So it's my pleasure to introduce Matan Ela from the Weissmann Institute, who's going to talk about rigidity of Romanian embeddings of discrete metric spaces. Okay. Hi everyone. I'm happy to be here. First of all, thank you for inviting me. And like Jacob said, I'm going to talk about rigidity of Romanian embeddings of discrete metric spaces. And this is, I'm Matan. I'm a PhD student from Weissmann. And I'm a student of Boaz. This is joint work with Boaz, my advisor. And okay, so let me start. So, first of all, our setup. So we have a manifold. Okay, so we have a manifold M and our manifold will always be completed connected. I'm going to say what what complete means in a second. With no boundary. Okay, and at some point I'm going to talk about manifolds with boundary but in general with no boundary. And we consider this manifold as a metric space. So, if I have two points in my manifold the distance is the infimum of length of curves connecting these two points. And the fact that the manifold is complete mean means that it's complete as a metric space so every Cauchy sequence converges. And by the hopefully not theorem, it's equivalent to saying that if I have a geodesic I can continue it indefinitely in both directions. So every geodesic is defined for all time. So both of these are equivalent and the fact that the manifold is complete. It implies that for any two points, or any two points in the manifold X, Y, I can connect them by minimizing geodesic. Minimizing geodesic and means that the length of this curve equals the distance between these two points. And this geodesic is not necessarily unique. For example, if you take the sphere, and you take two anti portal points so you have a lot of great circles connecting these two points, and each of these are minimizing. Okay, so this is a minimizing geodesic. And by the way I didn't say it but you can feel free to stop me if you have any questions or remarks so feel free to jump in. If you don't hear me or don't see anything. Okay. And we are interested in isometric embeddings of a metric space. We said so what does it mean to for a metric space X to be isometrically embedded inside the manifold. And we write it like this, this X is a is a metric space and isometrically embedded inside the manifold. It means that I have a map. This map from my metrics to the manifold that preserve distances. Okay, so the distance in the manifold between the images of the two points is the same as the distance between these two points in the metric space. Okay, this is isometric embedding. And it's something convenient to think about this isometric embedding as as when the metric space is just a subset of our many. So if I have some manifold, some part of the manifold it's without boundary but I have, I can think about the images of the X. So we will mainly talk about discrete metric spaces. And for each between and for each of each pair of points I know the distance between them in manifold intrinsic distance and the manifold is the same as their distance in the metric space. Okay, these two are equal. Okay, so sometimes more convenient to think about my metric space as a subset of the manifold. Any questions so far about what's isometric embedding. Okay, so let's talk a little bit about some result about a finite metric spaces. So what do we know about isometric embedding into finite. So I have a finite metric space. And I want to say something about the existence of an isometric embedding. So if I have a metric space metric space with at most three points I can always embed it inside our two problem I can embed these three points and just have a triangle inside my Euclidean plane. And if I have metric spaces for four points, even four points are summit spaces that I can't embed not only in Euclidean plane not only in space but any complete remaining manifold I can't admit it because they violate some some basic property that the remaining manifolds have. And these kinds of matrices are branching matrices. Okay, so what is a branching metric space so I have four points, four points x, y, z, w here, and I have this setting where x, y, and z is three points x, y, and z are collinear so why lies on the same geodesic between x and z so these three points are collinear and x, y, and w are also collinear. Okay, however, the four points are not collinear so why z and w are not collinear and why w and z are not collinear. Okay, so I have this picture. For example, this picture we can think of a tree with a root connected to three nodes here the distances of one. For example, this kind of metric space, I can't embed it inside any remaining manifold, because like I said it violates the property that geodesics cannot branch if I have a geodesic that overlap then they must coincide. I can't have this picture inside the remaining manifold so this kind of metric spaces, they do not embed isometric in any completely. Okay, but if I just lose this the generosity, and then then I can say things about about isometric embeddings. So, while then the rest of ski and show that if I have a non non branching so it's not the case that I just mentioned four point matrices I can isometrically embedded inside some constant curvature surface. SKN is just a complete simply when they mention manifold of constant curvature. Okay, if the curvature is positive sphere if it's zero then the space and if it's negative then it's the hyperbolic space. Okay, so if I have a four point matrices, I can always isometrically embedded inside some manifold. And they give also characterization in different cases, for example if three points are collinear, if all points are collinear they say what are the curvatures that that you can embed it in but I don't want to get into it but give a characterization of all of the, all of the cases. Okay, this is a four point metric spaces