Stokes for complex forms on a bounded C1 Euclidean domain

In standard oriented coordinates set $dV=d{x}_1\wedge \cdots \wedge d{x}_m$. Every complex $(m-1)$-form has a unique expression $\alpha ={\sum }_{j=1}^m(-1{)}^{j-1}{A}_{j}d{x}_1\wedge \cdots \wedge \widehat{d{x}_j}\wedge \cdots \wedge d{x}_m$ with $C^1$ complex coefficients $A_j$ up to the boundary. Directly differentiating these coefficients (the same coordinate formula as [F6], valid here because they are $C^1$) gives $d\alpha =({\sum }_j{\mathrm{\partial }}_j{A}_j)dV=(\mathrm{div}A)dV$. All terms with a repeated $d{x}_i$ vanish, and the sign $(-1{)}^{j-1}$ is canceled by moving $d{x}_j$ into its ordered volume position.

At a boundary point choose a positively oriented orthonormal tangent frame $t_1,\dots ,t_{m-1}$ so that $(\nu ,t_1,\dots ,t_{m-1})$ is positive, where $\nu$ is the outward unit normal. The displayed form is ${\iota }_{A}dV$. Its boundary trace evaluated on that frame is $dV(A,t_1,\dots ,t_{m-1})=(A\cdot \nu )dS(t_1,\dots ,t_{m-1})$, because the tangential component of $A$ repeats a tangent direction in the top-degree volume form. Thus the outward-normal-first boundary trace is exactly $(A\cdot \nu )dS$; this identity is complex-linear in $A$.

By [F1], full AC supplies the weaker ${\mathrm{A}\mathrm{C}}_{\omega }$ hypotheses in [F2] and the surface convention in [F4]. Apply [F3] separately to the real and imaginary vector fields $\mathrm{Re}A$ and $\mathrm{Im}A$. Adding the two equalities and using steps 1.1 and 1.2 gives ${\int }_{D}d\alpha ={\int }_{\mathrm{\partial }D}\alpha$. This is the only use of AC.