so by the
independence of chart it has the same value
whichever of the two charts is used to compute it. The right-hand side is therefore symmetric in the two partitions.
Running the identical argument starting from \(\sum_j \int_M \tilde\psi_j\, \omega\), and using \(\sum_i \psi_i = 1\) on
\(\operatorname{supp}\omega\), 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, let \(S \subseteq M\) be an oriented
immersed \(k\)-dimensional submanifold, and let \(\omega\) be 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, and it 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 the chart-based definition above requires
\(n \ge 1\). 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.
Integration here reduces to addition, and that degenerate case 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.
One further remark widens the class of integrands. Although differential forms on this site are smooth, nothing in the
construction has used more than continuity of the coefficients. Call an assignment \(p \mapsto \omega_p\) of an alternating
\(n\)-tensor on \(T_pM\) to each point a continuous \(n\)-form if its coefficient in every smooth chart is continuous. The
basic example is \(f\,\eta\), with \(f\) a continuous function and \(\eta\) a smooth \(n\)-form. Multiplication by a smooth
function, pullback by a diffeomorphism, and the formula for the
pullback of a top-degree form
are pointwise operations that apply to such forms unchanged. So the single-chart integral, its independence of the chart,
the integral over \(M\), its independence of the choices, and every result established below hold verbatim for
compactly supported continuous \(n\)-forms. This is what allows a continuous function to be integrated against a fixed
smooth form, as it will be against the Haar volume form at the end of this page and against the Riemannian volume form
when integration on Riemannian manifolds is developed.
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.
Call \(\omega\) everywhere non-negative with respect to the orientation if
\(\omega_p(v_1, \dots, v_n) \ge 0\) for every \(p \in M\) and every positively oriented basis \((v_1, \dots, v_n)\) of
\(T_pM\), and strictly positive at \(p\) if this value is \(\gt 0\) there. (Two positively oriented bases differ by a
matrix of positive determinant, which multiplies the value by that determinant, so neither condition depends on the
basis. For \(n = 0\), where \(\omega\) is a function, both conditions are read with the orientation sign at \(p\), as
\(\pm\omega(p) \ge 0\) and \(\pm\omega(p) \gt 0\).) If \(\omega\) is everywhere non-negative and strictly positive at
some point, then \(\int_M \omega \gt 0\). In particular, a positively oriented orientation form has positive integral.
(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.
For \(n = 0\) the integral is a finite signed sum, and (a) and (b) are immediate. For \(n \ge 1\), each property
reduces to forms supported in a single chart, where it is a statement about Euclidean integrals, and the partition of
unity carries it to all of \(M\) through 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. On \(M\), choose one family of charts covering
\(\operatorname{supp}\omega \cup \operatorname{supp}\eta\), and a partition of unity subordinate to it together with
the complement of this compact set. By the independence of the choices it computes the integrals of \(\omega\),
\(\eta\), and \(a\,\omega + b\,\eta\) simultaneously, and summing the pieces, 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 everywhere non-negative and strictly positive at \(p\). For \(n = 0\) the
integral is the signed sum \(\sum_q \pm\omega(q)\), whose terms are all \(\ge 0\) and one of which is \(\gt 0\). Let
\(n \ge 1\). In a positively oriented chart the coordinate frame \((\partial_1, \dots, \partial_n)\) is a positively
oriented basis at every point, so the coefficient of \((\varphi^{-1})^*\omega\), which is
\(\omega(\partial_1, \dots, \partial_n)\), is \(\ge 0\), and the single-chart integral takes the positive sign. In a
negatively oriented chart the coordinate frame is negatively oriented and \((-\partial_1, \partial_2, \dots, \partial_n)\)
is positively oriented, so the coefficient is \(\le 0\), and the compensating negative sign again makes the contribution
\(\ge 0\). Take charts and a partition of unity \(\{\psi_0, \dots, \psi_k\}\) as in the definition of the integral. Since
\(\psi_i \ge 0\), each \(\psi_i\,\omega\) is again everywhere non-negative, so each term \(\int_M \psi_i\,\omega\) is
\(\ge 0\) and \(\int_M \omega \ge 0\). The point \(p\) lies in \(\operatorname{supp}\omega\), where \(\psi_0\) vanishes,
so \(\psi_1(p) + \cdots + \psi_k(p) = 1\) and \(\psi_i(p) \gt 0\) for some \(i \ge 1\). The coefficient of
\(\psi_i\,\omega\) in the chart \(\varphi_i\) is continuous and nonzero at \(\varphi_i(p)\), hence nonzero, with the sign
just described, on a neighborhood of \(\varphi_i(p)\) in \(\varphi_i(U_i)\). Such a neighborhood, open in \(\mathbb{R}^n\)
or in \(\mathbb{H}^n\), has positive Lebesgue measure, so that term is strictly positive. Hence \(\int_M \omega \gt 0\).
