Kodaira dimension of algebraic fiber spaces over abelian varieties or projective surfaces
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Cao, Junyan
Let f:X→Y be a fibration between two projective manifolds. The Iitaka’s conjecture predicts that the Kodaira dimension of X is larger than the sum of the Kodaira dimension of X and the Kodaira dimension of the generic fiber. We explain a proof of the Iitaka conjecture for algebraic fiber spaces over abelian varieties or projective surfaces.
 Okay. Thanks a lot for the invitation. So today I will explain a joint work with Mihaly So it's essentially an application of the productivity of the image which he explained which he has explained before. So to begin with in the first part introduction, I will first explain a little bit what is the Itaka furniture and after that I will explain what's our main theorems. So to begin with I first record the definition of Kodaiyara dimension. So to begin with we assume that x is a projected manifold defined over c. It also holds in the compact color case. In this case for the Kodaiyara dimension, kappa x is defined as follows. If the k-ac cannot bundle of x is not creative, it's not creative. In this case we define the kappa x as equals to minus infinity. If not, if this one is a creative saying in this case the kappa x is defined as the largest number d such that once we calculate the dimension of the global section over x of the current bundle tensors of times divided by m. So in the case one this one is not equals to zero, so in this case we define the Kodaiyara dimension of x is the largest number d such that for the dimension it's really positive. So one remark here is that and so a very simple remark here is that so in this case by the monarchy it's easy to see that in this case the kappa x should take the values minus infinity 0, 1, 2 and 2 the dimension of x. And also another remark is yes that if we replace the kappa bundle x by a line bundle by a homomorphic line bundle, line bundle over x so by using the same definition here we can also define the kappa error with respect to error. We just replaced the kappa kappa by the line bundle error here. And so for this one we call it's the Itaka dimension of the line bundle error. So now I can explain what's the Itaka conjecture. So for the Itaka conjecture it says as follows. So letter p via fibrillation, so I explain later what the fibrillation means between two projective manifolds x and y. And so for the fibrillation here I just means that she has a projective and the fiber and the fibers are connected. So for the general fiber is so in this case since we have set it smooth so in this case so for the general fiber is a smoother connected manifold. So in this case the Itaka conjecture states that in this case we have the sub-adjectivity of the correct dimension. So in this case it's conjectured at kappa x here is much bigger than the kappa y plus the kappa f. So where the f is the general fiber is the general fiber of the fibrillation p. So that's the Itaka conjecture. So there are a lot of thoughts about the Itaka conjecture. So here I just so here I just just a little bit. So in the case one the dimension of y is equals to 1 in the case the base years of dimension one it's solved a long ago 40 years ago by Kamata. He proved that in this case the conjecture holds and also another extreme case is in the case one the fiber one the fiber f years of general type of general type. That means set the kappa f years the extremo in this case for this conjecture it's also solved by it's also solved by Kola and also there was another proof by Fieberg and also in the case one f years of log general type case it's of log general type case. So in this case it's also same this case for this conjecture it's also solved recently by Kovac and Pata. And yes so here yeah so there is also another very important relater which is due to Kamata. He says that f for f at the midter a good minimum model minimum model sets means that sets means that so for this fiber it's biogenally equivalent to your manifold so for the fiber is biogenally equivalent to a variety of such that it's kind of the window is some example in this case Kamata proves that it holds Kamata proves that the conjecture holds so in particular so in particular so in particular abundance conjecture so in particular abundance conjecture implies the attack conjecture. Yeah so so now I can explain our result so for our result we prove that yeah so maybe I need that also to say that there is also so there is also so there is also another more simple proof or prove that by a little better simple proof by a proportionella and also there is a former worker due to Chen and the Haken we can prove the following since and yes so if set up so later at P yeah fibrations yeah fibrations between between two projective manifolds and the letter and now we consider a KLT case that to be a KLT if at the advisor on X so in this case yeah for yeah for so in this case yeah for yeah for the base is a billion variety if it's a projective polys saying this case we can prove that for the high dimension of this one should be at least the the current dimension of the fiber plus the restriction of the pair on the general fiber plus kappa y and so in this case the same so in this case the same so why is a billion variety so we know that for this one it equals to zero in fact and yeah and the and and and the biasing and the biasing document and biasing document and the biasing that argument that developed so biasing document developed in our paper we can also prove another router so so we can prove another router yeah set if for the base yes if the base yes dimension equals to two so in this case we can have also this equality so you try improve a little bit the result of karma saying this yes no no no tenness as way so I see no no no so so so yeah for you in this case so in this case yeah for straight away in this father since ever since ever since ever since ever we take it to be the junk fiber so in this case from this one equals to zero so in this case by by so in this case by using our two windows that this one is much bigger than this one and if we add a pair it has more more sections yeah so so yeah so for the proof of our router so for the proof of our router it's based on the worker so for the proof of our for a lot so it's based on a lot of folks so for the proof of our reality it's based on the productivity of the relative kind of bundle which is already explained in the talks of me hi so in the second part I just record briefly so in the second part I just record briefly what he has explained in his course so single party is the productivity of the relative kind of window and also the productivity of the direct image so here we take a so so here we so for the end here we take a n to be a number and then maybe we assume certain and and maybe we assume set the m e as much as and true and and we assume set for the direct image is not equals to zero that's just a means to set a set that's just a means that a fall for sets means just set just means that a four generative point in why for this dark