Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 57 / 1

Let $M$ be an oriented smooth $m$-dimensional manifold with boundary and $\alpha$ a smooth compactly supported $(m-1)$-differential form. Give $\partial M$ the outward-normal-first orientation. A partition of unity subordinate to oriented interior and boundary charts reduces the proof to a compactly supported form on ${\mathbb{R}}^m$ or on a half-space. Indeed $\alpha ={\sum }_j{\chi }_j\alpha$, so $d\alpha ={\sum }_{j}d({\chi }_j\alpha )$; the sum of terms involving $d{\chi }_j$ is zero. Exterior derivative commutes with chart pullbacks, so each summand can be calculated in coordinates.

Write the local form asFor an interior chart, integrate each derivative using the fundamental theorem of calculus: its integral is zero because $f_j$ has compact support. For a boundary chart modeled on $x^m\ge 0$, all tangential derivatives again integrate to zero, while ${\int }_0^{\infty }{\partial }_m{f}_{m}d{x}^m=-f_m(x^{\prime },0)$. The outward normal is $-{\partial }_m$; its contraction with the volume form gives exactly this sign in the induced boundary integral of $\alpha$. Summing the chart identities proves the Generalized Stokes theorem:Compact support can be replaced by compactness of $M$; on a noncompact manifold, sufficient support or convergence conditions are necessary.

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.