Objects `dx`, `dy`, and `dz` in the `stokes` package

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%}

dx <- d(1)
dy <- d(2)
dz <- d(3)

Convenience objects dx, dy, and dz are discussed here. The conjugate basis vectors i, j and k are discussed in ijk.Rmd.

Spivak, in a memorable passage, states (p89):

Fields and forms

If $f\colon\mathbb{R}^n\longrightarrow\mathbb{R}$ is differentiable, then $Df(p)\in\Lambda^1(\mathbb{R}^n)$. By a minor modification we therefore obtain a $1$-form $df$, defined by $$df(p)(v_p)=Df(p)(v).$$ Let us consider in particular the $1$-forms $d\pi^i$ ^[Spivak introduces the $\pi^i$ notation on page 11: "if $\pi\colon\mathbb{R}^n\longrightarrow\mathbb{R}^n$ is the identity function, $\pi(x)=x$, then [its components are] $\pi^i(x)=x^i$; the function $\pi^i$ is called the $i^\mathrm{th}$ *projection function*"]. It is customary to let $x^i$ denote the _function_ $\pi^i$ (on $\mathbb{R}^3$ we often denote $x^1$, $x^2$, and $x^3$ by $x$, $y$, and $z$) $\ldots$ Since $dx^i(p)(v_p)=d\pi^i(p)(v_p)=D\pi^i(p)(v)=v^i$, we see that $dx^1(p),\ldots,dx^n(p)$ is just the dual basis to $(e_1)_p,\ldots, (e_n)_p$.

- Michael Spivak, 1969 (Calculus on Manifolds, Perseus books). Page 89

Spivak goes on to observe that every $k$-form $\omega$ can be written $\omega=\sum_{i_1 < \cdots < i_k}\omega_{i_1,\ldots i_k}dx^{i_1}\wedge\cdots\wedge dx^{i_k}$. If working in $\mathbb{R}^3$, we have three elementary forms $dx$, $dy$, and $dz$; in the package we have the pre-defined objects dx, dy, and dz. These are convenient for reproducing textbook results. We start with some illustrations of the package print method.

dx

This is somewhat opaque and difficult to understand. It is easier to start with a more complicated example: take $dx\wedge dy -7dx\wedge dz + 3dy\wedge dz$:

dx^dy -7*dx^dz + 3*dy^dz

We see three rows for the three elementary components. Taking the row with coefficient $-7$ [which would be $-7dx\wedge dz$], this maps $\left(\mathbb{R}^3\right)^2$ to $\mathbb{R}$ and we have

$$(-7dx\wedge dz)\left(\begin{pmatrix} u_1\u_2\u_3\end{pmatrix}, \begin{pmatrix}v_1\v_2\v_3\end{pmatrix}\right)= -7\det\begin{pmatrix}u_1&v_1\u_3&v_3\end{pmatrix}$$

Armed with this familiar fact, we can interpret $dx$ as a map from $\left(\mathbb{R}^3\right)^1$ to $\mathbb{R}$ with

$$dx\left(\begin{pmatrix} u_1\u_2\u_3\end{pmatrix} \right)= \det\begin{pmatrix}u_1\end{pmatrix}=u_1$$

or, in other words, $dx$ picks out the first component of its vector (as the print method gives, albeit obscurely). This is easily shown in the package:

as.function(dx)(c(113,3,6))

We might want to verify that $dx\wedge dy=-dy\wedge dx$:

dx ^ dy == -dy ^ dx

Elementary forms and the print method

The print method is configurable and can display kforms in symbolic form. For working with dx dy dz we may set option kform_symbolic_print to dx:

options(kform_symbolic_print = 'dx')

Then the results of calculations are more natural:

dx
dx^dy + 56*dy^dz

However, this setting can be confusing if we work with $dx^i,i>3$, for the print method runs out of alphabet:

rform()

Above, we see the use of NA because there is no defined symbol.

The Hodge dual

Function hodge() returns the Hodge dual:

hodge(dx^dy + 13*dy^dz)

Note that calling hodge(dx) can be confusing:

hodge(dx)

This returns a scalar because dx is interpreted as a one-form on one-dimensional space, which is a scalar form. One usually wants the result in three dimensions:

hodge(dx,3)

This is further discussed in the dovs vignette.

Other ways to create the elementary one-forms

It is possible to create these objects using package idiom:

d(1) == dx

Basis vectors

Package dataset

Following lines create dx.rda, residing in the data/ directory of the package.

save(dx,dy,dz,file="dx.rda")

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