The shortest path in a metric space with a given metric

Question

My questions seem to be very basic and intuitively correct but I can't formally prove them. Before learning metric spaces, for $\mathbb R^2$, we always define the distance between 2 points as $d_2 = ((x_1-y_1)^2+(x_2-y_2)^2)^{1/2}$ and we admit naturally (and intuitively) that the segment is the shortest path amongst those which connect 2 points. In $\mathbb R^1$, the distance between 2 points is define as $abs(x-y)$, which can be thought as a special case of $d_2$, where 2 points lie on 1 axis.

My questions are:

1. How to prove this statement (that is for $d_2$ in $\mathbb R^2$)

2. If we have another metric in another metric space, say for example: $d_n$ in $\mathbb R^n$, what is the shortest path and how to find it?

Answer 1

For $d_2$ in $\mathbb R^2$ we want to minimize the functional $ L[y]=\int_{x_2}^{x_1} \sqrt{1+ ( \frac{dy}{dx})^2} dx$. We can do this by looking for the critical points of $L[y]$. In other metrics we alter the functional that we want to minimize.

Answer 2

What you are looking for are geodesic curves. In that particular metric it's easy to prove it, even by linear algebra methods (the triangular inequality assures you that deviating from the straight line gets you more distance.

In more general metrics, a metric tensor is defined. Then you can use a number of methods to get the equations of geodesic curves. One is using parallel transport (google it). Another one is minimizing the functional that gives you the integrated length of a curve between two points: $\int_a^b dt\sqrt{g_{\mu\nu}\dot y^\mu\dot y ^\nu}$, where $y^\mu$ are the components of the curve your integrating to (parametrised with $t$), and so $\dot y^\mu$ are the components of it's tangent vector.

If you apply a variational principle to that you should arrive to very ugly expressions that give you the general equation for the geodesic under the metric $g_{\mu\nu}$. This equations are of the form: $$\ddot x^\mu+{\Gamma^\mu}_{\nu\lambda}\dot x^\nu\dot x^\lambda=0$$

where the $\Gamma$ objects are called Christoffel symbols, and are the second derivatives of the metric. Look here. Observe that if the metric is the euclidean one those are zero and you get equations for straight lines.

Solving those may be really hard if you include a more complex metric tensor. This may be difficult to see, I suggest you read differential geometry, this is covered in basically every textbook about it (and it's really cool in my opinion :) ).