gauge theory (alternative descriptions)
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r_n_tsai@yahoo.com - 14 Apr 2008 15:53 GMT I am trying to understand gauge theories in an algebraic setting (I hope to avoid arguments over what setting is "better",...).
It seems that the basic construct is a combination of two algebras :
one algebra is the commutative algebra generated by the differentials (d1,d2,d3,d4).
the second algebra is the "internal symmetry" algebra. For the strong force, this is "su(3)", but over what? not C, but the the ring of functions A(x,y,z,t) -> C?
the "covariant derivative" is formed by combining the generators of the two in what seems like a reasonable enough process. I'd like to see if there are alternative approaches of describing this. Again my preference is for an algebraic treatment.
Thanks and please excuse the vagueness of the post.
Stephen Blake - 17 Apr 2008 15:05 GMT r_n_t...@yahoo.com wrote:
> I am trying to understand gauge theories in an algebraic setting > (I hope to avoid arguments over what setting is "better",...). [quoted text clipped - 14 lines] > > Thanks and please excuse the vagueness of the post. Dear r_n_t,
I would also like to understand gauge theories and so here is my current understanding of what is going on.
A gauge theory is one in which the state space is too big; there are many states which correspond to each physical situation. If G is a group with elements g, then the big Hilbert space carries a representation U(g). The group G has an invariant subgroup H called the gauge group. Since H is an invariant subgroup, there is a quotient group G/H. The quotient group G/H will show up in the representations of G on the big Hilbert space as reps, U(g)=gH (eqn 1). The subspace of the big Hilbert space that transforms under the quotient rep (eqn 1) is spanned by states of the regular rep |gH> so that, U(g1)|gH>=|g1gH> (eqn 2). These states |gH> are the physical states. However, when we work in the big Hilbert space we don't know about the subspace which transforms as the quotient rep; we use some states |i> which span the whole of the big Hilbert space. We write a state as |psi>=\sum c_i|i> and we have an equation of motion F(|psi>)=0 which is satisfied when |psi> lies in the subspace which transforms as the quotient rep.
One often reads that "a gauge group is a group of transformations that leaves invariant every operator representing an observable." If h is an element of the gauge group H, then the quotient rep is U(h)=H and so, U(h)U(g)U(h^-1)=hgh^-1H=hgH=gh1H (for some h1 in H) and so, U(h)U(g)U(h^-1)=gH=U(g).
Stephen http://www.stebla.pwp.blueyonder.co.uk
r_n_tsai@yahoo.com - 18 Apr 2008 15:25 GMT Thanks for the responses Alexey and Stephen,
The sort of setting I'm after ideally doesn't get too intertwined with other subjects. I'd rather not deal with groups, bundles, cohomology, observables,....if that can be avoided.
I'll try to work out "u(1)" gauge algebra (electomagnetism) as an example; although keep in mind that there are gaps in the definitions that I'm not sure how to fill. Let's call this algebra X. I don't know if X should be over a simple field (R or C) or something more complicated (functions over R,...). X can be "generated" by 8 elements : X=<d1,d2,d3,d4,a1,a2,a3,a4> Here "generated" is along the lines of universal enveloping algebra (polynomials in the 8 generators). You can think of the di's as partial derivatives, and ai's as related to electromagnetic 4- potential; but that's only for motivation...strictly speaking these are just generators that obey certain multiplication rule. What are these rules? let the algebra multiplication be g1g2, then define a second multiplication (commutation) on the algebra g1*g2=g1g2-g2g1. Then I think all the rules you need are :
di*dj=0; ai*aj=0; di*aj=bij;
the first just says derivatives commute; the second is because this is an abelian gauge; the last one is a definition of 16 elements of the algebra; related to partials of 4-potentials. Higher derivatives can also be treated as definitions of other elements. I think that's it.
What can you do with this. Let's try to derive Maxwell's equations! First define "covariant derivative" :
Di=di+ai;
these are just 4 elements in the algebra
Next define the "field strength" :
Fij=Di*Dj;
These are just 16 elements in the algebra. Because g1*g2=-g2*g1, only 6 of these are significant; these can be identified with the electric and magnetic field (E1,E2,E2,B1,B2,B3).
Take the identity g1*(g2*g3)+g2*(g3*g1)+g3*(g1*g2)=0 and substitute Di's in it and with the right identification of Fij with E and B you get \Del.E=0 and \Del x E = \partial B / \partial t Note that these are 4 equations relating three components at a time.
