Natural Science Forum / Physics / Research / March 2008

I prefer to use an algebraic form, wherever possible. Express the
underlying Lie algebra within its universal enveloping algebra. Lie
brackets then become simple commutators.

Express the contraction with Lie covectors algebraically as a "trace"
which satisfies the appropriate cyclicity property.

The real point of brining in a vector bundle representation is that
the underlying gauge dynamics are not just applied to the gauge field
itself (via its principle bundle representation) but to other fields
that are acted on by the symmetries as well (via the appropriate
associated bundle representation). What you're calling the "vector
bundle" approach is just an encapsulation of the latter and is not
mutually exclusive wth the former, but rather complementary.

The simplest way to incorporate the associated bundle is to simply
represent the group action as an ordinary product on the left. A
further assumption may then  be made that the algebra extends to a *-
algebra. Then under the "adjoint", or *, operator left action becomes
right action on the adjoint field.

So, (Y_a) is a basis for the Lie algebra and (e_A) a basis for the
extra fields; with respective dual bases (Y^a), (e^A), then the
actions are given by:
  [Y_a,Y_b] = Y_a Y_b - Y_b Y_a = f^c_{ab} Y_c
  Y_a e_B = T^C_{aB} e_C
  Tr(Y^b Y_a) = delta^b_a
  Tr(e^B e_A) = delta^B_A
where f^c_{ab}, T^C_{aB} are the corresponding "structure
coefficients".

The action (Y_b Y^c - Y^c Y_b) is then mandated by the requirement
that cyclicity apply to its trace; i.e.,
  Tr( (Y_b Y^c - Y^c Y_a) Y_a)
  = Tr(Y^c Y_a Y_b) - Tr(Y^c Y_b Y_a)
  = Tr(Y^c (Y_a Y_b - Y_b Y_a)) = f^c_{ab}.
Thus, one has
  Y_b Y^c - Y^c Y_b = f^c_{ab} Y^a.
This is the "co-adjoint action" of the generator Y_b on Y^c.
Similarly, one has the co-adjoint action on the dual basis
  Tr(e^C Y_a) e_B) = Tr(e^C (Y_a e_B)) = T^C_{aB}.
Thus
  e^C Y_a = T^C_{aB} e^B.
This is the right action mentioned above.

There is also a kind of "co-adjoint" action of the (e_A) basis on the
dual (e^A),
  Tr((e_B e^C) Y_a) = Tr(e^C (Y_a e_B)) = T^C_{aB}.
Thus, one should also have
  e_B e^C = T^C_{aB} Y^a.
Ultimately it is out of this that gauge currents are constructed from
the external field and its adjoint.

All of the foregoing is meshed with form-valued objects by adopting
the convention that the products are formed component-wise by the
wedge product. Thus, for instance,
  (Y_a dx) (Y_b dy) = (Y_a Y_b) (dx ^ dy).
For the potential 1-form
  A = A^a_m Y_a dx^m,
one can then form the square
  AA = A^a_m A^b_n Y_a Y_b dx^m ^ dx^n.
Employing the anti-symmetry with respect to (mn), this can also be
written
  AA = 1/2 (A^a_m A^b_n - A^a_n A^b_m) Y_a Y_b dx^m ^ dx^n.
This transfers to an anti-symmetry in (ab), by
  AA = 1/2 (A^a_m A^b_n - A^b_m A^a_n) Y_a Y_b dx^m ^ dx^n
  = 1/2 A^a_m A^b_n (Y_a Y_b - Y_b Y_a) dx^m ^ dx^n
  = 1/2 f^c_{ab} A^a_m A^b_n Y_c (dx^m ^ dx^n).

Thus, combining with the exterior derivative, one gets a simple
algebraic form for the field strength
  F = dA + AA.
The Bianchi identity comes easily from this,
  dF = d(dA + AA) = d(dA) + dA A - A dA.
The Leibnitz rule was applied d(AA) = (dA)A - A(dA), noting that the
second minus sign came because A is a 1-form. Noting that d(dA) = 0,
the result is
  dF = (F - AA)A - A (F - AA) = FA - AAA - AF + AAA
or
  dF + AF - FA = 0.