(d) For \(n = 0\), \(F\) is a bijection that preserves every sign or reverses every sign, and
\(F^*\omega = \omega \circ F\), so the signed sum defining \(\int_N F^*\omega\) is a reindexing of the one defining
\(\pm\int_M \omega\). Let \(n \ge 1\). Writing \(\omega = \sum_i \psi_i\,\omega\) with a partition of unity as in the
definition, so that \(F^*\omega = \sum_i F^*(\psi_i\,\omega)\), linearity reduces the claim to the case where
\(\omega\) is supported in a single chart \((U, \varphi)\) of \(M\), whose sign we write \(\varepsilon = \pm 1\).
Then \(\bigl(F^{-1}(U), \varphi \circ F\bigr)\) is a smooth chart of \(N\) whose domain contains the compact set
\(\operatorname{supp} F^*\omega = F^{-1}(\operatorname{supp}\omega)\). Its sign is \(\varepsilon\) when \(F\) is
orientation-preserving and \(-\varepsilon\) when \(F\) is orientation-reversing, since \(dF\) carries positively oriented
bases to positively, respectively negatively, oriented ones. Its image is
\((\varphi \circ F)\bigl(F^{-1}(U)\bigr) = \varphi(U)\), and the coordinate representative of \(F^*\omega\) is
\[
\begin{align*}
\bigl((\varphi \circ F)^{-1}\bigr)^* F^*\omega
&= \bigl(\varphi^{-1}\bigr)^* \bigl(F^{-1}\bigr)^* F^*\omega\\\\
&= \bigl(\varphi^{-1}\bigr)^*\omega.
\end{align*}
\]
The Euclidean integrals defining \(\int_N F^*\omega\) and \(\int_M \omega\) are therefore the same integral, and only the
attached signs can differ. They agree when \(F\) is orientation-preserving and are opposite when it is
orientation-reversing.
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 the integral 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 \(n \ge 1\), 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 map \(F_i : \overline{D_i} \to M\) that is
smooth on
\(\overline{D_i}\), 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.
Both sides are linear in \(\omega\), the left by property (a) and the right because pullback and integration over each
\(D_i\) are linear. Write \(\omega = \sum_j \psi_j\,\omega\) with charts and a partition of unity as in the definition
of the integral. Each \(\psi_j\,\omega\) satisfies the hypotheses, because its support lies in
\(\operatorname{supp}\omega\). It therefore suffices to prove the formula when \(\operatorname{supp}\omega\) lies in the
domain of a single positively or negatively oriented chart \((U, \varphi)\), whose sign we write \(\varepsilon = \pm 1\).
The restriction \(F_i|_{D_i}\) is a diffeomorphism onto \(W_i\), so \(\bigl(W_i, (F_i|_{D_i})^{-1}\bigr)\) is a smooth
chart whose image \(D_i\) is open in \(\mathbb{R}^n\), an interior chart. Every point of \(W_i\) is therefore an interior
point of \(M\).
