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

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

To cite the stokes package in publications, please use Hankin (2022). 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).

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.

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”↩︎

Objects dx, dy, and dz in the stokes package