 Je pense qu'il y a un organisateur pour l'invitation, pour parler à cette conférence. C'est un honneur de donner une parole à cette conférence pour le Bazaar. Ce projet de la conférence sera un résultat qui est en commun avec Kashiwara et Shapira et il y aura une application facile de ces résultats. Le taux principal de la conférence est la notion de micro-support, qui nous rendra le lien entre les chiffres et le manifold sur la géométrie simplifique du Bazaar. C'était introduit par Kashiwara et Shapira dans les années 80. Je vous raconte la définition, mais avant peut-être que je vous explique très rapidement ce que l'idée est. Nous considérons un manifold réel M. Dans le taux, tout le manifold serait réel. Nous considérons un exemple. Nous considérons un manifold M. Nous considérons un constant shift de N. Et le micro-support considère la question suivante. Nous prenons un point X dans le M. Et un numéro U de ce point. Nous restriquons la homologie de F pour un plus grand objectif. Nous considérons la homologie de F. Quand nous faisons la première restriction, la homologie n'a pas changé. Nous trouvons qu'il n'y a pas de différence entre ce arc et celui-là. Quand nous faisons la seconde restriction, nous trouvons qu'il n'y a pas de différence entre ce arc et celui-là. C'est l'idée. La définition de Shapira dans les années 82 est la suivante. Nous considérons un chiffre ou un objectif dans la catégorie d'un M. Ici, il y a une question. Nous définirons le micro-support. Au début, c'était le single arc support. C'est un subset de la banque de la catégorie d'un M. C'est défini comme un follow-up. C'est le close-law du set d'une direction intéressante. L'intéressant signifie qu'on a un changement dans la homologie. Il y a un set de x, c'est-à-dire qu'il existe une fonction de test, c'est-à-dire qu'il existe un point x, c'est-à-dire que c'est x. Il y a un point integer i, c'est-à-dire que la homologie de F sur des petits boules n'est pas isomorphique à la homologie de F mais c'est sévélisé par un set de psi. Ici, on prend la limite sur le niveau de x. C'est-à-dire que le psi est dans le micro-support si ce n'est pas un isomorphisme. Ou en autres mots, si on fait la différence entre ces deux homologies, cela signifie que la homologie de F sur les petits boules n'est pas 0. Merci. Donc la définition est très simple, en fait. Et par définition, le micro-support est fermé en conic. J'ai juste donné un exemple. Si on suppose que u et n sont l'open set avec une bande douleur, le micro-support du constat-chif sur u c'est... C'est la bande auto-acronomique de la bande douleur de u. Et le micro-support du constat-chif de la vue est l'opposite. C'est-à-dire que le conic est plus clair sur la définition. Ce n'est pas clair en tout cas, et c'est un débat en fait, sur le micro-support. C'est-à-dire qu'il est prouvé qu'avec cette définition très générale, le micro-support est toujours assez utopique. Donc je vais venir bientôt sur ce C.O.M. et je vais dire ce qu'il veut dire. Donc il y a un lien entre le micro-support et la géométrie. Mais la motivation n'était pas pour étudier la géométrie du constat-chif. Mais pour étudier les modules. Donc pour exemple Shapir a mentionné l'invitivité C.O.M. pour les modules D. Et donc c'est relativement avec le C.O.M. parce qu'ils n'ont pas pu compter le micro-support de la solution complexe de la solution des modules D. Donc ici, X est un manifle complexe et on considère des modules des modules C.O.M. sur X. Donc nous pouvons définir la solution holomophique et prouver que le micro-support de la solution de la solution est equal à l'invitivité caractéristique de la module D. Donc de cette façon, le C.O.M. ici c'est l'éprouvé de l'invitivité C.O.M de la caractéristique de la solution de la solution de l'invitivité C.O.M. donc nous avons déjà éprouvé de l'auteur de Excusez-moi, excusez-moi, excusez-moi. Oui, c'est ça. C'est un peu carré de chouin. Il y a aussi une pure algebraique prouve par Gable. Mais ça donne une complétie de différentes prouves, qui ont un sens dans la ville réelle. OK. Mais au début, il n'y avait pas de lien avec les propriétés mondiales de la géométrie symphonique. Et pour cela, nous avons deux différents résultats. L'une, par Nedelin, sur Zaslo en 2006. On ne va pas vraiment expliquer ce qu'il veut. Nous considérons un compact manifold M. Et ils prouvent que la catégorie de chouin constructif sur M est l'équivalent de la catégorie de l'abandon de Cotangent. Et le correspondant est ici. Nous considérons la chouin constructif sur l'open set. C'est la chouin de micro-support. Si j'ai une photo en dimension 1. Si vous avez un intervalle en dimension 1. La chouin de micro-support sera comme ça. C'est la chouin de l'outre de l'économie. Et le correspondant de l'agrégion ici est même par le graphe de la fonction de U, qui va à l'infinité de l'abandon. Donc le graphe de la fonction de l'agrégion est une approcution de la micro-support. Et il y a un autre résultat, qui est entre les chouins de la géométrie. Donc en 2008, la chouin de la chouin de l'abandon de Cotangent a été considérée en 2 sub-sets A, B, Cotangent et London. Et après, ce sont des sub-sets connex. Ce sont des sub-sets non micro-support. Donc il y a un variable pour faire tout en connexant. Donc on considère, disons, F sur M x R, avec une chouin