Function dovs() function in the stokes package

Robin K. S. Hankin

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 \(\mathbb{R}\), 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:

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 \(\left(\mathbb{R}^4\right)^2\) to \(\mathbb{R}\):

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 \(\left(\mathbb{R}^5\right)^2\) to \(\mathbb{R}\):

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 \(a\colon\left(\mathbb{R}^5\right)^2\longrightarrow\mathbb{R}\), 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 \(\mathbf{v}\in\mathbb{R}^5\).

(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 \(e_5\) 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