 So I'll talk about, the title of my talk is the Permanent Kernel Approach, the Pogaine-Milmauian Equality and Helping Functionals, and this is some joint work with Professor Rubinstein. So I want to fix some notation before I begin, so just ignore this column because I don't think I'll have time to talk about functions. So, K is going to denote the convex body, S is going to be used to denote the symmetric K-Serve is the polar, so, you know, this is for all X and K missing there. Observe the volume of K with the non-to-volume, Pk the paracenter, Mk the mother volume, and Tk the two domains, so it's like Rn, cross, like Rn Cartesian for the Tk, B2n is the usual antimenson volume Rn, and K subscript superscript N is like the Cartesian product of K itself. Sorry, do you assume that origin belongs to the interior? I don't. You don't? No. But doesn't it that way, don't you, when you get that the polar of the polar is K again? Yeah, you will not. And you will not be, as you know, you don't mind it? Not yet, because I'm usually translated by the paracenter, so, for now it's just, yeah, and I've put the definitions here, so let's keep going. So, in particular, I want to emphasize that for the mother volume I'm going to use this N factorial here, because this way it's multiplicative, so the mother volume of K times L is just the product of the two volumes. So we don't have to worry about factorials. Alright, and so there's this famous conjecture by Mahler, like 1939 for a symmetric complex body, the mother volume is lower bounded below by 4N, which should be, in general, bounded below by this number there, which is the mother volume of this complex, like translated so that it's centered with the origin, and this is approximately asymptotically e to the n. And these are still open, but with this theorem first approved by Puget and Milman in 1987, that at least there is some constant that doesn't depend on dimension, so that the mother volume is bounded below by C and all complex bodies and all dimensions. And we have several proofs, like Milman gave another proof. Next year, like 1988, that by Cooper Burke, Nisarov, Janopoulos, Paulis and Pitu, and two proofs by Bernstein. And I'll focus on the Nisarov proof. So, alright, so the reason why I want to focus on the Nisarov proof is that all of these proofs work for symmetric complex bodies. So, they prove their theorem for all symmetric complex bodies and then using this symmetrization trick, they get a bound for all complex bodies, and this comes from a theorem of Rogers and Sephard, like from the 50s, that if we take a complex body k that contains the origin, and we consider this complex volume of k in negative k, then this gives a symmetric complex body that contains k, and the volume is bounded from above by two n times the volume of k. And so it follows as a corollary from that, but if we have a lower bound and the mother volume of symmetric bodies, then if we get a lower bound for all bodies, that we have to divide the constant by two. But Nisarov's proof, using the Bergman kernels, so Nisarov's proof disbound specifically by k over 16, which is approximately 1.9, and I think Professor Foske gave another proof of the same theorem, like by another method, but still using Bergman kernels. No, I don't think I'm really getting another proof, just some aspects of Nisarov. But it's the same bound. Alright, so I don't know. But Nisarov, in the final lines of his paper, predicts that the same proof would work for any complex body. And this would be without using the symmetrication tool, like someone would be able to carry out all the details in a similar manner and get this bound by 4, which is not as good as what one would get from the symmetrication trick, but still it's interesting. So I'll talk about two things, first of all, how to actually carry out the details. So we only encounter one difficulty doing this, and so I'll explain what the difficulty is and at least what we did to solve it. Then I want to talk about some nice definitions that came up from this work, and some helping variants of the model volumes that really give a better approximation. Alright, so Nisarov's proof can be realized in three steps. The first step is that we have a really good, a very explicit formula for the Bergman kernels of two domains. So Tk is a two-domain, it's like Rn times K, and it was ordered by the 6th and more explicitly in 2005. The formula for the Bergman kernels was explicitly computed. And so in particular on the diagonal for A into K, we get this integral, which is just illegal. The second step is to... So first of all, we would like to multiply the Bergman kernel with the volume of K squared and call this Bk, because it behaves a little bit better. So for example, Bk is in the final variant. And also we have this inequality connecting B and N, first given by Nisarov's first symmetric body, then by Hultre and a more general setting of functions that are centered at the origin. And then we reformulate it in terms of B and N at the upper bound. The upper bound is not interesting for the proof of the Bergman, but I think it's interesting in general, as we all see now. So this follows from Jensson's inequality. And the third step is to construct a homomorphic function on the two domains that evaluate at one, say at the one at the Poirier center on the diagonal, and then depending on what the setting is, we have a different constant, like for symmetric bodies, it should be like a pi cubed over 16, and in general, 5 over 4. And I'll explain where these two numbers come from. But this would give the desired bound, because this would give a bounded B, and by this lemma, the bounded B would give a bounded M, just by multiplying by... One observation here is that we have those truss terms, E