Orientations of Hypersurfaces
An orientation of a manifold is a choice that lives in the top exterior power of each tangent space,
recorded globally by a nowhere-vanishing
orientation form.
Suppose now that \(S\) sits inside an oriented manifold \(M\) as a hypersurface, a submanifold of
one dimension less.
One might hope that \(S\) simply inherits the ambient orientation, but the mechanism that records
orientation refuses to transfer directly. An orientation form on \(M\) is an \(n\)-covector field,
and pulling it back to a space of dimension \(n-1\) produces a covector of degree \(n\) on an
\((n-1)\)-dimensional space, which must vanish identically. The top-degree information has nowhere
to live once a dimension is removed.
The remedy is to supply the missing dimension externally. If we can single out, at each point of
\(S\), a tangent vector of \(M\) that is not tangent to \(S\), then prepending that vector to a
basis of \(T_pS\) recovers a basis of \(T_pM\). The ambient orientation can then adjudicate whether
that enlarged basis is positively oriented. A consistent choice of such a transverse vector along
all of \(S\) therefore transports the ambient orientation onto \(S\). The next result makes this
precise and identifies the resulting orientation form by a contraction.
Proposition (Orientation Induced by a Transverse Vector Field)
Let \(M\) be an oriented smooth \(n\)-manifold with or without boundary, with orientation
form \(\omega\), let \(S \subseteq M\) be an immersed hypersurface, that is, an
immersed submanifold
or an
immersed submanifold with boundary
of codimension one, with inclusion \(\iota_S : S \hookrightarrow M\),
and let \(N\) be a
smooth vector field along \(S\)
that is nowhere tangent to \(S\), meaning
\(N_p \in T_pM \setminus d(\iota_S)_p(T_pS)\) for every \(p \in S\). Then \(S\) has a unique
orientation with the following property: for every \(p \in S\), an ordered basis
\((E_1, \dots, E_{n-1})\) of \(T_pS\) is positively oriented if and only if
\(\bigl(N_p,\, d(\iota_S)_p E_1,\, \dots,\, d(\iota_S)_p E_{n-1}\bigr)\) is a positively
oriented basis of \(T_pM\). This orientation is determined by the nowhere-vanishing
\((n-1)\)-form
\[
\iota_S^{*}\bigl(N \lrcorner\, \omega\bigr),
\]
where \(N \lrcorner\, \omega\) denotes the
interior multiplication
of \(\omega\) by \(N\).
Proof.
Write \(\sigma = \iota_S^{*}(N \lrcorner\, \omega)\), where the pullback evaluates
\((N \lrcorner\, \omega)_p\) on the images under \(d(\iota_S)_p\) of vectors tangent to \(S\). This is a
smooth \((n-1)\)-form on \(S\). Indeed, near each point of \(S\) the field \(N\) agrees with a smooth
vector field \(\widetilde N\) on an open set \(V \subseteq M\). The form \(\widetilde N \lrcorner\, \omega\)
is smooth on \(V\), because its components in any chart are sums of products of the components of
\(\widetilde N\) and of \(\omega\), and \(\sigma = \iota_S^{*}(\widetilde N \lrcorner\, \omega)\) on the open
subset \(\iota_S^{-1}(V)\) of \(S\). We claim \(\sigma\) vanishes nowhere. Fix \(p \in S\) and a basis
\((E_1, \dots, E_{n-1})\) of \(T_pS\). To lighten notation, identify each \(E_i\) with its image
\(d(\iota_S)_p E_i \in T_pM\). Because \(\iota_S\) is an immersion, these images are linearly
independent and span an \((n-1)\)-dimensional subspace of \(T_pM\). Since \(N_p\) is not tangent
to \(S\), it lies outside that subspace, so \((N_p, E_1, \dots, E_{n-1})\) is a basis of
\(T_pM\). Using the defining relation of interior multiplication,
\[
\sigma_p(E_1, \dots, E_{n-1})
= (N \lrcorner\, \omega)_p(E_1, \dots, E_{n-1})
= \omega_p(N_p, E_1, \dots, E_{n-1}).
\]
The right-hand side is the value of a nonzero top-degree alternating tensor on a basis, hence
nonzero. Therefore \(\sigma_p \neq 0\), and as \(p\) was arbitrary, \(\sigma\) is
nowhere-vanishing. By the correspondence between nowhere-vanishing top forms and orientations,
\(\sigma\) determines a unique orientation of \(S\).