image here yeah yeah yes sorry so so so so for our proof so for now on so for now on so for now on we just assume set up for the pair data equals to zero so for the reason of simplicity so in the general case for the proof it's almost the same proof as before so far now for simplicity we just assume that the data equals to zero the general proof will be essentially the same proof so in this case so in this case it's not equal to zero the sets just means that for the gem here which is equals to this one yes not equals to zero we have at least one section for every for generic fiber same this case same this case so maybe I just recall thanks to Bob Watson and he has already explained in his course in this case we said that we can find a so-called data m so-called the m relative for m backman kernel type matrix kernel type matrix which we write here as the edge B on the relative kind of window such that once we calculate the curvature of it it might be a singular matrix in this case it's a curvature is semi-polytive for on the total space of x in the sense of current yeah I need to recall that's also so so so I just say a particular side x the recall a little bit of set up so I just explained a little bit of set up so by the definition here so the restriction of the dysmetric on the fiber on a generic fiber is constructed by yes constructed by by the sections yes constructed by the sections of this one yeah so for the so for the construction so I think that is already explained by me hi and I need also another router which is also proved by and which is also proved in his paper in his paper with both and that's a set it's proved by hi and Takayama still in this setting still in this setting and now we take a we just take air to be the m minus one take kind of window and on air we equipped it with the matrix as as the HB so HB is on the relative it should tensile m minus times say in this case say in this case say in this case let let y1 in y be the be the locally free free locus of the direct image plus air which is equals to the to this one because apioki for this dark image is just a coherent shifu so we just take the locally free locus of it so for the complementary is of could measuring at least two in y let y1 so let y1 be the locally free locus of it and and the vector y zero in y1 yeah we can prove so uh so yeah so yeah so so that so so that y zero be the smooth locus be the smooth locus of the vibration p that's just the main set of for mb fiber over y zero so for the fiber so so for the fiber is smoother and the by and the by and by the inverse of the shunher window center is in fact included in y1 so so so let y be this one say in this case say in this case they prove that the Ix is a possible so so the Ix is a possible single metric possible singular matrix which is coded which is related to the nashimha matrix on the direct image so here we just defined over the locally free locus on this one such set such set this one such that so for this possible single metric so for this possible single image here it's not necessarily smoother but it's in fact bounded but in fact it's bounded on y zero and and it's positively curved in the sense of a griffith y1 so so so for this one that's uh so for the productivity of this one I don't know if it's already explained the you you have already explained okay yeah so so yeah so so maybe I just explained a little bit of the construction of for h maybe I explained that a little bit the construction of h so under so so let y be a pointer in y zero say in this case so let y be a pointer in y zero so in this case so so for the dark image here we know that it's just equal to this one so therefore we want to define the metric on this space how we define it we take a section here say in this case you wish I'd have defined the metric h it's defined as the integration of the section s and with respect to hr yeah so in this case we know that for this one it's a volume form and by the construction of and by the construction of the backman chronotype metric it's easy to check set up for this one it's finite so therefore that's it's plain set up for this metric it's in fact bounded on y zero so and so and and so therefore we know that for this metric here it's in fact bounded on y zero and after set and after set we can prove that and after set we can prove that in fact we can extend the metric as a possible singular metric on y one and so and so for the notion of for weakly positive curve that's just means set and so for this one here's the weakly positive curve that's just means set on on y one that's just means set that's just means set for uh that's just means set uh that's just means a letter e be the dual yeah that's just a means that for every open set for every open set in y one every open set and for every open set uh for every open set uh for every topological open sets in y one And every section in the dual of this vector bundle, here we know that also for every section here, we see that it's weakly semi-partitive. So we see that it's weakly, we see that it's polydiped in the sense of gray surface. That just means that for every topological bundle set in Y1 and every section here, so in this case, we can consider the function log S. So here we take the dual of this one. So therefore, we should take the dual of this matrix. Yes, PSH function. Yeah. Yeah. Because we know that Y1, because we know that Y0, it might be a straight thing subset in Y1. So in this case, by the construction here, we know only that for the matrix H here, it's bounded on this set. It's bounded on this set. So on the complementary in Y0, it might be infinity. But for this definition, it still holds. And one more remark here is that if in the case H here, smooth, if the matrix is a smooth commission matrix, in this case, for this notation, it's your current to say that it's positively curved. It's positively curved. It's your current to say that it's gray surface semi-quality. It's gray surface semi-positive. Yeah. So that's the explanation of the result of Mihaly and Takayama. And as a color, by using a lot of roof feed, we know that in this case, actually, in use a matrix, we write as determinant H on the determinant of the draft image. And in this case, since it's of rank 1, we can take the double dual. It's a line bundle on the total space Y, such that once we calculate the curvature of it, it will be semi-positive in the sense of current on the total space Y. And so now I will explain two propositions, which will be crucial in the proof of our results. So for the first proposition, it is that still under this condition here, still under the construction here, one proposition is that later, let you be some open set in Y, not necessarily in Y1. Be some open set, topological open set in Y. Therefore, the curvature of the determinant of the draft image here is identically equals 2 0 on U. So in this case, in fact, we can prove that for the draft image here, with respect to H, it's Hermitian flat. It's a Hermitian flat vector bundle. Vect bundle on U, not only on U intersect with Y1. I think that I don't have enough time to explain the proof. So