Anyway u(1) gauge is probably the simplest example. It would be good to find a reference where a non-abelian gauge is treated along the lines of the above example...
Rock Brentwood - 25 Apr 2008 21:22 GMT On Apr 18, 9:25 am, r_n_t...@yahoo.com wrote:
> Anyway u(1) gauge is probably the simplest example. It would be > good to find a reference where a non-abelian gauge is treated > along the lines of the above example... Here: phi, A, E, B are Lie vector valued; D, H, J, rho are Lie co- vector valued. The permittivity epsilon is the gauge group metric (actually: epsilon c is).
In a universal enveloping algebra of the underlying Lie algebra B = curl A + A x A E = -grad phi - dA/dt + A phi - phi A div B + A.B - B.A = 0 curl E + AxE + ExA + dB/dt + phi B - B phi = 0
Extending the universal covering algebra to the dual space of the Lie algebra div D + A.D - D.A = rho curl H + AxH + HxA - dD/dt + phi D - D phi = J div J + A.J - J.A + d(rho)/dt + rho phi - phi rho = D.E - E.D + B.H - H.B.
The Lorentz force law Force density = rho E + J x B = rho_a E^a + J_a x B^a. The law for power density = J.E = J_a.E^a.
For a test charge, the force F and power P are dp/dt = F = e (E + v x B) = e_a (E^a + v x B^a) dT/dt = P = e (v.E) = e_a (v.E^a) where the charge (e_a) is a Lie co-vector and p, T represent the momentum and kinetic energy of the charge.
In first-order form: d(p + eA)/dt = -div (e(phi - v.A)) d(T + e phi)/dt = d/dt (e(phi - v.A)) where the d/dt on the right and the "div" on the right are taken as partial derivatives only with respect to the field coordinates in phi and A (i.e., derivatives taken with v and e constant).
The first order and second order form are consistent only if the charge is endowed with a time variability, itself, given by de/dt = e(phi - v.A) - (phi - v.A)e. The precession is the classical underpinning to "flavor-changing" interactions.
The most general Lorentz-invariant relations in 4-D (making the Lie algebra index explicit): D_a = epsilon_{ab} E^b + theta_{ab} B^b + k_{abc} E^b x B^c + 1/2 l_{abc} (E^b x E^c - c^2 B^b x B^c) H_a = epsilon_{ab} c^2 B^b - theta_{ab} E^b + 1/2 k_{abc} (E^b x E^c - c^2 B^b x B^c) + c^2 l_{abc} E^b x B^c where epsilon, theta are symmetric and k and l are completely anti- symmetric.
For Yang-Mills fields with a simple Lie group, theta = 0 and epsilon_{ab} c = K_{ab}/g^2 where K is the Killing metric and g the coupling coefficient. With a semi-simple Lie group, epsilon is the sum of the respective contributions from each simple subgroup (e.g. for U(1) x SU(2) x SU(3) the metric is defined by 3 "coupling coefficients").
The coupling coefficients, the gauge group metric, and the permittivity are all synonymous with one another.
For the SU(3) sector theta is taken to be a non-zero multiple of the SU(3) Killing metric in the standard model.
Nobody says anything about k_{abc} or l_{abc}.
If one takes the permeability to be the coefficients corresponding to the inverse relation (E,B) <- (D,H) then in the presence of theta, (epsilon mu) as a matrix product will be smaller than 1/c^2. Things get more complicated if k and l are present.
Stress tensors can be written down in a form that is independent of what the constitutive law (and Lagrangian) may be ... so it can be defined for Yang-Mills gauge fields or more general gauge fields. The total energy for a static point-like source -- on the assumption that the stress tensor is Lorentz covariant and the energy is well-defined -- can be written down in closed form ... independently of what the Lagrangian is. This result is not well-known ... and (in fact) not "known" at all (yet).
If the Lagrangian is homogeneous to the first degree in the quadratic Lorentz field invariants and independent of the cubic Lorentz field invariants, then the energy is the limiting value of 1/2 (q(r) phi(r)) as r -> 0, where q(r) is the flux of the D field. Otherwise, if I recall correctly, it's the limiting value of (q(r) phi(r)) as r -> 0 minus the Lagrangian, itself. I'll have the check my notes on this. But the argument is simple (and almost trivial!)
For the gauge field corresponding to gravity the canonical stress tensor is non-zero, while the symmetrized stress tensor is 0. The difference between the two ... in all cases for all gauge fields ... is related to the divergence of the angular momentum tensor.
