Fredholm theory and Deligne-Mumford spaces for witch balls
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Transkript: Englisch(automatisch erzeugt)
00:00
And thanks to the organizers for inviting me. It's a pleasure to be here.
00:21
So I'm talking about Fredholm theory and lean Mumford spaces for something you haven't heard of called witch balls. All of this is joint work with Katchen Verheim. And you can see what we've written so far in these two preprints. And so here's the plan. So I'm going to give you some motivation.
00:40
I'm going to tell you what quilts and correspondences are. I'll tell you the blueprint for the algebra that we've come up with for this thing called simp, this two-category-like thing. And finally, in the last section, I'll tell you a new result about Fredholm-ness for quilts with certain kinds of singularities. So four could be called the p's section.
01:05
OK. So let me recall for you that if we have a simplistic manifold, then under favorable circumstances, we
01:20
can define the Foucault category associated to it. So this is an A infinity category. The objects are certain kinds of Lagrangians inside of M.
01:40
Depending on your situation, they'll be required to satisfy some hypotheses. And if you take two morphisms, two Lagrangians, then as long as these guys intersect transversely, the morphisms are defined to be the free vector space generated by their intersections. And in general, it's equal to the Flir complex.
02:14
OK. And then the last thing I want to tell you about Fouc is that there are composition operations that eat d inputs for d at least 1, which is something
02:25
that any infinity category is required to do. And the d-ary composition operation is defined by counting pseudo-holomorphic d plus 1 gons.
02:46
So the images of these pseudo-holomorphic polygons in M look something like this.
03:01
OK, so this is sort of the main thing we care about in the talk. So before I move on, any questions about this? All right. So several years ago, Verheim and Woodward
03:20
defined something that they called quilted Flir theory. And the point of quilted Flir theory is to build functoriality into the Foucier category, so relate the Foucier categories of different symplectic manifolds.
03:52
So in particular, they had this idea that if you take a Lagrangian and a product
04:00
of symplectic manifolds, so L01 sits inside M0 minus times M1, the minus just means flip the sign of the symplectic form 1, M0, then that should give rise to an infinity functor FL01 from Fouc M0 to Fouc M1.
04:27
And if this looks funny to you, I can at least tell you some motivation for why the right notion of morphism from M0 to M1 should be a Lagrangian like this, which is Weinstein's symplectic creed. Everything is Lagrangian, and therefore,
04:42
morphisms between symplectic manifolds should be Lagrangians. And I'll tell you more about these Lagrangian correspondences in the next section. But for now, if you don't want to keep in mind, you can think of the graph of a symplectic morphism from M0 to M1.
05:00
Hey, Nate, can I ask just a real stupid question? So the floor chain complex here for the morphisms, how is that different if I switch L and L prime? I have to have a morphism, so morphism has to go between two objects, so it's directed. So what's the difference between CFL L prime and CFL prime L? They're closely related.
05:21
So in good circumstances, there's a duality between those things. OK, now it turned out when they tried to carry out this goal that this is really a non-trivial thing to do. On the object level, it's not so hard to understand how this functor should work.
05:42
Well, it's not trivial, but it's not as difficult as what happens on the morphism level where you're counting something pseudoholomorphic, but it's going to be these funny objects with singularities that will be the subject of today's talk. And Virheim-Woodwards' resolution was to study some related objects where the analysis was not so hard, but where the result was not quite this.
06:05
So the goals of Katrin and me are, first of all, to actually do the analysis for these singular objects. So let me call this goal star.
06:35
And then the second goal is to sort of enlarge this algebraic framework.
06:41
So their algebraic picture is Lagrangian correspondence should induce an infinity functor. And the bigger algebraic framework is that we actually expect there to be an A infinity 2 category.
07:01
So just think of this as some kind of A infinity version of 2 category, where the objects are some large class of symplectic manifolds, the morphisms are Lagrangian correspondences, and the two morphisms are Fleur code chains. So the picture is something like M0 and M1 are objects.
07:21
Lagrangian correspondences give you your morphisms, your one morphisms anyway, and your two morphisms are given by Fleur code chains. So I'll say more about this algebraic picture in the third section.
07:44
So what I'll do today is I'll, if I have time, recap the progress we've made. And then the thing that I'll definitely say is I'll explain a new result for Fred Holmness of the pseudoholomorphic objects used to define the structure maps.
08:01
Before I move on to section two, I want to mention the result that Verheim and Oder were able to prove. So like I said, they worked with some pseudoholomorphic things which are easier to deal with analytically. But they got this result, which was a little bit different than what they had originally aimed at. So if you assume that you're in the either monotone
08:22
or exact setting, then they showed that a Lagrangian correspondence gives rise to an A infinity functor, but not between Fukaya categories, between these things called extended Fukaya categories, which they invented for this purpose,
08:40
the objects of an extended Fukaya category not being Lagrangians, but by formally composable sequences of Lagrangian correspondences. Yes, thank you.
