Well, it's a participatory democracy kind of talk, so everyone is welcome to choose his own title.

The only thing is that this is joint work with Alexandru Oancha, and I would like to thank the organizer for the invitation. Ah, merci beaucoup.

So this started with two, well, let's say it's sources. The first one is a paper by Eli Ashberg and Hofer, which is called Towards the Definition

of Symplectic Boundary, and which asks the following question, well, if you're given,

I will be more specific on the assumption, but a symplectic manifold, so just for a second the symplectic manifold will be open, but such that it has a boundary, and the question is

does W omega determine sigma? So that's one, and actually the second one is a paper by, or, well a result, let's

say, by Fleur, Dusa, and Eli Ashberg, about, the following says that if you have m omega,

which is symplectically aspherical, yeah, do you mean the interior determines the boundary? Yeah, does the interior determine the boundary? That's what I mean. So it's, if m omega, and determining the boundary just means it's an abstract manifold?

No, as, well if it's a contact, if it's a contact boundary with the contact structure for example, what can you, what can you say about that?

With omega over pi two is zero, so symplectically aspherical, and if m omega is symplectomorphic to R2n with the standard symplectic form, then m is de-pheomorphic to R2n.

You haven't said that correctly, you mean it's symplectomorphic to that at infinity? At infinity, yes, I said that, I didn't write it down, but I said it, well, maybe.

But as I said, it's participatory democracy talk, so you're welcome to add your favorite assumptions to, or maybe the necessary assumptions to the statements, and also to modify the

conclusion if you like. Then m is de-pheomorphic to R2n, actually for n equal four, for n equal two, sorry, there's a stronger and older result by Gromov, which implies that m is actually symplectomorphic

to R4 with the standard symplectic form. But this is very specific to four-dimensional situation, and I will not talk about that, even though what I'm going to explain could be maybe adapted in dimension four to

get stronger results in this kind of spirit. So just one word about this, well first it adds, it uses modularized spaces of curves, which is the excuse for discussing this kind of problems here.

And the second is that the proof is actually in two parts. First you prove that the homology of m is zero, except in dimension zero, and then you prove that the pi one is actually zero, which is, if I'm correct, what Yasha proved.

So I will again here not be interested in pi one questions, even though it could be interesting to check that. So what I will, what you should remember about this theorem is that if m omega is like R2n with the standard structure at infinity, then the homology of m is like the homology

of R2n. And there's a slightly different way to rephrase that, which is the following, is that if

W omega, and I will again explain a little bit more about that, has contact boundary,

which is the sphere with the standard contact structure, then it's diffeomorphic to R2n,

or to the, sorry, first it's not m, it's W, and here let's say it's the same of course, but to the unit ball, it makes it more. But you do need omega.

We still, again, with omega over pi two. Well just because if you have such a manifold, you add the simplectization of the sphere, and then you get something which is, so you have your, something which bounds the sphere

here inside, and then you just have the simplectization and you get some manifold which is simplectomorphic to R2n with the standard structure at infinity. So this is the kind of question I will be interested in, and in fact, let me just

unfortunately, we have to write down some definitions which are pretty much standard. If you have sigma xi or contact manifold,

and if you have an embedding in a symplectic manifold, then this is a contact

embedding if there exists some form alpha near sigma such that xi is determined by alpha,

so it's the kernel of alpha, and the alpha is omega, well of course, near sigma since that's where it's defined, and the embedding is exact, or sometimes it's a restricted contact type,

if alpha extends to m, and of course, as a primitive of omega.

So that's definition one. Definition two, which is basically the same, is that the symplectic

filling of sigma xi is just a symplectic manifold m omega, and I will add with no closed component such that the boundary of w is sigma, when by slight abuse of language

I will say that sigma xi is contact embedded in w omega, well there's slight abuse of language

because w only exists on one side of sigma, not on the other side, so it's not a usual embedding, so the existence of alpha is only on one side, but I think it's pretty clear what it means. Can you in fact always extend by some such activation?

