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dx, dy,
and dz in the stokes package
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:
## [1] 2 5 702As 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 \(\mathbb{R}^3\).
## [1] 1 0 0## [1] 0 1 0## [1] 0 0 1We 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 1Above we see the conjugacy clearly [obtaining \(I_3\) as expected].
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:
## [1] -10Above we see the package correctly giving \(\det\begin{pmatrix}2&4\\3&1\end{pmatrix}=2-12=-10\).
Here I give some illustrations of the package print method.
## An alternating linear map from V^1 to R with V=R^1:
## val
## 1 = 1This is somewhat opaque and difficult to understand. It is easier to start with a more complicated example: take \(X=\mathrm{d}x\wedge\mathrm{d}y -7\mathrm{d}x\wedge\mathrm{d}z + 3\mathrm{d}y\wedge\mathrm{d}z\):
## An alternating linear map from V^2 to R with V=R^3:
## val
## 1 3 = -7
## 2 3 = 3
## 1 2 = 1We 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 \(\mathrm{d}x\wedge\mathrm{d}y=-\mathrm{d}y\wedge\mathrm{d}x\):
## [1] TRUEThe 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:
Then the results of calculations are more natural:
## An alternating linear map from V^1 to R with V=R^1:
## + dx## An alternating linear map from V^2 to R with V=R^3:
## + dx^dy +56 dy^dzHowever, this setting can be confusing if we work with \(\mathrm{d}x^i,i>3\), for the print method runs out of alphabet:
## 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^dNAAbove, we see the use of NA because there is no defined
symbol.
Function hodge() returns the Hodge dual:
## An alternating linear map from V^1 to R with V=R^3:
## +13 dx + dzNote that calling hodge(dx) can be confusing:
## [1] 1This 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:
## An alternating linear map from V^2 to R with V=R^3:
## + dy^dzThis is further discussed in the dovs vignette.
Package function d() will create elementary one-forms
but it is easier to interpret the output if we restore the default print
method
## An alternating linear map from V^1 to R with V=R^8:
## val
## 8 = 1Spivak 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”↩︎
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