`dovs()`

function in the `stokes`

package: the dimension of the underlying vector space

` dovs`

```
## function (K)
## {
## if (is.zero(K) || is.scalar(K)) {
## return(0)
## }
## else {
## return(max(index(K)))
## }
## }
## <bytecode: 0x560d83e4ede8>
## <environment: namespace:stokes>
```

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 \(V=\mathbb{R}^n\). 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:

```
set.seed(0)
rform(n=4,k=2)
a <- a
```

```
## 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 \(\mathbb{R}\):

```
as.function(a)
f <- matrix(1:8,4,2)) (M <-
```

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

`f(M)`

`## [1] -148`

However, `a`

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

```
as.function(a)
f <- matrix(c(1,2,3,4,1454,5,6,7,8,-9564),ncol=2)) # row 5 large numbers (M <-
```

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

`f(M)`

`## [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 \(\mathbf{v}\in\mathbb{R}^5\).

` cbind(c(0,0,0,0,1),runif(5))) (M <-`

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

`f(M)`

`## [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:

` dx`

```
## 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:

`dovs(dx)`

`## [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

`hodge(dx)`

`## [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:

`hodge(dx,3)`

```
## 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:

```
options(kform_symbolic_print="dx")
hodge(dx,3)
```

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