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: 0x560d8076fef8>
## <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 = 6Above, 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
## 8 7 1 = -1
## 8 1 7 = 1
## 7 8 1 = 1
## 7 1 8 = -1
## 1 8 7 = -1
## 1 7 8 = 1Above 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)]))## [1] -0.1645455 0.1645455Observing 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
## 2 2 3 = 1000
## 1 2 4 = 12Above 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
## 2 2 3 = 1000
## 1 2 4 = 12Alt(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 = 2Note 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
## 1 2 3 3 = 5
## 7 7 4 7 = 4
## 1 1 2 3 = 3
## 1 4 1 4 = 2
## 3 2 1 1 = 1Alt(S)## The zero linear map from V^4 to R with V=R^n:
## empty sparse array with 4 columnsWe 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 = 1AS <- Alt(S)
V <- matrix(rnorm(45),ncol=5) # element of (R^9)^5Then 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)## [1] 0.226959 0.226959c(as.function(AS)(V),as.function(AS)(V_odd)) # should differ in sign only (odd permutation)## [1] 0.226959 -0.226959as.function(AS)(V_rep) # should be zero## [1] -1.01644e-20Alt()In his theorem 4.3, Spivak proves the following statements:
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 = -7P == Alt(P)## [1] TRUEThe third point, idempotence is also easy:
P <- rtensor()*6 # the "6" avoids numerical round-off issues
Alt(Alt(P))==Alt(P) # should be TRUE## [1] TRUESpivak 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## A linear map from V^2 to R with V=R^3:
## val
## 2 3 = 7
## 3 2 = -7eta## A linear map from V^3 to R with V=R^7:
## val
## 1 4 7 = 2
## 3 2 1 = -1
## 4 6 5 = -3
## 1 7 4 = -2
## 4 5 6 = 3
## 7 1 4 = 2
## 4 1 7 = -2
## 4 7 1 = 2
## 6 5 4 = -3
## 6 4 5 = 3
## 5 4 6 = -3
## 1 3 2 = -1
## 2 1 3 = -1
## 1 2 3 = 1
## 2 3 1 = 1
## 7 4 1 = -2
## 3 1 2 = 1
## 5 6 4 = 3Above 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)## An alternating linear map from V^5 to R with V=R^7:
## val
## 2 3 4 5 6 = 2.1
## 1 2 3 4 7 = 1.4f <- 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)]))## [1] -0.852457 0.852457Above 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
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 = 1000Alt(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 columnsT <- rtensor()
c(is.zero(Alt(S %X% T)), is.zero(Alt(T %X% S)))## [1] TRUE TRUEsecondly, \(\operatorname{Alt}(\operatorname{Alt}(\omega\otimes\eta)\otimes\theta) =\operatorname{Alt}(\omega\otimes\eta\otimes\theta)= \operatorname{Alt}(\omega\otimes\operatorname{Alt}(\eta\otimes\theta))\):
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 = 1eta## A linear map from V^2 to R with V=R^5:
## val
## 5 4 = -30
## 4 5 = 30theta## A linear map from V^2 to R with V=R^7:
## val
## 7 6 = -7
## 6 7 = 7f1 <- 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))## [1] -3.154794 -3.154794 -3.154794Verifying 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))## [1] TRUE TRUE TRUEgive_kformFunction 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
## 7 3 6 4 7 = 120
## 8 1 3 4 4 = 240
## 2 6 4 5 7 = 360
## 8 4 2 8 8 = 840
## 7 1 3 5 1 = 480
## 1 6 4 7 6 = 600
## 6 7 3 3 4 = 1080
## 9 2 3 5 3 = 720
## 6 5 6 5 4 = 960S1 <- 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 866c13390582e2f7e562be3e046abda44be30079
##
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -3 -3 0 0 3 3
##
##
## Representative selection of index and coefficients:
##
## A linear map from V^5 to R with V=R^7:
## val
## 7 6 5 4 2 = 3
## 2 6 4 7 5 = -3
## 7 4 2 6 5 = 3
## 2 7 6 4 5 = -3
## 4 6 2 5 7 = -3
## 6 5 2 4 7 = -3Above, 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
## 2 4 5 6 7 = 3Verification 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
## 23.86928 23.86928 0.00000Above we see agreement to within numerical error.