It remains to verify the stated basis criterion. By definition, \((E_1, \dots, E_{n-1})\) is
positively oriented for the orientation determined by \(\sigma\) precisely when
\(\sigma_p(E_1, \dots, E_{n-1}) \gt 0\), which by the computation above means
\(\omega_p(N_p, E_1, \dots, E_{n-1}) \gt 0\). This last inequality holds exactly when
\((N_p, E_1, \dots, E_{n-1})\) is positively oriented for the ambient orientation determined by
\(\omega\). The equivalence is thus immediate.
When \(n = 1\) the hypersurface \(S\) is a \(0\)-manifold, so \(T_pS = \{0\}\) and the basis
\((E_1, \dots, E_{n-1})\) is empty. The proposition is then read pointwise. The contraction
\(N \lrcorner\, \omega\) reduces to the \(0\)-form \(\omega(N)\), and \(S\) receives the sign \(+1\)
at a point \(p\) when \(N_p\) is a positively oriented basis of \(T_pM\) and \(-1\) when it is
negatively oriented. With this reading the argument above goes through without change.
Interior multiplication by \(N\), the operation \(\eta \mapsto N \lrcorner\, \eta\), is an
antiderivation that lowers degree by one, and its behavior on bases is exactly what the proof
exploits. Its general algebraic properties are recorded in the
properties of interior multiplication.
The geometric content is worth isolating. Inducing an orientation on a hypersurface from the ambient
one is not an intrinsic operation but requires a transverse direction, and reversing \(N\) reverses
the induced orientation. More precisely, let \(N\) and \(\widetilde N\) be two smooth vector fields
along \(S\) that are nowhere tangent to \(S\), and fix \(p \in S\). Since \(N_p\) together with
\(d(\iota_S)_p(T_pS)\) spans \(T_pM\), we can write \(\widetilde N_p = \lambda N_p + v\) with
\(v \in d(\iota_S)_p(T_pS)\), and \(\lambda \neq 0\) because \(\widetilde N_p\) is not tangent to
\(S\). Replacing \(N_p\) by \(\widetilde N_p\) in the first slot of the enlarged basis is a change
of basis of determinant \(\lambda\), so the two fields induce the same orientation at \(p\) exactly
when \(\lambda \gt 0\).
The cleanest source of a transverse field is the position vector field on a sphere. Regard
\(S^n\) as the unit sphere in \(\mathbb{R}^{n+1}\), with \(\mathbb{R}^{n+1}\) carrying its
standard orientation. The vector field
\[
N = \sum_{i=1}^{n+1} x^i\, \frac{\partial}{\partial x^i}
\]
restricts along \(S^n\) to the outward radial direction, and since the radial direction is
orthogonal to the sphere it is nowhere tangent. The orientation it induces through the preceding
proposition is, by definition, the standard orientation of \(S^n\). For the
zero-sphere \(S^0 = \{-1, +1\} \subset \mathbb{R}\), the same prescription assigns the orientation
\(+1\) to the point \(+1\) and \(-1\) to the point \(-1\), consistent with the convention that a
zero-dimensional manifold is oriented by a choice of sign at each point.
A transverse field is also available whenever a hypersurface arises as a level set, which is the
most common way hypersurfaces appear in practice. Here the gradient supplies the normal direction,
once a metric is chosen to convert the differential into a vector.
Proposition (Regular Level Sets Are Orientable)
Let \(M\) be an oriented smooth manifold without boundary, let \(f : M \to \mathbb{R}\) be
a smooth function,
and let \(c\) be a
regular value
of \(f\). Then the
level set
\(S = f^{-1}(c)\) is an orientable hypersurface in \(M\).
Proof.
Choose a Riemannian metric \(g\) on \(M\), which
exists
on every smooth manifold. Let \(N = \operatorname{grad} f\big|_S\) be the restriction along \(S\) of the
gradient
of \(f\). Because \(c\) is a regular value, \(df_p \neq 0\) at every \(p \in S\). The gradient
characterization
\(g_p(\operatorname{grad} f_p, v) = df_p(v)\) shows \(\operatorname{grad} f_p \neq 0\) there as
well, since a vector whose inner product against every direction vanished would force
\(df_p = 0\).
