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For a finite dimensional Banach space $V$ with strictly convex and smooth norm (outside the origin), there is a natural way to induce a Riemannian structure on the unit sphere $S$ (see Alvarez-Paiva, "Some problems on Finsler Geometry"): consider $L=\|x\|^2/2 $, then $dL$ is a diffeomorphism from $V\backslash0$ to its dual $V^*\backslash0$. Now, for $q\in S$ we get $D_q(dL):T_qV \rightarrow T_qV^*$, and for $u,v\in T_qS$ take $g_q(u,v)=D_q(dL)(u)(v)$. What is known about this structure? What are the geodesics, for instance? What are their lengths?

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I don't know as much about this as I should, but you might be able to find some papers on this topic here: math.poly.edu/research/finsler. This site is maintained by Alvarez-Paiva. –  Deane Yang Nov 20 '10 at 5:07
    
Do you mean that the Riemann structure gives the usual intrinsic metric on $S$? –  Fedor Petrov Nov 20 '10 at 5:57
    
Can you point out more precisely where in Alvarez-Paiva's problem list this construction is discussed? –  Sergei Ivanov Nov 23 '10 at 5:06
    
The construction is described in section 2 of Alvarez's paper. –  Deane Yang Dec 4 '10 at 18:56

1 Answer 1

Maybe i'm wrong, but isn't $D_q(dL)$ the legendre transform of of the "Lagrangian"? In which case you should get the usual metric back. See for example Sternberg's book on differential geometry (the chapter on calculus of variations) or (i know, it sounds strange but) Mackey's Mathematical Foundations of Quantum Mechanics (the first chapter). If my memory doesn't fail me Abraham, Marsden should have something on that specific example too.

Edit: This applies in the common case where the norm is induced by an inner product.

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What is the "usual metric" ? This riemannian metric is something like the second fundamental form of $S$. –  BS. Nov 20 '10 at 10:18
    
Well, $dL(q)=\langle q,−\rangle$ which is linear in q so $D_q(dL)=dL$, meaning $g_q(u,v)=\langle u,v\rangle$, which when restricted to the tangent bundle of the sphere gives you the metric you would usually put on it –  Sky Nov 20 '10 at 11:00
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The norm $\| \|$ is not a Hilbert norm ($L$ isn't quadratic). –  BS. Nov 20 '10 at 11:42
    
I'm sorry, i was of the impression that the OP just cared of the fact that there was a natural (almost physical) way to get the metric in the usual case where the V is a Hilbert space, which lead to a very cool way to get the associated connection on S. I will edit the post to specify that. –  Sky Nov 20 '10 at 15:19

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