For each \(i\), set
\[
\begin{align*}
B_i &= U \cap W_i,\\\\
A_i &= (F_i|_{D_i})^{-1}(B_i),\\\\
C_i &= \varphi(B_i).
\end{align*}
\]
Then \(A_i \subseteq D_i\) is open, and \(C_i\) is an open subset of \(V = \varphi(U \cap \operatorname{Int} M)\),
which is open in \(\mathbb{R}^n\) as in the proof of the
independence of chart. The sets \(C_i\) are
pairwise disjoint, because the \(W_i\) are and \(\varphi\) is injective. The composite
\(G_i = \varphi \circ F_i : A_i \to C_i\) is a diffeomorphism between open subsets of \(\mathbb{R}^n\). It is
orientation-preserving if \(\varepsilon = 1\) and orientation-reversing if \(\varepsilon = -1\), since \(F_i\)
preserves orientation.
Next we show that the part of \(\operatorname{supp}\omega\) missed by the \(W_i\) is negligible. Let
\(q \in \overline{W_i} \setminus W_i\). Since \(M\) is first countable, there are points \(x_m \in D_i\) with
\(F_i(x_m) \to q\). The set \(\overline{D_i}\) is closed and bounded, hence compact, so a subsequence converges to some
\(x \in \overline{D_i}\), and continuity gives \(F_i(x) = q\), limits in the Hausdorff space \(M\) being unique. If
\(x\) were in \(D_i\), then \(q = F_i(x)\) would lie in \(W_i\), so \(x \in \partial D_i\). Thus
\(\overline{W_i} \setminus W_i \subseteq F_i(\partial D_i)\), and hypothesis 3, together with
\(\operatorname{supp}\omega \subseteq U\), gives
\[
\operatorname{supp}\omega \setminus (W_1 \cup \cdots \cup W_k) \subseteq U \cap \bigcup_{i=1}^{k} F_i(\partial D_i).
\]
The image of this uncovered part under \(\varphi\) is contained in the union of the sets
\((\varphi \circ F_i)\bigl(\partial D_i \cap F_i^{-1}(U)\bigr)\). Each map \(\varphi \circ F_i\) is smooth on the set
\(\partial D_i \cap F_i^{-1}(U)\). Near any point of this set, \(F_i\) has a smooth extension to an open subset of
\(\mathbb{R}^n\). After shrinking that open subset, the extension takes values in the open set \(U\), and composing it
with \(\varphi\) extends \(\varphi \circ F_i\) smoothly. Since \(\partial D_i\) has measure zero and
smooth maps preserve sets of measure zero,
each of these images has measure zero.
Now let \(f\) be the coefficient of \((\varphi^{-1})^*\omega\). It is continuous and vanishes outside the compact set
\(\varphi(\operatorname{supp}\omega)\), so it is bounded and integrable. As in the chart-independence argument,
\(\int_M \omega = \varepsilon \int_V f\, dV\). Within \(V\), the function \(f\) vanishes outside
\(\varphi(\operatorname{supp}\omega)\), and the part of that set not covered by the disjoint open sets \(C_i\) lies in
\(\varphi\bigl(\operatorname{supp}\omega \setminus (W_1 \cup \cdots \cup W_k)\bigr)\), which has measure zero. Hence
\[
\begin{align*}
\int_V f\, dV
&= \sum_{i=1}^{k} \int_{C_i} f\, dV\\\\
&= \sum_{i=1}^{k} \int_{C_i} \bigl(\varphi^{-1}\bigr)^*\omega.
\end{align*}
\]
Finally, apply the
diffeomorphism invariance on open sets
to \(G_i : A_i \to C_i\). Since \(G_i^*\bigl(\varphi^{-1}\bigr)^*\omega = F_i^*\omega\) on \(A_i\) by functoriality,
\[
\int_{C_i} \bigl(\varphi^{-1}\bigr)^*\omega = \varepsilon \int_{A_i} F_i^*\omega.
