This one confused me for a long time, so I figured I should write this down before I forgot again.
Let be an abstract smooth manifold. We want to define the notion of a tangent vector to at a point . With that, we can define the tangent space , which will just be the (real) vector space of tangent vectors at .
Geometrically, we know what this should look like for our usual examples. For example, if is a circle embedded in , then the tangent vector at a point should just look like a vector running off tangent to the circle.
Similarly, given a sphere , the tangent space at a point along the sphere would look like plane tangent to at .
However, the point of an abstract manifold is that we want to see the manifold as an intrinsic object, in its own right, rather than as embedded in . This can be thought of as analogous to the way that we think of a group as an abstract object in its own right, even though Cayley’s Theorem tells us that any group is a subgroup of the permutation group. (This wasn’t always the case! During the 19th century, a group was literally defined as a subset of or of . In fact Sylow developed his theorems without the word “group” Only much later did the abstract definition of a group was given, an abstract set which was independent of any embedding into , and an object in its own right.) So, we would like our notion of a tangent vector to not refer to an ambient space, but only to intrinsic properties of the manifold in question.
So how do we capture the notion of the tangent to a manifold referring just to the manifold itself? Well, the smooth structure of the manifold lets us speak of smooth functions . In the embedded case, we can thus think of taking a directional derivative along (i.e. some partial derivative). To give a concrete example, suppose we have a smooth function and a point . By the structure of a manifold, near the point , looks like a function on some neighborhood of the origin in So we are allowed to take the partial derivative of with respect to any of the vectors in .
For a fixed this partial derivative is a linear map . It turns out this goes the other way: if you know what does to every smooth function, then you can figure out which vector it’s taking the partial derivative of. This is the trick we use in order to create the tangent space. Rather than trying to specify a vector directly (which we can’t do because we don’t have an ambient space), we instead look at arbitrary derivative-like functions, and associate them with a vector. More formally, we have the following.
This is just a “product rule”. Then the tangent space is easy to define:
In fact, one can show that the product rule for is equivalent to the following three conditions:
- is linear, meaning .
- , where is the constant function on .
- whenever . Intuitively, this means that if a function vanishes to second order at , then its derivative along should be zero.
This suggests a third equivalent definition: suppose we define
to be the set of functions which vanish at (this is called the maximal ideal at ). In that case,
is the set of functions vanishing to second order at . Thus, a tangent vector is really just a linear map
In other words, the tangent space is actually the dual space of ; for this reason, the space is defined as the cotangent space (the dual of the tangent space). This definition is even more abstract than the one with derivations above, but it has the advantage (or so I’m told) that it can be transferred to other settings (like algebraic varieties).
EDIT (Oct 5 2015): Reproducing this Reddit comment by tactics, The beauty of the definition given in this blog post is stuffed away into an easily-forgotten sentence. This definition:
- Does not rely on any kind of parametrization, and
- It is defined only in terms of the ring of “regular functions” defined on the space.
The former is nice for philosophical reasons. The latter is nice because we can pull in a lot of intuition about manifolds into the study of algebraic varieties, and similarly, we can inject a lot of ring theory into the study of manifolds.
With all these equivalent definitions, the last thing I should do is check that this definition of tangent space actually gives a vector space of dimension . To do this it suffices to show verify this for open subsets of , which will imply the result for general manifolds (which are locally open subsets of ). Using some real analysis, one can prove the following result:
Thus, it follows that there is an isomorphism
and so the cotangent space, hence tangent space, indeed has dimension .