[proofplan] The curvature is defined globally as $F_A = d_A \circ d_A : \Omega^0(E) \to \Omega^2(E)$, and we compute its local form by directly applying the local formula for the extended connection twice and simplifying. The two applications of $d_A = d + \theta_\alpha \wedge$ produce four terms, of which $d^2 s_\alpha = 0$ by nilpotence of the de Rham differential, and two cross terms of the form $\pm \theta_\alpha \wedge ds_\alpha$ cancel exactly. The surviving terms are $d\theta_\alpha \wedge s_\alpha$ and $\theta_\alpha \wedge \theta_\alpha \wedge s_\alpha$, giving the claimed formula $(F_A)_\alpha = d\theta_\alpha + \theta_\alpha \wedge \theta_\alpha$ once the section $s_\alpha$ is stripped off. Since this expression is a fixed 2-form-valued endomorphism and not a differential operator, its action on sections is manifestly $C^\infty(M)$-linear, which proves the tensoriality claim. [/proofplan] [step:Recall the global definition of curvature and the local formula for the extended connection] The curvature of a connection $A$ on $E \to M$ is the composition \begin{align*} F_A = d_A \circ d_A : \Omega^0(E) \to \Omega^2(E), \end{align*} where $d_A: \Omega^i(E) \to \Omega^{i+1}(E)$ is the extension of the connection to $E$-valued forms constructed in the [Extension of a Connection theorem](/theorems/1539). In a trivialising chart $U_\alpha$ with local frame and connection matrix $\theta_\alpha \in \Omega^1(U_\alpha; \mathrm{End}(\mathbb{R}^r))$, that theorem gives \begin{align*} (d_A \omega)_\alpha = d\omega_\alpha + \theta_\alpha \wedge \omega_\alpha \end{align*} for $\omega \in \Omega^i(E)$ with local representative $\omega_\alpha$. [/step] [step:Apply $d_A$ twice to a local section and expand using the Leibniz rule] Let $s \in \Gamma(E)$ with local representative $s_\alpha \in C^\infty(U_\alpha; \mathbb{R}^r)$. From the previous step, \begin{align*} (d_A s)_\alpha = ds_\alpha + \theta_\alpha \wedge s_\alpha \in \Omega^1(U_\alpha; \mathbb{R}^r), \end{align*} treating $\theta_\alpha \wedge s_\alpha = \theta_\alpha \cdot s_\alpha$ since $s_\alpha$ is a 0-form. Applying $d_A$ again (now with $i = 1$): \begin{align*} (d_A^2 s)_\alpha = (d_A(d_A s))_\alpha = d\big((d_A s)_\alpha\big) + \theta_\alpha \wedge (d_A s)_\alpha = d(ds_\alpha + \theta_\alpha \wedge s_\alpha) + \theta_\alpha \wedge (ds_\alpha + \theta_\alpha \wedge s_\alpha). \end{align*} Expanding the first term using the graded Leibniz rule for $d$ (the form $\theta_\alpha$ has degree 1, so the sign is $(-1)^1 = -1$): \begin{align*} d(ds_\alpha + \theta_\alpha \wedge s_\alpha) = d^2 s_\alpha + d\theta_\alpha \wedge s_\alpha - \theta_\alpha \wedge ds_\alpha. \end{align*} Expanding the second term by distributivity of $\wedge$: \begin{align*} \theta_\alpha \wedge (ds_\alpha + \theta_\alpha \wedge s_\alpha) = \theta_\alpha \wedge ds_\alpha + \theta_\alpha \wedge \theta_\alpha \wedge s_\alpha. \end{align*} [guided] We compute $(d_A^2 s)_\alpha$ explicitly in a trivialising chart. The strategy is pure substitution and bookkeeping: apply the local formula $d_A = d + \theta_\alpha \wedge$ twice and expand. First application. For a section $s$ with local representative $s_\alpha$ (a 0-form-valued vector), \begin{align*} (d_A s)_\alpha = ds_\alpha + \theta_\alpha \wedge s_\alpha = ds_\alpha + \theta_\alpha s_\alpha \in \Omega^1(U_\alpha; \mathbb{R}^r), \end{align*} where $\theta_\alpha \wedge s_\alpha = \theta_\alpha s_\alpha$ because $s_\alpha$ is a 0-form and wedge with a 0-form is ordinary multiplication. Second application. Now $(d_A s)_\alpha$ is a 1-form-valued vector, so we apply $d_A: \Omega^1(E) \to \Omega^2(E)$ in its local form, giving \begin{align*} (d_A^2 s)_\alpha = d\big((d_A s)_\alpha\big) + \theta_\alpha \wedge (d_A s)_\alpha. \end{align*} Term 1: $d\big((d_A s)_\alpha\big) = d(ds_\alpha + \theta_\alpha \wedge s_\alpha) = d^2 s_\alpha + d(\theta_\alpha \wedge s_\alpha)$. The graded Leibniz rule for $d$ states $d(\eta \wedge \omega) = d\eta \wedge \omega + (-1)^{\deg \eta} \eta \wedge d\omega$. With $\deg \theta_\alpha = 1$, the sign is negative: \begin{align*} d(\theta_\alpha \wedge s_\alpha) = d\theta_\alpha \wedge s_\alpha - \theta_\alpha \wedge ds_\alpha. \end{align*} Term 2: $\theta_\alpha \wedge (d_A s)_\alpha = \theta_\alpha \wedge ds_\alpha + \theta_\alpha \wedge \theta_\alpha \wedge s_\alpha$, by distributivity of $\wedge$ over addition. Adding them together is the next step. The crucial observation is that in Term 1 we produced $-\theta_\alpha \wedge ds_\alpha$ (from the Leibniz sign), while Term 2 produced $+\theta_\alpha \wedge ds_\alpha$. These two cancel — this is a geometric manifestation of the fact that the failure of $d$ to commute with $\theta_\alpha \wedge$ is exactly the connection non-triviality. [/guided] [/step] [step:Collect terms using $d^2 = 0$ and cancel the cross terms] Adding the two expanded expressions from the previous step: \begin{align*} (d_A^2 s)_\alpha = \underbrace{d^2 s_\alpha}_{= 0} + d\theta_\alpha \wedge s_\alpha \underbrace{- \theta_\alpha \wedge ds_\alpha + \theta_\alpha \wedge ds_\alpha}_{= 0} + \theta_\alpha \wedge \theta_\alpha \wedge s_\alpha. \end{align*} Here $d^2 = 0$ is the nilpotence of the de Rham differential, applied component-wise to $s_\alpha$, and the $\pm \theta_\alpha \wedge ds_\alpha$ terms cancel exactly. Hence \begin{align*} (d_A^2 s)_\alpha = (d\theta_\alpha + \theta_\alpha \wedge \theta_\alpha) \wedge s_\alpha. \end{align*} [/step] [step:Identify the result as multiplication by a global $\mathrm{End}(E)$-valued 2-form] Since the expression $d\theta_\alpha + \theta_\alpha \wedge \theta_\alpha$ acts on the local representative $s_\alpha$ by wedge product (equivalently, by matrix multiplication, since $s_\alpha$ is a 0-form-valued vector), we have shown that $(F_A)_\alpha := d\theta_\alpha + \theta_\alpha \wedge \theta_\alpha$ is the local representative of the curvature: \begin{align*} (F_A)_\alpha = d\theta_\alpha + \theta_\alpha \wedge \theta_\alpha \;\in\; \Omega^2(U_\alpha; \mathrm{End}(\mathbb{R}^r)). \end{align*} Here $d\theta_\alpha$ is the component-wise de Rham differential of the matrix $\theta_\alpha$, and $\theta_\alpha \wedge \theta_\alpha$ is the wedge combined with matrix multiplication: $(\theta_\alpha \wedge \theta_\alpha)_{jk} = \sum_\ell (\theta_\alpha)_{j\ell} \wedge (\theta_\alpha)_{\ell k}$. Note that $\theta_\alpha \wedge \theta_\alpha \ne 0$ in general — the matrix multiplication breaks the usual skew-symmetry $\omega \wedge \omega = 0$ that holds for scalar 1-forms. [guided] Writing $(F_A)_\alpha = d\theta_\alpha + \theta_\alpha \wedge \theta_\alpha$ finishes the local computation, but two subtleties deserve attention. First, $(F_A)_\alpha$ is matrix-valued. The first summand $d\theta_\alpha$ is the component-wise exterior derivative: $(d\theta_\alpha)_{jk} = d((\theta_\alpha)_{jk})$, producing a matrix of 2-forms. The second summand combines wedge with matrix product: \begin{align*} (\theta_\alpha \wedge \theta_\alpha)_{jk} = \sum_{\ell=1}^r (\theta_\alpha)_{j\ell} \wedge (\theta_\alpha)_{\ell k}. \end{align*} Both summands live in $\Omega^2(U_\alpha; \mathrm{End}(\mathbb{R}^r))$, so their sum does too. Second, a common point of confusion: for a scalar 1-form $\eta$, one has $\eta \wedge \eta = 0$ (because $\wedge$ is skew on 1-forms). But $\theta_\alpha \wedge \theta_\alpha$ need not vanish, because the matrix multiplication interleaves the entries non-commutatively: \begin{align*} (\theta_\alpha \wedge \theta_\alpha)_{jk} = \sum_{\ell} (\theta_\alpha)_{j\ell} \wedge (\theta_\alpha)_{\ell k}, \end{align*} and the sum mixes different 1-forms. Equivalently, $\theta_\alpha \wedge \theta_\alpha = \frac{1}{2}[\theta_\alpha, \theta_\alpha]$ in the graded commutator of the matrix Lie algebra, which vanishes iff $\theta_\alpha$ is abelian-valued. Hence the curvature has genuine non-abelian content. Finally, the assignment $s \mapsto F_A \cdot s$ acts pointwise: at each $p \in U_\alpha$, $(F_A)_\alpha(p)$ is an $\mathrm{End}(\mathbb{R}^r)$-valued 2-tensor in $T^*_p U_\alpha$, and the action on $s_\alpha(p)$ is matrix--vector multiplication. This is what we exploit next for tensoriality. [/guided] [/step] [step:Deduce $C^\infty(M)$-linearity of $F_A$ from the pointwise nature of the local formula] Let $f \in C^\infty(M)$ and $s \in \Gamma(E)$. Since $(F_A)_\alpha$ acts on the local representative by wedge product — a $C^\infty(U_\alpha)$-linear operation — the action on $fs$ satisfies \begin{align*} (F_A \cdot (fs))_\alpha = (F_A)_\alpha \wedge (fs)_\alpha = (F_A)_\alpha \wedge (f \cdot s_\alpha) = f \cdot (F_A)_\alpha \wedge s_\alpha = f \cdot (F_A \cdot s)_\alpha, \end{align*} so $F_A \cdot (fs) = f \cdot F_A \cdot s$. By the tensor--hom correspondence (see the [Structure of the Space of Connections](/theorems/1538)), $C^\infty(M)$-linearity of $s \mapsto F_A \cdot s$ identifies $F_A$ with a global section of $\mathrm{End}(E) \otimes \Lambda^2 T^*M = \mathrm{End}(E) \otimes \Lambda^2(T^*M)$. Hence $F_A \in \Omega^2(\mathrm{End}(E))$ — i.e. $F_A$ is a genuine tensor, not a differential operator. Similarly, $F_A$ is $C^\infty(M)$-linear in its two 1-form arguments (the arguments that feed the wedge $\wedge^2 T^*M$) because $\wedge$ is $C^\infty(M)$-bilinear on 1-forms. [/step] [step:Conclude the local formula and global tensoriality] From Steps 2--4 we have the local formula \begin{align*} (F_A)_\alpha = d\theta_\alpha + \theta_\alpha \wedge \theta_\alpha \in \Omega^2(U_\alpha; \mathrm{End}(\mathbb{R}^r)) \end{align*} valid in every trivialising chart. Step 5 shows that the global operator $F_A = d_A^2: \Gamma(E) \to \Omega^2(E)$ is $C^\infty(M)$-linear, so by tensor--hom it is a global section of $\mathrm{End}(E) \otimes \Lambda^2 T^*M$, i.e. $F_A \in \Omega^2(\mathrm{End}(E))$. Together with the local formula, this establishes the theorem. [/step]