Objects dx, dy, and dz in the stokes package

Robin K. S. Hankin

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

To cite the stokes package in publications, please use Hankin (2022). This function monograph discusses convenience objects dx, dy, and dz, corresponding to elementary differential forms (basis vectors $e_1$, $e_2$, $e_2$ are discussed in vignette ex).

Spivak (1965), 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 $\mathrm{d}\pi^i$¹. 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))

## [1]   2   5 702

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)

## [1] 1 0 0

e(2,3)

## [1] 0 1 0

e(3,3)

## [1] 0 0 1

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)

##      [,1] [,2] [,3]
## [1,]    1    0    0
## [2,]    0    1    0
## [3,]    0    0    1

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)))

## [1] -10

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

## An alternating linear map from V^1 to R with V=R^1:
##        val
##  1  =    1

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)

## An alternating linear map from V^2 to R with V=R^3:
##          val
##  1 3  =   -7
##  2 3  =    3
##  1 2  =    1

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

## [1] TRUE

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

## An alternating linear map from V^1 to R with V=R^1:
##  + dx

dx^dy + 56*dy^dz

## An alternating linear map from V^2 to R with V=R^3:
##  + 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()

## An alternating linear map from V^3 to R with V=R^7:
##  +6 dy^dNA^dNA +5 dy^dNA^dNA -9 dNA^dNA^dNA +4 dx^dz^dNA +7 dx^dNA^dNA -3 dy^dz^dNA -8 dx^dNA^dNA +2 dx^dy^dNA + dx^dNA^dNA

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)

## An alternating linear map from V^1 to R with V=R^3:
##  +13 dx + dz

Note that calling hodge(dx) can be confusing:

hodge(dx)

## [1] 1

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)

## An alternating linear map from V^2 to R with V=R^3:
##  + dy^dz

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)

## An alternating linear map from V^1 to R with V=R^8:
##        val
##  8  =    1

Package dataset

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

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

References

Hankin, R. K. S. 2022. “Stokes’s Theorem in R.” arXiv. https://doi.org/10.48550/ARXIV.2210.17008.

Spivak, M. 1965. Calculus on Manifolds. Addison-Wesley.

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1. 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”↩︎