options(rmarkdown.html_vignette.check_title = FALSE) knitr::opts_chunk$set(echo = TRUE) library("stokes") set.seed(0)

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

Alt

Spivak, in a memorable passage, states:

A $k$-tensor $\omega\in{\mathcal J}(V)$ is called **alternating** if
$$
\omega(v_1,\ldots,v_i,\ldots,v_j,\ldots,v_k)=
-\omega(v_1,\ldots,v_j,\ldots,v_i,\ldots,v_k)\qquad\mbox{for all $v_1,\ldots,v_k\in V$.}
$$
$\ldots$ The set of all alternating $k$-tensors is clearly a subspace
$\Lambda^k(V)$ of ${\mathcal J}^k(V)$. Since it requires considerable
work to produce the determinant, it is not surprising that alternating
$k$-tensors are difficult to write down. There is, however, a uniform
way of expressing all of them. Recall that the sign of a permutation
$\sigma$, denoted $\operatorname{sgn}\sigma$, is $+1$ if $\sigma$ is
even and $-1$ if $\sigma$ is odd. If $T\in{\mathcal J}^k(V)$, we
define $\operatorname{Alt}(T)$ by
$$
\operatorname{Alt}(T){\left(v_1,\ldots,v_k\right)}=
\frac{1}{k!}\sum_{\sigma\in S_k}\mathrm{sgn}(\sigma)\cdot
T{\left(v_{\sigma(1)},\ldots,v_{\sigma(k)}\right)}
$$
where $S_k$ is the set of all permutations of numbers $1$ to $k$.

**- Michael Spivak, 1969** *(Calculus on Manifolds, Perseus books). Page 78*

For example, if $k=3$ we have

[\operatorname{Alt}(T)\left(v_1,v_2,v_3\right)= \frac{1}{6}\left(\begin{array}{cc} +T{\left(v_1,v_2,v_3\right)}& -T{\left(v_1,v_3,v_2\right)}\cr -T{\left(v_2,v_1,v_3\right)}& +T{\left(v_2,v_3,v_1\right)}\cr +T{\left(v_3,v_1,v_2\right)}& -T{\left(v_3,v_2,v_1\right)} \end{array} \right) ]

In the `stokes`

package, function `Alt()`

performs this operation on a
given $k$-tensor $T$. Idiom is straightforward:

S <- as.ktensor(rbind(c(1,7,8)))*6 # the "6" is to stop rounding error S

Above, object $S=6\phi_1\otimes\phi_7\otimes\phi_8$, where
$\phi_i\colon\mathbb{R}^8\longrightarrow\mathbb{R}$, with
$\phi_i(e_j)=\delta_{ij}$ [$e_j$ are basis vectors in $\mathbb{R}^8$].
We may calculate $\operatorname{Alt}(S)$ in the package using `Alt()`

:

```
Alt(S)
```

Above we see that
$\operatorname{Alt}(S)=\phi_1\otimes\phi_7\otimes\phi_8-\phi_1\otimes\phi_8\otimes\phi_7\pm\cdots
-\phi_8\otimes\phi_7\otimes\phi_1$, and we can see the pattern
clearly, observing in passing that the order of the rows in the R
object is arbitrary (as per `disordR`

discipline). Now, `Alt(S)`

is
an alternating form, which is easy to verify, by making it act on an
odd permutation of a set of vectors:

V <- matrix(rnorm(24),ncol=3) c(as.function(Alt(S))(V),as.function(Alt(S))(V[,c(2,1,3)]))

Observing that $\operatorname{Alt}$ is linear, we might apply it to a more complicated tensor, here with two terms:

(S <- as.ktensor(rbind(c(1,2,4),c(2,2,3)),c(12,1000)))

Above we have $S=12\phi_1\otimes\phi_2\otimes\phi_4+1000\phi_2\otimes\phi_2\otimes\phi_3$. Calculating $\operatorname{Alt}(S)$:

```
S
Alt(S)
```

Note that the `2 2 3`

term (with coefficient 1000) disappears in
`Alt(S)`

: the even permutations (being positive) cancel out the odd
permutations (being negative) in pairs. This must be the case for an
alternating map by definition. We can see why this cancellation
occurs by considering $T=\phi_2\otimes\phi_2\otimes\phi_3$, a linear
map corresponding to the `[2 2 3]`

row of `S`

:

[ T(v_1,v_2,v_3) = (v_1\cdot e_2)(v_2\cdot e_2)(v_1\cdot e_3)x ]

(where $x\in\mathbb{R}$ is any real number and "$\cdot$" denotes the dot product), and so

[ T(v_2,v_1,v_3) = (v_2\cdot e_2)(v_1\cdot e_2)(v_3\cdot e_3)x = T(v_1,v_2,v_3). ]

Thus if we require $T$ to be alternating [that is, $T(v_2,v_1,v_3) =
-T(v_1,v_2,v_3)$], we must have $x=0$, which is why the `[2 2 3]`

term
with coefficient 1000 vanishes (such terms are killed in the function
by applying `kill_trivial_rows()`

). We can check that terms with
repeated entries are correctly discarded by taking the
$\operatorname{Alt}$ of a tensor all of whose entries include at least
one repeat:

S <- as.ktensor(matrix(c( 3,2,1,1, 1,4,1,4, 1,1,2,3, 7,7,4,7, 1,2,3,3 ),ncol=4,byrow=TRUE),1:5) S Alt(S)

We should verify that `Alt()`

does indeed return an alternating tensor
for complicated tensors (this is trivial algebraically because
$\operatorname{Alt}(\cdot)$ is linear, but it is always good to
check):

S <- rtensor(k=5,n=9) S AS <- Alt(S) V <- matrix(rnorm(45),ncol=5) # element of (R^9)^5

Then we swap columns of $V$, using both even and odd permutations, to verify that $\operatorname{Alt}(S)$ is in fact alternating:

V_even <- V[,c(1,2,5,3,4)] # an even permutation V_odd <- V[,c(2,1,5,3,4)] # an odd permutation V_rep <- V[,c(2,1,5,2,4)] # not a permutation c(as.function(AS)(V),as.function(AS)(V_even)) # should be identical (even permutation) c(as.function(AS)(V),as.function(AS)(V_odd)) # should differ in sign only (odd permutation) as.function(AS)(V_rep) # should be zero

`Alt()`

In his theorem 4.3, Spivak proves the following statements:

- If $T$ is a tensor, then $\operatorname{Alt}(T)$ is alternating
- If $\omega$ is alternating [he writes $\omega\in\Lambda^k(V)$] then $\operatorname{Alt}(\omega)=\omega$
- If $T$ is a tensor, then $\operatorname{Alt}(\operatorname{Alt}(T))=\operatorname{Alt}(T)$.

We have demonstrated the first point above. For the second, we need
to construct a tensor that is alternating, and then show that `Alt()`

does not change it:

P <- as.ktensor(1+diag(2),c(-7,7)) P P == Alt(P)

The third point, idempotence is also easy:

P <- rtensor()*6 # the "6" avoids numerical round-off issues Alt(Alt(P))==Alt(P) # should be TRUE

Spivak defines the wedge product as follows. Given alternating forms $\omega\in\Lambda^k(V)$ and $\eta\in\Lambda^l(V)$ we have

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

So for example:

omega <- as.ktensor(2+diag(2),c(-7,7)) eta <- Alt(ktensor(6*spray(matrix(c(1,2,3,1,4,7,4,5,6),3,3,byrow=TRUE),1:3))) omega eta

Above we see that `omega`

is alternating by construction, and `eta`

is
alternating by virtue of `Alt()`

. Thus tensor $\omega\wedge\eta$ is
defined as per Spivak's definition, and we may calculate it directly:

Alt(omega %X% eta,give_kform = TRUE) f <- as.function(Alt(omega %X% eta))

Observe that the tensor $\omega\otimes\eta$ (which would be returned
if the default argument `give_kform=FALSE`

was sent to `Alt()`

) is
quite long, having $2\cdot 5!=240$ nonzero components, which is why it
is not printed in full. We may verify that `f()`

is alternating by
evaluating it on a randomly chosen point in
$\left(\mathbb{R}^7\right)^5$:

V <- matrix(rnorm(35),ncol=5) c(f(V),f(V[,c(2:1,3:5)]))

Above we see a verification that `f()`

is fact alternating: writing
the matrix `V`

in terms of its five vectors as
$V=\left[v_1,v_2,v_3,v_4,v_5\right]$, where $v_i\in\mathbb{R}^7$, we
see that
$f{\left(v_1,v_2,v_3,v_4,v_5\right)}=-f{\left(v_2,v_1,v_3,v_4,v_5\right)}$.