I just explained the idea of the proof of proposition 1. In the case 1, H is smooth. In this case, for the proof, it's very simple. If H is a smooth Hermitian matrix, say in this case, it's rather entry. So in this case, it's very easy to see. It's very elementary to see that for this draft image here. Yes, it's just equals to the trace of the curvature. We have this one on U intersect with Y1. And now we assume that for this one, it's identically equals 2 0. So here we know that in the case 1, it's smooth. So here in the case 1, H is smooth. We know that in the case 1, H is smooth. So we know that since to the root of my head, Takayama, we know that for the draft image curvature here, it's semi-prolative in the sense of graph. So for this one, it's semi-prolative matrix. And if the trace equals to 0, so that's where you imply set. And the source sets will imply set for the draft image here. Should there be Hermitian flat? Are you intersect with Y1? And after set by using a little bit more argument, we can prove that it's in fact a Hermitian flat on the total space U. Since here, it comes from the draft image here. So I have no time to explain the last one. And so to prove our result, we need also another proposition, which is essentially proved by Fievega. And also we follow a proof of 2G, which says that the following is so in the talk of Mihai, so we know that in the talk of Mihai, he proved that if the direct image of relative kind of kind of plane is not equal to 0, say in this case, he proved that set for the relative kind of plane, kx over y, it's considered effective. That means that we can find a possible singular matrix such that it's semi-prolative in the sense of current on x. But for the following proposition here, in fact, so in the proposition 2 here, in fact, we want to compare the productivity of the relative kind of bundle and the direct image of the relative kind of bundle. So the proposition 2 says that in fact, the probability 2 says that in fact, so for the relative kind of bundle here, it's in fact more positive than the direct image. So that's as follows. So still in this setting, say in this case, we can prove that we can find some small epsilon positive, which depends on m. And some divisor on x such set for such set of so forth, this divisor here, it's in fact very small. It's in fact, it's all for co-dimension, complex co-dimension at least a 2 at least a 2 in y. And more importantly, we can prove that for the relative kind of bundle plus e. It's much bigger than the direct image. That means that it's much bigger than the. So for the direct image, it comes from the base. So we need to take the determinant of this one. So now this one becomes a lambda on the base. And we prove back it on the total space x. So for here, it's a class in on x such that it's considered effective on the total space x. So in particular, for this one, it's a big line bundle on the base. So therefore, the progression implies that for the relative kind of bundle, model since modifies here, it will be straight and positive in the horizontal direction. And so that's the population too. Now I'll explain the proof of our loss. In fact, for the divisor e here, it's in fact harmless because of the factor set for its image on y is of co-dimension into when we prove the itaqa conjecture, it's either to get rid of this divisor e. So now for the third part, I will explain the proof of our theorem. So we talk about the results in two cases. So for the first case, we let P be a vibration between two projective manifolds. And in the case, we first explained the case 1, y is a Bayesian variety. So in this case, we want to show that, so in this case, we need to prove that for the co-dimension of x, y is much bigger than the colloidal dimension of the fiber. Because in this case, y is an Bayesian variety. So for the colloidal dimension of y is equals to 0. So that's namely, we want to construct a lot of sections in the direct image for this one, which is just equals to since y is a Bayesian variety, so we know that it's kind of complicated as trivial. So it's just equal to this one. So it's going to say that once we can find a lot of global section here, so for this one, it's just equal to x minus x. So it's therefore to prove the eta connoisseur, in fact, we just want to find a lot of global sections on the direct image here. So here, we have already, thanks to the lot of time, we have already no setter. We can find a simple matrix on it. So now we want to construct a lot of global sections. So for the step one, in fact, we need a simplification. In fact, we can assume that in fact, in the step one, we would like to show that we can assume that y is a simple torus. It is a simple torus. That's just a means that there is no structure sub torus in y. Why we can assume this one? So for the reading, you said in fact, for this inequality, it's in fact, in fact, for this inequality, it's in fact invariant after finite eta cover. That means that if we take some y theta, which is a finite eta cover, which is mean setter sub torus 1, it's a finite eta cover and no ramified locus. And here it's the product. If we can prove that inequality, for this vibration, say in this case, it will imply that we have also the inequality on this vibration. So for this fact, it's well known. So you want to increase the possibility? So we assume that it's a finite eta cover. Yeah, but you want to increase the possibility? No, no. So since there is no ramified locus, so for the collier dimension of this one, it's because all the collier dimension sequence sets. Yeah, so it's literature. Yeah, so by reading this route, so by using this route, we know that y is not simple. It's not a simple torus. So in this case, it will imply that, say in this case, by Bronga's reductive theorem, it will imply that after some finite eta cover, after some finite eta cover, so for example, once y is a torus, after some finite eta cover, it's also a torus. So for the collier dimension, it's sort of weird. We can assume, if it's not a simple torus, say in this case, after some finite eta cover, we know that y can be written as the product of two torus. There are two sub-torus in y. And so in fact, in this case, we can prove that, and once we have this route, so in this case, for the fibrous here, we have another projection to another torus setter. So in this case, for the inequality here, we can prove it by the induction on the dimension of the base. So we can prove the inequality here. Can be proved by induction on the dimension of y. Yeah. So for the advantage of this construction here, setter. So for the advantage of this construction here, setter. So the advantage of this construction here, setter, in fact, in this case, once y is a simple torus, say in this case, for every pursuit effect of lambda on y, there is only two possible cases. So the advantage of this one is setter. If y is a simple torus, in this case, we can easily prove that if y is pursuit effective for, say it will imply the setter y, if it's a pursuit effective lambda on y. So in this case, it will imply setter is y, where the topological true