In an arbitrary number of spacetime dimensions $D$, Maxwell's equations are \begin{align*} \mathrm{d}F &= 0, \\ \mathrm{d}(\star F) &= -J. \end{align*}
How many independent equations does this represent for arbitary $D$?
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1$\begingroup$ This is explained in my Phys.SE answer here. $\endgroup$– Qmechanic ♦Commented Dec 22, 2023 at 8:56
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I think the second equation is missing a $*$. The first one has n(n-1)(n-2)/6 independent equations (it's setting a 3 form equal to 0 and there's one equation for $dx_idx_jdx_k$ for each $i, j,k$ distinct and order doesn't matter). The second equation is one of 1 forms so it has $n$ independent equations. This yields a total of $(n^3-3n^2+8n)/6$ independent equations. For $n=4$, this yields 8 as expected.
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$\begingroup$ I think the second equation is missing a ∗. Wikipedia leaves out the second Hodge star on the inhomogeneous equation and considers the current density to be a 3-form. So how does that approach — with two 3-form equations — give the same counting for arbitrary dimension as one 3-form and one 1-form? $\endgroup$– GhosterCommented Dec 22, 2023 at 7:16
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$\begingroup$ it's not a 3 form - the dual of a 2 form is an n-2 form, d of this is an n-1 form ("pseudo vector"), the dimension of this space is also n. Under this convention, j is generally a pseudovector, not a 3 form. $\endgroup$ Commented Dec 22, 2023 at 7:22
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$\begingroup$ The second bullet point explicitly says “$J$ … is the current 3-form”. $\endgroup$– GhosterCommented Dec 22, 2023 at 7:23
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1$\begingroup$ Yes, it is a 3 form in 4 dimensions, but looking at the definition of J_a *dx_a (which generalizes), it's an n-1 form generally $\endgroup$ Commented Dec 22, 2023 at 7:25
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