 raised to the power n, various normalizations here. And actually, I want to put a zero here also, which again means that phi vanishes on the boundary. So if n is equal to 1, this is the integral of, if n is equal to 1, it's the integral minus phi Laplacian of phi, which then by integration by parts, since it vanishes on the boundary, is equal to this. It's the energy. But in higher dimensions, when you use the Montchampere energy, you have to assume that the function here is pleurisabharmonic in order for this to be reasonable, because you want this expression here to be positive. So it's important that it be pleurisabharmonic. And then you can ask the following question. Save it. It's time to ask the question. Yeah, it is a question. So problem. What question? Does the Pauliasega theorem hold for Montchampere energy? So is it true that, well, the energy omega phi is that greater than the energy over the ball of the same volume of phi hat? Do you have such an inequality for the Schwarz symmetrization? In particular, you need for this, in order for the question to be reasonable, you want your functions to be pleurisabharmonic. So one thing you need is that phi hat is also pleurisabharmonic. That's important for the problem to make sense. And unfortunately, this is not true. So the project somehow stops here, at least for a while. It doesn't really make sense, because it's not pleurisabharmonic anymore. But we add an extra hypothesis. So add hypothesis, namely that phi is S1 invariant. And that means, now, by definition, it's going to mean that phi of e to the i theta of c is equal to phi of c, when theta lies on S1. So this somehow eliminates the problem completely in one complex variable, because if you have S1 invariant, it is already radial. So of course, the problem is clear. But if you are in higher dimensions, then just one symmetry is not so much. And so you still have a lot of degrees of freedom. And you can ask the problem for such functions there. And then you can say something. So that's the first thing is that the problem makes sense. Assume also that you need to assume something about the domain. So assume omega is balanced. And that means, for me right now, that if lambda is a complex number smaller than 1, then it should imply, ah, and if c lies in omega, then I want lambda c to lie in omega. So it's a little bit stronger than saying just that it's S1 invariant. It was complete in a way. So it particularly contains the origin and things like that. And then we have the first theorem then. It says that if phi S1 invariant and omega balanced, for instance, of all, implies that phi hat is pleurisaparmonic. To the radial, the Schwarz symmetrization is still pleurisaparmonic. And so the problem makes sense. So that's theorem one. I'm going to say something about the proofs here later. Actually, I think I'm almost going to give the proofs. But let me give you the other result also then. The theorem two then, about the poly. I think it's an idea why we might even have expected that to be true, especially since pleurisaparmonics, this condition on planes. Yes, and it gets completely distorted, right? So no, I can only say that you wanted it to be true. And then you checked if it were true. Actually, there is a counterpart which I think is known. And that is that if you take a convex function in Rn and you take the Schwarz symmetrization, then it is still convex. I haven't really exactly found that statement. But I found a corresponding statement for convex sets in the book by Bondeson and Fenchel. So that's more or less equivalent, I think. I should say that this statement for convex functions is based on Brunminkowski inequality. And this will also, in the complex setting, it will also depend on the version of Brunminkowski theorem. Well, and then anyway, then we get back to the question there about the energy. And then you have that then, I can write it like this, that the answer to question is yes, if and only if. It's not always yes, but it depends on the domain. So it is yes if and only if omega is an ellipsoid. So in particular, it's true for the ball, which is not trivial. You might think it's trivial since it's already, but the function is not radial. So if you take any plurisobamonic function in the ball and you symmetrize, it's still plurisobamonic and the energy goes down, just like in the Polyaseketer. But you cannot say this about any domain. Actually, it's true if and only if the domain is an ellipsoid. That's the main theorem. OK, before I go on here and indicate the proofs, I would like to just give one little interlude here, which I find intriguing about theorem one here, that plurisobamonicity is preserved. So that's just a remark, which is the following. It's connected with the openness conjecture remark. So let me recall the openness conjecture in complex analysis for plurisobamonic functions. It was formulated by de Magie and Collard. And it says the following, if let phi be plurisobamonic, say, in the ball, it's not so important exactly to the domain there, and be such that the integral over the ball e to the minus phi is finite. So you should think of it as singular at the origin, but this is a measure of how singular it can be. The e to the minus phi is still finite. Then the openness conjecture, so that's the problem, still open, there is an r greater than 1, such that if you look at this integral, so you make it a little bit more singular, this thing, over the ball, then