lambda sets mean setter, if C1 plus equals to 1. Or there is ample. So for this one, it seems quite easy to check. But I have no time to explain it. So sets the advantage of this assumption. So now for step two, now we can use. So now for step two, we can use the result of behind the takayama. So in this case, by the result of takayama, we know setter for the black image of this one. By applying their result, we know that it's positively curved on y1. And also, we know setter by using the result of Rufi. We know setter for the determined window. It's semi-political in the sense of current on the total space y. And in particular, we know setter for this one, it's pursuit effective. And now by using this route, we know setter for this pursuit effective window, it should be whether it equals ample or its topological attribute. So setter for the determined. Setter for the determined. It's topological attribute or ample. So now there are essentially two cases we need to check. Now step three, we check the case of when it's topological attribute. If the determined window is c1 class equals to 0, if it's topological attribute, say in this case, since for this current, you see in the class of this determinant. And for this one, it's equals to 0. So for semi-political current, it's equals to 0. So in this case, it's identically equals to 0. So that means setter. So for this one, for the determinant, this one, it's better identically equals to 0 on the total space y. So now we are in a position to apply proposition 1. Now by using proposition 1, we know setter for the direct image here, it's in vector Hermitian flat. It's a Hermitian flat vector bundle on y. Once we know that it's Hermitian flat vector bundle on y, since y is an Abellion variety, we know setter. Since y is an Abellion variety, we know setter. So for the fundamental group of y, it is also Abellion. So in this case, once it's representation, it's in the unitary group. So therefore, since for the fundamental group, it's Abellion, we know setter. We know setter. We know setter in fact. For the direct image here, it can be split as the direct sum of numerical trival and bundle on y. So here, r is the rank of the direct image. So by using this variety, so here we use the factor setter. The fundamental group is Abellion. And so we are, they are numerical trival. So how many times I have? I'll take a minute. Yeah. So once we know this, we have to solve the setter. So by using a route of Kampana, Kampana and Paternela. So by using a route of Kampana and Paternela, which based on a route of Simpson, we can easily prove that for the, since for the direct image here, we know that it's in fact a numerical trival. And by using a route of Kampana and Paternela, we can prove that we can construct a lot of global sections. So setter will imply setter for this one. It's much bigger than this one. So I have no time to, so I'm sorry. So for the last set, I have not enough time to explain it. But it's essentially based on a route of Kampana and Paternela. And now I explain the last step. And so for the step four, we have already explained setter for the direct image here. Since we assume setter, why is it simple to us? So for the determinant of direct image, it's equals to a topological trival lavender or an ample lavender. So in the second part, we use that case when the direct image is ample. Therefore, the direct image is ample. So in this case, because I list also the second proposition, we are throwing the second case, so we need to apply the second proposition. So in this case, by using proposition two, setter will imply setter, it will imply setter for this one. It's it will be very critical. Well, yes, so in fact, as we explained before, so in fact, as we explained before, since the image here is of co-dimension at least two, since for this derisers, it's in fact a small derisers. So in this case, there are the standard, so in this case, there are the standard method to get rid of this small divisor. So for simplicity, we just assume that E is equals to 0. So in this case, that's imply setter. Since this one is ample, so that's where imply setter for this kind of window is political. Yes, it's really political in the horizontal direction. So setter will, and so setter for by using the relative for both and the power, and after setter and together with essentially by using the oscillatory gosh extension theorem, we can prove setter. So for this one, maybe another M, so for M large enough, we can prove setter by using the result and together with oscillatory gosh extension, we can prove setter for this further restructuring map is such a tip. So setter for setter will imply setter. kappa x is much bigger than kappa f. So I have one minute of, yeah, so in the last one minute, so maybe I explained that a little bit of the idea of the proof in case one y is a surface, and one y is a surface, and one y is a surface, and one y is a surface. In fact, in this case, it's quite easy to check setter for the collinear dimension. It's an imagined random number after birectional equivalent. So in this case, we can run the MMP on y. So therefore, in particular, we can assume setter for the kind of window of y is semi-ample. And in this case, in the case, if the collinear dimension is much bigger than when, then in this case, it's well known due to the relative of Flegger. So the only interesting case is in the case when the collinear dimension of base is equals to 0. So in this case, so after some finite a dot cover, since we assume that it's minimal, then in this case, we know that y should be a torus or k-3 surface. In the case 1, y is torus. It's already proved. And in the case 1, it's a k-3 surface. So in this case, we still, and so therefore, we still studied the direct image here. And we know that for this one, it's a pseudo-effective lambda on y. And in the case 1, y is a k-3 surface. So in this case, we know that if y is k-3 surface, it's k-3 surface. And 1, y is pseudo-effective by using the Savatsky decomposition and by using the Karmata-Flegger vanishing theorem and the He-Man-Rohlker. It's easy to check that for the collinear dimension of y, should be equals to the numerical dimension of y. And in the case 1, the collinear dimension of y is at least 1. So in this case, by following almost the same proof as here, we can prove the Idyac furniture. So the difficulty case is in the case 1, the collinear dimension equals 2, 0. So in this case, that just means that so for the collinear dimension of this one, equals 2, 0. So that's the difficult case. But in this case, by using the Savatsky decomposition, we know that for the direct image here, it should be equivalent to the sum of some exceptional locus. So it's an expression of divisor. So for this one, it's in fact the minus y rational curve. And by using a lot of compana, we can prove that we can in fact prove that for the fundamental group for outside the exceptional locus, we'll be very simple. So by following almost the same proof as in step 3, we can also prove the Idyac infinity. So that's the finished proof. Sick. Thank you.
