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A projective metric on a projective space is a metric on the underlying set such that shortest path with respect to this metric are parts of entire projective straight lines.

The 2-norm induces the Fubini-Study metric on a projective space (up to a real constant multiplier). Does any other p-norm also induce a projective metric on a projective space?


One idea is that a distance function like $d(x,y)=\operatorname{min}_{|\alpha|=1}||\frac{x}{||x||_q}-\alpha\frac{y}{||y||_q}||_p$ could work, either with $p=q$, or with $\frac{1}{p}+\frac{1}{q}=1$. But if one can show that neither distance function works, then that would also be valuable information.


This is a crosspost of this question from MSE. It currently has an open bounty saying

The linked Encyclopedia of Mathematics article about projective metric makes it sound like the problem of determining all projective metrics has been completely solved. So it seems to me that this question should have a definitive answer.

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  • $\begingroup$ I'm almost sure that no other (complex) norm on $\mathbb{C}^n$ $(n > 1)$ induces a "projective" Finsler metric on complex projective space. I'll try to get back to you on that soon. $\endgroup$ Mar 16, 2015 at 15:51
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    $\begingroup$ By the way, in the real case (if you ask which norms in $\mathbb{R}^n$ induce a projective Finsler metric in its projectivization) there is only the Euclidean norm(s). This is a nice exercise in classical variational calculus and convex geometry. $\endgroup$ Mar 16, 2015 at 16:00
  • $\begingroup$ Sorry, I still can't answer the question about the complex projective space. I wanted to mention that the question for real projective spaces was posed more than ten (perhaps twenty ...) years ago by Z. Shen in a list of questions about Finsler geometry (but I forgot where it is published, or even if it was published). It was this question---for which normed spaces are the geodesics on its unit sphere planar curves?---that Gautier Berck and I set to solve at some point with the dissapointing result that the solution was not "technical enough" to publish. $\endgroup$ Mar 18, 2015 at 14:19
  • $\begingroup$ Continuation: it suffices to consider the case of norms in $\mathbb{R}^3$ and after some basic variational calculus the key element is Blaschke's theorem stating that if the singular set of all two-dimensional projections of a three-dimensional convex body are planar curves, then the body must be an ellipsoid. The complex case looks a lot harder and more interesting. $\endgroup$ Mar 18, 2015 at 14:23

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