it's still finite. That's the problem. So this is a very elementary problem, but it's been open for like 15 years, so I don't know exactly. So it was proved in the case of n equal to 2 by Charles Favre and Matthias Jonsson. So the only remark here just to illustrate the theorem here is that the corollary of the theorem is that if the answer is yes, if the function phi is S1 invariant, it still doesn't make the problem completely evident that it's S1 invariant, because again, it's just one symmetry, and you can have a lot of variables here. In particular, functions that are associated to a matrix on line bundles over a projective space that will have this thing here, you can apply to them. So the proof here is so easy that I think I would like to give it. It's always good to give a real easy proof to inspire a sentiment of security in the audience. So let's see that. So the proof is just this, that integral e to the minus phi finite implies that integral e to the minus phi hat is finite, because they are the same. They are equidistributed. And now you can check the openness conjecture for radial functions. And that's a completely elementary exercise convex function. So it implies that there isn't some r such that e to the minus r phi hat is finite. I'm not going to do that, it's very elementary. And then since they are equidistributed, this is the same as integral e to the minus r phi. So it's finite, and that ends the proof. So it's very easy in that case. And of course, it would be nice if one could somehow, by some finagling, get the general result this way. But I don't know how. So I leave it to you too. That's a problem too, get the general result from there. OK, so now I'm going to say something about the proof of theorem one. So let's see what that. Yes. Yeah, so it uses a proposition. This is where bagman kernels appear. So the proposition is the following. Let d be a subset of Cn plus 1 and say it's pseudo-convex. Pseudo-convex. And you'll look at the slices. d tau is the set of all C in Cn such that tau C lies in d. So it's going to look something like this. Here's the tau axis. Here is the Cn axis. And here is a pseudo-convex domain d. And you look at the slices like this. Assume now that these functions are S1 invariant. Well, assume they are balanced, say. Well, say d tau, yeah, d tau balanced. Basically S1 invariant and connected. That's what I want. Then the proposition says that the logarithm of the volume of d tau is superharmonic. So the map that sends tau to this is superharmonic. So in particular, if it's independent of the imaginary part of tau, it's concave. So concave if independent of imaginary part. So that's the theorem. And this you should compare to the Brunminkowski theorem. So compare of Brunminkowski. So one way to formulate the Brunminkowski theorem is to say exactly the same thing in Rn. But this is convex then, instead. You take the slices. And then there's no assumption on S1 invariant or anything. And the conclusion is that this function is concave then. So this is, if you want to, with a complex version of the Brunminkowski inequality for pseudo-convex sets in Sia. Now the proof of this, it rests on property of bagman kernels that I'm not going to get into. So the starting point is a theorem of mine from a couple of years ago. It says that if you let k tau of 0 be the bagman kernel d tau at the point 0, at the origin. The origin is inside of d tau. Then logarithm of k tau 0 is subharmonic. That's the theorem that I'm not going to prove. We can use this that the variation of bagman kernels are subharmonic in the parameter there. So this rests on L2 estimates and d bar operators and such things. So you have that. And then because of the symmetry, so d tau balanced implies that you can actually easily compute the bagman kernel. And you find that it is equal to 1 over the volume of d tau. Because the bagman kernel is a holomorphic function with a reproducing property for 0 and a constant by the mean value property or for an S1 invariant set, a constant is such a reproducing holomorphic function so it must be the bagman kernel. So that ends the proof when you know the fact about the bagman kernel. So that's it. So this we have that the slices here, the volumes of those are superharmonic. And for us, they would be concave then if they don't depend on an imaginary part. OK, now you can look at proof of theorem 1 then. For theorem 1, remember that the defining property of the Schwarz symmetrization was that we had to look at the set where phi hat was going to be radial. So now I say it's a function of not of c but of logarithm or modulus of c. It's more convenient. So the defining property was that the set where phi hat smaller than t was going to be equal to the measure of this was going to be equal to the measure of the set where phi was smaller than t. Now, this is equal to f is going to be increasing. You can assume it's strictly increasing. So this is going to be equal to the set where c, where mod c is smaller than e to the f inverse of t. This is the same thing as that. So this just says that f of log c is smaller than that. You take the inverse of this increasing function and you get the f inverse there. And then you take the exponential, you get this. And you can compute this. This is the volume of a ball. So it's equal to some constant e to the n times f inverse of t. That's what it does. Now, this is equal to the set where phi minus the real part of tau is smaller than 