[math-fun] algebraic geometry query
Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a region R of the plane (possibly extending to infinity), I assert that C avoids R provided C avoids the boundary of R ; and C has no critical points within R . There must be a well-known theorem to this effect (unless, of course, it's actually false --- a situation by no means previously unknown). But I don't know a reference (or a counter-example) --- anybody? A straightforward way to locate the critical points seems to be to compute the discriminant g of f with respect to (say) x , then find the roots of g(y) = 0 . Is there a more respectable alternative? WFL
<< A straightforward way to locate the critical points ... >> Er, no --- that locates a zero, not a CP. WFL On 5/29/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a region R of the plane (possibly extending to infinity), I assert that C avoids R provided C avoids the boundary of R ; and C has no critical points within R .
There must be a well-known theorem to this effect (unless, of course, it's actually false --- a situation by no means previously unknown). But I don't know a reference (or a counter-example) --- anybody?
A straightforward way to locate the critical points seems to be to compute the discriminant g of f with respect to (say) x , then find the roots of g(y) = 0 . Is there a more respectable alternative?
WFL
Fred, You need some other hypothesis. For example the elliptic curve y^2 = x^3 - x has it's real locus two disconnected ovals (one of them passes through infinity so looks like an open oval). Since they're disconnected, you can surround one by a circle not encroaching on the other. Victor On Thu, May 29, 2014 at 1:01 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a region R of the plane (possibly extending to infinity), I assert that C avoids R provided C avoids the boundary of R ; and C has no critical points within R .
There must be a well-known theorem to this effect (unless, of course, it's actually false --- a situation by no means previously unknown). But I don't know a reference (or a counter-example) --- anybody?
A straightforward way to locate the critical points seems to be to compute the discriminant g of f with respect to (say) x , then find the roots of g(y) = 0 . Is there a more respectable alternative?
WFL
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(1) Dan has admonished me in private, and perfectly correctly, for confusing surfaces and curves with functions. In future I shall try to restrain my wayward imagination: fundamentally, it's functions we're discussing. In particular, my "critical points" --- where some relevant combination of differentials vanishes --- depend essentially on the coordinate frame (x,y) , the particular polynomial function f(x,y) , and whether I'm discussing: z = f(x,y) in 3-space, with CR's at df/dx = df/dy = 0 ; or y(x) or x(y) defined implicitly (and locally) by f = 0 in 2-space, with CR's at f = df/x = 0 , etc. (2) That confusion misled me into proposing a bogus counter-example to my 3-space algorithm, based on a "tilting cigar" surface meeting the z-plane in the boundary of the 2-space region of interest R , but rising to a maximum z at a point P which projects down to outside R . Intuitively (aha!) this is impossible for simply-connected R , since the surface would have to buckle under itself, and would no longer represent a single-valued function. (3) But as Warren's example of a circle of zeros within annular R shows, simple-connectivity is necessary (the surface can be a paraboloid). This means that any proof is going to involve more topological nous than I can currently muster. (4) Furthermore a "pedestrian" 2-space algorithm actually provides more information about what region R can be guaranteed free of zeros. (5) Andy's counter-example --- a straight line of zeros within an infinite strip R --- I did foresee, but ignored on the grounds that it can be fixed by compactification. Adjoin a complex point or projective line at infinity: the boundary of R then includes points at infinity where the line meets it. (6) I didn't follow Victor's reasoning concerning elliptic curves, I'm afraid; which may be well a problem of communication, given my own earlier muddle. In particular, at least two points on his interior oval will satisfy f = df/x = 0 , (7) But it's probably more constructive to work through an actual case illustrating my proposal in action: so I shall go prepare an example. Fred Lunnon On 5/30/14, Victor Miller <victorsmiller@gmail.com> wrote:
Fred, You need some other hypothesis. For example the elliptic curve y^2 = x^3 - x has it's real locus two disconnected ovals (one of them passes through infinity so looks like an open oval). Since they're disconnected, you can surround one by a circle not encroaching on the other.
Victor
On Thu, May 29, 2014 at 1:01 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a region R of the plane (possibly extending to infinity), I assert that C avoids R provided C avoids the boundary of R ; and C has no critical points within R .
There must be a well-known theorem to this effect (unless, of course, it's actually false --- a situation by no means previously unknown). But I don't know a reference (or a counter-example) --- anybody?
A straightforward way to locate the critical points seems to be to compute the discriminant g of f with respect to (say) x , then find the roots of g(y) = 0 . Is there a more respectable alternative?
WFL
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Fred, I took your reference to "critical points" as meaning singularities, especially when you mentioned that one way to find them was to find the discriminant, and then find the roots of that. However, on second though, if you mean that the discriminant has no real roots, then the real locus of an elliptic curve will have only one component, and thus not be a counter-example to what you hoped for. Now, of course you need your f(x,y) irreducible over the reals, for otherwise you could just take a product of a bunch of such and your locus would be the union of all the curves. Victor On Fri, May 30, 2014 at 10:52 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
(1) Dan has admonished me in private, and perfectly correctly, for confusing surfaces and curves with functions. In future I shall try to restrain my wayward imagination: fundamentally, it's functions we're discussing.
In particular, my "critical points" --- where some relevant combination of differentials vanishes --- depend essentially on the coordinate frame (x,y) , the particular polynomial function f(x,y) , and whether I'm discussing: z = f(x,y) in 3-space, with CR's at df/dx = df/dy = 0 ; or y(x) or x(y) defined implicitly (and locally) by f = 0 in 2-space, with CR's at f = df/x = 0 , etc.