but what about finite, maybe more than four non branching matrices. So, just some result I'm going to explain how to some reasoning about it. In a second, and is that any finite non branching metric space embeds isometrically in some complete Riemannian surface. So a two dimensional manifold I don't need more two dimension, but maybe it can have a large genus so let's see how you do it. So let's take for example, five point metric space, I have a five point metric space non branching, and I want to embed it inside the toes. Okay, so how do you do it, I take the complete graph. Okay, complete graph, every two complete graph of five vertices, every two vertices have an have an edge connecting. Okay, so I can't embed it in the plane, not a planar graph, but I can embed it on the toes just like I drew here. Okay, this is a topological embedding of this complete graph of five vertices. Okay topological embedding means that that vertices are points and edges are curves and two curves do not cross cross one. Okay, so I topologically embedded. And now what do I do, I define a Riemannian metric that be very, very large on the complement of these curves. Okay, here it will be very large almost infinite and assigns the proper value the proper length to this curve. Okay, so it makes you, if you want to go from one point to another, it forces you to go through this curve. This is how you can prescribe the proper distances between points and if you have more than five points then okay no problem maybe you can take very many many holes and you can topologically embed any graph in a sufficiently high genus surface and then you do the same procedure and you have this isometric embedding. Okay, any questions so far something. So, let's get to our case what about countable metric spaces okay so finite metric spaces we know some things and other results I didn't mention but what about countable spaces so let's take for example as a model this metric space so what do I have here I have. Okay, this is a subset of R3. Okay, this is a subset of R3, where I have in z equals zero floor this is the purple part. This is zero. Okay, so here I have z two. Okay, I take all the points in z two with integer, integer coordinates. Okay, and I add another point I add this point. Okay, and I treat this as a metric space for distances are just Euclidean so here in the distance between two adjacent points one here is one there is a square root of two and just Euclidean distance. Okay. Okay, so the question we asked ourselves our motive in question, can I take this metric space. It's countable. I can take any finite chunk and embed it like we just saw I can take any finite number of points and embed them in service. Right, but can I embed the entire and the entire metric space is countable in this inside the service to complete the money and service. And the answer is no answer our answer to this question is negative you can't, you can embed it in a three dimensional manifold of course R3 where it was taken from, but you can't embed it in any two dimensional and remind me for complete today. And why is that this is just because of theorem we prove that if you take a z two. Okay, just integer coordinates and you isometrically embedded in the surface. And this surface is, is, is the clean and plain it's isometric to do clean. Okay, I took this purple points all of these purple points and isometrically embedded them inside some manifold now I have a flat manifold I have just Euclidean play now I have no place to put the additional point and preserve distances right there is no point in our in Euclidean play that have distances that this point have from all the points of that to have no place to this point. This is the main result this is the main result and now I'm going to talk a little bit about generalizations because I don't need to I can have any lattice and even having every net is fine but this is maybe the picture that's nice to have in our heads of this z two that I embedded inside.  And so this many for must be isometric to Euclidean. Okay, so any questions about this result maybe something's not clear. Or is everything. Can I ask, did we. We're embedding it into a complete connected Romanian surface that means the dimension is to right. Yes, yes. Yes, always. Yes, the dimension is to I'm going to talk about the end dimension and always the. Okay, yes, the dimension here is to the dimension should always be the same otherwise of course I can embed it inside our three. Right. Okay, so like I said, and this proof and generalizes it with someone asked a question, why can't M and have a larger metric of the lattice and what do you mean, why can it have a larger. Okay, so I have. So, so I'm not sure what the question but the picture to have let's say I have this manifold. Okay, let's say it's manifold. Okay, now I took all the purple points, all the purple points of the two. And embedded like the tourist example okay just. This is this is the result I can the result is that I can't have, you would expect that maybe I could have like an infinite genius thing and connect them, but the result is that that you can that if you have you can embed all of these points. So that to here, many, many, many, many points. And you know the distances here, every distance here is clearly on all distance here you clearly on, then it flattens it flattens your, your, your manifold, somehow it enforces it to be isometric to. And, and this is, yeah, I just, I just do a chunk of course it's infinite. There's a lot many, many, many geodesics. Okay, so we will see I hope I hope you see some of the proofs. And, okay, so, and like I said, so like I said it generalized is easy to any lattice and apparently it holds for more generalized which are nets, and also quasi nets, but I'm not going to talk about it. So if you have any