[proofplan] We prove the formula locally in an arbitrary frame and then use $\operatorname{Ad}$-invariance to see that the resulting forms glue globally. Along the affine path of connections, the curvature varies by the covariant [exterior derivative](/theorems/1525) of the difference form, namely $\frac{\partial}{\partial t}F_{\nabla^t}=d_{\nabla^t}A$. Differentiating the polarized invariant polynomial gives the factor $k$, and the Bianchi identity plus infinitesimal $\operatorname{Ad}$-invariance identifies the derivative with an exterior derivative. Integrating the resulting identity over $t\in[0,1]$ gives the transgression formula. [/proofplan]

[step:Fix the polarized Chern-Weil convention in a local frame] Let $U\subset M$ be an [open set](/page/Open%20Set) over which $E$ admits a smooth frame \begin{align*} s=(s_1,\dots,s_r):U\times \mathbb C^r\to E|_U. \end{align*} In this frame, write \begin{align*} \Gamma_t\in \Omega^1(U;\mathfrak{gl}(r,\mathbb C)) \end{align*} for the connection matrix of $\nabla^t$, write \begin{align*} a\in \Omega^1(U;\mathfrak{gl}(r,\mathbb C)) \end{align*} for the matrix of $A$, and write \begin{align*} \Omega_t\in \Omega^2(U;\mathfrak{gl}(r,\mathbb C)) \end{align*} for the curvature matrix of $\nabla^t$. Thus \begin{align*} \Gamma_t=\Gamma_0+ta \end{align*} and \begin{align*} \Omega_t=d\Gamma_t+\Gamma_t\wedge \Gamma_t. \end{align*} The expression $P(\beta_1,\dots,\beta_k)$ for matrix-valued forms $\beta_j\in \Omega^{m_j}(U;\mathfrak{gl}(r,\mathbb C))$ denotes the form obtained from the symmetric polarization of $P$ by wedging the form parts and composing the matrix parts in the standard Chern-Weil symmetrized convention. Since $P$ is $\operatorname{Ad}$-invariant, this local expression is unchanged under change of frame, so the local forms $P(\Omega_t,\dots,\Omega_t)$ and $P(a,\Omega_t,\dots,\Omega_t)$ represent the global forms $P(F_{\nabla^t})$ and $P(A,F_{\nabla^t},\dots,F_{\nabla^t})$. [/step]

[step:Compute the curvature variation along the affine path]For an $\operatorname{End}(E)$-valued $m$-form $\beta$ represented locally by a matrix-valued form $b\in \Omega^m(U;\mathfrak{gl}(r,\mathbb C))$, the covariant exterior derivative $d_{\nabla^t}\beta$ is represented by \begin{align*} d_{\Gamma_t}b:=db+\Gamma_t\wedge b-(-1)^m b\wedge \Gamma_t. \end{align*} Since $a$ is independent of $t$, differentiating the local curvature formula gives \begin{align*} \frac{\partial \Omega_t}{\partial t}=da+a\wedge \Gamma_t+\Gamma_t\wedge a. \end{align*} Because $a$ has degree $1$, the preceding formula is precisely \begin{align*} \frac{\partial \Omega_t}{\partial t}=d_{\Gamma_t}a. \end{align*} Equivalently, in global notation, \begin{align*} \frac{\partial}{\partial t}F_{\nabla^t}=d_{\nabla^t}A. \end{align*}[/step]

