162 CHAPTER 3. MATHEMATICAL FOUNDATIONS
2) SU(N)tensor gab. The structure constantsλbca generated by the generatorsωain
(3.5.27) satisfy
(3.5.29) [ωb,ωc] =iλbcaωaA†, ∀A∈SU(N)then we can define another 2-covariant tensorgabonTASU(N)by the structure con-
stantsλbcaas follows(3.5.30) gab=1
4 N
λadcλcbd.The following theorem shows that bothGabandgabare symmetric second-orderSU(N)
tensors.
Theorem 3.31.The fieldsGab(A)and gabgiven by (3.5.28) and (3.5.30) are 2-order sym-
metric SU(N)tensors, and for any A∈SU(N), the generator basis (3.5.27) satisfies (3.5.29),
whereλbcaare the structure constants of SU(N)independent of A.
Proof.We first considergab. By the anti-symmetry of the structure constants:
λabc =−λbac,we deduce the symmetry ofgab:
gab=λadcλcbd=λdacλbcd=gba.Now we verify the symmetry ofGab. By (3.5.17) the basis(ω 1 ,···,ωK)ofTASU(N)
satisfy that
(3.5.31) A†ωa= (A†ωa)†, tr(A†ωa) = 0 , A†=A−^1 ,
NamelyA†ωaare Hermitian. LetA†ωa=τa, then
(3.5.32) τa=τa†, trτa= 0.
Thus we have
(3.5.33) Gab=
1
2
tr(ωaωb†) =1
2
tr(Aτaτ†bA†) =1
2
tr(τaτb†).Thanks to (3.5.32),(τ 1 ,···,τK)∈TeSU(N). Hence we have
(3.5.34) τaτb†=τbτ†a+iλabcτc, trτc= 0.
It follows from (3.5.33) and (3.5.34) that
Gab=1
2
tr(τaτ†b) =1
2
tr(τbτa†) =Gba.2.1.4 Symmetry.
Finally we prove that for anyA∈SU(N), each generator basis(ω 1 ,···,ωK)ofTASU(N)
satisfy (3.5.29). In fact, by (3.5.31),τa=A†ωa( 1 ≤a≤K)constitute a basis ofTeSU(N).
Thereforeτasatisfies (3.5.34), i.e.
A†[ωa,ωb]A=iλabcA†ωc.Hence we obtain (3.5.29). The proof is complete.