Sectional curvatures determine the Riemann tensor

The sectional quotient is intrinsic to a two-plane using only the given pair skews. Indeed, if its ordered basis $(X,Y)$ is changed by a matrix $A\in {\mathrm{G}\mathrm{L}}_2(\mathbb{R})$ with determinant $d$, the alternating-pair numerator $R(X,Y,Y,X)$ becomes $d^{2}R(X,Y,Y,X)$, while the Gram determinant becomes $\det (AG{A}^{\mathsf{T}})=d^2\det G$. Thus the quotient is basis-independent. For independent $X,Y$, equality of the two quotients and positivity of the common Gram denominator give $T(X,Y,Y,X)=0$. For dependent $X,Y$, the same equality follows from first-pair skewness.

Expanding $0=T(X+Y,Z,Z,X+Y)$ and using step 1.1 removes both diagonal terms. Pair interchange followed by the two pair skews identifies the two cross terms, so $2T(X,Z,Z,Y)=0$. Hence $T(X,Z,Z,Y)=0$ for all $X,Y,Z$.

Polarize step 2.1 in $Z$: expanding $0=T(X,Z+W,Z+W,Y)$ leaves $T(X,Z,W,Y)+T(X,W,Z,Y)=0$, so $T$ is also skew in its two middle slots. The Bianchi identity now gives $0=T(X,Y,Z,W)-T(Y,X,Z,W)-T(X,Z,Y,W)=3T(X,Y,Z,W)$ by first-pair and middle-slot skewness. Therefore $T=0$ and $R_1=R_2$.