to the O-N, that are sub-exponential to mention, and I want to explain the next slide why those things don't matter, and how to get rid of them. So first of all, B is multiplicative. So this is another similarity between B and M. And so this means that if we have a lower bound in B that is of the type C to the N, to the N times sub-exponential terms on dimension, then we can get rid of all those terms by the following trick, just by using this multiplicative progressive B. We have the same bound of KM, but B of KM is B of K to the N power. So we can take the M through and send M to infinity, and then those truss terms will just go away and we're left with a constant, with a C. Can you remind me? Yes. B is that the Bergman kernel? Yes. B is the Bergman kernel times the volume of K squared. And this is a final variant, like M is in a final variant, and also it has this multiplicative property. And so the idea for bounding B due to Nazarf was to use Hermander's del bar theorem to construct a holomorphic function that evaluates to one at the origin. So since this is symmetric, we know that the bar center is just at the origin and give this upper bound on the L to normal F, which has sometimes... So the first level is S squared, because B is not the Bergman kernel. B is the Bergman kernel times the volume of the body squared. So we want this term. This 16 over pi squared is the constant we're trying to prove for. Then we have a term e to the 2n delta times the constant plus 2n log sigma. But this is all delta in 0, close to 0, and then sigma close to 1. So we can take delta to 0 and sigma to 1. So this term goes away. And we have some truss terms that are sub-expansion to dimension which we just described how to get rid of. And finally we have this quotient big R over small r squared where big R is the radius of the ball containing S and small r is the radius of the ball containing S. And so the only term we want to deal with is this R over R squared. And to do this, Nazar refuses John's theorem. So since everything is in a finite variant, M is in a finite variant, B is in a finite variant, we don't really care about affine transformations. And so by John's theorem, by applying an affine transformation, we can make sure that this complex body is contained in two balls whose radii, the quotient of the radii is bounded above by squared N. And so in particular, this is also, this R over R here is also sub-expansion of the dimension. And so all the terms are here. And we're left with a bound we need for the symmetric problem. Just one more comment on Nazarov's proof is that to get this L to bound to F, he's applying Hermann's theorem on the two domain of some bunk functions I don't want to do more details about. And the weight function is given by this phi, small phi, where the big phi is the conformal map between the symmetric strip where the imaginary function between negative 1 and 1 and the unit ball. And the reason that this 16 over pi squared shows up in the proof is because exactly the derivative of this conformal map at the origin is pi over 4. So if you take pi over 4 and you square it, you get pi squared over 16. And then you invert that and you get 16 over pi squared. So if you want to do this in a non-symmetric setting because this unit product, is that a T will not longer lie this strip of 1 to negative 1. We know that 1 would be another bound still but negative 1 may not be the lower bound. So we have to go all the way to negative infinity. And so for the non-symmetric setting we would have to change this conformal map from the half space where the imaginary part is less than or equal to 1 to the unit ball and the derivative of the map would change this is where the different constant comes from. So in general we want to do the same thing. There is a problem. Like here I just explained that we could really do the same thing and get it bound like this way. But now we need to make sure that this quotient R over R has a good bound which in general we kind of do because John's theorem for non-symmetric body says that we can't find an affine transformation so that a case contained between two balls of radius R and small R and big R so that the quotient is less than or equal to n. But it doesn't guarantee that the body center will be contained in say the maximal ellipsoid. So if the body center is not contained in the maximal ellipsoid then taking the ballers would give really bad bounds. So this would destroy several parts of the proof. So that's the problem with the non-symmetric setting but we need to find a way to control this R over R. For us the solution was to just let R depend on k. So we actually don't need to use John's theorem. And the observation is that for powers of k we can choose the same R as rk and for the big R we can take squared m of rk because if k is contained in a ball of radius rk km will be contained in a ball of radius squared m. And so this shows how to choose R in products of k. And so having this observation in mind we can let R depend on k so that we get this ball. And as I described before we don't have 16 over pi squared here with 4 to the power of n. So this 4 to the power of n comes from this fact that we have to change the conformal map in general so that from the half space now instead of this symmetric string to the ball of radius 1. And this map has derivative of the origin equal to the half which means that at the bound for f it will show up as 1 half squared which is 1 over 4 and this inverted which is this thing. Alright, so this f because b is a Bergman kernel this gives a lower bound than b and gives this lower bound here and k squared is going because it was k squared is in the definition of b and the rest of the terms are there 1 over 4 to the n we need. The other term