 That's really a great pleasure for me to be here for Jean-Pierre Bourguignon was my advisor and I would Just want to to tell you about two two stories where he was very very influential for me So maybe I will speak about the field slayer laplacian and then about critical expandant of projective projective structures Well, it's especially of this year. It's about Lagrange. So when I came just to to see him I was very interested in calculus of variation and as a member of Arthur Bess team Immediately give me this book many followers of geodesics all of those geodesics are closed and Tell me you should try to understand the geometry included in this equation of variation It's what I did for four many years. So Thank you to Robert Bryant and this morning we have a discussion about Finstler geometry Finstler geometry with Indicatrix and the manifold and maybe some trajectories and This will stand for the same thing as the unit tangent bundle if you want it's a bit different because when you want to put several metrics you prefer the motion is bound and the equation that was written this morning and for geodesic equation is that you have a vector field Which is the vector field with flow It was running the geodesics above this morning Yeah, and yeah, but a is now the contact form so I'm in the smooth case, but I will leave the smooth case Soon though X is a red field So this is classical things and a is a Hilbert contact form Okay We had discussion for four years about finding a Laplace operator in Finstler geometry and There has been several attempts and this morning again Robert told us about Legendre transform, but legendre transform is not linear So if you want to transport gradient compare gradient and the differential function Then you get something which is not linear. So the not natural way to create Laplace operator You get something which is not linear So the idea I mean really was a discussion with Jean-Pierre. He told me one day, okay If you want to have a Laplace operator think you make a sum of differential second derivative partial derivatives But instead of making a sum do the partial derivative in the direction and make an average Then you get the usual Laplace operator So the question was with respect to which measure we could make the average Okay, but the idea is very simple because you have a contact form. So you have a contact volume above You have the fibers the space of directions you have the fibers So you have the conditional of the measure. So what you do you make the integration and then you get the Laplace operator It was written the first for the first time by your Thomas Bartelme, which was one of my former former students PhD students well, I can't resist to Continue in the way of you mean Let's consider cathodic zillermetric. What are cathodic zillermetric you continue for as to you consider the round the round sphere It's an ocean around ocean with some stream Exactly what's happening you have a drift on the trajectories and you keep only two Close geodesics only to close geodesics the equator in one direction in a quarter another direction Well, if you write it a bit more purly then It's a killing field V the stream and then you add to a miltonian the standard miltonian plus an epsilon and using the gen transform you go back to Lagrange and It was quite a surprise for Chen and me that we discovered independently that these are Non-reversible feature metric of curvature plus one in the sense of flat curvature these metrics and They have only two close geodesics In what which case when when epsilon is irrational with respect to pi Okay, so I asked my PhD student you can you compute the spectral data and at least find the Lowest eigenvalue indeed that you got lambda one with an expression very easy eight of a pi of a volume omega But is volume omega the volume with respect to the contact form. So we got that formula for this metric Let me recall you this HCRM Which is well known in the Riemannian setting and Compare it means that this matrix They fulfill the here the HCRM something should be done, but It still has to be done Okay, so a very short story because of ten minutes second story As second story, which is in the van of Jean-Pierre study the geometry of the equation So let me recall if you take a surface or higher dimensional Object closed for the moment. I restrict myself. You want to have a projective structure So you take charts you you might that locally into our P2 when you have a change of charts This is a restriction of an element of PSS Ria When you go to the universal cover then you have the developing map The image of that of the universal cover surface is nothing else, but in that case a strictly converse strictly convex proper open set omega You have seen one this morning for a belgemetry and in the same time what you get very interesting is that you get a representation of the fundamental group of the surface into SSR not SL2R as usually SSR higher Representation and the idea is that you have a group Can you say something about this group? For instance, can