Terence Tao: An integration approach to the Toeplitz square peg problem

I think it's a great honor to to be here for his birthday you know my when I when I first started doing research as a master's student my master's thesis was on q1 and here B theorems actually in Clifford algebra of all things but yeah you know I said I sort of you know I I started out learning about you know these papers of defeating journey and question and Sam's and so forth and I got this met Turkey as a postdoc in 2007 this famous MSRI program in harmonic analysis and I think the thing that struck me the most is you know you you as you know when you're when you're starting out you you read all these papers by these famous people and you know you know it's all very intimidating you know and then you meet them and he is he's the least intimidating person I know you know but it's a really good thing I think you you you need to know that you know sort of the leaders in your field are not scary people you know most of them anyway anyway so I will talk about a fun little problem it's it's a key prop which is it's a rare problem which is both open and and easy enough to state that even non non repetitions can at least understand the question so it's a question of topics from 19:11 and the question is faces a conjecture square peg problem wasp it's a big conjecture okay every simple closed curve gamma in the plane so simple means it doesn't cross itself and close resistive it starts inscribe to square and so if a simple closed curve in the plane so by inscribing a square I mean just an ordinary geometric square so in other words there exists a square with all vertices pull-pull vertices that on the curve okay so inscribe is not exactly the right word because you know that the square can I keep poke out side the curve but I just want the vertices to lie on the square okay so this this is a very simple question and it's still open so you can phrase it sort of you know sort of more analytically so the occult accordance per square I can call this point XK become a Y and this point let me just check my conventions at this point I'm gonna call X plus ay Y plus B and then this point is X sorry X py minus a and this point is a plus B what has P minus a okay so what you're asking really is can you solve an equation can you solve the system of four equations and four unknowns [Applause] [Music] so in other words okay you can one solve final solution X Y are your numbers and crucially you don't allow the squares your length okay then of course description okay so yeah you want to find for your numbers which solver so each gamma is a co dimension one curve so these are sort of four constraints for four unknowns it's not all over the terminal under determine it's a determined question okay so there's a there's a lot of partial results on this question so the conjecture is true if for example if gamma is a convex curve this is true this is so it was out of several people as a topless first claim this didn't quite give a proof but the gaps in the view for him and syndulla and 21 it's true for polygons applica no curve this is without habit 1914 it's true for piecewise analytic Kirstin 16 and this and it's true for c2 curves this is true slavish newman and 1944 it's true if gamma has a point symmetry or line symmetry so if it's symmetric by reflection around you've got line or point then it is only truths if it's not know something right and 99 Phi it's true if your time is locally monotone so I run every point there's a coordinate system where the camera is the graph of a monotone function and that's what sort of quest you know I and very recently it was stronger triple c1 curves and that's our other D and McCleary okay so there's a lot of classes of curves for which we know it to be true basically you know this other in some way if the curve is nice enough yeah it's it's true it's a little bit so you can draw an analogy with the adora line curve theorem but every simple curve close curve you know divides the plane into two regions and this is this is easy if your curve is nice like a polygon and the only authority in the jordan curve theorem is is that if you fearful if all you have a C naught hey okay so case occur so by curve I mean I see not then it's it's then the Jordan curve theorem is is is non-trivial yeah so okay so then II many was out some that they all have slightly different proofs but they always sort of have a topological flavor to them which is not surprising sort of given D there's a mr. Jordan curve theorem and more precisely there is they have a server homological flavor to them counting intersection numbers so as I said you're accounting for you're trying to solve four equations and four unknowns if you like and I was saying it is that you can look at that at the set of four tables on the curve so this is just a subset of our create okay on the one hand you have this format of all what's not really a manifold it's not smooth but it's a four-dimensional set thank you the set of forty books on this curve and you're trying to intersect it with this what is essentially a linear space which I my paper will squares so square by definition it's just a set of quadruples of varys's squared so you take X comma Y X plus a y plus B X is B Y minus a X plus a plus B y plus B minus a or X Y a B are wheels and a B is non zero okay so this this is almost a four dimensional linear subspace of of our eight except that you you remove a two dimensional subspace of that which is the degenerate squares so either this for T sub space you take up the two d sub space so that's a full manifold and what the so an equivalent way of operating this conjecture is is that given any simple closed curve this four-dimensional set and this four-dimensional set have non-empty intersection to intersection that's just the way we for milady so the homological approach basically is that the way you can prove something is not empty is that you just