set.seed(0) library("stokes") library("spray") # needed for spraycross() options(rmarkdown.html_vignette.check_title = FALSE) knitr::opts_chunk$set(echo = TRUE) knit_print.function <- function(x, ...){dput(x)} registerS3method( "knit_print", "function", knit_print.function, envir = asNamespace("knitr") )

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

wedge wedge2

To cite the `stokes`

package in publications, please use
@hankin2022_stokes. In a memorable passage, @spivak1965 states:

$\ldots$ we would like a theorem analogous to 4.1 [the dimensionality
of $k$-fold tensor products is $n^k$]. Of course, if
$\omega\in\Lambda^k(V)$ and $\eta\in\Lambda^l(V)$, then
$\omega\otimes\eta$ is usually not in $\Lambda^{k+l}(V)$. We will
therefore define a new product, the **wedge** product
$\omega\wedge\eta\in\Lambda^{k+l}(V)$ by
$$
\omega\wedge\eta=\frac{\left(k+l\right)!}{k!l!}\operatorname{Alt}(\omega\otimes\eta),\qquad\omega\in\Lambda^k(V),\eta\in\Lambda^l(V)
$$
(The reason for the strange coefficient will appear later).

**- Michael Spivak, 1969** *(Calculus on Manifolds, Perseus books). Page 79*

Function `wedge()`

returns the wedge product of any number of
$k$-forms. Function `wedge2()`

returns the wedge product of two
$k$-forms, although the idiom is somewhat opaque, especially the
"strange" combinatorial coefficient $(k+l)!/(k!l!)$.

`spraycross()`

Function `wedge()`

is essentially a convenience wrapper for
`spraycross()`

; the meat of `wedge2()`

is the last line:
`kform(spraycross(K1, K2))`

. Function `spraycross()`

is part of the
`spray`

package and gives a tensor product of sparse arrays,
interpreted as multivariate polynomials:

(a <- spray(matrix(1:4,2,2),c(2,5))) (b <- spray(matrix(c(10,11,12,13),2,2),c(7,11))) spraycross(a,b) spraycross(b,a)

Observe that `spraycross()`

(and by association `wedge()`

) is
associative and distributive but not commutative.

`wedge2()`

Function `wedge2()`

takes two kforms and we will start with a very
simple example:

(x <- as.kform(cbind(1,2),5)) (y <- as.kform(cbind(3,4,7),7)) wedge2(x,y)

It looks like the combinatorial term has not been included but it has.
We will express `x`

and `y`

as tensors (objects of class `ktensor`

)
and show how the combinatorial term arises.

tx <- as.ktensor(x) # "tx" = tensor 'x' (ty <- as.ktensor(y)) # "ty" = tensor 'y'

As functions, `y`

and `ty`

are identical:

M <- matrix(round(rnorm(21),2),7,3) # member of (R^7)^3 c(as.function(y)(M),as.function(ty)(M))

Both are equivalent to

7*( +M[3,1]*M[4,2]*M[7,3] -M[3,1]*M[4,3]*M[7,2] -M[3,2]*M[4,1]*M[7,3] +M[3,2]*M[4,3]*M[7,1] +M[3,3]*M[4,1]*M[7,2] -M[3,3]*M[4,2]*M[7,1] )

We can see that `y`

is a more compact and efficient representation of
`ty`

: both are alternating tensors but `y`

has alternatingness built in
to its evaluation, while `ty`

is alternating by virtue of including
all permutations of its arguments, with the sign of the permutation.

We can evaluate Spivak's formula (but without the combinatorical term)
for $x\wedge y$ by coercing to ktensors and using `tensorprod()`

:

(z <- tensorprod(as.ktensor(x),as.ktensor(y)))

Above, each coefficient is equal to $\pm 35$ (the sign coming from the sign of the permutation), and we have $2!3!=12$ rows. We can now calculate $\operatorname{Alt}(z)$, which would have $5!=120$ rows, one per permutation of $[5]$, each with coefficient $\pm\frac{12\times 35}{5!}=\pm 3.5$.

