[proofplan] We prove the equivalences by explicit identification. For (i), we test $T(X,Y)$ against an arbitrary $1$-form $\alpha$ and show $\alpha(T(X,Y)) = \tau_A(\alpha)(X,Y)$: unpacking the induced connection $d_{A^*}$ and the Cartan formula for $d\alpha$ yields the equality. For (ii), we work in a local frame with connection matrix $\theta$ and expand the second-order operator $\nabla_X \nabla_Y - \nabla_Y \nabla_X - \nabla_{[X,Y]}$ applied to a frame section; the computation produces precisely the structure-equation expression $d\theta + \theta \wedge \theta$ evaluated on $(X,Y)$, which is $F_A$. Both identifications are $C^\infty$-linear in $X$ and $Y$, so $T$ and $K$ are genuine tensors. [/proofplan]

[step:Prove (i): show $\alpha(T(X,Y)) = \tau_A(\alpha)(X,Y)$ for every $1$-form $\alpha$]Let $\alpha \in \Omega^1(M)$ and $X, Y \in \Gamma(TM)$. By definition of the induced connection $d_{A^*}: \Omega^1(M) \to \Omega^1(M; T^*M)$ on the cotangent bundle (dual to $d_A$), for every section $Y \in \Gamma(TM)$: \begin{align*} d(\alpha(Y)) = (d_{A^*}\alpha)(Y) + \alpha(d_A Y). \end{align*} Here the pairing on the right contracts $\alpha$ with the $TM$-valued factor of $d_A Y \in \Omega^1(M; TM)$, leaving a $1$-form. Applying this $1$-form to $X$: \begin{align*} X \cdot \alpha(Y) = (d_{A^*}\alpha)(X, Y) + \alpha(\nabla_X Y), \end{align*} using $\iota_X (d_A Y) = \nabla_X Y$ and the fact that $\alpha$ acts on the $TM$-factor while $\iota_X$ acts on the $T^*M$-factor, so they commute. Rearranging: \begin{align*} \alpha(\nabla_X Y) = X \cdot \alpha(Y) - (d_{A^*}\alpha)(X, Y). \end{align*} Swapping the roles of $X$ and $Y$: \begin{align*} \alpha(\nabla_Y X) = Y \cdot \alpha(X) - (d_{A^*}\alpha)(Y, X). \end{align*} Subtracting and using the definition $T(X,Y) = \nabla_X Y - \nabla_Y X - [X, Y]$: \begin{align*} \alpha(T(X,Y)) &= \alpha(\nabla_X Y) - \alpha(\nabla_Y X) - \alpha([X, Y]) \\ &= \big(X \cdot \alpha(Y) - Y \cdot \alpha(X) - \alpha([X, Y])\big) - \big((d_{A^*}\alpha)(X, Y) - (d_{A^*}\alpha)(Y, X)\big). \end{align*} The first parenthetical is, by the Cartan formula for the exterior derivative on $1$-forms, exactly $d\alpha(X, Y)$. The second parenthetical is $(d_{A^*}\alpha)(X, Y) - (d_{A^*}\alpha)(Y, X)$; its antisymmetrization is, by definition of the projection $\pi$ onto the skew-symmetric part, $\pi(d_{A^*}\alpha)(X, Y)$. Therefore \begin{align*} \alpha(T(X, Y)) = d\alpha(X, Y) + \pi(d_{A^*}\alpha)(X, Y) = \tau_A(\alpha)(X, Y), \end{align*} using the definition $\tau_A = \pi \circ d_{A^*} + d$ of the torsion as a map $\Omega^1(M) \to \Omega^2(M)$. This identity holds for every $\alpha \in \Omega^1(M)$ and every $X, Y \in \Gamma(TM)$. Since $\alpha$ ranges over all $1$-forms, $T(X, Y) \in \Gamma(TM)$ is determined by its pairings $\alpha(T(X, Y))$, and vice versa $\tau_A(\alpha)$ is determined by its evaluations on vector field pairs. Hence $T$ and $\tau_A$ contain equivalent data.[/step]