de micro-support en A. Donc cette chouin de la chouin de l'abandon de Cotangent en A x R, à un module de Cotangent M, c'est un Roi de X, T, XI, T. Donc, T et T sont les sujets de Thie et T star R. Et on n'oublie pas la variable T et on supprime la connexité. Donc ça se définit pour une positive T. Donc on est sûr que vous avez un felt ici avec cette propriété d'inclusion. En cas de l'un, g avec la règle du micro-sport de g obtenant b. Donc il prouve que si on a un homomorphisme non-vanécié, donc le homomorphisme est de f à g, mais on peut peut-être transmettre g verticalement. Donc si on a un homomorphisme non-vanécié comme ça, a et b sont non-disposibles. Donc tc, tc of xt, is a vertical translation. And non-disposible means that we cannot find either metonyne and isotope phi t, such that a and phi t or b are disjoint. And it applies this then to prove that the cliffhorn torus in cpn is non-disposible for itself and non-disposible from the diagonal result of this kind. So to understand Ney-dor Zaslow, we have to know the focaric category, but to understand the market result, it is enough to know the micro-support. So we try to understand his result. And so in building on this idea, we stated a result which says that, in some sense, a metonyne isotope phi over cotangent bundle acts on the category of shields. And I explain this. So consider some amyelton isotope phi. Of cotangent bundle, but I assume that it is homogeneous. For this, I remove the zero section. So this dot here means that I remove the zero section. So phi t is, of course, simplistic on r plus homogeneous. So by this, I mean that phi t of x lambda xi is lambda acting of phi t of xi, so positive lambda. OK, so in this situation, we prove the following. General n, some years ago. So there exists a unique shift k phi, the drive category of shields on m squared times i, which is locally bounded, I mean bounded on each complex set. So it's that, well, at time zero, so k phi restricted to m squared times zero is constant shift on the diagonal. Its micro-support is equal to the graph, so mu soup of k phi, is equal to the graph of lambda of phi. So by this graph, I mean conic Lagrangian subset of m squared times i. For, of course, the natural, so the obvious graph, we define d star m squared times i, but with no fiber over here. And there is a unique conic Lagrangian subset above this gamma phi. And this is the micro-support of our shift. So in particular, if I consider the composition with this shift k phi at times t, it will give an equivalent between the drive category on itself. So for all t, so the composition f maps to k phi t composed with f. So this is the direct image by the second projection of the transfer product of k phi t on the inverted image by the first projection. So p1, p2 go from m squared to m. And of course k phi t is restriction i at times t. So in particular, this function is an equivalent. It respects micro-support. I mean micro-support of the image is the image by the Hamiltonian isotope of the micro-support of f. Maybe one picture to give the geometric meaning of this. So a simple example is to consider the geodesic flow, the homogenous geodesic flow, which means phi t of x psi. So we are on rm, and we have the geodesic flow on the cotangential model, which is x psi is mapped to x translated in the direction given by psi. And psi is stable. And you can easily compute this example on finds that our kernel k phi t phi t is a constant shift on some neighborhood of the diagonal with a shift. So ut here is a set of x, y with distance of x on y less than t. So it's an open set. And if we want to understand the composition, so for example, take Lagrangian manifold given by the interior connormal bundle of some parabola. So say p is this closed subset, and we consider constant shift on p. So of course, if we apply the geodesic flow, so at the beginning, we obtain the connormal bundle of some smaller curve. But at some point, we will get a swallow tail or a singularity. And the shifts that we find here, so k phi t composed with f, will be, say, we start here with a constant shift in degree 0. So in the upper part, we still have the constant shift in degree 0. But here, in this small triangle, we shift it by 1. And the set where it is defined is a locally closed set. So the upper part, it is closed. And here, it is a locally closed. So the micro support is given like this. Oh, yes, yes, yes, yes. So here, n is 2. So in general, our shift k phi is of as follows. So assume we have the time here horizontally. But actually, we have r2n. So what we find is, excuse me, in this case, k phi is very simple. It is the inverse image of some shift on rn itself. If I make the difference between x and y. So my k phi is the inverse image of something. On this phi is a constant shift on some cone, which is open on the right, shifted by n. And here, it is a constant shift in degree 0 on a closed subset. And micro support is like this. So at the point here, of course, the singularity is informed. But the micro support is still smooth. So here's a picture of what happens. We gave you an application of the theorem above by recovering Arnold conjecture on the intersection of the zero section of some Cartesian bundle of compact manifold by its deformation by any emitting isotope. So there