Grisha got it!


by marcus
Tags: grisha
gvk
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#19
Aug28-06, 01:07 PM
P: 83
I want to understand why 3D case is totaly different from 1D-2D and nD,
n > 3 for which analoques of Poincare conjecture were proof a long time ago. Why the dimention 3D is really matter?
neutrino
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#20
Aug28-06, 01:38 PM
P: 2,048
Quote Quote by gvk
I want to understand why 3D case is totaly different from 1D-2D and nD,
n > 3 for which analoques of Poincare conjecture were proof a long time ago. Why the dimention 3D is really matter?
That question's been nagging my mind, too. I'm don't know about gvk, but I'm a complete outsider to pure maths. So all you math-types, keep your explanations simple.
waht
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#21
Aug28-06, 02:01 PM
P: 1,636
Supposedly, in 3d there isn't enough dimensions (only 3) to do their math stuff. Where as in higher dimensions you can utilize alot of different techniques because you have lots of room to move around.
selfAdjoint
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#22
Aug28-06, 03:20 PM
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Quote Quote by waht
Supposedly, in 3d there isn't enough dimensions (only 3) to do their math stuff. Where as in higher dimensions you can utilize alot of different techniques because you have lots of room to move around.

You got it. The one and two dimensional cases are trivial, and the n greater than three results, while difficult, were solved by Smale a long time ago. The reason three-manifolds are so interesting is that they are right on this boundary between being so limited they're trivial and being so unlimited they're trivial in a different way.

Back when I was a beginning topology student I was not very interested in three-manifolds (which had been an active subtopic since Poincare's day), thinking, "Well then, when they get that all resolved they'll just have to go on to four-manifolds, and then 5, 6, ..., google, ...". It seemed like a mug's game to me. But it ain't so; three is, even at the most abstract level, a very special, very important case.
gvk
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#23
Aug28-06, 04:18 PM
P: 83
Thanks, selfAjoint.
Now, more specific question: if the Ricci-flow technic works just fine for 3D case (on the boundary), it can work for nD case (n>3). Is that correct?
And what does mean the parameter t in Ricci-flow eq.: d (g_{i,j})/dt =-2*Rici? My guess: we issue any parametric curve on manifold from point P and take derivitive along it. Correct?
Sorry, I don't have any original Hamilton paper to look details.
Doodle Bob
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#24
Aug29-06, 03:43 AM
P: 255
Quote Quote by gvk
TAnd what does mean the parameter t in Ricci-flow eq.: d (g_{i,j})/dt =-2*Rici? My guess: we issue any parametric curve on manifold from point P and take derivitive along it. Correct?
Actually, the "t" is the variable for a parametrized curve of metrics on the manifold. I.e. for each fixed t, g_t is a Riemannian metric over the entire manifold.

In other words, they are looking at the (infinite-dimensional) space of all Riemannian metrics living on the given 3-manifold and studying paths in that space that satisfy the Ricci-flow equation, which is actually just a differential equation.


Regarding the strangeness of 3- versus higher-dimensions, dimensions 3 and 4 seem to be -- in general -- harder to work with than higher dimensions. 4-manifold theory is quite active these days, although it got a little "easier" when the Seiberg-Witten equations were discovered in the early 90s.
gvk
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#25
Aug30-06, 11:57 AM
P: 83
Thanks, Doodle Bob.
It seems that the application of Ricci-Flow equation to the dimentions n>3 is not straightforward because Riemann curvature for n>3 is not totaly defined by Ricci tensor, which is the right side of Ricci-Flow equation. Is that correct?
mathwonk
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#26
Aug30-06, 08:10 PM
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the two dimensional case follows from a clasification of all 2 manifolds S OBTAINED BY joining handles to spheres. the three dimensional case wouldfollow aND APPRENTLY DOES, FROM AN analogous conjecture of thurston on how to obtain three manifolds from joining various basic types of objects.

the definition of "trivial" is of cousre open to discussion, but m,y definition is nythign I MYSELF KNOW HOW TO DO, AND I CANNOT SAY I NKOWHOW TO PROVE THE CLKASIFICATION OF 2 MANIFOLDS.

help!! i should have taken typing asa high school studentbut in thsio days it was only offered to secretaries!!!@
mathwonk
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#27
Aug30-06, 08:17 PM
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recalling my ancient history courses in 2 manifolds, i seem to recall that clasificaion of 2 manifolds is ropecedded by a proof that all 2 manifolds are triangulable, itself perhaps not so easy.

or maybe there is an easier morse theory proof?
gvk
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#28
Sep2-06, 08:23 AM
P: 83
Quote Quote by gvk
Thanks, Doodle Bob.
It seems that the application of Ricci-Flow equation to the dimentions n>3 is not straightforward because Riemann curvature for n>3 is not totaly defined by Ricci tensor, which is the right side of Ricci-Flow equation.
The full answers to such trivia questions are available now in references here:
http://en.wikipedia.org/wiki/Grigori_Perelman
J77
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#29
Sep4-06, 03:21 AM
P: 1,157
A very impressive CV, that Tao guy's got.

It's amazing when you see people your own age with results like that!


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