Yeah, you can always extend by some small simplification, so there's no real, there's no issue there, and so it's exact, the filling is exact, if

again, as before, let me say, if it's exact as above, and the last thing I need is

can everyone see when I write here? You can't, but it's okay. Well it only says definition three.

And the filling of sigma xi is sine, well if it's a sine manifold, so if I have w

omega, the simplistic manifold, xi is a function, so xi is j pluricere harmonic,

and j times omega, sigma is xi minus one of zero, and xi is negative on w, so it's

w is a sub-level of pluricere harmonic, so as a consequence of being pluricere harmonic, sorry I missed something here, and xi is j star of deep psi. Well sorry, the kernel I should say. So the contact form is given by the kernel of j star

of deep psi, so if you have a pluricere harmonic function, all critical points have index less or equal to n, and it's time subcritical if and only if all critical points

of psi have index strictly less than n.

Are you asking what the j series is about? No, but no. So I will start on one of the blackboards. Let's start on this one.

By something which is an elementary remark based on the Fluhr-McDuff-El-Yasberg theorem,

so some of you may say very elementary remark maybe, so the theorem is the following,

so if sigma xi has the contact embedding in R2n with the standard symplectic form

and interior component z, so if you have an hypersurface in R2n, it has an inside and an outside, and the inside is called z. Now let w omega be a symplectic filling of sigma xi,

and let's say such that the second relative homology of w with respect to sigma vanishes, then the map from hp to w, well let's say one first,

to hp of sigma induced by inclusion is injective, and the second remark

is that any two such fillings have the same betty numbers, and the third remark

is that the pth betty number of sigma is the p betty number of w plus the 2n minus p minus 1

betty number of w. So I must say we are a little bit surprised to discover that, so it's not just

the sphere, and you will see that it's totally, I mean once I tell you it's totally elementary, you can actually prove it in two lines by yourself, but it's not just that the sphere

that has a unique contact filling, unique in the sense that the topology is completely determined by sigma, but you have a world class here, and in fact if you have a contact embedding of something into R2n, it's very easy to construct many more, so let's say it's very

hard if you give yourself a sigma, a contact manifold, it's very hard to embed it in R2n a priori, I mean unless it's already given something embedded in R2n. But once you have something embedded in R2n, it's very easy to construct new ones by just doing surgery

along handles of index strictly less than n. You have something like this, you can start with a sphere, and then you can add a handle like this just as much as you like, provided the index is less than n.

So you're adding handles to what? Well you're doing surgery, you're doing surgery I mean, so you're adding a Lagrangian, or actually isotropic because I say it's less than n, some isotropic... So you're modifying sigma. So you're modifying sigma, so once you have a sigma for example in the unit sphere,

then you can construct lots of things with quite different topologies, so the only thing that is delicate to touch is the n-dimensional homology. Can I ask about the definition of fulfilling, you don't distinguish between sigma being convex or a concave boundary?

Sorry, it's always a convex boundary. I should have said that from the beginning, that all my contact structures are oriented, and everything here is considered compatible with orientation. So indeed they're all convex boundaries.

I'm a little confused about this surgery. So you have sigma as a contact surface, and then you're modifying... Well you attach an isotropic handle and make surgery on this handle, because it's isotropic you can just put it in any way you want by still being embedded.

This is something you can find in a paper by... Well, an old paper by Lodenbach in the 80s. But why is it still embedded in autoread? Well because it's embedded, I mean, you start with something embedded, you add

an isotropic handle here, and basically the isotropic, but not Lagrangian, so isotropic in dimension i minus 1 at most satisfies H principle. So you can just realize that as something embedded and not crossing again the sphere,

and then you do the surgery around this. So you're finding an isotropic sub-angle of R2n which is going to be the core of the handle? Yeah, and you can always do that because you're in the dimension where you have H principle. Did Z enter into the centerware?

