[proofplan] The proof is a direct unfolding of the chapter's sign convention $R = -\nabla \circ \nabla$, where the second covariant derivative $\nabla^2 Z : \mathfrak{X}(M) \times \mathfrak{X}(M) \to \mathfrak{X}(M)$ is the $(1,3)$-tensor characterised by $(\nabla^2 Z)(X, Y) = \nabla_X \nabla_Y Z - \nabla_{\nabla_X Y} Z$. Antisymmetrising in $X, Y$ kills the symmetric part of the connection coefficients in the first slot and leaves a torsion-free combination involving the bracket. We compute $R(X, Y) Z$ from the antisymmetrisation, use the torsion-free identity $\nabla_X Y - \nabla_Y X = [X, Y]$ for the Levi-Civita connection, and cancel terms to obtain $R(X, Y) = \nabla_{[X,Y]} - [\nabla_X, \nabla_Y]$. [/proofplan]

[step:Recall the second covariant derivative and the chapter's sign convention] For a vector field $Z \in \mathfrak{X}(M)$, the second covariant derivative is the $\mathbb{R}$-bilinear map \begin{align*} \nabla^2 Z : \mathfrak{X}(M) \times \mathfrak{X}(M) &\to \mathfrak{X}(M), \\ (X, Y) &\mapsto \nabla_X \nabla_Y Z - \nabla_{\nabla_X Y} Z. \end{align*} The correction term $-\nabla_{\nabla_X Y} Z$ ensures tensoriality in $X$ and $Y$: it cancels the non-tensorial part of $\nabla_X \nabla_Y Z$ that arises from differentiating $Y$ along $X$. The chapter's curvature endomorphism is defined as the antisymmetrisation \begin{align*} R(X, Y) Z &:= -\bigl( \nabla^2 Z(X, Y) - \nabla^2 Z(Y, X) \bigr). \end{align*} This is the sign convention $R = -\nabla \circ \nabla$ stated at the start of the chapter; equivalently $R(X, Y) Z = \nabla^2 Z(Y, X) - \nabla^2 Z(X, Y)$. [/step]

[step:Expand $R(X, Y) Z$ in terms of single covariant derivatives] Substituting the definition of $\nabla^2 Z$ from Step 1, \begin{align*} R(X, Y) Z &= \nabla^2 Z(Y, X) - \nabla^2 Z(X, Y) \\ &= \bigl( \nabla_Y \nabla_X Z - \nabla_{\nabla_Y X} Z \bigr) - \bigl( \nabla_X \nabla_Y Z - \nabla_{\nabla_X Y} Z \bigr) \\ &= \nabla_Y \nabla_X Z - \nabla_X \nabla_Y Z + \nabla_{\nabla_X Y} Z - \nabla_{\nabla_Y X} Z. \end{align*} [/step]

[step:Combine the connection-correction terms using the torsion-free property]The Levi-Civita connection is **torsion-free**: for all $X, Y \in \mathfrak{X}(M)$, \begin{align*} \nabla_X Y - \nabla_Y X &= [X, Y]. \end{align*} By the linearity of the connection in the lower index, \begin{align*} \nabla_{\nabla_X Y} Z - \nabla_{\nabla_Y X} Z &= \nabla_{\nabla_X Y - \nabla_Y X} Z = \nabla_{[X, Y]} Z. \end{align*} Substituting into Step 2, \begin{align*} R(X, Y) Z &= \nabla_Y \nabla_X Z - \nabla_X \nabla_Y Z + \nabla_{[X, Y]} Z \\ &= \nabla_{[X, Y]} Z - \bigl( \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z \bigr) \\ &= \nabla_{[X, Y]} Z - [\nabla_X, \nabla_Y] Z. \end{align*}[/step]

[guided]We arrive at Step 3 with the expansion from Step 2, \begin{align*} R(X, Y) Z &= \nabla_Y \nabla_X Z - \nabla_X \nabla_Y Z + \nabla_{\nabla_X Y} Z - \nabla_{\nabla_Y X} Z, \end{align*} and we want to massage it into the operator form $\nabla_{[X,Y]} - [\nabla_X, \nabla_Y]$. What is the obstruction? Two non-tensorial pieces appeared: $\nabla_{\nabla_X Y} Z$ came from $\nabla^2 Z(X, Y)$, and $\nabla_{\nabla_Y X} Z$ came from $\nabla^2 Z(Y, X)$. Individually neither is tensorial in $X$ or $Y$ (each differentiates the field in the other slot via $\nabla$). But their *difference* depends only on the antisymmetric part of the map $(X, Y) \mapsto \nabla_X Y$, and the antisymmetric part of $\nabla$ is precisely the torsion. So if the connection is torsion-free, the difference should collapse to something tensorial. Let us make this precise. The Levi-Civita connection is by definition **torsion-free**, meaning that for all $X, Y \in \mathfrak{X}(M)$, \begin{align*} \nabla_X Y - \nabla_Y X &= [X, Y]. \end{align*} This is not a coincidence we need to prove here — it is one of the two defining properties of Levi-Civita (the other being metric compatibility). We invoke it directly from the connection's definition. Why is this exactly what we need? The connection $\nabla_W Z$ is $C^\infty(M)$-linear in its lower index $W$, so it is in particular $\mathbb{R}$-linear: \begin{align*} \nabla_{\nabla_X Y} Z - \nabla_{\nabla_Y X} Z &= \nabla_{\nabla_X Y - \nabla_Y X} Z = \nabla_{[X, Y]} Z. \end{align*} The two non-tensorial pieces have collapsed to a single tensorial term $\nabla_{[X, Y]} Z$, exactly as we hoped. This is the step where torsion-freeness is consumed; on a connection with non-zero torsion $T(X, Y) := \nabla_X Y - \nabla_Y X - [X, Y]$, the right-hand side would acquire an additional $\nabla_{T(X, Y)} Z$ correction. Now substitute back into the Step 2 expression: \begin{align*} R(X, Y) Z &= \nabla_Y \nabla_X Z - \nabla_X \nabla_Y Z + \nabla_{[X, Y]} Z. \end{align*} The remaining piece $\nabla_Y \nabla_X Z - \nabla_X \nabla_Y Z$ is, up to sign, the commutator of the operators $\nabla_X$ and $\nabla_Y$ acting on $Z$: \begin{align*} \nabla_Y \nabla_X Z - \nabla_X \nabla_Y Z &= -\bigl( \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z \bigr) = -[\nabla_X, \nabla_Y] Z. \end{align*} Regrouping, \begin{align*} R(X, Y) Z &= \nabla_{[X, Y]} Z - [\nabla_X, \nabla_Y] Z. \end{align*} The minus sign attached to $[\nabla_X, \nabla_Y]$ is a direct consequence of the sign convention $R = -\nabla \circ \nabla$ chosen in this chapter. Under the opposite convention $R = +\nabla \circ \nabla$ (used by Lee, do Carmo, and many others), the formula reads $R(X, Y) = \nabla_X \nabla_Y - \nabla_Y \nabla_X - \nabla_{[X, Y]}$, which differs only by an overall sign — the underlying calculation is identical.[/guided]

[step:State the operator identity] The result of Step 3 holds for every $Z \in \mathfrak{X}(M)$, so as operators on $\mathfrak{X}(M)$, \begin{align*} R(X, Y) &= \nabla_{[X, Y]} - [\nabla_X, \nabla_Y]. \end{align*} This completes the proof. [/step]