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The exterior algebra of a vector space $V$ is ${\Lambda }^{\ast }{V}^{\ast }={⨁}_{r=0}^{\dim V}{\Lambda }^r{V}^{\ast }$, whose degree-$r$ elements are alternating $r$-linear forms. Its exterior product satisfies $\alpha \wedge \beta =(-1{)}^{rs}\beta \wedge \alpha$ for degrees $r,s$. For example, one-forms anticommute, so $d{x}^i\wedge d{x}^i=0$. A differential form of degree $r$ on a smooth manifold is a smooth section of ${\Lambda }^r{T}^{\ast }M$; in coordinates it is an alternating sum of products of $r$ coordinate one-forms. Its exterior derivative increases degree by one, satisfies $d^2=0$, and obeysThe coordinate definition and pullback of a differential form are compatible, making differential forms suitable for coordinate-independent integration.

An orientation and a nondegenerate metric tensor define the Hodge star operator by $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩{\mathrm{vol}}_g$ for equal-degree forms. It maps degree $r$ to degree $d-r$. If the metric has $s$ negative directions, its square is ${\ast }^2=(-1{)}^{r(d-r)+s}$. Thus Hodge duality turns antisymmetric tensors into complementary-degree tensors, using the metric and orientation rather than merely the underlying vector space. The Generalized Stokes theorem says ${\int }_{M}d\eta ={\int }_{\partial M}\eta$, with the induced boundary orientation, for appropriate smooth forms and compactness or support assumptions. It generalizes the fundamental theorem of calculus, circulation version of Stokes theorem, and divergence theorem.

For the field calculation, choose signature $(-,+,+)$ and a fixed orientation on Minkowski spacetime. Write $a=\delta A$, so $\delta F=da$. Symmetry of the metric pairing and the Hodge star operator gives $\delta (F\wedge \ast F)=2da\wedge \ast F$. Applying the exterior derivative product rule,Since $F\wedge a=a\wedge F$, variation of the action isFix the boundary data or use compactly supported variations. Arbitrary interior one-forms $a$ then imply the bulk field equationWith $B=\ast F$, the Lorentzian Hodge star operator squares to $-1$ on both one- and two-forms, hence $F=-\ast B$. Thus the Maxwell–Chern–Simons one-form equation becomesThese two displayed equations are equivalent. Also $dF=0$ gives $d\ast B=0$. For $k\ne 0$ the latter follows from applying $d$ to the field equation; for $k=0$ it must be retained as the identity inherited from $F=dA$. Using the opposite metric-sign convention changes the corresponding ${\ast }^2$ sign; the signature-independent equation is $d\ast F+kF=0$.

Under the Abelian gauge transformation $A\mapsto A+d\Lambda$, the curvature $F=dA$ and the one-form $B$ are unchanged because $d^2\Lambda =0$. The field equation is therefore exactly gauge-invariant. The kinetic term is unchanged, while the other term changes byThe last equality uses $dF=0$ and the Generalized Stokes theorem. Thus the action density changes by a total derivative and the integrated action can change on a boundary. The source's statement that the action is not invariant needs this qualification: on a boundaryless spacetime, or when the boundary integral vanishes, the integrated action is invariant too. No such boundary restriction is needed for invariance of the bulk field equation.