variation of the field. We have:
p{x) = Tr(F A F)
Sp{x) = 2 Tr(F A SF)
1
F = dA + A A A =
d x ^ a dx"
TOPOLOGY OF GAUGE FIELDS AND RELATED PROBLEMS
95
(6.35)
(6.36)
SF dSA + A A SA F SA A A = V SA
(where F is a 2-form; d is called the exterior derivative). For pedagogical reasons let us repeat (6.35) in the ordinary notation:
P(^) = iW p Tr(F^vF^p) d^x
=
Tr(F^,^F^^) d^x
F fiv =
F
^v]
SF^, = d^SA, + [A^, (5/1J ~{p^v) = V^(5/l, - V,SA^
From these equalities we obtain:
Sp{x) = 2 Tr F A {dSA F A a SA
SA a A)
= 2 Tr(F A dSA) + 2 Tr(F a A - A a F) a SA
= 2 Tr F A d^/lH- 2 Tr dF A (5/1 = 2d(Tr F a SA)
=
Tr{F,,SA^)d^x
where we have used the Bianchi identity:
[V, F] = dF + /I A F - F A /I
=
dx'’ A dx“ ^ A dx^ = 0
(6.37)
(6.38)
We have:
Sp{x) = (d^Sji^ix)) d^x
SJf^{x) = e^^^^ Tr(F,,SA^)
(6.39)
Now we have to obtain the current
itself. To do this let us introduce
a parameter t: 0 < t < 1 and consider a family of gauge fields
A^{x, t) = xAJ^x). According to (6.39) we have:
dJi\x, t)
, /
dA^
(6.40)
> Tr{(r(a,/1, - d,AJ +
AJ) A^