(2) That confusion misled me into proposing a bogus counter-example to my 3-space algorithm, based on a "tilting cigar" surface meeting the z-plane in the boundary of the 2-space region of interest R , but rising to a maximum z at a point P which projects down to outside R .
Intuitively (aha!) this is impossible for simply-connected R , since the surface would have to buckle under itself, and would no longer represent a single-valued function.
(3) But as Warren's example of a circle of zeros within annular R shows, simple-connectivity is necessary (the surface can be a paraboloid). This means that any proof is going to involve more topological nous than I can currently muster.
(4) Furthermore a "pedestrian" 2-space algorithm actually provides more information about what region R can be guaranteed free of zeros.
(5) Andy's counter-example --- a straight line of zeros within an infinite strip R --- I did foresee, but ignored on the grounds that it can be fixed by compactification. Adjoin a complex point or projective line at infinity: the boundary of R then includes points at infinity where the line meets it.
(6) I didn't follow Victor's reasoning concerning elliptic curves, I'm afraid; which may be well a problem of communication, given my own earlier muddle. In particular, at least two points on his interior oval will satisfy f = df/x = 0 ,
(7) But it's probably more constructive to work through an actual case illustrating my proposal in action: so I shall go prepare an example.
Fred Lunnon
On 5/30/14, Victor Miller <victorsmiller@gmail.com> wrote:
Fred, You need some other hypothesis. For example the elliptic curve y^2 = x^3 - x has it's real locus two disconnected ovals (one of them passes through infinity so looks like an open oval). Since they're disconnected, you can surround one by a circle not encroaching on the other.
Victor
On Thu, May 29, 2014 at 1:01 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a region R of the plane (possibly extending to infinity), I assert that C avoids R provided C avoids the boundary of R ; and C has no critical points within R .
There must be a well-known theorem to this effect (unless, of course, it's actually false --- a situation by no means previously unknown). But I don't know a reference (or a counter-example) --- anybody?
A straightforward way to locate the critical points seems to be to compute the discriminant g of f with respect to (say) x , then find the roots of g(y) = 0 . Is there a more respectable alternative?
WFL
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<< for otherwise you could just take a product of a bunch of such and your locus would be the union of all the curves >> Don't follow --- why does that affect things? WFL On 5/30/14, Victor Miller <victorsmiller@gmail.com> wrote:
Fred, I took your reference to "critical points" as meaning singularities, especially when you mentioned that one way to find them was to find the discriminant, and then find the roots of that. However, on second though, if you mean that the discriminant has no real roots, then the real locus of an elliptic curve will have only one component, and thus not be a counter-example to what you hoped for. Now, of course you need your f(x,y) irreducible over the reals, for otherwise you could just take a product of a bunch of such and your locus would be the union of all the curves.
Victor
On Fri, May 30, 2014 at 10:52 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
(1) Dan has admonished me in private, and perfectly correctly, for confusing surfaces and curves with functions. In future I shall try to restrain my wayward imagination: fundamentally, it's functions we're discussing.
In particular, my "critical points" --- where some relevant combination of differentials vanishes --- depend essentially on the coordinate frame (x,y) , the particular polynomial function f(x,y) , and whether I'm discussing: z = f(x,y) in 3-space, with CR's at df/dx = df/dy = 0 ; or y(x) or x(y) defined implicitly (and locally) by f = 0 in 2-space, with CR's at f = df/x = 0 , etc.
(2) That confusion misled me into proposing a bogus counter-example to my 3-space algorithm, based on a "tilting cigar" surface meeting the z-plane in the boundary of the 2-space region of interest R , but rising to a maximum z at a point P which projects down to outside R .
Intuitively (aha!) this is impossible for simply-connected R , since the surface would have to buckle under itself, and would no longer represent a single-valued function.
(3) But as Warren's example of a circle of zeros within annular R shows, simple-connectivity is necessary (the surface can be a paraboloid). This means that any proof is going to involve more topological nous than I can currently muster.
(4) Furthermore a "pedestrian" 2-space algorithm actually provides more information about what region R can be guaranteed free of zeros.
(5) Andy's counter-example --- a straight line of zeros within an infinite strip R --- I did foresee, but ignored on the grounds that it can be fixed by compactification. Adjoin a complex point or projective line at infinity: the boundary of R then includes points at infinity where the line meets it.
(6) I didn't follow Victor's reasoning concerning elliptic curves, I'm afraid; which may be well a problem of communication, given my own earlier muddle. In particular, at least two points on his interior oval will satisfy f = df/x = 0 ,
(7) But it's probably more constructive to work through an actual case illustrating my proposal in action: so I shall go prepare an example.
Fred Lunnon
On 5/30/14, Victor Miller <victorsmiller@gmail.com> wrote:
Fred, You need some other hypothesis. For example the elliptic curve y^2 = x^3 - x has it's real locus two disconnected ovals (one of them passes through infinity so looks like an open oval). Since they're disconnected, you can surround one by a circle not encroaching on the other.
Victor
On Thu, May 29, 2014 at 1:01 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a region R of the plane (possibly extending to infinity), I assert that C avoids R provided C avoids the boundary of R ; and C has no critical points within R .
There must be a well-known theorem to this effect (unless, of course, it's actually false --- a situation by no means previously unknown). But I don't know a reference (or a counter-example) --- anybody?