with the delta and the sigma is the derivative of by taking delta to 0 and sigma to 1. And since this is the bound for all convex bodies in all dimensions we get the same bound for the product of k. And now we know how they transform when we take products. So we can replace rk n by rk and big rk n by squared m rk. And then in this last expression here we're going to take the nth root and this part is now prime star that doesn't depend on n and 1 over m is some exponential so everything will go to 1. So we don't really need John's theorem. In fact we can even let more things depend on k but I don't have much time to talk about it so let's keep this part. And I want to focus on some observation and that comes to this proof that we have an explicit formula for the Perkman kernel so we have an explicit formula for k. And in particular we can write p of this and we introduce this hk1 which is like the l1 version of the I mean this is like the Laplace transform but it looks like an l1 version of the support function then we write pk as like that which looks a lot like the mother volume because if we had the support function here this would be n factorial the volume of the polar which is exactly like the times k is exactly the mother volume. So b behaves we saw that b has similarities with m and now we wrote it in a form that really resembles m like the mother volume. The observation is that we can do this more generally so we can define those LP versions of the support function and then define some LP versions of the mother volume by this definition. So not for p equals infinity we just have the maximum over the inner products which give exactly what we've defined as the mother volume above. And the nice thing about this mp is that now we can generalize the previous lemma so we call that we had this inequality for p and m and now p is exactly a multiple of m1 and so we can restate this inequality using m1 and m as it's shown in the second example this inequality can be generalized for all p so that we get better approximations of m and the observation here is that as p goes to infinity we really get better and better on the voltage because this limit goes to of exponentias if we take the limit of those mp volumes then they converge to the mother volume. So what was the inequality on the previous page? In the limit let's p turn the infinity in this and the right-hand side is the mother volume and the left-hand side is the mother volume times one. So this generalizes the previous lemma. So it has many similarities to mp has many of the same properties m has so for example they are invariant under the action of gln and also if we take k and we move it so that it's centered to the origin then this expression here is a complete affine invariant even under translations but in general this is not invariant under polarity like the mother volume because this is no longer a duality brace those support functions. Okay so we lose duality but that's not a big deal to say something more that really resents the mother volume first of all all the translations of k by a point in the interior have finite volume but if we translate by something that is not in the interior then we get nothing. We can compute exactly what the derivatives of those mps are we can also compute the second derivatives and so that this mp is strictly complex and in particular this implies that I mean since they're strictly convex and outside of the complex set they are exactly infinity so they must have a unique minimizer so we get the existence of some kind of points for all those mp functionals so there exists a unique point in the interior that minimizes mp overall translations similarly to the mother volume and it's exactly the point where there is a bar center of the support function of the piece of the function it's the bar center for the function so some other point centralized to the setting and it's also multiplicative so this means that we can get rid of truss terms alright so maybe some examples one can explicitly compute I mean I don't know I'm not sure if this can be simplified for a large p but at least for p equals 1 there's a very nice formula that was also discovered by Professor Wozki in terms of Berkman kernels but here m1 which is reformulated formula for p equals 1 for higher p I don't know if the cube has a nice expression two balls we have an expression in terms of the modified Bessel functions so the modified Bessel function is given by this series and one can compute the mp of p to n for all p and all n in terms of an interval this function is a nice expression because you can put it in a computer and it can give you an estimate of what this would be and this is working properly but it should be the case that another interesting example is the simplex so the simplex but you have to move it so that it's centered at the origin and you can compute say m1 of the two-dimensional simplex was also computed in m1 of the three-dimensional but I did a part here and you can see that m1 of the simplex is less than m1 of the cube and here this is also work in progress but there is good indication that asymptotically the mp volume of the simplex should be given more this nice expression okay so I can make conjectures that for symmetric bodies those mp are minimized by the cube and in general they're minimized by the simplex and by the previous lemma if those conjectures are true then they imply the mother's conjecture if they're true for all p then we can take p to infinity this may be easier because there is only one conjecture minimizer namely the cube so people say maybe the existence of multiple minimizers make the problem harder the first conjecture for p equals to 1 was also conjectured by Professor Wojci in terms of Berkman-Carris previous paper