Misha Gromov - Novikov Conjecture and Scalar Curvature with Corners

[00:00:05][music] [music] Okay, I want to make a few remarks on the subject matter which is kind of obvious and historically you know that of about 30 or 40 years or two subject matter were going all along and Rosenberg here he can kind of one of the contributors to the big step that formally Some question on on scary question follows from the solution of strong version of of nical conjecture. Here is a lot of literature and so on. One of the question which arise in both situations is if a spherical manifolds have can admit and another if they satisfy the no of conjecture. Now when I speak about the no of conjecture I consider only the most elementary form of this which is amenable to various what I want to say generalization and relating it more directly to scient so on one hand the simplest case on the other hand is caring case sufficiently representative and uh so so because we don't we can't answer any of this question just make another question [clears throat] which kind of cheating And exactly if indeed the two things are equivalent. It's not strong conjecture. is usual no conjecture usual SC conjecture and it is if the fundamental class validate no conjecture then it must have positive scal and vice versa and so this kind of cheating but this kind of may I'm using to think about that we know we don't know answer to either of the two if they equivalent or not and um So I'm not certain that even was in the version Rosenberg it is exactly obvious in either direction but of course I emp emphasize it is usual no of conjecture was very very naive and what it says I forate here. So we have background manifold about which we care about of conjecture and consider two other man manifolds you map into there and smooth generic map and and take pull back of a generic point has some signature dimension is right and if the signature is hematotopic invariant of manifold being mapped take another manifold and at allotopic equivalence between the two and this of course this map related with this commutive diagram if the signature is invariant is doesn't change and the homotopus and so what's remarkable this is a follow from of novik of brower theory that if y is simply connected you can have any signature which you want ex except for it can vary change hematopy type from x X1 to X2 you may have any signature of this pullback except of course for some dimension what [clears throat] co dimension is just zero this is a and and on the other hand conjecturally if Y is a spherical manifold it's not so in and in in general if the core fundamental class of Y is kind of come from fundamental group you expect conjecture this is a kind of essentially a general form of this that this is invariant. So when you take another manifold mapped to your manifold take pull back of a point and then signature of this pullback if of course dimension with multiple of four istop equivalent to the manifold being domain man the domain man your source is fixed but conjecture can show this manifold y it must be true for all axis and all maps and this is more or less no conjecture it's some variation dimension is not multiple of four but you can reduce to that and just to this fundamentals to special case it's very special case on the other hand it is essentially everything reused to this case at least if you speak about rational conjecture okay so this is understood yeah yes what I say or I say quickly assuming we kind of know that yes >> yes >> does it make any difference whether we talk about homotopic equivalences or simple homotopic equivalences or is this the same Well, I'm not ready. I'm not ready to answer this. I believe it's a material here.

[00:04:58]>> Okay.

[00:04:58]>> I believe it the same. But you see because you stabilize see as much as you can. All this subtle point I believe of simple amount of disappear.

[00:05:07]>> Yeah. And it's rational anyway. So >> it's rational anyway. Yeah. Everything is rational anyway. Signature is rational. Very >> good.

[00:05:15]>> Okay. So and now I I I just I turn my major focus is concern manifold discon for this context seems to be good good structure and so this definition and manifold being stratified. So we divide it in the union of locally closed sets such that top dimensional set is just manifold itself in theory of the manifold. Everything else boundary and this and now they come the point that this locally this composition if you reduce it to small point will be this the same up toomorphism that what happens for convex polyhedra in the paper.

[00:06:01]There is some subtle point of convex and nonconvex by the way right because some angles you know equivalent and some not right when you pass 90 through pi yeah something happens but let's assume we understand it's locally everywhere to a convex polyhedra in the space or actually positive cone in the convex cone in the space so this is our object and this is standard terminology and as a remark which kind of now we have manifold with this property and something related to no conjecture when all faces are a spherical manifolds they're covering are contractable and when inject in embeddings from one face to another injective from their respective fundamental group this is kind of a creature and here you kind of mix topology mixed is chotoics and in fact ktoics carries secretly lots of topology even if we have a usual convex polyhedrin in the ukidian space secretly it carries a lot of topology for example if it's cube you reflect it around the faces then you have taurus and as you know was the first case where no conjecture was proven by no himself and this already non-trivial theorem and this is kind of topology of okmetoatoric of the cube and the question is how of that extends to other convex polyhedra both when it comes to scal curvature or to no conjunction.

[00:07:36]So here here is some reference which I bring here now. So what is no of conjecture here?

