Pulling Back Forms
We’ve just seen that smooth real-valued functions are differential forms with grade zero. We also know that functions pull back along smooth maps; if is a smooth function on an open subset and if is a smooth map, then is a smooth function — .
It turns out that all -forms pull back in a similar way. But the “value” of a -form doesn’t only depend on a point, but on vectors at that point. Functions pull back because smooth maps push points forward. It turns out that vectors push forward as well, by the derivative. And so we can define the pullback of a -form :
Here is a -form on a region , is a point in , and the are vectors in . Since the differential is a linear function and is a multilinear function on , is a multilinear function on , as asserted.
This pullback is a homomorphism of graded algebras. Since it sends -forms to -forms, it has degree zero. To show that it’s a homomorphism, we must verify that it preserves addition, scalar multiplication by functions, and exterior multiplication. If and are -forms in , we can check
so . Also if we can check
As for exterior multiplication, we will use the fact that we can write any -form as a linear combination of -fold products of -forms. Thus we only have to check that
Thus preserves the wedge product as well, and thus gives us a degree-zero homomorphism of the exterior algebras.