0, or the set of c such that this holds. So if you let the set of all tau and c such that phi minus real part of tau less than 0, if you call that d, this is a pseudo convex set. I can apply the theorem of the proposition over there. And this is a slide of those sets. And I find that the proposition implies that the logarithm of the volume here, of the volume where phi hat is smaller than t, is concave. So it says exactly that f inverse is concave. So f inverse is concave. And then it means that f is convex in the inverse. And so if f is convex, log c, this is precisely the condition that this function is plural subharmonic. And then we're done. That's the theorem. OK, so that was the first part here. And then, now let's see how you estimate the energy. So that's theorem 2, the inequalities for the energies. And that is, that uses the method of geodesics. So let me say something about that. It uses, now, so for theorem 2, we let phi, and I take then first, there are several directions, two directions. First, we take omega to be the ball. Omega to be a ball. So then we want to prove that the theorem is true, that the energy goes down. And now we let phi be the function then, which is plural subharmonic vanishing on the boundary of the ball. And now I'm going to compare it to another function for which I know that the theorem is true. So I take another function, phi 0, which is also going to be in there. So this is an auxiliary function. And it's going to satisfy that phi 0 is some function. It is already radial. So it's its own short cementization. Phi 0 is equal to phi 0 hat. So the theorem is certainly OK for phi 0. And I want to prove that it's OK for phi. And I connect them now with the geodesic. So connect with the geodesic. So we let phi equal to phi 1 and connect with geodesic. So I'm using abuse of language here. So there is a well-established notion of geodesics in the space of smooth, strictly plural subharmonic function. It goes back to Mabucchi, Sems and Donovson, while in the setting of manifolds and matrix on line bundles. I'm going to use it a little bit more sloppily here. So I'm going to say that it's a geodesic. It means that I have phi t of c. They are all the time going to lie in the same space. And the geodesic condition is that phi of, if I let it, if I think of it as depending on the complex parameter, but independent of the imaginary part, this function here is a function of, I could write it as, capital phi of tau and c. It should be plural subharmonic with respect to all the variables. And that's not a geodesic equation because it's just some inequality. But it should set the geodesic equation then is that the Monchland pair with respect to all of the variables of capital phi equals 0. So by pluripotential theory, it's enough that the function is bounded or something like that. It's enough to define this, and you can give a sense to this, even though it's a little bit of using. It's not quite properly geodesic. But that's what I want to do. And you can always, by some super construction, you can always connect those functions with geodesic in this way. It's a theorem of Chen in the complex setting that says that you have some regularity up to almost the second derivatives. But it's not so important here. So we can just take it in a generalized sense. So I connect them there. And then I look at, then we need a lemma first. So I'm going to look at the energy along the geodesic. And then it's a known fact, basically known that the energy for phi t is an affine function. It's affine. So the energy depends very nicely on variations. And now we take for each fixed value of t there. We look at, right at the lemma, we can look at the Schwarz symmetrization. So they are all, as we know, pluripotential. Yeah, they're going to be S1 symmetric. So if phi is S1 symmetric, and then the geodesic is going to be complete all the time, S1 symmetric. So those are pluripotential all the time. And we know, actually, it's a sub-geodesic. That's maybe my own word. So it means that only the first part here, that phi hat of real part of tau and C is pluripotentharmonic with respect to all the variables. No more shamperequation. It's just the inequality. So that's the cousin of this theorem. You prove it more or less the same way. It's a little bit more complicated, but basically you can verify that. And then it follows from this a known fact of the energy functional, that if you look at the energy of phi hat of t, it's concave. So then things start to look good, because you have a concave function, and you have an affine function, and you want to prove some inequality between them. So we're not quite done yet. But this is the first important thing that you have this. So let's see. So it looks like this. Let me draw a graph. Here is t-axis. Here is the energy. And this is the energy of phi t. And for t equal to 0, they are the same, since it was already radial. So you have a concave function. It looks something like maybe like this, or maybe like that. That would be E of phi t hat. The question is, which of those pictures is the right one, so to speak? You want to have an inequality there. And the final lemma that does the trick is that the derivative of E phi t at t equal to 0 is the same thing as the derivative t equal to 0 of E of phi t hat. You have the same derivative. And then, of course, we are in that picture there. In this picture. And this was not correct. And so theorem follows. So how do you