📂Geometry

Bianchi Identity

Theorem¹

Let’s call $R$ the Riemann curvature. Then, the following holds.

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

Proof

It is proven by a straightforward, though complex, calculation without any special techniques. By the definition of Riemann curvature,

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

Since $\nabla$ has symmetry due to the Riemannian connection, $\nabla_{X}Y - \nabla_{Y}X = [X, Y]$ holds. Therefore, to summarize,

$$ \begin{align*} &\ R(X, Y) Z + R(Y, Z) X + R(Z, X) Y \\ =&\ \nabla_{Y}[X, Z] + \nabla_{Z}[Y, X] + \nabla_{X}[Z, Y] + \nabla_{[X,Y]}Z + \nabla_{[Y,Z]}X + \nabla_{[Z,X]}Y \\ =&\ \nabla_{Y}[X, Z] + \nabla_{Z}[Y, X] + \nabla_{X}[Z, Y] - \nabla_{[Y,X]}Z - \nabla_{[Z,Y]}X - \nabla_{[X,Z]}Y \\ =&\ [Y, [X, Z]] + [Z, [Y, X]] + [X, [Z, Y]] \\ =&\ 0 \end{align*} $$

The last equality is due to the Jacobi identity.

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