you say something about the critical exponent of that group the critical exponent is the exponent? Which is linked to the Poincare series The press carry series diverges if you take something smaller than that Okay, it comes a growth of the group. It comes also the number of Closier desix smaller than a certain Shorter than a certain number so As to one of our former students could you compute this and he said yes? Yes, we can do it. I will say in two words. Why? The number of the this critical exponent of the group is also the topological entropy of the group and It's less in general. It's it's here and is to but in general is less or equal to n minus 1 Well in the for the remaining surfaces. It's not so I mean It's very interesting, but it is you have a quality So if you go on the Taylor space, you have a quality this number doesn't change But if you go to projective structures Real prejudice projective suture which are not remain then it's strictly smaller than n minus 1 to understand this oh I don't say much more, but you have this omega this set and you have a group acting on it whose quotient is close manifold But if it's remanions and that's a conic We know everything if it's not remanion by Benoit. We know that the boundary is not C2 So if it's not C2 I cannot compute the curvature as usually I mean there is no remanion metric well defined on a unit tangent bundle But but using the ideas of geometry The idea is that curvature is something you get when you go along the trajectories because the trajectories of Hilbert geometry are straight lines You can go along the lines So you can compute everything and what you get the curvature is minus one So I complete the list of this morning. There is a huge huge space of manifold of curvature minus one But not in the smooths category They are C1 plus epsilon And this is truly a diamond because you the specimen of this is as they mentioned for surfaces eight times the other the absolute value of the other characteristic What is embedded in it is the stationary space and What is the king gradient the king gradient is that is That's very surprising. So you have the same curvature. So you say, okay I sit down and familiar with Jacobi equation. It works and I look I understand everything But in the Jacobi equations, you have two ingredient the parallel transport and The parallel transport and the curvature But the problem is that the parallel transport is now is a global object It's not metric. It depends of what you see at the boundary at infinity And it has Lyapunov exponents So you change a Lyapunov exponents So you change the entropy and you get some information and you can go further only due to this parallel transport so this was very interesting to have the geometric inside and I want to finish and say thank you very much jump here for that. Good ideas How just by Saying a few remarks Jean-Pierre has been very influential this work and in fact We didn't talk too much about what we intended to say on these very short talks But I will tell you a story which is quite similar Well the difference between Patrick's talk would be that there will won't be much mathematics in my talk because I wanted to stick on the human side and to speak about how Jean-Pierre Supervised students and how this was very important in my own work and in my own way of trying to do something in mathematics. So The when I began my PhD studies and my PhD work and the Jean-Pierre supervision I was struck by the fact that physics was very important for him And probably that was one of the things that was attracting me and I went to to ask him if I could be a student and Well, I was lucky enough that he accepted and When I began One of the first things he suggested was that I should look at the problems in mathematical general relativity and Especially questions related to the mass of asymptotically flat manifolds. So to understand my story I should at least say a few words on what are these things. So the context of my story is The real of asymptotically flat manifolds. So asymptotically flat manifold is the way physicists describe an isolated gravitational system. So it's just remanin manifold and Very roughly speaking Remanin and forget about the Lorentz and say take remanin. Let's make the thing simple And and you just assume that Complement of a compact set is the morphic you can't come out of a bowl in our head and then ask Conditions on the metric and you you just assume something like g ij minus Delta ij behaves like well Are the geodesic distance in the Euclidean space to a power of minus tau for some tau Which is strictly positive and then you can also add conditions of the same type on derivatives of the metric Okay, and so this is rough definition of an asymptotically flat manifold and the special feature of This kind of manifold is that in fact, this is the definition of coordinates But whenever you take another chart of these types, so let's call this one Psi one and then you have another one psi two and Just assume that you have the same kind of testing it's on the