[proofplan] The proof is purely linear-algebraic and exploits the symmetries of a Riemann-type curvature tensor. Set $D := R' - R''$. Then $D$ is multilinear, satisfies all the symmetries (skew in the first pair, skew in the second pair, pair symmetry, and first Bianchi), and the hypothesis on sectional curvatures says $D(X, Y, X, Y) = 0$ for all $X, Y \in V$ (this uses pair symmetry to identify the sectional-curvature numerator with $D(X,Y,X,Y)$). The strategy is then to *polarise*: replace $X$ with $X + Z$, expand using multilinearity, cancel terms that vanish from $D(\cdot, Y, \cdot, Y) = 0$, and use the symmetries of $D$ to extract a smaller relation. Iterating the polarisation in the third slot (and then in the fourth) reduces to an identity of the form $3D(X, Y, Z, W) = 0$ when combined with the first Bianchi identity. We conclude $D = 0$, hence $R' = R''$. [/proofplan]

[step:Reduce to showing the difference $D := R' - R''$ vanishes] Define \begin{align*} D : V^{\otimes 4} &\to \mathbb{R}, \\ (X, Y, Z, W) &\mapsto R'(X, Y, Z, W) - R''(X, Y, Z, W). \end{align*} $D$ is multilinear (difference of multilinear maps) and satisfies all four symmetries of the Riemann curvature tensor (each symmetry holds for $R'$ and $R''$ separately, hence for their difference): \begin{align*} D(X, Y, Z, W) &= -D(Y, X, Z, W), \tag{S1} \\ D(X, Y, Z, W) &= -D(X, Y, W, Z), \tag{S2} \\ D(X, Y, Z, W) &= D(Z, W, X, Y), \tag{S3} \\ D(X, Y, Z, W) + D(X, Z, W, Y) + D(X, W, Y, Z) &= 0. \tag{B} \end{align*} We must show $D \equiv 0$. [/step]

[step:Translate the sectional-curvature hypothesis into $D(X, Y, X, Y) = 0$] The sectional curvature determined by a tensor $\tilde R$ satisfying the four symmetries above is, for a $2$-plane $\sigma = \mathrm{span}(X, Y) \subseteq V$ with $X, Y$ linearly independent, \begin{align*} K_{\tilde R}(\sigma) &= \frac{\tilde R(X, Y, X, Y)}{|X|^2 |Y|^2 - \langle X, Y \rangle^2}, \end{align*} where $\langle\cdot, \cdot\rangle$ is any inner product on $V$ used to interpret "sectional curvature". (The formula in the introduction of the chapter uses the metric on $V$; here we are working purely algebraically and the inner product is fixed throughout.) By hypothesis $K_{R'}(\sigma) = K_{R''}(\sigma)$ for all $2$-planes $\sigma$. The denominator is the same, so the numerators agree: \begin{align*} R'(X, Y, X, Y) &= R''(X, Y, X, Y) \qquad \text{for all linearly independent } X, Y. \end{align*} Both sides extend to all of $V \times V$ by multilinearity, and the equality continues to hold when $X, Y$ are linearly dependent (both sides vanish by $(\mathrm{S1})$ when $X = \lambda Y$). Therefore \begin{align*} D(X, Y, X, Y) &= 0 \qquad \text{for all } X, Y \in V. \tag{H} \end{align*} [/step]

[step:First polarisation: replace $X \to X + Z$ to eliminate the diagonal]Apply $(\mathrm{H})$ with $X$ replaced by $X + Z$ (and $Y$ unchanged): \begin{align*} 0 &= D(X + Z, Y, X + Z, Y). \end{align*} Expanding by multilinearity in the first and third slots, \begin{align*} 0 &= D(X, Y, X, Y) + D(X, Y, Z, Y) + D(Z, Y, X, Y) + D(Z, Y, Z, Y). \end{align*} The terms $D(X, Y, X, Y)$ and $D(Z, Y, Z, Y)$ vanish by $(\mathrm{H})$. Hence \begin{align*} 0 &= D(X, Y, Z, Y) + D(Z, Y, X, Y). \end{align*} By pair symmetry $(\mathrm{S3})$, $D(Z, Y, X, Y) = D(X, Y, Z, Y)$. Substituting, \begin{align*} 0 &= 2\, D(X, Y, Z, Y), \end{align*} so \begin{align*} D(X, Y, Z, Y) &= 0 \qquad \text{for all } X, Y, Z \in V. \tag{H'} \end{align*}[/step]

