MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The geodesic distance $d(p_1, p_2)$ on a geodesically complete, Riemanian manifold is a metric in the sense of a metric space metric. I'd like to know when other infinitesimal metric structures (e.g. finlser metrics) induce such a global metric and what the exact relation of the two things is.

What is the largest class of Lagrangians $\mathcal{L}: TM \rightarrow \mathbb{R}$ such that the action it assigns to its stationary curves defines a distance function in the sense of metric spaces.

I'm also interested in any procedure that recovers a local metric structure (i.e. Riemanian tensor/Finsler function) from a given global metric $d:M \times M \rightarrow \mathbb{R}$. Any references about this would be helpful.

A few sources state that the Fubini-Study metric on $\mathbb{C}P^N$ can be "easily" recovered from the distance function $d(|\psi \rangle, |\phi\rangle) = \arccos(| \langle \psi |\phi \rangle|)$, what is the procedure for this and are these ideas rigorous?

share|cite|improve this question
up vote 2 down vote accepted

I answer only your last question.

Take a finite-dimensional $\mathbb C$-linear space $V$ equipped with a positive-definite hermitian form $\langle-,-\rangle$. The tangent space $\text{T}_p{\mathbb P}_{\mathbb C}V$ at a $1$-dimensional linear subspace $p\subset V$ is naturally identified with $\text{Lin}_{\mathbb C}(p,V/p)$ which in turn is naturally identified with $\text{Lin}_{\mathbb C}(p,p^\perp)$ with respect to the orthogonal decomposition $V=p\oplus p^\perp$. Using this decomposition, we can assume that $\text{Lin}_{\mathbb C}(p,p^\perp)\subset\text{Lin}_{\mathbb C}(V,V)$ (extending by zeros). The hermitian form on $\text{Lin}_{\mathbb C}(V,V)$ given by the rule $\langle t_1,t_2\rangle=\text{tr}(t_1\circ t_2^*)$ induces on ${\mathbb P}_{\mathbb C}V$ the famous Fubini-Study hermitian structure. Naturally, the distance can be expressed in terms of the original hermitian form on $V$. The distance in the projective space between the points $0\ne p_1,p_2\in V$ is given by $\text{dist}(p_1,p_2)=\text{arccos}\frac{\langle p_1,p_2\rangle\langle p_2,p_1\rangle}{\langle p_1,p_1\rangle\langle p_2,p_2\rangle}$.

A similar approach works for many hyperbolic geometries (including the complex hyperbolic one), elliptic geometries, and many others that come from grassmannians.

Late extension. It is not necessary to require the form $\langle-,-\rangle$ to be positive-definite, and one can also take $p\subset V$ of a fixed dimension, say, $k$, thus dealing with the grassmannian $\text{Gr}_{\mathbb C}(k,V)$. As above, the loci of non-degenerate points get their pseudo-hermitian structure. Many objects of these geometries (such as geodesics, bisectors, etc) admit an easy linear/hermitian description. The degenerate points form the absolute whose geometry can also be described in the hermitian/linear terms. (It is also possible to take $\mathbb R$ or $\mathbb H$ (or even simetimes $\mathbb O$) in place of $\mathbb C$.)

share|cite|improve this answer
Are these things written down somewhere? The complex projective space pops up often enough as well as the Fubini-Study metric but I would love to see a comprehensive treatment including also other base-fields and Grassmanians. – Vít Tuček Feb 2 '14 at 0:07
@Vít Tuček Yes, you can read it here (For first year students, can be easier.) – Sasha Anan'in Feb 2 '14 at 0:27

I think the book on Petrunin's homepage might be of interest to you.

The part of the Finsler metric answer your question in the positive.

share|cite|improve this answer
More precisely, chapter 2 defines the notion of "Length space" and example 2.2.5 is on the lenght structure of a Finsler geometry. – user3487 Jan 31 '14 at 23:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.