coefficient of the metric Then you have this special feature that there always exists an isometry and you can do an isometry such that Psi one composed with psi two minus one equals the isometry plus some remainder which is Decaying to zero infinity So in fact it makes sense to speak about the limiting Euclidean metric at infinity So it's a well-defined geometric notion Okay, and then So this phenomenon has been removed by a number of people Robert Barthnick, maybe he wasn't sure who we are and Mass is a very important invariant of remaninvariant of asymptotically flat manifold and this Well the surprising part of the very Improving fact of mass is that it's an invariant defined by Using first derivatives of the metric So here is the formula you have It's defined this way you take the limit as the radius goes to infinity of an integral Overlarge spheres of something which is g ij It's a ij ij minus the i times the coordinates of your unit normals, and of course you sum over i's and j's In coordinates, but then it depends on the clock. Yes So wait a minute and And if you are a remanian geometry, this looks completely crazy This looks completely crazy because Every remanian geometry knows the motto of remanian geometry and the motto of remanian geometry is There is no content in the no geometry content in the first derivatives of the metric The difference here is that we are around infinity not around the point and It turns out that this makes sense So this integral exists, but not only it exists, but also it has geometry content and moreover If you compute this interval in any of these charts and if the tau here is large enough Then you always get the same number So it's a remanian Okay, I'm like very oscillating Because we won't be able to do that Just for one variable, don't do what you said You're making some conditions Maybe some kind of equation No, no, no, no The precise conditions are that you need this calculator to be l1 and tau Ah, still equation no one Ah, it's a mistake, okay But it's just a decay connection at infinity No, we see it, it's a very strong condition Not so strong Otherwise it's not true It depends on the condition Sure Otherwise it's not true for one way Sure, you need to take a little definition Right And then If you want to prove that energy Is independent Under these conditions Of the choice Of the chart It's just a computation And then something which is Well, maybe you can Go with Miraculous Because you do the computation So you have m psi 2 of g And then The difference between this And m psi 1 of g And then Well, do it explicitly And there are a lot of turns Which are very nice Because of course it's given by limit And there is an integral, of course And then there are a lot of turns Which are okay They go to zero at infinity As the radius tends to infinity But there is at least A series of turns Which has no reason To go to zero But then you look very carefully At how these turns How they are organized Then you realize that These are divergence Of something So they just cancel out Because you integrate over spheres Which are compact closed manifolds Okay, and This is the context of my story And then I was beginning my PG studies And one of the first questions Sir Jean-Claire asked me was Can you try to explain this And I was struck By this question because I believe That this is a really Very nice question Not the precise question But the way We mathematicians can use Physics To do mathematics Because You have to really look At what is behind Okay, well unfortunately I didn't solve that question But It turns out That I love this question Very much And so many years later I began to have PG students Myself And so I had the habit To ask them all the time the same Question And it turns out that one of the PG Students did solve it So let me give Just his name On the shelf And very unfortunately He stopped doing mathematics But he found the answer And in fact Improved that you can do Exactly the same kind of Transition for a very very general In a very very general setting So in fact this works For a very large class Of invariants And let me just say a few words I won't Because I don't have the time And I know that Usama is in trouble to talk But The idea is that Whenever you have A natural differential operator Let's say phi From the space of matrix And it can be even Larger in fact Into the Well sections of some Of the manifold Then you can do this Very simple thing I'm assuming that If you compute this capital phi To be in metric It's 0 And then you do this You just perform a tele-expansion So you start By writing this Take an asymptotically flat metric And compute phi of g And then of course you subtract 0 Which is Capital phi to be in metric And then you write The first derivative of the Capital phi Is to add the opinion metric Apply g minus the opinion metric And then of course There is some remainder Here And then This is with values in the Sections of vector bundles So you take this kind of product With some section of the vector bundle Over the whole manifold And