References to articles I have on-line:
The Gauge Field Equations in Maxwell Form http://federation.g3z.com/Physics/index.htm#MaxwellYangMills
The Maxwell Equations for Non-Abelian Gauge Fields http://federation.g3z.com/Physics/index.htm#Hehl2
The Anatomy of the Electroweak Field Supplementary article included under: http://federation.g3z.com/Physics/index.htm#Hehl1
The Gauge-Scalar Fields in Maxwell Form http://federation.g3z.com/Physics/index.htm#GaugeScalar (This one is going to be expanded)
The Constitutive Law in Gauge Theory http://federation.g3z.com/Physics/index.htm#Constitutive (What replaces/generalizes the Lorentz relations?)
r_n_tsai@yahoo.com - 26 Apr 2008 15:26 GMT I appreciate the effort that went into this detailed response, but I think this is still the "standard" treatment for gauge theory which I think I understand well enough. What I'm looking for, and perhaps I'm wrong in expecting to find any such thing, is an "alternative" description that merges the two algebras (the algebra of the differentials and the internal symmetry algebta) into a single algebra.
for example :
> phi, A, E, B are Lie vector valued; what does this mean?
> In a universal enveloping algebra of the underlying Lie algebra > B = curl A + A x A What's the "underlying Lie algebra" here? Let's pick the strong force gauge with su(3) as the internal symmetry algebra. If "A" is in su(3), then I can see that AxA would be in the enveloping algebra of su(3), but curl A involves differentials, so it has to be in something bigger, right?
> References to articles I have on-line:
>The Gauge Field Equations in Maxwell Form >http://federation.g3z.com/Physics/index.htm#MaxwellYangMills I've run accross this site before. A lot of interesting (and well written) material there. What's the story behind the "federation"? ..just curious.
Rock Brentwood - 06 May 2008 03:40 GMT On Apr 26, 9:26 am, r_n_t...@yahoo.com wrote:
> > phi, A, E, B are Lie vector valued; > > what does this mean? That its components are not numeric but reside in a Lie algebra.
Hence, if (Y_a: a=1,...,G) is the basis of the underlying Lie algebra, where G is its dimension, then phi = phi^a Y_a, A = A^a Y_a, E = E^a Y_a, B = B^a Y_a; where each B^a, E^a, A^a is, itself, a 3-vector. The x-component of B, for instance is the Lie vector B^x = B^{ax} Y_a, where B^{ax} is the x-component of B^a.
The fields Maxwell terms "electric displacement" (D) and "magnetic strength" (H) are, in contrast, dual Lie vectors. So the Lie index goes in the downstairs position, D_a, H_a. Likewise for the charge e_a. All this comes out of the Lorentz law. To contract e with E and B in e (E + v x B), you need e's index downstairs, while E and B's are upstairs. To make Gauss' law consistent (div D = e), you need D's index downstairs, like e's; similarly for the Maxwell equation (curl H - dD/dt = J).
Then you can see the true significance of the oft-neglected Lorentz relation D = epsilon E: epsilon is a Lie algebra metric, up to proportionality. It's lowering the index.
> What's the "underlying Lie algebra" here? Let's pick the strong > force gauge with su(3) as the internal symmetry algebra. If "A" > is in su(3), then I can see that AxA would be in the enveloping > algebra of su(3), You've just answered your own question here.
> but curl A involves differentials, so it has to be in something bigger, right? No. The a-component of curl A is the curl of A^a: (curl A)^a = curl (A^a).
In the language of differential forms, the x,y,z components of A fit together with phi into the 1-form: A_x dx + A_y dy + A_z dz - phi dt. The differential of this is just d(A_x dx + A_y dy + A_z dz - phi dt) = (dA_y/dx - dA_x/dy) dx^dy - (dA_x/dt + d(phi)/dx) dx^dt + (dA_z/dy - dA_y/dz) dy^dz - (dA_y/dt + d(phi)/dy) dy^dt + (dA_x/dz - dA_z/dx) dz^dx - (dA_z/dt + d(phi)/dz) dz^dt. In 3-vector form, using the notation dr = (dx,dy,dz); dS = (dy^dz,dz^dx,dx^dy) this becomes d(A.dr - phi dt) = (curl A).dS - (dA/dt + grad phi).dr^dt.
You can work out the quadratic terms, likewise, as per your indication above. The z-component of AxA, for instance, is (AxA)^z = A_x A_y - A_y A_x = [A_x, A_y]. The second equality is where we exploit the fact that we're working in the covering algebra.
