Commutative/ symmetric second covariant derivative

Question

Consider a smooth manifold $M$ together with an affine connection (or covariant derivative) $\nabla$ on the tangent bundle $TM$.

Is it possible to have an affine connection, possibly with non-zero torsion, such that $\nabla_{X,Y}^2 = \nabla_{Y,X}^2$? I.e. the second covariant derivative is symmetric/commutative, while the covariant derivative itself is not (since it might have torsion)?

How are such connections or manifolds called? If anyone can refer me to an article or reference, that would be great.

PS. Note that the connection is not necessarily curvature-free in the case that $\nabla_{X,Y}^2 = \nabla_{Y,X}^2$, but instead all the curvature is due to torsion. That is, $R(X,Y)Z = \nabla_{T(X,Y)}Z$.

Answer 1

If the second covariant derivative of every vector field $Z$ is symmetric in the sense that $\nabla(\nabla Z)$ (which is a section of $TM\otimes T^*M\otimes T^*M$) is a section of the sub bundle $TM\otimes S^2(T^*M)$, then the connection $\nabla$ is torsion-free and flat.

This follows directly from Cartan's structure equations (cf. Kobayashi and Nomizu). Here is the calculation. Let $\omega = (\omega^i)$ be a coframing on an open set in $M$ and let $\theta = (\theta^i_j)$ be the connection forms of $\nabla$ relative to the coframing $\omega$. The Cartan structure equation are $$ \mathrm{d}\omega^i = -\theta^i_j\wedge\omega^j +\tfrac12 T^i_{jk}\,\omega^j\wedge\omega^k \quad\text{and}\quad \mathrm{d}\theta^i_j = -\theta^i_k\wedge\theta^k_j + \tfrac12 R^i_{jkl}\,\omega^k\wedge\omega^l, $$ where $T^i_{jk} = - T^i_{kj}$ and $R^i_{jkl} = -R^i_{jlk}$ are, respectively, the components of the torsion and curvature of $\nabla$ in the coframing $\omega$. (Note the use of the Einstein summation convention both here and below.)

If $Z$ is an arbitrary vector field in the domain of $\omega$, with $\omega$-components $z^i = \omega^i(Z)$, then there are functions $z^i_j$ and $z^i_{jk}$ that satisfy the equations $$ \mathrm{d}z^i = -\theta^i_j\,z^j + z^i_j\,\omega^j \quad\text{and}\quad \mathrm{d}z^i_j = -\theta^i_k\,z^k_j + \theta^k_j\, z^i_k + z^i_{jk}\,\omega^k. $$ Here, $z^i_j$ and $z^i_{jk}$ are the components of the first and second covariant derivatives of $Z$ relative to the coframing $\omega$.

Suppose that $z^i_{jk}=z^i_{kj}$ for every vector field $Z$. (This is the component version of the condition that $\nabla(\nabla Z)$ be a section of $TM\otimes S^2(T^*M)$.) Use the above structure equations and the symmetry assumption to compute $$ 0 = \mathrm{d}(\mathrm{d}z^i) = \tfrac12\bigl(-R^i_{jkl}z^j + T^j_{kl}z^i_j\bigr)\,\omega^k\wedge\omega^l. $$ Thus, $-R^i_{jkl}z^j + T^j_{kl}z^i_j = 0$ for every vector field $Z$. (I.e., the relation $R(X,Y)Z = \nabla_{T(X,Y)}Z$ holds for every triple of vector fields $X$, $Y$, and $Z$.) Since the value of $Z$ and its first covariant derivative can be specified arbitrarily at any point of $M$, this relation can only hold for all vector fields $Z$ if and only if $T^j_{kl}$ and $R^i_{jkl}$ vanish identically, i.e., the connection $\nabla$ is both torsion-free and flat.