Sign problems in complex computations

Question

$\newcommand{\dd}{\mathrm{d}} \newcommand{\eg}{\epsilon} \newcommand{\mg}{\mu} \newcommand{\ng}{\nu} \newcommand{\rg}{\rho} \newcommand{\et}{\wedge} \newcommand{\lbar}{\overline} \newcommand{\ubar}{\underline} \newcommand{\dubar}[1]{\underline{\underline{#1}}} \newcommand{\pd}{\partial} \newcommand{\pa}[1]{\left(#1\right)} \newcommand{\xfr}{\frac}$ My Physics two professor has incredible passion for differential forms. With us having some knowledge of 1-forms and minimal knowledge of 2-forms, he bombarded us with lots of form equations involving both the differential operator $\mathrm{d}$ and the Hodge dual $\ast$. I tried to fill in the computation holes he left in his notes, and I'm having lots of sign problems. First of all, we are in the context of covariant notation for Maxwell's equations. We have Maxwell's stress tensor at hand: $$F^h=\left(\begin{array}{cccc} 0 & \frac{E_1}{c} & \frac{E_2}{c} & \frac{E_3}{c} \\ -\frac{E_1}{c} & 0 & B_3 & -B_2 \\ -\frac{E_2}{c} & -B_3 & 0 & B_1 \\ -\frac{E_3}{c} & B_2 & -B_1 & 0 \end{array}\right).$$ I call it $F^h$ both to distinguish it from the 2-form $F$ I will shortly introduce and to stress the fact that the indices are high, since lowering them involves turning $F^h$ to $F_l=\eta F^h\eta$, $\eta$ being the diagonal matrix with diagonal entries $(-1,1,1,1)$, yielding $F_l$ has changed signs where the electric field's components $E_i$ show up. In indices, the above reads: \begin{align*} F^{0i}=\frac{E_i}{c}, && F^{ij}=\epsilon_{ijk}B_k, \end{align*} $\epsilon_{ijk}$ being the Levi-Civita symbol. Next, we introduce: $$F=\frac12F_{\mu\nu}\mathrm{d}x^\mu\wedge\mathrm{d}x^\nu.$$ Yes, low indices. From there, I have checked that: $$F=-\mathrm{d}t\wedge E+\ast_3B,$$ $\ast_3$ denoting the 3-dimensional Hodge dual, to distinguish it from the 4-dimensional $\ast$. I have checked that $\mathrm{d}F=0$ is equivalent to two of Maxwell's equations, the one on the magnetic divergence and Faraday's law of induction, or to the law $\partial_{[\mu}F_{\nu\rho]}=0$ where that notation with square brackets in the subscripts is meant to compactify the sum of all cyclical permutations of the three indices, i.e. $\partial_{[\mu}F_{\nu\rho]}=\partial_\mu F_{\nu\rho}+\partial_\nu F_{\rho\mu}+\partial_\rho F_{\mu\nu}$. What I have problems with is proving that $\ast\mathrm{d}\ast F$ enters an equation which is equivalent to the other two Maxwell equations. First of all, 4-dimensional $\ast$ requires 4-index $\epsilon$s, which I only ever saw in this course. ALl I know about them is this phrase from the professor's notes: «We can define $\epsilon_{0123}=1$. Careful though: $\epsilon^0{}_{123}=-1$!». This alone sounds terribly confusion-inducing. From this phrase, I assumed that, having an $\epsilon$ with four indices, of which $k$ up and $4-k$ down, I first do as many index swappings (keeping the order of index heights the same, i.e. if I start with high low low low I always keep high low low low), changing sign at every swap, as necessary to get the sequence 0123, and then that $\epsilon$ is $(-1)^k$. For example: $$\epsilon^{03}{}_{21}=-\epsilon^{02}{}_{31}=\epsilon^{02}{}_{13}=-\epsilon^{01}{}_{23}=-1,$$ since with two indices up $\epsilon^{01}{}_{23}=1$. Under this assumption I performed the following calculations. First of all, $\ast F$. Here is what I did: \begin{align*} \ast F={}&\ast(F_{01}\dd x^0\et\dd x^1+F_{02}\dd x^0\et\dd x^2+F_{03}\dd x^0\et\dd x^3+{} \\ &{}+F_{23}\dd x^2\et\dd x^3+F_{31}\dd x^3\et\dd x^1+F_{12}\dd x^1\et\dd x^2)={} \\ {}={}&\eg_{01}{}^{23}F_{01}\dd x^2\et\dd x^3+\eg_{02}{}^{13}F_{02}\dd x^1\et\dd x^3+\eg_{03}{}^{12}F_{03}\dd