[00:00:06]
>> Ok. So we've proved this slicing inequality for functions. Carries a star board. In Narron. Af is an intra global function on k.. Then then to go forward case more the device. Will you marry should distance from k. to. The class of intersection boy does. Times the maximal section.

[00:01:34]
Times. Well him of get to the number and. And the class of intersection boy does. It was defined as the class of what is the star board is the. Sides that there exists a measure new d. on this fear. Satisfying. A quality the norm of d. the negative one.

[00:02:16]
On any continuous function. The same way is. The measure a new direct on their I don't transform of this function. Ok and then the quibble and statement is. And the origin symmetric. Star d. D n a r And. He's an intersection buddy. If and on leave. The norm of the to the negative one is positive definite distribution.

[00:03:17]
Which means that for any. Negative test function by. Their action of the free transfer of the norm to the negative one and by is no negative. So. First time shows some particular cases of this so so far doesn't have to be convex it's a new star but it doesn't have to be even symmetric nothing become Tuesday anything in the proof.

[00:04:02]
So 1st of all. Ellipsoids are intersection good is. Why because they clearly didn't bowl is the intersection but if itself constant all sections are equal and they feel confident in their transformation of an intersection but do the Fourier transform changes by the determinant to the value of the term and so they no transformation of an intersection but is an intersection budgies the.

[00:04:47]
Intersection bodies. And this means that if k. is. Symmetric convex. Then. The out the wm ratio distance from Kate to the class of intersection budgets is over smaller than the square root defense. Why because by joins there or. There exists another. Thing didn't. And case contained in this square to defend this of the so it so this will be d. d. it's an intersection but did.

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It contains k. and. This was the defense. So in particular this means that we can craft a constant care all of those 400 call convex metric on that board. Which doesn't depend on. The fan. Ok. Second. Let's look at the unit balls. Of subspaces of b. with speed between 0 and do.

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So. All of them are intersection buddies you need both a find and mention. Find dimensional subspace p. this p. between 0 and 2 are all intersection but this. So the. Distance from every side. Is just one. Ok. So this so called was the constant do this constant do is is not here you are and minus one time something that is really smaller than one starting from 3 or 4 or disappears.

[00:07:31]
So how do you prove this. There is a. Theorem of living. In the final dimensional case it's there I'm a poet living then Finot dimensional cases but then you are the queen of that increase in. That. Normal space. Find a dimensional are and of is the norm. Embeds a symmetrical in.

[00:08:18]
An LP. With speed between 0 and do if and done only if the exponential function of minus norm to p. is a positive definite function. So there is a difference between positive definite functions and positive definite distributions and the difference is that the post a different function is the food transform of a final measure on r. and Ok and the definite distribution is a Free Transform of a tempered measure gross not faster than some problem.

[00:09:14]
In particular all positive definite functions up positive definite distributions. So how do we use this as the proof is actually in the not very difficult it's kind of feeds of it I just don't have time for it here and it's the result of 19 something like that. So.

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What do we do after this. For every t. this function is also positive definite as a function of fx. For it transform changes by t. to negative one something that it's too positive and if integrated by t. from 0 to infinity. We get. The norm to the pound you get to one Ok just in the mentors up to dish.

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So if the norm belongs to subspace the phone be with b. between 0 and 2 and. No negative test function. If they apply this to shift at the facts it's the same as with this difficult to fix but integrate with this. Peak at effects and then integrate this was a spec the feet had to fix and that's in the integration goes inside it and this is the action of the Fourier transform of the exponential function on on fi by the definition of the 4 It turns so it's not a negative.

[00:11:20]
So this isn't a negative but this is the action of the Fourier transform of the norm to negative one on Siegel's So it's all was the negative this is a positive definite distribution Ok so what if p. is greater than 2. So fight a dimensional. Subspaces. Of b. would be greater than 2 then not necessarily intersection but does for example.

[00:12:09]
Be a small 5 dimensional the unit ball of p. 5 mins will this be great as and 2. Is not an intersection but I did. To prove it you just computed for transfer easily. And prove that it's not all of us was it so I'm not going to do it it takes 3 or 4 hours.

[00:12:39]
Ok. But 5 appears here for them so that for this intersection. Ok but. It was by and mine they were made in mn. The outer wall the American distance from the unit ball of every side shop space a full pig to the distance the intersection board is smaller than c. squared to be.

[00:13:22]
Sees an absolute constant. This is the unit ball of a subspace b. was big too. So we have this inequality with a constant depending on the on the. Square root of the. Now most interesting case unconditional. Convex but is. Convex buddies called unconditional if. Point x. one x. and belongs to k. if and on the if the point with positive chord and its belongs to k. so it symmetric with respect to planes.

