# 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: 0x7fc3a0c878e8>
## <environment: namespace:stokes>

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
## A linear map from V^3 to R with V=R^8:
##            val
##  1 7 8  =    6

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)
## A linear map from V^3 to R with V=R^8:
##            val
##  1 7 8  =    1
##  7 8 1  =    1
##  7 1 8  =   -1
##  8 1 7  =    1
##  1 8 7  =   -1
##  8 7 1  =   -1

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)]))
##  -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)))
## A linear map from V^3 to R with V=R^4:
##             val
##  1 2 4  =    12
##  2 2 3  =  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
## A linear map from V^3 to R with V=R^4:
##             val
##  1 2 4  =    12
##  2 2 3  =  1000
Alt(S)
## A linear map from V^3 to R with V=R^4:
##            val
##  1 2 4  =    2
##  2 4 1  =    2
##  1 4 2  =   -2
##  2 1 4  =   -2
##  4 1 2  =    2
##  4 2 1  =   -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 $$T=\phi_2\otimes\phi_2\otimes\phi_3$$, 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
## A linear map from V^4 to R with V=R^7:
##              val
##  3 2 1 1  =    1
##  1 4 1 4  =    2
##  1 1 2 3  =    3
##  1 2 3 3  =    5
##  7 7 4 7  =    4
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
##  9 7 3 4 5  =    1
##  9 2 6 4 7  =    6
##  6 6 8 9 1  =    3
##  7 3 3 8 6  =    2
##  9 8 6 9 9  =    4
##  8 2 7 7 7  =    5
##  7 6 8 7 3  =    7
##  6 1 1 1 2  =    9
##  8 6 7 6 6  =    8
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
##  6.776264e-21

# 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
##  2 1  =   -7
##  1 2  =    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
##  3 2  =   -7
##  2 3  =    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 2 3 3  =  1000
##  1 1 2 3  =  1001
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
##  1 2 3  =    1
##  2 1 3  =   -1
##  1 3 2  =   -1
##  2 3 1  =    1
##  3 1 2  =    1
##  3 2 1  =   -1
eta
## A linear map from V^2 to R with V=R^5:
##          val
##  4 5  =   30
##  5 4  =  -30
theta
## A linear map from V^2 to R with V=R^7:
##          val
##  6 7  =    7
##  7 6  =   -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. Here I verify that the two options return identical objects:

(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
##  9 2 3 5 3  =   240
##  8 4 2 8 8  =   360
##  6 7 3 3 4  =   600
##  6 5 6 5 4  =   480
##  1 3 4 4 5  =   840
##  3 6 4 7 4  =   720
##  6 4 5 7 1  =   960
##  1 3 5 1 5  =  1080
S1 <- Alt(rand_tensor)  # 120 terms, too long to print all of it
summary(S1)
## A ktensor object with 120 terms.  Summary of coefficients:
##
## a disord object with hash 8f7aa495692952e61f58fe758cdbb80768150a0b
##
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
##      -8      -8       0       0       8       8
##
##
## Representative selection of index and coefficients:
##
## A linear map from V^5 to R with V=R^7:
##                val
##  1 4 6 5 7  =   -8
##  4 6 5 1 7  =    8
##  1 7 4 5 6  =   -8
##  4 6 1 5 7  =   -8
##  1 4 7 6 5  =   -8
##  7 1 5 4 6  =   -8

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))
## An alternating linear map from V^5 to R with V=R^7:
##                val
##  1 4 5 6 7  =    8

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)
##           LHS           RHS          diff
## -5.637920e+00 -5.637920e+00  7.105427e-15

Above we see agreement to within numerical error.