 Okay. Thanks a lot for the invitation. So today I will explain a joint work with Mihaly  So it's essentially an application of the productivity of the image which he explained  which he has explained before. So to begin with in the first part introduction, I will  first explain a little bit what is the Itaka furniture and after that I will explain what's  our main theorems. So to begin with I first record the definition of Kodaiyara dimension.  So to begin with we assume that x is a projected manifold defined over c. It also holds in the  compact color case. In this case for the Kodaiyara dimension, kappa x is defined as follows. If  the k-ac cannot bundle of x is not creative, it's not creative. In this case we define the kappa x  as equals to minus infinity. If not, if this one is a creative saying in this case the kappa x is  defined as the largest number d such that once we calculate the dimension of the global section  over x of the current bundle tensors of times divided by m. So in the case one this one is not  equals to zero, so in this case we define the Kodaiyara dimension of x is the largest number  d such that for the dimension it's really positive. So one remark here is that and so a very simple  remark here is that so in this case by the monarchy it's easy to see that in this case the kappa x  should take the values minus infinity 0, 1, 2 and 2 the dimension of x. And also another remark is  yes that if we replace the kappa bundle x by a line bundle by a homomorphic line bundle,  line bundle over x so by using the same definition here we can also define the kappa  error with respect to error. We just replaced the kappa kappa by the line bundle error here.  And so for this one we call it's the Itaka dimension of the line bundle error. So now I can explain  what's the Itaka conjecture. So for the Itaka conjecture it says as follows. So letter p  via fibrillation, so I explain later what the fibrillation means between two projective manifolds  x and y. And so for the fibrillation here I just means that she has a projective and the fiber  and the fibers are connected. So for the general fiber is so in this case since we have  set it smooth so in this case so for the general fiber is a smoother connected manifold.  So in this case the Itaka conjecture states that in this case we have the sub-adjectivity of the  correct dimension. So in this case it's conjectured at kappa x here is much bigger than the kappa y  plus the kappa f. So where the f is the general fiber is the general fiber of the fibrillation p.  So that's the Itaka conjecture. So there are a lot of thoughts about the Itaka conjecture. So here I  just so here I just just a little bit. So in the case one the dimension of y is equals to 1  in the case the base years of dimension one it's solved a long ago 40 years ago by Kamata.  He proved that in this case the conjecture holds and also another extreme case is in the case  one the fiber one the fiber f years of general type of general type. That means set the kappa f  years the extremo in this case for this conjecture it's also solved by it's also solved by Kola and  also there was another proof by Fieberg and also in the case one f years of log general type case  it's of log general type case. So in this case it's also same this case for this conjecture it's  also solved recently by Kovac and Pata. And yes so here yeah so there is also another very  important relater which is due to Kamata. He says that f for f at the midter a good minimum model  minimum model sets means that sets means that so for this fiber it's biogenally equivalent to your  manifold so for the fiber is biogenally equivalent to a variety of such that it's kind of the  window is some example in this case Kamata proves that it holds Kamata proves that the  conjecture holds so in particular so in particular so in particular abundance conjecture so in  particular abundance conjecture implies the attack conjecture. Yeah so so now I can explain our  result so for our result we prove that yeah so maybe I need that also to say that there is also  so there is also so there is also another more simple proof or prove that by a little better  simple proof by a proportionella and also there is a former worker due to Chen and the Haken we can  prove the following since and yes so if set up so later at P yeah fibrations yeah fibrations  between between two projective manifolds and the letter and now we consider a KLT case that  to be a KLT if at the advisor on X so in this case yeah for yeah for so in this case yeah for yeah  for the base is a billion variety if it's a projective polys saying this case we can prove  that for the high dimension of this one should be at least the the current dimension of the fiber  plus the restriction of the pair on the general fiber plus kappa y and so in this case the same  so in this case the same so why is a billion variety so we know that for this one it equals  to zero in fact and yeah and the and and and the biasing and the biasing document and biasing  document and the biasing that argument that developed so biasing document developed in  our paper we can also prove another router so so we can prove another router yeah set if for the  base yes if the base yes dimension equals to two so in this case we can have also this equality  so you try improve a little bit the result of karma saying this yes no no no tenness  as way so I see no no no so so so yeah for you in this case so in this case yeah for straight away  in this father since ever since ever since ever since ever we take it to be the junk fiber so in  this case from this one equals to zero so in this case by by so in this case by using our  two windows that this one is much bigger than this one and if we add a pair it has more more  sections yeah so so yeah so for the proof of our router so for the proof of our router it's based  on the worker so for the proof of our for a lot so it's based on a lot of folks  so for the proof of our reality it's based on the productivity of the relative kind of  bundle which is already explained in the talks of me hi so in the second part I just  record briefly so in the second part I just record briefly what he has explained in his course  so single party is the productivity of the relative kind of window and also the productivity of the  direct image so here we take a so so here we so for the end here we take a n to be a number and  then maybe we assume certain and and maybe we assume set the m e as much as and true and and  we assume set for the direct image is not equals to zero that's just a means to set a  set that's just a means that a fall for sets means just set just means that a four  generative point in why for this dark image here yeah yeah yes sorry so so so so for our proof  so for now on so for now on so for now on we just assume set up for the pair data equals to zero  so for the reason of simplicity so in the general case for the proof it's almost the same proof as  before so far now for simplicity we just assume that the data equals to zero the general proof  will be essentially the same proof so in this case so in this case it's not equal to zero  the sets just means