sectional curvature determines Riemann curvature tensor

Theorem 1.

The sectional curvature operator $\mathrm{\Pi }\mathrm{\mapsto }K\mathit{}\mathrm{(}\mathrm{\Pi }\mathrm{)}$ completely determines the Riemann curvature tensor.

In fact, a more general result is true. Recall the Riemann $(1,3)$-curvature tensor $R:TM\otimes TM\otimes TM\to TM$ satisfies

$(x,y,z,t)+(y,z,x,t)+(z,x,y,t)$$=0\mathit{\hspace{1em}}\text{First Bianchi identity}$(1)

$(x,y,z,t)+(y,x,z,t)$$=0$(2)

$(x,y,z,t)-(z,t,x,y)$$=0,$(3)

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

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

Thus Theorem 1 is implied by

Theorem 2.

Let $V$ be a real inner product space, with inner product $\mathrm{⟨}\mathrm{-}\mathrm{,}\mathrm{-}\mathrm{⟩}$. Let $R$ and $R^{\mathrm{\prime }}$ be linear maps $V^{\mathrm{\otimes }\mathrm{3}}\mathrm{\to }V$. Suppose $R$ and $R^{\mathrm{\prime }}$ 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 $⟨-,-⟩$ in of $g(-,-)$.

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

Write

$(x,y,z,t)$$:=⟨R(x,y,z),t⟩$

${(x,y,z,t)}^{\prime }$$:=⟨R^{\prime }(x,y,z),t⟩.$

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

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

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

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

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

$$(x,y,z,t)={(x,y,z,t)}^{\prime }\mathit{\hspace{1em}}\forall x,y,z,t\in V$$

as desired.∎