The volume element

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.

References

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

Author

Robin K. S. Hankin

See also

Examples

dx^dy^dz == volume(3) 
#> [1] TRUE

p <- 1
for(i in 1:7){p <- p ^ as.kform(i)}
p
#> An alternating linear map from V^7 to R with V=R^7:
#>                    val
#>  1 2 3 4 5 6 7  =    1
p == volume(7)  # should be TRUE
#> [1] TRUE

o <- volume(5)
M <- matrix(runif(25),5,5)
det(M) - as.function(o)(M)   # should be zero
#> [1] 0


is.volume(d(1) ^ d(2) ^ d(3) ^ d(4))
#> [1] TRUE
is.volume(d(1:9))
#> [1] TRUE