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Keep the metric and hence the Hodge star fixed. Write $\alpha =\delta A$, so $\delta F=d\alpha$, and use the Bianchi identity for an Abelian p-form $dF=d^{2}A=0$. The bilinear pairing $U\wedge \star V$ on real four-forms is symmetric even for a fixed indefinite metric. Consequently the variation of the kinetic term is $d\alpha \wedge \star F$.

The two differentiated four-form factors in the cubic term have the same sign, since moving a four-form through another four-form or through a three-form contributes an even exponent. ThusThe graded Leibniz rule givesUsing the Generalized Stokes theorem, the complete first variation is thereforeFor compactly supported variations, a closed $M$, or boundary conditions removing this last term, the Euler-Lagrange equation and the Bianchi identity for an Abelian p-form areThe coefficient three follows from the normalization actually present in the action; inserting a differently normalized supergravity coupling would change it. This is the eleven-dimensional three-form action variation.

Under the gauge transformation $A\mapsto A+d\Lambda$, the four-form $F$ is unchanged because $d^2\Lambda =0$. The kinetic density is unchanged, and the cubic density changes byIt is an exact eleven-form, so the action changes only by a boundary integral. The Euler-Lagrange equations depend only on $F$ and are invariant even when this boundary integral is nonzero. Invariance of the action itself additionally requires suitable boundary or support conditions.