What is the stronger conjecture? Andy On Sat, May 31, 2014 at 8:20 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
I'm very relieved to hear that!
Now then, what about the (stronger) intrinsic conjecture? WFL
On 5/31/14, Andy Latto <andy.latto@pobox.com> wrote:
nnnI think this is really a theorem of differential geometry, not of algebraic geometry; the only place Warren's proof uses the fact that f is polynomial is when he needs it to be sufficiently smooth.
Since we can replace R by any smaller simply connected region that doesn't touch C, we may as well choose R to be diffeomorphic to a disc, and the theorem then becomes:
For any sufficiently smooth (C2?) function f on the unit disc, if f is nonzero on the unit circle, but has zeroes inside the disc, then it has a critical point inside the disc.
Proof: if f is negative on the unit circle, its maximum is a critical point; if f is positive on the circle, its minimum is a critical point.
This works in any dimension, and we don't really need the Perelman-Thurston classification theorem, the Jordan Curve theorem, or the classification of surfaces. All we need is the fact that the boundary of a compact simply connected set is connected; the theorem doesn't really require that R is simply connected, merely that it has connected boundary.
Andy
On Fri, May 30, 2014 at 1:44 PM, Warren D Smith <warren.wds@gmail.com> wrote:
STILL FURTHER CLARIFICATIONS/AMPLIFICATIONS TO PROOF:
1.note I also have implicitly used the Jordan-Schoenflies curve theorem (to see CC must have an "interior"... and R necessarily is "ball shaped..."), e.g. see http://en.wikipedia.org/wiki/Schoenflies_problem
2. also re my remark a bi-infinite spiral is not possible for a zero-curve of a finite-degree polynomial, if you are not willing to accept that as obvious, then, well..., a polynomial of degree d can only "wiggle" d times and only have d roots at most, along any line... anyhow this is known, indeed there is a known classification of the possible topologies of curves of degree=d for all small d, and partial but good enough classifications for every d...
3. and some of the "well known" theorems I used (in particular Jordan curve) are actually not so easy, in fact were historically proven incorrectly and eventually fixed by later authors... and you'd need to look them up & cite them, if you really wanted to do a solid job... they ought to be in standard textbooks... but anyhow, modulo such extra work I think it is clear I gave a real proof.
================
Now to move on, I will attempt to prove a 3-dimensional version of the Theorem.
CLAIM: 1. If f(x,y,z) is a real polynomial, then its zero set {x,y,z} such that f(x,y,z)=0 in general describe algebraic surfaces (although curves and isolated points can be possible) in 3-space; call this zero-set C. 2. Let R be a compact simply-connected region with connected interior in xyz space. 3. Then C avoids R provided: (A) C avoids boundary(R). (B) R does not contain a critical point of f, meaning a point where grad f = 0.
PROOF: The main tools we are going to use are
(i) Jordan-Brouwer Separation Theorem: Any imbedding of the n-1 dimensional sphere into n-dimensional Euclidean space, separates the Euclidean space into two disjoint regions.
(ii) Brouwer was unable to prove the 3D analog of the Jordan-Schoenflies theorem, that the inside and outside of such an imbedded sphere are homeomorphic to the inside and outside of the standard unit sphere |X|=1 in Euclidean 3-space. That is because this is false: there is well known counterexample called "J.W.Alexander's horned sphere." However, I believe it is known this analogue IS true for "non wild" surfaces, which are all that are possible for a finite-degree polynomial f. [Non-wild: surfaces homeomorphic to triangulated polyhedral surfaces with finite # faces.] We shall restrict attention to non-wild from now on.
(iii) Mobius's surface classification theorem: The only possible topologies for 2D compact surfaces are: spheres, h-holed tori for h=1,2,3,... and spheres with N disjoint circular disks surgically removed and replaced by "crosscaps", N=1,2,3,... These surfaces are orientable if and only if the number of crosscaps on them equals 0. They are simply connected if and only if the surface is a sphere. Surfaces with crosscaps are not embeddable in 3-space, (require more dimension, e.g. Klein bottle can be done in 4-space). For the orientable surfaces, if imbedded in 3-space, interiors and exteriors exist. (I'm not sure who proved the lattermost, Jodan-like, result... presumably it is known...) The interior, assuming it is connected, is simply connected only in the case of a sphere, then interior is ball.
(iv) Corresponding theorem for 1D compact curves with no endpoints: only possibilities are circles.
(v) Any continuous bounded function on a compact region has at least one max (also min).
(vi) Polynomials are infinitely differentiable and bounded on any compact region.
(vii) Any max (or min) of a differentiable function in the interior of a compact region is a critical point.
(viii) Any algebraic surface self-crossing, or curve, or isolated point, automatically is a critical point of f.
(ix) Poincare's conjecture (now theorem in every dimension d>1) Every simply connected, closed compact d-manifold is homeomorphic to the sphere. Actually, I think we'll only need the trivial case d=2 of Poincare which follows from Mobius classification. When d=3 it follows from the G.Perelman proof of W.P.Thurston classification. When d=4 M.Freedman proved it. When d>4 somobody else proved it (Donaldson?).
Sorry, I have to go now; I'll continue later with another post which will attempt to use these tools to provide the proof in 3D, following the line of thought in the 2D proof. It further ought to be possible (?) to prove it in 4D using Perelman-Thurston classification theorem (?).
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