Proposition 1. Let be a manifold, , and a curve of class , , such that . Let be defined by , which satisfies the properties
- is a linear mapping of into , and
- for all .
Then the set of mappings satisfying these conditions is a real vector space and: (i) The set of vectors at is a vector subspace of dimension , and (ii) the collection
is the basis for the set of vectors at .
Proof. Given two such mappings and along with constants , proving that the above space is a real vector space is straightforward as it suffices to show that and that satisfy the above conditions. Indeed, linearity in both cases is immediate upon the assumption that are linear as in condition (1). Also, and , whereby the claims will be immediate. It suffices to show that the vectors satisfy condition (i) and that the collection of partial differentials above form a basis as claimed in (ii). This will (essentially) be done in one step.
To that end, first let with neighborhood be arbitrary and choose a coordinate basis for . Let be an arbitrary curve for which for some , and let , , be the equations for in terms of . Let be arbitrary. Computing the derivative at of with respect to , it follows that
Therefore, any tangent vector at is immediately a linear combination of partial derivative operators , , with components . Moreover, given a linear combination , one can define a curve parametrically by for . Doing so and considering for arbitrary the quantity for the curve , , at yields that
The first term in the above product is because , and the second term reduces to since for all . Hence, the vector tangent to the curve , , at is precisely . The claim will thus be proven if it can be shown that the collection , , is linearly independent.
Suppose, then, that for all . In particular, considering the projection function, it would have to follow that
and because , it follows that the sum above is equal to for . Hence, the original sum equals zero precisely when the coefficients equal zero for all . Thus, linear independence is proved and so the collection of vectors at is a vector space of dimension with the basis claimed above.
Proposition 2. If , then the bracket is a vector field, where is the ring of functions on and where for all .
Proof. Let . To show that is a vector field, it suffices to show that assigns for all a vector to each point . To begin, define a local coordinate system and write and . Then per the definition of the bracket,
where denotes the partial differential operator with respect to . Combining the above expression, it follows that $
which is a vector whose components are given by . Hence, is a vector field.
Proposition 3. Let be a mapping of into and a point of .
- If is injectie, there exist a local coordinate system in a neighborhood of and a local coordinate system in a neighborhood of such that for and . In particular, is a homeomorphism of onto .
- If is surjective, there exist a local coordinate system in a neighborhood of and a local coordinate system of such that for and . In particular, the mapping is open.
- If is a linear isomorphism of onto , the defines a homeomorphism of a neighborhood of onto a neighborhood of and its inverse is also differentiable.