knitr::opts_chunk$set(echo = TRUE) options(rmarkdown.html_vignette.check_title = FALSE) library("stokes") set.seed(1)

![](`r system.file("help/figures/stokes.png", package = "stokes")`){width=10%}

To cite the `stokes`

package in publications please use
[@hankin2022_stokes]. Ordinary differential calculus may be
formalized and generalized to arbitrary-dimensional oriented manifolds
using the exterior calculus. Here I show how the `stokes`

package
furnishes functionality for working with the exterior calculus, and
provide numerical verification of a number of theorems. Notation
follows that of @spivak1965, and @hubbard2015.

Recall that a $k$-tensor is a multilinear map $S\colon V^k\longrightarrow\mathbb{R}$, where $V=\mathbb{R}^n$ is considered as a vector space; Spivak denotes the space of multilinear maps as $\mathcal{J}^k(V)$. Formally, multilinearity means

[ S{\left(v_1,\ldots,av_i,\ldots,v_k\right)} = a\cdot S{\left(v_1,\ldots,v_i,\ldots,v_k\right)} ]

and

[ S{\left(v_1,\ldots,v_i+{v_i}',\ldots,v_k\right)}=S{\left(v_1,\ldots,v_i,\ldots,x_v\right)}+ S{\left(v_1,\ldots,{v_i}',\ldots,v_k\right)}. ]

where $v_i\in V$. If $S\in\mathcal{J}^k(V)$ and $T\in\mathcal{J}^l(V)$, then we may define $S\otimes T\in\mathcal{J}^{k+l}(V)$ as

[ S\otimes T{\left(v_1,\ldots,v_k,v_{k+1},\ldots,v_{k+l}\right)}= S{\left(v_1,\ldots,v_k\right)}\cdot T{\left(v_1,\ldots,v_l\right)}. ]

Spivak observes that $\mathcal{J}^k(V)$ is spanned by the $n^k$ products of the form

[ \phi_{i_1}\otimes\phi_{i_2}\otimes\cdots\otimes\phi_{i_k}\qquad 1\leq i_i,i_2,\ldots,i_k\leq n ]

where $v_1,\ldots,v_k$ is a basis for $V$ and $\phi_i{\left(v_j\right)}=\delta_{ij}$; we can therefore write

[ S=\sum_{1\leq i_1,\ldots,i_k\leq n} a_{i_1\ldots i_k} \phi_{i_1}\otimes\cdots\otimes\phi_{i_k}. ]

The space spanned by such products has a natural representation in R
as an array of dimensions $n\times\cdots\times n=n^k$. If `A`

is such
an array, then the element `A[i_1,i_2,...,i_k]`

is the coefficient of
$\phi_{i_1}\otimes\ldots\otimes\phi_{i_k}$. However, it is more
efficient and conceptually cleaner to consider a *sparse* array, as
implemented by the `spray`

package. We will consider the case
$n=5,k=4$, so we have multilinear maps from
$\left(\mathbb{R}^5\right)^4$ to $\mathbb{R}$. Below, we will test
algebraic identities in R using the idiom furnished by the stokes
package. For our example we will define
$S=1.5\phi_5\otimes\phi_1\otimes\phi_1\otimes\phi_1+2.5\phi_1\otimes\phi_1\otimes\phi_2\otimes\phi_3+3.5\phi_1\otimes\phi_3\otimes\phi_4\otimes\phi_2$
using a matrix with three rows, one per term, and whose rows
correspond to each term's tensor products of the $\phi$'s. We first
have to load the `stokes`

package:

library("stokes")

Then the idiom is straightforward:

k <- 4 n <- 5 M <- matrix(c(5,1,1,1, 1,1,2,3, 1,3,4,2),3,4,byrow=TRUE) M S <- as.ktensor(M,coeffs= 0.5 + 1:3) S

Observe that, if stored as an array of size $n^k$, $S$ would have
$5^4=625$ elements, all but three of which are zero. So $S$ is a
4-tensor, mapping $V^4$ to $\mathbb{R}$, where $V=\mathbb{R}^5$. Here
we have
$S=1.5\phi_5\otimes\phi_1\otimes\phi_1\otimes\phi_1+2.5\phi_1\otimes\phi_1\otimes\phi_2\otimes\phi_3+3.5\phi_1\otimes\phi_3\otimes\phi_4\otimes\phi_2$.
Note that in some implementations the row order of object `S`

will
differ from that of `M`

; this phenomenon is due to the underlying `C`

implementation using the `STL map`

class; see the `disordR`

package
[@hankin2022_disordR] and is discussed in more detail in the `mvp`

package [@hankin2022_mvp].

