Is it valid to define the Levi-Civita connection as follows?
Start with a unit vector $\vec u$ at a point $x$. That defines a geodesic, call it $\gamma$. After moving along an infinitesimal displacement $d\vec r$ to $x'$, the parallel transport of $\vec u$ will be one that has the same angle as $\vec u$ relative to $d\vec r$. Out of these possibilities, we pick the vector $\vec u'$ whose geodesic $\gamma'$ has the minimum second derivative of its distance from $\gamma$, as measured at $x'$.
That seems to generalize the notion of parallel, since, in flat space, the second derivative of the distance between the two lines (really all derivatives) is zero, and for all other lines in the "same angle" plane (the plane parallel to $\gamma$) it is positive. Also, we should be able to interchange the "same angle" requirement with the statement that the first derivative of the distance is zero.
Is that or some variation thereof a valid definition, and if so, is there a reference that defines it that way? Thanks in advance.
On 2d surfaces it is a good intuition for the Levi-Civita connection: if you have a geodesic $\gamma$, then parallel transport of a unit tangent vector $u$ from $\gamma(t_0)$ to $\gamma(t_1)$ is done by requiring that $u$ and $\dot{\gamma}(t_0)$ have the same angle between them as the transported tangent vector $\tilde{u}$ and $\dot{\gamma}(t_1)$.
We can see this as follows: if $v:[t_0,t_1]\to TM$ is the curve that parallel-transports $u$ along $\gamma$, we have $\frac{d}{dt} g_{\mu\nu} \dot{\gamma}^\mu v^\nu = \nabla_{\dot{\gamma}} g_{\mu\nu}\dot{\gamma}^\mu v^\nu = (\nabla_{\dot{\gamma}}g_{\mu\nu})\dot{\gamma}^\mu v^\nu + g_{\mu\nu} (\nabla_{\dot{\gamma}}\dot{\gamma}^\mu)v^\nu + g_{\mu\nu}\dot{\gamma}^\mu\nabla_{\dot{\gamma}} v^\nu$. Now, since $\nabla$ is the Levi-Civita connection, the first term vanishes, since $\gamma$ is a geodesic, the 2nd term vanishes, and since $v$ is supposed to be the curve that parallel-transports $u$, the third term vanishes, so that $\frac{d}{dt}g_{\mu\nu} \dot{\gamma}^\mu v^\nu=0$, i.e. the inner product of the tangent vector at $\gamma$ and the transported version of $v$ stays constant (the lengths of $\dot{\gamma}$ and $v$ also stay constant, so that the angle stays constant too. If you sort-of take a derivative with respect to $t_1-t_0$, the "length" of the geodesic, you could recover your $\nabla$, so that might indeed work as a definition.
But while it is a good intuition, it doesn't really allow you to calculate anything or make any kind of meaningful statement. Furthermore, as soon as you look at manifolds of a higher dimension, tangent vectors stop being able to be identified solely by the angle they make with another vector: unless that angle is 0, there is a whole (hollow) cone of vectors that make that angle with your tangent vector (or if you restrict yourself to unit vectors, a whole $n-2$-sphere).