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Maurer-Cartan (use?)
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markwh04@yahoo.com
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PostPosted: Fri Jun 23, 2006 3:22 am    Post subject: Re: Maurer-Cartan (use?) Reply with quote

Cyberkatru wrote:
Quote:
How would you answer this:

What good are the Maurer-Cartan equations for a Lie group (other than just
sitting there being true)?

It's the formula for calculating the exterior derivative of the
Cartan-Maurer 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 Cartan-Maurer 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 Cartan-Maurer forms reside in the space
L x T_g*(G).

The Cartan-Maurer 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 1-form in T_g*(G).

The components dw(U,V) of the exterior derivative of the 1-form 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
tensor-index notation to allow arbitrary vectors as lower indices and
one-forms 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 curly-d 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
non-coordinate 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
non-coordinate 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 2-form
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 Cartan-Maurer 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 Lie-bracket + 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 2-form 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 Cartan-Maurer form associated with right-invariant vector
fields, everything follows through ultimately with a sign change in the
structure equation:
d(()/g) - [()/g, ()/g] = 0.
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Cyberkatru1
science forum beginner


Joined: 01 Jan 2006
Posts: 14

PostPosted: Thu Jun 22, 2006 6:40 am    Post subject: Maurer-Cartan (use?) Reply with quote

How would you answer this:

What good are the Maurer-Cartan equations for a Lie group (other than just
sitting there being true)?
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