Sorry? The interior component of Z. Well you could say that because they all have the same betty number here from 2, you could say that the betty numbers are the betty number of Z, if you want. So do you prove that the filling, you've got sigma fixed and you have HP mapping to, it's injected into HP sigma, is that, is the image of that determined?

Is that the primitive? Almost. You're not, you don't claim that quite. Almost except in dimension, sorry no, not in general.

Let me maybe state, yeah before I prove that which is, so here how many blackboards? Ah two only, okay.

Let me state a corollary, here you probably don't see. It's great, okay. Is there a blackboard on which you can't see?

Ah it's the one, okay behind, okay. Well I put myself in between the board and you. So the corollary is the following, is that if sigma xi has a contact embedding

in the standard R2N and

W omega is a Stein filling, then, and I need, actually I forgot why, but I need N to be greater than three, BP of sigma, so betty number of sigma is the same as the betty number of W

for P between zero and N minus two and BN minus one, so N minus one betty number of sigma, which is the same as the N betty number of sigma by Poincare duality,

is the sum of BN of W and BN minus one of W. So in particular, if W is Stein subcritical, then BN minus one of sigma is BN of minus one of W

and BN of sigma is BN.

No, I don't claim that. We have BN minus one of sigma and BN of sigma, does there have to be a plus sign between them? Here? Here, equal, on the left.

Here, equal, BN minus one and BN are the same because it's 2N minus one dimensional manifold and you have Poincare duality. Here it's a plus sign. The sum of the two and here is an equal sign. I think I missed the explanation as to why this elementary marker is elementary.

I haven't proved it is elementary yet. You're not going to do some sort of surgery on the... It's even more trivial, almost more trivial than that. So this room, you can't write in the corner like that?

Ah, here I am not supposed to write here, okay. Well it doesn't matter, it's elementary anyway. Okay, maybe I'll do my morning exercise then.

Maybe I want to send this one.

So the proof of the theorem is the following. Just look at W union R2N minus Z.

So you take R2N, you remove sigma, you get something with boundary, remove Z. You get something with boundary sigma and you plug in W. In the end you write major vietoris except that instead of doing that with R2N,

well it's the same R2N or the ball. So you get this sequence here, HP of R2N minus Z and goes to HP of sigma.

And then, well, there's no need to go back because this is going to be HP plus one of R2N

so this is zero. This is also zero, at least for most values of P. And so here you get that this is an isomorphism and so first you get that this map here is injective

and then you get, so this has to be injective, and then you get that BP of sigma is BP of W plus BP of R2N minus Z.

And this implies by Alexander duality that BP of sigma is BP of W plus B2N minus P minus one, sorry, of Z.

So I don't need Z in the theorem but I need Z.

And now what do we have? You can apply the same thing for W equals Z because it gives you a feeling

and so what you get is that BP of sigma is BP of Z plus B2N minus P minus one of Z.

And as a result, BP of Z is BP of W.

Well, this is actually true for P between zero and 2N minus one. You have to deal with the non-zero homology of R2N but I let you do that as an exercise. Okay, so I must, yeah?

Yes, of course. The fact that the last term, sorry, the last term is R2N. So it's the fact that the union, yes, sorry. So it's not only a major victory.

I would argue that that means it's not elementary. It's elementary from, it's an elementary step starting from a floor of McDorf-El-Yashburg.

That's what I meant. Because what? Because it's Yash. No, I think because the floor of McDorf proved that the homology is zero and then

after that El-Yashburg proved, am I correct? Proved that the pi one is zero. I mean we were, yeah. Is that correct? I think that's right. And the paper is actually your, in your paper. We were talking to MSRI and I was talking to Andreas and then Yasha came out and said, aha, right. Okay.

So we could say it's elementary mod Yasha's ya-ha-ha-ha. And that's if it uses, uses sort of J-homomorphic.

Yeah, J-homomorphic curve and modular spaces. That, yeah, that's the... And the fact that the only breaking can come from... Exactly. Is the condition on the relative H2 is what gives you symplecticly aspherical for...

