|
Revision difference : sectional curvature determines Riemann curvature tensor |
| Version 4 |
Version 3 |
| \begin{theorem}\label{thm:thm1} |
\begin{theorem}\label{thm:thm1} |
| The sectional curvature operator $\Pi\mapsto K(\Pi)$ completely determines the Riemann curvature tensor. |
The sectional curvature operator $\Pi\mapsto K(\Pi)$ completely determines the Riemann curvature tensor. |
| \end{theorem} |
\end{theorem} |
|
|
| 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 |
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 |
| \begin{align} |
\begin{align} |
| (x,y,z,t)+(y,z,x,t)+(z,x,y,t)&=0\quad\text{First Bianchi identity}\label{eqn:bianchi}\\ |
(x,y,z,t)+(y,z,x,t)+(z,x,y,t)&=0\quad\text{First Bianchi identity}\label{eqn:bianchi}\\ |
| (x,y,z,t)+(y,x,z,t)&=0\label{eqn:anti}\\ |
(x,y,z,t)+(y,x,z,t)&=0\label{eqn:anti}\\ |
| (x,y,z,t)-(z,t,x,y)&=0,\label{eqn:symm} |
(x,y,z,t)-(z,t,x,y)&=0,\label{eqn:symm} |
| \end{align} |
\end{align} |
| where $(x,y,z,t):=g(R(x,y,z),t)$, and the sectional curvature is defined by |
where $(x,y,z,t):=g(R(x,y,z),t)$, and the sectional curvature is defined by |
| \begin{equation} |
\begin{equation} |
|
K(\Pi=\SPAN\{x,y\})=\frac{g(R(x,y,x),y)}{g(x,x)g(y,y)-g(x,y)^2}.\label{eqn:sectcurv}
|
K(\PI=\SPAN\{x,y\})=\frac{g(R(x,y,x),y)}{g(x,x)g(y,y)-g(x,y)^2}.\label{eqn:sectcurv}
|
| \end{equation} |
\end{equation} |
| Thus Theorem~\ref{thm:thm1} is implied by |
Thus Theorem~\ref{thm:thm1} is implied by |
|
|
| \begin{theorem}\label{thm:thm2} |
\begin{theorem}\label{thm:thm2} |
| 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 |
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 |
| \begin{itemize} |
\begin{itemize} |
| \item Equations \eqref{eqn:bianchi},\eqref{eqn:anti},\eqref{eqn:symm}, and |
\item Equations \eqref{eqn:bianchi},\eqref{eqn:anti},\eqref{eqn:symm}, and |
| \item $K(\sigma)=K'(\sigma)$ for all $2$-planes $\sigma$, where $K,K'$ are defined by \eqref{eqn:sectcurv} using $\langle -,-\rangle$ in \PMlinkescapetext{place} of $g(-,-)$. |
\item $K(\sigma)=K'(\sigma)$ for all $2$-planes $\sigma$, where $K,K'$ are defined by \eqref{eqn:sectcurv} using $\langle -,-\rangle$ in \PMlinkescapetext{place} of $g(-,-)$. |
| \end{itemize} |
\end{itemize} |
| Then $R=R'$. |
Then $R=R'$. |
| \end{theorem} |
\end{theorem} |
|
|
| Write |
Write |
| \begin{align*} |
\begin{align*} |
| (x,y,z,t)&:=\langle R(x,y,z),t\rangle\\ |
(x,y,z,t)&:=\langle R(x,y,z),t\rangle\\ |
| (x,y,z,t)'&:=\langle R'(x,y,z),t\rangle. |
(x,y,z,t)'&:=\langle R'(x,y,z),t\rangle. |
| \end{align*} |
\end{align*} |
|
|
| \begin{proof}[Proof of Theorem~\ref{thm:thm2}] |
\begin{proof}[Proof of Theorem~\ref{thm:thm2}] |
| We need to prove, for all $x,y,z,t\in V$, |
We need to prove, for all $x,y,z,t\in V$, |
| \[ |
\[ |
| (x,y,z,t)=(x,y,z,t)'. |
(x,y,z,t)=(x,y,z,t)'. |
| \] |
\] |
|
|
| From $K=K'$, we get $(x,y,x,y)=(x,y,x,y)'$ for all $x,y\in V$. Expand $(x+z,y,x+z,y)=(x+z,y,x+z,y)'$ and use \eqref{eqn:symm}, we get |
From $K=K'$, we get $(x,y,x,y)=(x,y,x,y)'$ for all $x,y\in V$. Expand $(x+z,y,x+z,y)=(x+z,y,x+z,y)'$ and use \eqref{eqn:symm}, we get |
| \[ |
\[ |
| (x,y,x,y)+2(x,y,z,y)+(z,y,z,y)=(x,y,x,y)'+2(x,y,z,y)'+(z,y,z,y)'. |
(x,y,x,y)+2(x,y,z,y)+(z,y,z,y)=(x,y,x,y)'+2(x,y,z,y)'+(z,y,z,y)'. |
| \] |
\] |
| So $(x,y,z,y)=(x,y,z,y)'$ for all $x,y,z\in V$. |
So $(x,y,z,y)=(x,y,z,y)'$ for all $x,y,z\in V$. |
|
|
| Expand $(x,y+t,z,y+t)=(x,y+t,z,y+t)'$, we get |
Expand $(x,y+t,z,y+t)=(x,y+t,z,y+t)'$, we get |
| \[ |
\[ |
| (x,y,z,t)+(x,y,z,y)=(x,y,z,t)'+(x,t,z,y)'. |
(x,y,z,t)+(x,y,z,y)=(x,y,z,t)'+(x,t,z,y)'. |
| \] |
\] |
| Now use \eqref{eqn:anti} and \eqref{eqn:symm}, we get |
Now use \eqref{eqn:anti} and \eqref{eqn:symm}, we get |
| \begin{align*} |
\begin{align*} |
| (x,y,z,t)-(x,y,z,t)'&=(x,t,z,y)'-(x,t,z,y)\\ |
(x,y,z,t)-(x,y,z,t)'&=(x,t,z,y)'-(x,t,z,y)\\ |
| &=(z,y,x,t)'-(z,y,x,t)\\ |
&=(z,y,x,t)'-(z,y,x,t)\\ |
| &=(y,z,x,t)-(y,z,x,t)'. |
&=(y,z,x,t)-(y,z,x,t)'. |
| \end{align*} |
\end{align*} |
| 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 \eqref{eqn:bianchi}. So |
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 \eqref{eqn:bianchi}. So |
| \[ |
\[ |
| (x,y,z,t)=(x,y,z,t)'\quad\forall x,y,z,t\in V |
(x,y,z,t)=(x,y,z,t)'\quad\forall x,y,z,t\in V |
| \] |
\] |
| as desired. |
as desired. |
| \end{proof} |
\end{proof} |
|
|
|