prove that's the last part, then, to see that the derivatives are the same? That's what you need to prove. And that's where the ball is important. It's important that it is precisely the ball. So let's see how you can prove this. That's actually the punch line of the proof. So let's see why. Well, you know how to compute derivatives of energy functional. So we know that modularity and some technicalities, you know that the derivative of E of phi t at t equal to 0 is equal to minus, well, it's actually an n plus 1 here, phi dot 0, mon champere of phi 0 over the ball. That's a classical formula for the derivative of the mon champere of energy. And then, by the same token, you have the same thing of the energy of phi t hat, but with dot. Dot is the t derivative at 0, and you still have the mon champere of phi 0 over the ball. And you want to prove that the hat is the Schwartz. So this is the. But at the point 0, it was already radial, so it is its own Schwartz symmetrization. But still, so those two are the same, but those two are not the same. So how do you know that the two integrals are equal? Well, that's a funny thing, which turns out to be. It's a very trivial thing, but it's actually the heart of the matter, it turns out, very surprisingly, at least to me. And that's the following observation that if you look at the mon champere of phi 0, which is a radial function, this is a radial function times. The mon champere of a radial function is a radial thing. So let that sink in. I mean, this is completely obvious that if you take something that depends only on the radius, and you take mon champere of that function, it's still a radial function. So it means that it is equal to some function of phi 0 for some f. Because if it is radial, it's a function of phi 0 also. So it satisfies some sort of equation like this. And then by the fundamental theorem of calculus, this f has a primitive function, g prime. And now you can plug it in here. And you see that, well, this is equal to, well, let me write it here, d dt of the integral, or d dt of the energy of E of phi t t equal to 0 is equal to, well, minus n plus 1 integral phi t. And then you have g prime of phi t. This is f of phi t, sorry, 0. And here 0. g prime of that is f of this, which is the mon champere. So this is basically is equal to minus n plus 1 times the derivative of the integral g of phi t at t equal to 0. Now you use the defining property of Schwartz symmetrization again that this is the, since this is some function of phi t, this is the same thing as the derivative. I can replace phi t now by the Schwartz symmetrization, well, still minus n plus 1 d dt of the integral g of phi t hat. Because those integrals are the same, for any function, they are the same. And then this is the derivative of E of phi t hat. And so they are, and that finishes to prove them. So let me say something about why this holds only for the ball then. It turns out that it is this, it all centers around this observation, equations of that form. So let me say something about that. If you imagine that you try to prove the theorem for another domain, so let omega be a general balanced domain. Balanced, not necessarily the ball. You want to try to prove inequality like this over there for the energy. Then you would need to vary not only the function phi t, but also the domain, which is a bit complicated. So I don't want to do that. So I want to reformulate the problem so that it becomes a problem only for omega. So we can write it like this, that omega, since it is balanced, omega, you can think of it as a unit ball of something, unit ball for some norm, of norm n omega of c. Since it's a balanced thing, you can think of it as a norm of something. It's not convex anymore, so it's not a proper norm. But it's only pure sub-harmonic. But I'll put quotes there. And now I define a sort of a green function by taking the logarithm of that. So I let u omega be equal to the logarithm of the norm. So that's it. So that would be equal to, in the case of the ball, it would be my substitute for log mod c. Now I define omega symmetrization, not sword symmetrization, but omega symmetrization of phi. If I take a function phi, say, pure sub-harmonic or something in omega, I define the omega symmetrization by, I call it S omega phi. It's going to be equal to a function not of log c, but of a function of u omega instead. And it should still be equidistributed with phi. Equidistributed with phi. So I can define a symmetrization with respect to any domain like this by replacing logarithm of c by this function u that comes from the associated norm. OK, so that's my sword symmetrization. And then it's an easy exercise to see that it has the same energy, that's the omega symmetrization. And it's an easy exercise to see that it has the same energy as the sword symmetrization. So then E omega S omega of phi is equal to energy over the ball of phi hat of the sword symmetrization. So you can reformulate the problem then. The problem becomes, is it true that the energy over omega of phi is greater than the energy of its omega symmetrization? That's precisely equivalent. Well, you can pull the same proof. You can do the same proof. Try the same proof. You connect by a geodesic and everything works fine. So only problem is the last step then. Only problem is the last step. I want to find a comparison function. I want phi 0, which is a function, not radial now, but a function over u omega, such that the mon