[proofplan]
The determinant expansion expresses each Chern form as the Chern-Weil form associated to an $\operatorname{Ad}$-invariant polynomial on $\mathfrak{gl}(r,\mathbb C)$. Closedness follows by applying the [Chern-Weil closedness theorem](/theorems/7039) to this invariant polynomial and the curvature of the induced principal frame-bundle connection. To compare two connections, we join them by the affine path of connections and apply the [Chern-Weil homotopy formula](/theorems/9760), which shows that the corresponding Chern forms differ by an exact form. Therefore their de Rham classes are independent of the connection.
[/proofplan]
[step:Identify the determinant coefficients as invariant polynomials]
Let $\mathfrak{gl}(r,\mathbb C)$ denote the [Lie algebra](/page/Lie%20Algebra) of $GL(r,\mathbb C)$, and for each $k\in\{0,\dots,r\}$ define the [homogeneous polynomial](/page/Homogeneous%20Polynomial)
\begin{align*}
p_k:\mathfrak{gl}(r,\mathbb C)\to \mathbb C
\end{align*}
by requiring
\begin{align*}
\det\left(I_r+\frac{t}{2\pi i}X\right)=\sum_{k=0}^{r}p_k(X)t^k
\end{align*}
for every $X\in\mathfrak{gl}(r,\mathbb C)$.
For $g\in GL(r,\mathbb C)$ and $X\in\mathfrak{gl}(r,\mathbb C)$, multiplicativity of the determinant gives
\begin{align*}
\det\left(I_r+\frac{t}{2\pi i}gXg^{-1}\right)=\det\left(g\left(I_r+\frac{t}{2\pi i}X\right)g^{-1}\right)=\det\left(I_r+\frac{t}{2\pi i}X\right).
\end{align*}
Comparing coefficients of $t^k$ gives
\begin{align*}
p_k(\operatorname{Ad}_g X)=p_k(X).
\end{align*}
Thus $p_k$ is an $\operatorname{Ad}$-invariant homogeneous polynomial on $\mathfrak{gl}(r,\mathbb C)$. By the defining determinant identity for Chern forms,
\begin{align*}
c_k(E,\nabla)=p_k(F_\nabla).
\end{align*}
[guided]
The goal of this step is to put the Chern form into the exact input format required by Chern-Weil theory. The input is not the whole determinant at once, but each coefficient of that determinant.
Let $\mathfrak{gl}(r,\mathbb C)$ be the Lie algebra of $GL(r,\mathbb C)$. For each $k\in\{0,\dots,r\}$, define
\begin{align*}
p_k:\mathfrak{gl}(r,\mathbb C)\to \mathbb C
\end{align*}
as the coefficient of $t^k$ in the polynomial
\begin{align*}
\det\left(I_r+\frac{t}{2\pi i}X\right).
\end{align*}
Equivalently, $p_k$ is determined by
\begin{align*}
\det\left(I_r+\frac{t}{2\pi i}X\right)=\sum_{k=0}^{r}p_k(X)t^k
\end{align*}
for every $X\in\mathfrak{gl}(r,\mathbb C)$.
We must verify the invariant-polynomial hypothesis. Let $g\in GL(r,\mathbb C)$ and $X\in\mathfrak{gl}(r,\mathbb C)$. Since $gI_rg^{-1}=I_r$, we have
\begin{align*}
I_r+\frac{t}{2\pi i}gXg^{-1}=g\left(I_r+\frac{t}{2\pi i}X\right)g^{-1}.
\end{align*}
Taking determinants and using multiplicativity of $\det$ gives
\begin{align*}
\det\left(I_r+\frac{t}{2\pi i}gXg^{-1}\right)=\det(g)\det\left(I_r+\frac{t}{2\pi i}X\right)\det(g^{-1}).
\end{align*}
Because $\det(g)\det(g^{-1})=1$, this becomes
\begin{align*}
\det\left(I_r+\frac{t}{2\pi i}gXg^{-1}\right)=\det\left(I_r+\frac{t}{2\pi i}X\right).
\end{align*}
Comparing the coefficient of $t^k$ on both sides gives
\begin{align*}
p_k(\operatorname{Ad}_gX)=p_k(X).