First, we will define $E$ to be a random point in $V^k$ in terms of a matrix:

set.seed(0) (E <- matrix(rnorm(n*k),n,k)) # A random point in V^k

Recall that $n=5$, $k=4$, so $E\in\left(\mathbb{R}^5\right)^4$. We can evaluate $S$ at $E$ as follows:

f <- as.function(S) f(E)

Tensors have a natural vector space structure; they may be added and subtracted, and multiplied by a scalar, the same as any other vector space. Below, we define a new tensor $S_1$ and work with $2S-3S_1$:

S1 <- as.ktensor(1+diag(4),1:4) 2*S-3*S1

We may verify that tensors are linear using package idiom:

LHS <- as.function(2*S-3*S1)(E) RHS <- 2*as.function(S)(E) -3*as.function(S1)(E) c(lhs=LHS,rhs=RHS,diff=LHS-RHS)

(that is, identical up to numerical precision).

Testing multilinearity is straightforward in the package. To do this,
we need to define three matrices `E1,E2,E3`

corresponding to
points in $\left(\mathbb{R}^5\right)^4$ which are identical except for
one column. In `E3`

, this column is a linear combination of the
corresponding column in `E2`

and `E3`

:

E1 <- E E2 <- E E3 <- E x1 <- rnorm(n) x2 <- rnorm(n) r1 <- rnorm(1) r2 <- rnorm(1) E1[,2] <- x1 E2[,2] <- x2 E3[,2] <- r1*x1 + r2*x2

Then we can verify the multilinearity of $S$ by coercing to a function
which is applied to `E1, E2, E3`

:

f <- as.function(S) LHS <- r1*f(E1) + r2*f(E2) RHS <- f(E3) c(lhs=LHS,rhs=RHS,diff=LHS-RHS)

(that is, identical up to numerical precision). Note that this is
*not* equivalent to linearity over $V^{nk}$:

E1 <- matrix(rnorm(n*k),n,k) E2 <- matrix(rnorm(n*k),n,k) LHS <- f(r1*E1+r2*E2) RHS <- r1*f(E1)+r2*f(E2) c(lhs=LHS,rhs=RHS,diff=LHS-RHS)

Given two k-tensor objects $S,T$ we can form the tensor product $S\otimes T$, defined as

[ S\otimes T{\left(v_1,\ldots,v_k,v_{k+1},\ldots, v_{k+l}\right)}= S{\left(v_1,\ldots v_k\right)} \cdot T{\left(v_{k+1},\ldots v_{k+l}\right)} ]

We will calculate the tensor product of two tensors `S1,S2`

defined as follows:

(S1 <- ktensor(spray(cbind(1:3,2:4),1:3))) (S2 <- as.ktensor(matrix(1:6,2,3)))

The R idiom for $S1\otimes S2$ would be `tensorprod()`

, or `%X%`

:

```
tensorprod(S1,S2)
```

Then, for example:

E <- matrix(rnorm(30),6,5) LHS <- as.function(tensorprod(S1,S2))(E) RHS <- as.function(S1)(E[,1:2]) * as.function(S2)(E[,3:5]) c(lhs=LHS,rhs=RHS,diff=LHS-RHS)

(that is, identical up to numerical precision).

An alternating form is a multilinear map $T$ satisfying

[ T{\left(v_1,\ldots,v_i,\ldots,v_j,\ldots,v_k\right)}= -T{\left(v_1,\ldots,v_j,\ldots,v_i,\ldots,v_k\right)} ]

(or, equivalently, $T{\left(v_1,\ldots,v_i,\ldots,v_i,\ldots,v_k\right)}= 0$). We write $\Lambda^k(V)$ for the space of all alternating multilinear maps from $V^k$ to $\mathbb{R}$. Spivak gives $\operatorname{Alt}\colon\mathcal{J}^k(V)\longrightarrow\Lambda^k(V)$ defined by

[\operatorname{Alt}(T)\left(v_1,\ldots,v_k\right)= \frac{1}{k!}\sum_{\sigma\in S_k}\operatorname{sgn}(\sigma)\cdot T{\left(v_{\sigma(1)},\ldots,v_{\sigma(k)}\right)} ]

where the sum ranges over all permutations of $\left[n\right]=\left{1,2,\ldots,n\right}$ and $\operatorname{sgn}(\sigma)\in\pm 1$ is the sign of the permutation. If $T\in\mathcal{J}^k(V)$ and $\omega\in\Lambda^k(V)$, it is straightforward to prove that $\operatorname{Alt}(T)\in\Lambda^k(V)$, $\operatorname{Alt}\left(\operatorname{Alt}\left(T\right)\right)=\operatorname{Alt}\left(T\right)$, and $\operatorname{Alt}\left(\omega\right)=\omega$.