champere of phi 0 is equal to a function of phi 0. That's what I want. Not a radial function. I could find radial functions that satisfy this. Now I want to find a function which is a function of u omega that satisfies this. So why can I do this for the ball? Well, it's just because being radial means that you're invariant under the unitary group. But of course, for an arbitrary domain, there is nothing that corresponds to this. And actually, you can see that this works both ways so that the answer is yes if and only if you can find such a function. So the problem is actually equivalent to this. The problem is equivalent to the existence of such a function. Let me call this property a star or something. So the symmetrization inequality will hold if and only if there is a function satisfying this. So the problem is for which domain do we have such a function? Well, I should say one direction is that you could use the same proof and the other direction while it involves actually. But let me skip that. But it's a rather short argument saying that the problem is actually equivalent to this. And then the final result is this, and the proposition that completes the proof. The proposition is the following. That star holds if and only if omega is an ellipsoid. We can do it if and only if it's an ellipsoid. So let me say why that is true then to finish with that. The proof of this is the following. That proof u omega is log homogeneous. So it means that u of lambda c, since n is a norm or almost a norm, this is equal to logarithm of lambda plus u omega c. It satisfies this. Such functions are in one-to-one correspondence to metrics on a line bundle over a projective space. So this is the set of all u omega is in one-to-one correspondence with the set of metrics. Well, there is also pure subromanic, so the set of positively curved metrics on O1 over projective space. So you can reformulate the question here. Is there a function of u omega, such that, et cetera, et cetera, in terms of metrics over there? And then the surprise, well, at first glance, the surprising fact is that, and I'll write it here, then, so I call this something, so u omega corresponds to phi omega, which is a metric on O1. And then you can verify just by rather simple, explicit computation that u omega satisfies star if and only if phi, well, ddc of the metric. This is a calor metric on projective space. And it's precisely equivalent to this being K Einstein metric. It's K Einstein. This is just a computation that might look surprising. But when you write down the formula, it's not surprising anymore. You see that it's really this equation. And we know precisely which are the K Einstein metrics on projective space by special case of Mandel-Mabucchi theorem. So it implies that u omega, well, one case is, of course, logarithm of modulus of c. That's the standard round metric on projective space. And the other ones you get by applying an automorphism of projective space, which is a linear map in homogeneous coordinates. So it looks like this for some linear map. And that then says that omega is an ellipsoid. So that's the end of story. Not only bihologmorphic, but the linear image of a ball. So the problem is not the equivalent under bihologmorphic map, so it's a linear image of a ball. No, that could not be a complex ellipsoid, yeah. All right. In general case, do you have an estimate of the energy of the solicitation or something? Yes. So you mean to go back to a more saturated in equalities? No? To your question, if you recall. OK, OK. So if you take another one, that's the reason of a question I get. But I don't know. Because the method of proof with geodesics is something that it gives all or nothing, I think. So it's not conceivable that there could be a constant? I don't think so. But I can see no reason why there should be. Can just say that there is a, you can ask exactly the same problem for convex functions over convex domains in Rn. And then there is something like that. Then in order to get the reasonable problem, you have to multiply by a certain constant. That constant is the molar volume of the convex set. So then there is something in that line, I guess. But I'm not sure what happens here. Yeah? So you said the openness conjecture is proved in two dimensions? In two dimensions, yes. So where is this used? Where is it used? Yes. You mean applications of that? No, no, no. I mean the condition on the. Ah, OK, OK. So how do they use n equal to 2 in the proof? Well, the proof is actually very, very complicated. So it depends on, what I shouldn't say because I haven't read it in detail, but it depends on a theory that they developed independently. The local analysis of singularities of pleurisabarmonic functions. So they studied the singularities of pleurisabarmonic functions by taking successive blow-ups. And in case the pleurisabarmonic function is logarithm. If the singularities are analytic, so it's the logarithm of some holomorphic map or something, you can blow up. So many times and the singularities will disappear. But if it's general pleurisabarmonic function, you blow up infinitely many times. So you get something defined on some abstract space, a laburkovich space, or something like that. And then you get an object defined on something called the tree of valuations or something like that. And that theory seems to work only for n equal to 2. But it's not easy. I mean it's a book. No. Are there questions? Let's think through again.