[00:07:43]Now you speak about manifold which are cornet and then there is natural concept of a map of manifold corners and it's modeled by the situation when you have a generic map from some smooth manifold is boundary boundary goes to boundary but not manifold and this x manifold here x and y sometimes switch who goes to whom yeah and when you take pull genetically pull back of a corner structure you have a corner structure and more freedom of these corner structures and And now this how we define corresponding morphism to the category of corn and manifold just just faces go to faces and and this category you have a topic vence in everything and you can ask about no of conjecture exactly and because you can take pull back of generic point inside and and if there is no corners you have a ball then it's kind of simply connected situation and then pull back of a point by no means the signature is not homotopy invariant again the moment you have this maps you know what this category is you know what homotopy of maps what is hemattopic equivalence for many for Wisconsin right is nothing but it's of course this definition obvious but you you bring in discriminatorics which become kind of what in my view pleasant about that that you just you ask questions not about manif which fund the mental group but you ask about polyhedra and if you ask the question I don't really about you know triangles or simp because it had this topology secretly in it and conjecture is that if you take any kind of x polyhedrin it satisfies no of conjecture and I believe now follows from what people know like you go along and your guys probably can prove it easily for convex polyian But for general manifold this corners it's sticky. So when it is suppose it's foical of conjecture maybe slightly TK also I don't think it's horribly difficult when when all say all faces as biological cells but it is not and works polyhedron but some other kind of creature then it's not so obvious I don't see how to prove it instantaneously except when it happens to be for example all corners are 90° imagine you have So it's all corners I'm sorry 90° meaning they are said by isomeorphic to the corners of the cube so simply as possible corners then you can defle reflect this creature you have a reflection group reflection group is you know is subgroup of manifold of negative curvature and no conjure proof for this case and so you know the answer >> m here your high signature is like relative to the boundary >> no no yes signature for me is pulled back signature of pull back or a generic point inside.

[00:10:46]>> Oh okay. And that's you know super naive. You don't have even don't have to know by the way I emphasize what gology is cuz signature of manifold can be defined without knowing what homology is. All you have to know what are cycles and that's it right because what is the signature because the quadratic form you take all cycle on the middle dimension the cycle classes this quadratic form of the intersection and it has signature because it's has infinite dimensional kernel but modular kernel is fin dimensional and it has his signature you don't have to mention the word homology and in the proof often you don't need any homology it's much more elementary than homology theory right exactly nothing. It's just it looks kind of super elementary topology, right? And this was so nice in this setting when we define we don't have all this you know assembly map you don't need it just just just about in intersection of that it's remarkable thing that this such a thing existically invariant which you know not completely real requires a little touch of homology but really really underra undergraduate thing and um and and that's about So again for me there is this high signature your signature will pull back of manifold maps of the background manifold in this background if it's convex polyhedrin and your maps preserve faces and hematopy understood in this category then conjecturally it is hematopian invariant which is okay follows from usual no conjecture if all faces are say like of the cube or for any kind of difficult reflection group which so we can reflect this around faces formally to to make a manifold and this is coxic group yeah co reflection but this simple but if it's tricky for example if it is simply done because realize convex so galank I believe your techniques will easily bring this to to you just use instead of a manipulation with dra you operate the h operator and and check the signature behave okay this index of this operator is defined for many for the boundary and everything probably will work but I haven't tried to do that because you know there are people who know this better but just simp this is fun just for for simplex it is a triangle of course I can do it but I didn't try even for three for three simple but you see what's amusing because you can manipulate this thing with manifold you glue to manifolds along the face. We can reflect move this way and that way and all that because it's manifold this corners make kind of a category even with very simple corners you make kind of cabarism category because you take think like your manifold times interval and then out of them you can take next level time multiply by something and then you develop all this kind of categories like a Cabon categories or more than that but everything sits in this in this starting starting point just polyhedra all this kind of um cabon category kind of corresponds to cubes but simplic is already more complicated and we in the minute I formulate very specific one specific question I have a simplex I have another manifold which has combinatorial cornal structure simplex you can see the map of such manifold into the simplex faces go to faces is homotopia of this map preserving this property faces to faces is homotopic equivalence between many focus corners preserving the faces the question is if this pull back pull back of a point signature doesn't change certainly much more constrained than usual miracle conjection see it's more constraint so it might be easier it's easier therefore I believe if it is convex polyhedrin in Ukrainian space it's doesn't look horrible but for general prehedron you can make obvious obvious obvious conditions yeah it's not true for any but if sufficiently this corner structure sufficiently fine refine for example take more or less any structure then take something like division you make it finer then I'm pretty certain you have no conjecture but then it's interesting thing so how commenator will be most interesting to see interesting commenatorial characteristic where no conjecture is true and when is not. Okay, because this and and and this is a kind of interesting thing topology is kind of really unseparable.