In the stokes package, this is effected by the `Alt()`

function:

```
S1
Alt(S1)
```

Verifying that `S1`

is in fact alternating is straightforward:

E <- matrix(rnorm(8),4,2) Erev <- E[,2:1] as.function(Alt(S1))(E) + as.function(Alt(S1))(Erev) # should be zero

However, we can see that this form for alternating tensors (here
called $k$-forms) is inefficient and highly redundant: in this example
there is a `1 2`

term and a `2 1`

term (the coefficients are
equal and opposite). In this example we have $k=2$ but in general
there would be potentially $k!$ essentially repeated terms which
collectively require only a single coefficient. The package provides
`kform`

objects which are inherently alternating using a more
efficient representation; they are described using wedge products
which are discussed next.

This section follows the exposition of Hubbard and Hubbard, who introduce the exterior calculus starting with a discussion of elementary forms, which are alternating forms with a particularly simple structure. An example of an elementary form would be $dx_1\wedge dx_3$ [treated as an indivisible entity], which is an alternating multilinear map from $\mathbb{R}^n\times\mathbb{R}^n$ to $\mathbb{R}$ with

[ \left( dx_1\wedge dx_3 \right)\left( \begin{pmatrix}a_1\a_2\a_3\ \vdots\ a_n\end{pmatrix}, \begin{pmatrix}b_1\b_3\b_3\ \vdots\ b_n\end{pmatrix} \right)=\mathrm{det} \begin{pmatrix} a_1 & b_1 \ a_3 & b_3\end{pmatrix} =a_1b_3-a_3b_1 ]

That this is alternating follows from the properties of the determinant. In general of course, $dx_i\wedge dx_j\left( \begin{pmatrix}a_1\ \vdots\ a_n\end{pmatrix}, \begin{pmatrix}b_1\ \vdots\ b_n\end{pmatrix} \right)=\mathrm{det} \begin{pmatrix} a_i & b_i \ a_j & b_j\end{pmatrix}$. Because such objects are linear, it is possible to consider sums of elementary forms, such as $dx_1\wedge dx_2 + 3 dx_2\wedge dx_3$ with

[ \left( dx_1\wedge dx_2 + 3dx_2\wedge dx_3 \right)\left( \begin{pmatrix}a_1\a_2\ \vdots\ a_n\end{pmatrix}, \begin{pmatrix}b_1\b_2\ \vdots\ b_n\end{pmatrix} \right)=\mathrm{det} \begin{pmatrix} a_1 & b_1\ a_2 & b_2\end{pmatrix} +3\mathrm{det} \begin{pmatrix} a_2 & b_2\ a_3 & b_3\end{pmatrix} ]

or even $K=dx_1\wedge dx_2\wedge dx_3 +5dx_1\wedge dx_2\wedge dx_4$ which would be a linear map from $\left(\mathbb{R}^n\right)^3$ to $\mathbb{R}$ with

[ \left( dx_4\wedge dx_2\wedge dx_3 +5dx_1\wedge dx_2\wedge dx_4 \right)\left( \begin{pmatrix}a_1\a_2\ \vdots\ a_n\end{pmatrix}, \begin{pmatrix}b_1\b_2\ \vdots\ b_n\end{pmatrix}, \begin{pmatrix}c_1\c_2\ \vdots\ c_n\end{pmatrix} \right)=\mathrm{det} \begin{pmatrix} a_4 & b_4 & c_4\ a_2 & b_2 & c_2\ a_3 & b_3 & c_3 \end{pmatrix} +5\mathrm{det} \begin{pmatrix} a_1 & b_1 & c_1\ a_2 & b_2 & c_2\ a_4 & b_4 & c_4 \end{pmatrix}. ]

Defining $K$ has ready R idiom in which we define a matrix whose rows correspond to the differentials in each term:

M <- matrix(c(4,2,3,1,4,2),2,3,byrow=TRUE) M K <- as.kform(M,c(1,5)) K

Function `as.kform()`

takes each row of `M`

and places the elements in
increasing order; the coefficient will change sign if the permutation
is odd. Note that the order of the rows in `K`

is immaterial and
indeed in some implementations will appear in a different order: the
stokes package uses the `spray`

package, which in turn utilises the
STL map class of C++.

