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

In stokes: The Exterior Calculus

knitr::opts_chunk$set(echo = TRUE)
options(rmarkdown.html_vignette.check_title = FALSE)
library("stokes")
set.seed(0)

![](`r system.file("help/figures/stokes.png", package = "stokes")`){width=10%}

dovs

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)
a <- rform(n=4,k=2)
a

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))
f(M)

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
f(M)

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)))
f(M)

(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

Now, we know that dx is supposed to be a map from $\left(\mathbb{R}^3\right)^1$ to $\mathbb{R}$; but:

dovs(dx)

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)

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)

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

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

options(kform_symbolic_print = NULL)

stokes documentation built on Aug. 19, 2023, 1:07 a.m.