Wedge products and the exterior calculus

This section follows the exposition of Hubbard and Hubbard, who introduce the exterior calculus starting with a discussion of elementary forms. An elementary form is an object such as \(dx_1\wedge dx_3\), which is an alternating linear map from \(\mathbb{R}^n\times\mathbb{R}^n\) to \(\mathbb{R}\) with

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

That this is alternating follows from the properties of the determinant. Because such objects are linear, it is possible to consider more complicated 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 \(dx_1\wedge dx_2\wedge dx_3 +5dx_1\wedge dx_2\wedge dx_4\) which would be a linear map from \(\left(\mathbb{R}^n\right)^3\) to \(\mathbb{R}\) with

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

Taking the last form as an example, this has ready R idiom:

M <- matrix(c(4,2,3,1,2,4),2,3,byrow=TRUE)
M

##      [,1] [,2] [,3]
## [1,]    4    2    3
## [2,]    1    2    4

K <- as.kform(M,c(1,5))
K

##            val
##  2 3 4  =    1
##  1 2 4  =    5

Note that the order of the rows in K is immaterial and indeed in some implementations will appear in a different order: the wedge package uses the spray package, which in turn utilises the STL map class of C++.

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

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

and these products are called elementary forms: 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

##          val
##  1 2  =    1
##  1 3  =    2
##  2 3  =    3
##  1 4  =    4
##  2 4  =    5
##  3 4  =    6

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.

Formal definition of wedge product

Given two alternating forms \(\omega,\eta\), Spivak defines the wedge product \(\omega\wedge\eta\in\Lambda^{k+l}(V)\) as

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

and this is implemented in the package by function wedge(), or, more idiomatically, %^%:

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

##            val
##  3 4 5  =    2
##  1 4 6  =    7

M2 <- cbind(1:5,3:7)
K2 <- as.kform(M2,1:5)
K2

##          val
##  1 3  =    1
##  5 7  =    5
##  2 4  =    2
##  4 6  =    4
##  3 5  =    3

K1 %^% K2

##                val
##  1 4 5 6 7  =  -35
##  1 3 4 5 6  =  -21

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 but only two coefficients are nonzero out of a possible \({7\choose 5}=21\) terms. 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

##                    val
##  1 2 3 4 5 7 8  =   -5
##  1 3 4 5 6 7 8  =   -2
##  1 2 3 5 6 7 8  =   11
##  1 2 3 4 5 6 8  =    1
##  2 3 4 5 6 7 8  =    6
##  1 2 3 4 6 7 8  =    2
##  1 2 3 4 5 6 7  =    1
##  1 2 4 5 6 7 8  =   -5

F1 %^% (F2 %^% F3)

##                    val
##  1 2 3 4 5 6 7  =    1
##  1 3 4 5 6 7 8  =   -2
##  1 2 3 4 5 7 8  =   -5
##  1 2 3 4 6 7 8  =    2
##  1 2 3 4 5 6 8  =    1
##  1 2 3 5 6 7 8  =   11
##  2 3 4 5 6 7 8  =    6
##  1 2 4 5 6 7 8  =   -5

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:

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

## empty sparse array with 7 columns

Exterior derivatives

Hubbard and Hubbard 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 can be shown that

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

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))

##         val
##  1  =   0.4
##  2  =   0.1
##  3  =  -3.2
##  4  =   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

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

(where a hat indicates the absence of a term), and show analyically 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)

##              val
##  2 3 4 5  =    1
##  1 3 4 5  =    1
##  1 2 4 5  =    1
##  1 2 3 5  =    1
##  1 2 3 4  =    1

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)

##              val
##  1  =   4.05e-05
##  2  =  -2.84e-05
##  3  =   8.10e-06
##  4  =   2.03e-05
##  5  =  -5.67e-05

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

##                        val
##  1 2 3 4 5 6 7 8 9  =    0

##                        val
##  1 2 3 4 5 6 7 8 9  =    0

Stokes’s theorem

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)

##                            val
##  1 2 3 4 6 7 8 9  =  371423053
##  1 3 4 5 6 7 8 9  =  371423053
##  1 2 3 4 5 6 7 9  =  371423053
##  1 2 3 4 5 7 8 9  =  371423053
##  1 2 4 5 6 7 8 9  =  371423053
##  2 3 4 5 6 7 8 9  =  371423053
##  1 2 3 4 5 6 8 9  =  371423053
##  1 2 3 5 6 7 8 9  =  371423053
##  1 2 3 4 5 6 7 8  =  371423053

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)

##                              val
##  1 2 3 4 5 6 7 8 9  =  405071317

References

• J. H. Hubbard and B. B. Hubbard 2015. Vector calculus, linear algebra and differential forms: a unified aproach, Ithaca, NY.
• M. Spivak 1971. Calculus on manifolds, Addison-Wesley.