I think you are looking for the concept of a smooth functor. Essentially, any functorial construct on vector spaces, such as taking duals, tensor products and direct sums, can be applied to a vector bundle, if for all vector spaces V,W the map F:Hom(V,W)→Hom(F(V),F(W))
is smooth using the usual definition of what it means for a map between vector spaces to be smooth.
Thus if E and F both bundles over M with fibers V and W,, we can find a bundle denoted E⊗F, also over M, the fiber of which is V⊗W. So your first equality is true. In fact the proof is to do as you suggest and take the disjoint union of the fiber-wise tensor product. Then you only have to check that this gives a smooth bundle, but since the map (A,B)↦A⊗B is smooth for linear transformations A and B between finite-dimensional vector spaces, this is the case.
On the level of sections, you also have Γ(E⊗F)≅Γ(E)⊗Γ(F)
but the tensor product on the right is the tensor product of
C∞(M)-modules. There is not strict equality here, but the spaces are
naturally isomorphic. Here,
naturally means that you can construct an isomorphism that doesn't depend on arbitrary choices. (To clarify: you probably know that if
V and
W are vector spaces of the same dimension, you can find an isomorphism
V≅W by choosing bases for
V and
W. But this isomorphism depends on which bases you choose, so they are not
naturally isomorphic. However,
V⊗W and
W⊗V are
naturally isomorphic. Can you think of why?)
This post imported from StackExchange Physics at 2014-04-01 17:32 (UCT), posted by SE-user Robin Ekman