> I've run accross this site before. A lot of interesting (and well > written) material there. What's the story behind the "federation"? > ..just curious. It was supposed to be the research archive site adjoining the non- fiction book trilogy "The Federation Series", but has gotten a bit larger. Part of that is that the 2nd book, "The Fourth Wave" was going to delve a little into future physics; which is quite understandable, given what the central theme of the book is
http://federation.g3z.com/FedSeries/FourthWave.htm
Of course, it's difficult to accurately say what physics is going to be in the future ... um ... without making it happen, as a consequence! But either way whether that happens or not: of necessity, a description of future physics removed from the zeitgeist of the present and past entails a significant point of departure from the cliche'd points of view prevalent in the 20th and early 21st centuries; and a point of departure even from the cliche'd accounts that is commonly given of what future physics is going to be. Hence, the dissonance you might pick up reading the material is by design.
To answer your question directly: the third book forecasts the rise of the World Federation, which might be considered also as a culmination of Einstein's *other* dream. He was well-known as a staunch advocate of worldwide political unification and probably on his account (in part) several states in the US actually passed resolutions during his later life calling on such unification.
Both the archive and series are precursors to a bona fide interactive blog site dedicated to these issues and to the latest research in advanced and theoretical math, physics, computer science and other fields. In turn, the blog is a precursor (still a long ways off) to a foundation that will lie at the root of what will become known as the "World Federation Organization". In a way, this organization will be a continuation of what had been formerly known as the World Federalist Society (if memory serves me correctly), which Einstein was a member of and (I believe) even the leader of for a while.
The only affiliation I list in the set of papers I actually publish, for now (a non-empty set), is just "The Federation Archive" and its web address under my name. The personal stuff is going to be moved off to the side under a "personal profile" page, once the site opens up. There will probably be multiple people accessing and using it in a similar way with their own profile pages.
The invitation will be tendered to some of the best and the brightest to sign on and contribute when that time comes. But the research part of the site is meant to be neither a substitute for nor clone of either the Wikipedia or ArXiv.
r_n_tsai@yahoo.com - 08 May 2008 00:00 GMT > > > phi, A, E, B are Lie vector valued; > > what does this mean? [quoted text clipped - 4 lines] > x-component of B, for instance is the Lie vector B^x = B^{ax} Y_a, > where B^{ax} is the x-component of B^a. Let me linger on this point just a little bit longer. For each "a", B^{ax} is a function R^4 -> C. Let's call the ring of such functions "X1". The lie algebra we're dealing with is over "X1", let's move to it's univeral enveloping algebra and call that "X2"; phi = phi^a Y_a and B^x = B^{ax} Y_a, so both are *in* X2; (calling B^x a Lie vector is a little confusing). B=(B^x,B^y,B^z) is in a 3-dim vector space over "X2"; let's call that "X3". each B^a is also in "X3".
[interesting comments on electric displacement,contraction,... deleted] we can revsist this later.
> > but curl A involves differentials, so it has to be in something bigger, right?
>No. The a-component of curl A is the curl of A^a: (curl A)^a = curl >(A^a). [quoted text clipped - 10 lines] >this becomes > d(A.dr - phi dt) = (curl A).dS - (dA/dt + grad phi).dr^dt. You've jumped disciplines here. The properties of differentiation haven't been introduced yet in this setting. You have "curl" : a map "X3" -> "X3" "d/dt" : a map "X3" -> "X3" "grad" : a map "X2" -> "X3"
obviously these maps are not *in* "X2" so we're dealing with something bigger! component wise all these can be built with the 4 maps : "d/dx" : "X2" -> "X2" "d/dy" : "X2" -> "X2" "d/dz" : "X2" -> "X2" "d/dt" : "X2" -> "X2"
The "Y_a"'s are also maps "X2"->"X2". It is combining these G+4 in an alternative way that I've been looking for.
>You can work out the quadratic terms, likewise, as per your indication >above. The z-component of AxA, for instance, is > (AxA)^z = A_x A_y - A_y A_x = [A_x, A_y]. >The second equality is where we exploit the fact that we're working in >the covering algebra. the quadratic part is ok. A_x and A_y are in in "X2" so A_x A_y and A_y A_x are in "X2" too.
[interesting material on the federation deleted]
..dissonance you might pick up reading the material is by design.
aaah...that's the best part!
Rock Brentwood - 09 May 2008 07:47 GMT On May 7, 6:00 pm, r_n_t...@yahoo.com wrote:
> For each "a", B^{ax} is a function R^4 -> C. > Let's call the ring of such functions "X1". etc.
