# 47. Symmetries of the Curvature Tensor

We begin by investigating how the Levi-Civita connection behaves with respect to isometric maps.

In this lecture we then introduce a $(0,4)$ version of the curvature tensor of a connection $ \nabla$ on a Riemannian manifold $(M,g)$, which we denote the $ \mathcal{R}^{ \nabla}_g$:

$$ \mathcal{R}^\nabla_g(X,Y,Z,W) := \left \langle R^\nabla(X,Y)(Z),W, \right \rangle.$$

We show that:

- One always has $\mathcal{R}^{\nabla}_g(X,Y,Z,W)=-\mathcal{R}^{\nabla}_g(Y,X,Z,W)$. (Flip the first two variables.)
- If $\nabla$ is metric then in addition $\mathcal{R}^{\nabla}_g(X,Y,Z,W)=-\mathcal{R}^{\nabla}_g(X,Y,Z,W)$. (Flip the last two variables.)
- If $\nabla$ is torsion-free then $$\mathcal{R}^{\nabla}_g(X,Y,Z,W)+\mathcal{R}^{\nabla}_g(Y,Z,X,W)+\mathcal{R}^{\nabla}_g(Z,X,Y,W)=0.$$ (Fix the last variable and cyclically permute the other three.)
- If $\nabla$ is the Levi-Civita connection of $g$ then in addition to (2) and (3) one also has a “bonus” symmetry $$ \mathcal{R}^{ \nabla}_g(X,Y,Z,W) = \mathcal{R}^{\nabla}_m(W,Z,Y,X).$$ (Reverse all the variables.)

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