Chirality of the Electromagnetic Field Tensor

On can define the dual of a field strength tensor by $$ F^*_{\mu\nu}= \frac 12 \epsilon_{\mu\nu\alpha\beta} F_{\alpha\beta} $$ and one can impose the condition that $F^*_{\mu\nu} =F_{\mu\nu}$ (self dual) or $F^*_{\mu\nu}=-F_{\mu\nu}$. These conditions are preserved by Lorentz tranformations and correspend to the $(1,0)$ and $(0,1)$ representations.

See the table in the Wikipedia article on the "Representation theory of the Lorentz group"

It is possible to write Maxwells' "curl" equations so that they look like a spin one version of the the Weyl equation for chiral fermions: $$ i\hbar \frac{\partial \Psi_\pm}{\partial t}= \pm c(\Sigma\cdot {\bf P}) \Psi_\pm $$ where $\Psi_{\pm}$ are three component Riemann-Silberstein vectors $\Psi_\pm\equiv {\bf E}\pm i{\bf B}c $, ${\bf P}= -i\hbar \nabla$ and the spin-one matrices are $[\Sigma_i]_{jk}==-i\epsilon_{ijk}$. I've always assumed, but have not explicitly checked, that the $\Psi_\pm$ are the two chiral components described by the self-dual and anti-self-dual conditions.

Indeed under electromagnetic duality (and with $c=1$) we have $({\bf E,B})\mapsto ({\bf B,-E})$ so ${\bf E}+i{\bf B}\mapsto ({\bf B}-i{\bf E})=-i({\bf E}+i{\bf B})$ and ${\bf E}-i{\bf B}\mapsto i({\bf E}-i{\bf B})$. Note that $(F^*)^*=-F$, so that in Minkowski signature the eigenvalues of the duality transformation are $\pm i$. This means that real-valued EM fields cannot be self-dual in Lorenzian signature. They can be in Euclidean signature where the Lorentz group is replaced by ${\rm SO}(4)$.

There's a great existing answer, I just thought I'd check where the "rotation" comes from.

As you know, the electromagnetic field tensor decomposes under $SO(3)$ into two vectors, $\mathbf{E}$ and $\mathbf{B}$, which are preserved under rotation. In fact, any linear combination of $\mathbf{E}$ and $\mathbf{B}$ are preserved under rotations. Now if we add in the boosts, the specific combinations that are preserved under both rotations and boosts are $$\mathbf{E} = \pm i \mathbf{B}.$$ These correspond to the $(1, 0)$ and $(0, 1)$ irreps; they are called self-dual and anti-self-dual fields.

Here we're working with complex-valued electromagnetic fields, i.e. we have $$\mathbf{E} = \mathbf{E}_0 e^{ik \cdot x}, \quad \mathbf{B} = \pm i \mathbf{E}_0 e^{ik\cdot x}.$$ To get representative real-valued solutions, we may take the real part. For a wave propagating along $\hat{\mathbf{z}}$, guessing $\mathbf{E}_0 \propto (1, \pm i, 0)^T$, we find the self-dual and anti-self-dual fields correspond to light waves with clockwise and counterclockwise circular polarization, a clear manifestation of chirality. You can't boost or rotate a clockwise polarized wave into anything but a clockwise polarized wave.