You almost sound like you want to graft a theory of types on top of al this, a' la a typed programming language or typed logic. OK.
As a space-time function, B^{ax}: M -> R, where M is the spacetime manifold (which can be treated as R^4 when M is Minkowski space adapted to Cartesian coordinates). X1 = E(M,R) = E(M) where E(M,N) denotes the space of C^{infinity{ functions from manifold M to N. It's possible to expand the treatment to allow for singular functions, embedding E(M) into a larger space, but that's a digression here.
For each component, B^x: M -> L = Lie(G), where G is the gauge group and L its Lie algebra. So, for G = SU(3), L = su(3). So, your X2 is the tensor product E(M) x L of the algebras E(M) and L.
The 3-vector B: M -> L x R^3. So, X3 is E(M) x L x R^3.
The thing you're asking for is an amalgamation of the 2 algebras. This is gotten from treating the 1-form and 2-form, instead. The potential 1-form in Minkowski space in Minkowski coordinates is a = (A_x dx + A_y dy + A_z dz - phi dt). As a function it maps from M to L x Lambda^1(M), where Lambda^1(M) is the space of 1-forms over the manifold M. The 2-form f = (E_x dx + E_y dy + E_z dz)^dt + (B_x dy^dz + B_y dz^dx + B_z dx^dy) maps from M to L x Lambda^2(M). So they are, respectively, in a: E(M) x L x Lambda^1(M) f: E(M) x L x Lambda^2(M).
The amalgamation you seek out is the simple algebraic relation f = da + a^2, and its corresponding Bianchi identity df + af - fa = 0.
> You've jumped disciplines here. The properties of differentiation > haven't been introduced yet in this setting. You have > "curl" : a map "X3" -> "X3" > "d/dt" : a map "X3" -> "X3" > "grad" : a map "X2" -> "X3" All of what you're trying to get at here, and later in your reply, are subsumed by the considerations just raised.
But, this is actually NOT how the field is formalized mathematically. What's actually done is to extend the manifold M into a fibre bundle whose fibre is the group G. The fields are sections over the various spaces associated with the bundle, not merely functions.
This treatment is not equivalent to one expressed merely in terms of function spaces, but strictly more general (and robust). It can even survive the transition to discrete or non-commutative geometries mostly intact.
Rock Brentwood - 09 May 2008 22:41 GMT > The 3-vector B: M -> L x R^3. So, X3 is E(M) x L x R^3. ...
> As a function it maps from M to L x Lambda^1(M), where Lambda^1(M) is This needs to be fixed and more carefully done. The combined n-form algebra you're seeking takes place in Lambda(M) x L. This is what needs to be more carefully parsed.
So, first things first. Given a vector space V of n dimensions, the C(n,k) dimensional space Lambda^k(V) is the space formed by all k- vectors. The 2-vectors in Lambda^2(V) are, for instance, linear combinations of (u (x) v - v (x) u) for u, v in V, where (x) denotes tensor product.
The space-time manifold M may be coordinatized locally by (x^m) where the space-time index m ranges from 0 to n-1, and n is the dimension of M. At each point m in M, the tangent space T_m(M) has local coordinates (x-dot^m) and is equivalent, as a vector space, to R^n. Together, these form the "fibres" of the tangent space TM whose local coordinates are (x^m, x-dot^m).
The co-tangent space T_m*(M) is the dual of T_m(M) and has coordinates (p_m). It, too, is equivalent to R^n. This forms the co-tangent bundle T*(M) which has local coordinates (x^m, p_m).
Over each set of vector spaces one may define the spaces of anti- symmetric k-vectors: Lambda^k(T_m(M)) and Lambda^k(T_m*(M)). They are coordinatized locally by Lambda^k(T_m(M)): (v^{m1 ... mk}) and Lambda^k(T_m*(M)): (p_{m1 ... mk}), respectively. These give you the spaces, respectively, of k- multivectors and k-forms over the point m of M.
Together, these give you the spaces Lambda^k(T(M)) and Lambda^k(T*(M)), with respective coordinates given by (x^m, v^{m1...mk}) and (x^m,p_{m1...mk}).
The respective fields are smooth SECTIONS over these spaces. That is, a k-multivector field v: M --> Lambda^k(T(M)), and a k-form field omega: M --> Lambda^k(T*(M)) such that v(m) is in Lambda^k(T_m(M)) and omega(m) is in Lambda^k(T_m*(M)) for each m in M; the mappings are as C^{infinity} functions.
The spaces of k-form fields will be denoted as Lambda^k(M).
