Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A positive $1$-dimensional parametric integrand (short $1$-p.i.) is a continuous function on the tangent space of a manifold $F:TM\rightarrow \mathbb{R}_{\geq 0}$ that is positively homogeneous. This gives a variational problem involving $1$-dimensional manifolds (curves) or more generally $1$-integer multiplicity currents. A $1$-p.i. defined on TU, where U is an open set in R^n is called semi-elliptic if straight lines are locally minimizing. In the book of Krantz and Parks, Geometric integration Theory, it is proven that if $F$ is convex in the tangent variable for every $x\in U$ then it is semi-elliptic. Now take any Finsler norm on $TU=U\times \mathbb{R}^n$. It is strictly convex in the tangent variable (see Th. 1.2.2 in the book Bao+Chern+Shen). By definition the solutions of the associated variational problem are geodesics.

Combining the two results we get that straight lines are always geodesics. Where is my mistake?

share|improve this question

1 Answer 1

Probably you are misinterpreting some statement in Krantz and Parks, since the claim that if $F$ is convex in the tangent variable for every $x\in U$ then straight lines are locally minimizing is clearly false. Either you are misunderstanding their definition of 'semi-elliptic' or else you are missing some other assumption about $F$ in their sufficient condition for 'semi-ellipticity'. (I'm not familiar with their book, so I can't say which it is.) For example, the length integrand for most Riemannian metrics will give you counterexamples to the 'minimizers are straight lines' conclusion.

share|improve this answer
    
You are right Robert, thank you! The definition of semi-elipticity involves the constant $1$-p.i. $F_{x_0}(x,\omega)=F(x_0,\omega)$. It asks for the straight lines to be locally minimizing with respect to each one of these. I need bigger glasses! :) –  daniel Dec 8 '12 at 20:05

Your Answer

 
discard

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.