try to count the gnomon intersection points you had a discount you know so homology tozi be if you have two four dimensional objects and eight dimensions and then in some sense they're transverse if they're in general position they should intersect this and some finite number of times you should get some intersection number and maybe you can orient them and maybe he talked about oriented intersection number and he can try to count the domore intersections and so the the basic strategy of proof I mean it's not always freezing this language but but the basic strategy is to try to show that if you County the intersection between these two sets show that in turn this intersection is an odd number yeah because all things come up in this it has been one empty now it's so oh yeah it's a trivial thing but it but it's it it's a yeah now now yeah you have to be have a caveat that's all actually the way I've set it up this is never true because there's a symmetry the second it's always a mobile for because if you have a square root you can rotate it you can permute the the the vertices so yeah modulo symmetries is so this is there's a discrete symmetry of water for with you would you quotient output after you portion that out this happens to be all okay and roughly speaking though the reason why this is true is that you can just check one case and folks how we behave technical ifs you can check that for an ellipse which is not a key a circle there was this one there's one square up two symmetries that you can actually inscribed in in an ellipse and then you can just sort of use general homology theory arguments that you need to form [Music] continuously your curve and as long as you keep everything in general position you can you can show that the parody all this intersection never changes but that you know as these two curves as these two for manifolds or I mean this is always fixed as you perform this for manifold you can sometimes you can you can create a pair of intersections or some as two seconds can swap em disappear but you can't just create one okay so yeah maybe when you start a language you're using homology of z2 coefficients or something but all right and so the okay so this you can prove for sort of nice general position gamma so of course it's not quite true for certain curves for example gamma is if it's a circle and then there's an infinite number of squares okay so and that's because these two for minor forms then are not transverse okay degeneracy okay but as long it's a sort of general position then then then you can make this work answer and if things are smooth enough or piecewise political pellagra or something you can make this work and then to do all these other cases you try to approximate okay so then you know for more general gamma you just photo fox made by nice general position gamma PI by I so curves and take limits so get from some rough curve and you want to cut the square you can try to smooth it out no and bye-bye some nicer curve for which you can already do something find squares in your proximation and it takes you take take a limit and hope to you know you know you mumble compactness and then simmer and then is and then you should somehow get ask win a limit and this works in many cases again this is why we have won all these results but there is this key problem which is which is this set here the service all squares is not compact avoids not compacta mistakes not closed saying the more important thing because of this condition that you don't want your square to deteriorate to zero so so the main enemy is I said take a rough curve and your approximate saying smoother you know this was also you get a square but the square coming really tiny and as you take the limit the square we degenerate your point okay and then so all you conclude at the end is that you're setting your curtains got a point which is not very interesting and okay I mean that can't happen if your curve is nice enough you know so if the curve example is see to you okay it's not actually possible for all for birdie superstar but really tiny square to to vitamin C to occur I can even see one okay and and I thought that roughly is is what is behind all these restrictions on gamma okay so it you know if you freeze it this way it sounds like the the main difficulty of those problem is that you just have to be better at understanding rough curves which many people in this audience are very good at and just so don't understand understand limits better but it seems that that there's a limitation to this approach that is I don't think thank you you may be able to to me relax these conditions a little bit further but but I don't think you can prove the whole general conjecture this way and I'll try to give you some evidence of this later but but what I want to present is that this is a slightly different approach to to this problem which it's not so dependent on homology or counting intersections and it's instead more based on comb ology I work which is the fancy word for integrating well okay it's much fancier than that occiput okay but it but derive ecology at least since we're integrating and sort of in some sense you can think of no this is this is this is like the this problems I can't like a PDE except that there's no there's no D okay there's no differentiation but but these these are some of the system of equations and it turns up you'll say okay you can introduce the federal ID into the form and there is actually a conserved quantity like a conserved energy to this this problem which he can use to to have a different proof of a different style of proof which can handle some things that the homology an odd okay so so let me first give you a positive result okay okay so this was from last year I guess okay so all right so so this and I can prove the conjecture is true gammas union of two Lipsius graphs of Lipschitz constant less than 1 minus epsilon some positive Epsilon okay so if you