09:03
Okay. Oh, and final thing. Is Vivek in the audience? No, okay. Well anyway, for the algebraic geometers. Okay, great.
09:22
Well anyway, so if you're an algebraic geometer, then something that may have come to mind is Fourier-Mukai transforms. So there's this thing that at least formally seems very analogous in algebraic geometry. So if I take a object in dbCo of a product of smooth projective varieties x and y,
09:42
then it's this totally well-known construction that you can produce from that, a functor going from dbCo of x to dbCo of y. Really useful. And so the hope is that eventually we'll be able to understand how mirror symmetry
10:02
intertwines these functors FL01 and Fourier-Mukai transforms. So that's a really quite distant goal. All right, so before I move to the next section, any questions?
10:23
Does anyone know when I started? No? Okay, great, thank you. Okay, so let me tell you a little bit more about Lagrangian correspondences and give you some examples and that sort of thing.
10:43
So a bit of notation. From now on, I'm gonna write Lagrangian correspondence. Instead of writing it as a subset, I'm gonna write it as an arrow. So let me give you some examples.
11:04
If I take a Lagrangian in M0 and a Lagrangian in M1, then I can form their product and that will be a Lagrangian in M0 minus times M1. So it'll be a Lagrangian correspondence.
11:21
In particular, if M0 is a point, this tells you that Lagrangians induce Lagrangian correspondences from the point to whatever manifold they live inside. Okay, something I mentioned before is that if phi is a map from M0 to M1, let's say it's a symplectic morphism,
11:51
then its graph is a Lagrangian correspondence from M0 to M1. And the last example, which is the most non-trivial of these,
12:02
is that Hamiltonian group actions by Lie groups give rise to Lagrangian correspondences. So say that I have G acting on M in a Hamiltonian fashion. So the Hamiltonian-ness means that that comes with a map called mu from M to the dual of the Lie algebra.
12:23
There's this construction called symplectic quotient where you take the zero level set of mu and you quotient it out by G and you get another symplectic manifold. Out of some hypothesis. And it turns out that this zero level set is a Lagrangian correspondence
12:41
from M to the symplectic quotient. I'm sorry.
13:09
Now there's this important feature of Lagrangian correspondences, which is that you can compose them. So let's say that I take L01 from M0 to M1 and L12.
13:26
Then their composition is defined by pretending that they're graphs of functions and doing what we would do to compose those functions. So that is to say, we're first gonna form the fiber product of these guys over M1, and then we'll project down to M0 times M2.
13:50
And so an example of this is that the graph of phi and the graph of psi are gonna give you the graph of psi composed phi.
14:04
Okay, now in order for this to be sensible, we have to have some results saying that the compositions of Lagrangian correspondences are, again, Lagrangian. And it turns out that that's true under pretty general hypotheses. So this is the theorem of Gilman and Sternberg,
14:21
which says that if the fiber product is cut out transversely, which is to say that the intersection of the product with the diagonal in M1 is transverse, then pi zero two defines a Lagrangian immersion
14:42
of the fiber product into M0 minus times M2.
15:04
So important note, there is no similarly general hypothesis you can write down with the property that you'll get a Lagrangian embedding. So if you're going to be in the business of composing Lagrangian correspondences, you have to do stuff with immersed Lagrangians.
15:26
And all right, so before I move to the next section, any questions? Sorry, I guess this is the second half of the second section.
15:41
So I mentioned these pseudo holomorphic gadgets in the introduction. And now I'll tell you what they are. They're called pseudo holomorphic quilts. And the way you form these is you understand what role Lagrangian correspondences should play
16:00
when they interact with holomorphic curves. So we're all used to saying that if you have a Lagrangian, then that's defining a natural boundary condition for pseudo holomorphic curves. And what the theory of quilts tells us is that Lagrangian correspondences
16:24
define scene conditions. So I'll tell you what scene condition is and I'll also simultaneously define what a pseudo holomorphic quilt is.
16:52
So here's a Riemann surface. Let's give it some genus. And now let's divide it into two pieces
17:02
by drawing this circle here. Okay, and now let's label the chunks that it's been divided into, or the patches by symplectic manifolds. Let's label the boundary components by Lagrangians. And then let's label these one-dimensional submanifolds
17:22
by Lagrangian correspondences. So just as when you draw an unquilted normal Riemann surface and label it with m's and l's that represents PDE with boundary conditions, this represents the system of coupled PDEs.