questions at the end causing it is a little more inclusive than a net but not very. So, so what is a net and net. And so now it's in n dimension so I said, and now in is a net. If every point in our enemies of bound and distance from this set. Okay, I have some data. So every point x here, and I have some point why my net such that the distance here is at most that is called, let's say three so, and every point at some point from my set so of course, the two is a net, and every lattice is a net. So, it's a generalization and the theorem, same theorem but for next, if I have a manifold which is complete connected to the mention of manifold, and I was able to isometrically embed some net from our two inside it, then it must be isometric to, so just like in Z two or any lattice so it works for a net, and the same corollary that we had in our model, and just formulated in, in more general language so I have some set in our fluid it's not contained in any fine plane for example this said that we talked about so we have z two plus a point so it's not contained in any fact plan, but there is some fine plane, and such that intersects and intersection intersection is a net. So, of course this a fine plane is just the Z equals zero, and the intersection is a net inside it. So, the same reasoning works, and you can say that this text does not embed isometrically in any complete one. Okay, the same reasoning we said for the two plus a point. Once I embedded this lattice, I have no place to put the additional point that will preserve distances between it to every point. Okay. And this is some generalization and now we okay so I'm just going to say in a second but we don't know the n dimensional analog of this theorem we don't know that if we take a net from our end and embed it in an n dimensional complete it's not in a complete Romanian manifold so this manifold is an isometric to the Euclidean space that we don't know it. But we still think that maybe the ability to give a dimension to a metric space could be interesting so maybe maybe it's, it's not meaningful if we want for any dimension. So, that's simply we defined asymptotic Romanian dimension is the minimal dimension of the complete money and many fun in which the, the metric space and base so it's a dimension of a metric space. So for example, again, our metric space that we always talk about this Z two plus a point. So somehow this theorem was able to capture the fact that it's in some sense three dimensional.  Okay, so maybe this notion will be meaningful. And now a definition that maybe will help us formulate our theorem as more easily and connect the Romanian manifold so this plays the part of. So, how do I. Okay, see again that. Okay. So, here, and play the part of our tool in the theorem so if I have some n dimensional complete quantum manifold and I have some subset here it will be a net in the what we know and I have some subset. We say that it's metrically rigid. If, if whenever I embed it in another manifold. And also n dimensional complete connected the two must be isometric. Okay, so somehow if set is metrically rigid it somehow encodes the geometry of the. Okay, so the theorem says a simple way to formulate it is the net in our tool in the end plan are metrically metrically rigid so the, the geometry of the Euclidean plane, I can encode it in a net. Okay, just knowing all these distances is enough for me to recover all of the. And like I said we don't know that theorem and the first theorem generalizes to the end dimension setting or, in other words, we don't know the net in our and are metrically rigid, we don't know that. But in any dimension, we have different fees. And if I have a complete connected n dimensional the manifold, and again, I could I, I have some net in our end that embedded I don't know in two dimensions I know I know isometric. But here I have different movies. Okay, in n dimensions, I always have default. The first remark is that if I add some assumptions on my many fund here the assumptions are pretty weak only that it's completely connected. It's not a strongest assumptions but if I add more assumptions, for example, if the curvature tensor is completely supported, then, then I can have isometry, then, then this will turn into isometric. And this is just by reducing matters to some cases of Michelle's conjecture which I'm going to say what it is in a second, but maybe if you have any questions so. These are the two main theorems. The first one is, is the one when I have an isometry, and if I could embed a net inside a net from the plane inside a manifold and I have isometry. And here in the n dimensional, I know different. So if you have any questions about this before I move on to some some related results and hopefully as much of the proof is possible. I'm going to wait to questions. Okay, so boundary. So this is a some class of results and also Michelle's conjecture. So boundary duty is an inverse problem. So I have now I have a manifold with boundary. All the time I said that we don't have boundary but now we have some. Okay, let me do it. Maybe this is my manifold and I have a smooth boundary. And this manifold said to be boundary rigid. If the fact that I know distances between any two points on the boundary determines the metric inside. Okay, so if I have any other metric that agrees on the boundary in the sense of every two points of the sentences must be isometric to the metric G. Okay. And if, if that happens then the manifold is said to be boundary. Okay, and, okay, not all compact manifolds are boundary of course I can construct some metric and some metric on M and have some point here it's not that's very, very far from the boundary, very, very far from the boundary. So, I'm sorry to interrupt you, but I lost your sound now. I can't hear you. Maybe I'm alone about that. What