A straightforward way to locate the critical points seems to be to compute the discriminant g of f with respect to (say) x , then find the roots of g(y) = 0 . Is there a more respectable alternative?
WFL
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By critical points I assume that you mean points at which both partial derivatives vanish (that's compatible with your remark about the discriminant). Let R be a disk centered at the origin of radius 2. Let f(x,y) = (x^2 + y^2 - 1)((x-5)^2 + (y-5)^2 - 1). Then the locus of f(x,y) = 0 is the union of a circle of radius 1 centered at the origin and a circle of radius 1 centered at (5,5). This clearly avoids the boundary of R and has no critical points in R. Victor On Fri, May 30, 2014 at 11:34 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< for otherwise you could just take a product of a bunch of such and your locus would be the union of all the curves >>
Don't follow --- why does that affect things? WFL
On 5/30/14, Victor Miller <victorsmiller@gmail.com> wrote:
Fred, I took your reference to "critical points" as meaning singularities, especially when you mentioned that one way to find them was to find the discriminant, and then find the roots of that. However, on second though, if you mean that the discriminant has no real roots, then the real locus of an elliptic curve will have only one component, and thus not be a counter-example to what you hoped for. Now, of course you need your f(x,y) irreducible over the reals, for otherwise you could just take a product of a bunch of such and your locus would be the union of all the curves.
Victor
On Fri, May 30, 2014 at 10:52 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
(1) Dan has admonished me in private, and perfectly correctly, for confusing surfaces and curves with functions. In future I shall try to restrain my wayward imagination: fundamentally, it's functions we're discussing.
In particular, my "critical points" --- where some relevant combination of differentials vanishes --- depend essentially on the coordinate frame (x,y) , the particular polynomial function f(x,y) , and whether I'm discussing: z = f(x,y) in 3-space, with CR's at df/dx = df/dy = 0 ; or y(x) or x(y) defined implicitly (and locally) by f = 0 in 2-space, with CR's at f = df/x = 0 , etc.
(2) That confusion misled me into proposing a bogus counter-example to my 3-space algorithm, based on a "tilting cigar" surface meeting the z-plane in the boundary of the 2-space region of interest R , but rising to a maximum z at a point P which projects down to outside R .
Intuitively (aha!) this is impossible for simply-connected R , since the surface would have to buckle under itself, and would no longer represent a single-valued function.
(3) But as Warren's example of a circle of zeros within annular R shows, simple-connectivity is necessary (the surface can be a paraboloid). This means that any proof is going to involve more topological nous than I can currently muster.
(4) Furthermore a "pedestrian" 2-space algorithm actually provides more information about what region R can be guaranteed free of zeros.
(5) Andy's counter-example --- a straight line of zeros within an infinite strip R --- I did foresee, but ignored on the grounds that it can be fixed by compactification. Adjoin a complex point or projective line at infinity: the boundary of R then includes points at infinity where the line meets it.
(6) I didn't follow Victor's reasoning concerning elliptic curves, I'm afraid; which may be well a problem of communication, given my own earlier muddle. In particular, at least two points on his interior oval will satisfy f = df/x = 0 ,
(7) But it's probably more constructive to work through an actual case illustrating my proposal in action: so I shall go prepare an example.
Fred Lunnon
On 5/30/14, Victor Miller <victorsmiller@gmail.com> wrote:
Fred, You need some other hypothesis. For example the elliptic curve y^2 = x^3 - x has it's real locus two disconnected ovals (one of them passes through infinity so looks like an open oval). Since they're disconnected, you can surround one by a circle not encroaching on the other.
Victor
On Thu, May 29, 2014 at 1:01 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a region R of the plane (possibly extending to infinity), I assert that C avoids R provided C avoids the boundary of R ; and C has no critical points within R .
There must be a well-known theorem to this effect (unless, of course, it's actually false --- a situation by no means previously unknown). But I don't know a reference (or a counter-example) --- anybody?
A straightforward way to locate the critical points seems to be to compute the discriminant g of f with respect to (say) x , then find the roots of g(y) = 0 . Is there a more respectable alternative?
WFL
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On Fri, May 30, 2014 at 10:52 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
(5) Andy's counter-example --- a straight line of zeros within an infinite strip R --- I did foresee, but ignored on the grounds that it can be fixed by compactification. Adjoin a complex point or projective line at infinity: the boundary of R then includes points at infinity where the line meets it.
If you know the conjecture you're making is false, and have a patched-up version of the conjecture that's actually what you're asking about, I'd find it more useful if you actually stated the patched-up conjecture, rather than having us each guess what patching up you intend. So the question you were asking about R2, you are actually meaning to ask about some compactification of R2. Would that be the one-point compactification? or RP2, the real projective plane? Or some other compactification? Andy
In answer to Andy: it shouldn't matter which compactification is used --- so long as we don't forget about the necessity. "Or some other ..." --- trick question? There aren't any others! In answer to Victor: [see point (2) in my list]. If we reason about z = f(x,y) in 3-space, there is a critical point at (x,y,z) = (-0.98,-0.980,-45) approx, which projects into the disc near the origin: so my (3-space) algorithm reports that it cannot confirm that there is no zero of f within R . Note that the criterion is one-way: it can only establish that there is NO zero in the region! I'll post an example shortly ... WFL On 5/30/14, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, May 30, 2014 at 10:52 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
(5) Andy's counter-example --- a straight line of zeros within an infinite strip R --- I did foresee, but ignored on the grounds that it can be fixed by compactification. Adjoin a complex point or projective line at infinity: the boundary of R then includes points at infinity where the line meets it.