Let \(v \in T_pS\). The
tangent space of a regular level set
is the kernel \(T_pS = \ker df_p\), since \(f\) restricts to a submersion on the open set
\(\{df \neq 0\} \supseteq S\), of which \(S\) is a level set. Thus \(df_p(v) = 0\), hence
\(g_p(\operatorname{grad} f_p, v) = 0\). The gradient is therefore orthogonal to \(T_pS\). In
particular \(N_p = \operatorname{grad} f_p\) is nonzero and not tangent to \(S\), so \(N\) is a
smooth vector field along \(S\) that is nowhere tangent to \(S\). The previous proposition then
equips \(S\) with an orientation, so \(S\) is orientable.
The orientation produced here does not depend on the auxiliary metric. Fix \(p \in S\). For any
choice of \(g\), the gradient \(\operatorname{grad} f_p\) with respect to \(g\) satisfies
\(df_p(\operatorname{grad} f_p) = g_p(\operatorname{grad} f_p, \operatorname{grad} f_p) \gt 0\). Let
\(N\) and \(\widetilde N\) be the restrictions along \(S\) of the gradients of \(f\) for \(g\) and
for another metric \(\widetilde g\), and write \(\widetilde N_p = \lambda N_p + v\) with
\(v \in d(\iota_S)_p(T_pS) = \ker df_p\). Applying \(df_p\) gives
\(\lambda\, df_p(N_p) = df_p(\widetilde N_p) \gt 0\), so \(\lambda \gt 0\). By the comparison of
transverse fields above, \(N\) and \(\widetilde N\) induce the same orientation at \(p\). Since
\(p\) was arbitrary, they induce the same orientation of \(S\).
Replacing \(f\) by \(-f\), which cuts out the same set \(S = (-f)^{-1}(-c)\), replaces \(N\) by
\(-N\) and therefore reverses the induced orientation of \(S\). The same mechanism will reappear
once boundaries enter the picture, since a boundary is the zero set of a boundary defining function,
whose negated gradient is an outward-pointing field playing the role of \(N\).
Boundary Orientations and the Stokes Convention
The boundary of a manifold with boundary is the most important hypersurface of all, because it is
where the fundamental theorem relating differentiation and integration localizes. To make that
theorem hold with the correct signs, the boundary must carry a definite orientation derived from
the ambient one, and the recipe of the previous section applies once a transverse direction is
chosen. Two structural facts make boundaries an ideal setting. First, the boundary
\(\partial M\) is a properly embedded hypersurface,
so it is a genuine codimension-one submanifold to which the construction applies. Second, there is
a canonical side to which a transverse vector may point.
At a boundary point the tangent space of \(M\) splits, relative to the tangent space of the
boundary, into vectors that
point inward or outward,
and a boundary chart detects the distinction through the sign of the
last coordinate component.
An outward-pointing vector has negative \(x^n\)-component in any boundary chart.
There is no intrinsic notion of inward versus outward for an arbitrary hypersurface, but for a
boundary the direction leaving the manifold is unambiguous, and that is precisely what is needed to
pin down a transverse field up to its side. The next result shows that every smooth outward-pointing
field induces one and the same orientation.
Proposition (The Induced Orientation on a Boundary)
Let \(M\) be an oriented smooth \(n\)-manifold with boundary, \(n \geq 1\), with orientation
form \(\omega\). Then \(\partial M\) is orientable, and every smooth outward-pointing vector
field along \(\partial M\) induces the same orientation on \(\partial M\) through the construction
for transverse fields. If \(N\) is any such field and
\(\iota : \partial M \hookrightarrow M\) is the inclusion, the induced orientation has
orientation form
\[
\iota^{*}\bigl(N \lrcorner\, \omega\bigr).
\]
Proof.
Since an outward-pointing vector is by definition not tangent to the boundary, any smooth
outward-pointing vector field \(N\) along it is in particular nowhere tangent to \(\partial M\).
The construction for transverse fields therefore applies and equips \(\partial M\) with an
orientation whose orientation form is \(\iota^{*}(N \lrcorner\, \omega)\).
Such a field exists. Let \(f\) be a
boundary defining function
for \(M\), which
exists,
and let \(g\) be a Riemannian metric on \(M\), which also
exists.