x^1\et\dd x^2+{} \\ &{}+\eg_{23}{}^{01}F_{23}\dd x^0\et\dd x^1+\eg_{31}{}^{02}F_{31}\dd x^0\et\dd x^2+\eg_{12}{}^{03}F_{12}\dd x^0\et\dd x^3. \end{align*} Naturally, the indices of those $\epsilon$s all have the wrong height, but being four height changes in my assumptions no harm is done. Substituting the $\epsilon$s with their values, I get: \begin{align*} \ast F={}&F_{01}\dd x^2\et\dd x^3-F_{02}\dd x^1\et\dd x^3+F_{03}\dd x^1\et\dd x^2+{} \\ &{}+F_{23}\dd x^0\et\dd x^1-F_{13}\dd x^0\et\dd x^2+F_{12}\dd x^0\et\dd x^3. \end{align*} And here starteth the trouble, for my professor's notes have exactly the same, with the $\dd$s swapped in the second term, the indices swepped in the fifth, but most importantly, all indices lowered. That is a problem, because where $F$ has a zero index and a nonzero one, the sign changes, whereas when both are nonzero, it doesn't. Substituting the fields, I get: $$\ast F=-\frac1c\ast_3E+c\dd t\et B,$$ where I haven't said it yet but $E=E_i\dd x^i$ and $B=B_i\dd x^i$, summed over repeated indices. Coherently, the professor's notes have a minus on the first term. Let us now see what happens when I $\dd$ that: \begin{align*} \dd\ast F={}& \begin{aligned}[t] (\dd x^{\mg`}\et\pd_{\mg`})(&F_{01}\dd x^2\et\dd x^3+F_{02}\dd x^3\et\dd x^1+F_{03}\dd x^1\et\dd x^2+{} \\ &{}+F_{23}\dd x^0\et\dd x^1+F_{31}\dd x^0\et\dd x^2+F_{12}\dd x^0\et\dd x^3)={} \end{aligned} \\ {}={}&\ubar{\pd_0F_{01}\dd x^0\et\dd x^2\et\dd x^3}+\lbar{\pd_0F_{02}\dd x^0\et\dd x^3\et\dd x^1}+\dubar{\pd_0F_{03}\dd x^0\et\dd x^1\et\dd x^2}+{} \\ &{}+\pd_1F_{01}\dd x^1\et\dd x^2\et\dd x^3+\dubar{\pd_1F_{31}\dd x^1\et\dd x^0\et\dd x^2}+\lbar{\pd_1F_{12}\dd x^1\et\dd x^0\et\dd x^3}+{} \\ &{}+\pd_2F_{02}\dd x^2\et\dd x^3\et\dd x^1+\dubar{\pd_2F_{23}\dd x^2\et\dd x^0\et\dd x^1}+\ubar{\pd_2F_{12}\dd x^2\et\dd x^0\et\dd x^3}+{} \\ &{}+\pd_3F_{03}\dd x^3\et\dd x^1\et\dd x^2+\lbar{\pd_3F_{23}\dd x^3\et\dd x^0\et\dd x^1}+\ubar{\pd_3F_{31}\dd x^3\et\dd x^0\et\dd x^2}. \end{align*} Gathering the similarly "decorated" terms, I get: $$\dd\ast F=\pd_{\mg`}F_{\mg`3}\dd x^0\et\dd x^1\et\dd x^2-\pd_{\mg`}F_{\mg`2}\dd x^0\et\dd x^1\et\dd x^3+\pd_{\mg`}F_{\mg`1}\dd x^0\et\dd x^2\et\dd x^3-\pd_{\mg`}F_{\mg`0}\dd x^1\et\dd x^2\et\dd x^3.$$ Observing that, in the star, the $\epsilon$s will have three indices up and one down, and so $\epsilon^{012}{}_3=-1$, it is now easy to see that this stars to: $$\ast\dd\ast F=-\pd_{\mg`}F_{\mg`\ng`}\dd x^{\ng`}.$$ Again, we have a sign problem, since first of all the notes have $\ast\dd\ast F=\pd_{\mg}F^{\nu}_{\mu}\dd x^\nu$, where it is impossible to know which comes first, and one is low and one is high. If I swap indices I get $\ast\dd\ast F=\pd_\mu F_{\nu\mu}\dd x^\nu$, but the equation I have next states that if $J=c\rho\dd t+J_i\dd x^i$, then $\pd_\mu F^{\nu\mu}=J^\nu$, with indices up though, and raising the indices causes sign problems again. Going on with the fields, I get: \begin{align*} \dd\ast F={}&-\dd t\et\ast_3\dot{E}-\dd_3\ast_3E+(-\pd_1B_2+\pd_2B_1)\dd x^0\et\dd x^1\et\dd x^2+{} \\ &{}+(-\pd_1B_3+\pd_3B_1)\dd x^0\et\dd x^1\et\dd x^3+(-\pd_2B_3+\pd_3B_2)\dd x^0\et\dd x^2\et\dd x^3. \end{align*} Of course, the sign of the $E$ part is opposite to the one of the notes. The $B$ part -- it is easy to check -- is $-c\dd t\et\dd_3B$, as the notes say. Going on: $$\ast\dd\ast F=-\ast_3\dd_3\ast_3 E\dd t+\xfr{1}{c^2}\dot{E}+\pd_j\pa{\eg_{ijk}B_k}\dd x^i.$$ For some miracle, the sign of the $E$ part is now the same as that of the notes. And for another, far sadder miracle, the $B$ part now has the opposite sign, as it is $\ast_3\dd_3B$, and I would like it to be the opposite of that. So what am I doing wrong in here?

