transform.RdGiven a \(k\)-form, express it in terms of linear combinations of the \(dx_i\)
pullback(K,M)
stretch(K,d)Function pullback() calculates the pullback of a function. A
vignette is provided at pullback.Rmd.
Suppose we are given a two-form
$$ \omega=\sum_{i < j}a_{ij}\mathrm{d}x_i\wedge\mathrm{d}x_j$$
and relationships
$$\mathrm{d}x_i=\sum_rM_{ir}\mathrm{d}y_r$$
then we would have
$$\omega = \sum_{i < j} a_{ij}\left(\sum_rM_{ir}\mathrm{d}y_r\right)\wedge\left(\sum_rM_{jr}\mathrm{d}y_r\right). $$
The general situation would be a \(k\)-form where we would have $$ \omega=\sum_{i_1 < \cdots < i_k}a_{i_1\ldots i_k}\mathrm{d}x_{i_1}\wedge\cdots\wedge\mathrm{d}x_{i_k}$$
giving
$$\omega = \sum_{i_1 < \cdots < i_k}\left[ a_{i_1,\ldots, i_k}\left(\sum_rM_{i_1r}\mathrm{d}y_r\right)\wedge\cdots\wedge\left(\sum_rM_{i_kr}\mathrm{d}y_r\right)\right]. $$
The transform() function does all this but it is slow. I am not
100% sure that there isn't a much more efficient way to do such a
transformation. There are a few tests in tests/testthat and a
discussion in the stokes vignette.
Function stretch() carries out the same operation but for \(M\)
a diagonal matrix. It is much faster than transform().
The functions documented here return an object of class
kform.
S. H. Weintraub 2019. Differential forms: theory and practice. Elsevier. (Chapter 3)
# Example in the text:
K <- as.kform(matrix(c(1,1,2,3),2,2),c(1,5))
M <- matrix(1:9,3,3)
pullback(K,M)
#> An alternating linear map from V^2 to R with V=R^3:
#> val
#> 1 2 = -33
#> 1 3 = -66
#> 2 3 = -33
# Demonstrate that the result can be complicated:
M <- matrix(rnorm(25),5,5)
pullback(as.kform(1:2),M)
#> An alternating linear map from V^2 to R with V=R^5:
#> val
#> 3 5 = -0.98311942
#> 1 2 = -0.36990935
#> 2 4 = -0.20729935
#> 1 5 = -0.54937477
#> 2 5 = 0.96506752
#> 3 4 = 0.27027023
#> 4 5 = 0.06051901
#> 1 4 = 0.09481045
#> 2 3 = -0.94233222
#> 1 3 = 0.91326109
# Numerical verification:
o <- volume(3)
o2 <- pullback(pullback(o,M),solve(M))
max(abs(coeffs(o-o2))) # zero to numerical precision
#> [1] 4.718448e-16
# Following should be zero:
pullback(as.kform(1),M)-as.kform(matrix(1:5),c(crossprod(M,c(1,rep(0,4)))))
#> The zero alternating linear map from V^1 to R with V=R^n:
#> empty sparse array with 1 columns
# Following should be TRUE:
issmall(pullback(o,crossprod(matrix(rnorm(10),2,5))))
#> [1] TRUE
# Some stretch() use-cases:
p <- rform()
p
#> An alternating linear map from V^3 to R with V=R^7:
#> val
#> 1 3 6 = -8
#> 1 2 5 = -6
#> 2 6 7 = 5
#> 3 6 7 = -13
#> 1 2 6 = 3
#> 4 6 7 = 7
#> 2 4 7 = -2
#> 1 3 5 = -1
stretch(p,seq_len(7))
#> An alternating linear map from V^3 to R with V=R^7:
#> val
#> 1 3 5 = -15
#> 2 4 7 = -112
#> 4 6 7 = 1176
#> 1 2 6 = 36
#> 3 6 7 = -1638
#> 2 6 7 = 420
#> 1 2 5 = -60
#> 1 3 6 = -144
stretch(p,c(1,0,0,1,1,1,1)) # kills dimensions 2 and 3
#> An alternating linear map from V^3 to R with V=R^7:
#> val
#> 4 6 7 = 7