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I know that for a given metric tensor $g_{\mu \nu}$ on a differentiable manifold exists only one connection which:

  • Preserves the metric form
  • Has no torsion

That's the Levi-Civita connection, and it could be derived by $g_{\mu \nu}$ .

I'm wondering is if there is some intuitive/geometrical way to understand how we can obtain this connection starting from the metric form.

It is clear that a parallel transport compatible with the metric has to preserve angles and length, and we can see geometrically it very well. But that is not enough in order to define it unequivocally, so, we are asking also zero torsion.

How can we interpret it geometrically?

Writing it very roughly, if we have an ant walking straight ahead (i.e. on a geodesic) on a $\mathbb{R}^2$ plane with a metric fom: $ds^2 = a(x,y)dx^2 + b(x,y)dxdy+c(x,y)dy^2$ how does it moves? Let's say that it calculates the new direction at each (very little) step

EDIT: I add an image for making it clear: we have a mesh of the surface, and on each cell the metric form is just $G^i=a^idx^2 + b^idxdy+c^idy^2$ with $a^i,b^i,c^i \in \mathbb{R}$ linear coefficient.

The ant is walking from the cell $1$ to cell $2$: how does its direction change?

enter image description here

I think that there is an analogy with refraction of light, which could be considered as the case where the metric tensor is in the form $g_{\mu \nu}=\phi(x)\delta_{\mu \nu}$

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how does its direction change?

-- the direction is already determined by the geodesic.

Let me quote from this answer to a similar question,

"...pull a string tight enough and it will be a geodesic.

These in turn determine a class of connections, determined up to torsion: twist the string while parallel transporting a tangent vector along it, and you are changing the connection keeping the same geodesics"

Provided your ant has found a geodesic and is going to keep to it, her next job is not to twist her body, her whiskers, etc., as she goes, and carefully, most economically, transport them along thus realizing the torsion-free connection.

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  • $\begingroup$ Sorry, that's not an answer to my question. Clearly it follows a geodesic, and it realise the minimal length, but the question is how the ant can locally interpret it using just the metric tensor. $\endgroup$ – user84976 Oct 21 '17 at 16:11
  • $\begingroup$ that was my attempt at an "intuitive notion" using the "ant". For a more rigorous description please refer to the linked MO question and perhaps also mathoverflow.net/questions/20493/… $\endgroup$ – rych Oct 22 '17 at 0:51
  • $\begingroup$ Thanks, I've already read this mathoverflow question linked. Anywhere I appreciate your attempt, but it is not an answer to my question if you read it carefully $\endgroup$ – user84976 Oct 22 '17 at 8:34

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