If you know the conjecture you're making is false, and have a patched-up version of the conjecture that's actually what you're asking about, I'd find it more useful if you actually stated the patched-up conjecture, rather than having us each guess what patching up you intend.
So the question you were asking about R2, you are actually meaning to ask about some compactification of R2. Would that be the one-point compactification? or RP2, the real projective plane? Or some other compactification?
Andy
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Sorry, make that (x,y,z) = (-0.098076, -0.098076, -50) . WFL On 5/30/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
In answer to Andy: it shouldn't matter which compactification is used --- so long as we don't forget about the necessity. "Or some other ..." --- trick question? There aren't any others!
In answer to Victor: [see point (2) in my list]. If we reason about z = f(x,y) in 3-space, there is a critical point at (x,y,z) = (-0.98,-0.980,-45) approx, which projects into the disc near the origin: so my (3-space) algorithm reports that it cannot confirm that there is no zero of f within R .
Note that the criterion is one-way: it can only establish that there is NO zero in the region!
I'll post an example shortly ...
WFL
On 5/30/14, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, May 30, 2014 at 10:52 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
(5) Andy's counter-example --- a straight line of zeros within an infinite strip R --- I did foresee, but ignored on the grounds that it can be fixed by compactification. Adjoin a complex point or projective line at infinity: the boundary of R then includes points at infinity where the line meets it.
If you know the conjecture you're making is false, and have a patched-up version of the conjecture that's actually what you're asking about, I'd find it more useful if you actually stated the patched-up conjecture, rather than having us each guess what patching up you intend.
So the question you were asking about R2, you are actually meaning to ask about some compactification of R2. Would that be the one-point compactification? or RP2, the real projective plane? Or some other compactification?
Andy
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Mmm ... maybe the 3-space criterion IS two-way, after all ... counter-example, anybody? WFL On 5/30/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Sorry, make that (x,y,z) = (-0.098076, -0.098076, -50) . WFL
On 5/30/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
In answer to Andy: it shouldn't matter which compactification is used --- so long as we don't forget about the necessity. "Or some other ..." --- trick question? There aren't any others!
In answer to Victor: [see point (2) in my list]. If we reason about z = f(x,y) in 3-space, there is a critical point at (x,y,z) = (-0.98,-0.980,-45) approx, which projects into the disc near the origin: so my (3-space) algorithm reports that it cannot confirm that there is no zero of f within R .
Note that the criterion is one-way: it can only establish that there is NO zero in the region!
I'll post an example shortly ...
WFL
On 5/30/14, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, May 30, 2014 at 10:52 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
(5) Andy's counter-example --- a straight line of zeros within an infinite strip R --- I did foresee, but ignored on the grounds that it can be fixed by compactification. Adjoin a complex point or projective line at infinity: the boundary of R then includes points at infinity where the line meets it.
If you know the conjecture you're making is false, and have a patched-up version of the conjecture that's actually what you're asking about, I'd find it more useful if you actually stated the patched-up conjecture, rather than having us each guess what patching up you intend.
So the question you were asking about R2, you are actually meaning to ask about some compactification of R2. Would that be the one-point compactification? or RP2, the real projective plane? Or some other compactification?
Andy
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I'm getting confused. Can someone please state the latest version of the conjecture (say in the original dimensions)? Thanks. --Dan
Is there a standard name for the fractal G that's the *boundary* of Bill Gosper's Flowsnake? I've been calling it "the map-of-France". Not sure where that originated, but it's kind of apt, since the actual map of France is hexagonalish and has a fractalish boundary. Omitting technical details: One way to define it is to start with the 2D regular hexagon H_0 in the complex plane C. Inductively, define H_(n+1) as a) First taking the union of 6 translated copies of the previous stage H_n, along with H_n itself, so that the center of one of them lies at 2+tau, where tau := exp(2pi*i/6) so they all fit snugly together. To complete this step, and then b) shrink this union of 7 copies of H_n by the effect of the function z |-> z/(2+tau). The result is H_(n+1). Then we define H_oo := lim H_n n->oo It's not too hard to show this actually converges in the Hausdorff metric on compact subsets of the plane. Finally, define the fractal G := bd(H_oo), the boundary of H_oo. It has the lovely property that a rosette of 7 translated copies of it fitting snugly together forms a magnified and rotate copy of H_oo, as the regular hexagon doesn't quite do. It's easy to see the Hausdorff measure d of G is the solution to sqrt(7)^d = 3, so d = log_7(9) = 1.1291500681.... QUESTION: Does G have an official name? --Dan
* Dan Asimov <dasimov@earthlink.net> [May 31. 2014 10:27]:
Is there a standard name for the fractal G that's the *boundary* of Bill Gosper's Flowsnake?
I've been calling it "the map-of-France". Not sure where that originated, but it's kind of apt, since the actual map of France is hexagonalish and has a fractalish boundary.
Omitting technical details:
One way to define it is to start with the 2D regular hexagon H_0 in the complex plane C. [...]
More ways: 1) L-system Start: ++++++ Rules: + --> +-+ - --> --+ where + and - are turns by 60 degrees, a unit stroke implied after each turn. (use turns of 120 degrees to get the terdragon) 2) "unit rectangle" with numeration system using radix (5+sqrt(3)) / 2 and the seven digits {zero and all roots of unity solving z^6-1}.
