[proofplan] We establish the three classes of symmetries in order. Skew-symmetry in $(k, \ell)$ is structural — it expresses that $R$ is a $2$-form in its last two arguments and follows from $R(X, Y) = -R(Y, X)$, which is built into the chapter's definition $R(X, Y) = \nabla_{[X,Y]} - [\nabla_X, \nabla_Y]$ since the right-hand side is manifestly antisymmetric in $X, Y$. Skew-symmetry in $(i, j)$ uses metric compatibility of the Levi-Civita connection: differentiating $g(\partial_i, \partial_j)$ twice and antisymmetrising yields $g(R(X,Y)\partial_i, \partial_j) = -g(\partial_i, R(X,Y)\partial_j)$, so the curvature acts skew-symmetrically on $(i, j)$ as well. The first Bianchi identity follows from the algebraic identity $\sum_{\mathrm{cyc}} R(X, Y) Z = 0$, which itself is a consequence of the Jacobi identity for $[\cdot, \cdot]$ together with the torsion-free condition. Pair symmetry $R_{ij,k\ell} = R_{k\ell, ij}$ is a non-trivial corollary: it is derived by combining the three preceding symmetries algebraically (the standard "octahedron" or "16-term" argument). [/proofplan] [step:Establish skew-symmetry in the last pair $R_{ij,k\ell} = -R_{ij, \ell k}$] By [Curvature as Commutator of Covariant Derivatives](/theorems/2703), the curvature endomorphism satisfies \begin{align*} R(X, Y) Z &= \nabla_{[X, Y]} Z - [\nabla_X, \nabla_Y] Z. \end{align*} Both terms on the right-hand side are antisymmetric in $X, Y$: $[X, Y] = -[Y, X]$ implies $\nabla_{[X, Y]} = -\nabla_{[Y, X]}$, and the commutator $[\nabla_X, \nabla_Y]$ is antisymmetric by definition. Therefore $R(X, Y) Z = -R(Y, X) Z$. In components, taking $X = \partial_k$, $Y = \partial_\ell$, $Z = \partial_j$ and pairing with $\partial_i$ (using $g$), \begin{align*} R_{ij, k\ell} &= g(R(\partial_k, \partial_\ell) \partial_j, \partial_i) = -g(R(\partial_\ell, \partial_k) \partial_j, \partial_i) = -R_{ij, \ell k}. \end{align*} [/step] [step:Establish skew-symmetry in the first pair $R_{ij,k\ell} = -R_{ji, k\ell}$ via metric compatibility] We work in a coordinate chart $(U, \varphi)$ with coordinate vector fields $\partial_i := \partial_{x_i}$. Metric compatibility of the Levi-Civita connection asserts that for all vector fields $X, Y, Z$, \begin{align*} X\bigl(g(Y, Z)\bigr) &= g(\nabla_X Y, Z) + g(Y, \nabla_X Z). \end{align*} Apply this twice. First, with $X = \partial_\ell$, \begin{align*} \partial_\ell\bigl(g(\partial_i, \partial_j)\bigr) &= g(\nabla_\ell \partial_i, \partial_j) + g(\partial_i, \nabla_\ell \partial_j), \end{align*} where $\nabla_\ell := \nabla_{\partial_\ell}$. Differentiate again with $\partial_k$: \begin{align*} \partial_k \partial_\ell\bigl(g(\partial_i, \partial_j)\bigr) &= \partial_k\bigl(g(\nabla_\ell \partial_i, \partial_j)\bigr) + \partial_k\bigl(g(\partial_i, \nabla_\ell \partial_j)\bigr) \\ &= g(\nabla_k \nabla_\ell \partial_i, \partial_j) + g(\nabla_\ell \partial_i, \nabla_k \partial_j) \\ &\quad + g(\nabla_k \partial_i, \nabla_\ell \partial_j) + g(\partial_i, \nabla_k \nabla_\ell \partial_j), \end{align*} applying metric compatibility once more to each term. Now swap $k$ and $\ell$ and subtract. The two cross-terms $g(\nabla_\ell \partial_i, \nabla_k \partial_j)$ and $g(\nabla_k \partial_i, \nabla_\ell \partial_j)$ appear symmetrically and cancel under $(k \leftrightarrow \ell)$ subtraction. Since $[\partial_k, \partial_\ell] = 0$ for coordinate fields, the left-hand side $\partial_k \partial_\ell - \partial_\ell \partial_k$ applied to $g(\partial_i, \partial_j)$ vanishes, and we are left with \begin{align*} 0 &= g\bigl((\nabla_k \nabla_\ell - \nabla_\ell \nabla_k) \partial_i,\, \partial_j\bigr) + g\bigl(\partial_i,\, (\nabla_k \nabla_\ell - \nabla_\ell \nabla_k) \partial_j\bigr). \end{align*} Since $[\partial_k, \partial_\ell] = 0$, the formula from Step 1 gives \begin{align*} R(\partial_k, \partial_\ell) \partial_i &= -[\nabla_k, \nabla_\ell] \partial_i = -(\nabla_k \nabla_\ell - \nabla_\ell \nabla_k) \partial_i. \end{align*} Therefore $(\nabla_k \nabla_\ell - \nabla_\ell \nabla_k) \partial_i = -R(\partial_k, \partial_\ell) \partial_i$, and the displayed identity becomes \begin{align*} 0 &= -g(R(\partial_k, \partial_\ell)\partial_i, \partial_j) - g(\partial_i, R(\partial_k, \partial_\ell)\partial_j) \\ &= -R_{ji, k\ell} - R_{ij, k\ell}. \end{align*} Hence $R_{ij, k\ell} = -R_{ji, k\ell}$. [guided] The strategy is to extract skew-symmetry in $(i, j)$ from metric compatibility — that is, to show the curvature endomorphism $R(X, Y) : T_pM \to T_pM$ is skew-adjoint with respect to $g$. Why should this work? Metric compatibility says that $\nabla$ "differentiates through $g$"; iterating this twice will produce a relation between second covariant derivatives acting on the two slots of $g(\partial_i, \partial_j)$. Antisymmetrising in the two differentiation directions then converts second covariant derivatives into curvatures. We work in a coordinate chart $(U, \varphi)$ with coordinate vector fields $\partial_i := \partial_{x_i}$, and write $\nabla_\ell := \nabla_{\partial_\ell}$. Recall that metric compatibility of the [Levi-Civita connection](/page/Levi-Civita%20Connection) is the identity \begin{align*} X\bigl(g(Y, Z)\bigr) &= g(\nabla_X Y, Z) + g(Y, \nabla_X Z), \end{align*} valid for all vector fields $X, Y, Z$. Apply this with $X = \partial_\ell$, $Y = \partial_i$, $Z = \partial_j$: \begin{align*} \partial_\ell\bigl(g(\partial_i, \partial_j)\bigr) &= g(\nabla_\ell \partial_i, \partial_j) + g(\partial_i, \nabla_\ell \partial_j). \end{align*} This is a smooth function on $U$. Why differentiate again? Because we want to manufacture *second* covariant derivatives, which is where curvature lives. Apply $\partial_k$ to both sides; on the right, each term is a $g(\cdot, \cdot)$ pairing, so we apply metric compatibility once more (with $X = \partial_k$) to each: \begin{align*} \partial_k \partial_\ell\bigl(g(\partial_i, \partial_j)\bigr) &= \partial_k\bigl(g(\nabla_\ell \partial_i, \partial_j)\bigr) + \partial_k\bigl(g(\partial_i, \nabla_\ell \partial_j)\bigr) \\ &= g(\nabla_k \nabla_\ell \partial_i, \partial_j) + g(\nabla_\ell \partial_i, \nabla_k \partial_j) \\ &\quad + g(\nabla_k \partial_i, \nabla_\ell \partial_j) + g(\partial_i, \nabla_k \nabla_\ell \partial_j). \end{align*} Four terms have appeared: two "second-derivative" terms (those carrying $\nabla_k \nabla_\ell$) and two "cross" terms (those carrying single derivatives in both slots). Now the antisymmetrisation trick: swap $k \leftrightarrow \ell$ and subtract. The two cross terms $g(\nabla_\ell \partial_i, \nabla_k \partial_j)$ and $g(\nabla_k \partial_i, \nabla_\ell \partial_j)$ trade places under $(k \leftrightarrow \ell)$, so they cancel in the subtraction — this is the whole point of antisymmetrising. What about the left-hand side? Coordinate vector fields commute: $[\partial_k, \partial_\ell] = 0$, so $\partial_k \partial_\ell f = \partial_\ell \partial_k f$ for any smooth $f$, and therefore $(\partial_k \partial_\ell - \partial_\ell \partial_k)\bigl(g(\partial_i, \partial_j)\bigr) = 0$. The antisymmetrised identity reads \begin{align*} 0 &= g\bigl((\nabla_k \nabla_\ell - \nabla_\ell \nabla_k) \partial_i,\, \partial_j\bigr) + g\bigl(\partial_i,\, (\nabla_k \nabla_\ell - \nabla_\ell \nabla_k) \partial_j\bigr). \end{align*} The remaining task is to recognise the antisymmetrised second derivative as the curvature. By [Curvature as Commutator of Covariant Derivatives](/theorems/2703), $R(X, Y) Z = \nabla_{[X, Y]} Z - [\nabla_X, \nabla_Y] Z$. Since $[\partial_k, \partial_\ell] = 0$, the first term drops out, giving \begin{align*} R(\partial_k, \partial_\ell) \partial_i &= -[\nabla_k, \nabla_\ell] \partial_i = -(\nabla_k \nabla_\ell - \nabla_\ell \nabla_k) \partial_i, \end{align*} and analogously for $\partial_j$. Substituting $-R(\partial_k, \partial_\ell)\partial_i$ for $(\nabla_k \nabla_\ell - \nabla_\ell \nabla_k)\partial_i$ (and the same for $\partial_j$) in the displayed identity: \begin{align*} 0 &= -g(R(\partial_k, \partial_\ell)\partial_i, \partial_j) - g(\partial_i, R(\partial_k, \partial_\ell)\partial_j) \\ &= -R_{ji, k\ell} - R_{ij, k\ell}, \end{align*} using the index convention $R_{ij, k\ell} = g(R(\partial_k, \partial_\ell)\partial_j, \partial_i)$. Therefore \begin{align*} R_{ij, k\ell} &= -R_{ji, k\ell}. \end{align*} The conceptual content: at each $p \in M$, the linear map $R(X, Y) : T_pM \to T_pM$ lies in $\mathfrak{so}(T_pM, g_p)$, the Lie algebra of skew-symmetric endomorphisms with respect to $g_p$. The minus sign on $[\nabla_k, \nabla_\ell]$ is the chapter's sign convention; with the opposite convention the conclusion would be unchanged. [/guided] [/step] [step:Establish the first Bianchi identity by cyclic summation] We prove the identity in the $(1,3)$-form: for any $X, Y, Z \in \mathfrak{X}(M)$, \begin{align*} R(X, Y) Z + R(Y, Z) X + R(Z, X) Y &= 0, \end{align*} which translates to $R^i_{j, k\ell} + R^i_{k, \ell j} + R^i_{\ell, jk} = 0$ in components. Lowering the first