about the rest of you. No, no, it said that my internet is unstable now maybe it's okay. Oh, I hear you better. Yeah. Okay, so again I'm going to repeat it so I can construct a metric such that for some point. It's not it's very, very far from the boundary maybe it lies on a huge hill so I can change the metric there and it won't affect boundary. It won't affect boundary. So, so this kind of manifold is not boundary rigid and another simple example is the hemisphere so if I have the hemisphere. Okay, and I change the metric. I change the metric here in some somewhere that's far from the boundary. Okay, I make it larger than the standard metric. Also it won't affect boundary distance because any two points and can go around the boundary, and there is no point it won't affect so. There are also no but not boundary, but Michelle's conjecture said, I'm going to say, there are sets that are boundary rigid, and there are many folks that are boundary rich, and Michelle's conjecture was from 81 is that a manifold is boundary if it's simple. So what's simple as simple as like it says here two points, any two points are connected by unique geodesic and the boundary is tricky contacts, it means that if I have two points on the boundary.  Okay, this is not a simple one. Okay, if I, if the minimizing geodesic goes on the boundary, then this is not a, not a simple method. Okay, so for example strongly convex sets are, hey, if you take it this course it's any two points that are connected by a line the interior of it is inside the sphere and, and all of these strongly convex sets, and all of these strongly convex sets of constant curvature, and for example strongly convex sets of the hemisphere and of the clean and space and of symmetric space with constant negative space and all of these are bound and rigid. Okay, so if you have another metric and it preserves the boundary distances on the sphere on the ball, that's for example, you have the ball in our three, and you have another metric such that any two points are just the Euclidean distance, then this metric must go isometric to the Euclidean. Okay, and this, the hemisphere case is by Michelin 81 and the Euclidean by Gromov and the symmetric space with negative curvature is a so called wine galore, I hope I say the name is correct. And, and this conjecture is also known to be true when the manifold is two dimensional this, this was proven by Pestoven Ulman in 2005 so if I have a two dimensional simple manifold, then it's boundary region. Okay, so this is some, it's not very much related but some field of rigidity. Yeah. I wanted to discuss so any questions maybe about boundary rigidity I'm not an expert but you can try it. Do you have any, oh no, no you did it. I'm sorry. So let's talk about another class of results that it's more related to our proof and our results so this is about manifolds with no conjugate points. Okay, so what are conjugate points. So if I have a geodesic dam, I have two points connects two points P and Q. And if I have a non zeroed a copy field on this geodesic that vanishes at the end point, but maybe it's easier to think about it, and it's equivalent actually if I have a variation of geodesics of this. All of these are geodesics. Okay. And this variation agrees at the end points up to first order. It doesn't have to fix them points, but it agrees to first. For example, the portal points on the sphere if I have a sphere here. So anti portal points are conjugate. I have if I have a geodesic right I can take a variation of great circles, and it agrees on the end points. Just then points are fixed. Okay, so I rotated this geodesic and I created a variation should these two points are called conjugate. Okay. And, and one important fact that we use is that if I have a geodesic emanating from a point if I have a point P, and some geodesic from P, if it reaches a point Q. If these two points are conjugate, then the geodesic stops minimizing. So here, this geodesic is no longer minimized. Okay, if you think about the sphere to anti portal points connected in the circle so they are conjugate and indeed if I continue this great circle. It's no longer meaning I can go in around the other and the other way around. Okay. So these are, these, this is a fact, and it can stop minimizing in other cases also you can have the cylinder and stop minimizing in other cases but, but this is one fact and the carton carton of them are theorem, which is a non theorem in relation to this in this area is that if I have a manifold again a completely manifold with a non positive curvature. So for example the hyperbolic space or the previous space. So there are no conjugate points. No, if you take a point there is no other point that's conjugate to this point. And using the carton of them are sometimes formulated otherwise that the universal covers the film morphic to the Euclidean space. But this is just a result of this if you have no conjugate point that the exponential map is non singular so it's a covering map so you have this property. Okay, maybe I won't say anything more about this thing but and this these results talk about what you can learn about the manifold from the fact that it has no content. So the first of all, it's mentioned that the converse of Katana Damar is not you can have a manifold with no conjugate points, but still have some positive coverage. Okay, for example, you can take the, the hyperbolic, a hyperbolic space. Okay, let's say the model, and you can change the metric here in a very very small disk around the origin to have some positive curvature but you will not introduce any Okay, so the fact that you have no conjugate point it doesn't mean that the curvature is is non positive. Okay, we will see that you can do the same thing in the space in a second but and but what what can you learn about