`wedge()`

and related functions in the `stokes`

package

```
## function (x, ...)
## {
## if (nargs() < 3) {
## wedge2(x, ...)
## }
## else {
## wedge2(x, Recall(...))
## }
## }
## <bytecode: 0x55abc8bf5360>
## <environment: namespace:stokes>
```

```
## function (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))
## }
## <bytecode: 0x55abc3d20cf8>
## <environment: namespace:stokes>
```

[the meat of `wedge2()`

is `kform(spraycross(K1, K2))`

].

Function `wedge()`

returns the wedge product of any number of kforms, function `wedge2()`

returns the wedge product of two kforms. Spivak gives us:

\[ \omega\wedge\eta=\frac{\left(k+l\right)!}{k!l!}\operatorname{Alt}(\omega\otimes\eta),\qquad\omega\in\Lambda^k(V),\eta\in\Lambda^l(V) \]

Function `wedge2()`

implements this, although the idiom is somewhat opaque, especially the combinatorial coefficient \((k+l)!/(k!l!)\).

`spraycross()`

Function `wedge()`

is essentially a convenience wrapper for `spraycross()`

. Function `spraycross()`

it is part of the `spray`

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

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

```
## val
## 11 13 = 11
## 10 12 = 7
```

```
## val
## 1 3 10 12 = 14
## 1 3 11 13 = 22
## 2 4 10 12 = 35
## 2 4 11 13 = 55
```

```
## val
## 10 12 1 3 = 14
## 11 13 1 3 = 22
## 10 12 2 4 = 35
## 11 13 2 4 = 55
```

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:

```
## 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
```

```
## An alternating linear map from V^5 to R with V=R^7:
## val
## 1 2 3 4 7 = 35
```

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.

```
## 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 = 7
```

As functions, `y`

and `ty`

are identical:

`## [1] 15.23211 15.23211`

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]
)
```

`## [1] 15.23211`

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 `cross()`

:

```
## 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 = 35
```

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:

```
## An alternating linear map from V^5 to R with V=R^7:
## val
## 1 2 3 4 7 = 35
```

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 \(a\in\mathcal{R}\) 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 basis for \(\Lambda^k(V)\). Spivak tells us that, if \(v_1,\ldots,v_k\) is a basis for \(V\), then the set of all

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

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 = 5
```

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:

```
## An alternating linear map from V^6 to R with V=R^8:
## val
## 1 2 3 5 6 8 = -15
```

See how function `wedge()`

does the legwork.