here I'm just explaining the thing formally So you have to add every Check on convergence and so on But I will just hide this And Of course you integrate And then you integrate By parts here On this part Of the formula And what you get is Here The other term What you expect Of v Against g minus the opinion metric Plus something which is a boundary Contribution Something which I call U of v and g Plus Depends on phi Just finishing And then the remainder Okay So that's the main formula This Is the explanation of the construction Of mass because The idea is that you consider v's in the corner of v for a star And then this Construction here Gives you a map Of the kind of v for a star To the wheels of complex safety If this is a complex Vector bundles Whatever And this is Of the same time of the mass And in fact It doesn't depend On the choice of the chart And the reason for that is also included In that formula It's just another simple computation You just have to look At the dependence of this thing Here on the chart The dependence on the chart It comes from the fact that you have Here a reference opinion metric And when you change the chart at infinity You get another opinion metric Which Is the same at infinity But if you are at a finite distance There is a small difference Which goes to zero when you go to infinity So you have to look At the way this thing Varies When you change the opinion metric And it turns out that the important thing Is to look at this U of Some v In the kernel of info star Times and you want to differentiate This U With respect to the background Neuclear metric What happens here Is that you find terms like The derivative of the nuclear metric In the direction of any vector field And you want to Say something like this And then to get Contact with this formula The natural idea Just to take the divergence Then you Find again the integration by pot So you get something like Here The v Define Of This Thing Plus V and dot But still v is an element Of the kernel so this one is zero And remember that It's somewhere here That's five the nuclear metric is zero So this here Since five is a natural differential Operator This is The variation of five The nuclear metric so it's zero two Okay so you have Here assume It's a map It's a natural map From Sections Of vector bound rule It's x Into closed forms And moreover it's very easy To see that if x equals zero Then u Of this Trillion And then there is a nice very nice theorem Which is due to Robert Vald Which says that whenever you have a map From section from natural map From sections of a vector bound rule Of closed forms in fact and such that The image of the zero section Is zero then it has Values in exact forms And you're done Because The fact that this divergence here Is zero this means that This u is a divergence So you do exactly the same computation As the one I have done a few Minutes earlier And it shows that Any construction of that type If this thing is non-zero Then it needs to remain in the Variant of the asymptotic effect So let me start with the problem claim This is a well known claim We learned about this this morning Speeders And there are operators Powerful And elegant tools In mathematics And in physics Of course So this is something Jean-Claire Told me at the beginning What I started to work with him That he heard this from In 79 And so I might get an idea So then there were some Subclaims with this fact Which was my thesis Work This was Jean-Claire Subclaim That was about Conform of covariance with the Dirac operator Conform of Comparance The Dirac operator Should be relevant To study Specific sign Problems, geometric problems And he had A precise Problem in mind This was the Problem And to say that When we work in the conformal class Of matrix Then we can make a special choice Of This conformal class Making this caricature constant So In fact I was not able To prove this fact But I Really tried to Make use of the Conform of covariance with the Dirac operator Which was mentioned this morning by Sir Michael This was A Nigerian Contribution to prove that the dimension Of the space of our experience Is conformal So We have, we heard about some A nice formula Which we usually call Ditionalist formula But one should say that Starting here In 1932 And Ditionalist In 1963 Proved A very nice simple formula And I worked to In fact to write down this formula In the case of Compact manifold with boundary And so For a compact manifold So this is what I think Called the Buchner formula By Misha So I don't want to define The whole thing So let me write down this formula So if you integrate one over a compact manifold M So let me write this R for this current temperature 1 over 4 R side squared Plus Namely that side squared Minus Integrated against Dm So what are these Objects? So thereby this is the Standard Li-V-Civitta connection Which is coming from the The many manifold And D is the direct operator R is the scalar current Which was s In the total this morning So the The Schrodinger-Leschomitz formula The compact manifold Without