= Solution
Use anti-Hermitian $SU(2)$ <vector-bundle curvature> and the fundamental <matrix trace>, so
$$
k(E)=c_2(E)[X]=\frac1{8\pi^2}\int_X\operatorname{Tr}(F\wedge F)
$$
is 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 \b[the value of $\operatorname{CS}$ 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 $\widetilde a$ on $W$ and set $\widetilde\nabla_t=\widetilde\nabla+t\widetilde a$. The <vector-bundle curvature> derivative is $\dot F=d_{\widetilde\nabla}\widetilde a$. Invariance of the trace and the <Bianchi identity> give
$$
\begin{aligned}
\left.\frac d{dt}\right|_0\operatorname{Tr}(F_t\wedge F_t)
&=2\operatorname{Tr}(d_{\widetilde\nabla}\widetilde a\wedge F)\\
&=2d\operatorname{Tr}(\widetilde a\wedge F).
\end{aligned}
$$
The second equality follows from the covariant product rule; its other term contains $d_{\widetilde\nabla}F=0$. Integrating and applying <Stokes theorem>, with the given boundary orientation, yields
$$
\boxed{\left.\frac d{dt}\right|_0\operatorname{CS}(\nabla+ta)
=\frac1{4\pi^2}\int_{S^3}\operatorname{Tr}(F_\nabla\wedge a).}
$$
This 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 $\widetilde u$ on $W$, use $\widetilde u^{-1}\widetilde\nabla\widetilde u$ as the extension of the transformed boundary connection. <Vector-bundle curvature> is conjugated, so its trace square is unchanged pointwise. \b[$\operatorname{CS}(u\cdot\nabla)=\operatorname{CS}(\nabla)$.] 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.
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