\end{align*}
Thus $p_k$ is $\operatorname{Ad}$-invariant. Finally, the Chern form $c_k(E,\nabla)$ was defined as the same coefficient after substituting the curvature form $F_\nabla$ for $X$, so
\begin{align*}
c_k(E,\nabla)=p_k(F_\nabla).
\end{align*}
This is precisely the bridge from the determinant definition of Chern classes to the general Chern-Weil closedness and transgression results.
[/guided]
[/step]
[step:Apply Chern-Weil closedness to the Chern forms]
Let $\operatorname{Fr}(E)\to M$ be the principal $GL(r,\mathbb C)$-bundle of complex frames of $E$. The complex-linear connection $\nabla$ induces a principal connection $A_\nabla$ on $\operatorname{Fr}(E)$ whose curvature corresponds to $F_\nabla$ under the standard associated-bundle identification
\begin{align*}
\operatorname{Fr}(E)\times_{\operatorname{Ad}}\mathfrak{gl}(r,\mathbb C)\cong \operatorname{End}(E).
\end{align*}
The notation $p_k(F_\nabla)$ means the standard Chern-Weil evaluation of the invariant polynomial $p_k$ on the curvature form, using the symmetric polarisation of $p_k$ in degree $k$. Since $p_k$ is $\operatorname{Ad}$-invariant, [citetheorem:9757] applies to $A_\nabla$ and gives
\begin{align*}
d\,p_k(F_\nabla)=0.
\end{align*}
Using $c_k(E,\nabla)=p_k(F_\nabla)$, we obtain
\begin{align*}
d\,c_k(E,\nabla)=0.
\end{align*}
Therefore each Chern form is closed.
[/step]
[step:Connect two connections by an affine path]
Let $\nabla_0$ and $\nabla_1$ be two complex-linear connections on $E$. Define
\begin{align*}
B:=\nabla_1-\nabla_0\in \Omega^1(M;\operatorname{End}(E)).
\end{align*}
For $s\in[0,1]$, define the affine path of complex-linear connections
\begin{align*}
\nabla_s:=\nabla_0+sB.
\end{align*}
This path satisfies $\nabla_s=\nabla_0$ at $s=0$ and $\nabla_s=\nabla_1$ at $s=1$.
[/step]
[step:Use the homotopy formula to show the difference is exact]
First consider $k=0$. By the determinant identity, $c_0(E,\nabla_0)=1=c_0(E,\nabla_1)$, so
\begin{align*}
c_0(E,\nabla_1)-c_0(E,\nabla_0)=0.
\end{align*}
Thus the degree-zero Chern forms agree.
Now assume $k\ge 1$. Apply [citetheorem:9760] to the smooth path $\nabla_s$ and to the invariant polynomial $p_k$ from the determinant expansion. Its hypotheses are satisfied because $E\to M$ is a smooth complex vector bundle, $\nabla_0$ and $\nabla_1$ are smooth complex-linear connections, and $p_k$ is $\operatorname{Ad}$-invariant. The theorem gives a form
\begin{align*}
T_k(\nabla_0,\nabla_1)\in \Omega^{2k-1}(M;\mathbb C)
\end{align*}
such that
\begin{align*}
p_k(F_{\nabla_1})-p_k(F_{\nabla_0})=dT_k(\nabla_0,\nabla_1).
\end{align*}
Substituting $c_k(E,\nabla_j)=p_k(F_{\nabla_j})$ for $j\in\{0,1\}$ gives
\begin{align*}
c_k(E,\nabla_1)-c_k(E,\nabla_0)=dT_k(\nabla_0,\nabla_1).
\end{align*}
Thus, for every $k\in\{0,\dots,r\}$, the two Chern forms either agree or differ by an exact form.
[/step]
[step:Pass to de Rham cohomology]
Since $c_k(E,\nabla_1)-c_k(E,\nabla_0)$ is exact, the two closed forms determine the same de Rham cohomology class:
\begin{align*}
[c_k(E,\nabla_1)]=[c_k(E,\nabla_0)]\in H^{2k}_{\mathrm{dR}}(M;\mathbb C).
\end{align*}
Because $\nabla_0$ and $\nabla_1$ were arbitrary complex-linear connections, the class
\begin{align*}
c_k(E):=[c_k(E,\nabla)]
\end{align*}
depends only on the complex vector bundle $E\to M$ and not on the chosen connection. Together with the closedness proved above, this completes the construction of the Chern classes by Chern-Weil theory.
[/step]
