Symmetries of Riemannian curvature

hbghlyj

cis.upenn.edu/~cis6100/cis610-18-sl14.pdf
Proposition 4.6. Let $\bar{R}$ be such that it has the same properties as $R$ given in Proposition 4.3. Suppose that for all $p \in M$ and for all 2-dimensional subspaces $\sigma=\operatorname{Span}\{X, Y\} \subseteq T_p M$ we have that
\[
\bar{K}(\sigma)=\frac{\bar{R}(X, Y, Y, X)}{g(X, X) g(Y, Y)-g(X, Y)^2}=K(\sigma)
\]
Then $R=\bar{R}$.
This result can be paraphrased as "the sectional curvature determines the Riemann curvature".
Proposition 8.1. A Riemannian manifold $(M, g)$ has constant sectional curvature $K$ if and only if for all vector fields $X, Y, Z, W$ on $M$
\[
R(X, Y, Z, W)=K(g(X, W) g(Y, Z)-g(X, Z) g(Y, W))
\]
Proof. Suppose that $(M, g)$ has constant sectional curvature $K$. Define
\[
\bar{R}(X, Y, Z, W)=K(g(X, W) g(Y, Z)-g(X, Z) g(Y, W))
\]
Then
\[
R(X, Y, Y, X)=K(g(X, X) g(Y, Y)-g(X, Y)^2)=\bar{R}(X, Y, Y, X)
\]
How to check $\bar{R}$ has the same symmetries as $R$? Proposition 4.6 implies that $R=\bar{R}$.

hbghlyj

To apply Proposition 4.6, we must verify that the tensor
$$\bar{R}(X,Y,Z,W)=K(g(X,W)g(Y,Z)-g(X,Z)g(Y,W))
$$enjoys the same algebraic symmetries as an arbitrary Riemann curvature tensor $R$.  These are:

• Skew‐symmetry in the first two arguments: $\bar{R}(X,Y,Z,W)=-\bar{R}(Y,X,Z,W)$.
• Skew‐symmetry in the last two arguments: $\bar{R}(X,Y,Z,W)=-\bar{R}(X,Y,W,Z)$.
• Pair symmetry (interchange symmetry): $\bar{R}(X,Y,Z,W)=\bar{R}(Z,W,X,Y)$.
• First (algebraic) Bianchi identity: $\bar{R}(X,Y,Z,W)+\bar{R}(Y,Z,X,W)+\bar{R}(Z,X,Y,W)=0$.

Because each of these properties follows directly from the symmetries of the metric $g$ and the bilinearity of the expression $g(X,W)\,g(Y,Z)-g(X,Z)\,g(Y,W)$, it matches all the required symmetries of $R$.  By Proposition 4.6, having identical values on all sectional curvatures and sharing the same algebraic symmetries forces $\bar R=R$, completing the proof of Proposition 8.1.

1. Skew‐symmetry in the first two arguments
For any tensor $T$ of type $(0,4)$, skew‐symmetry in the first two slots means
$$T(X,Y,Z,W) = -\,T(Y,X,Z,W)\quad\Longleftrightarrow\quad T_{abcd}=-T_{bacd}.
$$The candidate $\bar R$ is
$$\bar R(X,Y,Z,W)
=K\bigl(g(X,W)g(Y,Z)-g(X,Z)g(Y,W)\bigr).
$$Swapping $X\leftrightarrow Y$ changes
$$g(X,W)g(Y,Z)-g(X,Z)g(Y,W)
\;\mapsto\;
g(Y,W)g(X,Z)-g(Y,Z)g(X,W)
= -\bigl[g(X,W)g(Y,Z)-g(X,Z)g(Y,W)\bigr],
$$so $\bar R(Y,X,Z,W)=-\bar R(X,Y,Z,W)$.

2. Skew‐symmetry in the last two arguments
Similarly, exchanging $Z\leftrightarrow W$ in the bracket
$$g(X,W)g(Y,Z)-g(X,Z)g(Y,W)$$introduces a minus sign by the same reasoning, giving
$\bar R(X,Y,W,Z)=-\bar R(X,Y,Z,W)$.

3. Pair (Interchange) Symmetry
Interchange symmetry asserts
$$R(X,Y,Z,W)=R(Z,W,X,Y)\quad\Longleftrightarrow\quad R_{abcd}=R_{cdab}.
$$Since each term $g(X,W)g(Y,Z)$ becomes $g(Z,Y)g(W,X)=g(Y,Z)g(X,W)$ when swapping $(X,Y)\leftrightarrow(Z,W)$, and similarly for the other term, the whole expression is invariant under exchanging the first pair $(X,Y)$ with the second pair $(Z,W)$.

4. Bianchi Identity
Write out the three terms
$$
\bar R(X,Y,Z,W)
=K[g(X,W)\,g(Y,Z)-g(X,Z)\,g(Y,W)].
$$
$$
\bar R(Y,Z,X,W)
=K[g(Y,W)\,g(Z,X)-g(Y,X)g(Z,W)].
$$
$$
\bar R(Z,X,Y,W)
=K[g(Z,W)\,g(X,Y)-g(Z,Y)\,g(X,W)].
$$
Thus
$$
\bar R(X,Y,Z,W)+\bar R(Y,Z,X,W)+\bar R(Z,X,Y,W)=0,
$$
verifying the Bianchi identity for $\bar R$.