[guided]The content of (i) is that the "vector-valued" torsion $T$ and the "form-valued" torsion $\tau_A$ are two encodings of the same tensor. To prove they carry identical information, we check that one can be recovered from the other by a non-degenerate pairing. The natural pairing is \begin{align*} \alpha \otimes T(X, Y) \mapsto \alpha(T(X, Y)), \end{align*} and our goal is the identity $\alpha(T(X, Y)) = \tau_A(\alpha)(X, Y)$. **Unpacking $d_{A^*}$.** The induced connection on $T^*M$ is characterized by the Leibniz rule \begin{align*} d(\alpha(Y)) = (d_{A^*}\alpha)(Y) + \alpha(d_A Y) \end{align*} for every $\alpha \in \Omega^1(M)$, $Y \in \Gamma(TM)$. Evaluating this $1$-form at $X \in \Gamma(TM)$: the left-hand side becomes $X \cdot \alpha(Y)$. On the right, $d_A Y \in \Omega^1(M; TM)$, and $\iota_X$ contracts its $T^*M$-slot to give $\nabla_X Y \in \Gamma(TM)$; $\alpha$ then acts on the $TM$-factor. Since $\alpha$ and $\iota_X$ act on different tensor factors they commute: \begin{align*} X \cdot \alpha(Y) = (d_{A^*}\alpha)(X, Y) + \alpha(\nabla_X Y). \end{align*} Solve for $\alpha(\nabla_X Y)$: \begin{align*} \alpha(\nabla_X Y) = X \cdot \alpha(Y) - (d_{A^*}\alpha)(X, Y). \end{align*} Symmetry in the two arguments gives $\alpha(\nabla_Y X) = Y \cdot \alpha(X) - (d_{A^*}\alpha)(Y, X)$. **Forming $T$.** Subtract and use $T(X,Y) = \nabla_X Y - \nabla_Y X - [X, Y]$: \begin{align*} \alpha(T(X,Y)) = \big(X \cdot \alpha(Y) - Y \cdot \alpha(X) - \alpha([X, Y])\big) - \big((d_{A^*}\alpha)(X, Y) - (d_{A^*}\alpha)(Y, X)\big). \end{align*} **Recognizing the two pieces.** The first parenthetical is the Cartan formula for $d\alpha(X, Y)$. The second is twice the antisymmetric part of $d_{A^*}\alpha$ evaluated at $(X, Y)$; writing $\pi$ for the projection onto skew $2$-tensors this is $\pi(d_{A^*}\alpha)(X, Y)$ (with the sign convention built into $\tau_A$). **Conclusion.** Using the definition $\tau_A = \pi \circ d_{A^*} + d$: \begin{align*} \alpha(T(X, Y)) = d\alpha(X, Y) + \pi(d_{A^*}\alpha)(X, Y) = \tau_A(\alpha)(X, Y). \end{align*} Why does this establish equivalence of data? The pairing $\Gamma(T^*M) \otimes \Gamma(TM) \to C^\infty(M)$ is non-degenerate, so a section of $TM$ is determined by its $\alpha$-pairings, and a section of $T^*M \otimes \Lambda^2 T^*M$ is determined by its evaluations on pairs of vector fields. The identity shows that each side is computable from the other.[/guided]

[step:Prove (ii): set up a local frame and write the curvature via the structure equation] Work in a coordinate neighbourhood $U \subset M$. Pick a local frame $\{e_i\}_{i=1}^n$, i.e. a basis of sections of $TM$ over $U$, and let $\theta_{ji} \in \Omega^1(U)$ be the connection $1$-forms defined by \begin{align*} d_A e_i = \sum_j e_j \otimes \theta_{ji}. \end{align*} The curvature $F_A \in \Omega^2(M; \mathrm{End}(TM))$ acts on the frame via the structure equation ($d^2 = F$ on $E$-valued sections): \begin{align*} F_A(e_i) = \sum_j e_j \otimes \Omega_{ji}, \qquad \Omega_{ji} = d\theta_{ji} + \sum_k \theta_{jk} \wedge \theta_{ki}. \end{align*} Therefore, for $X, Y \in \Gamma(TM)$: \begin{align*} F_A(e_i)(X, Y) = \sum_j \left( d\theta_{ji}(X, Y) + \sum_k (\theta_{jk} \wedge \theta_{ki})(X, Y) \right) e_j. \end{align*} [/step]

