Let $E$ be a vector bundle on a smooth manifold $X$ and $\nabla$ be a connection on $E$, by Chern-Weil theory, the Chern character of $(E,\nabla)$ could be construct as $$ ch(E,\nabla):=tr(\exp(-\nabla^2)). $$
It is well-known that the cohomology class of $ch(E,\nabla)$ is independent of the connection $\nabla$ and for two different connections $\nabla_1$ and $\nabla_2$, we can explicitly construct a form $tch(E,\nabla_1,\nabla_2)$, called the transgression form, such that $$ ch(E,\nabla_1)-ch(E,\nabla_2)=d (tch(E,\nabla_1,\nabla_2)). $$ Notice that the construction of $tch$ involves an integration $\int_0^1$. Of course a similar construction works for chern classes too.
Now we consider the same problem in another way: we could treat $E\oplus E$ as a super bundle and denote $$ \nabla=\begin{pmatrix}\nabla_1&0\\ 0& \nabla_2 \end{pmatrix} $$ and $$ D=\begin{pmatrix}0&id_E\\ id_E& 0. \end{pmatrix} $$ Then $\mathbb{A}_t:=\nabla+\sqrt{t}D$ is a super connection on $E\oplus E$ in the sense of Heat Kernels and Dirac Operators. Then by the argument of Heat Kernels and Dirac Operators Chapter 9, we could define a form $\alpha(t)$ such that $$ ch(\mathbb{A}_0)-\lim_{t\to \infty}ch(\mathbb{A}_t)=d[\int_0^{\infty}\alpha(t)dt]. $$ By definition it is clear that $ch(\mathbb{A}_0)=ch(\nabla_1)-ch(\nabla_2)$ and $\lim_{t\to \infty}ch(\mathbb{A}_t)=0$ since $E\overset{id}{\to}E$ is an isomorphism (it works for the chern character but not for chern classes). So it seems that we found another transgression formula $$ ch(\nabla_1)-ch(\nabla_2)=d[\int_0^{\infty}\alpha(t)dt]. $$
My question is : do these two transgression form constructed in two ways coincide given that one works for chern classes too while the second works for chern characters only?