[00:15:41]You don't know what you're talking about. But closer to no of conjecture, you can see the manifolds where all faces are themselves as spherical manifolds and all maps are injective and then of course is some inter interpolation but I'm thinking this category you can make interesting reduction of one case to another case but I don't know but this what I find amusing. So this is one first question but again coming to my general general question is because now I'll from this moment on I'll turn to I'll turn to SC but these are the question yeah the basic questions so how far you can go from because of conjecture is topological is quantitative aspect are kind of slightly not so clean yeah there is no numbers in there on the other And looking in polyhedra you can produce numbers having purely topological significance. And here they are it will be on my next slide. You see it's here where there Yes. And another yes just this is what I want to say. So now how we can make topology when you have exactly polyhedra is that kind of make topological invariant starting from discriminatorial data and incorporating scales because g topology you can think maybe interestingly to nical conjecture. You can consider all geometries on on this on on a given manifold with corners which satisfies our usual condition. It has scaly curvature bounded from from below say all faces I mean convex and angles bounded from above from by some numbers.

[00:17:45]So we have a set of numbers and we consider only those where this condition is satisfied there existic with these properties metric all angles kind of rather sharp and scal is rather big so there are particularly vector of numbers where it's possible and sometime it's impossible so I have a set of vectors in the equation space right it's really kind of shape which carries a lot of numeric information what it specific examples take a simple simplest example probably cube where it's probably easy but already simp maybe rather subtle. The question is what is the nature of that set and and if it has kind of significance topological significance which may allow you to refine know of conjecture right not just something true or not true but something about how combinatoric of the hidden tell you about this set because topological is there is a set convex set is carries topological about many it's not like yes yes So as an inovical conjecture it still kind of introduces numbers. So that's the question evaluate this set for a specific polyhedra and give you a polyhedral like simplex and give you some generate some set. So what kind of set is this? A very simple question. I didn't try to answer this but I just want to >> m your sigma is not necessarily positive right >> whom >> your sigma like the lower bound of scale coverage. No. Yeah. I I I don't have to say equal. I mean probably it's not absolutely clear the same. Probably the same. I can say I consider only metric when certain numbers are realized or which B number bound. I don't think that's fundamental. But I prefer say bounded.

[00:19:38]I prefer to say bounded. Yeah. But there is also think about the volume. In a second we see that volume careful about the volume and go to negative cvature.

[00:19:47]And now we say some about that and this is related to some theorem by Anderson and and and thirst that so what I'm certainly thirst is perman shown that you have hyperbolic manifold and you then change it metric then and keeping the the volume where it is then scaly curvature may only go down right as usual and uh and then it observed by by Anderson that it's not just volume but just this integral dimension two a minus 3 will satisfy this condition only goes down therefore this it follows then the same you see it wouldn't follow from perman stance but from the additional argument by Anderson the Same apply to polyhedra. If you take hyperbolic polyhedrin then in dimension three it will satisfy this kind of condition if it is reflection polyhedra.

[00:20:57]For example when it is cube or whatever which are many it satisfy this condition. If you take this negative part of the scaling curvature integrated in this thing with 3 over2 power then there will be this for this formula which is kind of cute right but it's open for general convex polyhedrin only for those IC of course again here we we know that this there are several in the proof the one is in inside of the argument by Anderson and other imper and I guess all of them I believe again without seriously thinking but on the surface of things they must be they apply it usually to to closed manifold sometime many manifold with boundary but I think everything applies to many focus points so you have to carry this richer flow and etc and again this usually it works I mean I have no serious mathematical reason for that but superficially looking from general perspect effective it must be so so this very plausible and but then next level what happened dimension bigger than three or three well I don't know how to prove it but there is a strategy in high dimension who knows and before I realize this using that give it for three I would never conjecture this in higher dimensions look so so strong and uh but who knows maybe true and if not true it might be easy to count maybe to find counting example this will be kind of polyhedra because makes sense for polyhedra it's nice you can check it for simplices for for small for big where there are many cases when this can go wrong right but this is true probably for any convex polyhedrin or or in fact for any hyperbolic manifold this corners doesn't have to be convex it may be really manifold you may have a lot of topology but it is all faces flat hyperbolic flat. So there is some corners and then there is and then if you vary the metric then this this integral can only go down only goes down negative part of scalation must increase positive kind of immaterial only can become more negative and uh and this is nice because kind of rather strong conjecture and may be wrong but if it's true of your fun and the dimension trees there are ways to say something okay so actually I more than halfway through then I say formulate some other conjectures which are uh which more more traditional this is also so okay so what is so maybe I just must say little interruption And I suggest I not actually your question but my your opinion what you think about that the presence here you Bernard and just for context. So if you uh formulate this conjecture in for the spherical case so with um on spheres. So do we then have the uh so the kind of counter examples um as in the um as for the menu conjecture or is this something completely different? M conjunction what what do you >> yeah I mean that we um have a kind of um extremality or kind of um result if you have um uh so extremity for the half sphere on the uh so when when you have a metric it's another story this exactly but my emphasis here >> first all this conjecture makes sense for closed ventil but the emphasis When there are corners.