is a lower bound for the number of intersection points, which is given by the sum of the better numbers. This can be recovered by using this on, of course, the many result on the micro support. But here, I wanted to give another example of application, which I like, which is to make the link between a deep result in symmetric geometry, well, in the beginning of symmetric geometry. So it's a result of the end of the 70s or the beginning of the 80s. It's gone off a l'HBL to GCOM. And I wanted to make the link between GCOM and invulnerability of the micro support. This is as follow. So the GLOM of l'HBL theorem is the following. It says that the set of symplectic diffeumorphism of a symplectic manifold is closed in the set of diffeumorphism for the C0 topology. And we can restate it in the following. And in fact, it is, in fact, we can prove a local statement. So we consider a symplectic vector space r2n, which is a standard form. And we consider a sequence of diffeumorphism phi n from some ball in e to e. So it is a symplectic morphism. So we respect the form omega. And we assume that phi n converges to some diffeumorphism phi infinity, but only in the C0 sense. Then in this case, phi infinity is also symplectic. And the proof with invulnerability of GCOM is not too difficult. Well, first up to shrinking the radius r, we can assume that phi n is in fact the time 1 of some Hamiltonian with compact support. So hn is some Hamiltonian function. We assume that our phi n is given by the time 1 of the flow. And well, of course, it is not conic. So we have to do the same trick as the time marking. We make everything conic by adding one variable. So we define hn tilde of xt psi tau just by tau hn of x psi mod tau. So this one is a homogeneous. And we can consider its flow by hn tilde and obtain some shift kn tilde on r2n times 1 variable. So r2n plus 2, well, times r. But I restrict at time 1. So this is k, phi, n tilde at time 1, OK? But it's not difficult to see that, in fact, because this variable t here does not appear in hn tilde. In fact, it is also the form p inverse of some kn with kn defined on r2n plus 1. On this kn as a property that its micro support is mapped to the graph of phi n by the map rho, the same map as in the time marking setting. So rho of mu sub of kn is our graph phi n. So this is a Lagrangian subset in e square. OK, now we define a kind of limit of this kn, but a stupid limit. We define k infinity by, well, it's the quotient of the product of this kn by the sum. And by standard properties of the micro support. So the micro support is well behaved with product on sum. We find that the rho of this micro support of this shift is contained in the graph of phi infinity. So if we check that this is not empty, so assume that for p given in the graph of phi infinity, there exists some p tilde in the micro support, which is mapped to p. Then in this case, the invulnerability theorem says that, so this statement, well, I didn't give the statement, but the statement is that the cone of our micro support, so I take the cone of the micro support, and take the synthetic orthogonal, and it is contained again in the cone. But there are two natural notions of cone for laboratory set. So here, cp plus of a is the limit of all lines of the union of all lines, say q, q pi r on the neighborhood of p, where this point q, q prime r in a intersected with this neighborhood. The other one is the same, but only the limit of the lines p, q prime. Of course, this will contain a synthetic orthogonal of rho minus 1 of k minus infinity, and this will be contained rho minus 1 of k minus infinity. So this implies that this inverse image of the graph is quite atopic, and it is not difficult to see, so this implies that the graph itself is quite atopic. So since it is a graph, it will be, of course, a Lagrangian. And this proves that phi infinity is a simplified morphism. So maybe I have to say a word why this fact is true. So we have to check that we have indeed micro support at this point, at any point of the graph. So to prove the fact, just to give a quick idea, we have to avoid the following, the fallacy, which could happen. In general, a limit of Lagrangian subset needs has no reason to be Lagrangian, because we can take, say, we consider a kind of length, say, in R2, and consider the conormal bundle. So it is a micro support of the constant shift in this lens. And if you make, say, here, we take radius epsilon on here, her height, epsilon squared. When epsilon goes to 0, this only converges to the half line of all points. So this is not Lagrangian, of course. So we have to avoid this kind of phenomena. And to do this, we make the following remark. So our gamma infinity, so it is a subset of the cotangent bundle of R2n. If needed, I can change the coordinate so that the map from gamma infinity to the base is local diffomorphism near my point, so my point p. And now I make an approximation. So I have this approximation of gamma infinity, which also graph this graph gamma m. And if I consider the same picture, but in the front of my manifold with one more variable, what I find is a shift with a micro support like this. So the micro support is conormal to my lines. So here I have drawn the front. And it is, in general, generally given by a local closed set