 So, this will be a joint work with Tai Wu Yan, and yeah, this also here, so it's, yeah. And yes, I think that the setup is the following. So, we look at the homophic Lambondo L over a complex manifold X. Up to now, we don't have any restriction on the manifold. So, not that it's really to be kinder or compact, so it's an arbitrary complex manifold. And so, phi, but we have to assume that the phi is parisaparmonic. So, it's a locally defined weight, so it's parisaparmonic. And we also have a polar function g, which is non-positive on the X. So, g is a globally defined function. So, g is a globally defined function. Then we can consider the phi plus lambda g. Yeah, so this also defines a metric on L, because g is globally defined. So, we also assume that this is also a fluid harmonic function. So, we have two, we have two metrics on L. So, one is phi, and the other one is p plus lambda g. Yeah, and this is, so this one is a singular metric, but this one is even more singular. So, we have two, and this lambda is a constant, which is bigger than lambda. Yeah, so this is a set up. And then we look at the, we look at the measure of... So, g to be, yeah, it's a 12-hour difference in the size of the nanometer. Yeah, that's a good question. So, g, up to now, g is only the difference of two-prisaparmonic functions. Yeah, so it's not, not unnecessary to be upseming continuous, even. Yeah, but it's definitely our one, local our one, yes. So, yes, this is a set up. And then we, then we can define the metric on the states of measurable sections of kx plus L. Yeah, but we are more interested in the homophic sections. So, we define this by h to zero. We can think of this as a homophic section, but it also has some kind of two-introduction with respect to phi. So, this is space. And then, and we want to do some estimate in this space. But for example, the Rosawa-Kaikou-Chikutian theorem and the related things. But usually it's very difficult. So, the idea is we consider a family of, we consider a family of norms. We introduce a deformation and then we embed the original norm into this family. So, such as when t is zero, then the t-norm is precisely equal to our original norm. So, we define this most complex function. The derivative of norm that means having some complex increasing function. So, which is zero here, from minus infinity to zero. But the derivative is controlled by some linear function with slope of lambda. So, lambda is constant bigger than one. Yes, we look at those parts of functions. And then, somehow, this is a reasonable deformation. I don't understand the figure here. What is the graph? Yeah, so, this should be the function. So, from minus infinity to zero is zero. And then from zero to infinity, the derivative is from zero to lambda. So, it runs at t, right there. So, here, if you're not correct, it's not like this. Yeah, essentially. Yeah, it should be like this. So, this is a fun function with slope of lambda. And then, yeah, it should be like this. Oh, yeah, yeah, yeah, yeah, yes. Yes, yes, yeah, yeah. It should be, in any case, eventually the linear derivative is, it is wisely linear. So, we introduce this deformation. And you can see that when phase zero, then this g, g minus t will be, will define the geodesic, the trivial geodesic in the space of matrix. Because if you compare the ddc to power of 1 plus 1 by zero, it's induced by a flow of the trivial homomorphic field. Please, yeah. So, somehow, 300, yeah, yeah. So, this is a deformation of the original matrix. Yeah, but now, we can consider the, we can consider the Bogeyman projection with respect to this new metric. And then, we look at the, we look at the whole, we put all the Bogeyman projections into a single map and then we divide it by p. This is also mentioned in Barbara's talk. This is a total Bogeyman projection map. Yes. And we want to study the variation of this map. I think this is a setup. You can see that h is a Hilbert space. And h is linear with also a Hilbert space. So, they turn out that it's very difficult to approach theory here. But it's very easy to compute a derivative in the sense of the Hilbert space. So, that's observation. That's observation. And then, yeah. So, you can compute a derivative, you can compute a derivative of a smooth map between two Hilbert spaces. Yes. And then, so what we have, what we are going to do is, so we can actually throw that p to smooth. We can actually throw that p to smooth. So, smooth means that you look at the map between two Hilbert spaces. So, this is a smooth map. Yes. And, yes. So, this is a readout. Yeah. I think, so, we know this, we knew this readout from Wu. Yeah. And I think it's very interesting because the original proof is to use a regularity theorem of Kong and Hunter, which is not trivial at all, I think. Yeah. I mean, in a very special case. Yeah. Yeah. Yeah. Yeah. Yes. So, you do a lot. We shouldn't think of it as a necessarily complex. We allow... No, no, it's only the two kinds. No. I think the X is a arbitrary complex manifold. Sorry, X is... Abitrate or complex manifold. In that case, of course, it would be an interesting case. Yeah, yeah. Of course, it would be a complex manifold. Yeah. But you basically don't need to use it for this. Yeah. Yeah. So, for this, we have three cases. But even... I think even in a compact case, it's... Yeah, yeah, yeah. In a compact case, it might also not be very easy, I guess. Yeah. Even when the... Even when it's zero, as you know, after the H is always infinity. The L2 and this one. This... This kind of question, because it's not a professional problem, this is always thinking of it. In urban spaces, it varies. Yeah. Yeah. It varies. Yeah, yeah, yeah. So we are thinking about whether we come through that... this PT depends mostly on T. Yeah. Yeah. But in anarchists, the result is true. And the truth is very... It's very... It's surprisingly simple, actually. Yeah. So you can... You can define this map. And the thing is, okay, if you look at this program projection map, this is, as I say, the zero box of the smallest function of the smallest map. This smallest map signifies that if you compute the direction of your view, where you look back to any zero, this is the identity map. So you can use that and