[00:25:04]>> Exactly.

[00:25:04]>> Yes, there are indeed some conjunction as a result of this >> type for these corners. But when you come to positive curvature, you have to impose typically condition on the size of the manifold.

[00:25:16]>> Right?

[00:25:17]>> Because your maps when curvature is positive must shrink somehow controlled by this positive part of SC negative doesn't happen. So this are rather coarse statement. they're much coarser because this doesn't enter the game. We don't know how to make such a stronger statement here. So this causes negative in a way kind of more superficial may be >> but you don't know how to prove it and of course even the sperman theorem the per first kind of result in this direction there is this sharp inequality there are other this some result involving spectrum of the manifold this I forgot the name of this lady who woman who proved it and uh it gave pretty good estimate for the relating scaly curvature and in the first value of second the value of the universal covering of the manif but this essentially constant and but for the volume in dimension three of course is the core of the experiment the I don't think there is anything of the kind remote in exactly conjecture disappears It's exactly completely fail. So and and and nothing you don't expect anything quantitative there because just only it must become much more quantitative to become interesting right of course if you look at the no conjecture for kind of geometric form when you have this map between map they pull back no matter how much this map from quiz isometry how far boundary from there you know some argument of the imply no of conjecture of this of the shape of your signature is still invariant if you your boundary kind of moved far away in the way your map your homotopic equivalences still keep away from the boundary right certain proofs have perfectly work there but they work but I don't know maybe you know there really sharp result there quantitatively asymtoically you can say what happens but I don't know if There are sharp quantitative results really inequality interesting inequality saying that the signature is invariant if but again next question what happens if you slightly viate that in in how much you have error in this in the signature right so I this was my first question what would be the correct quantitative version of it but I don't know if it's how plausible such thing is and here. Well, what I'm saying some question which I ask I you may expect reasonably kind of kind of comprehensive answer but but but of course I don't know but again just my general kind of thought was in scaly we go further and fine and final quantitative results relation geometry scal it doesn't quite work so no conjecture and The question still brought it there you can have to work just in terms of direct hip operator and how spectrum behaves it's another nice and varian spectrum of the h operating some spectrum properties related to something else that's one way to perceive or you can play topological game but there is I think some actually delay in topology and comics may you can suggest a way to go.

[00:29:09]At the moment it seems to me that you cannot ignore discriminatorial aspects of that maybe fundamental.

[00:29:17]If you don't understand them you they're not the end of the story but one of the steps you have to bypass. Now let me again repeat the same question with scal simplicial volume and again kind of essentially here kind of was emboldened by this observation by of of refinement of the perman by Anderson. Yeah, it's little simple remark. It's not complete trivial that you can get back with integrals and that is the question which again follows from dimension three from this anderson for closed manifold and when I speak about simplicial volume so you represent your cycle is combination of real combination of simplices as usual and then there's inequality and if it is mainly with corners you insist that you simply the boundary fits the bound the faces that your your stratification into the corners and and accordingly you have this which were closed manifold follows in dimension three from Anderson and uh and this is again nice because very general statement and again makes sense for it makes sense for polyhedra and uh it's everything of course about negative case because po this would imply of course that many pos have zero simply ial volume which we cannot prove but in this form we have more examples to work on when you have this convex polyhedrin that's one of the points you and and you they may special cases here may be quite interesting and amenable to what we know so it's another question and and and then there is another invariant where I don't know what the sharp constant should be and I just want to again repeat the definition which I think is a kind of nice which is a nice object besides so you introduce so what you say we have a manifold and you know the bound on the scalation and you're assuming it's negative otherwise this invariant is supposed to be zero and you can see the this your manifold and you can see the such maps very very high degree for other manifolds and you see by how much this the Morse number increases. So how many critical points you need for morse function. So what is overall kind of homological complexity of a manifold and then this is a again conjectural statement. I want to edit it because I didn't isolate this invariant. This is a count counterpart of of the simplicial volume but defined not by simplicities but cells on manifolds and they are related and then probably I don't know how and the only estimates which I know come from from uh which are known only for manufacture of course which have non zero characteristic if the if they zero characteristic this environment probably zero which is by the way I don't know how to prove this it's it's zero threedimensional hyperabolic manifold by way we know about this circle this is a vibration of the circle theorem so this invariant is zero but for the fact it's non zero for even dimensional hypotic is a well it's just your index kind argument and some moment you need some particular presentation some property of some final project you know something about this Golang probably the way you look you said kind of know more than I do oh u yeah that well Chhatu and I have been well we work on this conjecture yeah we don't really get exactly the result you wanted but >> you meaningfully this with this invariant. Yeah. How many the Mor number?

[00:33:38]>> Uh I think it's more the simplicial volume.