have two two lectures grass so one has to be a lot lie about the other okay so y'all so like GE t of t TF of T second T a and B okay so you need G to be point wise bigger than F I think well it doesn't really matter good but you have them take you to the bottom then yeah so so so any any curve which has this form you can you can make you can ascribe the curve now that this class is not quite covered by the previous examples it's close to being locally monitor because it's looking on everywhere except at the two end points what is this a problem so this curve you know and it's almost c1 you know it's Lipschitz but it's yeah but the the normal to here is that there are these two points where things can get singular so you know be the thing is is that to the risk of you they can make an awfully quite a bit as you approach the endpoint in particular you can get there are these infant decimals of near squares well it's quite possible to create a code like this which has Clausen inscribed squares which was true to a point yeah yeah you can't really you can't describe squeezed locally here because because your because of the Lipchitz condition but okay but you can inscribe a lot of interesting small squares here and so the the the standard strategy does not really work because if you try to approximate a nicer curve you can get these tiny tiniest square is impacting it infinitely many squares sort of starting to accumulate at at the endpoint and then and and and so parity becomes a a meaningless concept in the limit and yes I have a more concrete reason why this this this this class of curves is not be able to be covered by the strategy I will talk about that later okay so let me first prove this so it's a yeah so it's basically you push my parts [Music] all right so let me draw bigger picture of this curve all right so we have four equations and four unknowns so it's let's first forget about the the final constraint and just solve three equations in four nodes which is a lot easier so now the problem under determined and this very quickly your freedom so what you expect actually is that if you think any point on the bottom side if you pick any point on the bottom curve you should be able to inscribe an L okay should be able to find what speaking my conventions what am i doing okay so given any point on the on the bottom curve you should be able to find an L shape so so two line segments are perpendicular and have the same and the same length okay but you but you you don't require the fourth point so this is three points per square but the fourth point it doesn't have to lie on the curve okay so you should be able to do this and in fact you always can M is unique so so given any point on the on the bottom curve there is a unique point also on the bottom curve until the right and a unique point on the upper curve such that you could it's a shape I'm not taking very easy because what you do is that once you fix point you take this this lucious curve and you rotate it by 90 degrees around this this point here if you rotate this couple 90 degrees this will give you another curve which is literature ellipses graph of X a secular y-axis okay so this this is ellipses graph you know of the y coordinate respectively the x coordinate and when you rotate 90 degrees this this curve rotates to a ellipses cough which is look in the other direction and this this looks just but curves I'm assuming have gotten less than one so the slopes here less than 45 degrees and so if you think about it this forces this was a unique intersection between this limb Fisker which is basically horizontal and this copy is basically vertical basically thanking me hex okay so you and and at the intersection point that's exactly where and then if your data back that's to create your own shape okay so so for every point on the base there is a there was a unique L shape okay and then you've got the point yeah and the question is can you also forces but just this endpoint to to also hit the curve okay so what you can do is that you can so this is a free parameter okay so so this point you can vary along the curve so so so basically we have this curve of L shapes okay so okay you can introduce a time parameter okay and and just reversely and just let X T YT traverse the the bottom edge of this this bottom curve and then you know and as this point moves around this L shape moves around as well so we have a time variable okay and then this corner also moves around and so everything depends on some type I'm dirty and it's easy to show that that just no contraction back from that that all of these these functions X dat y DB T are literature's functions of T okay another fuse concert like one of Absalom actually something like that okay so um so yeah this curve of squares okay I at that time zero everybody the square has degenerated to a point okay so you have a point then as you move baby this help the square grows and grows as you keep moving these three points lot stay on the curve this point doesn't stay on the curve and then eventually your friends back to point again okay so you can track the this this fourth point so this both my traces another curve okay and basically what what you want to do okay so let me say so so gamma get consists of Tuco's gamma 1 and gamma 2 and then this is this is extra curve gamma 3 which is coming from which is the trajectory of the fourth point of this curve all shapes and basically what you want is that you want these two occurs to have a non-trivial intersection okay so they're going to intersect at the two end points but they think if you can make the meet anywhere in between then you've created the square okay and now comes the conserved quantity okay so given any curve given any say Lipschitz or rectifiable curve gamma you can talk about the area under the curve and by the area under the curve what I mean I don't want to integrate this this one for y DX okay so yeah so yeah so is this this