Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 22 / 4 / a

Use anti-Hermitian $SU(2)$ vector-bundle curvature and the fundamental matrix trace, sois an integer on a closed oriented four-manifold. The supplied formula's phrase “extending $W$” is a typographical error in the original PDF: the connection extends the boundary connection $\nabla$. An extension exists, for example on the trivial bundle over $B^4$, by extending a boundary connection through a collar and cutting it off farther inside.

For well-definedness, glue two choices $W_0$ and $-W_1$ along their identified boundary bundles. Connections may be made product connections on a collar while keeping their boundary values, and the transgression identity below shows that this does not alter the relevant integral. The glued bundle over the resulting closed four-manifold has an integer Second Chern number. Therefore the two extension integrals differ by an integer, proving the value of $\mathrm{CS}$ is independent of all choices in $\mathbb{R}/\mathbb{Z}$. In particular, different extensions on one fixed bundle with the same boundary value have zero difference by transgression and Stokes theorem.

To compute the derivative, extend the boundary variation $a$ to an adjoint-valued one-form $\tilde{a}$ on $W$ and set ${\tilde{\nabla }}_t=\tilde{\nabla }+t\tilde{a}$. The vector-bundle curvature derivative is $\dot{F}=d_{\tilde{\nabla }}\tilde{a}$. Invariance of the trace and the Bianchi identity giveThe second equality follows from the covariant product rule; its other term contains $d_{\tilde{\nabla }}F=0$. Integrating and applying Stokes theorem, with the given boundary orientation, yieldsThis derivative means the derivative of any local real lift of the circle-valued functional; changing that lift by an integer changes no derivative.

If a bundle gauge transformation $u$ extends to $\tilde{u}$ on $W$, use ${\tilde{u}}^{-1}\tilde{\nabla }\tilde{u}$ as the extension of the transformed boundary connection. Vector-bundle curvature is conjugated, so its trace square is unchanged pointwise. $\mathrm{CS}(u\cdot \nabla )=\mathrm{CS}(\nabla )$. This proof even preserves the chosen extension's real integral, before reducing modulo integers. The local Chern-Simons three-form transgresses the same characteristic form and is the boundary expression used in the compactness argument below.