The criterion that has mostly been discussed up to now I want to re-christen "EXTRINSIC", since it considers the surface z = f(x, y) embedded in 3-space: Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a simply-connected compact region R of the plane, C avoids R if C avoids the boundary of R , and there is no point (x, y) in R where df/dx = df/dy = 0 . I asked earlier whether "if" could be strengthened to "if and only if"; a counter-example is easy to find. However, the following "INTRINSIC" criterion -- which is anyway what I have actually been experimenting with -- seems more robust in this respect, besides (like the rane in Spane?) remaining firmly in 2-space: Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a simply-connected compact region R of the plane, C avoids R if and only if C avoids the boundary of R , and there is no point (x, y) in R where df/dx = f = 0 . Counter-examples in either direction please? If the weak ("if") or strengthened ("if and only if") version stands up, can Warren's proof -- if it stands up -- be modified to apply to either? I keep promising a worked example, but have tripped over a technical detail: it's easy to compute one component of the coordinate of a critical point -- say y above -- but rather trickier to pair with the corresponding x . For my application I haven't needed to take that final step; but omitting it is didactically rather unsatisfactory. Fred Lunnon
On Sat, May 31, 2014 at 8:05 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote: . Taking f to be x^2 + y^2 = 1, so C is the unit circle, and the only critical point is the origin, and negating both sides of your conjectured equivalence, since I find "intersects" easier to think about than "avoids", your conjecture in this case reduces to R contains the origin
I asked earlier whether "if" could be strengthened to "if and only if"; a counter-example is easy to find. However, the following "INTRINSIC" criterion -- which is anyway what I have actually been experimenting with -- seems more robust in this respect, besides (like the rane in Spane?) remaining firmly in 2-space:
Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a simply-connected compact region R of the plane, C avoids R if and only if C avoids the boundary of R , and there is no point (x, y) in R where df/dx = f = 0 .
Taking f to be x^2 + y^2 = 1, so C is the unit circle, and the only critical point is the origin, and negating both sides of your conjectured equivalence, since I find "intersects" easier to think about than "avoids", your conjecture in this case reduces to R intersects the unit circle if and only if The boundary of R intersects the unit circle or R contains the origin. One direction of this is true: If R intersects the unit circle, but the boundary of R does not, then since the unit circle is connected, R contains the entire unit circle, and since it is simply connected, it contains the entire unit disc. But for the converse, a small disc around the origin is an obvious counterexample. This is the second conjecture you've posted about all plane algebraic curves where the answer is "well, it's trivially false for the circle". Since you're presumably not just spewing out random conjectures, presumably you have some intuition about plane algebraic curves that you are trying to capture and failing. Maybe it would work better for you to post what this intuition is informally, rather than making failed attempts to formalize it and having math-fun play the "well, this is obviously false, let's try to guess what Fred meant to say" game. Andy
Some confusion here, I'm afraid. The origin is not a critical point for the intrinsic criterion: the critical points lie at (x,y) = (0,+1), (0,-1) . WFL On 5/31/14, Andy Latto <andy.latto@pobox.com> wrote:
On Sat, May 31, 2014 at 8:05 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote: .
Taking f to be x^2 + y^2 = 1, so C is the unit circle, and the only critical point is the origin, and negating both sides of your conjectured equivalence, since I find "intersects" easier to think about than "avoids", your conjecture in this case reduces to
R contains the origin
I asked earlier whether "if" could be strengthened to "if and only if"; a counter-example is easy to find. However, the following "INTRINSIC" criterion -- which is anyway what I have actually been experimenting with -- seems more robust in this respect, besides (like the rane in Spane?) remaining firmly in 2-space:
Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a simply-connected compact region R of the plane, C avoids R if and only if C avoids the boundary of R , and there is no point (x, y) in R where df/dx = f = 0 .
Taking f to be x^2 + y^2 = 1, so C is the unit circle, and the only critical point is the origin, and negating both sides of your conjectured equivalence, since I find "intersects" easier to think about than "avoids", your conjecture in this case reduces to
R intersects the unit circle
if and only if
The boundary of R intersects the unit circle
or
R contains the origin.
One direction of this is true:
If R intersects the unit circle, but the boundary of R does not, then since the unit circle is connected, R contains the entire unit circle, and since it is simply connected, it contains the entire unit disc.
But for the converse, a small disc around the origin is an obvious counterexample.
This is the second conjecture you've posted about all plane algebraic curves where the answer is "well, it's trivially false for the circle". Since you're presumably not just spewing out random conjectures, presumably you have some intuition about plane algebraic curves that you are trying to capture and failing. Maybe it would work better for you to post what this intuition is informally, rather than making failed attempts to formalize it and having math-fun play the "well, this is obviously false, let's try to guess what Fred meant to say" game.
Andy
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I don't understand what you mean by "the critical points for the intrinsic criterion"; given a function f(x,y), what is your definition of what you're calling the critical points for the intrinsic criterion? Points where the tangent line is horizontal? What about point where the tangent line does not exist? is the origin a critical point in your sense of y^2 = x^3? Of x^2 = y^3? Andy On Sat, May 31, 2014 at 11:13 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Some confusion here, I'm afraid. The origin is not a critical point for the intrinsic criterion: the critical points lie at (x,y) = (0,+1), (0,-1) . WFL
On 5/31/14, Andy Latto <andy.latto@pobox.com> wrote:
On Sat, May 31, 2014 at 8:05 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote: .