[00:14:38]
And. Kerry uses 2 facts 1st is the result of was another scheme. That defy Kaizen unconditional convex body. And then there exists a linear operator and. So that they mature of the q.. Which side to this is the do so b. infinity and. Is. The set of. The actors in r.n. such that each chord in it is less or equal to one.

[00:15:40]
So the 2. Is an inside of k.. And the same time Kazan side and times the image on to the same I pray to of that one and will. Ok this is written in the book. If you are interested it's 2 pages. It's in and 3 the argument.

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And the 2nd thing that you are going to use this is the one and is in the section as the unit ball of subspace of one. So be one and times be one and. So so sorry. To be one and has an intersection buddy. Then it's Linaria mage is also an intersection with dilated but it's still an intersection so this is the.

[00:17:06]
And now we want to prove that. The volume of Dia is not very much bigger than the volume of k.. To make. The volume of d.. K. to run over and. This will be bigger than the wm or a should distance so what is the volume of d.

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It's absolute value of the determinant of 50 times the volume of b. and which is 2 to the power and over and factorial. Now will raise this to the one number and. We get determinant of t 21 overran. And. Replace this by an over to the. And and d. includes and so it's times and so and cancels and what stays is 2.

[00:18:30]
Times 2. So now the problem is the determinant of t. but we know that. They much of the cube is inside of k.. So the volume of t. of be infinity. Which is the absolute value of the determinant of t. times the volume of b. infinity and which is due to the power n. is smaller than the volume of k..

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So determinant of t. to one number n. times 2 is smaller than volume of k. to one over and this is smaller than the. Key to one over and. So the ratio is smaller than. The out there well you may show distance from every unconditional kind that is but it didn't the section budget is it's smaller than the.

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So this is what fed the. Ok. And then there are many out of the examples. Came through a section board is ordered it. Will Iran's. Students doing this things this but. I suspect. Yeah yeah I know the question Do you know Ok. Well this. Site the distance is like some.

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Reverse in the quote of just how tight the quote. Like. Actually. Should in no because it's always a constant as far as I'm concerned so and all this examples you get a constant. So after looking at this so initially when I started doing this I wanted to construct a measure that contradicts.

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The fact that this is a constant but then when I was at 2 doing it was getting buzz different. Started thinking that maybe. But know. So. So it was proved by paperless clerk that. And later the logarithmic term was removed in. The paper of the and Alina. The dirt exists.

[00:22:17]
Convex but the. One. In our and there exists no negative function calls in to go lower T.'s one. But all sections of this function. Smaller than it was 1st was. Over this quote defend and then. Do the same construction genetical it was a different point of choice a friend of this was removed by Bose and.

[00:23:07]
So it's. Sealed with the square root offend. So they're all small. And the construction is. It's around of both dope and the function is. The guy was centered that every vertex then you take this. Then 1st you use the connection between their I don't transform into for a transform and then you have to choose proper the points.

[00:23:49]
So this is then does a story with this. This is so essentially except for our. Dependence of the function f. so and f. becomes. Constant goes down. The force through the and at least. What. What a fair phrase a different kind of function for example. But maybe there is something about the function if that is interesting so this is done the question that really remains.

[00:24:37]
Ok and I see what. 20 minutes. Doesn't come. Ok so the 3rd problem that I wanted to mention is the bosom and. So this will still need but we don't need. So. I thought about where to start with it was a month there are people who don't know what this is.

[00:25:23]
So I will start from the very beginning. So. K. and Al are. Origen symmetric. Convicts Why does. And suppose that. The central section of k in every direction small area than this section of the same direction. Does it follow. That. The volume of caves smaller than the volume of.

[00:26:30]
So I must say that I had to use this result when I was moving my son in his dorm room. Because the volume of furniture was bigger than him of the room but sections were smaller. The I extended the room to Dinant. Ok. So if the dimension is still a section says just diameters the diameter of k. in every direction it's smaller than the damage to K.'s and side says obviously yes.

[00:27:16]
And it was yes for 30 years. Everything was going in the positive direction then Lerman and drudgery. Prove the dancer is no if the dimension is going to requote to 12. K. and this was done by probably stick argument then ball. Of the dancer is no if the dimension is greater equal to 10 and that this is the most visible example if you can see dimension.

[00:28:03]
Because the smaller bodies the cube and the bigger boy it is the lead and. In dimension. Then gentle pose and donate to gain. The dancers know the dimension is going to trickle to 7 and then but deem it drag is. And later Gardner very different. To the dancers know is a dimension is good to equal to 5.

[00:28:58]
And starting from this point. On the results were dependent on the connection between this problem and intersects and what is that was found by. So Lou took. Proved 2 things 1st if case and intersection board in. This case. Then dancer is yes for every star where deal. And there is no convex attainable here.

[00:29:51]
On the other candidates. Is not an intersection. One can perturb it twice. So that to get the counterexamples. Little Bit is this and perturbed even more k. So this there is a counterexample. And deny 994 the problem was considered to be solved completely because 1st of all Gardner.

