Functions wedge() and
wedge2() in the stokes package
Robin K. S. Hankin
Source:vignettes/wedge.Rmd
wedge.Rmd
wedgefunction (x, ...)
{
if (nargs() < 3) {
wedge2(x, ...)
}
else {
wedge2(x, Recall(...))
}
}
wedge2function (K1, K2)
{
if (missing(K2)) {
return(K1)
}
if (is.ktensor(K1) | is.ktensor(K2)) {
stop("wedge product only defined for kforms")
}
if (!is.kform(K1) | !is.kform(K2)) {
return(K1 * K2)
}
if (is.empty(K1) | is.empty(K2)) {
return(zeroform(arity(K1) + arity(K2)))
}
kform(spraycross(K1, K2))
}To cite the stokes package in publications, please use
Hankin (2022). In a memorable passage,
Spivak (1965) 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. The idiom of wedge2()
is somewhat opaque, especially the “strange” combinatorial coefficient
(k+l)!/(k!l!), which is discussed in
detail below.
Digression: function 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:
## val
## 2 4 = 5
## 1 3 = 2## val
## 11 13 = 11
## 10 12 = 7
spraycross(a,b)## val
## 1 3 10 12 = 14
## 1 3 11 13 = 22
## 2 4 10 12 = 35
## 2 4 11 13 = 55
spraycross(b,a)## val
## 10 12 1 3 = 14
## 11 13 1 3 = 22
## 10 12 2 4 = 35
## 11 13 2 4 = 55Observe that spraycross() (and by association
wedge()) is associative and distributive but not
commutative.
Cut to the chase: wedge2()
Function wedge2() takes two kforms and we will start
with a very simple example:
## An alternating linear map from V^2 to R with V=R^2:
## val
## 1 2 = 5## An alternating linear map from V^3 to R with V=R^7:
## val
## 3 4 7 = 7
wedge2(x,y)## An alternating linear map from V^5 to R with V=R^7:
## val
## 1 2 3 4 7 = 35It 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'## A linear map from V^3 to R with V=R^7:
## val
## 7 4 3 = -7
## 7 3 4 = 7
## 4 7 3 = 7
## 4 3 7 = -7
## 3 7 4 = -7
## 3 4 7 = 7As 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))## [1] 15.23211 15.23211Both 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]
)## [1] 15.23211We 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 combinatorial term)
for x\wedge y by coercing to ktensors
and using tensorprod():
(z <- tensorprod(as.ktensor(x),as.ktensor(y)))## A linear map from V^5 to R with V=R^7:
## val
## 1 2 3 4 7 = 35
## 2 1 3 7 4 = 35
## 1 2 4 3 7 = -35
## 2 1 3 4 7 = -35
## 1 2 4 7 3 = 35
## 2 1 4 3 7 = 35
## 2 1 4 7 3 = -35
## 1 2 7 3 4 = 35
## 2 1 7 3 4 = -35
## 1 2 3 7 4 = -35
## 1 2 7 4 3 = -35
## 2 1 7 4 3 = 35Above, 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)## An alternating linear map from V^5 to R with V=R^7:
## val
## 1 2 3 4 7 = 35So to conclude, the combinatorial term is present in the R idiom, it is just difficult to see at first glance.
Algebraic properties
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). Spivak (1965) 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:
## An alternating linear map from V^3 to R with V=R^8:
## val
## 1 3 7 = 6
## 1 2 8 = 5In 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:
## An alternating linear map from V^6 to R with V=R^8:
## val
## 1 2 3 5 6 8 = -15See how function wedge() does the legwork.