 Solution

The exterior derivative is a linear map $d:{\Omega }^k(M)\to {\Omega }^{k+1}(M)$, agrees with the ordinary differential on functions, satisfies $d^2=0$, and obeys the graded Leibniz ruleIn local coordinates it differentiates the coefficient functions:Antisymmetry cancels the second partial derivatives, proving $d^2=0$ locally. These properties determine the coordinate-independent derivative.

For any smooth $\varphi :N\to M$, the pullback of a differential form is defined byThere is no requirement that $\varphi$ be invertible or an immersion. Pullback preserves wedge products and sends $df$ to $d(f\circ \varphi )$ by the chain rule. Locally every form is a sum of $fd{x}^{i_1}\wedge \cdots \wedge d{x}^{i_k}$; applying the preceding identities and $d^2{\varphi }^i=0$ proves

Describe the brane by an embedding $X:\Sigma \to M$, with nondegenerate induced worldvolume metric $h=X^{\ast }g$. Locally the Poincaré lemma gives $F=dA$ because $F$ is a closed differential form. For a timelike worldvolume, the geometric area and Wess-Zumino brane coupling give the action$T$ is the brane tension and $q$ the charge in this normalization. The geometric term is the higher-dimensional version of the Nambu–Goto action. Both integrals are invariant under changes of worldvolume coordinates.

The additive gauge transformation is $A\mapsto A+d\Lambda$. Its change of action isThe last step is the Generalized Stokes theorem. It vanishes for a closed worldvolume, or with suitable boundary restrictions. An open worldvolume can instead have a compensating boundary coupling, as in gauge invariance of an open-brane coupling.

More directly, for a variation vector $V$ of the embedding, Cartan's magic formula givesThus the interior force depends on $F$, not the gauge potential. Define the trace of the second fundamental form byThe area variation is $-\int \sqrt{|h|}g(V,K)$ plus a boundary term. For compactly supported interior variations the resulting equations areThey exhibit the field-strength force on a charged brane and are gauge invariant.

Two global qualifications are necessary. First, replacing $A$ by $d\Lambda$ alone, as printed, gives $d{A}^{\prime }=0$ and would erase a nonzero $F$; the intended transformation must be the additive shift above. Second, closedness of $F$ only guarantees a potential locally. If a global potential does not exist and the worldvolume bounds a filling $B$, its charged term can be defined as $q{\int }_{B}F$. Different fillings differ by a period of $F$; independence of the quantum phase requires $q{\int }_{C}F\in 2\pi \hslash \mathbb{Z}$ for the corresponding closed cycles. Without the required global potential or patching/filling data, the given closed form still defines the local variational equations, but not automatically a single globally defined real-valued action.