[guided]The curvature of $\nabla^t$ is easiest to differentiate after choosing a frame, because a connection then becomes a matrix of $1$-forms. In the chosen frame on $U$, the affine path is represented by \begin{align*} \Gamma_t=\Gamma_0+ta, \end{align*} where $a$ is the matrix of the globally defined form $A=\nabla^1-\nabla^0$. The curvature matrix is \begin{align*} \Omega_t=d\Gamma_t+\Gamma_t\wedge \Gamma_t. \end{align*} Differentiating with respect to $t$ is legitimate coefficientwise because $\Gamma_t$ depends smoothly and affinely on $t$. We obtain \begin{align*} \frac{\partial \Omega_t}{\partial t}=d a+a\wedge \Gamma_t+\Gamma_t\wedge a. \end{align*} The order of the two product terms matters because the wedge product also composes matrix entries. Now recall the local formula for the covariant exterior derivative on an $\operatorname{End}(E)$-valued $m$-form represented by $b$: \begin{align*} d_{\Gamma_t}b=db+\Gamma_t\wedge b-(-1)^m b\wedge \Gamma_t. \end{align*} Here $b=a$ has degree $m=1$, so $-(-1)^1=1$, and therefore \begin{align*} d_{\Gamma_t}a=da+\Gamma_t\wedge a+a\wedge \Gamma_t. \end{align*} This is exactly the derivative computed above. Translating out of the frame gives the global identity \begin{align*} \frac{\partial}{\partial t}F_{\nabla^t}=d_{\nabla^t}A. \end{align*}[/guided]

[step:Differentiate the invariant polynomial evaluated on curvature] For each $t\in[0,1]$, the polarized form of $P$ gives \begin{align*} P(F_{\nabla^t})=P(F_{\nabla^t},\dots,F_{\nabla^t}). \end{align*} Since $P$ is symmetric and $k$-linear, differentiation with respect to $t$ and the curvature variation formula give \begin{align*} \frac{\partial}{\partial t}P(F_{\nabla^t})=kP(d_{\nabla^t}A,F_{\nabla^t},\dots,F_{\nabla^t}). \end{align*} [/step]

[step:Convert the covariant derivative term into an exterior derivative]We claim that, for each $t\in[0,1]$, \begin{align*} dP(A,F_{\nabla^t},\dots,F_{\nabla^t})=P(d_{\nabla^t}A,F_{\nabla^t},\dots,F_{\nabla^t}). \end{align*} In the local frame on $U$, the Bianchi identity says \begin{align*} d_{\Gamma_t}\Omega_t=0. \end{align*} The covariant Leibniz rule for an invariant polynomial gives \begin{align*} dP(a,\Omega_t,\dots,\Omega_t)=P(d_{\Gamma_t}a,\Omega_t,\dots,\Omega_t)-\sum_{j=2}^{k}P(a,\Omega_t,\dots,d_{\Gamma_t}\Omega_t,\dots,\Omega_t). \end{align*} The minus sign before the sum occurs because $a$ has degree $1$ and each $\Omega_t$ has degree $2$. Each summand in the sum vanishes by the Bianchi identity. The remaining commutator terms in passing from $d$ to $d_{\Gamma_t}$ cancel by differentiating the $GL(r,\mathbb C)$-$\operatorname{Ad}$-invariance of $P$. Explicitly, for all $X,Y_1,\dots,Y_k\in\mathfrak{gl}(r,\mathbb C)$, \begin{align*} \sum_{j=1}^{k}P(Y_1,\dots,[X,Y_j],\dots,Y_k)=0. \end{align*} This identity is applied coefficientwise after expanding the matrix-valued forms in local coordinates, with $X$ supplied by the connection matrix component of $\Gamma_t$ and the $Y_j$ supplied by the matrix components of $a$ and $\Omega_t$. Hence \begin{align*} dP(a,\Omega_t,\dots,\Omega_t)=P(d_{\Gamma_t}a,\Omega_t,\dots,\Omega_t). \end{align*} Since this identity is invariant under change of frame, it is the asserted global identity.[/step]