[guided]We have the identity $(\mathrm{H})$: $D(X, Y, X, Y) = 0$ for all $X, Y \in V$. The function $X \mapsto D(X, Y, X, Y)$ is "quadratic" in $X$ (it has $X$ in two slots, with $Y$ held fixed). How do we extract a bilinear identity in $(X, Z)$ from a quadratic-in-$X$ identity? The standard tool is **polarisation**: substitute $X + Z$ for $X$, then the diagonal terms $D(X, Y, X, Y)$ and $D(Z, Y, Z, Y)$ vanish by hypothesis and only the cross terms survive. Apply $(\mathrm{H})$ with $X$ replaced by $X + Z$ (and $Y$ left alone — we are polarising in the first/third slots, not the second/fourth): \begin{align*} 0 &= D(X + Z, Y, X + Z, Y). \end{align*} Now expand by multilinearity. $D$ is multilinear in each of its four slots, so we expand in slot 1 (replacing $X+Z$ by $X+Z$) and in slot 3 (replacing $X+Z$ by $X+Z$). Each slot gives two terms, and distributing the four products yields four mixed terms: \begin{align*} 0 &= D(X, Y, X, Y) + D(X, Y, Z, Y) + D(Z, Y, X, Y) + D(Z, Y, Z, Y). \end{align*} The first and last terms — the "diagonal" terms in which the same vector ($X$ or $Z$) appears in both slot 1 and slot 3 — vanish by $(\mathrm{H})$. We are left with only the off-diagonal mixed terms: \begin{align*} 0 &= D(X, Y, Z, Y) + D(Z, Y, X, Y). \end{align*} These two terms are not obviously equal. To merge them, we invoke the pair-symmetry $(\mathrm{S3})$: $D(A, B, C, D) = D(C, D, A, B)$. Applied with $A = Z$, $B = Y$, $C = X$, $D = Y$ this gives $D(Z, Y, X, Y) = D(X, Y, Z, Y)$. (Why does this work? Because $(\mathrm{S3})$ swaps the first pair of slots with the second pair as a block — and after the swap, both pairs read $(X, Y)$ and $(Z, Y)$ in the same order they did before, just exchanged.) Substituting, \begin{align*} 0 &= 2\, D(X, Y, Z, Y), \end{align*} and since we are working over $\mathbb{R}$ (characteristic $0$, in particular $2 \neq 0$), we may divide by $2$ to obtain \begin{align*} D(X, Y, Z, Y) &= 0 \qquad \text{for all } X, Y, Z \in V. \tag{H'} \end{align*} The takeaway: the diagonal restriction $D(X, Y, X, Y) = 0$ was actually a symptom of the stronger statement that the bilinear form $(X, Z) \mapsto D(X, Y, Z, Y)$ vanishes identically. This is the standard payoff of polarisation — a quadratic vanishing implies the corresponding bilinear vanishing (over a field where $2$ is invertible). $(\mathrm{H}')$ is the first reduction toward proving $D \equiv 0$.[/guided]

[step:Second polarisation: replace $Y \to Y + W$ to extract the full tensor up to a Bianchi correction] Apply $(\mathrm{H}')$ with $Y$ replaced by $Y + W$: \begin{align*} 0 &= D(X, Y + W, Z, Y + W). \end{align*} Expanding by multilinearity in the second and fourth slots, \begin{align*} 0 &= D(X, Y, Z, Y) + D(X, Y, Z, W) + D(X, W, Z, Y) + D(X, W, Z, W). \end{align*} The terms $D(X, Y, Z, Y)$ and $D(X, W, Z, W)$ vanish by $(\mathrm{H}')$. Hence \begin{align*} D(X, Y, Z, W) + D(X, W, Z, Y) &= 0, \end{align*} i.e., \begin{align*} D(X, Y, Z, W) &= -D(X, W, Z, Y). \tag{$\dagger$} \end{align*} [/step]

[step:Combine $(\dagger)$ with the first Bianchi identity to conclude $D \equiv 0$]The first Bianchi identity $(\mathrm{B})$ for $D$ reads \begin{align*} D(X, Y, Z, W) + D(X, Z, W, Y) + D(X, W, Y, Z) &= 0. \end{align*} Apply $(\dagger)$ to convert the second and third terms. For the second: \begin{align*} D(X, Z, W, Y) &= -D(X, Y, W, Z) \qquad \text{(apply } (\dagger) \text{ with } Y \to Z, Z \to W, W \to Y\text{)} \\ &= D(X, Y, Z, W) \qquad \text{(apply } (\mathrm{S2})\text{)}. \end{align*} For the third: \begin{align*} D(X, W, Y, Z) &= -D(X, Z, Y, W) \qquad \text{(apply } (\dagger) \text{ with } Y \to W, Z \to Y, W \to Z\text{)} \\ &= D(X, Z, W, Y) \qquad \text{(apply } (\mathrm{S2})\text{)} \\ &= D(X, Y, Z, W) \qquad \text{(by the previous line)}. \end{align*} Substituting into the Bianchi identity, \begin{align*} 0 &= D(X, Y, Z, W) + D(X, Y, Z, W) + D(X, Y, Z, W) = 3\, D(X, Y, Z, W). \end{align*} Hence $D(X, Y, Z, W) = 0$ for all $X, Y, Z, W \in V$.[/step]