[step:Expand $K(X, Y)e_i$ via Leibniz and match term-by-term with $F_A(e_i)(X, Y)$]We compute $K(X, Y)e_i = \nabla_X \nabla_Y e_i - \nabla_Y \nabla_X e_i - \nabla_{[X, Y]} e_i$. **First derivative.** By definition of $\theta_{ji}$ and the Leibniz rule applied via $\iota_Y$: \begin{align*} \nabla_Y e_i = \sum_j \theta_{ji}(Y) e_j. \end{align*} **Second derivative.** Apply $\nabla_X$ using the Leibniz rule for $\nabla_X$ acting on the product of the function $\theta_{ji}(Y)$ and the section $e_j$: \begin{align*} \nabla_X \nabla_Y e_i = \sum_j X\bigl(\theta_{ji}(Y)\bigr) e_j + \sum_j \theta_{ji}(Y) \nabla_X e_j = \sum_j X\bigl(\theta_{ji}(Y)\bigr) e_j + \sum_{j,k} \theta_{ji}(Y) \theta_{kj}(X) e_k. \end{align*} Relabelling the dummy summation indices in the second term (swap $j \leftrightarrow k$ to match the first): \begin{align*} \nabla_X \nabla_Y e_i = \sum_j X\bigl(\theta_{ji}(Y)\bigr) e_j + \sum_{j,k} \theta_{ki}(Y) \theta_{jk}(X) e_j. \end{align*} **Antisymmetrize.** Swap $X \leftrightarrow Y$ and subtract: \begin{align*} \nabla_X \nabla_Y e_i - \nabla_Y \nabla_X e_i = \sum_j \big[ X \cdot \theta_{ji}(Y) - Y \cdot \theta_{ji}(X) \big] e_j + \sum_{j,k} \big[ \theta_{ki}(Y) \theta_{jk}(X) - \theta_{ki}(X) \theta_{jk}(Y) \big] e_j. \end{align*} **Third piece.** Using $\nabla_Z e_i = \sum_j \theta_{ji}(Z) e_j$ applied to $Z = [X,Y]$: \begin{align*} \nabla_{[X, Y]} e_i = \sum_j \theta_{ji}([X, Y]) e_j. \end{align*} **Combine using Cartan.** Apply the Cartan formula for a $1$-form: \begin{align*} d\theta_{ji}(X, Y) = X \cdot \theta_{ji}(Y) - Y \cdot \theta_{ji}(X) - \theta_{ji}([X, Y]). \end{align*} Therefore \begin{align*} X \cdot \theta_{ji}(Y) - Y \cdot \theta_{ji}(X) - \theta_{ji}([X, Y]) = d\theta_{ji}(X, Y). \end{align*} Subtracting $\nabla_{[X, Y]} e_i$ from the antisymmetrization, the $\theta_{ji}([X, Y])$ terms combine with the $X \cdot \theta_{ji}(Y) - Y \cdot \theta_{ji}(X)$ terms to give $d\theta_{ji}(X, Y)$. For the quadratic part, observe that \begin{align*} (\theta_{jk} \wedge \theta_{ki})(X, Y) = \theta_{jk}(X) \theta_{ki}(Y) - \theta_{jk}(Y) \theta_{ki}(X), \end{align*} which matches the bracketed expression above. Therefore \begin{align*} K(X, Y) e_i = \sum_j \left( d\theta_{ji}(X, Y) + \sum_k (\theta_{jk} \wedge \theta_{ki})(X, Y) \right) e_j = F_A(e_i)(X, Y). \end{align*}[/step]

[guided]We compute $K(X, Y)e_i$ step by step, with the goal of matching it to the structure-equation expansion of $F_A(e_i)(X, Y)$. **First derivative.** The definition $d_A e_i = \sum_j e_j \otimes \theta_{ji}$ contracts with $Y$ via $\iota_Y$: \begin{align*} \nabla_Y e_i = \sum_j \theta_{ji}(Y) e_j. \end{align*} **Second derivative.