📂Geometry

Symmetry of the Riemann Curvature Tensor

Definition¹

The Riemann curvature tensor $R$ is defined as follows for $R: \frak{X}(M) \times \frak{X}(M) \times \frak{X}(M) \times \frak{X}(M) \to \mathcal{D}(M)$.

$$ R(X, Y, Z, W) := g(R(X, Y)Z, W) = \left\langle R(X, Y)Z, W \right\rangle $$

Here, $\frak{X}(M)$ is the set of all vector fields defined on $M$, $\mathcal{D}(M)$ is the set of differentiable functions defined on $M$, and $g$ is the Riemannian metric.

Description

Note that notation is being used redundantly. This is because these two definitions are essentially the same.

$R$(Eq. 1) has the following symmetries, which stem from the symmetry and compatibility of the Levi-Civita connection.

Properties

$$ \begin{align} R(X, Y, Z, W) + R(Y, Z, X, W) + R(Z, X, Y, W) &= 0 \\ R(X, Y, Z, W) &= - R(Y, X, Z, W) \\ R(X, Y, Z, W) &= - R(X, Y, W, Z) \\ R(X, Y, Z, W) &= R(Z, W, X, Y) \end{align} $$

Proof

$(1)$

It holds by the Bianchi identity.

$$ \begin{align*} & R(X, Y, Z, W) + R(Y, Z, X, W) + R(Z, X, Y, W) \\ =& g(R(X, Y)Z, W) + g(R(Y, Z)X, W) + g(R(Z, X)Y, W) \\ =& g\left( R(X, Y)Z, W) + g(R(Y, Z)X, W) + g(R(Z, X)Y, W \right) \\ =& g(0, W) \\ =& 0 \end{align*} $$

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$(2)$

It holds by the definition of the curvature tensor $R$.

$$ \begin{align*} R(X, Y, Z, W) &= g\left( R(X, Y)Z, W \right)) \\ &= g\left( \nabla_{Y}\nabla_{X}Z - \nabla_{X}\nabla_{Y}Z + \nabla_{[X, Y]}Z, W \right) \\ &= - g\left( \nabla_{X}\nabla_{Y}Z - \nabla_{Y}\nabla_{X}Z + \nabla_{[Y, X]}Z, W \right) \\ &= - g\left( R(Y, X)Z, W \right) \\ &= - R(Y, X, Z, W) \end{align*} $$

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$(3)$

It is not straightforward as in $(2)$. First, let’s show the following claim.

Claim: $R(X, Y, Z, Z) = 0$

$$ \begin{align*} R(X, Y, Z, Z) &= g\left( \nabla_{Y}\nabla_{X}Z - \nabla_{X}\nabla_{Y}Z + \nabla_{[X, Y]}Z, Z \right) \\ &= g\left( \nabla_{Y}\nabla_{X}Z, Z \right) - g\left( \nabla_{X}\nabla_{Y}Z, Z \right) + g\left( \nabla_{[X, Y]}Z, Z \right) \end{align*} $$

Since the Levi-Civita connection is compatible, the following holds.

$$ \begin{align*} && Y g(\nabla_{X}Z, Z) &= g(\nabla_{Y}\nabla_{X}Z, Z) + g(\nabla_{X}Z, \nabla_{Y}Z) \\ \implies && g(\nabla_{Y}\nabla_{X}Z, Z) &= Y g(\nabla_{X}Z, Z) - g(\nabla_{X}Z, \nabla_{Y}Z) \end{align*} $$

Similarly, the following also holds.

$$ \begin{align*} && X g(\nabla_{Y}Z, Z) &= g(\nabla_{X}\nabla_{Y}Z, Z) + g(\nabla_{Y}Z, \nabla_{X}Z) \\ \implies && g(\nabla_{X}\nabla_{Y}Z, Z) &= X g(\nabla_{Y}Z, Z) - g(\nabla_{Y}Z, \nabla_{X}Z) \end{align*} $$

$$ \begin{align*} && [X, Y] g(Z, Z) &= g(\nabla_{[X, Y]}Z, Z) + g(Z, \nabla_{[X, Y]}Z) \\ \implies && g(\nabla_{[X, Y]}Z, Z) &= \dfrac{1}{2} [X, Y] g(Z, Z) \end{align*} $$