about the manifold from the fact it's not it's not useless information the fact that you have no content. So, most in headland they consider the case of a closed two dimensional. Okay, so this closed two dimensional money and manifold that has no conjugate points what can I say about. Okay, so this is a these surfaces these closed surfaces they have a classification, they are homomorphic to either the sphere, or the sum of Torah or the sum of projective play. Okay, so one thing that you can say if you have no conjugate points that it's impossible to be homomorphic to the sphere or to the projective play. This is one thing I'm not going to explain it's because of the universal cover but I don't have enough time to get into it but okay so I have a manifold. Two dimensional closed manifold without conjugate points I already know I can eliminate the possibility that it's the sphere homomorphic to the sphere or the projective plan. But what if it what if it's a homomorphic to the total so all the client but so these two services they have a little characteristic zero. Okay. So, okay so maybe it's a more frequent one of these and they conjectured and they show some with additional assumptions that if I have some metric on these on these without conjugate points that it must be flat. The flat metric metric. Okay. And they also considered other the other cases. Let's focus on this one so, and hopefully in 48 he proved it so he proved the show that if I have a closed remanial surface, close remanial surface, two dimensional surface always two dimensional without conjugate points, then the total curvature the integral integral curvature, must be negative or zero, and if it's zero, then the curvature vanishes. Okay, so why does this prove the contracture. So if I have a by the by the Gauss Bonnet I know that the curvature I don't have any boundaries. I know it's two pi times the other. This is Gauss Bonnet theorem. Right, so if I know that this is zero. Okay, so in the case of the person in the client model, then I know that the total curvature is zero and hope still tells me that the curvature is zero everywhere so it's flat. Okay. And so this is about the two dimensional tools. And what when you want to be 94 showed it for the end dimension. We have some metric on the end dimensional tools without contribute points then also must be flat. Okay. And they use different. Okay, so. Okay, so we know that the counter of Katana doesn't hold, I don't know a non positive curvature if I have no content one is not true. But sometimes it has this kind of a similar behavior so what's non positive curvature means that if I have a point in my main phone. I have a very, very small ball on this point and infinitesimal small ball and its volume will be larger than the Euclidean volume of the same of the of the volume of the Euclidean ball of the same way. Okay, this is negative. Okay, so sometimes the lack of contribute points have a similar behavior asymptotically if I take very, very large. So, these are two results and the second one we actually using our proof so the first one by croak in 92 is if I have a combat combat manual method without contribute points, always without contribute points. And I take this, I denote the volume of the ball of radius how this is the universal car.  So the volume of the ball. Around some point that projects to P. Okay, and epsilon n is just the volume of the unit ball in RL. So the volume of the ball. And in this was a cover it grows at least as fast as the employee number so so balls are basically asymptotically and a very, very large scale there they have a volume which is bigger. And this is kind of non positive curvature behavior, we said so we can learn some things about asymptotic growth of the fact that you have no control. Okay, and but an image I'm not sure that I'm pronouncing it right but in 2013. They showed and we use this result so that if I have some manifold that's different morphic to the plane to R2 to the Euclidean plan. I have a complete running metric on it without contribute points in the area roses least as fast. Again, the area grows at least as fast as the Euclidean area. Okay. And if I have a quality, then I know that the metric is flat. Okay, so how, how will we use this. I'm going to say it again in a second, how do we use this result. So the proof in our proof again, let me remind you. Okay, so maybe it's and remind, let's remind us of the setup and and see how we use this result, in order to prove our fear. Any questions so far about anything. Okay. We have some. And now we're back to our case. We have some set L which is a net inside our end which embeds isometrically. Again, this M. And I have some L here points of L, which I know the distances, the distances are just Euclidean. And sometimes I will treat this L as a subset of RN and sometimes I will treat it as a subset of M. Again, like I said I don't distinguish between point and its image so and I will use this for the characterization and and for convenience I assume that the the origin is inside can always translate the net and it doesn't change. Okay, so what will be our strategy, how will we use the band of an image so the first step is to investigate geodesic passing through some point in air so high. Just a second. So, sorry, so okay. So again I have this M. And I have these points L points L and I take one of these points. And I investigate geodesic passing through this point. So we will show that any, any geodesic passing through this point is minimizing all the way. Okay. And if it's minimizing if every geodesic in any direction is minimizing, then, well the exponential map is a different morphism and I have different morphism to are exactly what we stated in the, in theorem two. The dimension is n dimensional I didn't yet restrict to the two dimensionals. Okay, this