# What's the fundamental invariants in $1+d$ dim electromagnetism?

As we know in $1+3$ dim electromagnetism, there are only two independent gauge invariant scalars $$\frac12F^{\mu\nu}F_{\mu\nu}=\mathbf{B}^2-\mathbf{E}^2$$ $$\frac14F_{\mu\nu}\ {}^\ast F^{\mu\nu}=\frac14\epsilon^{\mu\nu\alpha\beta }F_{\mu\nu}F_{\alpha\beta}=\mathbf{B}\cdot\mathbf{E}.$$

The first one can be certainly generalized to any dimension. But the second one heavily depends on the $1+3$ dim.

As we know in $1+1$ dim $$F^{\mu\nu} = \begin{pmatrix}0 & -E \\ E & 0\end{pmatrix}$$ So the two invariants I can think is $$\epsilon_{\mu\nu}F^{\mu\nu}=-2E$$ and $$F^{\mu\nu}F_{\mu\nu}=-2E^2$$ So only one is independent.

So my question: In $1+d$ dim electromagnetic theory, how many independent gauge invariant scalars and what are they respectively? How to prove they are the unique independent gauge invariant scalars? i.e. Any other gauge invariant scalars can be constructed by them. I found a proof in $1+3$ dim, but this proof heavily depends on the physics in $1+3$ dim, and cannot be generalized to any dimension.

If you restrict your question to what are the Lorentz-invariant polynomials of $F_{\mu\nu}$, then the question has a relatively simple answer. $F_{\mu\nu}$ is an antisymmetric matrix, which is the same representation as the adjoint of the Lorentz group.

Invariant polynomials in the adjoint representation of an algebra $g$ form what is known as the center of the universal enveloping algebra $Ug$. It is further known that the center of $Ug$ for a simple $g$ is finitely generated by $n$ generators known as the Casimir elements. Here $n$ is equal to the rank of $g$.

In our case the algebra is $so(d)$ (or $so(d-1,1)$ but the analysis does not depend on the signature), which is the same as $D_n$ for $d=2n$ or $B_n$ for $d=2n+1$.

For $B_n$ the Casimir elements are just $\mathrm{Tr}\,F^k$ for even $k=2,4,\ldots, 2n$. Here by $F^k$ I mean the usual $k$-th matrix power and not $F\wedge F \wedge \ldots \wedge F$. (For odd $k$ the invariant $\mathrm{Tr}\,F^k$ vanishes -- why?)

For $D_n$ the Casimir elements are $\mathrm{Tr}\,F^k$ for $k=2,4,\ldots, 2(n-1)$, which give $n-1$ invariants. The last $n$-th invariant is given by Pfaffian of $F$, $$\mathrm{Pf}\,F \propto \epsilon^{\mu_1\nu_1\mu_2\nu_2\ldots \mu_n\nu_n}F_{\mu_1\nu_1}\cdots F_{\mu_n\nu_n}.$$

All other invariants polynomial in $F$ can be written as polynomials in the above basic invariants.

In four dimensions we have thus $\mathrm{Tr} F^2=F_{\mu\nu}F^{\nu\mu}$ and $\epsilon^{\mu\nu\sigma\rho} F_{\mu\nu}F_{\sigma\rho}$ as expected. In two dimensions this gives just $\epsilon^{\mu\nu}F_{\mu\nu}$. In three dimensions there is only $F_{\mu\nu}F^{\nu\mu}$.