No, what you actually need is that the union that I picture here, so W union R2N minus Z should be still symplectically aspherical. So this assumption on the H2 implies that, actually in the paper with Alex, we have other conditions that imply that. But you need some condition that will guarantee that this manifold is aspherical

and it's not enough that W and the W itself is aspherical. So let me make a number of remarks.

One remark, I don't know how much it fits here, but so according to some result by Mei-Lin Yao,

if sigma xi is contact and satisfies that the first turn class of xi is zero and W is time subcritical, then we have that the cylindrical contact homology

of sigma and xi is isomorphic to the homology of W relative to sigma tensor Well basically with the homology of Cp infinity I think shifted by two.

And it's interesting to compare the kind of result that you get. So here what you get is that if you know the cylindrical contact homology of sigma and xi,

then you know something, I mean from this you can extract the homology of W basically. So you basically know the homology of W and vice versa, provided you have this time subcritical condition.

This kind of statement is different. You ask for information about the contact structure by saying it can be embedded somewhere, in this case it's in R2n, and then from that just the differential or the homology

of sigma gives you information about the homology of W. But of course you can also somehow combine the two and say well if for example sigma has a contact embedding in R2n then you're going to know the relative homology of W sigma and therefore you're going to know something, I mean if you're

in the situation of the corollary where you have a subcritical Stein filling, then you know also the cylindrical contact homology. But if you look at the two statements somehow separately,

they're telling you different things. One starts really from this contact structure that you're supposed to know quite well to compute this and gives you information about this and the other one you have a sort of much coarser assumption by just saying well there's a contact embedding somewhere and then I know

the homology of sigma and from this I know the homology of W basically. So this result is for any dimension right? Maybe, I don't know, m is greater than something at least.

So let me just continue on similar questions and then...

So another obvious question somehow, well I mean the question is obvious, I mean the answer. Is the following take L to be a compact manifold? Look at the sphere-cotangent bundle of L with the standard contact form

and then the question is what are the possible feelings of this, of S T star lambda,

S T star L with lambda?

Of course the question is well at least homologically should they have the same homology as D, so is it D T star L?

Well we have the proposition again I think N has to be greater than three is that if L, which obviously follows from what, if L has a Lagrange embedding in R to N,

unfortunately I didn't say exact Lagrange, so you have a non-empty set of examples, W a feeling of S T star L

lambda such that the relative homology of W with respect to the boundary is zero, then from what follows W has the homology.

And you can construct of course many of these if you have L which has an immersion in R to N, then L times S1 for example has an embedding in R to N plus two and the assumption of having

an immersion in R to N is just an assumption of the tangent bundle, so if the tangent bundle, the complexified tangent bundle is trivial then you have such an immersion for example.

So what did I want to say? Okay so of course it's a bit frustrating if you say that all this comes from one, a well-known theorem, and two, Mayor-Vietoris, and then you think well maybe we can work a little bit more and get some...

Can you use this as one trick to upgrade the language? No, no because if you try to multiply by D2, so the problem is that

the boundary of DT star, this is this term which is okay but it concerns also this term

which is not okay because it's I mean it really comes from the from the feeling that you had. So the idea is well let's try to generalize this theorem, the

Fluhr-McDuff-Eliaschberg theorem and then apply again Mayor-Vietoris and maybe we get to a

class of examples. So the theorem is the following. Assume sigma xi admits

a contact embedding in a subcritical Stein manifold. Now let w be an aspherical feeling

such that the relative H2 of w and sigma vanishes. Then the map,

I hope this is correct, from Hg of w to Hg of sigma is injected for all j.

Well if sigma xi is just s minus one, s to n minus one with the standard sphere you essentially get, well you exactly get, well up to this pi one thing which

I must admit we were too lazy to investigate. We get the Fluhr-McDuff-Eliaschberg theorem. Of course s to n minus one with the standard

structure has a contact embedding in R2n so in a subcritical Stein. What this says is that if you have an aspherical feeling then the map from Hg of w to Hg of sigma is injected but this is zero for most values.