boundary This is 0 Schrodinger wrote a paper In 1932 in fact He worked locally And he found this formula He said this is the beautiful formula But he was not able There was no index to conclude What the Schrodinger-Leschomitz Has done afterwards Schrodinger wrote for the Not for men and men, for locally Yes, but I mean The formula is really written And this one The gentleman Yes In the rots and leons etc And this was Unlike Hoffman who found this In a Polish Journal So What I want to do today Is to say okay We have seen this morning We have seen that If you are imposed, if you have A compact manifold With a positive scalar curvature Then the kernel could not be Could not be 0 The kernel is empty, is 0 You can just read And now in fact If we consider A compact manifold with boundary Then I am going to write the boundary term And to define Some appropriate Boundary value conditions So that we can also Look at this And to relate this to the Last which has been defined by Mark Just now So this is Nothing but the integral over Sigma of Deep Psi Psi I will define this Data Minus Psi squared H You can imagine that Here R is the double trace Of the curvature tensor R here Ah, sorry Yes, you are right But what I want to do Is to put this inequality Under some conditions Now it is almost 0 If the manifold has boundary So in fact On a Riemannian manifold If I take M to be Consider M To be a Riemannian Spin Compact Manifold Then we have The Spinal Bongo SM This is the Spinal Bongo And the drop operator Acts of sections of the Spinal Bongo If we Take now Sigma and consider Sigma to be a boundary Of M So then we have to Look At the restriction of the Spinal Bongo To Sigma So I will define S-Sigma to be the restriction Of this The Spinal Bongo to Sigma So in fact here If what should I should assume Maybe take M to be M plus 1 dimensional So then Sigma is n dimensional And this Extrinsic spin Bongo Of the hyper surface of the boundary Is nothing Is Itomorphic to the intrinsic Spin Bongo of Sigma If n is even So let us Stick to this case So then there is some identification What n is even When n is even Which is an interesting case When n is equal to So Here, so what is D-slash This is the Extrinsic Dirac operator And also it is Isomorphic To the intrinsic Dirac operator Of Sigma H as for R This is the mean curvature This is the trace Up the constant of this Is that equal To the intrinsic or no Directional For Dirac It really Corresponds to the Intrinsic Dirac operator In this case The boundary copies Of the intrinsic Dirac operator Of the boundary So now With this Beautiful formula Let me Express Some Appropriate boundary conditions So that we can control the sign Of the boundary term So here As we have seen So Here we have Say n And the boundary Sigma And take n To be The normal Vector field to Sigma So if we consider Gamma n, gamma n is Clifford multiplication by n Over s Sigma Then The Mind when the About the spinors And Clifford algebra Is this idea of square root Of minus one or minus identity So gamma n Square is minus identity If we are Look at i gamma n Square Then this is identity And this will give us Projections In such a way For any phi In Sigma In s Sigma Phi will be composed Into phi plus Phi minus So that they equal Because you have h term Without h term it wouldn't be equal It's not through that external Equal to internal plus h But it's not really your term But it's not really your term That's a strict term It's a strict term Yes, exactly So now You see here I would like to Put Some conditions on r and on h So that we can get Some conclusions For example the sign of the boundary term We can change it now We can change it on the boundary No, here We can leave it exclusive Explain it Explain it We keep it like this So here if we assume that r Is not negative And h is positive Okay These two conditions So then I look at this I evaluate the problem I evaluate the boundary Sorry, I evaluate the boundary problem That we could write it this way So For any phi In gamma s Sigma There is a unique phi In gamma s Sigma Solution Of the system Deep side This is the ambient Dirac Equal 0.n And Equal Phi plus On sigma So now I use This solution of the system I Look at it in this boundary term It's Not straightforward to show That using the properties of these Appointments and this is less Or equal than 2 over n 4 over sigma Of 1 over h d Phi plus This is Phi plus Because Phi plus is equal to Phi plus Squared Minus n2 over 4 h Phi plus Squared So this is straightforward We show That we have Such inequality Under these conditions if r is Non-negative This is non-negative And with the p being solution Of this system We get To prove that this is Non-negative So this term here For any phi plus On the boundary At sigma we have This inequality Under these conditions you always have this inequality Yes I can forget everything Now we have this So what we can conclude We can do the same trick With the other projection Projection With the negative Part Then what we get Is that this is also true We add up these two inequalities