is the first part of our second one. Okay, so now we restrict to the two dimensional we want to prove theorem one that we have isometry. So we restrict to the two dimensional setting and we show that there are no conjugate points at all, not only that this point, the point of our net doesn't have any kind of an image of point or it doesn't have contributed but any two points. Any two points are not conjugate. Okay, and now that we know that they're not conjugate. Okay, so this is, we're in good shape we know that they are not conjugate. We know that I'm default morphic to R2. So I use this result, all I need to show is that I have equality right if I can show that I have equality. I know the metric is flat and I know that I have a zone. Okay, so. So the equality case and the fact that the large scale geometry is Euclidean this is not hard just if I have a net, then I have to show that the, the, the, the area of a large body is approximate Euclidean, it's, it's the, it's not trivial but it's not as difficult, I guess, then to show that that I have no conjugate. This is the major part in the two dimensional I think this is the hard part is to show that there are no conjugate. Okay, this is like an overall and scheme of the proof and now I'm going to try to show this theorem to someone just talk about the end dimensional setup try to show as much as I can that if I take a point in my net. I think about it as a point in my manifold, then every geodesic through it in any direction it's minimizing all the way. So maybe if you have any questions before I begin, this is, I guess it will be a bit more technical so maybe if you have any questions on the, on the holistic view of the proof and so let me, let me try and prove this theorem and show you maybe the tools that we use. So how do I do it so I fix a point okay I fix a point. Some point, again I'm thinking about it as a point in my manifold so I have some point in my manifold P. And that and for every direction in our end, or any direction V. I will define a minimizing geodesic. Okay, a minimum, a complete meeting with you as defined for all time. Okay, gamma PV, it's not going to be a time zero it passes through P. So some kind of some kind. I'm going to give it some notion of direction I didn't say how still. And, but if I can construct such a geodesic and show that this map. It maps any direction to the tangent vector of the geodesic or to the geodesic is onto, then I know that every geodesic passing through the point takes this form. I'm minimizing, and I have the default will fit the exponential just give me the different office. Okay. The cut locus is empty, empty in this case if I have every point that leaves P takes this form it's gamma PV for some V. And I could, if I can prove that it's minimizing, then every geodesic emanating from P is minimizing and the cut locus is empty, and the exponential map is the different morphism and will the tangent space is just the form of two array. So I have different morphism, which was what I wanted to show. So now I reduced our goal to show the showing that I can construct such a geodesic such a minimizing geodesic and prove that this map is on. Okay, so now this is my goal. Okay, so how do I do it. Okay, so I use limits of minimizing geodesic that that I'll take but just connecting this point to other point in my neck. So I take this point P here. Okay, this is the point P, and I connected to a sequence going to infinity. Okay, this is the sequence going to infinity. It's norm is going to infinity. These are all points in my net. So I know the distances between any of these two points. Okay. And I connected these two in a minimizing through a minimizing geodesic and I know that it exists because the manifold is complete like we said and there's no. Okay, so I denoted and you know just this in this lecture, I denoted the just like this this the segment the minimizing segment connecting P to PM PM again will always be some sequence going to infinity. We will give it direction now now we can go in a direction but soon give it the direction some sense. But this up early does not convert the sequence of geodesics but they all passed through a point so it has a converging some sequence. Okay, so I will. Well, maybe sometimes assumed that it converges maybe. Maybe not but but basically the fact that limits exist. It's not so hard and show. Since I have again this sequence tending to infinity, then I know that the limit that the limit segment this limit. It's a geodesic rate, right because I have points very very far very far I connect them. Eventually they will converge if the limit exists they will convert to geodesic rate right geodesic rate defined on all positive times. Okay. Just because the distance function is continuous I also know that this limit is minimizing. Okay, all of these geodesics all of these segments were minimizing I chose me so the limit is also minimizing so I have a minimizing geodesic rate, just by connecting this this point. Okay, so now, what I want to do is I want to give some sense of direction so these points can go in any direction but I want to say okay they go in direction V in some way okay so I want to take these points I can treat them as Euclidean. Right I have a points like I said I have this net. Sometimes I treat it as a subset of R and sometimes the subset of N, but these are just points if I think about the tool for example, or is it and so these are just points in my lattice so. So, and I can have some kind of direction just Euclidean treating as Euclidean there points. Okay, so we said that this sequence is drifting in direction V, V is some direction on the on the sphere, and we denote it like this. Okay, so they have two things they have to admit so first they tend to infinity like I said in the second is that their direction