17:44
So this represents the following system. We require a map u, which goes from this left patch into m0, called u0. A map u1 going from the right patch into m1.
18:08
Then for all points in the boundary circle here, they're supposed to be mapped by u1 to l1. And finally, the new interesting condition is that if we take any point in this seam
18:26
and we pair up its image under u0 and under u1, note that they're both defined at that point, then we land in l01. That's what I meant by seam condition. Is everyone okay with that?
18:52
So first of all, let me say an important thing I didn't say, which is that u and u1 are supposed to be J-holomorphic, of course.
19:03
I'm, no, I'm not imposing any condition like that. So the curve l01 can somehow go in any way on this surface? Oh, I'm sorry, you're talking about this seam. Oh, okay, yeah, there are conditions. And what you mean if you're talking about
19:22
like a non-singular quilt is you don't allow any, let's just say you don't allow any intersections. But for the purposes of this talk, I'm only gonna be considering really specific quilts, so we don't have to think about the question, what's a general well-behaved quilt? I can tell you more afterward.
19:41
Okay, great.
20:01
So now it's time to talk about figure eights and witch balls. So what a figure eight bubble is is, well, it's a new kind of singularity that Verheim and Woodward discovered
20:21
when they were studying strip shrinking. So the situation that they were looking at is they were considering sequences of maps whose domain looks like this.
20:41
So this is defining a quilt problem. And they noticed that, well, if you have a family of quilts, well, this is the domain, and if the width of the middle strip is shrinking, then this funky thing can happen.
21:03
So like usual, you can have bubbling wherever you like on this quilt. And in particular, it could happen in the middle of this shrinking strip. So let me try to draw what's gonna happen at an intermediate step. So we're puffing out a bubble.
21:21
But we'd better be carrying the seams along with us when we blow this bubble out. So on the base quilt, the seams now look something like that. But the bubble is puffing them out. And if the rate of bubbling is proportional to the rate of the delta going to zero,
21:41
you're gonna get something interesting happening in the limit. So in the limit, you'd expect to see a quilted sphere sitting on top of a double strip where the seams on the sphere look like this. So this sphere has two circles as seams.
22:03
This one here, this one here, and they meet at the South Pole tangentially. Does that make sense to everyone? All right. Okay, and now it's time for the philosophy portion of this talk.
22:22
So in the late 80s, Fluhr was studying Fluhr strips, though I suppose he just called them strips. And he noticed that you can have this funky thing happen where bubbles can occur. In particular, you can have disk bubbling on the boundary. And he said, whoa, let's see what hypotheses
22:40
we can put on our situation so that we can avoid that. But then Kenji Fukaya comes along and he says, well, Fluhr strips are pseudomorphic disks, more or less, or inhomogeneous ones with one input and one output. And disk bubbles are also disks with zero inputs and one output. So let's, instead of regarding them as a bugaboo,
23:02
let's study pseudomorphic disks with arbitrarily many inputs and one output. And out of that, he got this amazing algebraic structure called the Fukaya category. Right, so then as a naive graduate student, I thought, let's try to do this with a figure eight bubble and it turned out that an interesting algebraic structure
23:23
seems to emerge from that. And by the way, the reason that this is called the figure eight bubble is that if you look at it from the south pole, then the two seams look like a figure eight. All right, great.
23:42
So then the idea that comes out of that philosophy is to count quilts like the figure eight bubble, but let's call them figure eight quilts since they're no longer bubbles, they're really the primary object of study.
24:03
And just like Fukaya did, why don't we put some mark points on the seams? And since the south pole is what attaches to the base quilt over in the situation where the figure eight emerged, let's regard the south pole as the output. Okay.
24:23
Now, before we can do anything, we have to have some idea of what do the inputs want to eat? So what should this be defining a map between? And the way you can understand that is let's look at a little neighborhood of one of these mark points.
24:41
So we get this little disk mapping it to M one and M zero with these seam conditions, L zero one, L zero one prime. Okay, and then you can see, like more or less trivially, that if you fold this thing across the center line,
25:03
this is equivalent to a little half disk mapping into M zero minus times M one with honest Lagrangian boundary conditions in L zero one and L zero one prime. Sorry, like, corresponding to the picture,
25:22
shouldn't that just be M one and M two, this one? Like, just from the, from all the picture. There's no, like, from there. Oh, yeah. You know, I should have, I probably should have reversed these. I mean, it doesn't matter, but I'll get confused otherwise. Okay, and then we know that this sort of input
25:43
mark point should eat a Flir co-chain in CF L zero one, L zero one prime. Okay, so now we know what sort of input this guy should produce. And, well, a removal of singularity theorem
26:05
that I might not talk about today tells us that we'd expect that this output mark point would produce something in, well, I'd better give things names. Let's see, so this is L zero one prime. This is L one two prime prime,
26:20
L one two prime, L one two, L zero one. So this guy should spit something out in CF. L zero one compose L one two. L zero one prime compose L one two prime prime. And so this is a Flir co-chain group in M zero
26:42
minus times M two. At that point, you've got four Lagrangians playing. How do you decide to group them one way rather than another? Does it not matter, or is that? I think you decide to group it like that because you can prove that that's where the limit will live
27:01
in that Flir co-chain group. Does it have to do with the fact that there's this tangency condition? Yes, that's right. Yeah, and maybe I should give a zoomed-in view near the South Pole. So let me cut out a little disk near the South Pole and then look at it from below,
27:21
and I'll show you what it looks like.
