Let be a Lie group. Let be a smooth curve in passing through the unit element 1of , i.e., a smooth mapping from a neighborhood of 0 on the real line into with . , the tangent space of at 1, consists of all matrices of the form , or just in a less clumsy notation.
is the Lie algebra of . I will show in a moment that is a vector space over R, and I really should (but I won't) define a binary operation (the Lie bracket) on .
Proof that is a vector space over R: if is a smooth curve and , then set , . This is also a smooth curve and . So is closed under multiplication by elements of R. (Note this argument fails for complex .) Similarly, differentiating (with , as usual) shows that is closed under addition.
Historically, the Lie algebra arose from considering elements of
``infinitesimally close to the identity''. Suppose
very small, or (pardon the expression), ``infinitesimally small''. Then
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 be an ``infinite''
integer. Then if is an ordinary real number (not
``infinitesimal''), we can let and so
In fact, the following is true: for any Lie group with Lie algebra , we have a mapping from into such that , and , for any and .
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, . Notation: the Lie algebra of is . If you differentiate the condition and plug in , you will conclude that all elements of are anti-symmetric. Fact: the converse is true.
Example: , rotations in 2-space. All elements of have the
In the earlier discussion of , I set up a one-one correspondence
, rotations in 3-space. Elements of have the form:
Fact: the vector is the angular velocity vector for the above element of . What does this mean? Well first, let be some arbitrary vector; if is a curve in , and we set , then is a rotating vector, whose tip traces out the trajectory of a moving point. The velocity of this point at is . It turns out that equals the cross-product , which characterizes the angular velocity vector. The next few paragraphs demonstrate this equality less tediously than by direct calculation.
This verifies the equation for the special cases of , and . The general case now follows by linearity.
© 2001 Michael Weiss