\]
At a point \(x \in D_i \setminus A_i\) the image \(F_i(x)\) lies in \(W_i\) but not in \(B_i\), hence outside \(U\) and
outside \(\operatorname{supp}\omega\). Since \((F_i^*\omega)_x\) depends only on \(\omega\) at \(F_i(x)\), the form
\(F_i^*\omega\) vanishes on \(D_i \setminus A_i\). Its coefficient is continuous on \(\overline{D_i}\). Near each point
of \(\overline{D_i}\) it is computed from a smooth local extension of \(F_i\), and the derivatives of such an extension
at points of \(\overline{D_i}\) are limits of derivatives of \(F_i\) on the open set \(D_i\), so they do not depend on
the extension. The integral \(\int_{D_i} F_i^*\omega\) over the domain of integration \(D_i\) is therefore defined, its
Riemann and Lebesgue values agree, and it equals \(\int_{A_i} F_i^*\omega\). Combining these equalities, and using
\(\varepsilon^2 = 1\),
\[
\int_M \omega = \sum_{i=1}^{k} \int_{D_i} F_i^*\omega.
\]
The proposition says exactly what one would hope. To integrate over a region covered by a parametrization \(F\), we 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 \(\overline{\mathbb{B}}^3\), namely the outward-pointing one. 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 \(F\) is smooth on \(\overline{D}\), being the restriction of a smooth map on \(\mathbb{R}^2\). Spherical coordinates
\(G(r, \varphi, \theta) = r\,F(\varphi, \theta)\) define an injective smooth map on \((0, \infty) \times D\) with Jacobian
determinant \(r^2 \sin\varphi \gt 0\), hence a diffeomorphism onto an open subset of \(\mathbb{R}^3\). Since
\(\lvert G(r, \varphi, \theta) \rvert = r\), its restriction to \((0, 1] \times D\) is an orientation-preserving
diffeomorphism onto an open subset of \(\overline{\mathbb{B}}^3\), and it carries \(\{1\} \times D\) into \(S^2\). The
boundary parametrization criterion,
applied with \(F(\varphi, \theta) = G(1, \varphi, \theta)\), shows that \(F\) restricted to \(D\) is a diffeomorphism onto an
open subset of \(S^2\), namely the sphere minus the closed half-meridian \(\{y = 0,\ x \ge 0\}\), and that it preserves the
Stokes orientation. The closure of this open set is all of \(S^2\), so \(F\) 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,
\[
\begin{align*}
F^*dx &= \cos\varphi \cos\theta\, d\varphi - \sin\varphi \sin\theta\, d\theta,\\\\
F^*dy &= \cos\varphi \sin\theta\, d\varphi + \sin\varphi \cos\theta\, d\theta,\\\\
F^*dz &= -\sin\varphi\, d\varphi.
\end{align*}
\]
Wedge these in pairs and substitute the coordinate functions \(x = \sin\varphi\cos\theta\), and so on. Every term then
combines into a single multiple of \(d\varphi \wedge d\theta\).
Carrying out the algebra, we find that 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.
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.
On \(\mathbb{R}^3\), \(\omega\) is the form whose integral over an oriented surface returns the flux of the radial field
\((x, y, z)\) through that surface. On the unit sphere this field is the outward unit normal, so the 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. What it needed was
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 a natural class of manifolds on which a canonical integral exists, requiring no metric: the compact Lie groups. The
symmetry of such a group singles out a preferred top-degree form, unique once we fix a
left-invariant orientation
and ask that the total volume be one. Integrating against that form is the construction on which harmonic analysis on groups
is built. The representation theory behind equivariant neural networks,
developed in the pages ahead, rests on the same construction. 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:
\[
L_g^* A = A \quad \text{for all } g \in G.
\]
A left-invariant form is completely determined by its value at the identity \(e\), since left translation carries that value
to every other point. That rigidity makes left-invariant top-degree forms unique up to a constant multiple, since the
alternating tensors of top degree on \(T_eG\) form a one-dimensional space.
Proof.
