volume()
in the
Stokes packagefunction (n)
{
as.kform(seq_len(n))
}
To cite the stokes
package in publications, please use
Hankin (2022). Spivak (1965), in a memorable passage,
states:
The volume element
The fact that \(\operatorname{dim}\Lambda^n\left(\mathbb{R}^n\right)=1\) is probably not new to you, since \(\operatorname{det}\) is often defined as the unique element \(\omega\in\Lambda^n{\left(\mathbb{R}^n\right)}\) such that \(\omega{\left(e_1,\ldots,e_n\right)}=1\). For a general vector space \(V\) there is no extra criterion of this sort to distinguish a particular \(\omega\in\Lambda^n{\left(\mathbb{R}^n\right)}\). Suppose, however, that an inner product \(T\) for \(V\) is given. If \(v_1,\ldots,v_n\) and \(w_1,\ldots, w_n\) are two bases which are orthonormal with respect to \(T\), and the matrix \(A=\left(a_{ij}\right)\) is defined by \(w_i=\sum_{j=1}^n a_{ij}v_j\), then
\[\delta_{ij}=T{\left(w_i,w_j\right)}= \sum_{k,l=1}^n a_{ik}a_{jl}\,T{\left(v_k,v_l\right)}= \sum_{k=1}^n a_{ik}a_{jk}.\]
In other words, if \(A^T\) denotes the transpose of the matrix \(A\), then we have \(A\cdot A^T=I\), so \(\operatorname{det}A=\pm 1\). It follows from Theorem 4-6 [see vignettedet.Rmd
] that if \(\omega\in\Lambda^n(V)\) satisfies \(\omega{\left(v_1,\ldots,v_n\right)}=\pm
1\), then \(\omega{\left(w_1,\ldots,w_n\right)}=\pm
1\). If an orientation \(\mu\)
for \(V\) has also been given, it
follows that there is a unique \(\omega\in\Lambda^n(V)\) such that \(\omega\left(v_1,\ldots,v_n\right)=1\)
whenever \(v_1,\ldots,v_n\) is an
orthornormal basis such that \(\left[v_1,\ldots,v_n\right]=\mu\). This
unique \(\omega\) is called the
volume element of \(V\),
determined by the inner product \(T\)
and orientation \(\mu\). Note that
\(\operatorname{det}\) is the volume
element of \(\mathbb{R}^n\) determined
by the usual inner product and usual orientation, and that \(\left|\operatorname{det}\left(v_1,\ldots,v_n\right)\right|\)
is the volume of the parallelepiped spanned by the line segments from
\(0\) to each of \(v_1,\ldots,v_n\).
- Michael Spivak, 1969 (Calculus on Manifolds, Perseus books). Page 83
In the stokes
package, function volume(n)
returns the volume element on the usual basis, that is, \(\omega{\left(e_1,\ldots,e_n\right)}\). We
will take \(n=7\) as an example:
## An alternating linear map from V^7 to R with V=R^7:
## val
## 1 2 3 4 5 6 7 = 1
We can verify Spivak’s reasoning as follows:
## [1] 1
Above, we see that \(\omega{\left(e_1,\ldots,e_n\right)}=1\). To verify that \(V{\left(v_1,\ldots,v_n\right)}=\operatorname{det}(A)\), where \(A_{ij}=\left(v_i\right)_j\):
## LHS RHS diff
## 1.770074 1.770074 0.000000
Now we create \(w_1,\ldots,w_n\), another orthonormal set. We may verify by generating a random orthogonal matrix and permuting its rows:
M1 <- qr.Q(qr(matrix(rnorm(49),7,7))) # M1: a random orthogonal matrix
M2 <- M1[c(2,1,3,4,5,6,7),] # M2: (odd) permutation of rows of M1
c(f(M1),f(M2))
## [1] 1 -1
Above we see that the volume element of M1
and
M2
are \(\pm1\) to within
numerical precision.