The restriction \(N = -\operatorname{grad} f\big|_{\partial M}\) is a smooth vector field along
\(\partial M\). At each \(p \in \partial M\), the
characterization of the gradient
gives \(N_p f = df_p(N_p) = -g_p(\operatorname{grad} f_p, \operatorname{grad} f_p)\), which is negative
because \(df_p \neq 0\) forces \(\operatorname{grad} f_p \neq 0\). The proposition that provides \(f\)
also shows that a vector \(v \in T_pM\) with \(vf \lt 0\) is outward-pointing, so \(N\) is a smooth
outward-pointing vector field along \(\partial M\). In particular \(\partial M\) is orientable. It
remains to show that the induced
orientation does not depend on which smooth outward-pointing field is used.
Let \(N\) and \(\widetilde N\) be two smooth outward-pointing fields along \(\partial M\). Fix
\(p \in \partial M\) and a boundary chart \((x^1, \dots, x^n)\) at \(p\), in which the boundary
is \(\{x^n = 0\}\) and \(M\) lies in \(\{x^n \geq 0\}\). Write the components of the two fields
at \(p\) as \(N_p = \sum_i N^i(p)\,\partial_{x^i}\) and
\(\widetilde N_p = \sum_i \widetilde N^{\,i}(p)\,\partial_{x^i}\). Because both are
outward-pointing, their final components are negative: \(N^n(p) \lt 0\) and
\(\widetilde N^{\,n}(p) \lt 0\). Choose a positively oriented basis \((E_1, \dots, E_{n-1})\) of
\(T_p\partial M\) for the orientation induced by \(N\). The induced orientations of \(N\) and
\(\widetilde N\) agree at \(p\) precisely when the change of basis carrying
\((N_p, E_1, \dots, E_{n-1})\) to \((\widetilde N_p, E_1, \dots, E_{n-1})\) has positive
determinant. Since the two enlarged bases differ only in the first slot, and
\(E_1, \dots, E_{n-1}\) are tangent to the boundary (hence have zero \(x^n\)-component), that
determinant equals the ratio of the \(x^n\)-components,
\[
\frac{\widetilde N^{\,n}(p)}{N^n(p)} \gt 0,
\]
a quotient of two negative numbers. The orientations therefore agree at \(p\). As \(p\) was
arbitrary, they agree on all of \(\partial M\).
Definition (Stokes Orientation)
The orientation of \(\partial M\) determined by the preceding proposition is called the
induced orientation or the Stokes orientation on the
boundary. Unless stated otherwise, the boundary of an oriented manifold is always understood
to carry this orientation.
A first consequence settles the sphere. The closed unit ball
\(\overline{\mathbb{B}}^{n+1} \subseteq \mathbb{R}^{n+1}\) is a manifold with boundary \(S^n\), and
the outward radial field is outward-pointing, so the Stokes orientation that
\(\overline{\mathbb{B}}^{n+1}\) induces on \(S^n\) is exactly the standard orientation introduced
earlier. The signs are less innocent on a half-space.
Proposition (Stokes Orientation of the Half-Space Boundary)
Let \(\mathbb{H}^n = \{x^n \geq 0\}\), \(n \geq 1\), carry the standard orientation inherited
from \(\mathbb{R}^n\), and identify its boundary \(\partial \mathbb{H}^n = \{x^n = 0\}\) with
\(\mathbb{R}^{n-1}\) through the coordinates \((x^1, \dots, x^{n-1})\). Then the Stokes
orientation on \(\partial \mathbb{H}^n\) coincides with the standard orientation of
\(\mathbb{R}^{n-1}\) when \(n\) is even and is its reverse when \(n\) is odd. Equivalently,
the two orientations differ by the factor \((-1)^n\).
Proof.
The vector \(-\partial_{x^n}\) is outward-pointing, since it has negative final component,
so the Stokes orientation is the one for which \((E_1, \dots, E_{n-1})\) is positively
oriented exactly when \((-\partial_{x^n}, E_1, \dots, E_{n-1})\) is positively oriented in
\(\mathbb{R}^n\). Comparing this enlarged frame with the standard frame involves two sign
contributions: extracting the minus sign from \(-\partial_{x^n}\) contributes a factor
\(-1\), and moving \(\partial_{x^n}\) from the first slot past the \(n-1\) remaining vectors
to the last slot is an \((n-1)\)-fold transposition contributing \((-1)^{n-1}\). Their
product is \((-1)\cdot(-1)^{n-1} = (-1)^n\):
\[
\bigl[-\partial_{x^n}, \partial_{x^1}, \dots, \partial_{x^{n-1}}\bigr]
= (-1)^n \bigl[\partial_{x^1}, \dots, \partial_{x^{n-1}}, \partial_{x^n}\bigr].