`Alt()`

Spivak goes on to prove the following three statements. If $S,T$ are tensors and $\omega,\eta,\theta$ are alternating tensors of arity $k,l,m$ respectively, then

- If $\operatorname{Alt}(S)=0$, then $\operatorname{Alt}(S\otimes T)=\operatorname{Alt}(T\otimes S)=0$.
- $\operatorname{Alt}(\operatorname{Alt}(\omega\otimes\eta)\otimes\theta) =\operatorname{Alt}(\omega\otimes\eta\otimes\theta)= \operatorname{Alt}(\omega\otimes\operatorname{Alt}(\eta\otimes\theta))$
- $(\omega\wedge\eta)\wedge\theta = \omega\wedge(\eta\wedge\theta) = \frac{(k+l+m)!}{k!l!m!}\operatorname{Alt}(\omega\otimes\eta\otimes\theta)$.

Taking the points in turn. Firstly $\operatorname{Alt}(S\otimes T)=\operatorname{Alt}(T\otimes S)=0$:

(S <- as.ktensor(rbind(c(1,2,3,3),c(1,1,2,3)),1000:1001)) Alt(S) # each row of S includes repeats T <- rtensor() c(is.zero(Alt(S %X% T)), is.zero(Alt(T %X% S)))

secondly, $\operatorname{Alt}(\operatorname{Alt}(\omega\otimes\eta)\otimes\theta) =\operatorname{Alt}(\omega\otimes\eta\otimes\theta)= \operatorname{Alt}(\omega\otimes\operatorname{Alt}(\eta\otimes\theta))$:

omega <- Alt(as.ktensor(rbind(1:3),6)) eta <- Alt(as.ktensor(rbind(4:5),60)) theta <- Alt(as.ktensor(rbind(6:7),14)) omega eta theta f1 <- as.function(Alt(Alt(omega %X% eta) %X% theta)) f2 <- as.function(Alt(omega %X% eta %X% theta)) f3 <- as.function(Alt(omega %X% Alt(eta %X% theta))) V <- matrix(rnorm(9*14),ncol=9) c(f1(V),f2(V),f3(V))

Verifying the third identity $(\omega\wedge\eta)\wedge\theta = \omega\wedge(\eta\wedge\theta) = \frac{(k+l+m)!}{k!l!m!}\operatorname{Alt}(\omega\otimes\eta\otimes\theta)$ needs us to coerce from a $k$-form to a $k$-tensor:

omega <- rform(2,2,19) eta <- rform(3,2,19) theta <- rform(2,2,19) a1 <- as.ktensor(omega ^ (eta ^ theta)) a2 <- as.ktensor((omega ^ eta) ^ theta) a3 <- Alt(as.ktensor(omega) %X% as.ktensor(eta) %X% as.ktensor(theta))*90 c(is.zero(a1-a2),is.zero(a1-a3),is.zero(a2-a3))

`give_kform`

Function `Alt()`

takes a Boolean argument `give_kform`

. We have been
using `Alt()`

with `give_kform`

taking its default value of `FALSE`

,
which means that it returns an object of class `ktensor`

. However, an
alternating form can be much more efficiently represented as an object
of class `kform`

, and this is returned if `give_kform`

is `TRUE`

. Here
I verify that the two options return identical objects:

(rand_tensor <- rtensor(k=5,n=9)*120) S1 <- Alt(rand_tensor) # 120 terms, too long to print all of it summary(S1)

Above, we see that `S1`

is a rather extensive object, with 120 terms.
However, if argument `give_kform = TRUE`

is passed to `Alt()`

we get a
`kform`

object which is much more succinct:

(SA1 <- Alt(rand_tensor,give_kform=TRUE))

Verification that objects `S1`

and `SA1`

are the same object:

V <- matrix(rnorm(45),ncol=5) LHS <- as.function(S1)(V) RHS <- as.function(SA1)(V) c(LHS=LHS,RHS=RHS,diff=LHS-RHS)

Above we see agreement to within numerical error.

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