Little g as a tangent space

Tangent vectors are in little g

Statement of theorem

0.00 We now make our first steps towards the proof that, for a matrix group $G$ , the set $\mathfrak{g}=\{X\in\mathfrak{gl}(n,\mathbf{R})\ :\ \exp(tX)\in G\ \forall t\in% \mathbf{R}\}$ is a Lie algebra. We start by proving the following auxiliary result.

Theorem:

0.40 If $\gamma(s)$ is a path in $G$ such that $\gamma(0)=I$ then $\frac{d\gamma}{ds}(0)\in\mathfrak{g}$ .

Example:

1.20 Take $\gamma(s)=\begin{pmatrix}\cos(s)&-\sin(s)\\ \sin(s)&\cos(s)\end{pmatrix}$ . This is a path of rotation matrices. We have $\frac{d\gamma}{ds}=\begin{pmatrix}-\sin(s)&-\cos(s)\\ \cos(s)&-\sin(s)\end{pmatrix}$ , so $\frac{d\gamma}{ds}(0)=\begin{pmatrix}0&-1\\ 1&0\end{pmatrix}$ which is in the Lie algebra of the rotation group (which we saw is the space of antisymmetric matrices).

Remark:

2.30 The theorem gives us a way of producing elements of $\mathfrak{g}$ , which will be a crucial part of proving that $\mathfrak{g}$ is a Lie algebra.

Proof of theorem

Proof:

3.00 We need to prove that $\exp\left(t\frac{d\gamma}{ds}(0)\right)\in G$ for all $t\in\mathbf{R}$ . It's sufficient to prove that $\exp\left(\frac{d\gamma}{ds}(0)\right)\in G$ , because if we can prove this seemingly weaker statement for all $\gamma$ then we can apply it to the path $\delta(s)=\gamma(st)$ , which has $\frac{d\delta}{ds}(0)=t\frac{d\gamma}{ds}(0)$ , so we recover the stronger statement.

4.20 When $s$ is sufficiently close to zero, $\gamma(s)$ is close to the identity, so we can define its logarithm $h(s)=\log\gamma(s)$ . We have $\frac{dh}{ds}(0)=(d_{1}\log)\left(\frac{d\gamma}{ds}(0)\right)$ Recall that $d_{1}\log=(d_{0}\exp)^{-1}=\mathrm{id}$ . Therefore $\frac{dh}{ds}(0)=\frac{d\gamma}{ds}(0),$ so it's sufficient to prove that $\exp\left(\frac{dh}{ds}(0)\right)\in G$ .

6.30 By definition of the derivative, $\frac{dh}{ds}(0)=\lim_{\epsilon\to 0}\frac{h(\epsilon)-h(0)}{\epsilon}.$ We have $h(0)=0$ because $\gamma(0)=I=\exp(0)$ . Let's take the limit by using the sequence $\epsilon=1/n$ as $n\to\infty$ : $\frac{dh}{ds}(0)=\lim_{n\to\infty}\frac{h(1/n)}{1/n}=\lim_{n\to\infty}nh(1/n).$

7.45 We can get at the quantity inside the limit a different way. Since $\gamma(s)=\exp(h(s))\in G$ for all small $s$ , we get $\gamma(s)^{n}=\exp(nh(s))\in G$ 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 $\exp(h(s))^{n}=\exp(nh(s))$ , which is true because $h(s)$ commutes with itself.

8.50 Now take $s=1/n$ for large $n$ (so that $s$ is small). Then we get $\exp(nh(1/n))\in G$ for large $n$ . As $n\to\infty$ , this sequence converges to $\exp\left(\frac{dh}{ds}(0)\right)$ , but since $G$ is a topologically closed group of matrices, this limit lies in $G$ . This shows that $\exp(dh/ds(0))\in G$ , as required.

Tangent spaces

Tangent vectors

10.10 This construction of taking the derivative of a path at a point has a name:

Definition:

If $\gamma(s)$ is a path in $\mathbf{R}^{n}$ then $\frac{d\gamma}{ds}$ is called the tangent vector field along $\gamma$ .

The theorem tells us that $\mathfrak{g}$ contains all tangent vectors to paths in $G$ passing through the identity (called the tangent space of $G$ at the identity).

Example

Example:

11.40 Consider the unit circle as a path $\gamma(s)=e^{is}$ in the complex plane. We have $\frac{d\gamma}{ds}=ie^{is}$ . So $\frac{d\gamma}{ds}(0)=i$ , which points vertically. Indeed this is tangent to the unit circle at $\gamma(0)=1$ . Similarly, $\frac{d\gamma}{ds}(\pi/2)=ie^{i\pi/2}=-1$ , which points to the left, which is again tangent to the unit circle at $\gamma(\pi/2)=i$ .

Little g is the tangent space of G at the identity

13.20 We've now shown that the tangent space to $G$ at $I$ is contained in $\mathfrak{g}$ . In fact, it is equal to $\mathfrak{g}$ :

Lemma:

If $X\in\mathfrak{g}$ then there is a path $\gamma(s)\in G$ such that $\frac{d\gamma}{ds}(0)=X$ .

Proof:

Take $\gamma(s)=\exp(sX)$ .

Pre-class exercise

Exercise:

Why is the tangent vector to $\exp(sX)$ at $s=0$ equal to $X$ ?