isn't a weakness of just sense you you keep you parameterize gamma by some X of T y of T okay and then you you you take the Reimer stotras integral of Y of T against T of X of T and this doesn't depend on the parameterization and it and for Lipchitz the only these all ellipses curve so so this is this water fine yeah so f gamma was the graph this is just precisely the area under the curve but no if your curve is it's not a graph like that then it's you know example it could be you will be this is this area like this okay I think it could be negative if you camera this is underneath the x axis then this could list this to be negative but you know it's always a number okay so okay so this I'm gonna call the area under gamma okay interpreted sense okay and so there are sort of these four vertices per square they Traverse different curves so so the two bottom vertices Traverse gamma one this vertex versus gamma 2 and this vertex vertices gamma 3 and so you get these four areas and under curves and this relationship between them because of a simple identity okay so get these four curves so you can look at Y of T the X of T okay so so so on one hand if you integrate this from you form this MIMO just going to go from T time from A to B or whatever this is the area and become one okay if you integrate Y plus B you might still just XA this is also the area under gamma 1 so it's the same number because because this point here is also parametrized and as time varies this all this is a different characterization of the same curve gamma 1 so these are both the same number and then finally a y of t minus a of t oh yeah the necks of t minus bt it's the area under gamma 2 and the final thing Y of T must be of t minus a of t DX of T plus a of T ok so we have all this area under the curves and the local they all look kind of similar so you can actually do a lot cancellation if you take it now training some of these things if you add this subtract this subtract this and add this what what you find is that a lot of terms cancel and the only thing that's left actually is you get a being of TDP of t minus a of t you okay it's a small matter of algebra okay so okay so you can just believe that but see these are closed forms that so this is just the differential 1/2 B squared and B starts at 0 and ends at 0 so this is a total differential this is 0 this is 0 so in fact you have this conservative Nicko of motion the outlining some of these for integrals has to be 0 ok so that's that's just a nice geometric fact and so and these two areas were already equal so what I means is that is that the area under this curve has to equal the area under this curve here ok so there was actually constraint so the way I promised picture is actually inaccurate ok so the area under this blue curve actually has to be the executive seams under the white curve and that forces there to be in section point they must actually cross to actually prove that properly you need your own curfew oh yeah so yeah so yeah implicit in all this is the fact of this circuit is yes also so this blue curve gamma 3 turns out to be a simple a simple closed curve and that's it because of the ellipsis hypothesis and so if they if they were disjoint if these two codes are disjoint they will trace out a drawdown curve and the difference between the areas of between the two occurs would be the area up to a sign would be the area of the domain are they enclose and I would be nonzero ok so the picture must instead look like this ok so there has to be both positive and negative area between the two curves and that's that forces intersection point ok so so this is this is a non homo logical argument ok so it works even though yeah it works at the level versus regularity even though this this is possibly an infinite no you know so they you know these guys can it can intersect an infinite number of times but but they did the area between them is finite and so some other area is a more analytically robust concept than in the intersection number that it can work in slightly less regular context than an exact match here ok so actually positive result yeah so unfortunately it I don't know how to push it much further than this well I should mention that that this argument was also independently worker by care he didn't quite phrase it this way in this language basically what carrots have showed was that the conjecture is true as long as as long as you can't find two L shapes on your curve with it you can't find two L shapes on your curve whose fourth vertex fourth abilities agree so long with you you you can't drew a picture like this in your curve then this argument shows that there are no inscribed squares and and the Union to Lipsius graphs with constant less than one okay so all right so this conjecture has so so far sort of the you know the intuition that we telling you is that this is a problem which is easy for nice curves and hard for rough curves but it turns out that less sort of an illusion this it seems like really the the problem is understanding via the smooth case better and so I can truly say a periodic version of the square peg ball okay so voila having a simple closed curve in r2 we now take it to the kitchen let gamma 1 gamma 2 be too simple and just join cuz in the cylinder a winding number one so the pictures that you give a cylinder so it takes you one ah you identify the left and right they just so this is a cylinder and so I simple close curve of winding number one it's okay so the simplest case would just be a graph of a function from from consumption from from zero to one day the end of where it starts but that you could also imagine you you you you you are allowed to so backtrack and you know you go back and forth as long as your total white numbers one okay so yeah so there was a notion of a simple closed curve gamma 1 and gamma 2 and you assume that they destroyed and so thanks to you so one of the Mac user has to lie about the other