These two terms are non-negative The sum Will give The fact that Zero is equal to An integral over h d Phi squared Minus n squared over 4 h Phi squared This is Make a factor into n Here No, no It's really here Yeah, yeah I can explain this But this is just I control some square Which is non-zero So here So In fact what we can say That for any phi That is non-negative The mean curvature is positive Then we have this fact So what's now I come back to The conformal covariance of the Dirac operator Which says that This is exactly Integral over sigma Of d bar h I would say what I mean by this Phi of h Minus squared So Phi h squared d sigma h So I have to explain What I mean by this Mutation So here I take the conformal metric I multiply by h squared The remaining metric On the manifold Or on the boundary So here that's where we need H to be positive So here This is the Dirac operator Associated with this conformal metric So now We can say We can see that this Phi of h is in fact h to the power Minus n minus 1 over 2 Times 5 So here we have Many results which we can read On this inequality Say here for example we can say That the first eigenvalue Of this operator d slash h Is at least equal to n over 2 n over 2 is the first eigenvalue Of the Dirac operator of this sphere And if we have equality That we have Phi is a restriction Of the power of spinom This is very strict And also we can get back To this information here And It concludes I will just finish with this If sigma is moreover Sigma is in Under these two conditions R is non-negative h is positive And we have an immersion In some spin manifold m0 Of dimension n plus 1 With Parallel spinor And sigma h0 Being the mean temperature Of this immersion Then for what we could conclude Is that Because of the Gauss formula for spinors When we have a parallel spinor Then it gives A kind of eigenspinor So what We have If the method of psi is Zero Then we have d phi n over 2 h0 Phi If I take this immersion Okay So here if we Make such a choice In this Second in the first inequality Then what we get Is that integral d sigma Is Non-negative So this is a kind of Holographic principle For parallel spinors Which says that now Sigma has been The boundary of the spin manifold With Non-negative temperature And if at the same time sigma Is an immersion The immersion should be Precisely This immersion should be By the spin That means The spin structure Which we get by restriction Should be the same at that Then in this case We have an equality Between we have such an equality And in case we have equality Every Parallel spinor on m0 Gives rise to a parallel spinor on m itself And if we have many parallel spinors here With this is The characterization Of the Euclidean space Then the manifold Itself is The Euclidean space Okay so Before I stop I would like to thank Jean-Pierre Boubignon For really sharing With all of us His insight and his taste For mathematics Thank you Jean-Pierre Are there any questions? For the second lecture About the mass And some in the very into expression Which very into expression You apply for the mass If you want to get The classical construction for mass You can just classify it To be this character Well I have a question From both slides This looks very much related From the classical mass This is exactly Inspired by Wittensburg But just As you noticed If you complete The picture Then that's what Mark has discussed Then this is The boundary is the sphere And when you take the limit At infinity This is the mass Which is inspired by Wittensburg Any other questions? I have a comment I just I would like to thank all the people Who have moved to today's conferences I hope they have not moved For me really But rather for the conferences Which were all conceptual people I must say that I was very touched That they all make the effort To come here To take the pain Where is he? He's over there To make this short trip For these conferences I really appreciate that And I'm Well these two days Yesterday was an official Ceremony Of going from one director To the next which was also some kind Of a special experience I must say that the idea Of these conferences comes from Emanuel Ulmo and I'm extremely thankful To him for having the idea Of not only doing The administration but thinking Of my own science which as you can Imagine has suffered quite a lot From my being director here That's life, one has to make decisions From time to time And I hope I can Recover from this Some of you know that My recovery may be short But I think I'm going to Stanford for three months And I'm going to stay invited by Ashtai and Gashberg And I'm sure I will enjoy that very much I have to give a course Which the topic is not yet fully chosen Because I understood that some faculty members Want you to talk about the RAC operators The other ones want me to talk about Ritchie curvature so they have to Make up their minds and follow Their advice. Thank you so much For your income and everybody For the organization and For the speakers for the wonderful lectures Thank you so much