Theorem 1   The sectional curvature operator $\Pi\mapsto K(\Pi)$ completely determines the Riemann curvature tensor.

Theorem 2   Let $V$ be a real inner product space, with inner product $\langle-,-\rangle$ . Let $R$ and $R'$ be linear maps $V^{\otimes 3}\to V$ . Suppose $R$ and $R'$ satisfies

• Equations (1),(2),(3), and
• $K(\sigma)=K'(\sigma)$ for all $2$ -planes $\sigma$ , where $K,K'$ are defined by () using $\langle -,-\rangle$ in place of $g(-,-)$ .

Then $R=R'$ .

Proof. [Proof of Theorem 2] We need to prove, for all $x,y,z,t\in V$ , $$ (x,y,z,t)=(x,y,z,t)'. $$

From $K=K'$ , we get $(x,y,x,y)=(x,y,x,y)'$ for all $x,y\in V$ . The first step is to use polarization identity to change this quadratic form (in $x$ ) into its associated symmetric bilinear form. Expand $(x+z,y,x+z,y)=(x+z,y,x+z,y)'$ and use (3), we get $$ (x,y,x,y)+2(x,y,z,y)+(z,y,z,y)=(x,y,x,y)'+2(x,y,z,y)'+(z,y,z,y)'. $$ So $(x,y,z,y)=(x,y,z,y)'$ for all $x,y,z\in V$ .

Unfortunately, the form $(x,y,z,t)$ is not symmetric in $y$ and $t$ , so we need to work harder. Expand $(x,y+t,z,y+t)=(x,y+t,z,y+t)'$ , we get $$ (x,y,z,t)+(x,t,z,y)=(x,y,z,t)'+(x,t,z,y)'. $$ Now use (2) and (3), we get

So $(x,y,z,t)-(x,y,z,t)'$ is invariant under cyclic permutation of $x,y,z$ . But the cyclic sum is zero by (1). So $$ (x,y,z,t)=(x,y,z,t)'\quad\forall x,y,z,t\in V $$ as desired.