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The problem is an abstract from applied science.

Given an $n$ dimensional Riemann manifold with metric $\langle M, g\rangle$, we could define deformation of the metric $g(t)$ where $t\in [0,1]$, for example, Ricci flow is such an deformation. And thus we should be able to measure the deforming energy(as an analog with kinetic energy in physics) point by point, and the total defroming energy may be in the form as

$E(t)=\int_{M} \left|\dfrac{\mathrm{d}R(t)}{\mathrm{d}t}\right|^n dA$

where $R$ indicate the Ricci curvature and $\left| \cdot \right|$ is a "proper" norm, such that the energy enjoys scale-invariant property. I will appreciate if someone could provide me some references.

Furthermore, if above formulation of energy is possible, we may define distance between two manifold $\langle M_1, g_1\rangle$ and $\langle M_2, g_2\rangle$ by

$d(M_1,M_2)=\mathrm{inf}_{g(t)}\int_0^1E(t)\mathrm{d} t$

where $g(0)=g_1$ and $g(1)=g_2$.

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This earlier MO question is perhaps tangentially relevant: "Measures of the complexity of a metric" mathoverflow.net/questions/32527 . See especially Will Jagy's discussion of the Willmore functional. –  Joseph O'Rourke Mar 30 '11 at 15:38
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So, any one parameter family of flat tori has zero energy... Are you sure you want it this way? –  Anton Petrunin Mar 30 '11 at 15:47
    
@Joseph: considering the OP's earlier question mathoverflow.net/questions/59715/… I daresay he is somewhat familiar with the Willmore functional. –  Willie Wong Mar 30 '11 at 16:11
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To add to Anton's comment: the Riemannian Schwarzschild metrics form another one-parameter family of Ricci-flat that are not (locally) conformally isometric to each other, are you sure you want them to have distance zero between them? If you want the motivation from the Ricci flow, wouldn't it make more sense to actually consider the energy functional used there? terrytao.wordpress.com/2008/04/24/… –  Willie Wong Mar 30 '11 at 16:48
    
@Willie: My oversight! –  Joseph O'Rourke Mar 30 '11 at 20:48
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