that for the gem here which is equals to this one yes not equals to zero we  have at least one section for every for generic fiber same this case same this case so maybe I  just recall thanks to Bob Watson and he has already explained in his course in this case  we said that we can find a so-called data m so-called the m relative for m backman kernel  type matrix kernel type matrix which we write here as the edge B on the relative kind of window  such that once we calculate the curvature of it it might be a singular matrix in this case  it's a curvature is semi-polytive for on the total space of x in the sense of current yeah  I need to recall that's also so so so I just say a particular side x the recall a little bit of  set up so I just explained a little bit of set up so by the definition here so the restriction of  the dysmetric on the fiber on a generic fiber is constructed by yes constructed by by the sections  yes constructed by the sections of this one yeah so for the so for the construction so I think  that is already explained by me hi and I need also another router which is also proved by and  which is also proved in his paper in his paper with both and that's a set it's proved by hi and  Takayama still in this setting still in this setting and now we take a we just take air to be  the m minus one take kind of window and on air we equipped it with the matrix as as the HB so HB  is on the relative it should tensile m minus times say in this case say in this case say in this  case let let y1 in y be the be the locally free free locus of the direct image plus air which is  equals to the to this one because apioki for this dark image is just a coherent shifu so we just  take the locally free locus of it so for the complementary is of could measuring at least  two in y let y1 so let y1 be the locally free locus of it and and the vector y zero in y1 yeah  we can prove so uh so yeah so yeah so so that so so that y zero be the smooth locus  be the smooth locus of the vibration p that's just the main set of for mb fiber over y zero  so for the fiber so so for the fiber is smoother and the by and the by and by the inverse of the  shunher window center is in fact included in y1 so so so let y be this one say in this case  say in this case they prove that the Ix is a possible so so the Ix is a possible single  metric possible singular matrix which is coded which is related to the nashimha matrix on the  direct image so here we just defined over the locally free locus on this one such set such set  this one such that so for this possible single metric so for this possible single image here  it's not necessarily smoother but it's in fact bounded but in fact it's bounded on y zero and  and it's positively curved in the  sense of a griffith y1 so  so so for this one that's uh so for the productivity of this one I don't know if it's already explained  the you you have already explained okay yeah so so yeah so  so maybe I just explained a little bit of the construction of for h maybe I explained that  a little bit the construction of h so under so so let y be a pointer in y zero say in this case  so let y be a pointer in y zero so in this case so so for the dark image here  we know that it's just equal to this one  so therefore we want to define the metric on this space how we define it  we take a section here say in this case you wish I'd have defined the metric h it's defined as the  integration of the section s and with respect to hr yeah so in this case we know that for this one  it's a volume form and by the construction of and by the construction of the backman  chronotype metric it's easy to check set up for this one it's finite so therefore that's  it's plain set up for this metric it's in fact bounded on y zero so  and so and and so therefore we know that for this metric here it's in fact bounded on y zero  and after set and after set we can prove that and after set we can prove that in fact we can  extend the metric as a possible singular metric on y one and so and so for the notion of for  weakly positive curve that's just means set and so for this one here's the weakly  positive curve that's just means set on on y one that's just means set that's just means set for  uh that's just means set uh that's just means a letter e be the dual yeah  that's just a means that for every open set for every open set in y one every open set and  for every open set uh for every open set uh for every topological open sets in y one  And every section in the dual of this vector bundle,  here we know that also for every section here,  we see that it's weakly semi-partitive.  So we see that it's weakly, we see that it's  polydiped in the sense of gray surface.  That just means that for every topological bundle set in Y1  and every section here, so in this case,  we can consider the function log S. So here we take the dual  of this one.  So therefore, we should take the dual of this matrix.  Yes, PSH function.  Yeah.  Yeah.  Because we know that Y1, because we know that Y0,  it might be a straight thing subset in Y1.  So in this case, by the construction here,  we know only that for the matrix H here,  it's bounded on this set.  It's bounded on this set.  So on the complementary in Y0, it might be infinity.  But for this definition, it still holds.  And one more remark here is that if in the case H here,  smooth, if the matrix is a smooth commission matrix,  in this case, for this notation, it's  your current to say that it's positively curved.  It's positively curved.  It's your current to say that it's gray surface semi-quality.  It's gray surface semi-positive.  Yeah.  So that's the explanation of the result of Mihaly and Takayama.  And as a color, by using a lot of roof feed,  we know that in this case, actually,  in use a matrix, we write as determinant H  on the determinant of the draft image.  And in this case, since it's of rank 1,  we can take the double dual.  It's a line bundle on the total space Y,  such that once we calculate the curvature of it,  it will be semi-positive in the sense of current  on the total space Y.  And so now I will explain two propositions, which  will be crucial in the proof of our results.  So for the first proposition, it is  that still under this condition here,  still under the construction here,  one proposition is that later, let you  be some open set in Y, not necessarily in Y1.  Be some open set, topological open set in Y.  Therefore, the curvature of the determinant  of the draft image here is identically  equals 2 0 on U. So in this case, in fact,  we can prove that for the draft image here,  with respect to H, it's Hermitian flat.  It's a Hermitian flat vector bundle.  Vect bundle on U, not only on U intersect with Y1.  I think that I don't have enough time to explain the proof.  So I just explained the idea of the proof of proposition 1.  In the case 1, H is smooth.  In this case, for the proof, it's very simple.  If H is a smooth Hermitian matrix,  say in this case, it's rather entry.  So in this case, it's very easy to see.  It's very elementary to see that for this draft image here.  Yes, it's just equals to the trace of the curvature.  