17
$\begingroup$

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.)

$\endgroup$
  • 4
    $\begingroup$ So... did you pull that out of your hat or is there a reason you suspect this will have a geometric meaning? $\endgroup$ – Gunnar Þór Magnússon May 21 '15 at 17:28
  • 6
    $\begingroup$ @Gunnar Þór Magnússon Good question ;-) It arises naturally when considering a noisy version of the "heat equation" with domain $S^1$ and target manifold $M$. Here, $\nabla$ defines the heat equation and $g$ is determined by the noise. The vector field $h$ is then a logarithmically divergent drift the solution picks up, due to the interaction between the noise term and the heat equation. $\endgroup$ – Martin Hairer May 21 '15 at 18:07
  • $\begingroup$ I have only vague thoughts. The connection defines geodesics. If you start with a frame of tangent vectors at a point, you can study how that frame changes when you parallel transport it along the connection's geodesics. And maybe if you differentiate twice, you get something useful. Your vector field appears to involve a trace of the curvature (so a Ricci-like tensor), so maybe you want to look at how the determinant of the frame changes relative to the volume form of the Riemannian metric. Maybe this vector field measures the direction of maximal rate of change of this? $\endgroup$ – Deane Yang May 23 '15 at 20:34
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.