In the previous section we defined objects such as "$dx_1\wedge dx_6$" as a single entity. Here I define the elementary form $dx_i$ formally and in the next section discuss the wedge product $\wedge$. The elementary form $dx_i$ is simply a map from $\mathbb{R}^n$ to $\mathbb{R}$ with $dx_i{\left(x_1,x_2,\ldots,x_n\right)}=x_i$. Observe that $dx_i$ is an alternating form, even though we cannot swap arguments (because there is only one). Package idiom for creating an elementary form appears somewhat cryptic at first sight, but is consistent (it is easier to understand package idiom for creating more complicated alternating forms, as in the next section). Suppose we wish to work with $dx_3$:

dx3 <- as.kform(matrix(3,1,1),1) options(kform_symbolic_print = NULL) # revert to default print method dx3

Interpretation of the output above is not obvious (it is easier to understand the output from more complicated alternating forms, as in the next section), but for the moment observe that $dx_3$ is indeed an alternating form, mapping $\mathbb{R}^n$ to $\mathbb{R}$ with $dx_3{\left(x_1,x_2,\ldots,x_n\right)}=x_3$. Thus, for example:

as.function(dx3)(c(14,15,16)) as.function(dx3)(c(14,15,16,17,18)) # idiom can deal with arbitrary vectors

and we see that $dx_3$ picks out the third element of a vector. These are linear in the sense that we may add and subtract these elementary forms:

dx5 <- as.kform(matrix(5,1,1),1) as.function(dx3 + 2*dx5)(1:10) # picks out element 3 + 2*element 5

The wedge product maps two alternating forms to another alternating form; given $\omega\in\Lambda^k(V)$ and $\eta\in\Lambda^l(V)$, Spivak defines the wedge product $\omega\wedge\eta\in\Lambda^{k+l}(V)$ as

[ \omega\wedge\eta={k+l\choose k\quad l}\operatorname{Alt}(\omega\otimes\eta) ]

and this is implemented in the package by function `wedge()`

, or,
more idiomatically, `^`

:

M1 <- matrix(c(3,5,4, 4,6,1),2,3,byrow=TRUE) K1 <- as.kform(M1,c(2,7)) K1 M2 <- cbind(1:5,3:7) K2 <- as.kform(M2,1:5) K2

In symbolic notation, `K1`

is equal to $7dx_1\wedge dx_4\wedge dx_6
-2dx_3\wedge dx_4\wedge dx_5$. and `K2`

is $dx_1\wedge dx_3+
2dx_2\wedge dx_4+ 3dx_3\wedge dx_5+ 4dx_4\wedge dx_6+ 5dx_5\wedge
dx_7$. Package idiom for wedge products is straightforward:

K1 ^ K2

(we might write the product as $-35dx_1\wedge dx_4\wedge dx_5\wedge dx_6\wedge dx_7 -21dx_1\wedge dx_3\wedge dx_4\wedge dx_5\wedge dx_6$). See how the wedge product eliminates rows with repeated entries, gathers permuted rows together (respecting the sign of the permutation), and expresses the result in terms of elementary forms. The product is a linear combination of two elementary forms; note that only two coefficients out of a possible ${7\choose 5}=21$ are nonzero. Note again that the order of the rows in the product is arbitrary.

The wedge product has formal properties such as distributivity but by far the most interesting one is associativity, which I will demonstrate below:

F1 <- as.kform(matrix(c(3,4,5, 4,6,1,3,2,1),3,3,byrow=TRUE)) F2 <- as.kform(cbind(1:6,3:8),1:6) F3 <- kform_general(1:8,2) (F1 ^ F2) ^ F3 F1 ^ (F2 ^ F3)

Note carefully in the above that the terms in `(F1 ^ F2) ^ F3`

and
`F1 ^ (F2 ^ F3)`

appear in a different order. They are
nevertheless algebraically identical, as we may demonstrate by
calculating their difference:

(F1 ^ F2) ^ F3 - F1 ^ (F2 ^ F3)

Spivak observes that $\Lambda^k(V)$ is spanned by the $n\choose k$ wedge products of the form