27:43
Yeah, so anyway, yeah, the tangency is what makes the result B co-chain in here. And I'll come back to this, but I'll also say that the reason that the analysis of these figure-eight quilts is difficult
28:00
is exactly because of this local picture here. So if the seams were coming into the output point like that, then you'd have no trouble at all. And that's pretty much what Verheim, and Woodward, and Mao did. Okay, great. So what kind of algebraic structures should we get?
28:26
I'll do it over here. So there's some heuristics for gluing that tell us what we should expect.
28:42
And I don't think I'll get into the heuristics, but I'll tell you the result. Right, so based on what I was saying over here,
29:00
counting these figure-eight quilts of the precise type that I wrote down with two mark points on the left seam and one on the right, that we expect to define a map from, let's see, CF L1-2 prime, CF L1-2 prime prime tensor, CF L1-2 L1-2 prime tensor,
29:24
CF L0-1 L0-1 prime, two CF L0-1 compose L1-2 L0-1 compose L1-2 prime. And just to make it really clear what I mean when I say that counting these quilts
29:45
should give you this map, what I mean is that if I call this map C2, and if I feed in Y2, Y1 into the left seam and X1 into the right seam,
30:01
then I'm defining the output to be the sum over all Z in the intersection of the composed Lagrangians, the count over the zero-dimensional stratum
30:20
of figure-eight quilts passing through those specified co-chains, and that's the coefficient in front of C.
30:44
So putting all of the different maps that you get out of this process, you know, for however many mark points on the left seam and however many on the right, what you expect to get is something called C2, which has input, these two fukai categories, fuk M1- times M2, fuk M0- times M1,
31:10
and it maps to the fukai category of M0- times M2. So what do I mean by this? Well, what I mean is, on the object level, it sends a pair of Lagrangians to a single Lagrangian,
31:24
which when L0-1 and L1-2 have good composition is just the composition of them, and if they don't, then you are gonna have to do some work, and on the morphism level, your font size is getting smaller and smaller.
31:41
Okay, I'll try to remedy that. Thank you. On the morphism level, it's gonna input however many morphisms you like in each of the fukai categories and spit out a single morphism, defined exactly by this process. Right, and I said that we have some expected relations for these maps,
32:01
and in the case of C2, what that relation is is that C2 is expected to be an A infinity bifunctor, which is essentially the same thing as saying that it is expected to be an A infinity functor from the tensor product of these two fukai categories to this fukai category.
32:25
And let me draw the picture that gives us that expectation. So these are all fukai categories of immersed Lagrangian? Yes. Does saying something is an A infinity bifunctor mean that you have to know a whole lot
32:41
of other higher CIs to make that definition, or does it mean that already the structure tells you what the thing is? You don't need to know any higher CIs. So it's, to be an A infinity bifunctor, that's defined just in terms of the A infinity operations on the three A infinity categories involved.
33:01
And the CIs are not the same thing as those A infinity operations. I'll say it in a second, but the A infinity operations are the same thing as C1. Okay, right, okay, so here's an example
33:24
of that gluing heuristic that I mentioned. Let's consider a one-dimensional moduli space of figure eight quilts with two mark points on the left seam and zero mark points on the right seam.
33:41
So let's think about how this thing can degenerate, i.e. let's think about what the boundary of this should be. So one thing that can happen is these two mark points can come together. So when these two mark points come together, we'll get a two-patch sphere
34:00
sitting on a figure eight quilt. So the two-patch sphere is divided into two by this circle here, okay? Now other stuff can happen. For instance, the two seams can come together and collide.
34:23
And depending on where the mark points are, when that collision happens, you're gonna get different bubbles resulting. So for instance, if the two mark points do not come together as the seams collide, then you'll get two figure eight quilts
34:41
sitting on a two-patch sphere, whoops, or they could come together as the circles come together,
35:02
in which case you'll have a single figure eight bubble sitting on a two-patch sphere. Okay, have I missed anything? Oh yeah, I've missed two things,
35:21
which is that you can have floor breaking at either of these two mark points. So that is to say you can have energy concentration at either of these guys.
