Given a point in an -dimensional manifold , we have the vector space of tangent vectors at . Given a coordinate patch around , we’ve constructed coordinate vectors at , and shown that they’re linearly independent in . I say that they also span the space, and thus constitute a basis.
To see this, we’ll need a couple lemmas. First off, if is constant in a neighborhood of , then for any tangent vector . Indeed, since all that matters is the germ of , we may as well assume that is the constant function with value . By linearity we know that . But now since we use the derivation property to find
and so we conclude that .
In a slightly more technical vein, let be a “star-shaped” neighborhood of . That is, not only does contain itself, but for every point it contains the whole segment of points for . An open ball, for example, is star-shaped, so you can just think of that to be a little simpler.
Anyway, given such a and a differentiable function on it we can find functions with , and such that we can write
where is the th component function.
If we pick a point we can parameterize the segment , and set to get a function on the unit interval . This function is clearly differentiable, and we can calculate
using the multivariable chain rule. We find
We can thus find the desired functions by setting
Now if we have a differentiable function defined on a neighborhood of a point , we can find a coordinate patch — possibly by shrinking — with and star-shaped. Then we can apply the previous lemma to to get
with . Moving the coordinate map to the other side we find
Now we can hit this with a tangent vector
where we have used linearity, the derivation property, and the first lemma above. Thus we can write
and the coordinate vectors span the space of tangent vectors at .
As a consequence, we conclude that always has dimension — exactly the same dimension as the manifold itself. And this is exactly what we should expect; if is -dimensional, then in some sense there are independent directions to move in near any point , and these “directions to move” are the core of our geometric notion of a tangent vector. Ironically, if we start from a more geometric definition of tangent vectors, it’s actually somewhat harder to establish this fact, which is partly why we’re starting with the more algebraic definition.