Electromagnetic stress tensor is only traceless in 4D? The electromagnetic stress tensor $F_{\mu \nu}$ is as we all know traceless in 4 dimensions. With $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ and $A = (A_0,A_1,A_2,A_3)= (\phi, A_1, A_2, A_3 )$
In other dimensions this is not the case? If so how does one expand either the definition of $F_{\mu \nu}$ and A?
Edit: I was clearly wrong with the with terminology, I meant the electromagnetic stress-energy tensor (or electromagnetic energy-momentum tensor) which was pointed out in the comment section.
 A: I am not sure if you confused the terminology, but I'll address both interpretations of your question.
$F_{\mu \nu}$ is called the Faraday/electromagnetic/field strength tensor. Its antisymmetric property does not depend on the number of dimensions, so it is always traceless. See also the answer by @J. Murray above.
On the other hand, the electromagnetic stress-energy tensor (or electromagnetic energy-momentum tensor) is the tensor that you get by varying the Maxwell action with respect to the metric. In $d$ dimensions, it is given by:
$$T^\alpha_{\ \beta} = \frac{1}{4 \pi} \left( F^{\alpha \mu} F_{\beta \mu} - \frac{1}{4} \delta^\alpha_{\ \beta} F^{\mu \nu} F_{\mu \nu} \right)$$
Its trace is then:
$$T = \frac{1}{4 \pi}F^{\mu \nu}F_{\mu \nu} \left( \frac{4-d}{4} \right)$$
which is zero only for $d=4$. In $d=4$, then, free Maxwell theory is scale invariant and conformally invariant. For $d \neq 4$, it is only scale invariant.
A: The electromagnetic tensor $F$ is the exterior derivative of the one-form $A$.  This makes it a two-form, which is antisymmetric (and thus traceless) by definition, completely independent of the number of spacetime dimensions involved.