35:56
And now let me write down the algebraic expressions
36:01
that correspond to these sets. So if you count quilts like this, oh, and I didn't write the mark points on it. So this thing, so this sphere with two patches, if you fold across the seam, you see that this is just the same thing
36:21
as a pseudo-holomorphic disk mapping to M one minus times M two. So this corresponds to doing the C two operation to the product of two morphisms in M one minus times M two.
36:41
And this bar here divides the things that you're feeding into the left seam and the things you're feeding into the right seam. Now what's this? So this is gonna be mu two of C two Y two C two Y one.
37:07
This one here is mu two of C two of Y one Y two This is C two of Y two U one of Y one.
37:25
And then this one is C two of U one of Y two Y one.
37:43
So anyway, the fact that these guys arise as the boundary of this one dimensional modular space or they're expected to anyway, tells us that these algebraic expressions should sum to zero. And this is one of the infinitely many relations that must be satisfied in order for C two to be an A infinity bi-functor.
38:21
Okay, so I put which balls in the title, so what's a which ball? Well, the answer is figure eight quilts have two circles as seams. The structure maps in the Tokai category of a product, you can represent as counts of spheres with one circle as seams. So why don't we study circles
38:41
with arbitrarily many circles as seams? So what we call those quilts is which balls. D patch which balls. So, and counting them is gonna give rise
39:11
to some operation called C D going from Fook M D minus one minus times M D
39:26
through Fook M zero minus times M one to Fook M zero minus times M D. And I should say that it's only for D is equal to two that this thing should be a A infinity multi-functor
39:43
which is exactly analogous to the fact that the only A infinity operation which is a chain map is mu two. Great. Now, for every D we have,
40:01
or for some D we have predicted relations and we expect predicted relations for all D. We're just working on the algebra. So the first relation is that if you sum up all ways of sticking C one into itself,
40:21
you should get zero. Now C one goes from Fook M zero one minus times M one to itself and as it turns out, basically what I said before is that C one is exactly defined using the infinity structure maps in that Fukai category. So this is exactly equivalent to saying
40:42
that Fook M zero minus times M one is an A infinity category. I could see this by like taking the sphere and folding it into a disc across that one circle. Yes.
41:01
Okay, the second relation is equivalent to saying that C two is an A infinity bifunctor. So that you get by sticking C one into C two either on the left or the right
41:30
and also adding in the result of sticking C two into C one however many times as you like. So the dot dot dot here means fill up C one with C twos.
41:49
So we know what R one and R two are and we have an idea of what all the other expected relations should be, but exactly the algebra, exactly what it should be we're working out.
42:01
And the last thing I wanna say before I move on, let's see. Is that, well, let's see. So I'll say that this thing I mentioned in the title,
42:23
this expected A infinity two category is just, we plan to get it by using these CDs as a structure maps.
42:40
I'll tell you one special case of this whole construction which is when D is equal to two and let's see, M zero is equal to a point. In this case, the fact that C two should be an A infinity bifunctor means that if you fix L one two, then that
43:07
as a formal corollary of that expectation gives rise to an A infinity functor from Fuke M one to Fuke M two defined by counting quilted disks
43:25
that look like that. The reason that I say disks and not spheres is that a priori, it's defined using these figure eight quilts, these three patch spheres except one of the patches is mapping to a point. So you can just delete that patch,
43:42
flatten it out to a disk and this is what you get. And let me note now that what Mao, Berheim and Woodward did is they straightened out the seams in the neighborhood of this singular point.
44:01
So they studied quilted disks where the seam is this teardrop thing. So these seams come into the output point in this sort of transverse fashion. All right, and then really the last thing I'll say before I move on to the new result
44:22
is that in the Fukae category, you get the A infinity operations because your spaces of pseudo holomorphic curves live over the operative associated hetero. And the A infinity operations reflect algebraic structure of that operand. So we expect that these spaces of curves
44:43
live over another operand and these relations exactly come from that operand structure. Specifically the operand where a typical space is the moduli space of D patch which bolts.
45:03
So I'll show you an example. So one of these spaces called P three zero zero is the space of thrice quilted disks
45:21
with three boundary mark points. The example is an example of disks and knot spheres because they're easier to draw. And it turns out that what you get is, well, so take this picture, cut it out along the edges. There are certain edge identifications that you have to make that I haven't indicated in this figure.
45:42
But the point is that it'll glue up to a polytope inside of R three. And what that polytope represents is this compactified lean-mumpered space of quilted disks.
