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sectional curvature determines Riemann curvature tensor
Theorem 1 The sectional curvature operator $\Pi\mapsto K(\Pi)$ completely determines the Riemann curvature tensor.
In fact, a more general result is true. Recall the Riemann $(1,3)$ -curvature tensor $R\colon TM\otimes TM\otimes TM\to TM$ satisfies
First Bianchi identity
where $(x,y,z,t):=g(R(x,y,z),t)$ , and the sectional curvature is defined by
Thus Theorem
![[*]](http://images.planetmath.org/cache/objects/7924/js//usr/share/latex2html/icons/crossref.png "[*]"){kind=icon}
is implied by
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 (),(),(), and
-
$K(\sigma)=K'(\sigma)$ for all $2$ -planes $\sigma$ , where $K,K'$ are defined by () using $\langle -,-\rangle$ in place of $g(-,-)$ .
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Proof. [Proof of Theorem ] We need to prove, for all $x,y,z,t\in V$ ,$$ (x,y,z,t)=(x,y,z,t)'.$$
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 (). So$$ (x,y,z,t)=(x,y,z,t)'\quad\forall x,y,z,t\in V$$ as desired. 
] 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 (), 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 () and (), we get
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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 (
). So$$ (x,y,z,t)=(x,y,z,t)'\quad\forall x,y,z,t\in V$$ as desired.
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