transform kform objects

[RobinHankin]

I asked a question on wikipedia (mathematics helpdesk, 24 April 2019), about transforming kform objects to another coordinate system. The question asked how to express a kform like dx_1 ^ dx_2 ^ dx_3 in terms of dy_1,dy_2,dy_3 where we have linear relationships like dx_1=M[1,1]dy_1 + M[1,2]dy_2 + M[1,3]dy_3. I wanted to know if there was a conceptually pleasant way of doing this.

Well the best I can do is this:

f <-  function(omega,M){
Reduce(`+`,sapply(seq_along(value(omega)),function(i){do.call("wedge",apply(M[index(omega)[i,,drop=FALSE],,drop=FALSE],1,as.1form))*value(omega)[i]},simplify=FALSE))
}

which is pretty dense idiom. But:

> (omega <- rform())
           val
 2 4 6  =   -1
 1 5 6  =   -1
 2 5 7  =    1
 1 4 6  =    2
 1 3 6  =   -1
 1 2 6  =   -1
 2 3 7  =   -1
 2 3 4  =    1
> (M <- matrix(round(rnorm(49),2),7,7))
      [,1]  [,2]  [,3]  [,4]  [,5]  [,6]  [,7]
[1,] -0.75 -0.92 -1.30 -0.80 -0.10 -0.56 -0.39
[2,]  0.83 -0.76 -2.33 -1.80  0.12  1.54  0.56
[3,]  1.00 -1.47  0.09  0.22 -0.47  1.92  0.77
[4,]  0.72 -1.27 -0.31  0.74 -0.92 -0.43  0.43
[5,]  0.89 -0.36  2.49  1.35 -0.26  0.68  0.94
[6,] -0.03 -0.80 -0.37  0.03  0.73  0.65  0.39
[7,] -0.48 -1.77 -1.81  1.77  0.31 -0.37  1.92
> f(omega,M)
                 val
 1 2 3  =   0.080255
 1 3 6  =  -6.201454
 1 2 6  =   2.450020
 4 5 7  =  -4.005752
 2 4 7  =   3.807063
[snip]

and indeed we can transform and then transform back to the original coordinates with only small numerical error:

> max(abs(value(omega %>% f(M) %>% f(solve(M)) - omega)))
[1] 5.77316e-15
> 

But this takes a very long time to run and I can't help thinking that there might be some factorization or other trick or technique that would make the code run faster, and perhaps more transparently.

[RobinHankin]

but there is a bug in f():

> omega <- as.1form(1:7)
> f(omega,M)
Error in inherits(F2, "kform") : 
  argument "F2" is missing, with no default
>

This is because wedge() expects at least two arguments, I will deal with this separately.