Answer 1

One source of sign errors may be that you are lowering/raising the indices on $\epsilon_{\mu\nu\rho\sigma}$ incorrectly. The mantra is to change sign each time the $0$ index is raised or lowered, not each time any index is raised or lowered. Have a look at $$ \epsilon^{01}{}_{23} = \eta_{0\mu} \eta_{1\nu}\epsilon_{\mu\nu 23}= (-1)(1)\epsilon_{01 23}=-1 $$ but you have it as $+1$.

Answer 2

$\newcommand{\hg}{\eta} \newcommand{\w}[1]{\bigwedge_{#1}}$ Origin of the answer $\newcommand{\dd}{\mathrm{d}} \newcommand{\eg}{\epsilon} \newcommand{\et}{\wedge} \newcommand{\mg}{\mu} \newcommand{\ng}{\nu} \newcommand{\pd}{\partial} \newcommand{\rg}{\rho} \newcommand{\lbar}{\overline} \newcommand{\ubar}{\underline}$

As Mark pointed out in his answer, my only problem was that I misunderstood the rules governing the sign changes with index raising and lowering on $\eg$. I believed any index, when raised or lowered, changed the $\eg$'s sign. The correct assertion is that only 0 does that, whereas the other indices don't. That is because raising an index implies an $\hg$. In other words, $\eg^i{}_{jkl}=\hg^{im}\eg_{mjkl}$, summed over $m$, and similarly, $\eg_{ijkl}=\hg_{im}\eg^m{}_{jkl}$, summed, again, over $m$. Then I tried doing the computations again, and got stuck for the $n$th time. I browsed the web, and found this pdf. It is my aim in this answer to prove that what it says is, up to missing $c$ factors which are probably absorbed into different measure units, and also up to an extra $\ast_s$ in the $B$ part, all correct. I will restrict myself to the computations concerning $\ast\dd\ast F$, as the rest is already clear to me.

Starting assumptions and definitions

I will start from the following assumption, which I know to be true: $$F=\frac12F_{\mg\ng}\dd x^\mg\et\dd x^\ng=-\dd t\et E+\ast_3B,$$ where $F_{\mg\ng}$ are the entries of the Maxwell stress tensor with the signs changed coherently with the low indices and $\ast_3$ is the three-dimensional Hodge dual, contrasted with the 4-dimensional $\ast$. Taking a quick look at the pages just before my computation, I noticed he is treating the magnetic field as a 2-form, which explains the extra $\ast_s$. I will start from the following two definitions: \begin{align*} E=E_i\dd x^i=E_1\dd x^1+E_2\dd x^2+E_3\dd x^3, && B=B_i\dd x^i=B_1\dd x^+B_2\dd x^2+B_3\dd x^3, \end{align*} meaning my $B$ is the pdf's $\ast_sB$ (or, in my notation, $\ast_3B$) and viceversa. So all is well.