and it tends to be so if I take this point and I look and it's normalized direction of the sphere, then it tends to be just just I think the most trivial way to tend to direction. Okay. But now we introduce something that's stronger. But so now I have not only that I okay again is would always be points and drifting to infinity, but now I want that the additive L will tend to zero. So not only am I have this way to tend to V I also have this more the stronger way which we call narrow. So of course if I if I this condition implies this condition so. So it's a stronger notion of tending to infinity in direction V. Okay, so now what I'm going to do I'm going to connect my point P to points PM, they will tend to infinity narrowly or not, they will tend to infinity in direction V sometimes narrowly sometimes not but this. And these are the two ways they have. We introduced to tend to infinity in a direction. So where we use the fact that we have a net. Okay, so this, the main place we use in this part of the proof later in other parts we use it also for other things but the main place we use is just this that. The fact that it's a net, it means that in any direction, I have some sequence tending in direction V narrowly so if the directions you want I can choose some points that will tend in my net never in direction V one. This is the place so once I have this I can, I have all I need I don't need to use other the fact that I know the distance. This is where we use the fact that it's a net. Okay, in this part of the. So the lemma, which we use I hope as much as time permits is in any direction I have a sequence of points and direction choose in direction V. Okay, I have some sequence of points tending to infinity narrowly in this direction, and I'm going to connect my point points to these points and create a limit geodesic so what will be a strategy again trying to do that again reminding that what I want to do is I want to define gamma PV. So I have this point. And I want to define a minimizing geodesic passing to this point. And I want to show that this map is okay so how will I do this. I will choose some sequence tending to be narrow, some sequence tending to be narrowly, like we said, and I'm just going to take its limit. I'm going to define a gamma PV to be this limit. Okay, so I have my point P. And I have a sequence PM. P1 P2. I connect all of these and I take the limit. And this will be my gamma PV. And it's important that they will tend never. So, in any direction I have such a sequence so there is no problem. Okay, so how do I know that this limit exists. So, this is just this proposition if I have two sequences one tending narrowly one tending drifting maybe drifting is the term we use. So, if I have some p and I connected to sequence of points again, and I do this limit exists and it's always the same if I tend to, if I drifting direction be it's always the same. Okay, this is the first part that we will show where. Maybe I won't have time. This is. And, but this is what we show and, and we define gamma PV for any positive time like I said this limit will give me a ray. So for positive time I have this gamma PV. Okay, what do I do for negative time. I extend it just by looking at the opposite direction. Okay, what do I do for negative time. I extended just by looking at the opposite direction. So I just gamma PV for negative time is just gonna be of minus V at the minus of T. So this gives me a definition for a for all of all of our. This is to minimize the geodesic, because of this proposition so if I have a P, and I connected to a sequence in tending narrowly in direction V, and another sequence tending drifting drifting narrowly in direction minus V. And if I take the limit and you find it for positive time to be this geodesic for this geodesic will be complete minimizing. Okay, so, again, what will be our logic so now I defined gamma PV, and it's a complete minimizing geodesic. And, and, and now I'm going to show this map is onto. It's just by showing that it's continuous and odd, I won't have that this is what we do it. Okay, the fact that it's obvious right just because of this definition. So of course this map is odd. Right, and we also showed it's continuous and it's not so difficult this bar of instruction and therefore it's on. This is the point of the sphere and this is the point that's not on the sphere but. Something will move. Okay, so maybe I don't have much more time so maybe I'll just talk a little bit about how we do this so not getting to just what we do is we use a leapshitz function so leapshitz function leapshitz functions are in some ways well too close so if a curve gives me an upper bound on the distance is by the definition so the leapshitz one give me a lower bound. Okay, so, and we use something called a transport curve so if I have some geodesic gamma. It's a transport curve of some leapshitz function if it's this function grows in speed one. Okay, and what's useful for us is a few properties that we use on transport curves and I wanted time but and the most important. It's minimizing geodesic okay this is not very hard to see and the fact that if I have a transport curves and of some leapshitz functions that intersect intersect in an interior.  Okay, so this will help us in some way show that. If I have this sequence can be narrowly direction being direction minus V, then this will give me the limit will give me the connection will give me just a minimizing good. So, this is a mainly the tools that we use this one leapshitz functions maybe I don't have enough time to talk about. Now we exactly prove and but this, this is the basic tool that allows us to prove these two things so and the fact that I can stretch this. This geodesics and it will give me a complete minimizing good, so I hope some of this was I think I'm out of time, and I hope it was a somewhat clear but. If you have any questions.


