Vanishing divergence implies closed Hodge dual

Fix a point $p$ and use [F4], making the chart positively oriented by reversing one spatial coordinate if necessary. The Christoffel symbols and first metric derivatives vanish at $p$. Write $ϵ$ for the metric volume form of [F5]. Its components are $\sqrt{\mathrm{\mid }\det g\mathrm{\mid }}$ times the alternating symbol. Since the determinant is a polynomial in the metric entries and the square-root function is smooth at $\mathrm{\mid }\det g(p)\mathrm{\mid }=1$, the vanishing of their first derivatives gives $\mathrm{\partial }ϵ(p)=0$. For distinct indices $a,b,c$ let $j$ be the missing fourth index. The Hodge formula gives $({\star }_{g}U{)}_{bc}={ϵ}_{bcaj}{U}^{aj}$, with no sum over $a,j$, since the two complementary terms in the half-sum agree.

At $p$, cyclically adding derivatives and using $\mathrm{\partial }ϵ(p)=0$ gives $(d{\star }_{g}U{)}_{abc}={ϵ}_{abcj}({\mathrm{\partial }}_a{U}^{aj}+{\mathrm{\partial }}_b{U}^{bj}+{\mathrm{\partial }}_c{U}^{cj})={ϵ}_{abcj}{\mathrm{\nabla }}_k{U}^{kj}$ for distinct $a,b,c$, with only $k$ summed in the last expression; the missing $k=j$ term is zero by antisymmetry. Put $V^j={\mathrm{\nabla }}_k{U}^{kj}$. Metric compatibility gives $V^j=g^{jb}{\mathrm{\nabla }}^a{U}_{ab}$, with $a,b$ summed. Thus vanishing divergence implies $V=0$, and the displayed identity makes every distinct-index component of $d{\star }_{g}U$ zero. Repeated-index components vanish by alternation. Since $p$ was arbitrary, $d{\star }_{g}U=0$ globally.

Conversely, suppose $d{\star }_{g}U=0$. For each $j$ at $p$, choose the other three indices as $a,b,c$. The identity in step 2.1 gives $0={ϵ}_{abcj}{V}^j$, with no summation here. By [F5] and the normal metric at $p$, ${ϵ}_{abcj}=\pm 1$, so $V^j=0$ for each $j$. Lowering the free index in $V^j=g^{jb}{\mathrm{\nabla }}^a{U}_{ab}$ using the nondegenerate metric yields ${\mathrm{\nabla }}^a{U}_{ab}=g_{bj}{V}^j=0$, with the contracted indices summed. The point was arbitrary, proving the converse globally and hence the asserted equivalence.