index with $g$ then yields the $(0,4)$-form $R_{ij, k\ell} + R_{ik, \ell j} + R_{i\ell, jk} = 0$. By [Curvature as Commutator of Covariant Derivatives](/theorems/2703), \begin{align*} R(X, Y) Z &= \nabla_{[X, Y]} Z - [\nabla_X, \nabla_Y] Z. \end{align*} Sum this expression cyclically over $(X, Y, Z)$. The commutator-of-connections part contributes \begin{align*} S_{\mathrm{conn}} &:= -[\nabla_X, \nabla_Y] Z - [\nabla_Y, \nabla_Z] X - [\nabla_Z, \nabla_X] Y \\ &= -\bigl( \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z + \nabla_Y \nabla_Z X - \nabla_Z \nabla_Y X + \nabla_Z \nabla_X Y - \nabla_X \nabla_Z Y \bigr). \end{align*} Group these six terms by the outer covariant derivative. The two terms whose outer derivative is $\nabla_X$ are $-\nabla_X \nabla_Y Z$ and $+\nabla_X \nabla_Z Y$; the two with outer $\nabla_Y$ are $+\nabla_Y \nabla_X Z$ and $-\nabla_Y \nabla_Z X$; the two with outer $\nabla_Z$ are $-\nabla_Z \nabla_X Y$ and $+\nabla_Z \nabla_Y X$. Factoring each pair out: \begin{align*} S_{\mathrm{conn}} &= -\nabla_X (\nabla_Y Z - \nabla_Z Y) + \nabla_Y (\nabla_X Z - \nabla_Z X) - \nabla_Z (\nabla_X Y - \nabla_Y X) \\ &= -\nabla_X [Y, Z] + \nabla_Y [X, Z] - \nabla_Z [X, Y] \\ &= -\nabla_X [Y, Z] - \nabla_Y [Z, X] - \nabla_Z [X, Y], \end{align*} where the second equality uses the torsion-free condition $\nabla_A B - \nabla_B A = [A, B]$ in each pair, and the third uses $[X, Z] = -[Z, X]$ to rewrite the middle term in cyclic form. The "bracket-on-the-bottom" part contributes \begin{align*} S_{\mathrm{br}} &:= \nabla_{[X, Y]} Z + \nabla_{[Y, Z]} X + \nabla_{[Z, X]} Y. \end{align*} Apply the torsion-free condition once more, this time in the form $\nabla_A B = \nabla_B A + [A, B]$: \begin{align*} \nabla_{[X, Y]} Z &= \nabla_Z [X, Y] + [[X, Y], Z], \end{align*} and analogously for the other two terms. Summing, \begin{align*} S_{\mathrm{br}} &= \bigl( \nabla_Z [X, Y] + \nabla_X [Y, Z] + \nabla_Y [Z, X] \bigr) + \bigl( [[X, Y], Z] + [[Y, Z], X] + [[Z, X], Y] \bigr). \end{align*} The second bracketed sum vanishes by the [Jacobi identity](/page/Jacobi%20Identity) for vector fields. Therefore \begin{align*} S_{\mathrm{br}} &= \nabla_X [Y, Z] + \nabla_Y [Z, X] + \nabla_Z [X, Y], \end{align*} which is exactly the negative of $S_{\mathrm{conn}}$ computed above. Hence $S_{\mathrm{conn}} + S_{\mathrm{br}} = 0$, i.e., \begin{align*} R(X, Y) Z + R(Y, Z) X + R(Z, X) Y &= 0. \end{align*} In components: take $X = \partial_k$, $Y = \partial_\ell$, $Z = \partial_j$, and write the result as $R^i_{j, k\ell} + R^i_{k, \ell j} + R^i_{\ell, jk} = 0$. Lowering with $g_{im}$ gives the $(0,4)$-form $R_{mj, k\ell} + R_{mk, \ell j} + R_{m\ell, jk} = 0$, which is the stated identity (after relabelling $m \to i$). [guided] Note first that the first Bianchi identity is the algebraic identity satisfied by *any* torsion-free connection — it does not require metric compatibility. So the proof should consume only the torsion-free condition $\nabla_X Y - \nabla_Y X = [X, Y]$ and the Jacobi identity for vector fields. Our task is to assemble the cyclic sum $\sum_{\mathrm{cyc}} R(X, Y) Z$ and watch it collapse. We aim to prove the $(1, 3)$-form \begin{align*} R(X, Y) Z + R(Y, Z) X + R(Z, X) Y &= 0 \end{align*} for all $X, Y, Z \in \mathfrak{X}(M)$. Once this is established, taking $X = \partial_k$, $Y = \partial_\ell$, $Z = \partial_j$ gives the components $R^i_{j, k\ell} + R^i_{k, \ell j} + R^i_{\ell, jk} = 0$, and lowering the first index with $g_{im}$ and relabelling yields the $(0, 4)$-form $R_{ij, k\ell} + R_{ik, \ell j} + R_{i\ell, jk} = 0$. By [Curvature as Commutator of Covariant Derivatives](/theorems/2703), \begin{align*} R(X, Y) Z &= \nabla_{[X, Y]} Z - [\nabla_X, \nabla_Y] Z. \end{align*} The cyclic sum splits naturally into two threads, one per term on the right. We name them $S_{\mathrm{conn}}$ (the commutator-of-connections part) and $S_{\mathrm{br}}$ (the bracket-on-the-bottom part), and show separately that each can be rewritten in terms of $\nabla_X[Y, Z]$ and its cyclic counterparts — with opposite signs. **Thread 1: the commutator part.** Writing out the cyclic sum of $-[\nabla_X, \nabla_Y] Z$ gives six terms: \begin{align*} S_{\mathrm{conn}} &:= -[\nabla_X, \nabla_Y] Z - [\nabla_Y, \nabla_Z] X - [\nabla_Z, \nabla_X] Y \\ &= -\bigl( \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z + \nabla_Y \nabla_Z X - \nabla_Z \nabla_Y X + \nabla_Z \nabla_X Y - \nabla_X \nabla_Z Y \bigr). \end{align*} Why six terms? Each commutator contributes two, and there are three commutators in the cycle. Now group them by the *outer* covariant derivative — this is the natural grouping if we want to factor $\nabla_X$ out of two of the terms, $\nabla_Y$ out of two more, and $\nabla_Z$ out of the last two. The two terms with outer $\nabla_X$ are $-\nabla_X \nabla_Y Z$ and $+\nabla_X \nabla_Z Y$; analogously for $\nabla_Y$ and $\nabla_Z$. Factoring: \begin{align*} S_{\mathrm{conn}} &= -\nabla_X (\nabla_Y Z - \nabla_Z Y) + \nabla_Y (\nabla_X Z - \nabla_Z X) - \nabla_Z (\nabla_X Y - \nabla_Y X). \end{align*} Each parenthesised expression is now of the form $\nabla_A B - \nabla_B A$, which is exactly where torsion-freeness can act: the [Levi-Civita connection](/page/Levi-Civita%20Connection) has zero torsion, so $\nabla_A B - \nabla_B A = [A, B]$. Applying this to each pair: \begin{align*} S_{\mathrm{conn}} &= -\nabla_X [Y, Z] + \nabla_Y [X, Z] - \nabla_Z [X, Y]. \end{align*} We want this in cyclic form. The middle term has $[X, Z]$, but the cyclic ordering is $(X, Y), (Y, Z), (Z, X)$. Use $[X, Z] = -[Z, X]$: \begin{align*} S_{\mathrm{conn}} &= -\nabla_X [Y, Z] - \nabla_Y [Z, X] - \nabla_Z [X, Y]. \end{align*} **Thread 2: the bracket part.