# 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

 $\displaystyle(x,y,z,t)+(y,z,x,t)+(z,x,y,t)$ $\displaystyle=0\quad\text{First Bianchi identity}$ (1) $\displaystyle(x,y,z,t)+(y,x,z,t)$ $\displaystyle=0$ (2) $\displaystyle(x,y,z,t)-(z,t,x,y)$ $\displaystyle=0,$ (3)

where $(x,y,z,t):=g(R(x,y,z),t)$, and the sectional curvature is defined by

 $K(\Pi=\operatorname{span}\{x,y\})=\frac{g(R(x,y,x),y)}{g(x,x)g(y,y)-g(x,y)^{2}}.$ (4)

Thus Theorem 1 is implied by

###### Theorem 2.

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

• Equations (1),(2),(3), and

• $K(\sigma)=K^{\prime}(\sigma)$ for all $2$-planes $\sigma$, where $K,K^{\prime}$ are defined by (4) using $\langle-,-\rangle$ in of $g(-,-)$.

Then $R=R^{\prime}$.

Write

 $\displaystyle(x,y,z,t)$ $\displaystyle:=\langle R(x,y,z),t\rangle$ $\displaystyle(x,y,z,t)^{\prime}$ $\displaystyle:=\langle R^{\prime}(x,y,z),t\rangle.$
###### Proof of Theorem 2.

We need to prove, for all $x,y,z,t\in V$,

 $(x,y,z,t)=(x,y,z,t)^{\prime}.$

From $K=K^{\prime}$, we get $(x,y,x,y)=(x,y,x,y)^{\prime}$ 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 $(x+z,y,x+z,y)=(x+z,y,x+z,y)^{\prime}$ and use (3), we get

 $(x,y,x,y)+2(x,y,z,y)+(z,y,z,y)=(x,y,x,y)^{\prime}+2(x,y,z,y)^{\prime}+(z,y,z,y% )^{\prime}.$

So $(x,y,z,y)=(x,y,z,y)^{\prime}$ 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 $(x,y+t,z,y+t)=(x,y+t,z,y+t)^{\prime}$, we get

 $(x,y,z,t)+(x,t,z,y)=(x,y,z,t)^{\prime}+(x,t,z,y)^{\prime}.$

Now use (2) and (3), we get

 $\displaystyle(x,y,z,t)-(x,y,z,t)^{\prime}$ $\displaystyle=(x,t,z,y)^{\prime}-(x,t,z,y)$ $\displaystyle=(z,y,x,t)^{\prime}-(z,y,x,t)$ $\displaystyle=(y,z,x,t)-(y,z,x,t)^{\prime}.$

So $(x,y,z,t)-(x,y,z,t)^{\prime}$ is invariant under cyclic permutation of $x,y,z$. But the cyclic sum is zero by (1). So

 $(x,y,z,t)=(x,y,z,t)^{\prime}\quad\forall x,y,z,t\in V$

as desired. ∎

Title sectional curvature determines Riemann curvature tensor SectionalCurvatureDeterminesRiemannCurvatureTensor 2013-03-22 15:55:09 2013-03-22 15:55:09 kerwinhui (11200) kerwinhui (11200) 10 kerwinhui (11200) Theorem msc 53B21 msc 53B20