\]
This factor of \((-1)^n\) is exactly the sign that the boundary term of the integration theorem
acquires on a half-space when it is written in the standard coordinates of \(\mathbb{R}^{n-1}\),
and tracking it correctly is the whole point of fixing the convention.
To compute with the boundary orientation in practice, one parametrizes the boundary by restricting a
parametrization of a collar neighborhood and asks whether the restriction preserves orientation. The
following lemma reduces a boundary computation to an ambient one.
Lemma (Boundary Parametrization Criterion)
Let \(M\) be an oriented smooth \(n\)-manifold with boundary, and let
\(F : (a, b] \times U \to M\) be a diffeomorphism onto an open subset of \(M\), where
\(U \subseteq \mathbb{R}^{n-1}\) is open, \((a, b] \times U \subseteq \mathbb{R}^n\) carries the
standard orientation, and \(F\) carries \(\{b\} \times U\) into \(\partial M\). Define
\(f : U \to \partial M\) by \(f(x) = F(b, x)\). Then \(f\) is a
diffeomorphism onto an open subset of \(\partial M\), and \(f\) is an
orientation-preserving
parametrization of its image for the Stokes orientation on \(\partial M\)
if and only if \(F\) is orientation-preserving for \(M\).
Proof.
The standard orientation of \((a, b] \times U\) is the one for which
\((t, x^1, \dots, x^{n-1})\), with \(t\) the coordinate on \((a, b]\), is positively oriented.
Its boundary is \(\{b\} \times U\), and the outward-pointing direction there is \(+\partial_t\),
since \(t\) increases up to the endpoint \(b\). Here the outward field \(+\partial_t\) already
sits in the first slot of the positively oriented frame
\((\partial_t, \partial_{x^1}, \dots, \partial_{x^{n-1}})\), so no reordering and no sign
factor arise. The Stokes orientation on \(\{b\} \times U\) is by definition the one for
which \((x^1, \dots, x^{n-1})\) is positively oriented. (This is the collar analogue of the
half-space computation,
with the sign trivial because the outward direction is the leading coordinate rather than
the negated final one.) Thus the standard orientation of \(U\) is the boundary
orientation.
Next we examine \(F\) at a point \((b, x)\) of the boundary of the model, writing
\(p = f(x) = F(b, x) \in \partial M\). Since \(f\) takes values in \(\partial M\), in any
boundary chart \((y^i)\) of \(M\) around \(p\) the last coordinate of \(f\) vanishes
identically, and the remaining coordinates \((y^1, \dots, y^{n-1})\) restrict to a chart of
\(\partial M\) for the smooth structure that makes it a
properly embedded hypersurface.
Hence \(f\) is smooth as a map into \(\partial M\), and \(df_x : T_xU \to T_p\partial M\) is the
restriction of \(dF_{(b,x)}\) to the subspace \(T_xU\) spanned by
\(\partial_{x^1}, \dots, \partial_{x^{n-1}}\). Because \(F\) is a diffeomorphism onto an open
subset, \(dF_{(b,x)}\) is an isomorphism onto \(T_pM\), so \(df_x\) is injective, and as both
\(T_xU\) and \(T_p\partial M\) have dimension \(n-1\), it is an isomorphism. Thus \(f\) is a
local diffeomorphism, and being injective it is a diffeomorphism onto an open subset of
\(\partial M\). Two consequences follow. First, \(dF_{(b,x)}(\partial_t)\) does not lie in
\(T_p\partial M = dF_{(b,x)}(T_xU)\), since \(dF_{(b,x)}\) is injective and
\(\partial_t \notin T_xU\). Second, \(dF_{(b,x)}(\partial_t)\) is outward-pointing: the curve
\(\gamma(s) = F(b - s, x)\), defined for \(s \in [0, b - a)\), lies in \(M\), satisfies
\(\gamma(0) = p\), and has velocity \(\gamma'(0) = -\,dF_{(b,x)}(\partial_t)\), which is not
tangent to \(\partial M\). By the definition of
inward-pointing vectors,
\(-\,dF_{(b,x)}(\partial_t)\) is inward-pointing, and, by the
last-coordinate criterion,
a vector is inward-pointing exactly when its negative is outward-pointing.