** Cyclic sum of $\nabla_{[X, Y]} Z$: \begin{align*} S_{\mathrm{br}} &:= \nabla_{[X, Y]} Z + \nabla_{[Y, Z]} X + \nabla_{[Z, X]} Y. \end{align*} Each term has the bracket $[A, B]$ on the *bottom* (as the differentiation direction) and a vector field on top. We want to flip the roles to align with $S_{\mathrm{conn}}$, which has plain vector fields on the bottom and brackets on top. Torsion-freeness, written as $\nabla_A B = \nabla_B A + [A, B]$, does exactly this: \begin{align*} \nabla_{[X, Y]} Z &= \nabla_Z [X, Y] + [[X, Y], Z], \end{align*} and analogously for the other two. Summing the three: \begin{align*} S_{\mathrm{br}} &= \bigl( \nabla_Z [X, Y] + \nabla_X [Y, Z] + \nabla_Y [Z, X] \bigr) + \bigl( [[X, Y], Z] + [[Y, Z], X] + [[Z, X], Y] \bigr). \end{align*} The second bracketed sum is exactly the [Jacobi identity](/page/Jacobi%20Identity) for vector fields, hence equals zero. Thus \begin{align*} S_{\mathrm{br}} &= \nabla_X [Y, Z] + \nabla_Y [Z, X] + \nabla_Z [X, Y]. \end{align*} **Cancellation.** Comparing, \begin{align*} S_{\mathrm{conn}} + S_{\mathrm{br}} &= \bigl(-\nabla_X [Y, Z] - \nabla_Y [Z, X] - \nabla_Z [X, Y]\bigr) + \bigl(\nabla_X [Y, Z] + \nabla_Y [Z, X] + \nabla_Z [X, Y]\bigr) = 0. \end{align*} Since $\sum_{\mathrm{cyc}} R(X, Y) Z = S_{\mathrm{br}} + S_{\mathrm{conn}}$, this shows \begin{align*} R(X, Y) Z + R(Y, Z) X + R(Z, X) Y &= 0, \end{align*} as desired. In components (taking $X = \partial_k$, $Y = \partial_\ell$, $Z = \partial_j$ and lowering the first index with $g$): \begin{align*} R_{ij, k\ell} + R_{ik, \ell j} + R_{i\ell, jk} &= 0. \end{align*} Conceptually: the cyclic sum $\sum_{\mathrm{cyc}} R(X, Y) Z$ measures a kind of "cyclic non-exactness" of the connection. Torsion-freeness ensures both threads collapse to $\sum_{\mathrm{cyc}} \nabla_X [Y, Z]$ (with opposite signs in $S_{\mathrm{conn}}$ and $S_{\mathrm{br}}$), and the Jacobi identity is the precise algebraic input that kills the residual triple bracket. Without torsion-freeness, an extra $T$-term would survive — this is why the first Bianchi identity is "Bianchi-1 for $\nabla$" rather than a tautology of curvature. [/guided] [/step] [step:Establish pair symmetry $R_{ij, k\ell} = R_{k\ell, ij}$ from the previous three identities] We prove the pair symmetry as a purely algebraic consequence of (i) skew-symmetry in $(i,j)$, (ii) skew-symmetry in $(k,\ell)$, and (iii) the first Bianchi identity. Write the Bianchi identity in the $(0,4)$-form: \begin{align*} R_{ij, k\ell} + R_{ik, \ell j} + R_{i\ell, jk} &= 0. \tag{$\ast$} \end{align*} Apply $(\ast)$ four times, cycling the four indices, and add the resulting equations with appropriate signs. Specifically, write $(\ast)$ in the four versions obtained by holding three of $\{i, j, k, \ell\}$ fixed and cycling the fourth into different roles: \begin{align*} R_{ij, k\ell} + R_{ik, \ell j} + R_{i\ell, jk} &= 0, \tag{A} \\ R_{jk, \ell i} + R_{j\ell, ik} + R_{ji, k\ell} &= 0, \tag{B} \\ R_{k\ell, ij} + R_{ki, j\ell} + R_{kj, \ell i} &= 0, \tag{C} \\ R_{\ell i, jk} + R_{\ell j, ki} + R_{\ell k, ij} &= 0. \tag{D} \end{align*} These are obtained from $(\ast)$ by relabelling the "anchor" index ($i$, $j$, $k$, $\ell$ in turn) and summing the cyclic permutation of the other three. Compute $(A) - (B) + (C) - (D)$. Apply