[00:33:40]>> No, no, but that's another story. With this volume, it's another story. And this volume you see the point is we know the simplicial volume non zero for all manifolds for all locally symmetric spaces for most manifold in any sense and here we don't this is not true. So the meaning of that is not that this corresponding wall group of non zero fundamental class. So essentially what it says you take this fundamental class is element in the wall group in this dimension no conjecture says it's non zero right >> but this says if you take a multiple of this class the rank may when represented goes linearly to infinity it's much stronger so this is much stronger not sing of this variant is a significant strengthening of the no of conjunction in terms of the L groups the number of the cell essentially at the rank of the so we have quadratic form and you and which quadratic form and this all this environment you iterate them right you don't have in the naive way you have some algebra and take some weed group on the all this algebra so it's quadratic form which appear there what is the kind of what dimension non-trivial problem but how it grows when you iterate and amazingly it may know simple cases goes proportionally but but with three remains it's not so it's not zero but it goes very fast to zero because because of the vibrations there so I think it's very amusing and variant and the fact topologically even the first example where it's not trivial when it is for for human surfaces it's kind of easy if you take even surface map one to another with degree D then genus roughly multiplies by D at least by D right and it's kind of clear but if you A product of surfaces is not so clear right if you map any manifold with high degree D then the number of cell grows proportional to D. So you have product of two rem surfaces of of posive genus and you map in this product you map many photo of the same dimension with degree d then the minimal number of of um cells in the in the cell de composition grows at least as fast as the up to constant I don't know what the constant is of course the proof looks to give a reasonable constant because rather sharp proof using is index argument and there is no aspire I remember there is no kind of rather everything goes out sharply but still it is kind of interesting interesting interesting thing what happened there in in general if you think about many false locally symmetric spaces and what the homology of they covering already there not speaking about anybody map any degree it's very subtle arithmetic question right you have JL and Z and You take some group of arithmetic subgroups.

[00:36:38]What the homology of those? Who knows how they grow up? Especially when you look at with coefficients, people like choice probably have something to say about that. So it is a very interesting thing that is still present topological fact but there is no simple topological proof of it. It depends on existence very known example very particular representation of this group and usual proof of conjecture as we know we really need some some flat bundle with some quadratic form and so it's just imitating very closely the original and which does not generalize as far as I can see for example if you take um general even dimensional manifold of negative curvature we don't know whether it's true or not and of course I don't know if there are I think there was a kind of funny example that's odd dimensional manifold of constant negative cure being vibrated over something I thought recently some papers came like that which suggest this environment will be zero so naturally conjecture that for manifold of odd dimensional manifold possible of any kind of odd dimensional meaning for this environment always zero but but for the even dimensional say manifold of negative curvature I guess it is non zero but we have absolutely no inkling how to prove it and what I wrote here with scal curvish is kind of well kind of opposite g bound on So this is now let's turn to the last topic but Misha saw this um statement about the scala curvature this where you have also a control of mapping degrees this vaguely reminds me of something that you also mentioned here in your old book um spectral gaps and higher signatures where you know this is why I I I realized there. I mentioned it there and then >> proved what was needed for because before I only could do it for kind of simple example from the product. There was some simple but then he proved representation where you expect it to be the right representation and classical some classical representation of certain properties and and but then there is a related question about characteristic numbers may or may not be related to what we say but now come back to skovish now the following question that in some sense that certain environments can be multiply multiply main for scal scal adds and you and see there are many cases when it's not particularly scary adds but extreme of scaly cover also adds but extremely you have to define what it means right so you can using topology as a multip property of many property you define some extreal number the extremal scalation this might be additive and just specific question which I want to address which is we discussed with with Charlie and where also you can prove something when there are reflection and no reflection and here what's amusing the case look rather hopeless even for triangles we don't quite know in general but the first nice symmetric case about pentagon. So we know something true for regular triangle something true for regular square but we don't know it for pentagon. So this will formulate the question and uh we again we start with convex polyhedrin and but now we are concerned not scal curvature on this but scaly curvature on this polyhedrin cross s2 and geome and we compare geometry of this product it may be slightly more general but that's good enough for exposition which has positive mean curvature as usual and has all corners greater than greater than sorry smaller than the corners of your polyhedrin and moreover there is a map from your manifold to this polyhedrin. All angles go down and the map is distance decreasing, right?

[00:41:42]And you you want and the scaling curvature will be greater than that of a sphere of a sphere.

