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 \(T_pM\), and the ambient orientation can 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 orientation form \(\omega\), let
\(S \subseteq M\) be an immersed hypersurface with inclusion \(\iota_S : S \hookrightarrow M\),
and let \(N\) be a
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 \(\sigma = \iota_S^{*}(N \lrcorner\, \omega)\); this is a smooth \((n-1)\)-form on \(S\),
since \(\omega\) is smooth, \(N\) is smooth along \(S\), and interior multiplication followed
by pullback preserves smoothness. 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
\(\sigma_p(E_1, \dots, E_{n-1}) > 0\), which by the computation above means
\(\omega_p(N_p, E_1, \dots, E_{n-1}) > 0\); and 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. Independence of the choice of \(p\) within a
connected piece follows because \(N\), \(\omega\), and the immersion vary smoothly, so the
sign of \(\sigma\) cannot jump.
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.
The contraction \(N \lrcorner\, \omega\) is an antiderivation that lowers degree by one, and its
behaviour 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: orienting a hypersurface is not an intrinsic operation
but requires a transverse direction, and reversing \(N\) reverses the induced orientation. Two
transverse fields that point to the same side of \(S\) induce the same orientation, a fact we will
use repeatedly.
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, 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 \(M\); one exists by a
partition of unity
argument. 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 \(p \in S\), and 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
\(df_p = 0\). For any \(v \in T_pS\), the tangent space of a regular level set is the kernel
\(T_pS = \ker df_p\), so \(df_p(v) = 0\), hence \(g_p(\operatorname{grad} f_p, v) = 0\): the
gradient is orthogonal to
\(T_pS\). In particular \(N_p = \operatorname{grad} f_p\) is nonzero and not tangent to \(S\),
so \(N\) is a nowhere-tangent vector field along \(S\). The previous proposition then equips
\(S\) with an orientation, so \(S\) is orientable.
The orientation produced here depends on the auxiliary metric only through the side to which the
gradient points, and reversing the sign of \(f\) reverses it; the orientability of \(S\), however,
is intrinsic and metric-independent. This result will reappear in concrete form once boundaries
enter the picture, since a boundary is the level set of a defining function and its outward
direction plays 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 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 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
\[
\iota^{*}\bigl(N \lrcorner\, \omega\bigr).
\]
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.
Proof.
Since \(\partial M\) is an embedded hypersurface, any outward-pointing vector field \(N\)
along it is in particular nowhere tangent to \(\partial M\): an outward-pointing vector is by
definition not tangent to the boundary. The construction for transverse fields therefore
applies and equips \(\partial M\) with an orientation whose orientation form is
\(\iota^{*}(N \lrcorner\, \omega)\). It remains to show this orientation does not depend on
which outward-pointing field is used.
Let \(N\) and \(\widetilde N\) be two 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:
\(N^n(p) < 0\) and \(\widetilde N^{\,n}(p) < 0\). 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,
\[
\frac{\widetilde N^{\,n}(p)}{N^n(p)} > 0,
\]
a quotient of two negative numbers. The orientations therefore agree at \(p\); as \(p\) was
arbitrary and everything varies continuously, they agree on all of \(\partial M\). In
particular \(\partial M\) is orientable.
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, and the discrepancy that appears there is
the source of the alternating sign in the boundary term of the fundamental integration theorem.
Consider the upper half-space \(\mathbb{H}^n = \{x^n \geq 0\}\) with its standard orientation
inherited from \(\mathbb{R}^n\). Its boundary is \(\partial \mathbb{H}^n = \{x^n = 0\}\),
naturally identified with \(\mathbb{R}^{n-1}\) through the coordinates \((x^1, \dots, x^{n-1})\).
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 \((-1)\cdot(-1)^{n-1} = (-1)^n\):
\[
\bigl[-\partial_{x^n}, \partial_{x^1}, \dots, \partial_{x^{n-1}}\bigr]
= (-1)^n \bigl[\partial_{x^1}, \dots, \partial_{x^{n-1}}, \partial_{x^n}\bigr].
\]
Hence 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. This factor of
\((-1)^n\) is exactly what the boundary term of the integration theorem carries, 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 neighbourhood 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 an orientation-preserving diffeomorphism onto an open subset
of \(M\), where \(U \subseteq \mathbb{R}^{n-1}\) is open and \(F\) carries \(\{b\} \times U\)
into \(\partial M\). Define \(f : U \to \partial M\) by \(f(x) = F(b, x)\). Then \(f\)
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.
Give \((a, b] \times U\) the orientation 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
product orientation of \(U\) is the boundary orientation. Now \(F\) maps this model with its
outward direction
\(+\partial_t\) to \(M\) with an outward direction along \(\partial M\), and because \(F\) is
a diffeomorphism it carries the boundary orientation of \(\{b\} \times U\) to the boundary
orientation of \(\partial M\) exactly when it preserves the ambient orientations. Restricting
\(F\) to \(\{b\} \times U\) is the map \(f\), so \(f\) preserves the Stokes orientation if and
only if \(F\) preserves the orientation of \(M\).
As an illustration, spherical coordinates orient the two-sphere correctly. Parametrize a
neighbourhood in \(\mathbb{R}^3 \setminus \{0\}\) by
\(F(\rho, \varphi, \theta) = (\rho \sin\varphi \cos\theta, \rho \sin\varphi \sin\theta,
\rho \cos\varphi)\), whose Jacobian determinant is \(\rho^2 \sin\varphi > 0\) on the relevant
range, so \(F\) is orientation-preserving. Viewing \(S^2\) as the boundary reached as \(\rho\)
increases to \(1\), 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 measurement of size on such oriented manifolds, through a distinguished volume form
attached to a metric, is what remains to assemble the integral that the boundary orientation was
designed to serve.