[math-fun] curves and critical points -- proof of corrected conjecture
So the current repaired conjecture is, or anyhow I presume should be, this: CONJECTURE: Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a simply-connected compact region R of the plane with connected interior C avoids R provided 1. C avoids the boundary of R, and 2. f(x,y) has no critical points (meaning grad f = 0) within R. HERE'S MY PROOF: The only way C can fail to avoid R is (i) C hits boundary(R). But that case is already handled in the theorem statement. (ii) A connected component CC of C lies entirely within R's interior.
From now on we only consider case (ii) since that is all we need to consider. Note that any self-crossing points, or isolated points, or end-points of the curve automatically are critical points. (Actually I do not think end-points are possible, but their possibility or impossibility will not matter.) Hence wlog we need only consider the case where CC has no self-crossing or isolated or end points. Since CC is by assumption entirely contained in a compact region with no endpoints and no crossing points (and it s not just an isolated point) it necessarily either closes on itself, or spirals forever, but the latter option is impossible because f is a finite-degree polynomial. So CC is a closed compact curve of the same topological type as a circle.
Now such a CC necessarily contains a critical point of f(x,y) inside it, because f(x,y)=0 on the closed circle-like curve CC and is (wlog) >0 everytwhere that is infinitesimally inside it, and is continuous and differentiable, hence must have a maximum on the compact region inside CC, which necessarily is >0, and which necessarily is a critical point. Q.E.D. GENERALIZATIONS: It seems to me this conjecture ought to be true not only for algebraic curves in the xy plane, but also for algebraic (D-1)-dimensional surfaces in D-dimensional space. Anyhow when D=3. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
SLIGHT CORRECTION/CLARIFICATION/AMPLIFICATION: Change: Now such a CC necessarily contains a critical point of f(x,y) inside it, because f(x,y)=0 on the closed circle-like curve CC and is (wlog) >0 everytwhere that is infinitesimally inside it, and is continuous and differentiable, hence must have a maximum on the compact region inside CC, which necessarily is >0, and which necessarily is a critical point. Q.E.D. To: Now such a CC necessarily contains a critical point of f(x,y) inside it, because f(x,y)=0 on the closed circle-like curve CC and (wlog) f>0 everywhere that is just infinitesimally inside CC -- i.e. within epsilon of CC and inside CC, for all epsilon<eps0, for some positive eps0; we here are taking advantage of the known fact |grad f|>0 everywhere on CC and continuity of grad f since f is polynomial, which forces (grad f).(inward normal vector to c)>delta>0 everywhere on CC for some delta>0 -- and f is necessarily bounded, continuous and differentiable throughout the interior of (and on) CC, hence must have a maximum on the compact region inside-and-on CC, which necessarily is >0 and strictly inside CC, and which necessarily is a critical point. [We are here using the well known theorem that any continuous bounded function on a compact set has a max; and the other well known theorem that if the function also is differentiable, then grad f =0 at any max and any min. Also we'd previously implicitly used the same theorem in 1 less dimension using the fact CC itself is a compact 1-dimensional set and grad f is continuous and differentiable. Q.E.D. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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 (?). -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
OK, now to complete the proof of the 3D version of the theorem; I will use tools (i) (ii)... (ix) from my previous post: 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:
From tools listed plus simple-connected compact assumption, R must be topologically a ball.
The only way C can fail to avoid R is (i) C hits boundary(R). But that case is already handled in the theorem statement. (ii) A connected component CC of C lies entirely within R's interior. So from now on we only consider case (ii) since that is all we need to consider. Note that any surface-surface or curve-surface self-crossing points, or isolated points, curve end-points, curves, surface-boundary points automatically are critical points of f. (Actually I do not think some of those are possible, but their possibility or impossibility will not matter.) Hence wlog we need only consider the case where CC has none of those things. Since CC is by assumption entirely contained in a compact region with no endpoints and no crossing points (and it is not just an isolated point) it necessarily is a boundaryless manifold. Furthermore, we claim it must be compact and closed. Why? Because any kind of infinite-spiraling-surface confined within a compact region of 3-space, could not arise from a finite degree polynomial. [Really there should be some lemma here, that any connected component of an algebraic surface, in any dimension D, which happens to lie entirely within a compact subregion of D-space, automatically has finite measure and finitely bounded geodesic distance from anyplace on it to anyplace else.] So CC is a closed compact boundaryless 2-manifold embeddable in 3-space, hence by Mobius is an h-holed torus for h=0,1,2,3,...m where m is finitely bounded by some function of the degree of polynomial f. And CC has an interior. Now since f=0 on CC and wlog f>0 for all points (x,y,) interior to CC and within distance epsilon from CC (for all epsilon with 0<epsilon<eps0, for some finite positive eps0) and since interior(CC) U CC is compact, we see f must have a max which must be positive and must be located strictly interior to CC. This max is a critical point. Now given that CC is wholy contained inside R, and R is ball-shaped, it follows that interior(CC) also is wholy contained inside R ("interior" meaning the finite measure side -- exterior has infinite measure). Hence the critical point lies in R. Q.E.D. ================== Now one may ask -- can this result be generalized to any dimension, not just 2D and 3D? I'm not sure. I think my proof will prove it in any dimension D where the necessary ingredient-tool-lemmas are available in that dimension. I think we need these tools: 1. finite number of possible topologies for any surface curve compact connected component CC of finite algebraic degree. 2. measure and geodesic diameter automatically bounded for such a CC. 3. R automatically must be homeomorphic to ball. 4. Any kind of surface self-crossing, lower dimensionality than D-1, etc automatically yields a critical point so we can wlog assume CC is a non-wild boundaryless compact (D-1)-manifold. 5. CC "interior" and "exterior" then automatically always exist. We can wlog demand f<0 in exterior and f>0 in interior, at least if we stay within distance epsilon of CC for all small-enough epsilon. 6. If ball-shaped R contains CC then R automatically also contains interior(CC). If those tools are valid in some dimension then I think the theorem follows in that dimension. The only ones I don't feel entirely confident about are #3 and #5 (I feel confident about them in 2D and 3D, but in dimensions higher than 4 they are not so obvious... although likely the answers are known...).
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 (?).
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
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 (?).
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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 (?).
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
"if and only if" WFL On 5/31/14, Andy Latto <andy.latto@pobox.com> wrote:
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.
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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 (?).
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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participants (3)
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Andy Latto -
Fred Lunnon -
Warren D Smith