[00:41:54]And and then you want to say that it contains inside a sphere which has codic curvature as big as of the usual sphere which of course not true literally but what is to what you expect it is inequality. So there is a sphere inside which represent fundamental. So again you have a manifold which is kind of looks expected to be smaller than this this this sphere but you don't map to the sphere. You map but you map to this polyhedrin. So sphere IP is a fiber right and so there's decent decreasing map all angles sharper than for the actually metric product mean then you have this conclusion that there is a surface inside which homologous to zero multiple of non-zero class it's area smaller than area of the of the visual sphere in the way it is kind of smaller than usual sphere but it's not and and the proof if it work if the background were not a polyhedrin that was honestly closed manifold then it was theorem by by you know zoo pronounce his names correctly then it is it's done by producing this minimum surface there we produce minimum surface in in this class and it has area less for pi but Here if your domain where you map reflection polyhedrin for example maybe cube or some nice triangle whatever then you reflect it again you reduce to this case but if it's not you cannot reduce and usually split argument we use the cha completely dies for triangles maybe this argument by which you discuss it may work with triangles for quadruples it wouldn't work because And this problem runs into very technical thing about capillaries at the corners. So if you know if you knew that capillaries at the corners were okay then it would be pruned for triangles.

[00:44:02]That's it. No no more than that. But but if you look at this case this pentagon absolutely there is no visible approach to the problem kind of absolutely we can prove nothing which was none of the method tell you. Of course the dimension two is special. The same question exists when we multiply by sphere of any dimension. But then it's more technical statement. So what happens? Again you we say all the same words but now we don't claim this sphere of small area but there is a sphere of small kind of generalized kind of k area or some some generalized still small sphere where again if it were closed manifold in dimension now must be careful about which dimension dimension below seven or eight or something actually the recent improvement and some recent improvement we probably up to mention 10 probably but not more not as much as 11 or 12 but we you really need honest minimal subariety not some cheating by blow up from passia then you can say that combining this with some indexic argument you can prove there are spheres and this has something to do with this question many for which cannot be immersed into bounds domain and ucleian space for small curvature the same kind of problem but but this is a kind of key question which is for kind of you see difficulty of this problem because none of of known methods can you give this conclusion that if you have manifold it's sphere cross something and it's polyhedrin there is a small distance decreasing map to the to this polyhedral domain like this simplic so and curvature is positive then you you expect that This cannot be big in this direction, right?

[00:45:58]Because station this direction on the sphere po kind of model cure goes to zero. So but we cannot prove it for of course it's if you know it's something for square we have have trivial proof for that some kind of problem rough rough estimate maybe I'm not certain by the way but this is a what's so nice this is a very clearcut conjecture just about pentagon something we don't understand about pentagon and uh even if you say okay not poish say zero sky say zero kovish She's a free domain. It's already unclear what to do. It doesn't help. So I can say is this a domain in in in in the ukidian space maybe which had this kind of features when maybe is maybe not in a ukian space but sphere crossbow or something to avoid irrad problems or can of course sphere by a disc and then your condition will be slightly more delicate. But in principle when this poly very simple polyhed there is a problem.

[00:47:06]So your theorem about angles you see now has this stronger version right if you multiply it by a sphere it still leaves this conjecture. So it's again this philosophy that all thing about scal up leave dimension you multiply them by something and properties still remain.

[00:47:26]So it's they behave very nicely under the product of manifolds and again that doesn't have strictly a product but somebody which is mapped to a product as usual but the same by the way conjecture with the spheres you say instead of this polyhedrin you take closed manifold which say spherical and then we have the spherical manif and the stronger conjecture will multiply by sphere You map it some way and there is some fiber. This fiber must be kind of morally small and this is a refinement and generalization of any theorem you have about and my issue is once you even accept it and imag conjecture because you see we mix on the bottom you say something which is kind of no conjecture maybe true maybe not in the fire sphere was opposite to no conjunction but it's still clear opposite and then the product is properly formulated. Still remember that in scalature world what happens in no conjecture world and then exactly if you start doing that may help you to to to to make to to make some progress here.

[00:48:40]Right. So so this is this is my my my message that looking at main visas you have so many question opened and and and you have to go go from one domain to another and see what you can do. Maybe find a how to crawl inside and see what happens. Of course many of this conjecture make completely false but making count example kind of rather difficult. So far I haven't seen any convincing example with no con simple variation of what we know but I haven't seen construction very unexpected construction of many photo or somebody potentially breaks some version don't have that we optimistic and in this okay so I'm through I don't think they have anything else to So and I expect your reaction again >> because I >> thank you very much for the talk that's great something like that >> very interesting >> so if you don't again we don't understand the two issues ah many focus corners in this domain no truly understanding relation with the domain so I have kind of two edges open. What is the third? So we have lots of lots of unknown here. But I love it. But the last about pentagon for triangle say maybe even for quadruples if you believe in some regularity for capillary things which is kind of plausible but who knows it maybe on the other hand this capillarity may break down and then this may be not true even for triangles. For regular triangles it's okay. It's called by by reflection by reflection.