Applying these, we obtain the following.

$$ \begin{align*} &R(X, Y, Z, Z) \\ =& Y g(\nabla_{X}Z, Z) - g(\nabla_{X}Z, \nabla_{Y}Z) - X g(\nabla_{Y}Z, Z) + g(\nabla_{Y}Z, \nabla_{X}Z) + \dfrac{1}{2} [X, Y] g(Z, Z) \\ =& Y g(\nabla_{X}Z, Z) - X g(\nabla_{Y}Z, Z) + \dfrac{1}{2} [X, Y] g(Z, Z) \end{align*} $$

Similarly, as $\nabla$ is compatible, the following holds.

$$ YXg(Z, Z) = Yg(\nabla_{X}Z, Z) + Yg(Z, \nabla_{X}Z) \implies Yg(Z, \nabla_{X}Z) = \dfrac{1}{2}YXg(Z, Z) $$

Therefore, we get the following.

$$ \begin{align*} R(X, Y, Z, Z) &= \dfrac{1}{2}YXg(Z, Z) - \dfrac{1}{2}XYg(Z, Z) + \dfrac{1}{2} [X, Y] g(Z, Z) \\ &= \dfrac{1}{2}(YX- XY)g(Z, Z) + \dfrac{1}{2} [X, Y] g(Z, Z) \\ &= \dfrac{1}{2}[Y, X]g(Z, Z) + \dfrac{1}{2} [X, Y] g(Z, Z) \\ &= - \dfrac{1}{2}[X, Y]g(Z, Z) + \dfrac{1}{2} [X, Y] g(Z, Z) \\ &= 0 \end{align*} $$

Then, by the necessary and sufficient condition for an alternating function,

$$ R(X, Y, Z, W) = -R(X, Y, W, Z) $$

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$(4)$

First, the following holds by the Bianchi identity.

$$ R(X, Y, Z, W) + R(Y, Z, X, W) + R(Z, X, Y, W) = 0 $$

Cyclically changing these four variables, we similarly obtain the following three equations by the Bianchi identity.

$$ \begin{align*} R(Y, Z, W, X) + R(Z, W, Y, X) + R(Z, X, Y, X) &= 0 \\ R(Z, W, X, Y) + R(W, X, Z, Y) + R(X, Z, W, Y) &= 0 \\ R(W, X, Y, Z) + R(X, Y, W, Z) + R(Y, W, X, Z) &= 0 \end{align*} $$

Adding all four equations, following the proven symmetry leads to the elimination as follows.

$$ \begin{align*} &0 \\ =& {\color{red}\cancel{\color{black}R(X, Y, Z, W)}} + {\color{green}\cancel{\color{black}R(Y, Z, X, W)}} + R(Z, X, Y, W) \\ & + {\color{green}\cancel{\color{black}R(Y, Z, W, X)}} + {\color{orange}\cancel{\color{black}R(Z, W, Y, X)}} + R(W, Y, Z, X) \\ & + {\color{orange}\cancel{\color{black}R(Z, W, X, Y)}} + {\color{purple}\cancel{\color{black}R(W, X, Z, Y)}} + R(X, Z, W, Y) \\ & + {\color{purple}\cancel{\color{black}R(W, X, Y, Z)}} + {\color{red}\cancel{\color{black}R(X, Y, W, Z)}} + R(Y, W, X, Z) \\ =& R(Z, X, Y, W) + R(W, Y, X, X) + R(X, Z, W, Y) + R(Y, W, X, Z) \\ =& (-1)(-1)R(X, Z, W, Y) + (-1)(-1)R(Y, W, X, Z) + R(X, Z, W, Y) + R(Y, W, X, Z) \\ \end{align*} $$

Therefore, we obtain the following.

$$ \begin{align*} && 2R(Y, W, X, Z) + 2R(X, Z, W, Y) &= 0 \\ \implies && R(Y, W, X, Z) &= -R(X, Z, W, Y) \\ && &= R(X, Z, Y, W) \\ \end{align*} $$

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