[ dx_{i_1}\wedge dx_{i_2}\wedge\ldots\wedge dx_{i_k}\qquad 1\leq i_i<i_2<\cdots <i_k\leq n ]

where these products are the elementary forms (compare
$\mathcal{J}^k(V)$, which is spanned by $n^k$ elementary forms).
Formally, multilinearity means every element of the space
$\Lambda^k(V)$ is a linear combination of elementary forms, as
illustrated in the package by function `kform_general()`

. Consider
the following idiom:

Krel <- kform_general(4,2,1:6) Krel

Object `Krel`

is a two-form, specifically a map from
$\left(\mathbb{R}^4\right)^2$ to $\mathbb{R}$. Observe that
`Krel`

has ${4\choose 2}=6$ components, which do not appear in any
particular order. Addition of such $k$-forms is straightforward in R
idiom but algebraically nontrivial:

K1 <- as.kform(matrix(1:4,2,2),c(1,109)) K2 <- as.kform(matrix(c(1,3,7,8,2,4),ncol=2,byrow=TRUE),c(-1,5,4)) K1 K2 K1+K2

In the above, note how the $dx_2\wedge dx_4$ terms combine [to give ```
2
4 = 113
```

] and the $dx_1\wedge dx_3$ term vanishes by cancellation.

Although the spray form used above is probably the most direct and natural representation of differential forms in numerical work, sometimes we need a more algebraic print method.

U <- ktensor(spray(cbind(1:4,2:5),1:4)) U

we can represent this more algebraically using the `as.symbolic()`

function:

```
as.symbolic(U)
```

In the above, `U`

is a multilinear map from
$\left(\mathbb{R}^5\right)^2$ to $\mathbb{R}$. Symbolically, `a`

represents the map that takes $(a,b,c,d,e)$ to $a$, `b`

the map that
takes $(a,b,c,d,e)$ to `b`

, and so on. The asterisk `*`

represents
the tensor product $\otimes$. Alternating forms work similarly but
$k$-forms have different defaults:

K <- kform_general(3,2,1:3) K as.symbolic(K,d="d",symbols=letters[23:26])

Note that the wedge product $\wedge$, although implemented in package
idiom as `^`

or `%^%`

, appears in the symbolic representation as an
ascii caret, `^`

.

We can alter the default print method with the `kform_symbolic_print`

option, which uses `as.symbolic()`

:

options(kform_symbolic_print = "d") K

This print option works nicely with the `d()`

function for elementary
forms:

(d(1) + d(5)) ^ (d(3)-5*d(2)) ^ d(7) options(kform_symbolic_print = NULL) # restore default

Given a $k$-form $\phi\colon V^k\longrightarrow\mathbb{R}$ and a
vector $\mathbf{v}\in V$, the *contraction* $\phi_\mathbf{v}$ of
$\phi$ and $\mathbf{v}$ is a $k-1$-form with

[ \phi_\mathbf{v}{\left(\mathbf{v}^1,\ldots,\mathbf{v}^{k-1}\right)} = \phi{\left(\mathbf{v},\mathbf{v}^1,\ldots,\mathbf{v}^{k-1}\right)} ]

if $k>1$; we specify $\phi_\mathbf{v}=\phi(\mathbf{v})$ if $k=1$. Verification is straightforward:

(o <- rform()) # a random 3-form V <- matrix(runif(21),ncol=3) LHS <- as.function(o)(V) RHS <- as.function(contract(o,V[,1]))(V[,-1]) c(LHS=LHS,RHS=RHS,diff=LHS-RHS)

It is possible to iterate the contraction process; if we pass a matrix
$V$ to `contract()`

then this is interpreted as repeated contraction
with the columns of $V$:

as.function(contract(o,V[,1:2]))(V[,-(1:2),drop=FALSE])

If we pass three columns to `contract()`

the result is a $0$-form:

```
contract(o,V)
```

In the above, the result is coerced to a scalar; in order to work with
a formal $0$-form (which is represented in the package as a `spray`

with a zero-column index matrix) we can use the `lose=FALSE`

argument:

contract(o,V,lose=FALSE)

Suppose we are given a two-form $\omega=\sum_{i<j}a_{ij}dx_i\wedge dx_j$ and relationships $dx_i=\sum_rM_{ir}dy_r$, then we would have

[ \omega = \sum_{i<j} a_{ij}\left(\sum_rM_{ir}dy_r\right)\wedge\left(\sum_rM_{jr}dy_r\right). ]

The general situation would be a $k$-form where we would have

[ \omega=\sum_{i_1<\cdots<i_k}a_{i_1\ldots i_k}dx_{i_1}\wedge\cdots\wedge dx_{i_k} ]

giving

[\omega = \sum_{i_1<\cdots <i_k}\left[ a_{i_1<\cdots < i_k}\left(\sum_rM_{i_1r}dy_r\right)\wedge\cdots\wedge\left(\sum_rM_{i_kr}dy_r\right)\right]. ]

So $\omega$ was given in terms of $dx_1,\ldots,dx_k$ and we have expressed it in terms of $dy_1,\ldots,dy_k$. So for example if

[ \omega= dx_1\wedge dx_2 + 5dx_1\wedge dx_3]

and

[ \left( \begin{array}{l} dx_1\ dx_2\ dx_3 \end{array} \right)= \left( \begin{array}{ccc} 1 & 4 & 7\ 2 & 5 & 8\ 3 & 6 & 9\ \end{array} \right) \left( \begin{array}{l} dy_1\ dy_2\ dy_3 \end{array} \right) ]

then

[ \begin{array}{ccl} \omega &=& \left(1dy_1+4dy_2+7dy_3\right)\wedge \left(2dy_1+5dy_2+8dy_3\right)+ 5\left(1dy_1+4dy_2+7dy_3\right)\wedge \left(3dy_1+6dy_2+9dy_3\right) \ &=&2dy_1\wedge dy_1+5dy_1\wedge dy_2+\cdots+ 5\cdot 7\cdot 6dx_3\wedge dx_2+ 5\cdot 7\cdot 9dx_3\wedge dx_3+\ &=& -33dy_1\wedge dy_2-66dy_1\wedge dy_3-33dy_2\wedge dy_3 \end{array} ]

Function `pullback()`

function does all this:

options(kform_symbolic_print = "dx") # uses dx etc in print method pullback(dx^dy+5*dx^dz, matrix(1:9,3,3)) options(kform_symbolic_print = NULL) # revert to default

However, it is slow and I am not 100\% sure that there isn't a much
more efficient way to do such a transformation. There are a few tests
in `tests/testthat`

. Here I show that transformations may be inverted
using matrix inverses:

(o <- 2 * as.kform(2) ^ as.kform(4) ^ as.kform(5)) M <- matrix(rnorm(25),5,5)

Then we will transform according to matrix `M`

and then transform
according to the matrix inverse; the functionality works nicely with
magrittr pipes:

o |> pullback(M) |> pullback(solve(M))

Above we see many rows with values small enough for the print method
to print an exact zero, but not sufficiently small to be eliminated by
the `spray`

internals. We can remove the small entries with `zap()`

:

o |> pullback(M) |> pullback(solve(M)) |> zap()

See how the result is equal to the original $k$-form $2dy_2\wedge dy_4\wedge dy_5$.

Given a $k$-form $\omega$, Spivak defines the differential of $\omega$ to be a $(k+1)$-form $d\omega$ as follows. If

[ \omega = \sum_{ i_1 < i_2 <\cdots<i_k} \omega_{i_1i_2\ldots i_k} dx^{i_1}\wedge dx^{i_2}\wedge\cdots\wedge dx^{i_k} ]

then

[ d\omega = \sum_{ i_1 < i_2 <\cdots<i_k} \sum_{\alpha=1}^n D_\alpha\left(\omega_{i_1i_2\ldots i_k}\right) \cdot dx^{i_1}\wedge dx^{i_2}\wedge\cdots\wedge dx^{i_k} ]

where $D_if(a)=\lim_{h\longrightarrow 0}\frac{f(a^1,\ldots,a^i+h,\ldots,a^n)-f(a^1,\ldots,a^i,\ldots,a^n)}{h}$ is the ordinary $i^\mathrm{th}$ partial derivative (Spivak, p25). Hubbard and Hubbard take a conceptually distinct approach and define the exterior derivative $d\phi$ (they use a bold font, $\mathbf{d}\phi$) of the $k$-form $\phi$ as the $(k+1)$-form given by