basically function zero to do this. I think it's very interesting. Yeah. You can stretch that. This is... I think you can... You can come through this without. Which basically... Yeah. I think it's... Yeah. This proof is due to Boo. I think. Very clever observation, I think. Yeah. Because then we can totally avoid the complex... I mean the very complex variable, already zero. Yeah. So... On the other hand, it's more general, yes. So then... Yeah. In any case, this is a starting point. Yeah. So once we have this without, then you can imagine that you can do calculus in a cube-based sense. And then everything would flow very easily. So you can... You can show that we have a cube-based space. You can see all this as a program space. And then you fix a bounded linear functional on this space. You fix the element to f. And then the Smoleson's theorem in priset, and you consider the norm of this functional, but now it's related to the T-norm. And you'll see now there's a function of T. So this function, then the Smoleson's theorem immediately in priset, this functional depends mostly on the parameter T. So here the point is, how do you fix the only normal change? I mean the... Yeah. Because it's a family of the human norms defined on the same cube-based space. Only the norm changes. So this is the definition of the norm of the radar. So we can prove that it's Smoleson, and then you can compute a variation of this norm. So it will give you some coverage of the formula. A variation of the norm. And then the result, I think that this result will be proved by the formula. So... Yeah. So when you compute a curvature, one part comes from the L2 bundle, and the other part comes from the second fundamental form. And we can use the Hormand L2 estimate to control the second fundamental form. This is also a boost idea, I think. So in that case, once you come up with Hormand L2 estimate, Hormand L2 type estimate, then you have proved a certain convex state. So you compute... You compute the norm. And then you compute the second one, the O2, where you have to... How do you do that for a quantum loop carrier? Yeah, yeah, yeah. You have to have it straightly. Yeah, yeah, yeah. That's a good question. Yes, that's a... Well, because... Because if you first solve the Laplacian equation, and then you want to see that the Laplacian equation will reduce to the divide equation, and you need a family of kind of functions. And the original definition of complex carrier is we have a family of kind of functions, let's say gj. And then the path that is equal to 1 eventually will cover the whole space x. And then about the gradient, the show frame model is... I mean, smaller, after gj will go to 0. The gradient is arbitrarily small when gj goes in painting. And the part that is equal to 1 will go to eventually this side will go to the whole space x. So this is a definition. But now the the plastic complex carrier you can actually introduce that definition. You can change the formula into gj. Such as gj will be decreasing anyway before to 1. So this is a definition for plastic complex carrier. Yeah, but the advantage is because here the metric is not smooth. When you solve the divide equation you need to use the mind of the summation to regularize the to regularize the single metric. But... But then you only get the smooth metric outside an exam set. So you have to really need that. You really need to consider the complex carrier, the complement, the territory open set of the complex carrier. But this is in general not complex carrier. Obviously we cannot prove that it's complex carrier. Yeah, I guess it's quite difficult to prove. Yeah, but if you look at if you adjust the definition of complex carrier which is a more general definition and then, so we call this as a plastic complex carrier then you can easily show that if actually a plastic complex carrier then the target open set is also a plastic complex carrier. It's stable when you take the target open set. I can explain to you later but it's not an inventory, I think. Because I have a very basic question. Do you have family or metrics that you have on T? Do you have them on T? Yeah. Are there some other ones you won't be able to see? Yeah, that's a good question. Yes, exactly. Because we have this assumption now you go back to the... Because the derivative is controlled by this from 0 to 100. So this combination is somehow some kind of combination of this we can think and think as a quantity. Yeah, so it's a positive combination and you can... you can compute the... you can compute the identity by... So it is... It is, for example, when you go back to the Z and the... So if everything was smooth and compact then this would be the result. Yeah. Yeah, but if everything is then in that case it would fall off for most of the system. Yeah. Yeah. But this one is a commander of personally so it's... it's nothing... Yeah, but the idea is... Of course... But the singularity is that different? Yeah. So it's a certain altruismate with a singular weight. Yeah. So this is the most general result that we can... And then how to apply this result? So... we have the... the issue here of space between the work and space and we look at... we look at subspace. Yeah. So this is very problematic. The recurrence is quite accurate. When I look at the question model of the literature is also politically correct. The literature is also politically correct. So we can come through that every form of exception a lot of the moment you become X. Then if you come through the behavior when... when you go to mice and pain in the... you can have some interesting property and then go to the... So they go to the pool, which means they go to the region here, or the acid. So you look at the relation. So you look at the pool bundle, because it has to be perturbed to the law of how much acid you're using on the water. And then if you come from a pool that long-distance, you look at the basic programming. And then go to the pool, that would be crazy. So this is here. I think it's here. There's a lot of paper, but there's a lot of data on those two. Yes. So I don't think I would give the details here, but the idea is like this. And then, so we can get the decreasing property. Yes. Yes, so we can