TITLE: Vector field built from connection and metric
QUESTION [17 upvotes]: Consider a smooth finite-dimensional manifold $M$ with metric $g$ and connection $\nabla$. For some local coordinate system, denote by $g^{\alpha \beta}$ the inverse of the metric tensor and by $\Gamma^\alpha_{\beta\gamma}$ the Christoffel symbols for $\nabla$. The connection $\nabla$ is assumed to be torsion free, but it does in general not equal the Levi-Civita connection for $g$. Denote also by $R^\alpha_{\beta\gamma\eta}$ the curvature tensor built from $\nabla$. We can then define a vector field $h$ by
$$
h^\alpha = R^\alpha_{\beta\gamma\eta} g^{\gamma\zeta}\nabla_\zeta g^{\eta\beta}\;.
$$
Is there a "hands-on" geometric interpretation for $h$? (Does it maybe even have a name?) 
In particular, 
is there a geometric way of "seeing" when $h$ vanishes? (Of course there are the trivial cases where $\nabla$ is flat or Levi-Civita for $g$, but there must be others.)

REPLY [2 votes]: This is not a complete answer to your question. However, there is one fairly general case in which $h$ will vanish, which is when $(M,g, \nabla)$ is a statistical manifold. A statistical manifold 
is a Riemannian manifold $(M,g)$ with an affine connection $\nabla$ satisfying 
$$ (\nabla_X g)(Y,Z) =(\nabla_Y g)(X,Z) $$
for any vector fields $X,Y$ and $Z$. This is the natural geometry associated with a parametrized family of probability distributions, so these spaces appear in information geometry (which is where the name comes from). However, statistical manifolds are a well defined notion independent of this application. 
To see why this is sufficient for the vector field to vanish, it's helpful to first rewrite $h$.
In particular, we use the fact that
$$\nabla_\zeta g^{\eta\beta} = -(\nabla_\zeta g_{\mu \nu}) g^{\eta\mu} g^{\nu \beta} $$ so that
$$h^\alpha = -R^{\alpha\mu\zeta\nu} \nabla_\zeta g_{\mu \nu},$$
where we have written the curvature in terms of covectors 
(this might be off by a sign).
The definition of a statistical manifold forces the derivative terms to be totally symmetric in $\mu,\zeta$ and $\nu$ while the Bianchi identity (which holds for any torsion-free connection) shows that the cyclic sum of the curvature term vanishes. Taken together, these identities force $h$ to vanish.  It's worth noting that being statistical is not necessary for $h$ to vanish, but it is a fairly general condition which is sufficient.