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
is
very small, or (pardon the expression), ``infinitesimally small''. Then
is approximately
, or
(remembering
)
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
form
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.
Let
This verifies the equation
for the special
cases of
, and
. The general case now follows by
linearity.
© 2001 Michael Weiss