Explicit Chern–Simons transgression between two connections

By [F3], $A$ is a global endomorphism-valued one-form; for a complex bundle [F7] ensures that the difference remains complex-linear. In every supplied $G$-frame its matrix $a={\omega }_1-{\omega }_0$ is $\mathfrak{g}$-valued. If $g_{\beta \alpha }$ is a transition matrix, then $a_{\beta }=g_{\beta \alpha }^{-1}{a}_{\alpha }{g}_{\beta \alpha }$ and ${\omega }_{t,\beta }=g_{\beta \alpha }^{-1}{\omega }_{t,\alpha }{g}_{\beta \alpha }+g_{\beta \alpha }^{-1}d{g}_{\beta \alpha }$; hence ${\omega }_t={\omega }_0+ta$ is a compatible connection for every $t\in [0,1]$. The curvature transforms by conjugation, so [F2] makes ${\beta }_t=P_k(A,{\mathrm{\Omega }}_t,\dots ,{\mathrm{\Omega }}_t)$ agree in all frames. Its coefficients are smooth in $(x,t)$, hence integrating on the compact interval defines a global smooth form of degree $1+2(k-1)=2k-1$.

In a fixed $G$-frame, [F1] gives ${\mathrm{\Omega }}_t=d{\omega }_t+{\omega }_t\wedge {\omega }_t$. Differentiating this expression in $t$ yields ${\dot{\mathrm{\Omega }}}_t=da+a\wedge {\omega }_t+{\omega }_t\wedge a=D^{{\mathrm{\nabla }}_t}A$, since for a one-form $a$ the local covariant derivative is $D^{{\mathrm{\nabla }}_t}a=da+{\omega }_t\wedge a+a\wedge {\omega }_t$.

Set ${\alpha }_t=P_k({\mathrm{\Omega }}_t,\dots ,{\mathrm{\Omega }}_t)$. Symmetry of $P_k$ and the even degree of every curvature factor give ${\dot{\alpha }}_t=k{P}_k({\dot{\mathrm{\Omega }}}_t,{\mathrm{\Omega }}_t,\dots ,{\mathrm{\Omega }}_t)$. Applying [F4] to $(A,{\mathrm{\Omega }}_t,\dots ,{\mathrm{\Omega }}_t)$ leaves $d{\beta }_t=P_k(D^{{\mathrm{\nabla }}_t}A,{\mathrm{\Omega }}_t,\dots ,{\mathrm{\Omega }}_t)$: every other term contains $D^{{\mathrm{\nabla }}_t}{\mathrm{\Omega }}_t=0$ by [F5]. Here the local $D^{{\mathrm{\nabla }}_t}$ in [F4] is the covariant exterior derivative $d^{{\mathrm{\nabla }}_t}$ in [F5]. Therefore ${\dot{\alpha }}_t=kd{\beta }_t$.

Integrating the identity from step 3.1 gives ${\alpha }_1-{\alpha }_0=k{\int }_0^{1}d{\beta }_{t}dt=d\left(k{\int }_0^1{\beta }_{t}dt\right)$. To justify the last equality, write ${\beta }_t={\sum }_I{b}_I(x,t)d{x}^I$ in a chart; each $b_I$ is smooth, and each coordinate derivative commutes with its integral over compact $[0,1]$, so the equality holds coefficient by coefficient. On boundary charts the same calculation restricts from local extensions by [F6]. For $k=0$ both endpoint forms are the same constant and their difference is zero. No axiom of choice is used. $\mathrm{\square }$