We're now going to tackle the first part of the question and that is they want us to do constant volume combustion and determine the maximum pressure. So constant volume combustion. And for that what we'll be doing is we'll be using the form of the first law for fixed mass system for constant volume. So if you recall when we looked at the different forms of the first law for the adiabatic flame temperature we said that for constant volume the boundary work was zero and with that when we write out our formulation for the first law we had that they're dealing with adiabatic so heat is gone there's no boundary work and so with that we were able to rewrite the first law like so and so we have that equation for the first law. Now what we're going to do is we want to be able to determine the temperature of the product stream coming out of this and one of the challenges that we have here well what we'll do I'll begin by writing out the table. We'll do a table very much like we did earlier when we applied the first law to a reacting system. However the challenge that we're going to have is we don't know the temperature of the product so that will remain an unknown and that's what makes these questions a lot of fun for professors and a lot of stress for students. I think professors probably dream about giving these types of questions on final exams because there's so many iterations required as we will soon see. Okay so what we'll do we'll write out the different substances that we have so we have this equation or the table here now notice we do not have enthalpy for our products or the product temperature and the reason is because we do not know that temperature that's what we're trying to solve here that's the adiabatic flame temperature and that means that we have a big unknown here for each of these different substances and so that will be the way that we will formulate our first law so let's proceed to do that and what I'll do is I'll write in the values that we know and for the ones that we don't we will leave them as an unknown and I will show the value for the unknown as you'll see in the next equation that I'll put together. So that's the equation that we get by subbing in the values from the table into the first law for constant volume combustion and adiabatic flame temperature now what I'll do is I'll rewrite that and we end up with the following equation. So this is the equation that we get and what you'll notice is we have the enthalpy values for carbon dioxide, water vapor and nitrogen these are all of the things on the product side as well as the temperature of the product and it's equated to some number. Now we don't know the adiabatic flame temperature which is the temperature of the product and if we knew it we could then go into the tables and get the value of the enthalpy here the enthalpy there and the enthalpy there but we don't know that. Only way to solve this equation is by trial and error and what we need to do is we need to guess a couple of temperatures and with those temperatures we can then interpolate between the answers and help us converge on the final adiabatic flame temperatures. So I'm going to guess two temperatures and then we will go ahead and read the values from the table and see what we get and then we'll do an interpolation and so I'll show you that in the next section here. So there we get our results and if I go back a page you can see the right hand side was 734 387 with my two guesses I get one let me put that value it was 734 387 so 734 387.56 is what it should be and so you can see I was lucky with my guesses and that's partially because I do know what the answer is and that helps in solving this but by placing the guesses that I did and looking up the values we see that on this side we are above and on this side we are below so that means that our final temperature is going to be somewhere between 2850 and 2800 how do we determine that well the way that I'll do that is by doing an interpolation and that will get us close enough to the final adiabatic flame temperature so with that we get this for being the temperature of the products that is our adiabatic flame temperature when we have constant volume combustion with that now the question asked what is the maximum pressure if we have constant volume combustion so in order to answer that we need to take this temperature and use the ideal gas equation and that will enable us to get the pressure so let's proceed to do that so we have the ideal gas equation for our reactants and we have it also for the products I'm going to combine those enabling me then to isolate for the pressure of the products so when we do that we get this for being our final pressure 983 kPa or about 142.6 psi and this would simulate what we would have for the auto cycle if we were able to get the adiabatic flame temperature inside of the auto cycle if you recall the auto cycle looks something like this we started at state one we move up to a state two we then went into constant volume combustion taking us to state three and then that brought us down to state four and this was where we had our heat addition so with this calculation that would tell us it would provide us with an estimate of the maximum pressure that we may see in an auto cycle assuming that we have no heat transfer from the engine which we in reality do but that's assuming that we don't so that would be the max pressure that you could get for this particular case for example of the auto cycle so that provides us with a demonstration of how to do this for a max pressure of constant volume in the last segment what I'll do is I'll look what would the volume be if we were to restrict the pressure and have constant pressure so we'll proceed and do that in the next segment