So it means that this is zero for basically all values. Subcritical Stein manifold does not mean that somehow it's a subset of R2n already? Subcritical Stein means that, well implies that it's a product of C by our theorem by Chilliback. It's the product of C and I think a Stein manifold.

Okay and so in particular subcritical Stein implies that the symplectic homology,

so let's say this one is m omega subcritical Stein, the symplectic homology of m, well the plus part let's say and constant part vanishes.

So let me state the corollary of this.

If sigma is a rational or just a homology sphere and if sigma psi

embeds in subcritical Stein then any symplectically aspherical feeling

will be a rational homology ball.

Well I had plans to describe a little bit the proof but I think that was overly optimistic.

Oh the clock is fast and we started five minutes late so you have till 40. I have till 40. On this clock. On this clock. So 15 minutes. Ah yes, my says 20. Passed.

Okay then so I was about to skip many things so I will just skip half of what I plan to skip. Still I don't think I want to embark on the idea of the proof.

Can you tell us what the ingredients are? What the ingredients are? They're basically the same as, well let's say

no, for this no. So what we use is that, so the first thing is that is a result by Chilliback which says that subcritical Stein is the product of

something which I will denote by m and c.

So that's how I start. The second point is resolved by Liska and Matej which tells me

that one can close m to a manifold p and then the idea is to work in p times s2

and now the idea is essentially the same as the one but requires more work in the theorem by Fleur MacDuff and Eli Ashberg at looking at

rational curves which are in the homology class of s2. So what actually we do is to replace, and then consider v as, well z was embedded in

m times c here, so becomes embedded now in p times s2, so you take p times s2, remove z and then you add over, you glue over sigma your manifold w.

And then you somehow count holomorphic curves which are, so somehow at infinity you have the curves in p times s2. You use some analog of hyperplane section to sort of normalize the holomorphic curve by fixing three points so that you don't have to deal with the reparameterization.

And then you have also some homological argument but basically by kind of continuation you can fill the space v by this holomorphic curve. That's exactly the same idea, exactly.

z is the interior of, no for once it was not on the board. Everybody asked what in all previous statements there was a z and since you complained I dropped

it but now you see it's useful. So it embeds in a subset, in a subcritical Stein and the interior is z. So here you would say with interior component equal to z.

No here you have no choice on which thing to make to s2 but I mean that's not so different. And then the sort of homological argument in the end is slightly different but basically the

ideas are essentially the same. So is it an assumption that there's an interior and just

being completely simple? Sorry, why is there an interior? Is it an assumption? It says no because if it's subcritical Stein h2n minus one is zero and then by Alexander duality the hypersurface will have an inside and an outside. So it follows from the from the hypothesis here.

So there are two more things I wanted to discuss about the problem, about this kind of problem. The first one are symplectic, let's say symplectic homology

obstructions. Because so far except for, I mean there's a sort of starting point which uses

rational curves in the symplectic manifold but apart from that it's all ordinary homology

somehow. I mean there's no floor homology, here there's a hidden gram of wheat and invariant but there's not much. So what can we say if we use more about symplectic homology? Well

there was the statement I quoted before by Mei-Lin Yao but let me say a few more things. So I will say, actually I don't know what this that w omega is,

well I feel I would like to say floor aspherical but if it satisfies that the symplectic homology of w vanishes. I thought of symplectically, no not aspherical sorry

meant acyclic. I don't know if there's, yes, p is the manifold you get from m which was probably,

so you start from the subcritical stein, you write it as m times c. So in the original situation of Fleur, Macduff and Eliashberg the subcritical stein is just r to n so it's

easy to write it as a product of something and c. So it's going to be cn minus one times c and then the cn minus one, you close it as a compact symplectic manifold and this is p.

In the original paper it's closed as a torus, sorry as a torus, as a product of spheres but it doesn't matter what you close it as. Then you have that sigma is embedded, of course it was embedded in m times c so it's embedded in p times s2

and then you make surgery over p times s2 by removing this z which is the interior and gluing w instead and see what you get from there by looking at a toromorphic curves.

