Does every connection admit a parallel volume form?

Question

Let $E$ be a smooth orientable vector bundle of rank $k$ over a manifold $M$. Let $\nabla$ be a connection on $E$. Does there always exist a non-zero parallel section of $\Lambda_k(E)$?

What about $E=TM$? (when $M$ is orientable)

Answer 1

No, in general this is not true. The curvature $R^k$ of the connection $\nabla$ induces on $\Lambda^k E$ can be identified with $(X, Y) \mapsto \operatorname{tr}R^{\nabla}(X, Y) \,\cdot\,$. So, if $\xi$ is a parallel section of $\Lambda^k E$, differentiating $0 = \nabla \xi$ and skewing gives $0 = R^k(X, Y)(\xi) = (\operatorname{tr}R(X, Y)) \xi$. Thus, the following are equivalent:

• $\Lambda^k E$ (locally) admits parallel sections;
• the curvature $R^{\nabla}$ of $\nabla$ takes values in $\Lambda^2 T^*M \otimes \mathfrak{sl}(E)$;
• the curvature $R^k$ of the induced connection vanishes.

It follows that if $\dim M = 1$ or $k = 1$ then $\Lambda^k E$ always (locally) admits parallel sections.

There need not even be a parallel volume form even in the special case $E = TM$---which is the subject of this converse question---even if we restrict to torsion-free connections. In that setting, applying the (First) Bianchi identity to $\operatorname{tr}R$ implies that the curvature of the induced connection coincides up to a nonzero multiple with the skew-symmetrization $(X, Y) \mapsto \frac{1}{2}[\operatorname{Ric}(X, Y) - \operatorname{Ric}(Y, X)]$ of the Ricci tensor of $\nabla$.

To see that $R^k$ need not automatically vanish, it's enough to carry out a quick local computation. For the torsion-free connection $\nabla$ with Christoffel symbols $\Gamma_{ab}^c = \Gamma_{ba}^c$ in the standard chart on $\Bbb R^2$, the curvature of the induced connection is $$(\partial_x \Gamma_{xy}^x - \partial_y \Gamma_{xx}^x + \partial_x \Gamma_{yy}^y - \partial_y \Gamma_{xy}^y) \,dx \wedge dy ,$$ and obviously one can choose the $\Gamma_{ab}^c$ so that this quantity is not zero.

As observed in the comments, if $\nabla$ preserves some metric $g$, then it preserves both of its volume forms, because they are constructed from $g$ invariantly (or, just as well, because the curvature of $\nabla$ takes values in $\Lambda^2 T^*M \otimes \mathfrak{so}(g) \subseteq \Lambda^2 T^*M \otimes \mathfrak{sl}(TM)$).

If one is only interested in the (unparameterized) geodesics of a connection $\nabla$ on $TM$, i.e., in the projective structure $[\nabla]$, this is strictly a global phenomenon: One can always choose a connection $\nabla'$ with the same geodesics for which $M$ locally admits parallel volume form. In the projective differential geometry literature, such connections are sometimes called special. See my answer to the question linked above for more about this picture.