# Symmetry of the (Riemann) curvature tensor of the Riemannian connection: $\nabla_{\nabla_Y Z}W = \nabla_Y(\nabla_Z W)$?

My book is Connections, Curvature, and Characteristic Classes by Loring W. Tu (I'll call this Volume 3), a sequel to both Differential Forms in Algebraic Topology by Loring W. Tu and Raoul Bott (Volume 2) and An Introduction to Manifolds by Loring W. Tu (Volume 1).

I refer to Theorem 22.6 in Section 22.4. While the term "Riemann curvature tensor" is not introduced until Section 22.6, I believe what Theorem 22.6 refers to is the Riemann curvature tensor.

For proving Theorem 22.6, I'm trying to understand what it means when Tu says that adding (22.5) and (22.6) cancels out the $$\langle R(Y,Z)X,W \rangle$$ term.

I think the canceling out is in that $$\langle R(Y,Z)X,W \rangle + \langle R(Y,Z)W,X \rangle = 0$$, which I can show assuming $$\nabla_{\nabla_Y Z}W = \nabla_Y(\nabla_Z W)$$ for all $$Y,Z,W \in \mathfrak X(M)$$ is true.

Question if $$\nabla_{\nabla_Y Z}W = \nabla_Y(\nabla_Z W)$$ is true: Why is $$\nabla_{\nabla_Y Z}W = \nabla_Y(\nabla_Z W)$$ true?

Question if $$\nabla_{\nabla_Y Z}W = \nabla_Y(\nabla_Z W)$$ is false: How else do I approach this?

No, $$\nabla_{\nabla_Y Z}W = \nabla_Y(\nabla_Z W)$$ is not true in general.

All Tu means is that $$\langle R(Y,Z)X,W \rangle$$ in (22.5) will cancel with $$\langle R(Y,Z)W,X \rangle$$ in (22.6), due to the skew-symmetry (Prop 12.5).

• Oh, I missed this "As an algebraic consequence of these three properties" Thanks!
– user636532
Oct 7 '19 at 2:51