Volume element

Volume element of a linear subspace

Consider the linear subspace of the n-dimensional Euclidean space Rⁿ that is spanned by a collection of linearly independent vectors $${\displaystyle X_1,\dots ,X_k.}$$ To find the volume element of the subspace, it is useful to know the fact from linear algebra that the volume of the parallelepiped spanned by the ${\displaystyle X_i}$ is the square root of the determinant of the Gramian matrix of the ${\displaystyle X_i}$: $${\displaystyle \sqrt{det(X_i\cdot X_j{)}_{i,j=1\dots k}}.}$$

Any point p in the subspace can be given coordinates ${\displaystyle (u_1,u_2,\dots ,u_k)}$ such that $${\displaystyle p=u_1{X}_1+\cdots +u_k{X}_k.}$$ At a point p, if we form a small parallelepiped with sides ${\displaystyle \mathrm{d}{u}_i}$, then the volume of that parallelepiped is the square root of the determinant of the Grammian matrix $${\displaystyle \sqrt{det{\left((d{u}_i{X}_i)\cdot (d{u}_j{X}_j)\right)}_{i,j=1\dots k}}=\sqrt{det(X_i\cdot X_j{)}_{i,j=1\dots k}}\mathrm{d}{u}_1\mathrm{d}{u}_2\cdots \mathrm{d}{u}_k.}$$ This therefore defines the volume form in the linear subspace.

Volume element of manifolds

On an oriented Riemannian manifold of dimension n, the volume element is a volume form equal to the Hodge dual of the unit constant function, ${\displaystyle f(x)=1}$: $${\displaystyle \omega =\star 1.}$$ Equivalently, the volume element is precisely the Levi-Civita tensor ${\displaystyle ϵ}$. In coordinates, $${\displaystyle \omega =ϵ=\sqrt{\left|detg\right|}\mathrm{d}{x}^1\wedge \cdots \wedge \mathrm{d}{x}^n}$$ where ${\displaystyle detg}$ is the determinant of the metric tensor g written in the coordinate system.

Area element of a surface

A simple example of a volume element can be explored by considering a two-dimensional surface embedded in n-dimensional Euclidean space. Such a volume element is sometimes called an area element. Consider a subset ${\displaystyle U\subset {\mathbb{R}}^2}$ and a mapping function $${\displaystyle \phi :U\to {\mathbb{R}}^n}$$ thus defining a surface embedded in ${\displaystyle {\mathbb{R}}^n}$. In two dimensions, volume is just area, and a volume element gives a way to determine the area of parts of the surface. Thus a volume element is an expression of the form $${\displaystyle f(u_1,u_2)\mathrm{d}{u}_1\mathrm{d}{u}_2}$$ that allows one to compute the area of a set B lying on the surface by computing the integral $${\displaystyle \mathrm{Area}(B)={\int }_{B}f(u_1,u_2)\mathrm{d}{u}_1\mathrm{d}{u}_2.}$$

Here we will find the volume element on the surface that defines area in the usual sense. The Jacobian matrix of the mapping is $${\displaystyle J_{ij}=\frac{\mathrm{\partial }{\phi }_i}{\mathrm{\partial }{u}_j}}$$ with index i running from 1 to n, and j running from 1 to 2. The Euclidean metric in the n-dimensional space induces a metric ${\displaystyle g=J^{T}J}$ on the set U, with matrix elements $${\displaystyle g_{ij}=\sum _{k=1}^n{J}_{ki}{J}_{kj}=\sum _{k=1}^n\frac{\mathrm{\partial }{\phi }_k}{\mathrm{\partial }{u}_i}\frac{\mathrm{\partial }{\phi }_k}{\mathrm{\partial }{u}_j}.}$$

The determinant of the metric is given by $${\displaystyle detg={\left|\frac{\mathrm{\partial }\phi }{\mathrm{\partial }{u}_1}\wedge \frac{\mathrm{\partial }\phi }{\mathrm{\partial }{u}_2}\right|}^2=det(J^{T}J)}$$

For a regular surface, this determinant is non-vanishing; equivalently, the Jacobian matrix has rank 2.

Now consider a change of coordinates on U, given by a diffeomorphism $${\displaystyle f:U\to U,}$$ so that the coordinates ${\displaystyle (u_1,u_2)}$ are given in terms of ${\displaystyle (v_1,v_2)}$ by ${\displaystyle (u_1,u_2)=f(v_1,v_2)}$. The Jacobian matrix of this transformation is given by $${\displaystyle F_{ij}=\frac{\mathrm{\partial }{f}_i}{\mathrm{\partial }{v}_j}.}$$

In the new coordinates, we have $${\displaystyle \frac{\mathrm{\partial }{\phi }_i}{\mathrm{\partial }{v}_j}=\sum _{k=1}^2\frac{\mathrm{\partial }{\phi }_i}{\mathrm{\partial }{u}_k}\frac{\mathrm{\partial }{f}_k}{\mathrm{\partial }{v}_j}}$$ and so the metric transforms as $${\displaystyle \tilde{g}=F^{T}gF}$$ where ${\displaystyle \tilde{g}}$ is the pullback metric in the v coordinate system. The determinant is $${\displaystyle det\tilde{g}=detg{\left(detF\right)}^2.}$$

Given the above construction, it should now be straightforward to understand how the volume element is invariant under an orientation-preserving change of coordinates.

In two dimensions, the volume is just the area. The area of a subset ${\displaystyle B\subset U}$ is given by the integral $${\displaystyle \begin{array}{cc}\text{Area}(B)& ={\iint }_B\sqrt{detg}\mathrm{d}{u}_1\mathrm{d}{u}_2\\ & ={\iint }_B\sqrt{detg}\left|detF\right|\mathrm{d}{v}_1\mathrm{d}{v}_2\\ & ={\iint }_B\sqrt{det\tilde{g}}\mathrm{d}{v}_1\mathrm{d}{v}_2.\end{array}}$$

Thus, in either coordinate system, the volume element takes the same expression: the expression of the volume element is invariant under a change of coordinates.

Note that there was nothing particular to two dimensions in the above presentation; the above trivially generalizes to arbitrary dimensions.

Example: Sphere

For example, consider the sphere with radius r centered at the origin in R³. This can be parametrized using spherical coordinates with the map $${\displaystyle \varphi (u_1,u_2)=(r\cos u_1\sin u_2,r\sin u_1\sin u_2,r\cos u_2).}$$ Then $${\displaystyle g=\left(\begin{array}{cc}{r}^2{\sin }^2{u}_2& 0\\ 0& r^2\end{array}\right),}$$ and the area element is $${\displaystyle \omega =\sqrt{detg}\mathrm{d}{u}_1\mathrm{d}{u}_2=r^2\sin u_2\mathrm{d}{u}_1\mathrm{d}{u}_2.}$$