We have this one on U intersect with Y1.  And now we assume that for this one,  it's identically equals 2 0.  So here we know that in the case 1, it's smooth.  So here in the case 1, H is smooth.  We know that in the case 1, H is smooth.  So we know that since to the root of my head,  Takayama, we know that for the draft image curvature here,  it's semi-prolative in the sense of graph.  So for this one, it's semi-prolative matrix.  And if the trace equals to 0, so that's where you imply set.  And the source sets will imply set for the draft image here.  Should there be Hermitian flat?  Are you intersect with Y1?  And after set by using a little bit more argument,  we can prove that it's in fact a Hermitian flat  on the total space U. Since here,  it comes from the draft image here.  So I have no time to explain the last one.  And so to prove our result, we need also another proposition,  which is essentially proved by Fievega.  And also we follow a proof of 2G, which says that the following  is so in the talk of Mihai, so we know that in the talk of Mihai,  he proved that if the direct image of relative kind of  kind of plane is not equal to 0, say in this case,  he proved that set for the relative kind of plane,  kx over y, it's considered effective.  That means that we can find a possible singular matrix  such that it's semi-prolative in the sense of current on x.  But for the following proposition here,  in fact, so in the proposition 2 here,  in fact, we want to compare the productivity of the relative  kind of bundle and the direct image  of the relative kind of bundle.  So the proposition 2 says that in fact,  the probability 2 says that in fact,  so for the relative kind of bundle here,  it's in fact more positive than the direct image.  So that's as follows.  So still in this setting, say in this case,  we can prove that we can find some small epsilon  positive, which depends on m.  And some divisor on x such set for such set of so forth,  this divisor here, it's in fact very small.  It's in fact, it's all for co-dimension,  complex co-dimension at least a 2  at least a 2 in y.  And more importantly, we can prove that for the relative  kind of bundle plus e.  It's much bigger than the direct image.  That means that it's much bigger than the.  So for the direct image, it comes from the base.  So we need to take the determinant of this one.  So now this one becomes a lambda on the base.  And we prove back it on the total space x.  So for here, it's a class in on x such  that it's considered effective on the total space x.  So in particular, for this one, it's a big line bundle  on the base.  So therefore, the progression implies  that for the relative kind of bundle,  model since modifies here, it will  be straight and positive in the horizontal direction.  And so that's the population too.  Now I'll explain the proof of our loss.  In fact, for the divisor e here,  it's in fact harmless because of the factor set for its image  on y is of co-dimension into when we prove the itaqa  conjecture, it's either to get rid of this divisor e.  So now for the third part, I will explain the proof  of our theorem.  So we talk about the results in two cases.  So for the first case, we let P be a vibration between two  projective manifolds.  And in the case, we first explained the case 1,  y is a Bayesian variety.  So in this case, we want to show that, so in this case,  we need to prove that for the co-dimension of x,  y is much bigger than the colloidal dimension of the fiber.  Because in this case, y is an Bayesian variety.  So for the colloidal dimension of y is equals to 0.  So that's namely, we want to construct a lot of sections  in the direct image for this one, which is just  equals to since y is a Bayesian variety,  so we know that it's kind of complicated as trivial.  So it's just equal to this one.  So it's going to say that once we can find a lot of global  section here, so for this one, it's just equal to x minus x.  So it's therefore to prove the eta connoisseur,  in fact, we just want to find a lot of global sections  on the direct image here.  So here, we have already, thanks to the lot of time,  we have already no setter.  We can find a simple matrix on it.  So now we want to construct a lot of global sections.  So for the step one, in fact, we need a simplification.  In fact, we can assume that in fact, in the step one,  we would like to show that we can assume that y is a simple  torus.  It is a simple torus.  That's just a means that there is no structure  sub torus in y.  Why we can assume this one?  So for the reading, you said in fact,  for this inequality, it's in fact,  in fact, for this inequality, it's  in fact invariant after finite eta cover.  That means that if we take some y theta, which  is a finite eta cover, which is mean setter sub torus 1,  it's a finite eta cover and no ramified locus.  And here it's the product.  If we can prove that inequality, for this vibration,  say in this case, it will imply that we have also  the inequality on this vibration.  So for this fact, it's well known.  So you want to increase the possibility?  So we assume that it's a finite eta cover.  Yeah, but you want to increase the possibility?  No, no.  So since there is no ramified locus,  so for the collier dimension of this one,  it's because all the collier dimension sequence sets.  Yeah, so it's literature.  Yeah, so by reading this route, so  by using this route, we know that y is not simple.  It's not a simple torus.  So in this case, it will imply that,  say in this case, by Bronga's reductive theorem,  it will imply that after some finite eta cover,  after some finite eta cover, so for example,  once y is a torus, after some finite eta cover,  it's also a torus.  So for the collier dimension, it's sort of weird.  We can assume, if it's not a simple torus,  say in this case, after some finite eta cover,  we know that y can be written as the product of two torus.  There are two sub-torus in y.  And so in fact, in this case, we can prove that,  and once we have this route, so in this case,  for the fibrous here, we have another projection  to another torus setter.  So in this case, for the inequality here,  we can prove it by the induction on the dimension of the base.  So we can prove the inequality here.  Can be proved by induction on the dimension of y.  Yeah.  So for the advantage of this construction here,  setter.  So for the advantage of this construction here, setter.  So the advantage of this construction here,  setter, in fact, in this case, once y is a simple torus,  say in this case, for every pursuit effect of lambda  on y, there is only two possible cases.  So the advantage of this one is setter.  If y is a simple torus, in this case,  we can easily prove that if y is  pursuit effective for, say it will imply  the setter y, if it's a pursuit effective lambda on y.  So in this case, it will imply setter  is y, where the topological true lambda  sets mean setter, if C1 plus equals to 1.  Or there is ample.  So for this one, it seems quite easy to check.  But I have no time to explain it.  So sets the advantage of this assumption.  