`volume()`

in the Stokes package

` volume`

```
## function (n)
## {
## as.kform(seq_len(n))
## }
## <bytecode: 0x7fc3a58dc838>
## <environment: namespace:stokes>
```

Spivak, in a memorable passage, states (p83):

**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 vignette`det.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
**- Michael Spivak, 1969** *(Calculus on Manifolds, Perseus books). Page 89*

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:

`<- volume(7)) (V `

```
## 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:

```
<- as.function(V)
f f(diag(7))
```

`## [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 \(A_{ij}=\left(v_i\right)_j\):

```
<- matrix(rnorm(49),7,7)
A <- f(A)
LHS <- det(A)
RHS c(LHS=LHS,RHS=RHS,diff=LHS-RHS)
```

```
## 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 permutiting its rows:

```
<- qr.Q(qr(matrix(rnorm(49),7,7))) # M1: a random orthogonal matrix
M1 <- M1[c(2,1,3,4,5,6,7),] # M2: (odd) permutation of rows of M1
M2 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.