by shift constant on a local closed set. And if I restrict my shift kn to some line x times r, so I fix a point x in the base, it is not difficult to see that a shift on a line is given by a sum of constant shift on interval, well, at least if k is a field. So this is the consequence of Gabriel's theorem. So what I want to avoid is that I want to avoid the fact that the intervals have a length going to 0. And this can be seen as follows. The first remark is I have at least one interval with one end in the right part here and one end in the green part. I mean that the map gamma n, let's say, we call gamma n prime here, the intersection of gamma n near a box, with a box near the graph of phi infinity on gamma n double prime the other part. So the map gamma n prime to the base is of degree 1. So this implies that there is at least one interval i with ends of i, both in the projection of gamma n prime and the projection of gamma n double prime. So the micro support is contained in two disjoint sets. One set associated with this gamma n prime where I approximate gamma infinity on another set far away from gamma infinity, which is added because I made some approximation. And now I can use, so there is a way to give direct sum to decompose a shift according to its micro support, which is called a cutoff lemma. So this is also given by Kashiwai and Shapia, at least in the convex case. And it can be rather a solo. Suppose you have some ball, say, of radius r in some vector space aik, a shift f on this ball with micro support contained in two disjoint subsets. So it is contained b times gamma 1 union gamma 2 disjoint union. So gamma 1 and gamma 2 are closed conic subsets in a k dual. Then I can find a smaller ball such that the restriction of f to this smaller ball is decomposed as f1 plus f2 modulo locally constant shift, where the micro support of fI is the micro support of f intersected with gamma i. So this cutoff lemma says that if I have a decomposition of the micro support into disjoint close subsets, then up to restricting everything to a smaller ball, I can also decompose my shift. So using this, no, so the first way, yes. So if the gamma 1 and gamma 2 are convex conic, then this is given in Kassurian Shapira's book. But using the theorem on isotope, we can extend it to arbitrary closed subsets. So the idea is maybe very quick. Very quickly is the idea. If you assume that, so we can choose some direction. So we are r k minus 1 here and r here. First assume that our two sets are contained. First assume that they contain one in the variable here t2. One contains t positive and the other one t negative. Then we can apply the projection given by the marking, which is convolution with a constant shift on r plus. And the convolution with shift will split automatically our shift. So we want to change a general situation. So my gamma 1, these are conic sets. So my gamma 1 here and gamma 2 like this. So what I will do is find some Hamiltonian isotope, which keeps gamma 1 fixed or move gamma 2 in the negative direction with respect to tau. So this I can do by choosing some Hamilton function on multiplying some burn function with t. Because the flow is treated with t is d tau. So it will move gamma 2 to the negative direction. And phi is some burn function, which is phi 0 near gamma 1, and 1 near gamma 2, not 1, because it needs to be homogeneous. So near the hypersurface t equals 0, this will give exactly what I want. So it doesn't move the base variable, and it only moves the farther variable, and we get to some situation like this. And of course, if I move away from t equals 0, this is no longer 2, because I can move far away from my set. This is why I have to restrict to small r here. And r is more or less r times the distance between gamma 1 et non gamma 2. C'est sur la phi. But anyway, so once we can cut our sheath in a ball, this means that, for example, an interval like this, which starts in the gamma n prime and gamma n double prime, has to be of lengths bigger than the small r. Because this interval is not split even up to a local constant shift. And the constants here are independent of gamma n. They only depend of the box, which gives a good approximation. So we can find some interval with non-validation in length. And well, at least these proofs, as we avoid this one, it is not difficult to see that. Indeed, this proves that we must have micro support at a point near one end of the interval. So of course, we have to prove this fact. But this only standard proposition is theory of micro support. So the main result to obtain a chromophilic acid in this sitting is a rise of the theorem. So the deep result, which implies the chromophilic acid theorem, is this inviolativity of micro support. OK, I have 10 minutes. OK, so I will just state another quick application. So another easy application, which looks more like, it looks like an exercise about curves on the sphere. It is one conjecture of Arnold, but among 100 conjectures about sanctuaries of caustics, or something like that, it is a three-clubs conjecture, which is the following. So we take, we consider the sphere as two, and we consider the contact manifold given by the projectivization of the cotangent mandrel. So this is contact manifold. In this contact manifold, we consider the fiber over