there is a Jordan curve theorem in this periodic setting but whenever you're twisting for closed curves then the Union should inscribe a square okay I wear to the vertices line gamma one with two adjacent vertices in gamma 1 & 2 Lisa versus in gamma 2 so whenever you have two curves like this you should be able to find a square with where two bases are two adjacent versus the square lie on the bottom curve and two adjacent vertices found on the top curve okay okay so square okay yeah so by square I mean points XY y plus B is a good point David okay my - a okay so we're ex I guess it is periodic why is real and a and B are real numbers and I guess yeah it's no longer so important but but we can speakin to enforce their nonzero yes so I got lost squares we have lengths bigger than one and so you saw this I'm not gonna kind of joy but but you can find squeeze in the center don't look like squares because they wrap around and yes so even the true cuz a really far apart if you like you can yeah and one way to think about it if you extend them periodically that the two infinite curves in the plane or inscriber squares and all the way of thinking about it so even if they're very power part then a square of a square okay so this is another conjecture you can make sake related to that so you can't force these two between disjoint to be degenerate but actually um you don't need these to be disjoint and it's not important that that the square has your death anymore you can just take two curves you can allow them to cross well then it's easy actually okay if they cost but okay but you can eke you can you can you can allow you can also pose the problem allowing now the curve to to to degenerate it to zero length and that's an equivalent conjecture because if you have two curves that are not disjoint as David said you can just shift one up by a very large distance and then if you do it large enough they become just joint and then any square that they inscribe it has to be has to have known as your length and then you can take a square there and then you can be you can push it back down so so in contrast to the non pure existence from you can remove this this this non closed constraint and now the problem comes closed and because of this yes also on quality of this now is that for this problem if you can handle this if you can prove this this for smooth curves or polygons occurs or just nice curves you get it for free for rough curves so for this problem the smooth and rough cases are kind of equivalent whereas in the original problem there is really this subtle to think in salute cases and the rough cases and this so and this conjecture is assists to open even safe or polygonal coves oh you can epoxy here so so if you could do a couple ego curves then yeah okay so yeah then yeah so the second thing is open okay because yeah if you could have put clicking occurs then the approximation I got worse because there's no longer this this this open condition and it's it is true for general cases okay now one of these invites open is that is that you can try to run the same homological argument and you could you could account they're all squares in its problem and you could do the homology the intersection number without even and that's useless yes that thought yeah it's yeah the number of squares this is now an even number and and and you just can't make any conclusions out of that now it turns out the two problems are related okay so if okay so which which direction do I want to say yeah so yeah so if if if if if the original square problem is true if you eat okay drink is true then actually the periodic is good lecture is true let's say this something is bugging me but what I'm saying is this doesn't quite fit with what I just said so oh right yeah it's actually not completely true what I said that oh yeah yeah knock knock you to look the way I phrased it ah okay so there's actually okay they all basically fight there's actually two versions of the square peg Purex group ikon so so in this version i'm forcing two two vertices to lie on the bottom curve and to persist plan on the top curve you could talk maybe about yeah and the memo code is there's also sort of a weak version this conjecture where you don't make that requirement so you also allowing squares to you also allow two example all four vertices to lie on a bottom curve or three vertices that Monica but one define the top coat that's a weaker conjecture now for that week of conjecture it now can't easily get rid of the condition that the the square degenerates because now nobody even if you translate these two cursory file paths it's still possible that you can also be a bit obsolete small squares okay so for the weak conjecture there was no it was again a distinction between nice curves then and rough coat so careful a strong conjecture there isn't okay yeah so actually the okay the connection is that if you know that the scraper conjecture is true then the weak version of this conjecture where you you allow the squares to talaiya on just one of the two curves like this true first nice good sign nice I won't quite defined but for example polygons would suffice so this is a key this fairly simple observation well what it does tell you what a strongly suggests is that if you want to prove the ocean or conjecture you must at some point deal with you must also somehow deal with at least a week burden of this conjecture and and this is a conjecture in which the whole the whole module argues is do not work and so this kind of tells you then you would not expect to be able to prove this conjecture entirely about how logical methods whereas this integration argument does keep working or talk about that in just a little bit in mystery otic setting so why is this true so it's a second like he just explained by a picture so if you have two nice purity curves time on gamma - okay and you want to find a square inscribed in this union you can just extend