A couple remarks

Let us first of all explicit $F$: \begin{align*} F=\frac12F_{\mg\ng}\dd x^\mg\et\dd x^\ng={}&F_{01}\dd x^0\et\dd x^1+F_{02}\dd x^0\et\dd x^2+F_{03}\dd x^0\et\dd x^3+{} \\ &{}+F_{12}\dd x^1\et\dd x^2+F_{13}\dd x^1\et\dd x^3+F_{23}\dd x^2\et\dd x^3. \end{align*} If you are wondering where the $\frac12$ has ended up, well, let me make an example: I have e.g. $\frac12F_{01}\dd x^0\et\dd x^1+\frac12F_{10}\dd x^1\et\dd x^0$. Swapping the "factors" of a wedge product produces a minus, as does swapping $F$'s indices since $F$ is an antisymmetric tensor (should I indicate it differently, e.e. as $\lbar F^h$ -- h for high indices -- perhaps?), so those two terms are identical and summing them removes the $\frac12$.

By the way, my assumption is also in the pdf, since $-\dd t\et E=E\et\dd t$ by virtue of wedge anticommutation, since $E$ is "hiding" only one "leg" per term so bringing $\dd t$ to the right is producing only one swapping of wedge product "factors", and my $\ast_3B$ is the pdf's $B$. Another point in favour of the pdf.

The first star: developing an abbreviation for long wedge products

Anyway, time to star this. Let us first do everything in terms of the tensor. So $\ast F$ is: \begin{align*} \ast F={}&\ast(F_{01}\dd x^0\et\dd x^1+F_{02}\dd x^0\et\dd x^2+F_{03}\dd x^0\et\dd x^3+{} \\ &{}+F_{12}\dd x^1\et\dd x^2+F_{13}\dd x^1\et\dd x^3+F_{23}\dd x^2\et\dd x^3)={} \\ {}={}&F_{01}\eg^{01}{}_{23}\dd x^2\et\dd x^3+F_{02}\eg^{02}{}_{13}\dd x^1\et\dd x^3+F_{03}\eg^{03}{}_{12}\dd x^1\et\dd x^2+{} \\ &{}+F_{12}\eg^{12}{}_{03}\dd x^0\et\dd x^3+F_{13}\eg^{13}{}_{02}\dd x^0\et\dd x^2+F_{23}\eg^{23}{}_{01}\dd x^0\et\dd x^1. \end{align*} I really have to develop abbreviations for wedges. I hereby decide I will introduce the following notation: $\bigwedge_{i}=\dd x^i,\bigwedge_{ij}=\dd x^i\et\dd x^j,\bigwedge_{ijk}=\dd x^i\et\dd x^j\et\dd x^k,\bigwedge_{ijkl}=\dd x^i\et\dd x^j\et\dd x^k\et\dd x^l$, and so on, but I should need no more than 4 "legs". In this notation, the above reads: $$\ast F=F_{01}\eg^{01}{}_{23}\bigwedge_{23}+F_{02}\eg^{02}{}_{13}\w{13}+F_{03}\eg^{03}{}_{12}\w{12}+F_{12}\eg^{12}{}_{03}\w{03}+F_{13}\eg^{13}{}_{02}\w{02}+F_{23}\eg^{23}{}_{01}\w{01}.$$ Let us get rid of those $\eg$s. Now the 0 is up, so 0123 has a negative sign. Thus for example the first term has a minus. The second one has a plus, since it is 0213, which via swapping of 2 with 1 (onw swap, one minus) produces 0123. But we produce a minus by swapping the wedge factors, i.e. by $\w{13}=-\w{31}$. That is because we are going to find an $\ast_3E$, so we want coherent signs first of all, and secondly that $\ast_3$ produces $\w{31}$ from $\w2$, and not $\w{13}$. The third term is 1203, which by swapping 1 and 0 gives a minus and 0213, and by swapping 2 and 1 gives another minus and 0123, so all in all it is a minus. From now on, I will simply say $1203=-0213=0123=-1$, meaning the epsilons above with those index sequences and the 0 up, and the rest of the indices up or down as you wish. But back to work. The fourth term has $1203=-1023=0123=-1$, so another minus. By now, we expect all minuses. The fifth terms has $1302=-1032=0132=-0123=+1$, a plus? What a surprise. Anomalous, isn't it? We will see what happens when we get to the fields, if something doesn't figure, back to the calcs here and check this sign. Sixth and last term: $2301=-1230=0123=-1$, another minus. And yes: with four indices, cyclical permutations equate to three swaps, so three sign change, which means a minus, unlike with 3 indices where they equate 2 swaps, so no minus. But WRONG: the second triplet of terms has no up zero, so the signs are all the opposites of what I said above. With all this, we have: $$\ast F=-F_{01}\w{23}-F_{02}\w{31}-F_{03}\w{12}+F_{12}\w{03}-F_{13}\w{02}+F_{23}\w{01}.$$ Raising the indicese eats up the minuses in the first three terms, since there is a 0 being raised, and leaves the rest untouched. Thus: $$\ast F=F^{01}\w{23}+F^{02}\w{31}+F^{03}\w{12}+F^{12}\w{03}+F^{31}\w{02}+F^{23}\w{01}.$$ I took advantage of $F^{13}=-F^{31}$ to get the same sign for all terms with no zeros on $F$, and also because $F^{31}=B_2$ whereas $F^{13}$ is minus that.

