TITLE: Geodesic on a Riemannian manifold with a random metric tensor
QUESTION [6 upvotes]: Given a metric tensor $g_{\mu\nu}$ on a Riemannian manifold, it's possible to write the geodesic equations using:
$$\frac{d^2x^a}{ds^2} + \Gamma^{a}_{bc}\frac{dx^b}{ds}\frac{dx^c}{ds} = 0$$ where:
$$\Gamma^a_{bc} = \frac{1}{2} g^{ad} \left( g_{cd,b} + g_{bd,c} - g_{bc,d} \right)$$ are the Christoffel symbols. In the euclidean metrics, the tensor $g_{\mu\nu}$ is given by the identity matrix $I.$ Suppose we have:
$$g_{\mu\nu} =
 \begin{pmatrix}
  1 & \eta_1(t) &  \eta_2(t) \\
  0 & 1 &  \eta_3(t) \\
    \\
  0 & 0 & 1
 \end{pmatrix}$$
with:
$\eta_k(t)$ random variables normally distribuited with variance: $\sigma_1,\sigma_2,\sigma_3$, and zero means, how can I write down the geodesic equations? Thanks in advance.

REPLY [2 votes]: Geodesic is a shortest path in metric space. If $g$ is not symmetric, then triangle inequality does not hold : 
$$ g=I + E_{12}$$ where $E_{12}$ is $2\times 2 $ matrix whose only nonzero element is $(1,2)$-entry. And its value is $1$. Then $$\overline{(0,0)(1,1)} > \overline{(0,0)(1,0)} + \overline{(1,0)(1,1)} $$