Taking f to be x^2 + y^2 = 1, so C is the unit circle, and the only critical point is the origin, and negating both sides of your conjectured equivalence, since I find "intersects" easier to think about than "avoids", your conjecture in this case reduces to
R contains the origin
I asked earlier whether "if" could be strengthened to "if and only if"; a counter-example is easy to find. However, the following "INTRINSIC" criterion -- which is anyway what I have actually been experimenting with -- seems more robust in this respect, besides (like the rane in Spane?) remaining firmly in 2-space:
Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a simply-connected compact region R of the plane, C avoids R if and only if C avoids the boundary of R , and there is no point (x, y) in R where df/dx = f = 0 .
Taking f to be x^2 + y^2 = 1, so C is the unit circle, and the only critical point is the origin, and negating both sides of your conjectured equivalence, since I find "intersects" easier to think about than "avoids", your conjecture in this case reduces to
R intersects the unit circle
if and only if
The boundary of R intersects the unit circle
or
R contains the origin.
One direction of this is true:
If R intersects the unit circle, but the boundary of R does not, then since the unit circle is connected, R contains the entire unit circle, and since it is simply connected, it contains the entire unit disc.
But for the converse, a small disc around the origin is an obvious counterexample.
This is the second conjecture you've posted about all plane algebraic curves where the answer is "well, it's trivially false for the circle". Since you're presumably not just spewing out random conjectures, presumably you have some intuition about plane algebraic curves that you are trying to capture and failing. Maybe it would work better for you to post what this intuition is informally, rather than making failed attempts to formalize it and having math-fun play the "well, this is obviously false, let's try to guess what Fred meant to say" game.
Andy
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Now I have studied Andy's generalisation and proof of the extrinsic criterion, I take off my hat to a model of clarity and concision, making everything look easier than it is (as earlier discovered to my cost). However, I do not understand the necessity for C^2 --- why would not C^1 suffice? In contrast my attempts to cast light by comparing "extrinsic" with "intrinsic" side-by-side have evidently backfired in spades, resulting in dismissive refutations of conjectures which I remain unaware of having proposed in the first place. Anyway moving on, in the spirit of his neat one-liner, there follows a proposed more general intrinsic criterion for real 2-space zero-detection: Theorem: Given a continuously differentiable function f(x, y) , and a compact region R of the plane with f nonzero on the boundary of R : f has a zero within R if and only if there is some (intrinsic critical) point (x, y) in R where *** df/dx = f = 0 ***. Proof (offered tentatively): By the implicit function theorem, the constraint f(x, y) = 0 defines a function x(y) single-valued in any interval of y where df/dx <> 0 . Such an interval meeting R is finite since R is compact, and its endpoints lie in the interior of R since f <> 0 on the boundary; therefore df/dx = 0 at those endpoints. The converse is trivial. QED. The result may well generalise to an m-dimensional curve in n-space, but I prefer to take one step at a time. It is stronger than the earlier extrinsic criterion because nec. and suff., so should be applicable to situations such as solution of equations and curve-tracing, besides computational theorem-proving. The driver behind its practical utility is that for f polynomial there exists a constructive algorithm to detect critical points: compute discriminant g(y) of f(x, y) with respect to x , then solve g = 0 to yield their (real) y-components. Locating corresponding x-components is rather more tedious [ Gröbner bases prove helpful here in practice, despite theoretical complications ] though for some applications these may not be required. In particular if y_0 denotes the maximum critical y-component, we may rest assured --- pace compactness and boundary condition --- that there are no zeroes in the region y > y_0 ; and similarly (exchanging variables) for x . Fred Lunnon On 6/1/14, Andy Latto <andy.latto@pobox.com> wrote:
I don't understand what you mean by "the critical points for the intrinsic criterion"; given a function f(x,y), what is your definition of what you're calling the critical points for the intrinsic criterion? Points where the tangent line is horizontal? What about point where the tangent line does not exist? is the origin a critical point in your sense of y^2 = x^3? Of x^2 = y^3?
Andy
On Sat, May 31, 2014 at 11:13 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Some confusion here, I'm afraid. The origin is not a critical point for the intrinsic criterion: the critical points lie at (x,y) = (0,+1), (0,-1) . WFL
On 5/31/14, Andy Latto <andy.latto@pobox.com> wrote:
On Sat, May 31, 2014 at 8:05 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote: .
Taking f to be x^2 + y^2 = 1, so C is the unit circle, and the only critical point is the origin, and negating both sides of your conjectured equivalence, since I find "intersects" easier to think about than "avoids", your conjecture in this case reduces to
R contains the origin
I asked earlier whether "if" could be strengthened to "if and only if"; a counter-example is easy to find. However, the following "INTRINSIC" criterion -- which is anyway what I have actually been experimenting with -- seems more robust in this respect, besides (like the rane in Spane?) remaining firmly in 2-space:
Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a simply-connected compact region R of the plane, C avoids R if and only if C avoids the boundary of R , and there is no point (x, y) in R where df/dx = f = 0 .
Taking f to be x^2 + y^2 = 1, so C is the unit circle, and the only critical point is the origin, and negating both sides of your conjectured equivalence, since I find "intersects" easier to think about than "avoids", your conjecture in this case reduces to
R intersects the unit circle
if and only if
The boundary of R intersects the unit circle
or
R contains the origin.
One direction of this is true:
If R intersects the unit circle, but the boundary of R does not, then since the unit circle is connected, R contains the entire unit circle, and since it is simply connected, it contains the entire unit disc.
