Riemannian Volume Form

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The Volume Form and Its Coordinate Expression Hypersurfaces and Boundaries

The Volume Form and Its Coordinate Expression

An orientation tells us which ordered frames count as positive, but it says nothing about size. Two oriented frames are interchangeable as far as orientation is concerned, however much they differ in length or angle. A Riemannian metric supplies exactly the missing notion of size, assigning a length to every tangent vector and an angle to every pair.

The combination of an orientation and a metric singles out, at each point, a unique top-degree alternating tensor that evaluates to \(1\) on any positively oriented orthonormal frame, and such a frame therefore spans a unit of \(n\)-dimensional volume. Globalized into a nowhere-vanishing \(n\)-form, this object is what allows functions on the manifold to be integrated against a geometrically meaningful measure.

This construction is the geometric foundation on which invariant integration rests, and the present page builds the form itself rather than the analytic theory that follows from it.

Let \((M, g)\) be an oriented Riemannian \(n\)-manifold, with or without boundary, with \(n \geq 1\). Oriented orthonormal frames exist near every point. Given \(p \in M\), there is a smooth orthonormal frame \((E_1, \dots, E_n)\) on a neighborhood \(U\) of \(p\), by the existence of local orthonormal frames. Shrinking \(U\), we may take it connected, for instance the preimage under a chart of a small ball or half-ball. If \(\omega\) is an orientation form for \(M\), the function \(\omega(E_1, \dots, E_n)\) is continuous and nowhere zero on \(U\), hence of constant sign, and replacing \(E_1\) by \(-E_1\) if necessary makes the frame positively oriented.

Proposition (Existence and Uniqueness of the Volume Form)

Let \((M, g)\) be an oriented Riemannian \(n\)-manifold, with or without boundary, with \(n \geq 1\). There is a unique smooth orientation form \(\omega_g \in \Omega^n(M)\), called the Riemannian volume form, satisfying \[ \omega_g(E_1, \dots, E_n) = 1 \] for every local oriented orthonormal frame \((E_1, \dots, E_n)\) for \(M\).

Proof.

For uniqueness, suppose such a form \(\omega_g\) exists. If \((E_1, \dots, E_n)\) is any local oriented orthonormal frame on an open set \(U \subseteq M\) and \((\varepsilon^1, \dots, \varepsilon^n)\) is the dual coframe, then on \(U\) we may write \(\omega_g = f\, \varepsilon^1 \wedge \cdots \wedge \varepsilon^n\) for a smooth function \(f\). Evaluating on the frame and using the defining condition forces \(f = 1\), so \[ \omega_g = \varepsilon^1 \wedge \cdots \wedge \varepsilon^n \] on \(U\). Since every point lies in the domain of some such frame, as shown above, this determines \(\omega_g\) uniquely.

For existence, define \(\omega_g\) near each point by the displayed formula with respect to a chosen oriented orthonormal frame, and check that the result is independent of the choice. Let \((\widetilde E_1, \dots, \widetilde E_n)\) be another oriented orthonormal frame, defined on an open set meeting the first, and compare the two on the overlap. Give it the dual coframe \((\widetilde\varepsilon^1, \dots, \widetilde\varepsilon^n)\), and set \(\widetilde\omega_g = \widetilde\varepsilon^1 \wedge \cdots \wedge \widetilde\varepsilon^n\). Write \[ \widetilde E_i = A_i^{\,j} E_j \] for a matrix \((A_i^{\,j})\) of smooth functions. Because both frames are orthonormal, the matrix \((A_i^{\,j}(p))\) lies in the orthogonal group at each \(p\), so its determinant is \(\pm 1\). Both frames are positively oriented, so the determinant must be \(+1\). Using the expression of a top-degree alternating tensor as a determinant, \[ \omega_g(\widetilde E_1, \dots, \widetilde E_n) = \det\!\bigl(\varepsilon^j(\widetilde E_i)\bigr) = \det(A_i^{\,j}) = 1 = \widetilde\omega_g(\widetilde E_1, \dots, \widetilde E_n). \]

Hence \(\omega_g = \widetilde\omega_g\), and the locally defined forms patch into a single global smooth \(n\)-form. It is nowhere vanishing, since it evaluates to \(1\) on a frame, and by construction satisfies the defining condition for every oriented orthonormal frame.

The frame description is conceptually clean but rarely the form in which one computes. In coordinates the volume form acquires the square-root-of-determinant factor familiar from the change-of-variables formula in multivariable calculus, now intrinsic to the geometry.

Proposition (Coordinate Expression of the Volume Form)

Let \((M, g)\) be an oriented Riemannian \(n\)-manifold, with or without boundary, with \(n \geq 1\). In any oriented smooth coordinates \((x^i)\), the Riemannian volume form has the local expression \[ \omega_g = \sqrt{\det(g_{ij})}\, dx^1 \wedge \cdots \wedge dx^n, \] where \(g_{ij}\) are the components of \(g\) in these coordinates.