[ {d}\phi
\left({v}*i,\ldots,{v}*{k+1}\right)
=
\lim_{h\longrightarrow 0}\frac{1}{h^{k+1}}\int_{\partial
P_{x}\left(h{v}*1,\ldots,h{v}*{k+1}\right)}\phi
]

which, by their own account, is a rather opaque mathematical idiom. However, the definition makes sense and it is consistent with Spivak's definition above. The definition allows one to express the fundamental theorem of calculus in an arbitrary number of dimensions without modification.

It can be shown that

[ d{\left(f\,dx_{i_1}\wedge\cdots\wedge dx_{i_k}\right)}= df\wedge dx_{i_1}\wedge\cdots\wedge dx_{i_k} ]

where $f\colon\mathbb{R}^n\longrightarrow\mathbb{R}$ is a scalar
function of position. The package provides `grad()`

which, when given
a vector $x_1,\ldots,x_n$ returns the one-form

[ \sum_{i=1}^n x_idx_i ]

This is useful because $df=\sum_{j=1}^n\left(D_j f\right)\,dx_j$. Thus

grad(c(0.4,0.1,-3.2,1.5))

We will use the `grad()`

function to verify that, in
$\mathbb{R}^n$, a certain $(k-1)$-form has zero work function.
Motivated by the fact that

[ F_3=\frac{1}{\left(x^2+y^2+z^2\right)^{3/2}} \begin{pmatrix}x\y\z\end{pmatrix} ]

is a divergenceless velocity field in $\mathbb{R}^3$, H\&H go on to define [page 548, equation 6.7.16]

[ \omega_n= d\frac{1}{\left(x_1^2+\ldots +x_n^2\right)^{n/2}}\sum_{i=1}^{n}(-1)^{i-1} x_idx_1\wedge\cdots\wedge\widehat{dx_i}\wedge\cdots\wedge dx_n ]

(where a hat indicates the absence of a term), and show analytically that $d\omega=0$. Here I show this using R idiom. The first thing is to define a function that implements the hat:

f <- function(x){ n <- length(x) as.kform(t(apply(diag(n)<1,2,which))) }

So, for example:

f(1:5)

Then we can use the `grad()`

function to calculate $d\omega$,
using the quotient law to express the derivatives analytically:

df <- function(x){ n <- length(x) S <- sum(x^2) grad(rep(c(1,-1),length=n)*(S^(n/2) - n*x^2*S^(n/2-1))/S^n ) }

Thus

df(1:5)

Now we can use the wedge product of the two parts to show that the exterior derivative is zero:

x <- rnorm(9) print(df(x) ^ f(x)) # should be zero

We can use the package to verify the celebrated fact that, for any
$k$-form $\phi$, $d\left(d\phi\right)=0$. The first step is to define
scalar functions `f1(), f2(), f3()`

, all $0$-forms:

f1 <- function(w,x,y,z){x + y^3 + x*y*w*z} f2 <- function(w,x,y,z){w^2*x*y*z + sin(w) + w+z} f3 <- function(w,x,y,z){w*x*y*z + sin(x) + cos(w)}

Now we need to define elementary $1$-forms:

dw <- as.kform(1) dx <- as.kform(2) dy <- as.kform(3) dz <- as.kform(4)

I will demonstrate the theorem by defining a $2$-form which is the sum of three elementary two-forms, evaluated at a particular point in $\mathbb{R}^4$:

phi <- ( +f1(1,2,3,4) ^ dw ^ dx +f2(1,2,3,4) ^ dw ^ dy +f3(1,2,3,4) ^ dy ^ dz )

We could use slightly slicker R idiom by defining elementary forms
`e1,e2,e3`

and then defining `phi`

to be a linear sum, weighted with
$0$-forms given by the (scalar) functions `f1,f2,f3`

:

e1 <- dw ^ dx e2 <- dw ^ dy e3 <- dy ^ dz phi <- ( +f1(1,2,3,4) ^ e1 +f2(1,2,3,4) ^ e2 +f3(1,2,3,4) ^ e3 ) phi

Now to evaluate first derivatives of `f1()`

etc at point
$(1,2,3,4)$, using `Deriv()`

from the Deriv package:

library("Deriv") Df1 <- Deriv(f1)(1,2,3,4) Df2 <- Deriv(f2)(1,2,3,4) Df3 <- Deriv(f3)(1,2,3,4)

So `Df1`

etc are numeric vectors of length 4, for example:

Df1

To calculate `dphi`

