Effect of Hodge star operator on electromagnetic field From looking at similar questions, it seems that I'm far from the only one having problems with the notation of the effects of the Hodge star operator.
My question is quite specific, to page 92 of John Baez' excellent book -- Gauge Fields, Knots and Gravity.
The top of the page has the field represented by a  $4 \times 4$ matrix, on $\mathbb{R}^4$ using the Minkowski metric.
This matrix is then operated on by the Hodge star operator producing another $4 \times 4 $ matrix, with the $E$'s and $B$'s changed into each other and moved to different positions.
I can't see how the Hodge star operator does this.  I'd like an explicit calculation for one or two of the entries of the first matrix showing how the $E$'s and $B$'s get shifted around and changed into each other in the second matrix
 A: If we write stuff in coordinates, the Hodge star can be seen to be a glorified $\epsilon$ symbol. In particular, you have $(\star F)^{\mu\nu} = \frac{\sqrt{|g|}}{2} \epsilon^{\mu\nu\rho\sigma} F_{\rho\sigma}$. Now if you take your metric to be Minkowski, you have up to minuses $F_{0i} = E_i$ and $F_{ij} = \epsilon_{ijk}B_k$. If we now inspect $(\star F)^{0i}$ we find $(\star F)^{0i} = \frac{1}{2}\epsilon^{0i\rho\sigma}F_{\rho\sigma}$. Since $\epsilon$ is antisymmetric, it is only nonzero if $\rho,\sigma$ are neither 0 nor i nor equal, but this turns out exactly to be the indices where $F$ takes the value $B_i$, so that $(\star F)^{0i}=B_i$ (up to minuses).
Similarly, if you look at $(\star F)^{ij}$ then $\rho,\sigma$ have to be the other two indices for the $\epsilon$ not to vanish. One of them in particular has to be 0, so we have an electric field strength in that entry.
Another way to think about it is that if we forget about the time axis, components $F^{0i}$ sort-of behave like (polar) vectors i.e. can be written as 1-forms, like $\vec{E}$, while the components $F^{ij}$ like $\vec{B}$ behave more like bivectors (or axial vectors), i.e. 2-forms. Now the thing about $\star$ is that it changes the degree of a form from $k$ to $3-k$ (3 since we forgot about time for the moment), so in particular it changes vectors into bivectors and vice-versa. This makes it only natural that it should switch electric into magnetic fields an vice-versa.