But for the converse, a small disc around the origin is an obvious counterexample.
This is the second conjecture you've posted about all plane algebraic curves where the answer is "well, it's trivially false for the circle". Since you're presumably not just spewing out random conjectures, presumably you have some intuition about plane algebraic curves that you are trying to capture and failing. Maybe it would work better for you to post what this intuition is informally, rather than making failed attempts to formalize it and having math-fun play the "well, this is obviously false, let's try to guess what Fred meant to say" game.
Andy
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-- Andy.Latto@pobox.com
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Now that I finally understand what Fred was asking for :-), I think this is related to the hyperplane section theorem of Lefschetz: http://en.wikipedia.org/wiki/Lefschetz_hyperplane_theorem Basically this theorem relates the Homology of sections of a surface (or an n-fold in more generality) to the Homology of the whole surface. The curve that Fred is talking about is but one section. Victor On Sun, Jun 1, 2014 at 9:27 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Now I have studied Andy's generalisation and proof of the extrinsic criterion, I take off my hat to a model of clarity and concision, making everything look easier than it is (as earlier discovered to my cost). However, I do not understand the necessity for C^2 --- why would not C^1 suffice?
In contrast my attempts to cast light by comparing "extrinsic" with "intrinsic" side-by-side have evidently backfired in spades, resulting in dismissive refutations of conjectures which I remain unaware of having proposed in the first place.
Anyway moving on, in the spirit of his neat one-liner, there follows a proposed more general intrinsic criterion for real 2-space zero-detection:
Theorem: Given a continuously differentiable function f(x, y) , and a compact region R of the plane with f nonzero on the boundary of R : f has a zero within R if and only if there is some (intrinsic critical) point (x, y) in R where *** df/dx = f = 0 ***.
Proof (offered tentatively): By the implicit function theorem, the constraint f(x, y) = 0 defines a function x(y) single-valued in any interval of y where df/dx <> 0 . Such an interval meeting R is finite since R is compact, and its endpoints lie in the interior of R since f <> 0 on the boundary; therefore df/dx = 0 at those endpoints. The converse is trivial. QED.
The result may well generalise to an m-dimensional curve in n-space, but I prefer to take one step at a time. It is stronger than the earlier extrinsic criterion because nec. and suff., so should be applicable to situations such as solution of equations and curve-tracing, besides computational theorem-proving.
The driver behind its practical utility is that for f polynomial there exists a constructive algorithm to detect critical points: compute discriminant g(y) of f(x, y) with respect to x , then solve g = 0 to yield their (real) y-components. Locating corresponding x-components is rather more tedious [ Gröbner bases prove helpful here in practice, despite theoretical complications ] though for some applications these may not be required. In particular if y_0 denotes the maximum critical y-component, we may rest assured --- pace compactness and boundary condition --- that there are no zeroes in the region y > y_0 ; and similarly (exchanging variables) for x .
Fred Lunnon
On 6/1/14, Andy Latto <andy.latto@pobox.com> wrote:
I don't understand what you mean by "the critical points for the intrinsic criterion"; given a function f(x,y), what is your definition of what you're calling the critical points for the intrinsic criterion? Points where the tangent line is horizontal? What about point where the tangent line does not exist? is the origin a critical point in your sense of y^2 = x^3? Of x^2 = y^3?
Andy
On Sat, May 31, 2014 at 11:13 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Some confusion here, I'm afraid. The origin is not a critical point for the intrinsic criterion: the critical points lie at (x,y) = (0,+1), (0,-1) . WFL
On 5/31/14, Andy Latto <andy.latto@pobox.com> wrote:
On Sat, May 31, 2014 at 8:05 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote: .
Taking f to be x^2 + y^2 = 1, so C is the unit circle, and the only critical point is the origin, and negating both sides of your conjectured equivalence, since I find "intersects" easier to think about than "avoids", your conjecture in this case reduces to
R contains the origin
I asked earlier whether "if" could be strengthened to "if and only if"; a counter-example is easy to find. However, the following "INTRINSIC" criterion -- which is anyway what I have actually been experimenting with -- seems more robust in this respect, besides (like the rane in
Spane?)
remaining firmly in 2-space:
Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a simply-connected compact region R of the plane, C avoids R if and only if C avoids the boundary of R , and there is no point (x, y) in R where df/dx = f = 0 .
Taking f to be x^2 + y^2 = 1, so C is the unit circle, and the only critical point is the origin, and negating both sides of your conjectured equivalence, since I find "intersects" easier to think about than "avoids", your conjecture in this case reduces to
R intersects the unit circle
if and only if
The boundary of R intersects the unit circle
or
R contains the origin.
One direction of this is true:
If R intersects the unit circle, but the boundary of R does not, then since the unit circle is connected, R contains the entire unit circle, and since it is simply connected, it contains the entire unit disc.
But for the converse, a small disc around the origin is an obvious counterexample.
This is the second conjecture you've posted about all plane algebraic curves where the answer is "well, it's trivially false for the circle". Since you're presumably not just spewing out random conjectures, presumably you have some intuition about plane algebraic curves that you are trying to capture and failing. Maybe it would work better for you to post what this intuition is informally, rather than making failed attempts to formalize it and having math-fun play the "well, this is obviously false, let's try to guess what Fred meant to say" game.
