a new transformation interpretation of vector functions

A new transformation interpretation of vector functions This is somehow the missing link in my tensor wedge document. I need to get it nailed down solid. Three transformations. Think of three transformations A,B and C applied to a given starting position. We shall use the word rotate and rotation as conceptual aids, but we imply more general transformations. The "starting position" is an "experiment". There is an apparatus (the experiment) and there is an observer (the basis vectors). In transformation A, the apparatus is rotated while the basis vectors stay put. In transformation B, the basis vectors are rotated while the apparatus stays put. In transformation C, the apparatus and basis vectors are both rotated by the same amount. The apparatus is described by a set of tensors. In transformation A we find for rank-1 and rank-2 tensors, v'i = Rijvj T'ij = RiaRjbTab ei, eiej etc all stay put Thus, vector v is "actively" rotated into vector v', and tensor Tij is actively rotated into T'ij, and similarly for all higher rank tensors. Transformation A is our transformation of interest. Vector. Consider in more detail what happens to a vector v under transformation A. vi = (ei v) starting position vi' = (ei v') ending position after the action of transformation A In transformation A, the basis vectors ei stay put. Although (ei v) is the dot product of two vectors, it does not transform as a scalar under transformation A. It looks like a scalar, but in fact it transforms under transformation A as the component of a vector. It would transform as a scalar under transformation C, but that is not our transformation of interest. The expansions going with the above lines are these v = Σiviei starting position v' = Σiv'iei ending position after the action of transformation A Rank-2 tensor. We can generalize these expansions to a rank-2 tensor. The expansions are T = ΣabTab eaeb starting position T' = ΣabT'ab eaeb ending position after the action of transformation A The projections corresponding to ** are then Tab = (eaeb ) T starting position T'ab = (eaeb ) T' ending position after the action of transformation A Rank-1 tensor function Now we return to the vector case we define a function fi(v) as follows, fi(v) ≡ vi = (ei v) starting position Plan 1 Then after transformation A we have fi(v') = v'i = (ei v') ending position after the action of transformation A The function fi picks off component i of its vector argument, before and after transformation A. Now since transformation A does this to vector v (this is a standard-issue covariant vector component transformation), v'i = Rijvj we find that fi(v') = v'i = Rijvj = Rij fj(v) This function fi(v) is an example of a "rank-1 tensor function". We have just shown that the transformation rule for such a function is this: fi(v') = Rij fj(v) where v'i = Rijvj Had we started off with the dual basis vectors ei we would have defined fi(v) = vi = (ei v) starting position fi(v') = v'i = (ei v') ending position after the action of transformation A (ei stay put) fi(v') = Rij fj(v) transformation rule, where v'i = Rijvj This is exactly the situation with the dual space V* basis functional λi where we just replace fi by λi, λi(v) = vi = (ei v) starting position λi(v') = v'i = (ei v') ending position after the action of transformation A λi(v') = Rij λj(v) transformation rule, where v'i = Rijvj Using the notation v' = Rv we repeat the above lines λi(v) = vi = (ei v) starting position λi(Rv) = (Rv)i = (ei Rv) ending position after the action of transformation A λi(Rv) = Rij λj(v) transformation rule We repeat again replacing v by v1 where 1 is a label, not a component, λi(v1) = vi = (ei v1) starting position λi(Rv1) = (Rv1)i = (ei Rv1) ending position after the action of transformation A λi(Rv1) = Rij λj(v1) transformation rule Plan 2 Repeat the above arguing that after the transformation, you should have f'(v') We then get f'i(v') = Rij fj(v) and then fi(v) transforms as a vector field! Then we also have λ'i(v') = Rij λj(v) so λi(v) also transforms as a vector field. Now, what happens with a general functional in V* acting on V? Plan 1A α(v) = Σiαiλi(v) starting position α(v') = Σiαiλi(v') ending position after the action of transformation A Here I have made the assumption that αi does not change in transformation A. That is a crucial issue! This is saying that αi is part of the observer, and stays put like ei . And at the same time I assume Plan 1 above for λi . Then we find α(v') = Σiαiλi(v') = ΣiαiΣj Rij λj(v) = Σj [ Σi αi Rij] λj(v) = Σj [βj] λj(v) = Σi βiλj(v) = β(v) βj = Rijαi = (R-1)ji αi β = R-1α This says that α(v') = β(v) and that just leaves me cold. It is some other functional. Plan 1B Now suppose αi is part of the apparatus, and transforms the same way that vi transforms, as a vector, under transformation A. Then we get, still assuming Plan 1 for λ, α(v) = Σiαiλi(v) starting position α'(v') = Σiα'iλi(v') = Σi [Rijαj] λi(v') Now we use our earlier result that λi(v') = Rik λk(v) and we then have α'(v') = Rijαj [Rik λk(v)] = δjkαj λk(v) = αj λj(v) = α(v) and in Plan 2 we find that α(v) transforms as a scalar field, just as φ'(x') = φ(x) in tensor doc. Plan 2A α(v) = Σiαiλi(v) starting position α(v') = Σiαiλ'i(v') ending position after the action of transformation A This gives the same result as Plan 1A. Plan 2B α(v) = Σiαiλi(v) starting position α'(v') = Σiα'iλ'i(v') ending position after the action of transformation A This gives the same result as Plan 1B.