46:00
And the different faces represent co-dimension one degenerations which are indicated here. So the edges are co-dimension two degenerations and so forth. And it's now time to play a game. Will someone name two adjacent faces?
46:27
Good choice. The fact that they're adjacent should mean that alpha and beta have a common co-dimension one degeneration. So let's see if I can figure out what that is. I promise I haven't practiced.
46:43
Okay, so the common degeneration is if the seam in this component here
47:01
gets larger and larger until it hits the boundary. So it's gonna force two patch disks to bubble off. And here what's happening is the two attachment points of these guys are gonna come together. So what you'll get is, I hope that that's convincing.
47:50
Any questions about? Can you just explain that again? You're kind of blocking the view. Oh yeah, I'm sorry. Okay, so what's happening is the degeneration of alpha is that this seam here, so the border of this green disk
48:03
is expanding until it hits the boundary of the disk. The mark points in this degeneration have not come together simultaneously. So by definition of this space, when that happens, when that collision of the seam with the mark points occurs, you're forced to bubble off a quilted disk. So that's what these guys are.
48:23
And after this degeneration happens, this formerly quilted disk becomes an unquilted disk. The way that that happens with beta is that these two attachment points have moved together and so you're forced to bubble off a disk. And that's those two guys. Okay.
48:44
So I don't have a proof, but I expect that it will always be a polytope, which, a convex polytope. And I can say that in particular, you can specialize it in two different ways to the associated hedera.
49:01
You can also specialize it to the multiple hedera and furthermore to the bi-associated hedera and bi-multiple hedera. So it seems to be a pretty rich object. It's sort of like a two-categorical version of the associated hedera. Okay, so any more questions before I move on?
49:23
So in your picture here, it looks like you've got some points with valence higher than three. Am I reading that right? Yes. Is there a simple example that could be done where you can see, because you said this was supposed to be a convex polytope in R3, so I was just curious if I could see an example of a corner which had more than three faces
49:42
coming in to define it. There is a simple example, but I better show you afterward because it'll take up too much of the time. But that even happens with the multiple hedera and I claim that these are generalizations of the multiple hedera, so you're forced into that. I have a question. Usually, the number of nodes gives you the degeneracy index.
50:05
So why do we have three nodes on the boundary face? Right, okay, suggestive question. Yeah, so this has to do with these gluing heuristics that I alluded to earlier.
50:21
And the answer is that there's this funny thing that happens with figure eight bubbling. So suppose that we have this picture,
50:44
so a disc with two patches, and let's look at the degeneration when the seam gets larger and larger until it hits the boundary. So we'll get two quilted discs stuck on to an unquilted disc like that.
51:02
So you can ask what's the co-dimension of that degeneration. So that is to say how many gluing parameters are there for this situation. And you might think that there are two gluing parameters. There's one for this node and one for this node. And therefore, this stratum should not show up
51:20
in the algebra. But as it turns out, that's wrong. The reason it's wrong is that there's no way to only glue this node based on the seam structure. So the right way to think about it is in order to glue up this picture to something smooth, you're first going to have to sort of glue
51:41
this unquilted disc to something that's quilted with a very thin outer patch. So to something like that. And then you're gonna have to glue the bubbles in. And it turns out that there's a relation between those three gluing parameters, the two neck lengths at the nodes and the width of this very thin patch
52:02
that you've introduced. And in fact, that width determines what the neck lengths must be. Satisfactory answer, can't you? Okay, great, yeah. So your co-dimension is, I mean, you have a sort of central bubble and then it's the number of things going out to the band.
52:22
I mean, those two are sort of, because they're on the boundary, the same bubble have the same gluing parameter. Exactly right, yeah, so. Yeah. Each kind of seam has its own. Yeah, that's right. And so it's sort of like the usual situation
52:40
that you're used to is that if you have some nodal curve, then the different nodes don't know about each other. You can do whatever you want to any one of them locally. But here, if you have a bunch of different bubbles sitting on a seam, the same seam, then they all know about each other. Which, you know, that seems simple enough, at least in this example, but when you have many different seams and many mark points, it turns out to,
53:02
you know, you have to think carefully to figure out which bubbles know about which other bubbles. Any other questions? All right, so time to move on to the next section.
53:22
Which I had intended as a sort of, no, no, no, no, no, I wanted to get, I thought I was gonna have 20 minutes. You know, the theme of this conference is sort of especially educational, so I wanted to do a little bit of a summary of a standard technique
53:41
and then note how it gets adapted to this situation. So we'll see if I have time to do all of that. Right, so this section is about thread wholeness. So first of all, let me mention
54:01
the general strategy for proving thread wholeness of a linearized del bar operator. And let's say that this linearized del bar operator goes from H1 to H0.
