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Let $d:\mathbb R^n\times \mathbb R^n\to\mathbb R^n$ be a distance which induces the Euclidean topology and with which $\mathbb R^n$ is a length space. A continuous curve $\gamma:[a,b]\to\mathbb R^n$ is called a geodesic if it locally minimizes the length. Furthermore, we say that $(\mathbb R^n,d)$ is a projective space if straight lines are geodesics.

I'm wondering if, in a projective space $(\mathbb R^n,d)$, straight lines also globally minimize length, i.e. for every points $x,y\in\mathbb R^n$ we have that $d(x,y) = l(\overline{xy})$, where $l(\overline{xy})$ is the length of the line segment $\overline{xy}$ which connects $x$ and $y$.

I know that in general it is false that local geodesics are also global ones without the hypothesis of projective space, but I can't find a counterexample in this particular case.

My question originates studying a simple version of Hilbert's fourth problem and in particular reading the article Hilbert's fourth problem in two dimension. It seems to me that for the variational approach it is assumed the local definition of the geodesics, whereas for the geometric point of view (see for example Busemann, Alexander and Ambartzumian) segments are assumed to be globally minimizing. However I can't find a relation between the two definitions.

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  • $\begingroup$ If your distance function comes from a metric structure, the Hopf Rinow-Theorem tells you that the minimizing curve between any two points (globally) is a geodesic. Since at least locally geodesics are unique, that minimizing curve has to be locally everywhere a line, hence is a line. I however do not know wether every distance function is induced by a metric structure. $\endgroup$ Commented Jan 22, 2016 at 16:21
  • $\begingroup$ @GeorgLehner Consider the distance such that $d(x,y) = \max |x_i-y_i|$, where $x=(x_1,\dots,x_n)$ and $y=(y_1,\dots,y_n)$. This distance respects all the hypothesis, but geodesics aren't locally unique. $\endgroup$ Commented Jan 22, 2016 at 16:41
  • $\begingroup$ Yes, but the $d_\infty$-norm also fails to be induced by a riemannian metric. (And, it also doesn't have all straight lines as geodesics). For every riemannian metric, geodesics are locally unique. $\endgroup$ Commented Jan 22, 2016 at 19:19
  • $\begingroup$ I've been too concise... If I'm not mistaken, $d_\infty$-norm have all straight lines as global geodesics. In fact for $x,y,z$ collinear points (in this order) we have $d_\infty(x,y)+d_\infty(y,z)=d_\infty(x,z)$ and this easily implies that all segments are global geodesics. However $d_\infty$ is not induced by a riemannian metric, moreover its geodesics aren't locally unique. Therefore we can't apply your line. $\endgroup$ Commented Jan 23, 2016 at 9:37

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