Back to the fields for just a second

In terms of the fields, the above reads: $$\ast F=\frac1c\ast_3E+\dd x^0\et B=\frac1c\ast_3E-B\et\dd x^0.$$ The pdf confirms, and we are happy.

The $\dd$

Let us now take $\dd$ of $\ast F$. I will denote $\dd_3=\dd x^i\et\pd_i$ and $\dd=\dd x^\mg\et\pd_\mg$. With that: \begin{align*} \dd\ast F={}&\dd(F^{01}\w{23}+F^{02}\w{31}+F^{03}\w{12}+F^{12}\w{03}+F^{31}\w{02}+F^{23}\w{01})={} \\ {}={}&\widetilde{\pd_0F^{01}\w{023}}+\pd_1F^{01}\w{123}+\ubar{\pd_0F^{02}\w{031}}+\pd_2F^{02}\w{231}+\lbar{\pd_0F^{03}\w{012}}+\pd_3F^{03}\w{312}+{} \\[1em] &{}+\ubar{\pd_1F^{12}\w{103}}+\widetilde{\pd_2F^{12}\w{203}}+\widetilde{\pd_3F^{31}\w{302}}+\lbar{\pd_1F^{31}\w{102}}+\lbar{\pd_2F^{23}\w{201}}+\ubar{\pd_3F^{23}\w{301}}. \end{align*} I have given similar accents to terms with similar wedges, so you can more easily convince yourselves of what follows: \begin{align*} \dd\ast F={}&(\pd_0F^{03}+\pd_1F^{13}+\pd_2F^{23})\w{012}+(-\pd_0F^{02}-\pd_1F^{12}-\pd_3F^{32})\w{013}+{} \\ &{}+(\pd_0F^{01}+\pd_2F^{21}+\pd_3F^{31})\w{023}+(\pd_1F^{01}+\pd_2F^{02}+\pd_3F^{03})\w{123}. \end{align*} You can easily check that all signs match the pdf.

The second star

Let us star this at once. $\w{012}$ stars to $\eg^{012}{}_3\dd x^3=-\dd x^3$, $\w{013}$ stars to $\eg^{013}{}_2\dd x^2=+\dd x^2$. $\w{023}$ stars to $\eg^{023}{}_1\dd x^1=-\eg^0{}_1{}^{32}\dd x^2=\eg^0{}_1{}^{23}\dd x^1=-\dd x^1$. $\w{123}$ stars to $\eg^{123}{}_0\dd x^0=-\eg_0{}^{123}\dd x^0=-\dd x^0$. WIth that we get: \begin{align*} \ast\dd\ast F={}&-(\pd_0F^{03}+\pd_1F^{13}+\pd_2F^{23})\dd x^3-(\pd_0F^{02}+\pd_1F^{12}+\pd_3F^{32})\dd x^2-{}& \\ &{}+(\pd_0F^{01}+\pd_2F^{21}+\pd_3F^{31})\dd x^1-(\pd_1F^{01}+\pd_2F^{02}+\pd_3F^{03})\dd x^0={} \\ {}={}&(\pd_0F^{30}+\pd_1F^{31}+\pd_2F^{32})\dd x^3+(\pd_0F^{20}+\pd_1F^{21}+\pd_3F^{23})\dd x^2+{} \\ &{}+(\pd_0F^{10}+\pd_2F^{12}+\pd_3F^{13})\dd x^1-(\pd_1F^{01}+\pd_2F^{02}+\pd_3F^{03})\dd x^0={} \\ {}={}&\pd_\mg F_\ng{}^\mg\dd x^\ng. \end{align*} You can, again, check that the signs all agree with the pdf.

