Theorem:
If gamma of s is a path in G such that gamma of zero equals the identity then d gamma by d s at s = 0 is in little g.
Little g as a tangent space
Tangent vectors are in little g
Statement of theorem
We now make our first steps towards the proof that, for a matrix group G, the set little g equals the set of X in little g l n R such that exp (t X) is in G for all t in R is a Lie algebra. We start by proving the following auxiliary result.
Example:
Take gamma of s equals the matrix cos s, minus sin s, sin s, cos s. This is a path of rotation matrices. We have d gamma by d s equals minus sin s, minus cos s, cos s, minus sin s, so d gamma by d s at s = 0 equals the matrix 0, -1, 1, 0 which is in the Lie algebra of the rotation group (which we saw is the space of antisymmetric matrices).
Remark:
The theorem gives us a way of producing elements of little g, which will be a crucial part of proving that little g is a Lie algebra.
Proof of theorem
Proof:
We need to prove that exp of t d gamma by d s at 0 for all real numbers t. It's sufficient to prove that exp of d gamma by d s zt zero is in G, because if we can prove this seemingly weaker statement for all \gamma then we can apply it to the path delta of s equals gamma of s times t, which has d delta by d s at 0 equals t times d gamma by d s at 0, so we recover the stronger statement.
When s is sufficiently close to zero, gamma of s is close to the identity, so we can define its logarithm h of s equals log gamma of s. We have d h by d s at 0 equals the derivative at the identity of log applied to d gamma by d s at 0 Recall that the derivative of log at the identity equals the inverse of the derivative of exp at 0, which equals the identity. Therefore d h by d s at 0 equals d gamma by d s at 0 so it's sufficient to prove that exp of d h by d s at 0 is in G.
By definition of the derivative, d h by d s at 0 equals the limit as epsilon goes to 0 of h at epsilon minus h at zero all over epsilon. We have h at zero equals zero because gamma at zero equals the identity equals exp of 0. Let's take the limit by using the sequence epsilon equals 1 over n as n goes to infinity: d h by d s at 0 equals the limit as n goes to infinity of h at 1 over n, divided by 1 over n, that is the limit as n goes to infinity of n times h at 1 over n
We can get at the quantity inside the limit a different way. Since gamma of s equals exp of h of s is in G for all small s, we get (gamma of s) to the n equals exp of n times h of s is in G for all small s and for any integer n. This is because an element of G raised to any power is still in G; we have also used the fact that (exp of h of s) all to the n equals exp of n times h of s, which is true because h of s commutes with itself.
Now take s equals 1 over n for large n (so that s is small). Then we get exp of n times h of 1 over n is in G for large n. As n goes to infinity, this sequence converges to exp of d h by d s at 0, but since G is a topologically closed group of matrices, this limit lies in G. This shows that exp of d h by d s at 0 is in G, as required.
Tangent spaces
Tangent vectors
This construction of taking the derivative of a path at a point has a name:
Definition:
If gamma of s is a path in R n then d gamma by d s is called the tangent vector field along \gamma.
The theorem tells us that little g contains all tangent vectors to paths in G passing through the identity (called the tangent space of G at the identity).
Example
Example:
Consider the unit circle as a path gamma of s equals e to the i s in the complex plane. We have d gamma by d s equals i e to the i s. So d gamma by d s at zero equals i, which points vertically. Indeed this is tangent to the unit circle at gamma at 0 equals 1. Similarly, d gamma by d s at pi over 2 equals i times e to the i pi over 2, which equals minus 1, which points to the left, which is again tangent to the unit circle at gamma at pi over 2 equals i.
Little g is the tangent space of G at the identity
We've now shown that the tangent space to G at I is contained in little g. In fact, it is equal to little g:
Lemma:
If X is in little g then there is a path gamma of s in G such that d gamma by d s at 0 equals X.
Proof:
Take gamma of s equals exp of s X.
Pre-class exercise
Exercise:
Why is the tangent vector to exp of s X at s = 0 equal to X?