into its physical and pure-gauge parts, simiar to Chen et al. [@Chen08],[@Chen09]. Different from their treatment, however, we impose only the following quite general conditions : $$\begin{aligned}
 F^{\mu \nu}_{pure} \ \equiv \ \partial^\mu \,A^\nu_{pure} \ - \ 
 \partial^\nu \,A^\mu_{pure} \ - \ i \,g \,[A^\mu_{pure}, A^\nu_{pure}]
 \ = \ 0, \label{gcond1}\end{aligned}$$ and $$\begin{aligned}
 A^\mu_{phys} (x) &\rightarrow& U(x) \,A^\mu_{phys} (x) \,U^\dagger (x), \label{gcond2} \\
 A^\mu_{pure} (x) &\rightarrow& U(x) \,\left(\,A^\mu_{pure} (x) \ + \ 
 \frac{i}{g} \,\,\partial^\mu \,\,\right) \,U^\dagger (x) . \label{gcond3}\end{aligned}$$ The first is the pure-gauge condition for $A^\mu_{pure}$, while the second are the gauge transformation properties for these two components. (These transformation properties indicates that the physics is basically contained in the physical part $A^\mu_{phys}$, while the pure-gauge part $A^\mu_{pure}$ carries unphysical gauge degrees of freedom.) Actually, these conditions are not enough to fix gauge uniquely. However, the point of our theoretical scheme is that we can [*postpone*]{} a complete gauge fixing until later stage, while accomplishing a gauge-invariant decomposition of $M^{\mu \nu \lambda}$ based on the above conditions alone. Still, we find the way of gauge-invariant decomposition is not unique and are left with two possibilities.

We start with the decomposition (II) given in the form : $$M^{\mu \nu \lambda} 