volume: The volume element

Description

The volume element in $n$ dimensions

Usage

volume(n)
is.volume(K)

Arguments

Dimension of the space

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.

References

Spivak

Examples

Run this code

# NOT RUN {
as.kform(1) %^% as.kform(2) %^% as.kform(3)  == volume(3)  # should be TRUE

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


# }