# The Alt() function in the stokes package Alt
## function (S, give_kform = FALSE)
## {
##     if (is.kform(S)) {
##         return(S)
##     }
##     if (give_kform) {
##         return(kform(S)/factorial(arity(S)))
##     }
##     out <- kill_trivial_rows(S)
##     if (nrow(index(out)) == 0) {
##         return(S * 0)
##     }
##     ktensor(include_perms(consolidate(out))/factorial(ncol(index(out))))
## }
## <bytecode: 0x55abbff4b510>
## <environment: namespace:stokes>

Function Alt() converts a $$k$$-tensor to alternating form (see kform.Rd). Given a $$k$$-tensor $$T$$, we follow Spivak [p78] and define $$\operatorname{Alt}(T)$$ as follows:

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

where $$S_k$$ is the set of all permutations of numbers $$1$$ to $$k$$ and $$\operatorname{sgn}\sigma$$ is the sign of permutation $$\sigma$$ for any $$\sigma\in S_k$$. Thus for example if $$k=3$$ we have

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

Package idiom is straightforward:

S <- as.ktensor(rbind(c(1,7,8)))*6  # the "6" is to stop rounding error
S
## A linear map from V^3 to R with V=R^8:
##            val
##  1 7 8  =    6
Alt(S)
## A linear map from V^3 to R with V=R^8:
##            val
##  8 7 1  =   -1
##  8 1 7  =    1
##  7 8 1  =    1
##  7 1 8  =   -1
##  1 8 7  =   -1
##  1 7 8  =    1

and we can see the pattern clearly, observing in passing that the order of the rows is arbitrary. 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)]))
##  -0.1645455  0.1645455

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))
S
## A linear map from V^3 to R with V=R^4:
##             val
##  2 2 3  =  1000
##  1 2 4  =    12
Alt(S)
## A linear map from V^3 to R with V=R^4:
##            val
##  4 2 1  =   -2
##  4 1 2  =    2
##  2 4 1  =    2
##  2 1 4  =   -2
##  1 4 2  =   -2
##  1 2 4  =    2

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 $$T$$, a linear map corresponding to the second row of S above, that is, [2 2 3]. We have

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

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

$T(v_2,v_1,v_3) = (v_1\cdot e_2)(v_1\cdot e_1)(v_1\cdot e_3)x = T(v_1,v_2,v_3).$

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
## A linear map from V^4 to R with V=R^7:
##              val
##  1 2 3 3  =    5
##  7 7 4 7  =    4
##  1 1 2 3  =    3
##  1 4 1 4  =    2
##  3 2 1 1  =    1
Alt(S)
## The zero linear map from V^4 to R with V=R^n:
## empty sparse array with 4 columns

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
## A linear map from V^5 to R with V=R^9:
##                val
##  6 1 1 1 2  =    9
##  8 6 7 6 6  =    8
##  7 6 8 7 3  =    7
##  8 2 7 7 7  =    5
##  9 8 6 9 9  =    4
##  6 6 8 9 1  =    3
##  9 2 6 4 7  =    6
##  7 3 3 8 6  =    2
##  9 7 3 4 5  =    1
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)
##  0.226959 0.226959
c(as.function(AS)(V),as.function(AS)(V_odd))    # should differ in sign only (odd permutation)
##   0.226959 -0.226959
as.function(AS)(V_rep)                          # should be zero
##  -1.01644e-20

# Further properties of 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
## A linear map from V^2 to R with V=R^2:
##          val
##  1 2  =    7
##  2 1  =   -7
P == Alt(P)
##  TRUE

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
##  TRUE

# Wedge product

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(rtensor(2))*30
omega
## A linear map from V^2 to R with V=R^3:
##          val
##  2 3  =    7
##  3 2  =   -7
eta
## The zero linear map from V^3 to R with V=R^n:
## empty sparse array with 3 columns
f <- as.function(Alt(omega %X% eta))

(the tensor itself is quite long, having 60 nonzero components). We may verify that f() is in fact alternating:

V <-  matrix(rnorm(35),ncol=5)
c(f(V),f(V[,c(2:1,3:5)]))
##  0 0

## Further further properties of 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)) 
## A linear map from V^4 to R with V=R^3:
##               val
##  1 1 2 3  =  1001
##  1 2 3 3  =  1000
Alt(S)  # each row of S includes repeats
## The zero linear map from V^4 to R with V=R^n:
## empty sparse array with 4 columns
T <- rtensor()
c(is.zero(Alt(S %X% T)), is.zero(Alt(T %X% S)))
##  TRUE TRUE

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
## A linear map from V^3 to R with V=R^3:
##            val
##  3 2 1  =   -1
##  3 1 2  =    1
##  2 3 1  =    1
##  2 1 3  =   -1
##  1 3 2  =   -1
##  1 2 3  =    1
eta
## A linear map from V^2 to R with V=R^5:
##          val
##  5 4  =  -30
##  4 5  =   30
theta
## A linear map from V^2 to R with V=R^7:
##          val
##  7 6  =   -7
##  6 7  =    7
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))
##  -8.948642 -8.948642 -8.948642

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))
##  TRUE TRUE TRUE

# Argument 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:

(rand_tensor <- rtensor(k=5,n=9)*120)
## A linear map from V^5 to R with V=R^9:
##                 val
##  1 6 4 7 6  =   120
##  6 7 3 3 4  =   600
##  1 3 5 1 5  =  1080
##  9 2 3 5 3  =   240
##  8 4 2 8 8  =   360
##  6 5 6 5 4  =   480
##  3 6 4 7 4  =   720
##  1 3 4 4 5  =   840
##  6 4 5 7 1  =   960
S1 <- Alt(rand_tensor)  # 720 terms, too long to print
(SA1 <- Alt(rand_tensor,give_kform=TRUE))
## An alternating linear map from V^5 to R with V=R^7:
##                val
##  1 4 5 6 7  =    8

Verification:

V <- matrix(rnorm(45),ncol=5)
c(as.function(S1)(V),as.function(SA1)(V)) # should match
##  -5.63792 -5.63792