markwh04@yahoo.com science forum Guru Wannabe
Joined: 12 Sep 2005
Posts: 137

Posted: Fri Jun 23, 2006 3:22 am Post subject:
Re: MaurerCartan (use?)



Cyberkatru wrote:
Quote:  How would you answer this:
What good are the MaurerCartan equations for a Lie group (other than just
sitting there being true)?

It's the formula for calculating the exterior derivative of the
CartanMaurer form.
It helps substantially to remove the cobwebs from the underlying theory
with some convenient conventions. A Lie group G has a continuous
differentiable product (g,h) > gh. So, it's natural to extend this 
by differentiation  to a definition of the product with respect to
tangents, such that
({g} x T_h(G)) > T_{gh}(G)
and
(T_g(G) x {h}) > T_{gh}(G).
Thus, if g(t) is a curve, then d/dt (g(t)h) will just be  in this
notation g'(t) h; and d/dt (hg(t)) will be h g'(t), with g'(t) being a
tangent vector (by definition, in fact) of T_{g(t)}(G).
The tangent spaces are all identical, since each can be derived from
the others by multiplying, e.g. T_g(G) = (g/h) T_h(G) = T_h(G) (h\g),
where I'm using g/h to denote g h^{1} for convenience and h\g = h^{1}
g.
They are all copies of the tangent space L = T_e(G) at the identity e,
which in turn is just the Lie algebra.
The left and right invariant vector fields are just
L(g,v) = gv; R(g,v) = vg
for Lie vector v, w in L. They're invariance is gL(h,v) = g(hv) = (gh)v
= L(gh,v), and similarly R(g,v)h = R(gh,v).
The CartanMaurer forms are just the quotients g\() and ()/g. Thus,
g\(gv) = v and (vg)/g = v.
They are also cast as tensor products. Taking a basis {Y_1,Y_2,...,Y_n}
in L, you can write
g\() = sum Y_i [x] (g Y^i),
where the dual basis is defined by
(gY^i).(gv) = v^i.
Thus,
g\(gv) = sum Y_i (gY^i).(gv) = sum Y_i v^i = v.
Similarly
()/g = sum Y_i [x] (Y^i g).
This is a one form at g, residing in the dual T_g*(G) of the tangent
space T_g(G); and a vector field at e, residing in the tangent space L
= T_e(G). Thus, both CartanMaurer forms reside in the space
L x T_g*(G).
The CartanMaurer equations give you their differentials. The formula
is just a rendition of the general formula in differential geometry for
the exterior derivative of a vector field. The differential is taken
with respect to the second factor, since that's a 1form in T_g*(G).
The components dw(U,V) of the exterior derivative of the 1form w, in
"ordinary" notation are written as:
dw(U,V) = U w(V)  V w(U)  w([U,V]).
This becomes more familiar to Physicists if we extend the usual
tensorindex notation to allow arbitrary vectors as lower indices and
oneforms as upper indices. Also, instead of writing the vector U() as
its own differential operator, write it as @_U(). (I'm using @ as an
ASCII replacement for the curlyd partial derivative operator). Then
the above formula becomes
dw_{UV} = @_U w_V  @_V w_U  f^w_{UV}.
The structure coefficients are defined by
f^w_{UV} = w_{[U,V]} = w([U,V]) = [U,V]^w
(illustrating the convention, in the process).
For a coordinate basis, [@_m, @_n] = 0 and f^w_{mn} = 0, so the above
would become
dw_{mn} = @_m w_n  @_n w_m,
which is the usual familar form for the differential. For a
noncoordinate basis, {Y_a}, in a Lie algebra, [Y_a,Y_b] = f^c_{ab}
Y_c, and the f's are constant. For a general manifold with a general
noncoordinate frame {Y_a}, the f's would be variable, so that
distinguishes Lie groups from other manifolds.
If you take w(g) = gY^c, then the components are just (gY^c) =
delta^c_b (gY^b), with respect to the basis {gY^c}. So the formula
simplies, here, to
dw_{ab} = @_a w_b  @_b w_a  f^w_{ab}
= 0  0  f^c_{ab}.
Thus, as a 2form
dw = 1/2 dw_{ab} (gY^a) ^ (gY^b) = 1/2
f^c_{ab} (gY^a) ^ (gY^b).
Substutiting this into the expression for the CartanMaurer form g\(),
you get
d(g\()) = sum Y_c [x] d(g Y^c),
= 1/2 sum Y_c [x] f^c_{ab} (gY^a) ^
(gY^b).
= 1/2 sum f^c_{ab} Y_c [x] (gY^a) ^
(gY^b)
= 1/2 sum [Y_a,Y_b] [x] (gY^a) ^
(gY^b).
At this point, various notations are adopted to represent this
simultaneous Liebracket + exterior product.
One convention is to define the notation
[h(g)^k(g)] = [h^a(g) Y_a ^ k^b(g) Y_b]
= (h^a(g) ^ k^b(g)) [x] [Y_a, Y_b]
= f^c_{ab} (h^a(g) ^ k^b(g)) [x] Y_c.
Then, you can write
d(g\()) + 1/2 [g\() ^ g\()] = 0.
Another notation defines an extended commutator as the 2form with the
components
[h,k]_{UV} = [h_U,k_V].
So, then (using the generalized index notation):
[h_U, k_V] = [h^a_U Y_a, k^b_V Y_b]
= h^a_U k^b_V [Y_a, Y_b]
= h^a_U k^b_V f^c_{ab} Y_c.
This is related to the combined wedge/bracket product since
(h^a ^ k^b f^c_{ab} Y_c)_{UV}
= ((h^a [x] k^b  k^b [x] h^a) f^c_{ab} Y_c)_{UV}
= (h^a_U k^b_V + k^a_U h^b_V) f^c_{ab} Y_c
using the antisymmetry of the structure coefficients
f^c_{ab} = f^c_{ba}.
Thus,
[h^k] = [h,k] + [k,h].
The structure equation may therefore also be written as:
d(g\()) + [g\(), g\()] = 0.
For the CartanMaurer form associated with rightinvariant vector
fields, everything follows through ultimately with a sign change in the
structure equation:
d(()/g)  [()/g, ()/g] = 0. 