, or $d\phi$, we can use function `grad()`

:

dphi <- ( +grad(Df1) ^ e1 +grad(Df2) ^ e2 +grad(Df3) ^ e3 ) dphi

Now work on the differential of the differential. First evaluate the Hessians (4x4 numeric matrices) at the same point:

Hf1 <- matrix(Deriv(f1,nderiv=2)(1,2,3,4),4,4) Hf2 <- matrix(Deriv(f2,nderiv=2)(1,2,3,4),4,4) Hf3 <- matrix(Deriv(f3,nderiv=2)(1,2,3,4),4,4)

rownames(Hf1) <- c("w","x","y","z") colnames(Hf1) <- c("w","x","y","z")

For example

Hf1

(note the matrix is symmetric; also note carefully the nonzero diagonal term). But $dd\phi$ is clearly zero as the Hessians are symmetrical:

ij <- expand.grid(seq_len(nrow(Hf1)),seq_len(ncol(Hf1))) ddphi <- # should be zero ( +as.kform(ij,c(Hf1)) +as.kform(ij,c(Hf2)) +as.kform(ij,c(Hf3)) ) ddphi

as expected.

In its most general form, Stokes's theorem states

[ \int_{\partial X}\phi=\int_Xd\phi ]

where $X\subset\mathbb{R}^n$ is a compact oriented $(k+1)$-dimensional manifold with boundary $\partial X$ and $\phi$ is a $k$-form defined on a neighborhood of $X$.

We will verify Stokes, following 6.9.5 of Hubbard in which

[ \phi= \left(x_1-x_2^2+x_3^3-\cdots\pm x_n^n\right) \left( \sum_{i=1}^n dx_1\wedge\cdots\wedge\widehat{dx_i}\wedge\cdots\wedge dx_n \right) ]

(a hat indicates that a term is absent), and we wish to evaluate $\int_{\partial C_a}\phi$ where $C_a$ is the cube $0\leq x_j\leq a, 1\leq j\leq n$. Stokes tells us that this is equal to $\int_{C_a} d\phi$, which is given by

[ d\phi = \left( 1+2x_2+\cdots + nx_n^{n-1}\right) dx_1\wedge\cdots\wedge dx_n ]

and so the volume integral is just

[ \sum_{j=1}^n \int_{x_1=0}^a \int_{x_2=0}^a \cdots \int_{x_i=0}^a jx_j^{j-1} dx_1 dx_2\ldots dx_n= a^{n-1}\left(a+a^2+\cdots+a^n\right). ]

Stokes's theorem, being trivial, is not amenable to direct numerical verification but the package does allow slick creation of $\phi$:

phi <- function(x){ n <- length(x) sum(x^seq_len(n)*rep_len(c(1,-1),n)) * as.kform(t(apply(diag(n)<1,2,which))) } phi(1:9)

(recall that `phi`

is a function that maps $\mathbb{R}^9$ to 8-forms.
Here we choose $\left(1,2,\ldots,9\right)\in\mathbb{R}^9$ and
`phi(1:9)`

as shown above is the resulting 8-form. Thus, if we write
$\phi_{1:9}$ for `phi(1:9)`

we would have
$\phi_{1:9}\colon\left(\mathbb{R}^9\right)^8\longrightarrow\mathbb{R}$,
with package idiom as follows:

f <- as.function(phi(1:9)) E <- matrix(runif(72),9,8) # (R^9)^8 f(E)

Further, $d\phi$ is given by

dphi <- function(x){ nn <- seq_along(x) sum(nn*x^(nn-1)) * as.kform(seq_along(x)) } dphi(1:9)

(observe that `dphi(1:9)`

is a 9-form, with
$d\phi_{1:9}\colon\left(\mathbb{R}^9\right)^9\longrightarrow\mathbb{R}$).
Now consider Spivak's theorem 4.6 (page 82), which in this context
states that a 9-form is proportional to the determinant of the
$9\times 9$ matrix formed from its arguments, with constant of
proportionality equal to the form evaluated on the identity matrix
$I_9$ [formally and more generally, if $v_1,\ldots,v_n$ is a basis for
$V$, $\omega\in\Lambda^n(V)$ and $w_i=\sum a_{ij}v_j$ then
$\omega\left(w_1,\ldots,w_n\right) =
\det\left(a_{ij}\right)\cdot\omega\left(v_1,\ldots v_n\right)$].
Numerically:

f <- as.function(dphi(1:9)) E <- matrix(runif(81),9,9) f(E) det(E)*f(diag(9)) # should match f(E) by Spivak's 4.6

i <- 0 j <- 0 k <- 0 rm(i) rm(j) rm(k)

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