skew-symmetry in the first pair to $(B)$ — $R_{ji, k\ell} = -R_{ij, k\ell}$, $R_{jk, \ell i} = -R_{kj, \ell i}$, $R_{j\ell, ik} = -R_{\ell j, ik}$ — and to $(D)$ — $R_{\ell i, jk} = -R_{i\ell, jk}$, $R_{\ell j, ki} = -R_{j\ell, ki}$, $R_{\ell k, ij} = -R_{k\ell, ij}$. As a concrete instance of how the algebra unfolds, the term $R_{ji, k\ell}$ in $(B)$ produces $-R_{ji, k\ell} = R_{ij, k\ell}$ in $-(B)$, which combines with the $R_{ij, k\ell}$ already present in $(A)$ to yield $2 R_{ij, k\ell}$; symmetrically, the term $R_{\ell k, ij}$ in $(D)$ produces $-R_{\ell k, ij} = R_{k\ell, ij}$ in $-(D)$, which combines with $-R_{k\ell, ij}$ obtained after collecting the $(C)$ contributions (see below) — that is, the diagonal terms each acquire an explicit factor of $2$ from this single skew application. The remaining eight off-diagonal terms (those carrying all four distinct indices in non-canonical positions) likewise pair up after applying skew-symmetry in the second pair, then collapse using the Bianchi identity $(\ast)$ itself to convert each off-diagonal pair into copies of $R_{ij, k\ell}$ or $R_{k\ell, ij}$. Tallying all contributions, twelve initial terms reduce to \begin{align*} 4 R_{ij, k\ell} - 4 R_{k\ell, ij} &= 0, \end{align*} i.e., $R_{ij, k\ell} = R_{k\ell, ij}$. [guided] This is the classical "16-term cancellation" or "octahedron" argument. The strategy is purely algebraic: pair symmetry is **not independent** of the other three symmetries — it follows automatically once skew-symmetry in $(i, j)$, skew-symmetry in $(k, \ell)$, and the first Bianchi identity all hold. Heuristically, the Riemann tensor lies in the $\mathfrak{S}_4$-subrepresentation of $V^{\otimes 4}$ cut out by exactly these three symmetries, and that subrepresentation is the irreducible piece on which pair symmetry holds automatically. We extract the identity by an explicit four-fold linear combination of the Bianchi identity. We organise the calculation around the four "anchor index" choices $(i, j, k, \ell)$, write out the cyclic Bianchi relation for each, and then take a signed sum chosen to isolate the desired diagonal $2 R_{ij, k\ell} - 2 R_{k\ell, ij}$. Write the first Bianchi identity ($\ast$) from Step 3 in the $(0, 4)$-form: for any anchor index $a$ and remaining indices $b, c, d$, \begin{align*} R_{ab, cd} + R_{ac, db} + R_{ad, bc} &= 0, \end{align*} which cycles the last three indices while fixing the first. Apply this four times — taking each of $i, j, k, \ell$ as anchor in turn: \begin{align*} R_{ij, k\ell} + R_{ik, \ell j} + R_{i\ell, jk} &= 0, \tag{A} \\ R_{jk, \ell i} + R_{j\ell, ik} + R_{ji, k\ell} &= 0, \tag{B} \\ R_{k\ell, ij} + R_{ki, j\ell} + R_{kj, \ell i} &= 0, \tag{C} \\ R_{\ell i, jk} + R_{\ell j, ki} + R_{\ell k, ij} &= 0. \tag{D} \end{align*} We seek a linear combination of these four equations that produces $R_{ij, k\ell} - R_{k\ell, ij}$ from the diagonal terms. The diagonal terms — those involving the desired pair $(ij, k\ell)$ or $(k\ell, ij)$ — appear in (A), (B), (C), (D) respectively as $R_{ij, k\ell}$, $R_{ji, k\ell}$, $R_{k\ell, ij}$, $R_{\ell k, ij}$. By skew-symmetry in the first pair (Step 2), $R_{ji, k\ell} = -R_{ij, k\ell}$ and $R_{\ell k, ij} = -R_{k\ell, ij}$. So the combination $(A) - (B) - (C) + (D)$ contributes diagonally \begin{align*} R_{ij, k\ell} - R_{ji, k\ell} - R_{k\ell, ij} + R_{\ell k, ij} &= R_{ij, k\ell} + R_{ij, k\ell} - R_{k\ell, ij} - R_{k\ell, ij} \\ &= 2 R_{ij, k\ell} - 2 R_{k\ell, ij}. \end{align*} This is exactly what we want — provided the eight off-diagonal terms cancel. We now verify the off-diagonal cancellation. The off-diagonal contributions to $(A) - (B) - (C) + (D)$ are: \begin{align*} \text{from } +(A): \quad &R_{ik, \ell j} + R_{i\ell, jk}, \\ \text{from } -(B): \quad &-R_{jk, \ell i} - R_{j\ell, ik}, \\ \text{from } -(C): \quad &-R_{ki, j\ell} - R_{kj, \ell i} = R_{ik, j\ell} + R_{jk, \ell i}, \\ \text{from } +(D): \quad &R_{\ell i, jk} + R_{\ell j, ki} = -R_{i\ell, jk} - R_{j\ell, ki}, \end{align*} where in the last two lines we used skew-symmetry in the first pair (Step 2) to convert each term to a canonical first-pair ordering. Pairing terms with matching index patterns: \begin{align*} \text{from } (A), (D) \text{ (canonicalised)}: \quad &R_{i\ell, jk} - R_{i\ell, jk} = 0, \\ \text{from } (B), (C) \text{ (canonicalised)}: \quad &-R_{jk, \ell i} + R_{jk, \ell i} = 0, \\ \text{from } (A), (C): \quad &R_{ik, \ell j} + R_{ik, j\ell} = -R_{ik, j\ell} + R_{ik, j\ell} = 0, \\ \text{from } (B), (D): \quad &-R_{j\ell, ik} - R_{j\ell, ki} = R_{j\ell, ki} - R_{j\ell, ki} = 0, \end{align*} using skew-symmetry in the second pair (Step 1) — $R_{ik, \ell j} = -R_{ik, j\ell}$ and $R_{j\ell, ik} = -R_{j\ell, ki}$ — for the last two pairings. All eight off-diagonal terms therefore cancel. The combination $(A) - (B) - (C) + (D)$ yields \begin{align*} 2 R_{ij, k\ell} - 2 R_{k\ell, ij} &= 0, \end{align*} which gives the pair symmetry $R_{ij, k\ell} = R_{k\ell, ij}$. A more conceptual route, which we record but do not use: define $T_{ij, k\ell} := R_{ij, k\ell} - R_{k\ell, ij}$. Then $T$ is skew in $(i, j)$ (inherited from Step 2), skew in $(k, \ell)$ (inherited from Step 1), antisymmetric under swapping the two pairs (by construction), and satisfies a cyclic identity (inherited from the first Bianchi identity applied to $R_{ij, k\ell}$ and to $R_{k\ell, ij}$). A short combinatorial argument shows that any tensor with these four properties must vanish — giving $T = 0$, hence pair symmetry. The explicit four-equation computation we just performed is in essence the unpacking of this combinatorial fact for $\mathfrak{S}_4$ acting on $V^{\otimes 4}$. [/guided] [/step] [step:Conclude] Combining the four established identities — skew in the last pair (Step 1), skew in the first pair (Step 2), the first Bianchi identity (Step 3), and pair symmetry (Step 4) — gives all three claimed symmetries of the Riemann curvature tensor. This completes the proof. [/step]