First Bianchi identity

From Diffgeom
Jump to: navigation, search


Let \nabla be a torsion-free linear connection. The Riemann curvature tensor R of \nabla satisfies the following first Bianchi identity or algebraic Bianchi identity:

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

for any three vector fields X,Y,Z.

Notice that since this proof is applicable for any torsion-free linear connection, it in particular holds for the Levi-Civita connection arising from a Riemannian metric or pseudo-Riemannian metric.

Related facts


Using repeated simplication and the Jacobi identity

Let us plug the definition of the Riemann curvature tensor:

\nabla_X\nabla_YZ - \nabla_Y\nabla_XZ + \nabla_Y\nabla_ZX - \nabla_Z\nabla_YX + \nabla_Z\nabla_XY - \nabla_X\nabla_ZY - \nabla_{[X,Y]}Z - \nabla_{[Y,Z]}X - \nabla_{[Z,X]}Y

This can be regrouped as:

\nabla_X(\nabla_YZ - \nabla_ZY) + \nabla_Y(\nabla_ZX - \nabla_XZ) + \nabla_Z (\nabla_XY - \nabla_YX) - \nabla_{[X,Y]}Z - \nabla_{[Y,Z]}X - \nabla_{[Z,X]}Y

Now, since \nabla is torsion-free, we have \nabla_YZ - \nabla_ZY = [Y,Z] and similar simplifications yield:

\nabla_X[Y,Z] + \nabla_Y[Z,X] + \nabla_Z[X,Y] - \nabla_{[X,Y]}Z - \nabla_{[Y,Z]}X - \nabla_{[Z,X]}Y

again using the fact that \nabla is torsion-free, this simplifies to:

[X,[Y,Z]] + [Y,[Z,X]] + [Z,[X,Y]]

this becomes zero by the Jacobi identity.

Using the differential Bianchi identity

Fill this in later