If \(\dim G = 0\), the group is compact and discrete, hence a finite set of \(k\) points. An orientation is a choice of
sign at each point, and it is left-invariant when every \(L_g\) preserves these signs. Left translations act
transitively, so all points carry one common sign \(s = \pm 1\). A left-invariant \(0\)-form is a function with
\(f(gh) = f(h)\) for all \(g, h\), that is, a constant \(c\), and it is positively oriented when \(s\,c \gt 0\). Its
integral, the signed sum over the points, is \(k\,s\,c\), so the normalization forces \(c = s/k\), and \(\omega_G = s/k\)
is the unique such form. 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
\(E_1, \dots, E_n\), a basis of the Lie algebra. Replacing \(E_1\) by \(-E_1\) if necessary, we may take
\((E_1|_e, \dots, E_n|_e)\) to be positively oriented. The frame is then positively oriented at every point \(g\),
because \(E_j|_g = d(L_g)_e\, E_j|_e\) and \(L_g\) preserves the given orientation, which is left-invariant.
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 tangent space at
every point, they are equal. Hence \(\varepsilon^i\) is left-invariant.
Now set \(\omega_0 = \varepsilon^1 \wedge \cdots \wedge \varepsilon^n\). Pullback commutes with the wedge product, so
\[
\begin{align*}
L_g^* \omega_0
&= \bigl(L_g^*\varepsilon^1\bigr) \wedge \cdots \wedge \bigl(L_g^*\varepsilon^n\bigr)\\\\
&= \varepsilon^1 \wedge \cdots \wedge \varepsilon^n = \omega_0,
\end{align*}
\]
so \(\omega_0\) is left-invariant. It is nowhere zero, because \(\omega_0(E_1, \dots, E_n) = 1 \gt 0\). Hence 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, and among the left-invariant forms that are positively oriented for
a fixed left-invariant orientation it is unique up to a positive constant multiple. (Two such forms agree at the
identity up to a positive scalar, since alternating \(n\)-tensors on \(T_eG\) form a one-dimensional space and both
forms are positive on \((E_1|_e, \dots, E_n|_e)\), hence they agree everywhere by left-invariance.)
Compactness enters only now. Because \(G\) is compact, \(\omega_0\) is compactly supported, and since it is a
positively oriented orientation form, part (c) of the
properties of the integral makes
\(\int_G \omega_0\) a positive number. Rescaling by its reciprocal, the form
\[
\omega_G = \left( \int_G \omega_0 \right)^{-1} \omega_0
\]
is again left-invariant, positively oriented, and now satisfies \(\int_G \omega_G = 1\). Any left-invariant positively
oriented form is a positive multiple of \(\omega_0\), and the normalization fixes that multiple, so \(\omega_G\) 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 follows from part (d) of the
properties of the integral. Indeed
\(L_g^*(f\,\omega_G) = (f \circ L_g)\, L_g^*\omega_G = (f \circ L_g)\,\omega_G\), and \(L_g\) is an orientation-preserving
diffeomorphism, since the orientation is left-invariant. This translation-invariance allows one to average any continuous
function over the whole group in a way that respects the group structure, and the same average applies to objects built
from a representation of the group. A normalized average of total mass \(1\) behaves like a probability measure on \(G\).
The Haar integral 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. First, the existence of a left-invariant orientation form, unique
up to scale, holds for every Lie group. Compactness is used only to make the form compactly supported, so that its total
volume is defined, and hence to normalize the scale to \(1\). On a noncompact group the form still exists but cannot be
normalized this way. An orientation form vanishes nowhere, so its support is all of \(G\), which is not compact, and its
integral is not defined.
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 integral is in fact invariant under right translation as well, a property
called unimodularity. The form itself is right-invariant up to a sign, the sign recording whether the right translation
preserves the orientation. That two-sided invariance of the integral is what the representation theory of compact
groups ultimately requires. The refinement is
established where it is needed, in the
representation-theoretic development that comes later. Here we record only the left-invariant normalized form and its
integral, which is the foundation on which the rest is built.