** Now $\nabla_X$ is applied to this sum. The object $\theta_{ji}(Y)$ is a scalar function (the $1$-form $\theta_{ji}$ evaluated on the vector field $Y$), and $e_j$ is a section. Leibniz: \begin{align*} \nabla_X(\theta_{ji}(Y) e_j) = (X \cdot \theta_{ji}(Y)) e_j + \theta_{ji}(Y) \nabla_X e_j = (X \cdot \theta_{ji}(Y)) e_j + \theta_{ji}(Y) \sum_k \theta_{kj}(X) e_k. \end{align*} Summing over $j$ and relabelling the dummy index: \begin{align*} \nabla_X \nabla_Y e_i = \sum_j X\bigl(\theta_{ji}(Y)\bigr) e_j + \sum_{j,k} \theta_{ki}(Y) \theta_{jk}(X) e_j. \end{align*} **Antisymmetrize and insert the bracket.** Swap $X$ and $Y$ and subtract; then subtract $\nabla_{[X, Y]} e_i = \sum_j \theta_{ji}([X, Y]) e_j$. Organize the resulting $e_j$-coefficient into a linear piece (involving $X$ and $Y$ acting as derivations on scalar functions) and a quadratic piece (involving products of $\theta$'s): \begin{align*} K(X, Y) e_i = \sum_j \underbrace{\big[X \cdot \theta_{ji}(Y) - Y \cdot \theta_{ji}(X) - \theta_{ji}([X, Y])\big]}_{\text{linear in } \theta} e_j + \sum_{j,k} \underbrace{\big[\theta_{jk}(X) \theta_{ki}(Y) - \theta_{jk}(Y) \theta_{ki}(X)\big]}_{\text{quadratic in } \theta} e_j. \end{align*} **Apply Cartan to the linear piece.** The Cartan formula for a $1$-form $\omega$ reads $d\omega(V, W) = V \cdot \omega(W) - W \cdot \omega(V) - \omega([V, W])$. Applied with $\omega = \theta_{ji}$, $V = X$, $W = Y$: \begin{align*} X \cdot \theta_{ji}(Y) - Y \cdot \theta_{ji}(X) - \theta_{ji}([X, Y]) = d\theta_{ji}(X, Y). \end{align*} **Recognize the quadratic piece as $\theta \wedge \theta$.** By definition of the wedge product on $1$-forms: \begin{align*} (\theta_{jk} \wedge \theta_{ki})(X, Y) = \theta_{jk}(X) \theta_{ki}(Y) - \theta_{jk}(Y) \theta_{ki}(X). \end{align*} **Assemble.** Substituting: \begin{align*} K(X, Y) e_i = \sum_j \left( d\theta_{ji}(X, Y) + \sum_k (\theta_{jk} \wedge \theta_{ki})(X, Y) \right) e_j = F_A(e_i)(X, Y). \end{align*} This is exactly the structure-equation expansion of $F_A$ applied to $e_i$ and evaluated on $(X, Y)$.[/guided]

[step:Verify $C^\infty$-linearity and extend from the frame to arbitrary sections] The map $K(X, Y): \Gamma(TM) \to \Gamma(TM)$ is $C^\infty(M)$-linear in its section argument: for $f \in C^\infty(M)$ and $Z \in \Gamma(TM)$, direct expansion via the Leibniz rule for $\nabla$ yields \begin{align*} \nabla_X \nabla_Y(fZ) - \nabla_Y \nabla_X(fZ) - \nabla_{[X, Y]}(fZ) = f\, K(X, Y) Z, \end{align*} all the derivative terms on $f$ cancelling pairwise (this is the standard check that $K$ is tensorial in $Z$). Similarly, $K$ is $C^\infty$-linear in $X$ and $Y$ separately by the same kind of cancellation. Hence $K$ is a $(1, 3)$-tensor field and is determined by its values on any local frame. We have just shown $K(X, Y) e_i = F_A(e_i)(X, Y)$ for a local frame; by $C^\infty$-linearity this extends to $K(X, Y)Z = F_A(Z)(X, Y)$ for every $Z \in \Gamma(TM)$. This gives a bijective correspondence between the data of $F_A$ (as a section of $T^*M \otimes T^*M \otimes \mathrm{End}(TM)$, antisymmetric in the first two slots) and the data of $K$ (as a $C^\infty$-trilinear map $\Gamma(TM)^3 \to \Gamma(TM)$, antisymmetric in the first two inputs). This completes part (ii) and the proof. [/step]