both vertically okay and so now you have you have two infinite curves in the plane now this is not yet this is not a simple closed curve but it's very close to one and so you can sort of pinch it off of the edges to make it simply closed curve so so what what a Q do is that I pick a very large number n and I apply a map and it doesn't really matter what the math is okay but it turns out that this map works okay so what this map does is that it maps that the whole play are to cross our two conformally or not almost conforming to assess trip I think minus n times pi / 2 or something in PI over 2 okay this is just the derivative of this maybe I don't have these this sorry this is actually I'm not a part in it which we here but but ok what this does okay the precise form is not so important what it does okay what it does is it it will turn this this will turn the union of these two curves into a simple closed curve and if I did it correctly you were doing it and we just almost conformal okay that angles are almost preserved locally in the limit they will be preserved and so you you yeah you squeeze it to a to a simple closed curve and it'll be a nice curve except at the end points where it oscillates a lot and if you know the square peg problem is true for this sort of curves you can find you can find a square you can find a square inside this this curve and then it turns out that that you can take limits okay so you can go infinity you can sort of renormalize the square just sort of bpp somewhere in the middle and if you take limits you'll get so fat and almost square inscribed in in this ahir and you take limits as long as the curves are nice enough so that you can get obviously these infinitesimal almost squares as long as it's nice enough you can show the blue motoki non-degenerate square contained in the original curve so you you can you can apply the square pick problem to this compactified curve and then there's not compactness in the problem I feel good nice enough that you that you can recover it here okay so I won't detail that okay so are you so basically the periodic problem is is is a sub problem a sub case of the of the square of the square case and it is a dual so this instance or the first case of a square paper bottom where when there really is sort of an infinite amount of oscillation going on so so all the previous results positive results for the problem and sort of involve the case where the only a finite finite amount of solution that sort of roughly what all these conditions are saying but now yeah once I mean from our oscillation you can't distinguish mean even in audience sections basically and and is this the some sort of the technical term is swindled versus real so no going on where where you can indicate not or Dunedin parity here all right so the homogeneous don't work anyone of your deck setting but the integration argument still works exactly the same argument that okay perforce so it gives you that that the pure like square problem in strong form it's true if the truth is demo 1 and gamma 2 are 1 minus epsilon with this class okay so if your two curves like graphs of Lipsius functions and the slope is never bigger than 45 degrees minus epsilon then you exactly the same I can basically works yeah we can show that that you can sort of find ocean area under curve and so forth yeah yeah by integrating you stole this from Y DX this don't make sense ok now there's actually so this becomes so this is in some sense a bilinear version officer the square picking clique of as a linear problem in some sense yeah it's like you linear in but it has just one curve that's important this is where hey he bilinear version without the linear because it's these two curves but there's actually a Quadra linear version which I think should be true so in fact i conjecture a stronger version so instead of having two curves at the shiitake be a version of four curves okay so that gamma 1 gamma 2 gamma 3 gamma 4 before those curves so no whining number one and with one constraint that's the area in the flick area under gamma 1 minus K 1 a gamma 2 gamma 3 so I need this particular I would make some integrals to be zero then there exist a square XY say must be sleepless B minus a this B when I say for the first guy lies in gamma one second album camera to the car nice and kind of $3 to demo for so given any four curves now but where this area can strength so the the opening some of the areas must must must sum to 0 so gamma 1 number 2 can feel a force like that you can always find a square okay no gamma 4 gamer theaters better yes yeah you can you can always find a square where the first bit exciting gamma 1 Sigma I got my camera to Kirk Alyn came of three in the fourth guy ice in front of fourth yeah when traversing in it say guys or whatever okay so some condition like this is necessary so like for example in the degenerate case where if you give all four curve there just cause on two lines then it's it's easy to see that that and you can ascribe a script precisely when the condition is true because if the fall y coordinate of a square there how things always you okay and so this is in such as a perturbation of that claim so if you can prove this conjecture which I believe to be true that would imply these would be the strong version of a pure X square peg problem that cost must the casement to the Kurds are equal and the other two beakers are equal and so I think conditions is or likely satisfied so this I believe you true the same arguments are for is that this conjecture is true before for curves Lipschitz constant 1 is epsilon okay so if you if you have nice enough curves this is this is a true statement as assess the moment you have this this this quality you're forced to have a square but I can't prove it in general and again it's open even for put peligro curbs if you can prove before clicking on curves you can prove it in general a limiting argument and in fact it's