[00:30:45]
Prove the dancer is yes today in dimension 3 yes and dimension 3. And then there was a paper and then I was of mathematics that said that dancers know in Dimension 4 because the cube is not an intersection. And dimension 4 which which I disproved at some point the problem became open again.

[00:31:22]
You know and dimension for done so was no. He read my book. So and then in 1909 it was finally so the opposite way the cubit there to be and to sex. And dancer and dimension fours Yes. And this was done by John and at the same time and paper of his garden.

[00:32:07]
And slump ranked. So this paper. Was about about dimension for one and paper was solution in all dimensions the same time so what is. The point. We go back to the formula that. Like so much already what is the form of the so much. The derivative of the parallel section function.

[00:32:56]
0. Is negative one over to. By and minus minus one. For it transform of. The norm of the negative one. This is unstable. Negative and plus plus one. Point c.. So this is the from no that proved. So what if and this for. This is for we are interested in the norm of k. to the power negative one because if the forward transform is positive this in the section body if not positive it is not so take.

[00:34:04]
2. Then here we get the norm to the negative one. And this is 2 so this is negative minus. And then case symmetric convex this is the place work on the x. it is used. Symmetric on Vegas board because in every direction the Maximals section at the center.

[00:34:45]
By boons. So never direction the maximal section perpendicular to the center. So this function a cake she. Has maximum adhere. And the 2nd derivative is. Negative. So negative negative so this is all of us positive in Dimension 4 forever symmetric and that's what is so every semantic and that is void in Dimension 4 is an intersection body and then by little connection.

[00:35:25]
The answer is yes. Now what if the dimension is 5. In principle there is a version of this formula for. This is even. Right. There is a version of this formula but I'm not going to mention it let's just see what happens if and it's 5 has to be 3.

[00:35:58]
Right so we are talking about the 3rd do Eva 2 of the titles section function but convexity controls only the 2nd derivative doesn't control the 3rd period which she used to construct a counterexample the counterexample is not very short I'm not going to do it but this is the idea so this farm you are actually solves the.

[00:36:27]
Completely the problem so. I want to. Show you is a little a little disconnected in one direction at least but I will show it not for this problem. But for the isomorphic. Was a month so since dancer is all there is no starting from Dimension 5. We consider.

[00:37:11]
A different problem which is called isomorphic. It was an unpaid to problem so can do I still symmetric on rigs by descent there is a sections are still smaller. Does it follow. That. The volume of Kate minus one over n. is smaller than c Well you know of minus one over and.

[00:38:02]
Sees an absolute constant. So I can write it without and minus one over and but prefer the this way it's the same thing. And I'm going to prove that. Not blue books connection but. I'm going to prove that if we have this kind dition then well you know of gate then minus one over and is smaller than the will you marry should distance from Kate to the class offender section would this times well enough and this.

[00:39:01]
Is. No. Mean when pressure will come later if I have another. What this is this is skirted. Sometimes this is done cert. Question So the true the it should be the 1st to defend it at least. Because this is equivalent to. But for conditional context what does For example this is an absolute constant.

[00:39:51]
So now I'm going I'm just not a new result it's looks connection and a different safe gaze an intersection body this is one so well you move case smaller than the well being of. So they get the connection. So I. Say something about this and still. So let's use the.

[00:40:29]
Before you have. This this moves. The history. So. For your formulas. There is of sections it's the 4 it transform. Of k. a minus cent plus 1. 1 over but the constant and the same constant. So this is the condition of the theorem. Now let's take the best possible d. So we're taken at the section body d..

[00:41:43]
That gun taints k.. And the volume of d. to the number n. is the wm ratio distance. But once again it has to be lines was don't and delta goes to 0 but just. So which means that. The norm of get to the power negative one cat is positive we can approximate it by an infinity smooth intersection and.

[00:42:29]
Then this is a function so we multiply by this no negative function both sides and integrate this. Case get into Google with this here. Smaller than the same base. Show the right. And now we can us by surprise for me so if we have to come up genius functions on this here because degrees of common a to add to minus.

[00:43:31]
We can remove the forward transform So this is called spherical Parcival formula. So we remove. That's. Now a. Country. So their idea is served the is bigger than their ideas. Ok so this is bigger than norm of Kate as a minus and which is and times k.. So the norm of d. to minus one is greater than the norm of Kate a man's.

[00:44:53]
Carry years in the quote is. So it's norm. X. and the minus. Minus one over and and the norm of d. to the minus. But this is an volume. And cancels with this. And goes to the power and minus one over and over and over and cancels. So this says and volume.

[00:45:36]
And this is and William Deane so we get that k. is smaller than well you know of minus one over. Time swelling of d. to the one over and disease. The out there will you may show distance from Kate intersection good this. Time swelling of Kate divine over and bowing of Kate to one over and cancels and get this.

