phifunction (n)
{
ktensor(spray(n, 1))
}To cite the stokes package in publications, please use
Hankin (2022b). Function
phi() returns a tensor dual to the standard basis of V=\mathbb{R}^n. Here I discuss
phi() but there is some overlap between this vignette and
the tensorprod vignette.
In a memorable passage, Spivak (1965) states (theorem 4.1):
Integration on chains
Let v_1,\ldots,v_n be a basis for V, and let \phi_1,\ldots,\phi_n be the dual basis, \phi_i(v_j)=\delta_{ij}. Then the set of all k-fold tensor products
\phi_{i_1}\otimes\cdots\otimes\phi_{i_k}\qquad 1\leqslant i_1,\ldots,i_k\leqslant n
is a basis for \mathcal{J}(V), which therefore has dimension n^k.
- Michael Spivak, 1969 (Calculus on Manifolds, Perseus books). Page 75
Function phi() returns a very simple tensor:
phi(4)## A linear map from V^1 to R with V=R^4:
## val
## 4 = 1First we will verify the properties of phi(), using
V=\mathbb{R}^5, specifically
\phi_i(e_j) = \delta_{ij} = \begin{cases} 1, & i=j\\ 0 & i\neq j. \end{cases}
(package idiom is to use e() for basis vectors as
opposed to Spivak’s v). As numerical
verification, we will check that \phi_3(e_2)=0, \phi_3(e_3)=1, \phi_3(e_4)=0:
## [1] 0 1 0A more severe test might be
aa <- function(n){
outer(seq_len(n), seq_len(n),
Vectorize(function(i, j){as.function(phi(i))(as.matrix(e(j, n)))}))
}
aa(5)## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 0 0 0
## [2,] 0 1 0 0 0
## [3,] 0 0 1 0 0
## [4,] 0 0 0 1 0
## [5,] 0 0 0 0 1Above, we see that the matrix is I_5, as expected. Further:
## [1] TRUEThe objects created by phi() may be multiplied together
using tensorprod() or its binary operator
%X%:
## A linear map from V^3 to R with V=R^5:
## val
## 4 3 5 = 1If we want to create arbitrary tensor products of \phi objects the most natural way would be to
use tensorprod() repeatedly:
## A linear map from V^5 to R with V=R^8:
## val
## 4 5 6 7 8 = 1However, function phi() simply takes a vector:
## A linear map from V^3 to R with V=R^5:
## val
## 2 5 1 = 1This creates an element of the basis set, in this case \phi_2\otimes\phi_5\otimes\phi_1. Verification is straightforward:
(v <- sample(9))## [1] 9 4 7 1 2 6 3 8 5## [1] TRUEWe will consider an element X of \mathcal{J}^{2}(V) where V=\mathbb{R}^3 and construct an explicit basis for it along the lines of Spivak’s observation above.
## A linear map from V^2 to R with V=R^3:
## val
## 3 1 = 3
## 2 1 = 2
## 1 2 = 1Thus X=\phi_1\otimes\phi_2 +2\phi_2\otimes\phi_1 +3\phi_3\otimes\phi_1. Spivak asserts that \mathcal{J}^{2}(V) has dimension n^k=3^2=9.
## A linear map from V^2 to R with V=R^3:
## val
## 2 1 = 2
## 1 2 = 1
## 3 1 = 3With a little effort, we can create all 3^2=9 elements of a basis as follows:
apply(expand.grid(rep(list(seq_len(3)),2)),1,phi)## [[1]]
## A linear map from V^2 to R with V=R^1:
## val
## 1 1 = 1
##
## [[2]]
## A linear map from V^2 to R with V=R^2:
## val
## 2 1 = 1
##
## [[3]]
## A linear map from V^2 to R with V=R^3:
## val
## 3 1 = 1
##
## [[4]]
## A linear map from V^2 to R with V=R^2:
## val
## 1 2 = 1
##
## [[5]]
## A linear map from V^2 to R with V=R^2:
## val
## 2 2 = 1
##
## [[6]]
## A linear map from V^2 to R with V=R^3:
## val
## 3 2 = 1
##
## [[7]]
## A linear map from V^2 to R with V=R^3:
## val
## 1 3 = 1
##
## [[8]]
## A linear map from V^2 to R with V=R^3:
## val
## 2 3 = 1
##
## [[9]]
## A linear map from V^2 to R with V=R^3:
## val
## 3 3 = 1Or it might be logically better to use ellipsis constructs to pass multiple arguments:
## A linear map from V^3 to R with V=R^6:
## val
## 3 4 6 = 1Then we could have
1*s(1,2) + 2*s(2,1) + 3*s(3,1)## A linear map from V^2 to R with V=R^3:
## val
## 2 1 = 2
## 1 2 = 1
## 3 1 = 3
1*s(1,2) + 2*s(2,1) + 3*s(3,1) == X## [1] TRUEDistributivity
The tensor product is left- and right distributive. To illustrate this we can use the package to calculate, say, (2\phi_1+3\phi_2)\otimes(5\phi_3+7\phi_4):
## A linear map from V^2 to R with V=R^4:
## val
## 2 4 = 21
## 1 4 = 14
## 2 3 = 15
## 1 3 = 10Above we see package form for the result which is 10\phi_1\phi_3 + 14\phi_1\phi_4 + 15\phi_2\phi_3 + 21\phi_2\phi_4 in algebraic notation.
Reconstruction of a given tensor
Consider the following tensor
## A linear map from V^2 to R with V=R^7:
## val
## 4 4 = 8
## 7 3 = 9
## 3 5 = 7We may express b as the sum of its three terms, each with a coefficient:
## A linear map from V^2 to R with V=R^7:
## val
## 4 4 = 8
## 3 5 = 7
## 7 3 = 9Above, observe that the order of the terms may differ between the two
methods, as per disordR discipline (Hankin
2022a), but they are algebraically identical:
## [1] TRUEFunction Alt()
Function Alt() returns an alternating tensor as
documented in the Alt vignette in the package. It works
nicely with phi():
phi(1:3)## A linear map from V^3 to R with V=R^3:
## val
## 1 2 3 = 1## 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 = 1References
disordR Package.” https://arxiv.org/abs/2210.03856; arXiv. https://doi.org/10.48550/ARXIV.2210.03856.