[guided]The goal of this step is to replace the covariant expression produced by differentiating curvature with an ordinary exterior derivative, because the theorem claims that the difference of Chern-Weil forms is exact. Work in the same local frame. The curvature matrix $\Omega_t$ satisfies the Bianchi identity \begin{align*} d_{\Gamma_t}\Omega_t=0. \end{align*} This identity applies because $\Omega_t$ is the curvature of the connection represented by $\Gamma_t$. Now apply the covariant Leibniz rule to the invariant polarized expression \begin{align*} P(a,\Omega_t,\dots,\Omega_t)\in \Omega^{2k-1}(U). \end{align*} The first input $a$ has degree $1$, and every curvature input $\Omega_t$ has degree $2$. Therefore the sign accumulated when the derivative passes the first input is negative, while passing curvature factors introduces no additional odd-degree sign. The rule gives \begin{align*} dP(a,\Omega_t,\dots,\Omega_t)=P(d_{\Gamma_t}a,\Omega_t,\dots,\Omega_t)-\sum_{j=2}^{k}P(a,\Omega_t,\dots,d_{\Gamma_t}\Omega_t,\dots,\Omega_t). \end{align*} The sum vanishes term by term because $d_{\Gamma_t}\Omega_t=0$. Thus only the derivative of the first input remains. Why is the left side the ordinary exterior derivative $d$ rather than a covariant derivative of a matrix-valued form? The expression $P(a,\Omega_t,\dots,\Omega_t)$ is scalar-valued after applying the invariant polynomial. The difference between expanding with $d$ and expanding with $d_{\Gamma_t}$ consists exactly of commutator terms involving the connection matrix $\Gamma_t$. We now spell out the cancellation. Since $P$ is invariant under the adjoint action of $GL(r,\mathbb C)$, its symmetric polarization satisfies the infinitesimal identity \begin{align*} \sum_{j=1}^{k}P(Y_1,\dots,[X,Y_j],\dots,Y_k)=0 \end{align*} for every $X,Y_1,\dots,Y_k\in\mathfrak{gl}(r,\mathbb C)$. This follows by differentiating at $\lambda=0$ the identity $P(\operatorname{Ad}_{\exp(\lambda X)}Y_1,\dots,\operatorname{Ad}_{\exp(\lambda X)}Y_k)=P(Y_1,\dots,Y_k)$. Applying this formula coefficientwise to the matrix components of $\Gamma_t$, $a$, and $\Omega_t$ cancels precisely the commutator contributions. Hence \begin{align*} dP(a,\Omega_t,\dots,\Omega_t)=P(d_{\Gamma_t}a,\Omega_t,\dots,\Omega_t). \end{align*} Because $P$ is $\operatorname{Ad}$-invariant, the same identity holds after changing frames, so the local calculation represents the global formula \begin{align*} dP(A,F_{\nabla^t},\dots,F_{\nabla^t})=P(d_{\nabla^t}A,F_{\nabla^t},\dots,F_{\nabla^t}). \end{align*}[/guided]

[step:Integrate the derivative identity over the path of connections] Combining the previous two steps gives, for every $t\in[0,1]$, \begin{align*} \frac{\partial}{\partial t}P(F_{\nabla^t})=k\,dP(A,F_{\nabla^t},\dots,F_{\nabla^t}). \end{align*} Integrating this identity over $[0,1]$ with respect to one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) gives \begin{align*} P(F_{\nabla^1})-P(F_{\nabla^0})=\int_{[0,1]}\frac{\partial}{\partial t}P(F_{\nabla^t})\,d\mathcal L^1(t). \end{align*} Substituting the derivative identity yields \begin{align*} P(F_{\nabla^1})-P(F_{\nabla^0})=k\int_{[0,1]}dP(A,F_{\nabla^t},\dots,F_{\nabla^t})\,d\mathcal L^1(t). \end{align*} The exterior derivative commutes with integration over the compact parameter interval because the integrand is a smooth family of differential forms on $M$. Therefore \begin{align*} P(F_{\nabla^1})-P(F_{\nabla^0})=d\left(k\int_{[0,1]}P(A,F_{\nabla^t},\dots,F_{\nabla^t})\,d\mathcal L^1(t)\right). \end{align*} This is the claimed Chern-Weil homotopy formula. [/step]