[00:46:17]
Ok. So. I want to talk about measures just for. I'm not going to prove what I wanted to prove. Because Google is once to speak. So. We can do. So one of which. Extended this to arbitrary measures I mean the original was a month. Because finding a strictly positive continuous function on r..

[00:47:16]
And k. and. Symmetric convex. Budgets. And suppose they enter go of. Every section a central section of a smaller than the in to go. Over every section of. Doesn't follow. The Into go over all the call k. if f. is small then then to go to the. Hell of it and I'm sure is exactly the same as the original problem for every such function than Series yes.

[00:48:05]
In dimensions 4 and smaller and no in dimensions and 55 and greater. So for every positive density we can find care in. Providing this a months or. So of course one immediately wants. And isomorphic version and isomorphic version was proved in paper was. So they scanned dition. This kind dition and implies that.

[00:48:47]
They enter go over k. is smaller than the distance from Kate to intersection but does. Time sensible. So that similar result unfortunately. The distance is not out of 0 in the ratio but by now. So the banner come as a distance. Is the infimum. Fall a there exists.

[00:49:29]
An intersection. Which is contained in k. and. Contains gave. So the condiction that this condition remains then clues and but then doubt the volume ratio distant requires just there a show for him so this doesn't have to be inside. So this is a worse distance so this is a much worse distance very much.

[00:50:02]
So the 1st question that is open can vary move this country place this by the out of the image distance. You. That's exactly what I'm asking what if you have special densities but then Ginger in general in general is it true in general that William should. Now why do we want this because.

[00:50:39]
I Some are fic was a man take a problem implies the slicing problem with this is exactly the same constant the constant is even better there I'm not going to show it it's so the reason is with they could. Take here they clearly didn't pull. And suppose because among.

[00:51:06]
Some of the you get the volume of the they didn't bow and make make them equal here to the maximum so. Say this. Yes no. Ok so would take care and. Time Stickley Denbeaux. So that every section of this is the Maximo section of k.. And then compute tough and put it in this formula here and you get the slicing problem with the same constant.

[00:51:47]
Was measures this just doesn't work because you cannot make this constant if you take the cli didn't have a density. But we still want. Did use this slicing in a quote this one. From some kind of was a month to problem and we did did just recently was good glorious and Tim.

[00:52:28]
So. What do you have to do is consider different bodies and different functions. So. This is the. Kind of free dickless because it doesn't have anything to do that. Except for that is the bigger. So. So there is this suppose so f. and j. are negative functions. Locally into Google.

[00:53:37]
Kane Del Irenaeus star book does just nothing. So. Suppose. The maximum of Jesus 0 and is equal to one then we can actually get there is out surprisingly then to go for a k. is smaller than the volume ratio distance. Kate intersection but this. Times. The power run over on.

[00:54:19]
Time Center go. To the power and minus one over and. And there is the correct distance. First of all and 2nd. What to really need this just do equal to one so of J's one. We get. If G.'s identical to one this becomes. Well you must minus one over and.

[00:55:00]
Then this inequality if you take. So the cunt dition is now. This becomes volume. So while you miss a constant. And if you use this inequality you get precisely. Ok so this was. Ok and I will stop here thinking. Everything was so clear. To me. So. You. Can buy one.

[00:56:27]
You know you're just going to see. What you feel if there's one shark. Southeast. You have that. Much taller speaking section but I'm just. Being Me and you know the problem is that to construct a counterexample to. This bosom unfit to pull them or to you need to really work so you have to look at those proofs and see if you can get anything about the distance because this is a very special cases where and the bosom and the poem doesn't work.

[00:57:23]
But the about to. Be. So just so. It's not this problem for us to. Always distance from of the business and from o.p. you cover both sex and not on this until 6 so it's a different thing. But I don't know it's it's a nice question to look at the all of this but this counter-examples and try to find.

[00:58:01]
The William ratio of distance that appears there so the standard way to construct a counterexample with the poor to transform the norm to minus one is negative so it's not an intersection would you take a function of its support to peer they could transform and use that to perturb the norm.

[00:58:24]
And then all of a sudden it works. Or it doesn't give the. It's the you know so you don't know how to mate now in p 5 you can write precisely the transfer. Using the stable density and then use the moments of the stable density to prove that it's negative so.

[00:58:51]
There may be possible because but then you have this awful function stable density the transform of it to minus t. to the. So there is an expansion and. If somebody wants to do this. With a great. Question. You tell us so much. No no it's. Different it's minus one over and.

[00:59:32]
Especially. The. You want to number 26. But you. Explain. That's question for you number 27. So why do you see. What. You just. Constant changes from the square are different to the. Yeah but you. Know. I don't know. Next time next country. But. Thanks so. Much.