If is a path in such that then .
Tangent vectors are in little g
Statement of theorem
We now make our first steps towards the proof that, for a matrix group , the set is a Lie algebra. We start by proving the following auxiliary result.
Take . This is a path of rotation matrices. We have , so which is in the Lie algebra of the rotation group (which we saw is the space of antisymmetric matrices).
The theorem gives us a way of producing elements of , which will be a crucial part of proving that is a Lie algebra.
Proof of theorem
We need to prove that for all . It's sufficient to prove that , because if we can prove this seemingly weaker statement for all then we can apply it to the path , which has , so we recover the stronger statement.
When is sufficiently close to zero, is close to the identity, so we can define its logarithm . We have Recall that . Therefore so it's sufficient to prove that .
By definition of the derivative, We have because . Let's take the limit by using the sequence as :
We can get at the quantity inside the limit a different way. Since for all small , we get for all small and for any integer . This is because an element of raised to any power is still in ; we have also used the fact that , which is true because commutes with itself.
Now take for large (so that is small). Then we get for large . As , this sequence converges to , but since is a topologically closed group of matrices, this limit lies in . This shows that , as required.
This construction of taking the derivative of a path at a point has a name:
If is a path in then is called the tangent vector field along .
The theorem tells us that contains all tangent vectors to paths in passing through the identity (called the tangent space of at the identity).
Consider the unit circle as a path in the complex plane. We have . So , which points vertically. Indeed this is tangent to the unit circle at . Similarly, , which points to the left, which is again tangent to the unit circle at .
Little g is the tangent space of G at the identity
We've now shown that the tangent space to at is contained in . In fact, it is equal to :
If then there is a path such that .
Why is the tangent vector to at equal to ?