dovs

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 ℝ, where V = ℝⁿ (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:

set.seed(0)
a <- rform(n=4,k=2)
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 (ℝ⁴)² to ℝ:

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

##      [,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 (ℝ⁵)² to ℝ:

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

##      [,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: (ℝ⁵)² → ℝ, we observe that row 5 is ignored: e₅ = (0, 0, 0, 0, 1)^T maps to zero in the sense that f(e₅, v) = f(v, e₅) = 0, for any v ∈ ℝ⁵.

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

##      [,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 e₅ in the second column with the same result. Alternatively we see that the k-form a, evaluated with e₅ 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 (ℝ³)¹ to ℝ; but:

dovs(dx)

## [1] 1

So according to stokes, dx : (ℝ¹)¹ → ℝ. This does not really matter numerically, until we consider the Hodge star operator. We know that ⋆dx  = dy  ∧ dz , but

hodge(dx)

## [1] 1

Above we see the package giving, correctly, that the Hodge star of dx  is the zero-dimensional volume element (otherwise known as “1”). To get the answer appropriate if dx  is considered as a map from (ℝ³)¹ to ℝ [that is, dx : (ℝ³)¹ → 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