Turning to the currents

We are very happy, for this form can be easily written if we define a 1-form from the charge and current densities $\rg,\vec J$. Namely, we will have: $$J=J_\mg\dd x^\mg,$$ where $J_0=-c\rg=-J^0$. We have proven equations saying that $\pd_\mg F^{\ng\mg}=\mg_0J^\ng$. Can we apply them here to simply the expression of $\ast\dd\ast F$? Yes. Let us first rewrite things: $$\ast\dd\ast F=\pd_\mg F_\ng{}^\mg\dd x^\ng=-\pd_\mg F^{0\mg}\dd x^0+\pd_\mg F^{i\mg}\dd x^i=-\mg_0J^0\dd x^0+\mg_0J^i\dd x^i=\mg_0J.$$ The signs agree with the pdf, the $\mg_0$ is another vanishing factor.

Field $\dd$

Let us now write $\dd\ast F$ and its star in terms of the fields. Let us first recall the form of those: \begin{align*} \dd\ast F={}&(\pd_0F^{03}+\pd_1F^{13}+\pd_2F^{23})\w{012}+(-\pd_0F^{02}-\pd_1F^{12}-\pd_3F^{32})\w{013}+{} \\ &{}+(\pd_0F^{01}+\pd_2F^{21}+\pd_3F^{31})\w{023}+(\pd_1F^{01}+\pd_2F^{02}+\pd_3F^{03})\w{123}, \\ \ast\dd\ast F={}&(\pd_0F^{30}+\pd_1F^{31}+\pd_2F^{32})\dd x^3+(\pd_0F^{20}+\pd_1F^{21}+\pd_3F^{23})\dd x^2+{} \\ &{}+(\pd_0F^{10}+\pd_2F^{12}+\pd_3F^{13})\dd x^1-(\pd_1F^{01}+\pd_2F^{02}+\pd_3F^{03})\dd x^0. \end{align*} Next, we recall the relationships between $F^{\mg\ng}$ and the fields, and that of $\pd_0$ to $\pd_t$ and $\dd x^0$ to $\dd t$: \begin{align*} F^{0i}=\frac1cE_i, && F^{ij}=\eg_{ijk}B_k, \\ \dd x^0=c\dd t, && \pd_0=\frac1c\pd_t. \end{align*} With this in mind, we tackle $\dd\ast F$: \begin{align*} \dd\ast F={}&(\pd_0F^{03}+\pd_1F^{13}+\pd_2F^{23})\w{012}+(-\pd_0F^{02}-\pd_1F^{12}-\pd_3F^{32})\w{013}+{} \\ &{}+(\pd_0F^{01}+\pd_2F^{21}+\pd_3F^{31})\w{023}+(\pd_1F^{01}+\pd_2F^{02}+\pd_3F^{03})\w{123}={} \\ {}={}&(\frac1c\pd_0E_3-\pd_1B_2+\pd_2B_1)\w{012}+(-\frac1c\pd_0E_2-\pd_1B_3+\pd_3B_1)\w{013}+{} \\ &{}+(\frac1c\pd_0E_1-\pd_2B_3+\pd_3B_2)\w{023}+(\pd_1E_1+\pd_2E_2+\pd_3E_3)\w{123}. \end{align*} All signs agree with the pdf. What are those messes in terms of the 1-forms $E,B$ we definied at the beginning of the answer? The last term I know to be $\dd_3\ast_3E$. For example, taking $E_1\dd x^1$ and starring it yields $E_1\dd x^2\et\dd x^3$, which under $\dd_3$ gives $\pd_1E_1\dd x^1\et\dd x^2\et\dd x^3$, which we have above. Let us now consider the electric terms. $\pd_0=\frac1c\pd_t$, so we get a $\frac{1}{c^2}$. Apart from the coefficient, the first electric term is $\pd_tE_3\dd x^0\et\dd x^1\et\dd x^2=\dd x^0\et\ast_3(\pd_tE_3\dd x^3)$. We can thus infer these terms, collected, will give $\frac{1}{c^2}\ast_3\dot E$, where by $\dot E$ I have denoted the time derivative, as is customary. In other words, $\pd_tE\equiv\dot E$. The second electric term is $-\pd_0E_2\dd x^0\et\dd x^1\et\dd x^3$. If we swap the wedge factors 1 and 3 we eat up a minus, and obtain $\dd x^0\et\ast_3(\pd_tE_2\dd x^2)$, plus the factor. The last term is $\pd_0E_1\dd x^0\et\dd x^2\et\dd x^3=\frac1c\dd x^0\et\ast_3(\pd_tE_1\dd x^1)$. So the terms collected together are $\dd x^0\et\frac{1}{c^2}\ast_3\dot E$. Of course, we can restore $\pd_0$ by eating up a $\frac1c$, but then you may say, how did the $\dd x^0$ end up on the left without changing sign? Well, $\ast_3\dot E$ gives a 2-form, i.e. a form with two wedge factors per term, so bringing the $\dd x^0$ to the right equates to two swaps of wedge factors, two sign changes, and thus no sign change. So another way to write the electric terms, modulo the missing $\frac1c$, is precisely $\pd_0\ast_3 E\et\dd x^0$, as the pdf goes. Great. Let us gather what we have discovered: \begin{align*} \dd\ast F={}&\dd_3\ast_3E+\frac{1}{c^2}\ast_3\dot E\et\dd x^0+{} \\ &{}+(-\pd_1B_2+\pd_2B_1)\w{012}+(-\pd_1B_3+\pd_3B_1)\w{013}+(-\pd_2B_3+\pd_3B_2)\w{023}. \end{align*} Knowing that we must get $\dd_3B$, we evaluate that to see how it relates to the remaining magnetic terms: \begin{align*} \dd_3B={}&\dd_3(B_1\dd x^1+B_2\dd x^2+B_3\dd x^3)={} \\ {}={}&\pd_2B_1\w{21}+\pd_3B_1\w{31}+\pd_1B_2\w{12}+\pd_3B_2\w{32}+\pd_1B_2\w{13}+\pd_2B_3\w{23}={} \\ {}={}&(\pd_1B_2-\pd_2B_1)\w{12}+(\pd_1B_3-\pd_3B_1)\w{13}+(\pd_2B_3-\pd_3B_2)\w{23}. \end{align*} We can now very easily recognize that: $$\dd\ast F=\dd_3\ast_3E+\frac{1}{c^2}\ast_3\dot E\et\dd x^0-\dd x^0\et\dd_3B.$$ Like before, bringing $\dd x^0$ to the right of $\dd_3B$ changes nothing. Where is that $\ast_3$, you will say. Well, remember the pdf has $B$ defined as we have defined to be $\ast_3B$, so everything is fine.

