Are higher order covariant derivatives symmetric tensors?

Question

Suppose the tangent bundle $TM$ of a smooth manifold $M$ is equipped with a torsion-free connection. Then for a function $f$ on $M$, one can define the $k$-th order derivative $\nabla^k f $ iteratively to be the covariant derivative of $\nabla^{k-1} f$ and $\nabla^1 f = df$. It is known that the second derivative $\nabla^2 f$ is a symmetric tensor. Is $\nabla^k f$ a symmetric tensor for all $k$? Any references?

Answer 1

The symmetry of second covariant derivative is because of Ricci's identity for any section $S$: $$\nabla^2_{X,Y}S -\nabla^2_{Y,X}S=R(X,Y)S$$ where $$R(X,Y)S=\nabla_X(\nabla_Y)S-\nabla_Y(\nabla_X)S-\nabla_{[X,Y]}S$$ and if you substitute a smooth function $f$ for section $S$ we get $R(X,Y)f=0$ and explains the symmetry of second covariant derivative.

For third order covariant derivative we have Bianchi's identity which states: $$\nabla^3_{X,Y,Z}S -\nabla^3_{Y,X,Z}S=R(X,Y)(\nabla_ZS)-\nabla_{R(X,Y)Z}S$$ and if you substitute a smooth function $f$ for section $S$ I don't think the RHS vanishes. So at least for first two parameters it is not symmetric.