We define $x\wedge y$ to be $\frac{5!}{3!2!}\operatorname{Alt}(z)$, so
each coefficient would be $\pm\frac{5!}{3!2!}\cdot\frac{12\times
35}{5!}=35$. We know that $x\wedge y$ is an alternating form. So to
represent it as an object of class `kform`

, we need a `kform`

object
with *single* index entry `1 2 3 4 7`

. This would need coefficient
35, on the grounds that it is linear, alternating, and maps
$\begin{pmatrix}
1&0&0&0&0\
0&1&0&0&0\
0&0&1&0&0\
0&0&0&1&0\
0&0&0&0&0\
0&0&0&0&0\
0&0&0&0&1
\end{pmatrix}$ to $35$; and indeed this is what we see:

```
wedge(x,y)
```

So to conclude, the combinatorial term is present in the R idiom, it is just difficult to see at first glance.

First of all we should note that $\Lambda^k(V)$ is a vector space
(this is considered in the `kform`

vignette). If
$\omega,\omega_i\in\Lambda^k(V)$ and $\eta,\eta_i\in\Lambda^l(V)$ then

\begin{eqnarray} (\omega_1+\omega_2)\wedge\eta &=& \omega_1\wedge\eta+\omega_2\wedge\eta\ \omega\wedge(\eta_1+\eta_2) &=&\omega\wedge\eta_1 + \omega\wedge\eta_2\ \end{eqnarray}

(that is, the wedge product is left- and right- distributive); if $a\in\mathbb{R}$ then

\begin{equation} a\omega\wedge\eta = \omega\wedge a\eta=a(\omega\wedge\eta) \end{equation}

and \begin{equation} \omega\wedge\eta = (-1)^{kl}\eta\wedge\omega. \end{equation}

These rules make expansion of wedge products possible by expressing a general kform in terms of a basis for $\Lambda^k(V)$. @spivak1965 tells us that, if $v_1,\ldots,v_k$ is a basis for $V$, then the set of all

\begin{equation} \phi_{i_1}\wedge\phi_{i_2}\wedge\cdots\wedge\phi_{i_k}\qquad 1\leq i_1 < \cdots < i_k\leq n \end{equation}

is a basis for $\Lambda^k(V)$ where $\phi_i(v_j)=\delta_{ij}$. The package expresses a $k$-form in terms of this basis as in the following example:

(omega <- as.kform(rbind(c(1,2,8),c(1,3,7)),5:6))

In algebraic notation, `omega`

(or $\omega$) would be
$5\phi_1\wedge\phi_2\wedge\phi_8+6\phi_1\wedge\phi_3\wedge\phi_7$ and
we may write this as $\omega=5\phi_{128}+6\phi_{137}$. To take a
wedge product of this with $\eta=2\phi_{235}+3\phi_{356}$ we would
write

\begin{eqnarray} \omega\wedge\eta &=& (5\phi_{128}+6\phi_{137})\wedge (2\phi_{235}+3\phi_{356})\ &=& 10\phi_{128}\wedge\phi_{235} + 15\phi_{128}\wedge\phi_{356} + 12\phi_{137}\wedge\phi_{235} + 18\phi_{137}\wedge\phi_{356}\ &=& 10\phi_1\wedge\phi_2\wedge\phi_8\wedge\phi_2\wedge\phi_3\wedge\phi_5 + 15\phi_1\wedge\phi_2\wedge\phi_8\wedge\phi_3\wedge\phi_5\wedge\phi_6\&{}&\qquad + 12\phi_1\wedge\phi_3\wedge\phi_7\wedge\phi_2\wedge\phi_3\wedge\phi_5 + 18\phi_1\wedge\phi_3\wedge\phi_7\wedge\phi_3\wedge\phi_5\wedge\phi_6\ &=& 0+ 15\phi_1\wedge\phi_2\wedge\phi_8\wedge\phi_3\wedge\phi_5\wedge\phi_6+0+0\ &=& -15\phi_1\wedge\phi_2\wedge\phi_3\wedge\phi_5\wedge\phi_6\wedge\phi_8 \end{eqnarray}

where we have used the rules repeatedly (especially the fact that
$\omega\wedge\omega=0$ for *any* alternating form). Package idiom
would be:

eta <- as.kform(rbind(c(2,3,5),c(3,5,6)),2:3) wedge(omega,eta)

See how function `wedge()`

does the legwork.

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