The same applies to exterior fields. Denoting the field by q, one can
define the field velocity by
  v = dq + Aq.
This embodies the assumption, also, as you made reference to, that the
fields q are form-valued. I'm not sure who goes beyond degree 2 forms
for q .. at least in 3+1 dimensions. Spinors would be incorporated
simply as scalar-valued forms that possess a Lorentz gauge symmety. A
spin 3/2 field could be either a 1-form valued field with a spin 1/2
Lorentz representation for its components or a scalar valued field
with a spin 3/2 Lorentz gauge symmetry. This requires, in turn, that
Lorentz degrees of freedom be present in whatever comprises the gauge
field A. So, if fermions are present, the Lie group underlying the
potentials A has to include both the "internal" symmetries of whatever
forces are present, plus a representation for local Lorentz
symmetries.

So ... going back to your original query, this is the "algebraic" form
I prefer to the exterior covariant derivative D = d + A.

Manipulation with it is easy. Taking a second derivative, one gets as
a result
  dv = d(dq) + (dA) q - A (dq) = (F - AA) q - A (v - Aq) = Fq - AAq -
Av + AAq
Thus
  dv + Av = Fq.

Adjoints are brought in at this point. As mentioned above, the
assumption made here is that a *-operator can be defined with the
property
  (ab)* = b* a*; (a+b)* = a* + b*; a** = a.
So, at the outset, one is taking the "enveloping algebra", first
expanding it to a larger algebra that operates on the bases (Y^a),
(e_A), (e^A) and then further expanding it to a "enveloping *-
algebra".

It's also assumed that this works in conjunction with the differential
forms so that the product reversal property remains intact; e.g.
  ((a dx) ^ (b dy))* = (ab dx ^ dy)* = b* a* dy ^ dx = b* dy ^ a* dx.
Then one can immediately write down the adjoint of the field laws
above,
  A* d* + A* A* = F*; F* d* + F* A* - A* F* = 0
  q* d* + q* A* = v*; v* d* + v* A* = q* F*.

The convention adopted here is that the adjoint of the exterior
derivative d* is simply the same -- but operating to the left from the
right. Then one can write
  A* d*  = -d A*.
Thus, in order for (dA + AA = F) to be consistent under the action of
the adjoint, one adopts the convention that A* = -A. Then it follows
that F* = F. Component-wise,
  F = 1/2 F_{mn} dx^m ^ dx^n
so that
  F = *F = 1/2 (F_{mn})* dx^n ^ dx^m = -1/2 (F_{mn})* dx^m ^ dx^n,
so that
  (F_{mn})* = -F_{mn}.

If q is a form-valued field of degree m, then the adjoint relations
can be rewritten as
  (-1)^m d q* - q* A = v*
  (-1)^{m+1} d v* - v* A = q* F.

This background makes it easier to write down a Dirac-like Lagrangian
  L = Tr(q* a v + v* b q + q* c q).

Here, you can get a better fix on a size limit. The Lagrangian, L,
here is a 4-form. The action integral is just
  S = integral(L).
In order for this to work, one has to take (a,b) each to be forms of
degree (3-2m); while (c) is of degree (4-2m). In 3+1 dimensions this
limits you to m = 0, 1 for this particular form of the Lagrangian.

But the general framework may be of particular use in writing down
higher-dimensional field theories. I'll assume, below, that the
dimension of the underlying space is n; which generalizes from n = 4.

The field equation is relatively easy to develop from this Lagrangian.
For instance, one has the variational
  delta(q* a v) = delta(q*) a v + q* a delta(v).
Running the delta through the exterior derivative, one gets
  q* a delta(dq + Aq)
  = q* a d(delta q) + q* a delta(A) q + q* a A delta(q)
Integrating by parts, one gets
  q* a d(delta q) = (-1)^{n-m} d(q* a delta q) - (-1)^{n-m} d(q* a)
delta(q)

Under the trace operator, using cyclicity, one gets
  Tr(q* a delta(A) q) = (-1)^{n-q-1} Tr(delta(A) q q* a).
Thus, associated with the gauge field is a current
  J_q = (-1)^{n-q-1} q q* a.
A second contribution comes from the b term
  delta(v* b q) = delta(v*) b q + ...
  = delta((-1)^m dq* - q* A) b q + ...
  = ... - q* delta(A) b q + ...
  = -(-1)^{m(n-m)} delta(A) b q q* + ...
Thus, the second contribution is given by
  J_q* = (-1)^{m(n-m)} b q q*.

I can't answer specifically to that. But the general picture I painted
here may show the advantages of simply putting everything down in
algebraic form instead of dealing with it via degrees of indirection
(e.g. by making explicit reference to "adjoint action", "left
actions", "right actions", etc. and the specialized notations
developed for these).