How are the Maxwell's Electromagnetism equations traceless?

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I was reading this:

However, which only coincides with his final (correct) equation if the stress-energy tensor T (and hence also R) is traceless, i.e. that the sum of the elements on the main diagonal of the matrix trace are zero), which is true for Maxwell’s electrodynamics.
 
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The trace of the energy-momentum tensor of the electromagnetic field is ##{T^i}_i = \dfrac{1}{4\pi} \left\{{F^i}_j{F_i}^j - \dfrac{1}{4} \eta^i_i F_{jk} F^{jk} \right\}##. Since ##\eta^i_i = \delta^i_i = 4## and also ##{F^i}_j{F_i}^j = \eta^{ik} \eta_{il} F_{kj} F^{lj} = F_{lj} F^{lj}##, the bit inside the curly brackets vanishes.
 
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vanhees71 said:
The reason is scale invariance of the free Maxwell equations. This symmetry is, however, very fragile since it's anomalously broken when quantizing the theory.
So then it would seem that QED theory has a quandary; no wonder so little progress has been made.
 
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vanhees71 said:
What do you mean by "QED theory has a quandary"? I've no clue. The claim that "so little progress has been made" is just ridiculous particularly in the context of QED.
Because the symmetry is broken when quantizing?
 
What's the problem with scale invariance broken? It's broken anyway as soon as you have a single massive particle in the game (and the electrons and positrons in standard minimal QED are such particles).

Some anomalies are very helpful. E.g., the trace anomaly in QCD explains most of the mass of the hadrons consisting of light quarks (u, d, s) or the axial anomaly comes to the rescue for the neutral-pion decay.