stokes (version 1.2-0)

volume: The volume element

Description

The volume element in n dimensions

Usage

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

Value

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

Arguments

n

Dimension of the space

K

Object of class kform

Author

Robin K. S. Hankin

Details

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

If V has dimension n, it follows that ^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 ^n(V). it is not surprising to find it in the following theorem:

Let v_1,...,v_nv_1,...,v_n be a basis for V and let ^n(V).. If w_i=_j=1^n a_ijv_j. then

(w_1,...,w_n)=(a_ij)(v_1,... v_n)omitted; see PDF

(see the examples for numerical verification of this).

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

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))