The space Lambda(M) will then be defined as the sum of the Lambda^k(M) over all k = 0, 1, ..., n.
The 0-forms are scalar functions over M. This may be conveniently taken as the space E(M), the space of C^{infinity} functions mapping M to R. This is equivalent to Lambda^0(M). The potential 1-forms and curvature 2-forms reside, componentwise in Lambda^1(M) and Lambda^2(M). Since they are both Lie valued, then their spaces are the tensor products Lambda^1(M) (x) L and Lambda^2(M) (x) L where L is the Lie algebra of the underlying gauge group.
When L is treated as residing in a universal covering algebra, then one can form products over Lambda(M) (x) L. This leads to a product defined by (a (x) u) (b (x) v) = (ab) (x) (uv) where a, b are in Lambda(M) and u, v are in L. The Lie bracket then extends to a graded bracket over the combined algebra, given by [a(x)u, b(x)v] = (ab) (x) [u,v]. Expressed in terms of the product, this is just [a(x)u, b(x)v] = (a(x)u) (b(x)v) - (-1)^{AB} (b(x)v)(a(x)u) where A, B are the degrees of the forms a and b, respectively.
That's the combined algebra you're looking for.
Note, in particular, for 1-forms a, b you have [a,b] = ab + ba, not ab- ba. So, as a special case, for a 1-form a, you have [a,a] = 2a^2.
An author like Bleecker (in his 1980 tome on the math underlying gauge theory and principal bundles) only defines the bracket. He doesn't use the device of putting everuything inside a universal covering algebra, so things get obscured. In place of the equation f = da + a^2, he can only write f = da + 1/2 [a,a].
So, in n-dimensional spacetime, if A is the potential 1-form, and F the field strength 2-form (using upper case now), their Lie components are A^a, F^a which are, respectively, 1-form and 2-form fields residing in Lambda^1(M) and Lambda^2(M). The field law becomes (dA + A^2 = F), the Bianchi identity (dF + AF - FA = 0). Expressed in terms of the graded Lie bracket, these become dA + 1/2 [A,A] = F; dF + [A,F] = 0.
The conjugate fields (corresponding to what Maxwell calls D and H) are Lie co-vectors, and are captured by the (n-2) form which we'll call G. Thus, for Minkowski spacetime, you have G = (D^x dy^dz + D^y dz^dx + D^z dx^dy) - (H_x dx + H_y dy + H_z dz) ^ dt. These come out as derivatives of the field Lagrangian L. So if the Lagrangian L is a n-form (the kernel of the action integral S = integral(L)), then the total differential in terms of A and F would be dL = dA ^ J - dF ^ G. The current (n-1) form is, in Minkowski space, the 3-form J = rho dx^dy^dz - (J_x dy^dz + J_y dz^dx + J_z dx^dy)^dt.
To write the field law and conservation law algebraically, you need to also bring the dual Lie algebra L* within the universal covering algebra. First, the bracket may be extend to an operation over L and L* by [w,u].v = w.[u,v] where w is in L*, and u,v in L; with ().() denoting the contraction. Taking a basis (Y_a) for L, defining the structure coeffcients by [Y_a,Y_b] = f^c_{ab} Y_c, this gives you a product [Y^c,Y_a] = f^c_{ab} Y_b.
Assume this has been incorporated into the covering algebra in such a way that [Y^c,Y_a] = Y^c Y_a - Y_a Y^c.
Then one can define a graded bracket that for (Lie-valued) and (Lie- covalued) forms. That is, if u is in Lambda^A(M) x L and v is in Lambda^B(M) x L*, then the bracket [u,v] will be in Lambda^{A+B}(M) x L*, such that [u,v] = uv - (-1)^{AB} vu.
The field law then becomes dG + [A,G] = dG + AG - (-1)^n GA = J.
>From this, the conservation law follows dJ + [A,J] = [F,G] --> dJ + AJ + (-1)^n JA = FG - GF.
Alexey Popov - 17 Apr 2008 15:05 GMT > I am trying to understand gauge theories in an algebraic setting > (I hope to avoid arguments over what setting is "better",...). [quoted text clipped - 12 lines] > see if there are alternative approaches of describing this. Again > my preference is for an algebraic treatment. Necessary ingredient is external derivarive. Namely, main agrebra A is graded commutative algebra with nilpotent derivation d of degree +1. (this is algebra of differential form in total space of bundle). Lie group action induces action of lie algebra on A. This is way on which equivariant cohomology are constructed. May be this can help you.
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