2 The proof
We now prove of Theorem 12. The case is trivial, so assume .
Recall that we have to show that any natural transformation
is (uniquely) a linear combination of transformations of the form
The uniqueness part is easy to see since we can extract by evaluating for of dimension large enough. So we have to show every is of this form.
The idea of the proof is to first use naturality to show that for , the form depends only on the -jet of at for some large but finite number (of course, a posteriori, suffices). Once we know this, the problem is reduced to one of finite dimensional linear algebra and invariant theory.
For and , the value of at depends only on the -jet of at for some . In fact, suffices.
Suppose and have identical -jets at . Then there are functions vanishing at and such that
The first step is to replace the with more easily understood coordinate functions. Consider the maps
Let be the pullbacks of the corresponding forms under , and the standard coordinates on . Then are the pullbacks of under the first map.
So it suffices to show that and agree as -forms at .
The point now is that by multilinearity of a -form, it suffices to evaluate these -forms on -tuples of standard basis basis vectors (after choosing a chart for ), and there is at least one for which the is not in the list. So by naturality we can perform this evaluation in the submanifold defined by , in which these two -forms agree.□
By naturality, we may assume is a vector space and is the origin. The value of at the origin is given by a map
where is the space of -jets of elements of . This is a finite dimensional vector space, given explicitly by
Under this decomposition, the th piece captures the th derivatives of . Throughout the proof, we view as a quotient of , hence every function on is in particular a function on .
At this point, everything else follows from the fact that is functorial in , and in particular -invariant.
is a polynomial function.
This lemma is true in much greater generality — it holds for any set-theoretic natural transformation between “polynomial functors” . Here a set-theoretic natural transformation is a natural transformations of the underlying set-valued functors. This is a polynomial version of the fact that a natural transformation between additive functors is necessarily additive, because being additive is a property and not a structure.
We think of these as a functor (with fixed). The point is that for , the functions are polynomial in . This together with naturality will force to be polynomial as well.
To show that is polynomial, we have to show that if , then is a polynomial function in . Without loss of generality, we may assume each lives in the th summand (so that the summand has tensor powers of ).
Fix a number such that for all . We first show that is a polynomial function in the 's.
Let be the map that multiplies by on the th factor, and be the sum map. Consider the commutative diagram
Let be the image of under the inclusion of the th summand. Then gets sent along the top row to . On the other hand, is some element in , and whatever it might be, the image along the bottom row gives a polynomial function in the , hence in the . So we are done.
We now know that for any finite set , we can write
We claim each is a multiple of (if the corresponding is non-zero). Indeed, if we set , then the result must be a polynomial in the and as well, since it is of the form . But is polynomial in if and only if .
Now by taking th roots, we know is polynomial in the when . That is, it is polynomial when restricted to the cone spanned by the 's. But since the 's are arbitrary, this implies it is polynomial everywhere.□
Any non-zero -invariant linear map has and is a multiple of the anti-symmetrization map. In particular, any such map is anti-symmetric.
For convenience of notation, replace with . Since the map is in particular invariant under , we must have . By Schur's lemma, the second part of the lemma is equivalent to claiming that if we decompose as a direct sum of irreducible representations, then appears exactly once. In fact, we know the complete decomposition of by Schur–Weyl duality.
Let be the set of irreducible representations of . Then as an -representation, we have
where is either zero or irreducible, and are distinct for different . Under this decomposition, corresponds to the sign representation of .□
So we know is a polynomial in , and is anti-symmetric in the . So the only terms that can contribute are when or . In the case, it has to factor through . So is polynomial in . This exactly says is given by wedging together and (and pairing with elements of ).