[00:50:36]I mean what I find interesting fascinating is that uh so usually so when you talk about the nov conjecture then you have the simply connected case okay and then you also have the non-simply connected case so then you have higher signatures and all these these things and but now you kind of generalize it into a different direction right you you do not no longer talk about smooth manifolds closed smooth manifolds but manifolds with corners and the fundamental group part is no longer so important it's kind of orthogonal somehow it seems to me Yeah, it's kind of kind of orthogonal, right?

[00:51:07]>> Yeah. Mhm. Very >> but again secretly this but maybe on the other hand the structure of this meatoric it secretly generalizes what fundamental group is >> yes >> when it's reflection polyhedron you can say that's is fundamental group but if not it's something else it's some object and I don't know what it is some interesting object mathematical object and of course this may be not true maybe if you start making count example about maybe completely will collapse is nothing there. But if if it doesn't, it suggests interesting interesting development in and >> are these objectives more accessible if you assume that the homotopic equivalences that you consider are of a certain kind of maximum complexity so that you say okay it's you have some control on the homotop equivalence that you are looking at so that you have >> no but exactly this hemattopic equivalence in the category of of a category of corn manifold.

[00:52:13]>> Oh okay.

[00:52:13]>> Mhm.

[00:52:14]>> Of course. Yeah. It is the whole point otherwise if you don't say it it's just nothing is true.

[00:52:20]>> But then this corner structure doesn't interfere and you and the simplest case you take this manifold multiply it by a sphere and just ask about this how it changes my topype. And if your manifold is really kind of cornered manifold itself is product of somebody by sphere probably S1 or will not work S2 I'm not certain say by3 and then it behave again like in the simplicet case this my guess it will kill this this positive scary will kill will kill topology kind of interesting right and you connected so products play essential at all. Another issue immediately arising. We take this finite product. We don't know if if there is a reasonable way to go to infinite products, right? Or unspecified product or something and corresponding because every time we apply particular zero cooperator adapted to this dimension but we can do it without efficient way to with unspecified dimension. might there may be some way to do it I guess but I don't know how to how even to formulate properly and of course you can imagine some formulation but I don't know good enough formulation that's another issue so again with with kish my feeling you have to maximally try to generalize go to the edges what you know and and this may bring you to new new new vision what happens right because we a little bit stuck And uh we have this problem we don't know how to solve it. some of course like irregularity of minimal services right I think all believe eventually will be solved by pertubation or something but that see in principle we don't know what to do and one of them with spin here brings another one we don't know what to do for fourdimensional manifold and it is pentagon cross sphere really nice bridge manifold if it's hexagon it's okay but for pentagon We don't know this. He agrees strange you know maybe it is indeed something hidden there which maybe not exactly but this is amusing in my view that we have this problem and another again with no conjecture emphasize we don't have to know even what homology is and this is not so stupid we think in those terms maybe some secret some secret word because there are any way you may enter it and develop some general theory of completely different kind of conction of course spectrum of the operator how it behaves in in how influences topology and geometry but this I must don't don't quite have this feeling it's there is no spin here nothing But the questioner among us with us and so forth and again for triangle you know we spoke to to explain well this was what he proves himself dimension three and here is dimension four and this kind of subtle regularity theorem for you take minimum surface which is and boundary has a corner What happened on the corner? And maybe in so bad way you just get it. You can't do anything.

[00:56:10]This is when dimension three it's okay with dimension four is if you try to apply against this kind of shyo argument. But this argument become kind of artificial here because it's not a product. You see you go from cubes to general polyhedra. Maybe in principle it's not true. I mean it's not an issue and but some estimates need to be must be there.

[00:56:39]Okay. But so I have nothing to add what I >> was all I can all I can say about it.

[00:56:49]>> Okay.

[00:56:50]>> Myself I know I'm too old to think about I just can offer this to somebody who may may find it interesting.

[00:56:57]>> Yeah. Yeah, indeed. Very interesting.

[00:57:01]>> It's intriguing to have such a conjecture. Yeah.

[00:57:04]>> Yeah, definitely something for the future 50 years. [laughter] >> So, Misha, did you have some write up some notes for this?

[00:57:13]>> To to say what?

[00:57:15]>> Oh, do you have some notes like >> No, but this what this what's what I have here.

[00:57:21]>> That's all I have.

[00:57:22]>> Yeah, they are probably on your homepage, right? I mean on your homepage there are many notes. I I am not certain I put it on my page.

[00:57:28]>> Yeah, maybe you can do that.

[00:57:29]>> Maybe I have to. I thought about looking at this book more carefully and I probably put it on my page.

[00:57:36]>> It's very interesting to look on your page anyway.

[00:57:38]>> I Yeah, I doubt >> I recommend that to everybody.

[00:57:41]>> Yeah, I don't think it is. I don't [laughter] >> Can I ask a question?

[00:57:46]>> Yes.