Andy
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I don't understand how these results about homology relate to numerical questions about locations of function zeros. WFL On 6/1/14, Victor Miller <victorsmiller@gmail.com> wrote:
Now that I finally understand what Fred was asking for :-), I think this is related to the hyperplane section theorem of Lefschetz:
http://en.wikipedia.org/wiki/Lefschetz_hyperplane_theorem
Basically this theorem relates the Homology of sections of a surface (or an n-fold in more generality) to the Homology of the whole surface. The curve that Fred is talking about is but one section.
Victor
On Sun, Jun 1, 2014 at 9:27 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Now I have studied Andy's generalisation and proof of the extrinsic criterion, I take off my hat to a model of clarity and concision, making everything look easier than it is (as earlier discovered to my cost). However, I do not understand the necessity for C^2 --- why would not C^1 suffice?
In contrast my attempts to cast light by comparing "extrinsic" with "intrinsic" side-by-side have evidently backfired in spades, resulting in dismissive refutations of conjectures which I remain unaware of having proposed in the first place.
Anyway moving on, in the spirit of his neat one-liner, there follows a proposed more general intrinsic criterion for real 2-space zero-detection:
Theorem: Given a continuously differentiable function f(x, y) , and a compact region R of the plane with f nonzero on the boundary of R : f has a zero within R if and only if there is some (intrinsic critical) point (x, y) in R where *** df/dx = f = 0 ***.
Proof (offered tentatively): By the implicit function theorem, the constraint f(x, y) = 0 defines a function x(y) single-valued in any interval of y where df/dx <> 0 . Such an interval meeting R is finite since R is compact, and its endpoints lie in the interior of R since f <> 0 on the boundary; therefore df/dx = 0 at those endpoints. The converse is trivial. QED.
The result may well generalise to an m-dimensional curve in n-space, but I prefer to take one step at a time. It is stronger than the earlier extrinsic criterion because nec. and suff., so should be applicable to situations such as solution of equations and curve-tracing, besides computational theorem-proving.
The driver behind its practical utility is that for f polynomial there exists a constructive algorithm to detect critical points: compute discriminant g(y) of f(x, y) with respect to x , then solve g = 0 to yield their (real) y-components. Locating corresponding x-components is rather more tedious [ Gröbner bases prove helpful here in practice, despite theoretical complications ] though for some applications these may not be required. In particular if y_0 denotes the maximum critical y-component, we may rest assured --- pace compactness and boundary condition --- that there are no zeroes in the region y > y_0 ; and similarly (exchanging variables) for x .
Fred Lunnon
On 6/1/14, Andy Latto <andy.latto@pobox.com> wrote:
I don't understand what you mean by "the critical points for the intrinsic criterion"; given a function f(x,y), what is your definition of what you're calling the critical points for the intrinsic criterion? Points where the tangent line is horizontal? What about point where the tangent line does not exist? is the origin a critical point in your sense of y^2 = x^3? Of x^2 = y^3?
Andy
On Sat, May 31, 2014 at 11:13 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Some confusion here, I'm afraid. The origin is not a critical point for the intrinsic criterion: the critical points lie at (x,y) = (0,+1), (0,-1) . WFL
On 5/31/14, Andy Latto <andy.latto@pobox.com> wrote:
On Sat, May 31, 2014 at 8:05 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote: .
Taking f to be x^2 + y^2 = 1, so C is the unit circle, and the only critical point is the origin, and negating both sides of your conjectured equivalence, since I find "intersects" easier to think about than "avoids", your conjecture in this case reduces to
R contains the origin
I asked earlier whether "if" could be strengthened to "if and only if"; a counter-example is easy to find. However, the following "INTRINSIC" criterion -- which is anyway what I have actually been experimenting with -- seems more robust in this respect, besides (like the rane in
Spane?)
remaining firmly in 2-space:
Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a simply-connected compact region R of the plane, C avoids R if and only if C avoids the boundary of R , and there is no point (x, y) in R where df/dx = f = 0 .
Taking f to be x^2 + y^2 = 1, so C is the unit circle, and the only critical point is the origin, and negating both sides of your conjectured equivalence, since I find "intersects" easier to think about than "avoids", your conjecture in this case reduces to
R intersects the unit circle
if and only if
The boundary of R intersects the unit circle
or
R contains the origin.
One direction of this is true:
If R intersects the unit circle, but the boundary of R does not, then since the unit circle is connected, R contains the entire unit circle, and since it is simply connected, it contains the entire unit disc.
But for the converse, a small disc around the origin is an obvious counterexample.
This is the second conjecture you've posted about all plane algebraic curves where the answer is "well, it's trivially false for the circle". Since you're presumably not just spewing out random conjectures, presumably you have some intuition about plane algebraic curves that you are trying to capture and failing. Maybe it would work better for you to post what this intuition is informally, rather than making failed attempts to formalize it and having math-fun play the "well, this is obviously false, let's try to guess what Fred meant to say" game.
Andy
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On Fri, May 30, 2014 at 4:28 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
In answer to Andy: it shouldn't matter which compactification is used --- so long as we don't forget about the necessity. "Or some other ..." --- trick question? There aren't any others!
Not true; the Stone-Cech compactification is much, much, larger than RP2. Or without using such heavy machinery, consider the sequences (0,n) and (1, n). In a compactification, they each have to have a convergent subsequence. But there's no reason that these two convergent subsequences have to converge to the same point, as they do in the two compactifications you mention. Andy _____________
participants (5)
-
Andy Latto -
Dan Asimov -
Fred Lunnon -
Joerg Arndt -
Victor Miller