Connections can change a representative without changing its class

By [F6], $E={\mathbb{R}}^2\times \mathbb{C}$ is the product line; give its fibres the standard complex structure and $h(z,w)=z\bar{w}$. In the global frame, write ${\mathrm{\nabla }}_{a}s=ds+as$, where $a_0=0$ and $a_1=ixdy$. Each operator is complex-linear and satisfies ${\mathrm{\nabla }}_a(fs)=dfs+f{\mathrm{\nabla }}_{a}s$, so it is a connection. For either $a$, $h({\mathrm{\nabla }}_{a,X}s,t)+h(s,{\mathrm{\nabla }}_{a,X}t)=X(s\bar{t})+(a(X)+\overline{a(X)})s\bar{t}$. Here $a_0=0$ and $\overline{a_1}=-a_1$, so this equals $Xh(s,t)$; both connections are Hermitian for the stated metric, with the compatibility convention of [F3].

The structure equation [F2] gives ${\mathrm{\Omega }}_0=0$. For $a_1=ixdy$, $d{a}_1=idx\wedge dy$ and $a_1\wedge a_1=-x^{2}dy\wedge dy=0$, so ${\mathrm{\Omega }}_1=idx\wedge dy$.

Expanding the degree-one term of the determinant in the Chern-form definition [F1] gives $c_1=-\mathrm{\Omega }\mathrm{/}(2\pi i)$ for this rank-one bundle. Consequently $c_1({\mathrm{\nabla }}_0)=0$ and $c_1({\mathrm{\nabla }}_1)=-dx\wedge dy\mathrm{/}(2\pi )$. The latter is nonzero, since its value on $({\mathrm{\partial }}_x,{\mathrm{\partial }}_y)$ is $-1\mathrm{/}(2\pi )$; hence the representative forms are not equal.

Put $\eta =-xdy\mathrm{/}(2\pi )$. Direct differentiation gives $d\eta =-dx\wedge dy\mathrm{/}(2\pi )=c_1({\mathrm{\nabla }}_1)-c_1({\mathrm{\nabla }}_0)$. This is also the degree-one transgression in [F4]: its polynomial is $P_1(B)=-B\mathrm{/}(2\pi i)$ and $A=ixdy$, so $T_{P_1}({\mathrm{\nabla }}_0,{\mathrm{\nabla }}_1)=P_1(A)=\eta$. Thus the endpoint forms are unequal while [F5] identifies their de Rham classes; [F7] ensures these closed forms represent classes. The example uses only the displayed product bundle and connections; it requires no choice axiom. [F1, F4, F5, F7, step 1.2, step 2.1, given, algebra] $\mathrm{\square }$