get this. And then, so I should say that in some special case, when we look at the decreasing property, in case this section, in case X is the human pool, we'll consider the trivial property function to be of one. And here, that's the openness. Actually, this actually gives you the sharp openness result. And this is due to a book. And then, yes, so this is the first interesting case. And then later, it's in a bunch of other paper, they can consider the G is special. The G has a local, along the side of the line. And in that case, in that case, you will consider the case where the face is zero, and you'll like the G. You'll have to come to the lambda, go to the infinity. And then you come to the certain or some other type, zero. Yes. And also, yes, and also when I say compact, you can use the spectral theorem and the net reduction trick also by Booge. You can also get this decreasing theorem, which is, yeah, in a number of paper, a number on those paper. So this is, I think I mentioned this really loud. Yeah. But the proof is, it's, and then people ask that whether you come from this without the actual method to be compact. And even in the case when you're doing it all. Because if you come through this, then you'll get a very clear proof of the shortness without, with sharp estimate. Yeah. But, if you come through this, then, if you come through the uniform case, what I'm showing you, you'll get the shortness without with sharp estimate. But your question is to prove that you're doing the human-bondled analysis things. But they cannot do that. But they can use common that also has a very good notion. I'm telling you, guys, that it's 2 minus 2. It's 2 minus 1 plus actual t. This is because, and then we kind of have to go through here where we can categorically. Yeah. So we cheat a little bit. We don't really do human analysis. And we can use common to do that. So this is, yeah. But also, Was the conclusion on that you get this from open step way, or? Yeah, yeah, yeah. So do you know that action on both zeroes? Yeah. So this is the next patient. Yeah. So you can categorize sharp. So open is type of result. Yes. Yeah. So you can consider general F. And so here, also, actually, arbitrary, and you can estimate the, you can estimate the, the, the integrability index. You can, yeah. So in the, in the answer, we are going to, so here I think plus R for G means the, it means L2 to the E2 minus 3 plus R for G in the label flow of K. So then you can define the, the, the, the index. I think when K is equal to X, this index is also mentioned in when K is actually equal to X. So the solveness is when K is a part of L. And then I easily get from directly from the SNS. Yeah. So that's, yeah. So that's one thing. And in this, if the G itself is a result of only the kind of line-up of the infinity, and then you can get to the previous without through the by one, through by the chain one. Yeah. So, so, yeah. And of course the solveness is a result of G. So this is the first application. Yeah. And the other one is that I want to mention is that I push a lot. So, yeah, because it's decreasing and you can simply, you can simply control the, you can simply control the limit. I wouldn't say that's a trigger, but you can get such estimate from the estimate at a certain level and you will realize that you can do it. So now, so how to integrate this without? It means that, it means that if you have, if you have a whole discussion over the support of this coaching shift. Yeah. So, if that's it, you can, how to understand how people have this. So if this most, and the G is time thoughtful, this is precisely a reflection to the, to the, to the, to the minus and green day blocks of B. And so, so, so, if we have a whole discussion here, and we want to look at the, the auto-minimal dictation. We want to look at the auto-minimal dictation. And this is about being able to estimate the auto, for the auto-minimal dictation. Yes. And you can get it, you can get it directly from this, because, because eventually this one will go to the, maybe it will, will be, will be some constant times auto-minimal dictation. So we go to the, the auto-minimal dictation. Yes. So this is, yeah. But you can see that in order to do this, you have to show that this section already had auto-minimal dictation. So this is a problem. But, but, and we don't know how to, how to prove the dictation theorem without this assumption. So, so we really need to assume that of course it, if it, if it has a Stein characteristic open set, then of course we can do that. But in general, we don't know how to do that. So, yeah. So we're sure that this is a right question to ask a lot, but, but we think it might be possible to use a verbal bundle approach to also prove the existence of the dictation and that the, that the, the possible theorem will be probably this one. Yeah. So this is a redoubt that already been proved by, first by Damayi and then by Chao Damayi Machumbura, yeah, a few years ago. But in the most general version it's due to. So, yeah. So the redoubt is x is, it is yet a proper homophobic mapping to see it. So, so, so x into the space of a vibration over a Stein, over a Stein, over a Stein space. Yeah, proper homo, proper, proper vibration over a Stein space. And then we, then we actually have the, then we actually have the, the dictation redoubt. And essentially, when you see what the theorem is, it's not the dictation of the doubt. Because it's subjective. If you have a section here, then you can have a dictation. Do you require that from x to the Stein would be submersive? Or is submersive to be a proper homophobic? Yeah, that, that, that vibration is just a homophobic mapping. It's no, no sound motion. Yeah, yeah. Uh, close. Actually, I have a plan to see it. Yeah, that is of course true because it's Stein. They don't look at the trivial map to embed it to see it. Yeah, yeah, yeah, yeah, yeah. Yeah, this is a, yeah, I think this is a very nice redoubt. So, the question is whether we can use a global model approach to prove this redoubt. And we don't know how to do that. That would be for q equal to 0. Yeah, for q, yeah, yeah, at least, yeah, yeah, yeah. I think the most interesting case would be the q equal to 0 case. Yeah, I just know.