Proof.

In an oriented chart \((U, (x^i))\), write \(\omega_g = f\, dx^1 \wedge \cdots \wedge dx^n\) for a positive coefficient function \(f\), positive because both \(\omega_g\) and the coordinate volume element are positively oriented. To compute \(f\), let \((E_i)\) be a smooth oriented orthonormal frame on a neighborhood, let \((\varepsilon^i)\) be its dual coframe, and express the coordinate frame in terms of it as \[ \frac{\partial}{\partial x^i} = A_i^{\,j} E_j. \] Evaluating \(\omega_g = \varepsilon^1 \wedge \cdots \wedge \varepsilon^n\) on the coordinate frame gives \[ f = \omega_g\!\left(\frac{\partial}{\partial x^1}, \dots, \frac{\partial}{\partial x^n}\right) = \det\!\bigl(\varepsilon^j(\tfrac{\partial}{\partial x^i})\bigr) = \det(A_i^{\,j}). \]

On the other hand, since \((E_k)\) is orthonormal, \[ g_{ij} = \left\langle \frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j} \right\rangle_g = \bigl\langle A_i^{\,k} E_k,\, A_j^{\,l} E_l \bigr\rangle_g = \sum_k A_i^{\,k} A_j^{\,k}, \] which is the \((i, j)\)-entry of \(A A^{\top}\) for \(A = (A_i^{\,j})\). Therefore \[ \det(g_{ij}) = \det(A A^{\top}) = (\det A)^2, \] so \(f = \det A = \pm\sqrt{\det(g_{ij})}\). Both frames being oriented forces the positive sign, so \(f = \sqrt{\det(g_{ij})}\).

The coefficient \(\sqrt{\det(g_{ij})}\) is precisely the factor by which the metric distorts coordinate volume, and its appearance here explains why the same expression governs integration in curvilinear coordinates. With a single global volume form in hand, the next step is to see how it induces volume forms on hypersurfaces and, in particular, on boundaries, where it produces the surface measure that pairs with the Stokes orientation.

Hypersurfaces and Boundaries

When a hypersurface sits inside a Riemannian manifold, it inherits a metric by restriction, and that induced metric has its own volume form. There is a pleasant formula relating the two. The surface volume form is the ambient volume form contracted against a unit normal, restricted to the hypersurface. This is the same contraction that orients the hypersurface, now carrying metric information as well as orientation.

Proposition (Volume Form of a Hypersurface)

Let \((M, g)\) be an oriented Riemannian manifold, with or without boundary, let \(S \subseteq M\) be an immersed hypersurface, with or without boundary, with inclusion \(\iota_S\), and let \(\widetilde g = \iota_S^{*} g\) be the induced metric on \(S\), the pullback metric determined by the inclusion. Suppose \(N\) is a smooth unit normal vector field along \(S\). With respect to the orientation of \(S\) determined by \(N\), the volume form of \((S, \widetilde g)\) is \[ \omega_{\widetilde g} = \iota_S^{*}\bigl(N \lrcorner\, \omega_g\bigr). \]

Proof.

At each \(p \in S\), the unit vector \(N_p\) is nonzero and orthogonal to \(d(\iota_S)_p(T_pS)\), hence not tangent to \(S\), so \(N\) is a smooth vector field along \(S\) that is nowhere tangent to \(S\). By the transverse-field construction, the \((n-1)\)-form \(\iota_S^{*}(N \lrcorner\, \omega_g)\) is the orientation form for the orientation of \(S\) determined by \(N\), so it is nowhere vanishing and positively oriented. To identify it as the volume form of the induced metric, it suffices to show it takes the value \(1\) on every oriented orthonormal frame for \(S\).

Let \((E_1, \dots, E_{n-1})\) be such a frame at a point \(p \in S\). Since \(N_p\) is a unit vector orthogonal to \(T_pS\), the enlarged frame \((N_p, E_1, \dots, E_{n-1})\) is orthonormal in \(T_pM\), and it is positively oriented for the ambient orientation, by the very definition of the orientation \(N\) induces on \(S\). Therefore \[ \iota_S^{*}(N \lrcorner\, \omega_g)(E_1, \dots, E_{n-1}) = \omega_g(N_p, E_1, \dots, E_{n-1}) = 1, \] the last equality because \(\omega_g\) evaluates to \(1\) on any oriented orthonormal frame. A form that gives \(1\) on every oriented orthonormal frame of \((S, \widetilde g)\) is its volume form, which proves the claim.

