Finding lie algebra of a group by using exp map and tangent space

Question

I'm studying Lie groups and I am in trouble with finding lie algebras of the classical groups. How can I calculate $\mathfrak{sp}(n,\mathbb{C})$ or $\mathfrak{so}(n,\mathbb{R})$ using exp map and tangent space of these groups?

Answer 1

There are several ways.

I'll do one for $SO(3)$ and I guess you can figure out the general way of proceeding.

OUTLINE: We will start from a general form for an element of $SO(3)$ then we'll consider all the curves passing from the identity and we will derive effectively the tangent space in the identity. We'll find a base and that will be the Lie Algebra. If you exponentiate the tangent vector at the tangent space at the identity you'll get newly the element of the Lie Group $SO(3)$

Let's suppose that you know that every rotation can be seen as a rotation of an angle $\theta$ around a vector $\mathbf{v}=x\mathbf{e_x}+y\mathbf{e_y}+z\mathbf{e_z}$.

Then every element of $SO(3)$ can be parametrized as $$R_{\mathbf{v}}\left(\theta\right)=\mathbb{1}+\sin\theta X+\left(1-\cos\theta\right)X^{2},$$ where $X$ is this matrix $$X=\left(\begin{array}{ccc} 0 & -z & y\\ z & 0 & -x\\ -y & x & 0 \end{array}\right).$$ If you're smart you already understood that I'm cheating (because this parametrization of $SO(3)$ is already a sort of exponential... but let's continue...

So given the parametrization, a general rotation matrix is given by $$R\left(\theta,\,x,\,y,\,z\right)=\left(\begin{array}{ccc} x^{2}+\cos\theta\left(y^{2}+z^{2}\right) & \left(1-\cos\theta\right)xy-z\sin\theta & \left(1-\cos\theta\right)xz+y\sin\theta\\ \left(1-\cos\theta\right)xy+z\sin\theta & y^{2}+\cos\theta\left(x^{2}+z^{2}\right) & \left(1-\cos\theta\right)yz-x\sin\theta\\ \left(1-\cos\theta\right)xz-y\sin\theta & \left(1-\cos\theta\right)yz-x\sin\theta & z^{2}+\cos\theta\left(y^{2}+x^{2}\right) \end{array}\right),$$

Now take a regular curve $\theta\left(t\right)$ such that $\theta\left(0\right)=0$ and let $\mathbf{v}$ be a vector $\mathbb{R}^{3}$ with coordinates $\left(x,\,y,\,z\right)$. The tangent vector of the curve is then given by $$\frac{d}{dt}R_{\mathbf{v}}\left(\theta\right)=\left(\begin{array}{ccc} -\left(y^{2}+z^{2}\right)\sin\theta & xy\sin\theta-z\cos\theta & xz\sin\theta+y\cos\theta\\ xy\sin\theta+z\cos\theta & -\left(x^{2}+z^{2}\right)\sin\theta & yz\sin\theta-x\cos\theta\\ xz\sin\theta-y\cos\theta & yz\sin\theta+x\cos\theta & -\left(y^{2}+x^{2}\right)\sin\theta \end{array}\right)\cdot\dot{\theta}\left(t\right)$$. Then in $t=0$ $$\frac{d}{dt}R_{\mathbf{v}}\left(\theta\right)|_{t=0}=\left(\begin{array}{ccc} 0 & -z & y\\ z & 0 & -x\\ -y & x & 0 \end{array}\right)\cdot\dot{\theta}\left(t\right).$$ So let's now consider the axis $\mathbf{x}=\left(1,\,0,\,0\right)$, $\mathbf{y}=\left(0,\,1,\,0\right)$ and $\mathbf{z}=\left(0,\,0,\,1\right)$ Then you have $$ E_{1}=\frac{d}{dt}R_{\mathbf{x}}\left(\theta\right)|_{t=0} =\left(\begin{array}{ccc} 0 & 0 & 0\\ 0 & 0 & -1\\ 0 & 1 & 0 \end{array}\right)\cdot\dot{\theta}\left(t\right),$$ $$E_{2}=\frac{d}{dt}R_{\mathbf{y}}\left(\theta\right)|_{t=0} =\left(\begin{array}{ccc} 0 & 0 & 1\\ 0 & 0 & 0\\ -1 & 0 & 0 \end{array}\right)\cdot\dot{\theta}\left(t\right),$$ $$E_{3}=\frac{d}{dt}R_{z}\left(\theta\right)|_{t=0} =\left(\begin{array}{ccc} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0 \end{array}\right)\cdot\dot{\theta}\left(t\right).$$ Then you have a base for the Lie Algebra of $SO(3)$ and a general vector $X$ in the tangent space of $SO(3)$ at the identity can be written as $$X=xE_{1}+yE_{2}+zE_{3},$$ If you exponentiate the general vector of the Lie Algebra you get the general rotation in the exponential form $$R_{\mathbf{v}}\left(\theta\right)=\exp\left[\theta X\right],$$ with $$X=\left(\begin{array}{ccc} 0 & -z & y\\ z & 0 & -x\\ -y & x & 0 \end{array}\right)$$ which in fact was the parametrization we were starting from. Hope gave you some hints on how to proceed. The other classic and more easy way to proceed it's just to do a taylor expansion, but I think this one it really explicit.