`dovs()`

function in
the `stokes`

package```
function (K)
{
if (is.zero(K) || is.scalar(K)) {
return(0)
}
else {
return(max(index(K)))
}
}
```

To cite the `stokes`

package in publications, please use
Hankin (2022). Function
`dovs()`

returns the dimensionality of the underlying vector
space of a \(k\)-form. Recall that a
\(k\)-form is an alternating linear map
from \(V^k\) to \(\mathbb{R}\), where \(V=\mathbb{R}^n\) (Spivak 1965). Function `dovs()`

returns \(n\) [compare
`arity()`

, which returns \(k\)]. As seen above, the function is very
simple, essentially being `max(index(K))`

, but its use is not
entirely straightforward in the context of `stokes`

idiom.
Consider the following:

```
## An alternating linear map from V^2 to R with V=R^4:
## val
## 2 4 = 9
## 1 4 = 8
## 2 3 = 1
## 1 3 = -3
## 3 4 = -2
## 1 2 = 2
```

Now object `a`

is notionally a map from \(\left(\mathbb{R}^4\right)^2\) to \(\mathbb{R}\):

```
## [,1] [,2]
## [1,] 1 5
## [2,] 2 6
## [3,] 3 7
## [4,] 4 8
```

`## [1] -148`

However, `a`

can equally be considered to be a map from
\(\left(\mathbb{R}^5\right)^2\) to
\(\mathbb{R}\):

```
## [,1] [,2]
## [1,] 1 5
## [2,] 2 6
## [3,] 3 7
## [4,] 4 8
## [5,] 1454 -9564
```

`## [1] -148`

If we view \(a\) [or indeed
`f()`

] in this way, that is \(a\colon\left(\mathbb{R}^5\right)^2\longrightarrow\mathbb{R}\),
we observe that row 5 is ignored: \(e_5=\left(0,0,0,0,1\right)^T\) maps to zero
in the sense that \(f(e_5,\mathbf{v})=f(\mathbf{v},e_5)=0\),
for any \(\mathbf{v}\in\mathbb{R}^5\).

```
## [,1] [,2]
## [1,] 0 0.3800352
## [2,] 0 0.7774452
## [3,] 0 0.9347052
## [4,] 0 0.2121425
## [5,] 1 0.6516738
```

`## [1] 0`

(above we see that rows 1-4 of `M`

are ignored because of
the zero in column 1; row 5 is ignored because the index of
`a`

does not include the number 5). Because `a`

is
alternating, we could have put \(e_5\)
in the second column with the same result. Alternatively we see that the
\(k\)-form `a`

, evaluated
with \(e_5\) as one of its arguments,
returns zero because the index matrix of `a`

does not include
the number 5. Most of the time, this kind of consideration does not
matter. However, consider this:

```
## An alternating linear map from V^1 to R with V=R^1:
## val
## 1 = 1
```

Now, *we* know that `dx`

is supposed to be a map
from \(\left(\mathbb{R}^3\right)^1\) to
\(\mathbb{R}\); but:

`## [1] 1`

So according to `stokes`

, \(\operatorname{dx}\colon\left(\mathbb{R}^1\right)^1\longrightarrow\mathbb{R}\).
This does not really matter numerically, until we consider the Hodge
star operator. We know that \(\star\operatorname{dx}=\operatorname{dy}\wedge\operatorname{dz}\),
but

`## [1] 1`

Above we see the package giving, correctly, that the Hodge star of
\(\operatorname{dx}\) is the
zero-dimensional volume element (otherwise known as “1”). To get the
answer appropriate if \(\operatorname{dx}\) is considered as a map
from \(\left(\mathbb{R}^3\right)^1\) to
\(\mathbb{R}\) [that is, \(\operatorname{dx}\colon\left(\mathbb{R}^3\right)^1\longrightarrow\mathbb{R}\)],
we need to specify `dovs`

explicitly:

```
## An alternating linear map from V^2 to R with V=R^3:
## val
## 2 3 = 1
```

Actually this looks a lot better with a more appropriate print method:

```
## An alternating linear map from V^2 to R with V=R^3:
## + dy^dz
```

Hankin, R. K. S. 2022. “Stokes’s Theorem in
R.” arXiv. https://doi.org/10.48550/ARXIV.2210.17008.

Spivak, M. 1965. *Calculus on Manifolds*. Addison-Wesley.