54:23
You know, you can do, you can prove the same thread wholeness results for lots of different function spaces, but this is the easiest situation. So the procedure has three steps. The first step is to prove that DU is semi-thread-wholen,
54:45
which is by definition saying that the kernel is finite dimensional and the range is closed. Okay, and then the next step is to identify
55:04
the co-kernel of DU with weak solutions of DU star is equal to zero. So this is the formal adjoint that Chris mentioned. And then the final thing is you have to prove
55:21
a regularity result to show that you can identify weak solutions of DU star with strong solutions. So here's an example of number one. So let's say that sigma is a closed Riemann surface.
55:46
Then the standard elliptic estimates tell you that you have an inequality of the form, the H1 norm of C is bounded up to a constant by the H0 norm of DU C plus the H0 norm of C itself.
56:09
Okay, and now, I should have said this, we're gonna observe that there's a theorem saying that semi-thread-wholeness is equivalent to the condition that there exists an operator K going from H0 to E,
56:25
which is compact such that you can bound C H1 by DU C H0 norm plus the norm of K C.
56:48
Sorry for the font size. So anyway, this exactly satisfies that condition because the embedding of H1 into H0 on sigma,
57:01
which is closed, is compact. So that's great. So that's how you prove semi-thread-wholeness on a closed Riemann surface. Now, I'm gonna skip some stuff
57:23
because I don't have so much time, but let me just say that another example of proving semi-thread-wholeness is the case that sigma is not closed, it's equal to the cylinder.
57:41
So let's call that cylinder C. Then there's this problem. The problem is you still have elliptic estimates. So you have that the H1 norm of C is bounded by the H0 norm of DU C plus the H0 norm of C.
58:04
So that holds, you know, these function spaces are on C. But the problem is that Relec's theorem no longer applies because we're not on a compact Riemann surface. And therefore, H1 embedding into H0 is not compact.
58:24
So we don't yet know that this operator is semi-thread-wholen. The reason it's not compact if you haven't seen it before is because you can just think of a sequence of bump functions which are scooting off to infinity. Right, so let me just say that the resolution
58:46
is to study the asymptotic operator, which is the limit as s goes to plus or minus infinity
59:02
the linearized del bar operator. And then the asymptotic operator is, of course, s-invariant, and so you can argue that there's an injectivity estimate for it. It's actually an isomorphism. And that's what allows you to fix this problem. You can therefore replace this H0 norm on C
59:21
with an H0 norm on a compact sub-cylinder. Right, so rather than saying anything more about that, about what you do about that resolution in the quilted case, it turns out you can make it work but you can't make it work trivially. You have to do some work. Instead of doing that, I think I'll just say
59:42
what the analog of the elliptic estimates are in the quilted case. So here's what I mean by the quilted case. I apologize in advance for going about three minutes over. So in the quilted case,
01:00:02
Let's say we want to understand the figure eight quilt. Why don't we put some mark points on there and we would
01:00:26
We would like to know that this defines a Fredholm problem. So the linearized delbar operator in this case is Fredholm. In order for that to be true, we certainly need to have some kind of transversality conditions on the correspondences, but Let me not say anything about that. It's exactly what you would expect.
01:00:45
So to get the argument started We need elliptic estimates
01:01:01
Which you can get locally so Away from this bad point You can get these elliptic estimates just from the standard unquilted elliptic estimates But then you have this bad point down here. So let's figure out what happens there So let's cut out a disk center to that point and let's go into cylindrical coordinates
01:01:28
On that disk, so then what you get is a quilted cylinder with these non straight seams
01:01:55
Right, so how the heck are you going to get Elliptic estimates on this thing given this weird structure of the seams and the answer is
01:02:06
Well, look at my thesis Let's look at these chunks. So these are constant width constant height rectangles. We're looking at their translates and
01:02:23
We can note that you can you can try straightening out the seams in each of these rectangles so you know what I mean by that is choosing diffeomorphisms of these rectangles with Pictures that look like that So what you see is a sequence of quilted squares
01:02:58
With three patches where the width of the middle patch is shrinking to zero
01:03:04
and So Verheim Woodward came up with an estimate for quilts with domains like that and I Upgraded that in my thesis to in particular include the case when the domain complex structure is not standard Which will happen here because you're using these diffeomorphisms which are going to be tweaking the complex structure
01:03:25
All right, so That's all I'll say about Fred Holness. I'll just finish by saying that We we now know as a theorem that
01:03:40
In this case du is Fredholm as an operator from h1 to h0. Yeah No, I'm not done. Um Yeah, I thought people were starting at fun right, so I the last thing I want to do is I just want to want to mention the laundry list of
01:04:01
Analysis goals that we started with and I'll tell you which ones we've checked off
01:04:21
So the first analysis goal was a removal of singularity For figure eight quilts and more generally which balls? so That's done. The next thing is a