The second star and the fields

Now for the last step: the second star. Let us calculate the stars of the three terms separately. \begin{align*} \ast\dd_3\ast_3E={}&\ast(\vec\nabla\cdot\vec E)\dd x^1\et\dd x^2\et\dd x^3=(\vec\nabla\cdot\vec E)\eg^{123}{}_0\dd x^0={} \\ {}={}&(\vec\nabla\cdot\vec E)\eg_{1230}\dd x^0=-(\vec\nabla\cdot\vec E)\dd x^0=-(\ast_3\dd_3\ast_3E)\dd x^0. \end{align*} The sign agrees. I don't like putting a wedge between $\dd x^0$ and a term with no $\dd x$s. It just doesn't make sense to me. \begin{align*} \ast\left(\frac{1}{c^2}\ast_3\dot E\et\dd x^0\right)={}&\frac{1}{c^2}\ast(\dot E_1\w{230}+E_2\w{310}+E_3\w{120})={} \\ {}={}&\frac{1}{c^2}\left(\eg^{230}{}_1\dot E_1\dd x^1+\eg^{310}{}_2\dot E_2\dd x^2+\eg^{120}{}_3\dot E_3\dd x^3\right)=-\frac{1}{c^2}\dot E, \end{align*} with the right sign, the usual missing $\frac1c$ and an additional missing $\frac1c$ which is concealed in the $\pd_0=\frac1c\pd_t$ used on the pdf. Last term: \begin{align*} \ast(-\dd_3B\et\dd x^0)={}&-\ast\left[(\pd_1B_2-\pd_2B_1)\w{120}+(\pd_1B_3-\pd_3B_1)\w{130}+(\pd_2B_3-\pd_3B_2)\w{230}\right]={} \\ {}={}&-\eg^{120}{}_3(\pd_1B_2-\pd_2B_1)\dd x^3-\eg^{130}{}_2(\pd_1B_3-\pd_3B_1)\dd x^2-{} \\ &{}+\eg^{230}{}_1(\pd_2B_3-\pd_3B_2)\dd x^1={} \\ {}={}&\ast_3\dd_3B. \end{align*} So all signs are fine.

