Integrating (6.40) on t we obtain:
jr^(x) =
1 )
=
- d,A, -f f[4 „ 4 J ).4 ,)
^Tr(F*F) = d^jr%x)
(*^v =
Of course, after the answer (6.41) and (6.42) is known, it can be checked
by direct computation.
Substitution of (6.42) into (6.34) gives:
96
GAUGE FIELDS AND STRINGS
(6.41)
(6.42)
1 d)-;r,dV
(6.43)
5*3
where we integrate over a large S^. At these distances
= 0, and
(6.41) can be replaced by:
^^(x) =
Tr(d,A, - d,A, F lA,A,)A^
-
Tr(A,A,A^)
- ^ 6^^^^Tr(L,L,LJ
(6.44)
This proves the equivalence of (6.32) and (6.34).
Our aim now is to find an instanton solution with ^ = 1. As in the
case of the n-field we can avoid solving the Yang-Mills equations
themselves, by considering instead the “square root” of them. Let us use
an identity:
S =
1
4eo
Tr
d^x
= ¿ 1
J Tr((F,„ - *FJ^)
+ ¿2 j Tr(F,. d^x
We see that if we find a solution of the “duality” equation
F = *F
fiv
fiV
(6.45)
(6.46)
then the action for a fixed q will be minimal. Actually, it is trivial to
check that if the first-order equations (6.46) are satisfied, then the
Yang-Mills equations
(6.47)
y^F'^y = 0