If $\gamma (s)$ is a path in $G$ such that $\gamma (0)=I$ then $\frac{d\gamma}{ds}(0)\in \U0001d524$ .

# 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 $$\U0001d524=\{X\in \U0001d524\U0001d529(n,\mathbf{R}):\mathrm{exp}(tX)\in G\forall t\in \mathbf{R}\}$$ is a Lie algebra. We start by proving the following auxiliary result.

Take $\gamma (s)=\left(\begin{array}{cc}\hfill \mathrm{cos}(s)\hfill & \hfill -\mathrm{sin}(s)\hfill \\ \hfill \mathrm{sin}(s)\hfill & \hfill \mathrm{cos}(s)\hfill \end{array}\right)$ . This is a path of rotation matrices. We have $\frac{d\gamma}{ds}=\left(\begin{array}{cc}\hfill -\mathrm{sin}(s)\hfill & \hfill -\mathrm{cos}(s)\hfill \\ \hfill \mathrm{cos}(s)\hfill & \hfill -\mathrm{sin}(s)\hfill \end{array}\right)$ , so $\frac{d\gamma}{ds}(0)=\left(\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)$ 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 $\U0001d524$ , which will be a crucial part of proving that $\U0001d524$ is a Lie algebra.

### Proof of theorem

We need to prove that $\mathrm{exp}\left(t\frac{d\gamma}{ds}(0)\right)\in G$ for all $t\in \mathbf{R}$ . It's sufficient to prove that $\mathrm{exp}\left(\frac{d\gamma}{ds}(0)\right)\in G$ , 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.

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

By definition of the derivative, $$\frac{dh}{ds}(0)=\underset{\u03f5\to 0}{lim}\frac{h(\u03f5)-h(0)}{\u03f5}.$$ We have $h(0)=0$ because $\gamma (0)=I=\mathrm{exp}(0)$ . Let's take the limit by using the sequence $\u03f5=1/n$ as $n\to \mathrm{\infty}$ : $$\frac{dh}{ds}(0)=\underset{n\to \mathrm{\infty}}{lim}\frac{h(1/n)}{1/n}=\underset{n\to \mathrm{\infty}}{lim}nh(1/n).$$

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

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

## Tangent spaces

### Tangent vectors

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

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 $\U0001d524$ contains all tangent vectors to paths in $G$ passing through the identity (called the tangent space of $G$ at the identity).

### Example

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

We've now shown that the tangent space to $G$ at $I$ is contained in $\U0001d524$ . In fact, it is equal to $\U0001d524$ :

If $X\in \U0001d524$ then there is a path $\gamma (s)\in G$ such that $\frac{d\gamma}{ds}(0)=X$ .

Take $\gamma (s)=\mathrm{exp}(sX)$ .

## Pre-class exercise

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