[00:57:47]>> Yeah. So uh this uh area question uh for this sphere fibers is very this is somewhat analogous to the euro n minus two width question for scalar curvature because instead of saying that you have a map to a in this case prespecified pentagon >> for the fiber of the area is small there you're asking >> I map to the pentagon preserving this corner structure and the map is a decent decreasing >> so so so is this formal uh similarity just formal or there's uh some connection here. No, but see this question already if you map to a sphere and this decreasing map and and if this pullback will be small it's also unclear what happens but swear maybe more difficult even than this because for the Taurus you know and for for cubes we we know how how it works right and so right this is a kind of if you have many photo of certain positivity of SC negative cv contractioning then the fibers ought to be small whenever you can apply of course Shenya method you have enough room to apply it but this exactly exactly what you you don't have here because it's not cube yeah is as a split your in in this usual situation multiply by something which contains inside the cube it's cube or something bigger than cube and so you can splitting but here triangle you can split it but you can reflect it and then split Seems quite absurd. Yeah.

[00:59:22]>> No, my question was is there this relation some relationship between core dimension to uron width and this uh very uh restricted co- dimension to area of fibers.

[00:59:34]>> No, excuse me. Can you repeat it? You can quit.

[00:59:38]>> So your your your pentagon conjecture is that you have this map to a pentagon or in general a map uh and then the area of the uh core dimension two fibers is small because here core dimension two fibers are just spheres. So >> but no but this may be the question here indeed because I mentioned two fibers I'm in two dimensional fibers in general I may have high dimensional polyhedrin and multiply it by sphere in map to this polyhedrin and if this polyhedrin is refraction polyhedrin then what I said said is true okay it's it's not so much dimension two it just accidentally happened to be it's about two dimensional fiber if it's higher dimensional it's threedimensional fiber the statement of the theorem become more technical yeah I don't want to exactly formulate it but it says that that it's conjecturally that certain homology class is is must be small and this is formulated in some kind of specific terms and proven with index theorem here you don't need index theorem but you can do it also the index theorem for families but it's not dimension two it's a dimension two so in princ in general believe both big dimension big dimension you take any kind of polyhedron of certain kind you multiply it say by sphere of certain kind and you know all this and the map to a polyhedron distance contracting angles are right positive curvature big and and mean convex then the axis of this by which you multip multiply this manifold must be small which case when you just multiply you take your polyhedral like you multiply by the sphere and you see this equalities you know what are angles what all things are you saying you cannot and map is decreasing keeping this distance decreasing you cannot simultaneously make co positive angle smaller without contracting mass in dimension of your manifold you shrink this closed manifold of course you you you makes may make things smaller and nothing changes. So this obvious conjecture there are actual geometric products in saying the extremal situation is response this extreal solution for this question.

[01:02:04]Okay.

[01:02:05]>> And this can be proven in some cases when the basis is say Taurus, right?

[01:02:10]When exactly use Shyao, you split it and come to this small small fiber. But if it's not a Taurus, something which far from being split, we don't know what to do.

[01:02:21]>> So DJ, you mentioned the uh eison with uh so what what do you have in mind there? So it's >> I was I was I was essentially asking if you can ask the same question with you uh urone n minus two width. Uh so right now it is the uh area of uh dimension two fibers which is >> no but you need all fibers to be small.

[01:02:46]Yeah I just say all fibers must be small.

[01:02:49]>> I see.

[01:02:49]>> No no it's it's different. No of course you don't see it's not property of the map. The map only tells you give you bound up lower bound on your manifold. It's lower bound not upper bound. You see it's pinch scal positive it's pinch in one direction therefore expected to be and invo positive and pinch in another direction.

[01:03:15]So this this is exactly a play of one against another it nothing to do with but at the surface of things it's not related to but this is a I'm saying if the manifold you map into just hyperbolic manifold or just any spherical manifold by the way if it's four dimensional spherical manifolds you know a spherical manifold doesn't have so you expect the same be true for any spherical manifold conjecture that the spherical mean have no position have this generalization. So if you multiply it by sphere and map is decent decreasing time you have a small fiber and and the same remains for high dimension. This is st this product feature. But but this you the only way to approach it at the moment for higher code dimension you have to combine. Another of course interesting point that if you whenever case you can prove it except for spheres you have to combine both yo minimal surfaces and index for families. You cannot prove it by either one method which also very muted and and then you go beyond it and then you cannot prove it at all. You don't know where to start. This what I not just you don't know how to prove you don't know even where to start. That's that's what's so nice.

[01:04:43]Maybe you know how to make example.

[01:04:45]Maybe you can start learning how to make this manifest.

[01:04:53]So all all kind of conjectures here formulated in a maximally strong form.

[01:04:58]So just to make it easiest to to disprove experience showed it's very hard this conjecture hard to use proofs.