[guided]At this point we have two relations on $D$: the first Bianchi identity $(\mathrm{B})$ (a *built-in* symmetry of any curvature-type tensor) and the new relation $(\dagger): D(X, Y, Z, W) = -D(X, W, Z, Y)$ (a *consequence* of the sectional-curvature hypothesis). The plan is to combine them: write down $(\mathrm{B})$, use $(\dagger)$ together with the standard symmetries to rewrite each summand as a multiple of $D(X, Y, Z, W)$ itself, and observe that the result is $3 D(X, Y, Z, W) = 0$. The first Bianchi identity $(\mathrm{B})$ for $D$ reads \begin{align*} D(X, Y, Z, W) + D(X, Z, W, Y) + D(X, W, Y, Z) &= 0. \end{align*} The first summand is already $D(X, Y, Z, W)$ — what we are trying to extract. We must rewrite the second and third summands. **Rewriting the second summand $D(X, Z, W, Y)$.** We want to apply $(\dagger)$, which schematically reads "swapping the second and fourth slots flips sign". Compare $D(X, Z, W, Y)$ to $(\dagger)$'s left-hand side $D(X, Y, Z, W)$ — to match the patterns, relabel $(\dagger)$ via the substitution $Y \mapsto Z$, $Z \mapsto W$, $W \mapsto Y$: \begin{align*} D(X, Z, W, Y) &= -D(X, Y, W, Z) \qquad \text{(}(\dagger)\text{ with } Y \to Z, Z \to W, W \to Y\text{)}. \end{align*} Now we have $D(X, Y, W, Z)$, which differs from our target $D(X, Y, Z, W)$ by a swap of slots 3 and 4. That swap is exactly the second standard symmetry $(\mathrm{S2})$, which says $D$ is skew in the second pair of slots: \begin{align*} -D(X, Y, W, Z) &= D(X, Y, Z, W) \qquad \text{(by }(\mathrm{S2})\text{)}. \end{align*} Combining the two, \begin{align*} D(X, Z, W, Y) &= D(X, Y, Z, W). \end{align*} **Rewriting the third summand $D(X, W, Y, Z)$.** Apply $(\dagger)$ again, this time with the substitution $Y \mapsto W$, $Z \mapsto Y$, $W \mapsto Z$ to match the pattern $D(X, ?, ?, ?)$ to the target slots: \begin{align*} D(X, W, Y, Z) &= -D(X, Z, Y, W) \qquad \text{(}(\dagger)\text{ with } Y \to W, Z \to Y, W \to Z\text{)}. \end{align*} Now apply $(\mathrm{S2})$ to swap the third and fourth slots of $D(X, Z, Y, W)$: \begin{align*} -D(X, Z, Y, W) &= D(X, Z, W, Y) \qquad \text{(by }(\mathrm{S2})\text{)}. \end{align*} But we just computed $D(X, Z, W, Y) = D(X, Y, Z, W)$ in the previous step, so chaining, \begin{align*} D(X, W, Y, Z) &= D(X, Y, Z, W). \end{align*} **Substituting into Bianchi.** Both rewritten summands equal $D(X, Y, Z, W)$, so the Bianchi identity becomes \begin{align*} 0 &= D(X, Y, Z, W) + D(X, Y, Z, W) + D(X, Y, Z, W) = 3\, D(X, Y, Z, W). \end{align*} We are working over $\mathbb{R}$, where $3 \neq 0$, so dividing, \begin{align*} D(X, Y, Z, W) &= 0 \qquad \text{for all } X, Y, Z, W \in V. \end{align*} The combinatorial heart of the argument: the Bianchi identity is a sum over a cyclic permutation of three slots. The relation $(\dagger)$, combined with $(\mathrm{S2})$, lets us collapse each cyclic summand back into the single tensor value $D(X, Y, Z, W)$. The cyclic sum thus becomes $3 D(X, Y, Z, W)$, and the condition $3 \neq 0$ in our base field forces $D = 0$. The role of characteristic: over $\mathbb{F}_3$ the final division by $3$ would be illegal — and indeed in characteristic $3$, sectional curvature does not determine the curvature tensor. Over $\mathbb{R}$ (or any field of characteristic neither $2$ nor $3$, since we also divided by $2$ in the previous step), the argument goes through.[/guided]

[step:Conclude] By Step 5, $D \equiv 0$ as a multilinear map $V^{\otimes 4} \to \mathbb{R}$. Therefore $R' = R''$ on all of $V^{\otimes 4}$. This completes the proof. [/step]