So now for step two, now we can use.  So now for step two, we can use the result of behind the takayama.  So in this case, by the result of takayama,  we know setter for the black image of this one.  By applying their result, we know  that it's positively curved on y1.  And also, we know setter by using the result of Rufi.  We know setter for the determined window.  It's semi-political in the sense of current  on the total space y.  And in particular, we know setter for this one,  it's pursuit effective.  And now by using this route, we know setter for this pursuit  effective window, it should be whether it  equals ample or its topological attribute.  So setter for the determined.  Setter for the determined.  It's topological attribute or ample.  So now there are essentially two cases we need to check.  Now step three, we check the case  of when it's topological attribute.  If the determined window is c1 class equals to 0,  if it's topological attribute, say in this case,  since for this current, you see in the class of this  determinant.  And for this one, it's equals to 0.  So for semi-political current, it's equals to 0.  So in this case, it's identically equals to 0.  So that means setter.  So for this one, for the determinant,  this one, it's better identically equals to 0  on the total space y.  So now we are in a position to apply proposition 1.  Now by using proposition 1, we know setter for the direct image  here, it's in vector Hermitian flat.  It's a Hermitian flat vector bundle on y.  Once we know that it's Hermitian flat vector bundle on y,  since y is an Abellion variety, we know setter.  Since y is an Abellion variety, we know setter.  So for the fundamental group of y, it is also Abellion.  So in this case, once it's representation,  it's in the unitary group.  So therefore, since for the fundamental group,  it's Abellion, we know setter.  We know setter.  We know setter in fact.  For the direct image here, it can be split as the direct sum  of numerical trival and bundle on y.  So here, r is the rank of the direct image.  So by using this variety, so here we use the factor setter.  The fundamental group is Abellion.  And so we are, they are numerical trival.  So how many times I have?  I'll take a minute.  Yeah.  So once we know this, we have to solve the setter.  So by using a route of Kampana, Kampana and Paternela.  So by using a route of Kampana and Paternela,  which based on a route of Simpson,  we can easily prove that for the,  since for the direct image here, we  know that it's in fact a numerical trival.  And by using a route of Kampana and Paternela,  we can prove that we can construct a lot of  global sections.  So setter will imply setter for this one.  It's much bigger than this one.  So I have no time to, so I'm sorry.  So for the last set, I have not enough time to explain it.  But it's essentially based on a route of Kampana and Paternela.  And now I explain the last step.  And so for the step four, we have already explained setter  for the direct image here.  Since we assume setter, why is it simple to us?  So for the determinant of direct image,  it's equals to a topological trival  lavender or an ample lavender.  So in the second part, we use that case  when the direct image is ample.  Therefore, the direct image is ample.  So in this case, because I list also the second proposition,  we are throwing the second case, so we need to apply  the second proposition.  So in this case, by using proposition two,  setter will imply setter, it will imply setter for this one.  It's it will be very critical.  Well, yes, so in fact, as we explained before,  so in fact, as we explained before,  since the image here is of co-dimension at least two,  since for this derisers, it's in fact a small derisers.  So in this case, there are the standard,  so in this case, there are the standard method  to get rid of this small divisor.  So for simplicity, we just assume that E is equals to 0.  So in this case, that's imply setter.  Since this one is ample, so that's  where imply setter for this kind of window is political.  Yes, it's really political in the horizontal direction.  So setter will, and so setter for by using the relative for both  and the power, and after setter and together with essentially  by using the oscillatory gosh extension theorem,  we can prove setter.  So for this one, maybe another M, so for M large enough,  we can prove setter by using the result  and together with oscillatory gosh extension,  we can prove setter for this further restructuring map  is such a tip.  So setter for setter will imply setter.  kappa x is much bigger than kappa f.  So I have one minute of, yeah, so in the last one minute,  so maybe I explained that a little bit  of the idea of the proof in case one y is a surface,  and one y is a surface, and one y is a surface,  and one y is a surface.  In fact, in this case, it's quite easy to check setter  for the collinear dimension.  It's an imagined random number after birectional equivalent.  So in this case, we can run the MMP on y.  So therefore, in particular, we can assume setter  for the kind of window of y is semi-ample.  And in this case, in the case, if the collinear dimension is  much bigger than when, then in this case,  it's well known due to the relative of Flegger.  So the only interesting case is in the case  when the collinear dimension of base is equals to 0.  So in this case, so after some finite a dot cover,  since we assume that it's minimal,  then in this case, we know that y  should be a torus or k-3 surface.  In the case 1, y is torus.  It's already proved.  And in the case 1, it's a k-3 surface.  So in this case, we still, and so therefore, we still  studied the direct image here.  And we know that for this one, it's  a pseudo-effective lambda on y.  And in the case 1, y is a k-3 surface.  So in this case, we know that if  y is k-3 surface, it's k-3 surface.  And 1, y is pseudo-effective by using the Savatsky  decomposition and by using the Karmata-Flegger vanishing  theorem and the He-Man-Rohlker.  It's easy to check that for the collinear dimension of y,  should be equals to the numerical dimension of y.  And in the case 1, the collinear dimension of y  is at least 1.  So in this case, by following almost the same proof as here,  we can prove the Idyac furniture.  So the difficulty case is in the case 1,  the collinear dimension equals 2, 0.  So in this case, that just means that so for the collinear  dimension of this one, equals 2, 0.  So that's the difficult case.  But in this case, by using the Savatsky decomposition,  we know that for the direct image here,  it should be equivalent to the sum of some exceptional locus.  So it's an expression of divisor.  So for this one, it's in fact the minus y rational curve.  And by using a lot of compana, we can prove that  we can in fact prove that for the fundamental group  for outside the exceptional locus,  we'll be very simple.  So by following almost the same proof as in step 3,  we can also prove the Idyac infinity.  So that's the finished proof.  Sick.  Thank you.
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