Such a manifold without a c factor? What, it's a critical stein? No, I mean a fully acyclic manifold which is not something plus c.

So every flexible Weinstein domain is where acyclics is but they're not all subcritical, so subcritical is equivalent to this product. So critical is equivalent to the product by Chelyabak's theorem.

And okay, thanks. So let me state the proposition here.

So let sigma xi bound a subcritical stein w omega.

Let m omega be a floor acyclic and

sigma be embedded in such a way that you have a separating embedding

and sigma be separating in m with interior z and then the homology are the same.

The homology of z is the same as the homology of w. And well, if you want to know the ingredients, well there's a result by MacLean which says that when you're embedded,

where you have an exact embedding in fluorocyclic, you're again fluorocyclic, there's the, you use the long exact sequence in symplectic homology

and then you use the result by Bourgeois, sorry this is the second,

Barbourgois and Wanczia which says that sigma psi determines the positive part of w.

And there are some assumptions you need that there are no conlesander orbits on the boundary of index I think less than 3 minus n and you can prove that it's satisfied here if sigma bounds a subcritical style. Okay and so the last thing I wanted to say are applications

which maybe in a way is most fun and also maybe asks,

so somewhere here what's hidden is the idea that if you have a cobordism, so if you have a contact manifold which is concave on one side

and another which is convex on the other side, one of them is more complicated than the other. There should be a sort of increasing complexity in this way. And so there are sort of obvious questions that come up about complex singularities,

so for example let f be a complex polynomial and 0 be an isolated singularity,

then you can look at this singular hypersurface here, you can look at the intersection with a small sphere and this is actually,

well here it's going to be let's say plus, so a complex polynomial on Cn plus 1, so that this now is in s to n minus 1.

This has a contact structure by taking the maximal complex subspaces here and there's something which is called a symplectic manifold which is the Milner fiber

here which essentially you get by instead of looking at f minus 1 of 0, you just move a little bit so you have your singularity here, you have a sphere here where you would like to look at this filling here but you don't, you actually look at something which is just below here and so w will just be this,

so this is, this thing here is w. So w is a symplectic manifold that bounds sigma and w is a wedge of spheres

and it's a wedge of mu spheres and mu is the Milner number. And so here the idea is that the biggest mu, the more complicated singularity you have,

I mean if you have no singularity actually mu is equal to 2, and so the proposition I would like to state is the following, and then I'm true, so again for n greater than 3, so sigma psi is called the link

of a singularity such that the intersection form on w is non-zero,

then sigma psi has no embedding, so no contact embedding of course, in a subcritical style.

And the second theorem is that if you're looking at Briskorn manifolds and mu

is at least equal to 2, there are no subcritical, for example, Stein embeddings.

And the idea is that, I mean this is somehow the result for mu equal 1, says that if you have a trivial singularity, then w is essentially cn and so you cannot have a larger singularity, so something with a larger mu, with some assumption here, that will embed of course in cn. What you would like to say is that if you look

at two singularities with mu strictly less than mu prime and you take the manifolds w corresponding to mu and to mu prime, you cannot embed the more complicated one into the simplest one. That's what you would like to know, but I have absolutely no idea on how to... Can you write the statement if you would like to know?

Well, if you have sigma 1 w1 with mu 1, sigma 2 w2 with mu 2, so this one comes with psi 1

and this comes with omega 1, this comes with psi 2, and this one with omega 2, and then you would say that something like there's no contact embedding or exact contact embedding, I don't know, sigma 2 psi 2 in w1 omega 1, if mu 1 is less than mu 2.

So in this sense sigma 2 is more complicated than sigma 1.

Do you know if one of them degenerates into the other one? Does any of it avoid itself? No. No, the only thing I know is this, so with this extra assumption about the intersection, and it's essentially the trivial case, I mean it says you cannot embed in a subcritical style.

Okay, my time is more than up now. Thank you very much for your attention.