The dovs() function in the stokes package: the dimension of the underlying vector space

Robin K. S. Hankin

dovs

## function (K) 
## {
##     max(index(K))
## }
## <bytecode: 0x55abc1bff3f0>
## <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 \(n\). The function is, as seen above, very simple 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(1:10,5,2))

##      [,1] [,2]
## [1,]    1    6
## [2,]    2    7
## [3,]    3    8
## [4,]    4    9
## [5,]    5   10

f(M)

## [1] -185

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 \(e_5\), that is \(\left(0,0,0,0,1\right)^T\), maps to zero:

(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

(and of course, because a is alternating, we could have put \(e_5\) in the second column with the same result). The \(k\)form a returns zero because the index matrix of a does not include the number 5.

Most of the time, this 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

R gives, correctly, that the Hodge star of \(\operatorname{dx}\) is the zero-dimensional volume element (otherwise known as ‘’1’’). To get the right answer, 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