volume {stokes}

The volume element

Description

The volume element in n dimensions

Usage

volume(n)
is.volume(K,n=dovs(K))

Arguments

n

Dimension of the space

K

Object of class kform

Details

Spivak phrases it well (theorem 4.6, page 82):

If V has dimension n, it follows that \Lambda^n(V) has dimension 1. Thus all alternating n-tensors on V are multiples of any non-zero one. Since the determinant is an example of such a member of \Lambda^n(V) it is not surprising to find it in the following theorem:

Let v_1,\ldots,v_n be a basis for V and let \omega\in\Lambda^n(V). If w_i=\sum_{j=1}^n a_{ij}v_j then

\omega\left(w_1,\ldots,w_n\right)=\det\left(a_{ij}\right)\cdot\omega\left(v_1,\ldots v_n\right)

(see the examples for numerical verification of this).

Neither the zero k-form, nor scalars, are considered to be a volume element.

Value

Function volume() returns an object of class kform; function is.volume() returns a Boolean.

Author(s)

Robin K. S. Hankin

References

• M. Spivak 1971. Calculus on manifolds, Addison-Wesley

See Also

zeroform,as.1form,dovs

Examples

dx^dy^dz == volume(3) 

p <- 1
for(i in 1:7){p <- p ^ as.kform(i)}
p
p == volume(7)  # should be TRUE

o <- volume(5)
M <- matrix(runif(25),5,5)
det(M) - as.function(o)(M)   # should be zero


is.volume(d(1) ^ d(2) ^ d(3) ^ d(4))
is.volume(d(1:9))