Lie Algebras

Let $G$ be a Lie group. Let $x(t)$ be a smooth curve in $G$ passing through the unit element 1of $G$, i.e., a smooth mapping from a neighborhood of 0 on the real line into $G$ with $x(0)={\bf 1}$. $T(G)$, the tangent space of $G$ at 1, consists of all matrices of the form $\left.\frac{dx(t)}{dt}\right\vert _{t=0}$, or just $x'(0)$ in a less clumsy notation.

$T(G)$ is the Lie algebra of $G$. I will show in a moment that $T(G)$ is a vector space over R, and I really should (but I won't) define a binary operation $[x,y]$ (the Lie bracket) on $T(G)\times T(G)$.

Proof that $T(G)$ is a vector space over R: if $x(t)$ is a smooth curve and $x(0)={\bf 1}$, then set $y(t)=x(kt)$, $k\in{\bf R}$. This is also a smooth curve and $y'(0)=kx'(0)$. So $T(G)$ is closed under multiplication by elements of R. (Note this argument fails for complex $k$.) Similarly, differentiating $z(t)=x(t)y(t)$ (with $x(0)=y(0)={\bf 1}$, as usual) shows that $T(G)$ is closed under addition.

Historically, the Lie algebra arose from considering elements of $G$ ``infinitesimally close to the identity''. Suppose $\epsilon\in{\bf R}$ is very small, or (pardon the expression), ``infinitesimally small''. Then $x'(0)$ is approximately $\frac{x(\epsilon)-x(0)}{\epsilon}$, or (remembering $x(0)={\bf 1}$)

$$x(\epsilon)\approx {\bf 1}+\epsilon x'(0)$$

Historically, $x(\epsilon)$ is a so-called infinitesimal generator of $G$.

Robinson has shown how this classical approach can be made rigorous, using non-standard analysis. Even without this, the classical notions provide a lot of insight. For example, let $n$ be an ``infinite'' integer. Then if $t\in{\bf R}$ is an ordinary real number (not ``infinitesimal''), we can let $\epsilon=t/n$ and so

$$x(\epsilon)^n \approx \left({\bf 1}+\frac{tx'(0)}{n}\right)^n \approx e^{tx'(0)}$$

Assume that the left hand side is an ordinary ``finite'' element of $G$. Write $v$ for $x'(0)$, an arbitrary element of the Lie algebra $T(G)$. This suggests there should be a map $(t,v)\mapsto e^{tv}$ from ${\bf R}\times T(G)$ into $G$.

In fact, the following is true: for any Lie group $G$ with Lie algebra $T(G)$, we have a mapping $\exp$ from $T(G)$ into $G$ such that $\exp({\bf0})={\bf 1}$, and $\exp\left((t_1+t_2)v\right)=\exp(t_1v)\exp(t_2v)$, for any $t_1,t_2\in{\bf R}$ and $v\in T(G)$.

It also turns out that the Lie algebra structure determines the Lie group structure ``locally'': if the Lie algebras of two Lie groups are isomorphic, then the Lie groups are locally isomorphic. Here, the Lie algebra structure includes the bracket operation, and of course one has to define local isomorphism.

Now for our standard example, $SO(n)$. Notation: the Lie algebra of $SO(n)$ is $so(n)$. If you differentiate the condition $A^tA={\bf 1}$ and plug in $A(0)={\bf 1}$, you will conclude that all elements of $so(n)$ are anti-symmetric. Fact: the converse is true.

Example: $SO(2)$, rotations in 2-space. All elements of $so(2)$ have the form

$$\left[ \begin{array}{cc} 0 & -c \\ c & 0 \end{array} \right]$$

An element of $so(2)$ can be thought of as an angular speed.

In the earlier discussion of $SO(2)$, I set up a one-one correspondence

$$\left[ \begin{array}{cc} a & -b \\ b & a \end{array} \right] \leftrightarrow a+ib$$

between matrices and complex numbers. (We had the restriction $a^2+b^2=1$, but we can drop this and still get an isomorphism between matrices of this form and C.) Then the displayed element of $so(2)$ corresponds to the purely imaginary number $ic$, and we have a map $(t,ic)\mapsto e^{ict}$ mapping ${\bf R}\times so(2)$ onto the unit circle, which in turn is isomorphic to $SO(2)$.

$SO(3)$, rotations in 3-space. Elements of $so(3)$ have the form:

$$\left[ \begin{array}{ccc} 0 & -c & b \\ c & 0 & -a \\ ... ...& 0 \end{array} \right] \leftrightarrow (a,b,c)\in {\bf R}^3$$

(We'll see the reason for the peculiar choice of signs and arrangement of $a$, $b$, and $c$ shortly.)

Fact: the vector $(a,b,c)$ is the angular velocity vector for the above element of $so(3)$. What does this mean? Well first, let $v_0\in{\bf R}^3$ be some arbitrary vector; if $A(t)$ is a curve in $SO(3)$, and we set $v(t)=A(t)v_0$, then $v(t)$ is a rotating vector, whose tip traces out the trajectory of a moving point. The velocity of this point at $t=0$ is $A'(0)v_0$. It turns out that $A'(0)v_0$ equals the cross-product $(a,b,c)\times v_0$, which characterizes the angular velocity vector. The next few paragraphs demonstrate this equality less tediously than by direct calculation.

Let

$$\mbox{$\bf\hat{x}$}= \left[ \begin{array}{ccc} 0 & 0 & 0\\ ... ...}{ccc} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0 \end{array}\right]$$

so the general element of $so(3)$ can be written $a\mbox{$\bf\hat{x}$}+b\mbox{$\bf\hat{y}$}+c\mbox{$\bf\hat{z}$}$. And $\bf\hat{x}$, $\bf\hat{y}$, and $\bf\hat{z}$ are simply the elements of $so(3)$ corresponding to unit speed uniform rotations about the x, y, and z axes, respectively-- as can be seen by considering their effects on the standard orthonormal basis.

This verifies the equation $A'(0)v_0 = (a,b,c)\times v_0$ for the special cases of $A'(0)=\mbox{$\bf\hat{x}$},\mbox{$\bf\hat{y}$}$, and $\bf\hat{z}$. The general case now follows by linearity.

© 2001 Michael Weiss

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