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

In stokes: The Exterior Calculus

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

knitr::include_graphics(system.file("help/figures/stokes.png", package = "stokes"))

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

To cite the stokes package in publications, please use @hankin2022_stokes. Convenience objects dx, dy, and dz, corresponding to elementary differential forms, are discussed here (basis vectors $e_1$, $e_2$, $e_2$ are discussed in vignette ex.Rmd).

@spivak1965, in a memorable passage, states:

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 $\mathrm{d}f$, defined by $$\mathrm{d}f(p)(v_p)=Df(p)(v).$$ Let us consider in particular the $1$-forms $\mathrm{d}\pi^i$ ^[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 $\mathrm{d}x^i(p)(v_p)=\mathrm{d}\pi^i(p)(v_p)=D\pi^i(p)(v)=v^i$, we see that $\mathrm{d}x^1(p),\ldots,\mathrm{d}x^n(p)$ 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 $\omega=\sum_{i_1 < \cdots < i_k}\omega_{i_1,\ldots, i_k}\mathrm{d}x^{i_1}\wedge\cdots\wedge\mathrm{d}x^{i_k}$. If working in $\mathbb{R}^3$, we have three elementary forms $\mathrm{d}x$, $\mathrm{d}y$, and $\mathrm{d}z$; in the package we have the pre-defined objects dx, dy, and dz. These are convenient for reproducing textbook results.

We conceptualise dx as "picking out" the $x$-component of a 3-vector and similarly for dy and dz. Recall that $\mathrm{d}x\colon\mathbb{R}^3\longrightarrow\mathbb{R}$ and we have

$$ dx\begin{pmatrix}u_1\u_2\u_3\end{pmatrix} = u_1\qquad dy\begin{pmatrix}u_1\u_2\u_3\end{pmatrix} = u_2\qquad dz\begin{pmatrix}u_1\u_2\u_3\end{pmatrix} = u_3. $$

Noting that $1$-forms are a vector space, we have in general

$$(a\cdot\mathrm{d}x + b\cdot\mathrm{d}y +c\cdot\mathrm{d}z) \begin{pmatrix}u_1\u_2\u_3\end{pmatrix} = au_1+bu_2+cu_3 $$

Numerically:

v <- c(2,3,7)
c(as.function(dx)(v),as.function(dx+dy)(v),as.function(dx+100*dz)(v))

As Spivak says, dx, dy and dz are conjugate to $e_1,e_2,e_3$ and these are defined using function e(). In this case it is safer to pass n=3 to function e() in order to specify that we are working in $\mathbb{R}^3$.

e(1,3)
e(2,3)
e(3,3)

We will now verify numerically that dx, dy and dz are indeed conjugate to $e_1,e_2,e_3$, but to do this we will define an orthonormal set of vectors $u,v,w$:

u <- e(1,3)
v <- e(2,3)
w <- e(3,3)
matrix(c(
    as.function(dx)(u), as.function(dx)(v), as.function(dx)(w),
    as.function(dy)(u), as.function(dy)(v), as.function(dy)(w),
    as.function(dz)(u), as.function(dz)(v), as.function(dz)(w)
),3,3)

Above we see the conjugacy clearly [obtaining $I_3$ as expected].

Wedge products

The elementary forms may be combined with a wedge product. We note that $\mathrm{d}x\wedge\mathrm{d}y\colon\left(\mathbb{R}^3\right)^2\longrightarrow\mathbb{R}$ and, for example,

$$ (\mathrm{d}x\wedge\mathrm{d}y)\left( \begin{pmatrix}u_1\u_2\u_3\end{pmatrix}, \begin{pmatrix}v_1\v_2\v_3\end{pmatrix} \right) = \det\begin{pmatrix}u_1&v_1\u_2&v_2\end{pmatrix} $$

and

$$ (\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z) \left( \begin{pmatrix}u_1\u_2\u_3\end{pmatrix}, \begin{pmatrix}v_1\v_2\v_3\end{pmatrix}, \begin{pmatrix}w_1\w_2\w_3\end{pmatrix} \right) = \det\begin{pmatrix}u_1&v_1&w_1\u_2&v_2&w_2\u_3&v_3&w_3\end{pmatrix} $$

Numerically:

as.function(dx ^ dy)(cbind(c(2,3,5),c(4,1,2)))

Above we see the package correctly giving $\det\begin{pmatrix}2&4\3&1\end{pmatrix}=2-12=-10$.

The print method

Here I give 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 $X=\mathrm{d}x\wedge\mathrm{d}y -7\mathrm{d}x\wedge\mathrm{d}z + 3\mathrm{d}y\wedge\mathrm{d}z$:

(X <- dx^dy -7*dx^dz + 3*dy^dz)

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

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

The other two rows would be

$$(3\mathrm{d}y\wedge\mathrm{d}z)\left( \begin{pmatrix}u_1\u_2\u_3\end{pmatrix}, \begin{pmatrix}v_1\v_2\v_3\end{pmatrix} \right) = 3\det\begin{pmatrix}u_2&v_2\u_3&v_3\end{pmatrix}$$

and

$$(1\mathrm{d}x\wedge\mathrm{d}y)\left( \begin{pmatrix}u_1\u_2\u_3\end{pmatrix}, \begin{pmatrix}v_1\v_2\v_3\end{pmatrix} \right) = \det\begin{pmatrix}u_1&v_1\u_2&v_2\end{pmatrix} $$

Thus form $X$ would be, by linearity

$$ X\left( \begin{pmatrix}u_1\u_2\u_3\end{pmatrix}, \begin{pmatrix}v_1\v_2\v_3\end{pmatrix} \right) = -7\det\begin{pmatrix}u_1&v_1\u_3&v_3\end{pmatrix} +3\det\begin{pmatrix}u_2&v_2\u_3&v_3\end{pmatrix} +\det\begin{pmatrix}u_1&v_1\u_2&v_2\end{pmatrix}. $$

We might want to verify that $\mathrm{d}x\wedge\mathrm{d}y=-\mathrm{d}y\wedge\mathrm{d}x$:

dx ^ dy == -dy ^ dx

Configuring 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 $\mathrm{d}x^i,i>3$, 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.

Creating elementary one-forms

Package function d() will create elementary one-forms but it is easier to interpret the output if we restore the default print method

options(kform_symbolic_print = NULL)
d(8)

Package dataset

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

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

References

stokes documentation built on April 4, 2025, 1:48 a.m.