Now compare orientations at \(x\). Let \((E_1, \dots, E_{n-1})\) be a basis of \(T_xU\). By the
first paragraph, it is positively oriented for the boundary orientation of \(\{b\} \times U\)
exactly when \((\partial_t, E_1, \dots, E_{n-1})\) is positively oriented in
\(T_{(b,x)}\bigl((a, b] \times U\bigr)\). On the other side, write
\(N_p = dF_{(b,x)}(\partial_t)\). The pointwise comparison in the proof of the
induced orientation on a boundary
used nothing about the two outward-pointing fields beyond their values at \(p\), so the Stokes
orientation of \(T_p\partial M\) is the one for which a basis \((v_1, \dots, v_{n-1})\) is
positively oriented exactly when \((N_p, v_1, \dots, v_{n-1})\) is positively oriented in
\(T_pM\), the outward-pointing vector \(N_p\) being as good as any other. Applying this to
\(v_i = df_x E_i\),
\[
\bigl(N_p,\, df_x E_1,\, \dots,\, df_x E_{n-1}\bigr)
= dF_{(b,x)}\bigl(\partial_t, E_1, \dots, E_{n-1}\bigr),
\]
so \(df_x\) carries a positively oriented basis of \(T_xU\) to a positively oriented basis of
\(T_p\partial M\) exactly when \(dF_{(b,x)}\) carries the positively oriented basis
\((\partial_t, E_1, \dots, E_{n-1})\) to a positively oriented basis of \(T_pM\). That is,
\(f\) preserves orientation at \(x\) if and only if \(F\) preserves orientation at \((b, x)\),
and likewise for reversal.
It remains to pass from the boundary points \(\{b\} \times U\) to all of the model. Choose an
orientation
form \(\omega\) for \(M\). Since \(F\) is a local diffeomorphism, the
pullback
\(F^{*}\omega = h\, dt \wedge dx^1 \wedge \cdots \wedge dx^{n-1}\) is nowhere vanishing, so the
coefficient \(h\) is continuous and never zero, and \(F\) preserves orientation at a point
exactly when \(h\) is positive there. The sign of \(h\) is therefore constant on each connected
component of \((a, b] \times U\). Every such component has the form \((a, b] \times U_0\) with
\(U_0\) a connected component of \(U\), and it contains the boundary points
\(\{b\} \times U_0\). Hence \(F\) is orientation-preserving on all of \((a, b] \times U\) if and
only if it is orientation-preserving at every point of \(\{b\} \times U\), which by the
preceding paragraph holds if and only if \(f\) is orientation-preserving at every point of \(U\).
As an illustration, spherical coordinates orient the two-sphere correctly. Regard the closed unit ball
\(\overline{\mathbb{B}}^3\), with the standard orientation of \(\mathbb{R}^3\), as an oriented manifold with
boundary \(S^2\), and define \(F : (0, 1] \times (0, \pi) \times (0, 2\pi) \to \overline{\mathbb{B}}^3\) by
\(F(\rho, \varphi, \theta) = (\rho \sin\varphi \cos\theta, \rho \sin\varphi \sin\theta, \rho \cos\varphi)\).
This map is injective, its Jacobian determinant \(\rho^2 \sin\varphi\) is positive throughout the domain,
and its image is the open subset of \(\overline{\mathbb{B}}^3\) complementary to the closed set
\(\{y = 0,\ x \geq 0\}\). Hence \(F\) is an orientation-preserving diffeomorphism onto an open subset of
\(\overline{\mathbb{B}}^3\) that carries \(\{1\} \times (0, \pi) \times (0, 2\pi)\) into \(S^2\). Since \(S^2\) is
reached as \(\rho\) increases to its endpoint \(1\), with the radial direction outward, the lemma shows
that the angular parametrization
\(X(\varphi, \theta) = (\sin\varphi \cos\theta, \sin\varphi \sin\theta, \cos\varphi)\) preserves the
standard orientation of \(S^2\). The positivity of the spherical Jacobian is thus the same fact,
seen from the boundary, as the standard orientation of the sphere.
With the Stokes orientation fixed, the boundary of an oriented manifold is itself a fully oriented
manifold, and the apparatus of orientation forms developed for hypersurfaces transfers without
change. The integral that the boundary orientation was designed to serve, the integral of
differential forms over oriented manifolds, needs no further structure. Measuring size on such
manifolds is a separate matter, for which a metric supplies a distinguished volume form.