01:04:44
Compactness theorem for modulized bases of such things. Okay. The next thing is Fred Holness Are you working in the powerful context and you said
01:05:04
That's the goal but we haven't achieved that yet So when I said Fred home, I meant classically for classically for the standard h1 Yes We would really like h0 interested We would really like something like hk intersect wk minus one four to the analog one
01:05:25
Level of differentiation down, I think we can get that and then there's still some gluing stuff We'll have to do in order to put that into the polyfolds context so this is sort of Partway done, you know just because of what I was talking about with Dusa that we'd like this in the polyfolds context
01:05:44
We'd like it with slightly different function spaces and Then the final thing that's really not done is Gluing to undo strip-shrinking so
01:06:08
Well, no, I'd say that we need to figure out four in order to be able to put it into a polyfold setting Yeah once we have four done I I
01:06:20
Think we're really ready to turn the polyfold crank. So anyway four isn't it next up on the docket Thanks for your attention Just really quick on this last point you need classical gluing to undo strip-shrinking is what is what's next on the docket
01:06:46
Polyfolds gluing. Okay, so you can do it classically It's just a way to try it's a way to try and turn what you can do with a sort of hands-on application into the polyfold Something like that. So Virheim and Woodward did classical gluing in one situation. I don't know if you could
01:07:03
generalize that classical gluing to sort of which balls and I Don't know how general you could do that But yeah, the thing we need to do is polyfocal in which we have not so much of an idea of how to do I didn't understand what was the importance of having the actual which balls or this teardrop
01:07:24
If you use the teardrop, then what you'll get is maps between extended Fukai categories So these are Fukai categories where your objects are formal sequences of Lagrangian correspondences which is formally somewhat easier to deal with but is quite a bit further removed from geometry
01:07:41
So that's why we want these things amongst Fukai categories for it Furthermore, even if you work with the teardrop if you have bubbling the bubble is not going to have that teardrop structure It's gonna have this which ball structure So you would only be able to try to do things with teardrops when you could exclude all kinds of every kind of bubbling that happens
01:08:00
When strip shrinking also happens if you use this teardrop map and then you use the Same you actually compose those things using your map do you get the same as using your map to begin with? I I Probably but it's a little bit of a perverse thing to do because in order to prove something like that
01:08:21
You have to completely understand the which balls and therefore why are you thinking about the teardrops? You have a question do so. Yeah a question about I can't quite see the colors So some of those you had bubblings both in the level of blue and of the level of green as far as I would understand Then that would be sort of code mention to
01:08:43
So you're talking about Sigma first something I can't as I say I can't quite see but Yeah, or like what's the simplest example of this? Well, okay. Yeah, let's look at row So here we have a whole bunch of nodes and they're not all nodes of the same type I
01:09:02
Can't answer that very well in the time I have but the idea is that The lean muffled space I wrote down was not the lean muffled space that you would write down on your first try the thing you'd write down on your first try it would have some strata which are It's not clear how to cast them in terms of algebra so we've stuck on some additional cells in order to
01:09:21
Make every stratum correspond to an algebraic expression and the ones that you're confused about are exactly the new faces that you get And a pleasing thing about this picture so the picture I wrote down is exactly the polytope that tells you that C3 will define a homotopy between the functors corresponding to different Lagrangian correspondences
01:09:44
and Katrin and Chris Woodward They and and sick man him out proved such homotopy statements though They had to do some funny analysis with delay functions because they hadn't done this resolution So a pleasing thing about this resolution is that you won't have to do such funny mucking around with delay functions
01:10:08
an idea of the flavor of geometric applications that this sort of machinery might allow you to tackle or As opposed to the usual infinity structure. What other things might you be able to do?
01:10:24
I think an interesting goal would be take one of these correspondences coming from Symplectic quotient so going from M to M mod G and then This is expected to give you a functor between Fook of those two symplectic manifolds So how can you relate Fook of M to Fook of M mod G?
01:10:47
Which is My understanding is that's that's an interesting question in algebraic geometry, so you'd think it'd be interesting here, too More generally I have some ideas, but I'd rather not speculate in public
01:11:01
For instance Rookier has endless two categorical representations of this You have ideas about what some kind of manifold you might use to Get those out of some kind of job said Rookier has what I mean, you know, he makes two categories Out of the representation theory of the upgrades representation period at one category level up
01:11:24
and so in place of Adaptations you get a two category categorical representations I I do not have anything sensible to say about that. We should talk afterward
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