I'll address your first question, in some computational detail; although I cannot do better than the answers which have already been given in terms of what is conceptually going on.
Let G be a 1-connected Lie group with Lie algebra g. We assume that the group admits a bi-invariant metric or, equivalently, that the Lie algebra admits an ad-invariant inner product, denoted Tr by the traditional abuse of notation in the Physics literature. Let M be an orientable 3-manifold which, for simplicitly, I will take without boundary. Then every principal G-bundle on M is trivial and choosing a trivialisation once and for all we can identify connections with g-valued one-forms on M. Let A be one such connection.
In this language, the Chern-Simons action for A can be written as
I[A]=k4π∫MTr(A∧dA+23A∧A∧A).
There is of course some abuse of notation here. Both
A and
dA are
g-valued, whence the first term in the action is the composition of their wedge product with the ad-invariant inner product on
g. This results in a 3-form, which can be integrated on
M. The second term is perhaps best understood if one rewrites
23TrA∧A∧A=13Tr[A,A]∧A,
where
[A,A] is the composition of wedging and the Lie bracket on
g, hence it is again a
g-valued 2-form, and the second term in the action is (up to the factor of one-third) again the composition of the wedge product of
[A,A] and
A with the ad-invariant inner product.
(Notice that
[A,A] is not zero, because although
[−,−] is antisymmetric, so is
∧ between two one-forms.)
Now let us vary I:
δI=k4π∫MTr(δA∧dA+A∧dδA+2δA∧A∧A).
Using the fact that
d is an odd derivation, we have that
Tr(A∧dδA)=−dTr(A∧δA)+Tr(dA∧δA)
whence,
δI=k2π∫MTr(δA∧(dA+A∧A))+k4π∫∂MTr(δA∧A)
The boundary term drops out because we have assumed
∂M=∅. In summary, the Euler-Lagrange equation is the flatness of the curvature
F=dA+A∧A, which can also be written as
F=dA+12[A,A].
Under a gauge transformation g:M→G,
A↦Ag=gAg−1−dgg−1.
Again there's some abuse of notation in that one is assuming that
g is a matrix Lie algebra, but this only notational and it's perfectly possible to write this in a way that makes sense in general; namely,
Ag=AdgA+g∗θ
where
θ is the
right-invariant Maurer-Cartan form on
G.
In order to vary the action is perhaps best to write it in an equivalent way; namely,
I[A]=k4π∫MTr(A∧F−13A∧A∧A),
since the curvature transforms covariantly:
Fg=gFg−1.
The next calculation is a little long and I will not reproduce it here. It is not difficult: just plug in the expressions for Ag and Fg into the action I[Ag] and expand using the fact that Tr is Ad-invariant. One must also use the Maurer-Cartan structure equation dθ=−θ∧θ. The result of the calculation is
I[Ag]−I[A]=k4π∫MTr(d(g−1dg∧A)+13(dgg−1)3)
where the first term vanishes by Stokes's theorem since
M has no boundary, leaving
I[Ag]−I[A]=k12π∫MTr((dgg−1)3)
If G is a compact simple group, then one can rescale the inner product Tr in such a way that
112π∫MTr((dgg−1)3)∈Z
whence the Chern-Simons path integral is gauge-invariant provided that
k∈Z as well, in units where
ℏ=1.
This post imported from StackExchange MathOverflow at 2014-06-30 07:41 (UCT), posted by SE-user José Figueroa-O'Farrill