Timeline for Functions whose gradient-descent paths are geodesics
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Oct 19, 2010 at 3:34 | comment | added | Will Jagy | I'm starting to get this. Instead of a square in Pietro's construction as the set on which $f$ is zero, take an ellipse. Instead of the distance or the squared distance, start with the fourth power of the distance. If that works out alright, take $g(d) = d^4 e^{-1/d}$ to get a $C^\infty$ example. | |
| Oct 18, 2010 at 22:03 | comment | added | Ian Agol | I think Q4 should imply that $f$ is rotationally symmetric. Imagine using a function $h$ so that $h\circ f$ is nearly flat. Then the gradient flow lines of $f$ on $S$ are nearly planar geodesics, so the gradient flow lines of $f$ in the $xy$ plane should be planar geodesics (by taking a geometric limit). Then the level sets of $f$ (in the plane) should be circles. I haven't thought about how to make this heuristic rigorous yet. | |
| Oct 18, 2010 at 19:57 | comment | added | Joseph O'Rourke | @Pietro: Thanks for spelling out the consequences for me! I am starting to see (sorry for being slow) ... | |
| Oct 18, 2010 at 19:46 | comment | added | Pietro Majer | (sorry, I changed notation: $d_C(x)=\mathbb{dist}(x,C)$). Another way to characterize the functions $f$ should be, that sublevel sets {f<c} are uniform neighborhoods of $C$ (or of any sublevel set {f<b}, with b<c). | |
| Oct 18, 2010 at 19:31 | comment | added | Pietro Majer | (I guess) $C$ could be any subset, even not smooth, for the distance function from $C$ is 1-Lipschitz, thus differentiable a.e., and $|\nabla d_C|=1$, so in any case one gets a solution. A smooth, convex $C$ should give solutions defined everywhere. Note that $C$ a point gives the surfaces of revolutions you mentioned in the questions. | |
| Oct 18, 2010 at 19:21 | comment | added | Joseph O'Rourke | @Pietro: And $C$ is ... ? | |
| Oct 18, 2010 at 19:18 | comment | added | Pietro Majer | Two functions share the same level sets iff the values of each of them only depend on the values of the other. So Willie's condition writes as an eikonal equation $|\nabla f(x)|^2=h(f(x)),$ and I guess that solutions are all of the form $f(x)=g(\mathbb{dist}(x,C)),$ at least locally ($g$ and $h$ being related by $1/h= (g^{-1})^')$. Here $\mathbb{dist}(x,C)$ is the point-set Euclidean distance from $C$. | |
| Oct 18, 2010 at 18:23 | comment | added | Joseph O'Rourke | @Willie: That is a beautiful characterization, that $f$ and the square gradient share the same level sets! I do not yet see what this implies in terms of the global geometric shape of $f$, but it certainly is a succinct encapsulation. Thanks! | |
| Oct 18, 2010 at 14:26 | history | answered | Willie Wong | CC BY-SA 2.5 |