true even so it's open all right so it's true also if to the curves are horizontal okay so that's a key comes the jordan curve top-50 if you have two horizontal lines in there tune on horizontal curves okay you can inscribe it's quite precisely at the moment you have this condition I don't know whose true for even just one I was on the line I said so so it's opening one could so I was on so if you have one Co has also hot off and three arbitrary curves then I don't know whether this is true or not there's an equivalent form which is kind of appealing so maybe so in that special case which I think so this I think is a cue the correct mortal color you want to solve this problem what a problem which is special case of previous conjecture okay that if you have three curves gamma 1 and 2 gamma 3 or three coats a simple post in a cylinder sitting waiting that went and won and now the area condition one is that that the area is some 0 then they must exist x y1 y2 and y3 so X is a and I'm RT y 1 y 2 y 3 and the y sum to 0 such an x comma y 1 lies in gamma 1 X 1 Y 2 I think Emma to next my Y 3 is'nt gamma 3 okay so the spiritual picture here if there's somehow a very fancy version that you can be devalued with um I can't prove it so on a cylinder if you have three curves gamma 1 gamma 2 gamma 3 and the area of the signed area that underworld hookahs sum to 0 then somewhere there's a point X and a point XY 1 XY 2 XY 3 where the y coordinate is 0 at that point so if the area is sum to 0 then there was at some point where locally I think some to 0 so if all three of these curves were actually grass then this just follows an intermediate value theorem but yeah the problem is is that is that the because a large a backtrack and there's some there's some comment work is going on here which I don't understand and well I can reef I can phrase it I can okay I can so the problem is a key purely comentario in some sense actually um in my paper I have a complicated equivalent a cometary formulation of this question that has nothing to curves just has to do sets of numbers that service and their sign paths which sums are zeroed which has a positive or something negative so these questions are here this is equivalent to so the previous conjecture do I still have it yeah so if you take this previous conjecture and you sent one of the curves to be to be horizontal this conjecture is actually equivalent to to this this conjecture after a linear change of variables there's a simple linear you yeah you have to share two of the curves by 45 degrees and and and and this this problem is actually an equivalent form a special case of this problem so this I think is really the more important and again it suffice it to understand this for you know like like Pocono paths in fact even paths that are just vertical and horizontal lines really that's that that's really if you can do that case you can do everything and that boat that's why there's a common core but I mean you can phrase this as a purely combinatorial statement very complicated but koozie kono statement and yeah but this I think is is the weight remaining you can solve this problem then there's a real charge you can do a whole thing but anyway okay so thank you very much [Applause] I think just kill all the questions that I've been just a comment so Terry you're is is well aware of this this this notion of a of two Lipschutz graphs with the constant less than or equal to one anyway has has a name it's called a lip domain oh yes okay Creek Chris birds II okay introduced it in connection with that spots cancellation it aside because been and he drew a picture and in one of his papers of some lips yes they're shaped just like that anyway I thought that was a good sociological manic you may be possible to remove the epsilon and then you could really do the whole thing for the tomatoes that yeah but I didn't try to do that okay okay I didn't know that actually so yeah number three is also lip cheese yes yes the Lipschitz constant is like one of absolute it's a Lipschitz curve it's no longer a graph it can go backward it can go backwards but it it is parameterize by functions whose Lipschitz constant is like one over epsilon so they did it is to rectifiable and crucially it is simple yeah maybe I possibly T's of the generalization hope you see like taking triangles instead of squares or I don't know what a pretty good so does the stick T's can apply right yeah so it's it's important that the number of equations are number of constraints are the same if there's too many equations then it's basically if it's gonna be false and there's too few then it's easy so like it's it's you know if you want to discover rectangle puts out words it's almost trivial yes so there was a survey article of maybe yeah so in the notices of the AMS is a super Michael by Bayer Benjamin's have matched it I should have said this earlier if you're interested in this question I know about two or three years ago which has a nice survey up this question and various related questions yeah so this one seems to in some sense critical if you were placed squares if I have the shapes either the question was too easy or false Trudy oh I have not thought yeah I just have not thought about yeah I mean in three dimensions yeah yeah well for any such company you can again count how many equations how many unknowns there are yeah there may be generalizations but I would suspect that that really bottom is much harder than the 2d problem which is very hard so I just haven't thought about their questions there is no other questions props you can thank this bigger again [Applause]

Channel: Centre International de Rencontres Mathématiques

Views: 120,347

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Keywords: Cirm, CNRS, SMF, Mathematics, mathématiques, Marseille, Luminy, Centre international de rencontres mathématiques

Id: RGxnWgy5i80

Length: 58min 43sec (3523 seconds)

Published: Fri Oct 13 2017