When \(n = 1\) the hypersurface \(S\) is a \(0\)-manifold, to which the first proposition does not apply. We adopt the convention that the volume form of an oriented \(0\)-manifold is the function assigning to each point its orientation sign \(\pm 1\). The formula then holds without change. Here \(\iota_S^{*}(N \lrcorner\, \omega_g)\) is the function \(p \mapsto \omega_g(N_p)\). Since \(N_p\) is a unit vector in the one-dimensional space \(T_pM\), it equals \(\pm E_1\) for an oriented orthonormal basis \(E_1\) of \(T_pM\), so \(\omega_g(N_p) = \pm 1\). The sign is \(+1\) exactly when \(N_p\) is positively oriented, which is exactly when the orientation determined by \(N\) assigns \(+1\) to \(p\). Once integration over an oriented \(0\)-manifold is defined as a signed sum, integrating a function \(h\) against this volume form returns the unsigned sum \(\sum_p h(p)\), as a measure of size should.

The case that matters most is a boundary, where the ambient manifold itself supplies a canonical unit normal. For an arbitrary hypersurface a unit normal involves a choice of side and might not even exist globally, but a boundary has an unambiguous outward direction, and the metric promotes that direction to a distinguished unit field.

Proposition (The Outward Unit Normal)

Let \(M\) be a Riemannian manifold with boundary. There is a unique smooth outward-pointing unit normal vector field \(N\) along \(\partial M\).

Proof.

For uniqueness, fix \(p \in \partial M\). The orthogonal complement \((T_p \partial M)^{\perp} \subseteq T_pM\) is one-dimensional, so there are exactly two unit vectors at \(p\) normal to \(\partial M\). A unit normal is nowhere tangent to \(\partial M\), so it has nonzero final component in any smooth boundary chart. Of the two unit normals, exactly one has negative final component, which is the same as pointing outward. Thus the outward-pointing unit normal is unique at each point.

For existence, let \(f : M \to [0, \infty)\) be a boundary defining function, which exists, and set \[ N = -\,\frac{\operatorname{grad} f}{|\operatorname{grad} f|_g} \] restricted to \(\partial M\). Because \(df \neq 0\) at points of \(\partial M\), the gradient is nonzero there. The quotient above is a smooth vector field of unit length on the open set \(\{\operatorname{grad} f \neq 0\}\), which contains \(\partial M\), so \(N\) is a smooth unit vector field along \(\partial M\). It is normal to \(\partial M\). For any \(w \in T_p \partial M\), the gradient satisfies \(g_p(\operatorname{grad} f_p, w) = df_p(w)\), where \(w\) is identified with its image under the inclusion \(\iota_{\partial M}\), and \(df_p(w) = d(f \circ \iota_{\partial M})_p(w) = 0\) because \(f \circ \iota_{\partial M} \equiv 0\). Hence \(\operatorname{grad} f_p\), and with it \(N_p\), is orthogonal to \(T_p\partial M\).

Applying \(N\) to \(f\), we obtain \[ N f = -\,\frac{g(\operatorname{grad} f, \operatorname{grad} f)}{|\operatorname{grad} f|_g} = -\,|\operatorname{grad} f|_g \lt 0 \] at every point of \(\partial M\). The proposition that provides \(f\) also shows that a vector \(v\) at a boundary point with \(vf \lt 0\) is outward-pointing. Hence \(N\) is outward-pointing, and by uniqueness it is the outward unit normal, independent of the choice of defining function.

Combining the surface volume formula with the outward unit normal yields the boundary measure in its final form. The result, stated in the corollary below, is the volume form of the boundary, positively oriented for the Stokes orientation. Integrating against it gives the surface measure used for the boundary terms of Riemannian integration theorems such as the divergence theorem.

Corollary (Volume Form of a Boundary)

Let \((M, g)\) be an oriented Riemannian manifold with boundary, and let \(\widetilde g\) be the induced Riemannian metric on \(\partial M\). Then the volume form of \(\widetilde g\), taken with respect to the Stokes orientation, is \[ \omega_{\widetilde g} = \iota_{\partial M}^{*}\bigl(N \lrcorner\, \omega_g\bigr), \] where \(N\) is the outward unit normal vector field along \(\partial M\).

Proof.

The boundary is an embedded hypersurface, the outward unit normal is a smooth unit normal field along it, and the orientation it induces is by definition the Stokes orientation. Applying the hypersurface volume formula with this normal gives the stated expression.

The volume form thus assembled is the geometric measure that turns an oriented Riemannian manifold into a space on which functions can be integrated. Its contraction with the outward unit normal, pulled back to the boundary, is matched to the induced orientation so that the boundary integral and the interior integral fit together. On a positive-dimensional Lie group carrying a left-invariant metric and a left-invariant orientation, each left translation \(L_h\) is an orientation-preserving isometry. It therefore carries oriented orthonormal frames to oriented orthonormal frames, so \(L_h^{*}\omega_g\) takes the value \(1\) on every such frame, and uniqueness gives \(L_h^{*}\omega_g = \omega_g\). The invariant integral on a compact Lie group, which underlies harmonic analysis on the group, is integration against a normalized left-invariant top form of exactly this kind.