a point, over this point x naught. So this is the circle in my contact manifold. And I consider some Hamiltonian, no, some contact isotope of x. OK, and I think that's all. I denote by CT, the projection of the image of lambda t, where lambda t is, of course, the image of lambda 0. So this is a curve on, well, it's a set on S2, and we have the following result. When this set is a curve, with only curves on double points as singularities, je le baisse, à compter. So I assume that CT is generic. Then it has at least, then CT has at least three curves. This was proved by, so for R2 instead of S2, it was proved by check enough on Pushkar some long time ago in 2005. So the idea is to use a radon transform to turn this lambda nought into the jet bundle of S1. And then they can make a deep study of such not using the composition, oh, I forgot how it is called. Well, but anyway, so there is a proof using the contact geometry in dimension 3. And here is the proof, and the idea is, so let's follow. So of course, I will leave my phi t as some, say, phi t tilde, which is a Hamiltonian isotope in T star S2, again with zero section removed. And this one is not only r plus homogenous, it is homogenous. So this means that phi t of x lambda c is lambda phi t of x psi. So this time for all non zero lambda, not only positive lambda. So this will give you, by applying the theorem above, this will give you some equivalence between the shift with micro support, only over one point, on the shift with micro support on lambda t. But since it is r homogenous, in fact, this equivalence respects duality. I mean, it commutes with duality. We can also see that the first category has only countably many simple objects. So of course, I have to explain simple. So simple means, so we call the definition of the micro support. Micro support is a closure of points where some convergent group does not vanish. And in fact, Kashiwa and Shapiro prove that, at least if the micro support is lagrangian and smooth, this non-validation term is almost well defined. I mean that in our case, so since micro support of f is a smooth lagrangian, this is well defined in the sense that if I take some other function psi, I have an isomorphism between the two, up to some shift, a shift given by a local muscle of index. So top psi is the muscle of index, so defined by Kashiwa, of the three natural lagrangian space that we have here. So I consider point P in my micro support. So I have the tangent space of the micro support at this point. So my micro support is smooth. I have the tangent space of the graph of my test function psi. And I have the tangent space of the vertical fiber. When I have three lagrangian space in the Cartesian bundle, I have an index defined by Kashiwa. And so this is standard. And simple means that this is simple means that it is K in some degree. So now we can prove the theorem follow from the following proposition. Assume that I have an object F in my category of shifts micro support lambda t, which is simple on self-dual and with odd order number. Then I have many simple objects. So they exist. So there are more than a quantity of c star object in this category. And this number in fact correspond to some monodromes. So I don't give a proof, but I just make a picture. So this can happen in the following situation. Well, maybe first the generic. If I make a small deformation of the fiber in the contact manifold t star p2, what I get is a conomal bundle of some asteroid like this. And it's not difficult to find a shift with micro support like this. And this shift is of long two insight. And if I cut near cusp, so it is a shift on a line. And it is the sum of two constant shifts on half closed interval. And I can also cut above. I can glue, I can make three copies of the same picture and glue them. And then I can obtain a picture. No. Sorry. It seems like it's sad. It's OK. It seems to work. Great. I can obtain a curve with only one cusp here. And with a shift obtained by gluing three copies of this one. And it will satisfy this. But if I have such a shift, so of course it's zero here and here. So I can also define similar shifts by taking the transfer product with a locally constant shift of long one around this point, for example. This will give many objects in my category. So well, this is not a proof, of course, back to one. So thank you. Are there questions? Yeah. David? A group that acts on the category of sheaves that includes both Hamiltonian, a few morphisms of the cotangent bundle, and a C0 homeomorphisms of the base manifold. No. That's a good question. But it's going to be C. No, repeat. The base manifold that would induce no map on the cotangent bundle, is there something like discontinues flow that would act on the category? No, you don't know. That's a good question. OK, I have a small question. When you spoke of the micro-local cut of lemma, you gave the example of two cones, which are adjacent. But if you have one inside the other, it works, also? Yes, because you can, I mean, it doesn't have to be, I mean, something like this. Yeah. You can do this. So there is some space here. So you can take a band function, which is 0 here, and which is 1 here, and it will still work. Yes, it will work. OK, so no other question. Then we come back in 20 minutes. Thank you.