Inhomogeneous covariant Maxwell and the 2-form $F$

Now we will derive Gauss's law and Ampère's law from here. Let us first summarize what we have: $$J_\mg\dd x^\mg=\ast\dd\ast F=-\ast_3\dd_3\ast_3E\dd x^0-\frac{1}{c^2}\ast_3\dot E+\ast_3\dd_3B.$$ Now the first term has $\dd x^0$ and the others don't, so we will deduce two equations: $$\left\{\begin{array}{@{}l@{}} -\ast_3\dd_3\ast_3E\dd x^0=\mg_0J_0\dd x^0=-\mg_0c\rg\dd x^0, \\ -\frac{1}{c^2}\ast_3\dot E+\ast_3\dd_3B=J_i\dd x^i. \end{array}\right.$$ One can easily check, with a lot more boring calcs which I will not do here, that pairing all the $\dd x$s separately gives four equations, of which the 0th is Gauss's law (which we can already see from here -- notice how I lost a $\frac1c$ somewhere very high up :, which should fix the fact there is an extra $\frac1c$ on the RHS of that 0th equation above), and the others are the 3 components of Ampère's law with the displacement current. So my calcs are finally done.

Final bonus

As an addendum, since I've done so many computations, I will do a few more short computations to show something I have found out all by myself. First, let me show you what has led me to suspect what I then proved. FIrstly, we know that $\dd F=0$ is the same as Faraday's law plus the vanishing of the magnetic divergence. So $F$ is a closed form. Secondly, in terms of the potential four-vector $A^mg$ whose entries are $A^0=\frac{\varphi}{c}$ and the entries of the vector potential, we have proved $F_{\mg\ng}=\pd_\mg A_\ng-\pd_\ng A_\mg$. THis reminds us of a $\dd$. Could $F$ possibly be exact? Oh yes indeed. Let me prove this. Firstly, we define the 1-form $A=A_\mg\dd x^\mg$, which we call potential 1-form. We then set out to show $F=\dd A$. So let us calculate $\dd A$. But first, let us remember what the potentials are: \begin{align*} \vec B=\vec\nabla\times\vec A, && \vec E=-\pd_t\vec A-\vec\nabla\varphi, \\ \text{or in components:} \\ B_i=\eg_{ijk}\pd_jA_k, && E_i=-\pd_tA_i-\pd_i\varphi. \end{align*} With that, we get: $$\dd A=-\frac1c\dd_3\varphi\et\dd x^0-\dot A\et\dd x^0+\dd_3A_3,$$ $A_3$ being the spacial potantial 1-form $A_3=A_i\dd x^i$. Let us expand that: \begin{align*} \dd A={}&-\pd_1\varphi\w{10}-\pd_2\varphi\w{20}-\pd_3\varphi\w{30}-\pd_tA_1\w{10}-\pd_tA_2\w{20}-\pd_tA_3\w{30}+\dd_3A_3={} \\ {}={}&(-\pd_1\varphi-\pd_tA_1)\w{10}+(-\pd_2\varphi-\pd_tA_2)\w{20}+(-\pd_3\varphi-\pd_tA_3)\w{30}+\dd_3A_3={} \\ {}={}&E_1\w{10}+E_2\w{20}+E_3\w{30}+\dd_3A_3=E\et\dd x^0+\dd_3A_3, \end{align*} except I had canceled a $\frac1c$ when planning to pass from $\dd x^0$ to $\dd t$ and then forgot to do the passage, so this $\dd x^0$ is actually a $\dd t$, as appears in the assumption at the beginning. So all we need to show is $\dd_3A_3=\ast_3B$. Let us work on this then: \begin{align*} \dd_3A_3={}&\dd_3(A_1\dd x^1+A_2\dd x^2+A_3\dd x^3)={} \\ {}={}&\pd_2A_1\w{21}+\pd_3A_1\w{31}+\pd_1A_2\w{12}+\pd_3A_2\w{32}+\pd_1A_3\w{13}+\pd_2A_3\w{23}={} \\ {}={}&(\pd_1A_2-\pd_2A_1)\w{12}+(\pd_3A_1-\pd_1A_3)\w{31}+(\pd_2A_3-\pd_3A_2)\w{23}={} \\ {}={}&\ast_3[(\pd_1A_2-\pd_2A_1)\dd x^3+(\pd_3A_1-\pd_1A_3)\dd x^2+(\pd_2A_3-\pd_3A_2)\dd x^1]={} \\ {}={}&\ast_3[\eg_{3jk}\pd_jA_k\dd x^3+\eg_{2jk}\pd_jA_k\dd x^2+\eg_{1jk}\pd_jA_k\dd x^1]=\ast_3(B_3\dd x^3+B_2\dd x^2+B_1\dd x^1)={} \\ {}={}&\ast_3(B_i\dd x^i)=\ast_3B, \end{align*} as we wished it to be. So we have shown that: $$F=\dd A.$$ Naturally, the above terms with the $\eg$s are summed over $j,k$, each of them. It should anyway be natural to understand the sum, since those indices up there are repeated :).