A while ago, we interviewed Mindy Zempback, who works as the supervising casting producer from Shark Tank. During her conversation, we asked her what her most memorable moment from the show was. Without hesitation, she said that her most memorable moment was the pitch from the founders of Cupboard Pro that left everyone, sharks included, in tears. Cupboard Pro was invented by firefighter and 9-11 first responder, Guy Theon. He was named a two-time Boudin Network Chomp Champion and he invented his product to simplify the cooking prep process that always takes way longer than expected. Less time prepping means more time with your family. Before Keith could get some serious traction with his new product, tragedy struck. Keith's wife, Beth, passed away at 47 from breast cancer and it understandably crushed the whole family. And then, a couple years later, Keith passed away from a 9-11 related cancer, orphaning his three young children. It was their dad's dream to get Cupboard Pro into homes all around the world. So to carry on his legacy, Kaylee, Christian and Kira all went on Shark Tank seeking an investment to carry on his legacy. But now we're guessing you can see why their pitch had enough raw emotional power to leave the sharks in tears and create one of the most memorable moments in Shark Tank history. In today's episode, we chat with Kaylee and Kira about how they were manifesting the deal moments before making their pitch in front of the sharks, launching the cutting board into William Sonoma and what they've been doing now after this life-changing deal. All right, guys. Welcome to the podcast on today's show. We're talking to Cupboard Pro. Thank you guys for joining. People who don't know, what does your company do? Thanks for having us. Our company is the Cupboard Pro and it's a cutting board invention, a two-sided cutting board with a meat side and a prep side with a removable cup that inserts into your board. Our dad invented it when I was in high school and unfortunately we lost our dad to a 9-11 related cancer in 2018 and my siblings and I took it onto Shark Tank where we got a deal with all five sharks and a licensing deal with William Sonoma filed that so we could find the Cupboard Pro and William Sonoma's all around the country. It's a pretty amazing story. We had Mindy from Casting Who You Guys Know obviously and when we spoke to her on the podcast, it was really, I was trying to save her in Instagram DMs basically because apparently she gets DMed all the time of like, how do I get on the show? We wanted to create this episode where we tell people how to get on it and what to do in the process and so hopefully that avoids her from getting all these crazy people to hit her up. At the end of the interview, I had asked her what were her like in her head top memories and so not even companies but just like her best memories and you guys were your number one on the list and I think for obvious reasons. It's such an emotional story.  I would say the whole process from the time our dad invented it to getting onto Shark Tank was like a 10 year journey. Our dad probably started working on it in 2007, 2008, and it became just one of his little hobbies or he was carving into boards, trying to figure it out. He loved to cook. He was a chef. He wrote his own cookbook, cooking with the firehouse chef.  Unfortunately, we lost our mom August of 2022. And after that, our dad was asked to be on chopped and that kind of gave him the confidence to pursue the Cupboard Pro again. So we really started to get into it, had our prototypes made, had everything shipped from China in December of 2015.  So once again, everything kind of kind of got put on hold. He really fought his cancer head on. But as he was doing that, he still was applying to Shark Tank. He took his interview video, but because he didn't have any sales at the time, he didn't get on to the show. And we lost our dad in March of 2018. And I'll never forget, it was the night before my dad's one month anniversary. I'm sitting in class in the college and I get an email and it's from Mindy saying that she heard our story and if we'd like to apply to Shark Tank. So then that started the whole process. And three months after our dad passed away, we were sitting standing in front of all the sharks. That's incredible. And so gut gut wrenching at the same time. I lost my father when I was young. And in some way, when I hear your story, the legacy piece is so inspiring to some extent, where it's like this thing that you guys probably saw your dad tinkering with for so long. It could take on a life of its own in a different way and provide like this amazing reminder, any reminder is a nice reminder. But it can be just something that you could all rally behind also. But I can imagine that's also probably and maybe yes, maybe no, but for you guys, incredibly difficult to like have to take over a business or take over an idea and then go on Shark Tank and share such a personal story. What was that like for you guys? Unfortunately, I guess I had some experience in just taking over a business after parent passing when my mom passed away.  So I did that for seven years and just makes me laugh because every time she says that she's 19, I picture myself having to take over a business like by myself at 19. Why do that? We're too old. That's pretty wild. It's and I was alone. That's whack. Kira's 20 just turned 20. So I was her age. Yeah, teaching crazy Pilates dance class. Yeah, it's wild when I say that. And I look at Kira like that was me 10 years ago. That's crazy. Yeah. Did anything need to be finished before you guys went on Shark Tank as it relates to the product or did you already have it and all you had to do? Like you had the website, were you guys selling it prior to Shark Tank? So prior to Shark Tank and prior to like my dad getting sick, when I was in high school, he went to like the Chicago cooking like kitchen convention and I would go to those with him. So he just like always had me a part of the process. I was the one making his website for him, his logos. So it's kind of like really a part of it with him throughout high school, which was really cool. He never like didn't include us in all the things he was doing. And then when my dad was sick, even when he was laying in bed on hospice, he'd be like, Kaley, I just got an order on Amazon. Like, can you run it to UPS and I would be taking it to UPS for him sending it out. And then after he passed, it was just something that was too good not to continue his legacy. We saw how much work he put into that. And it was his dream to take it on to Shark Tank. So it was really cool to be able to do that. Yeah. And then walk us through. So you get on Shark Tank and the most amazing thing of all time happens. Or I guess the craziest thing happens. Did you have any I can't even imagine. Let's talk about let's just in a world where you're going on Shark Tank and are you guys thinking in your head, we'd like to get a specific shark. You know, you have a favorite. Is there someone you're specifically targeting that you think and take this product to the next level?  Like, pretty much, I feel like it feels like 24 hours, like 48 hours. You're pitching on and I remember we were me, my brother, my sister were getting ready in our trailer and I start writing thank you notes. Wait, that was so weird. I forgot about I start writing thank you notes to each one of the sharks to lead cutting their filming all day and to leave a cutting board in their trailer and my brother says to me, like, what are you doing? We're only going to get a deal if we get a deal only be from one. I'm like, I'm just putting it out there and saying thank you to all of them now.  That's amazing. That's actually pretty amazing. It was pretty cool. OK, so no favorites, you're writing thank you cards to them and then you get on the show and everyone's in tears. Yeah, we're standing there and it was wild. They asked us to step out for a minute and Kira is 14 at the time. And she's I was like rapping and like singing but they obviously didn't put that in. But I was like, what's going on? Oh, my God. Yeah, I was definitely nervous because you don't know like what they're talking about. If they're saying like these kids are young, they're going through a hard time right now, maybe it's like you have no idea what they're talking about. But they said like the camera crew never like they didn't ever have an entrepreneur on set and then the sharks asked them to leave. So they weren't really like they didn't think that we would be asked to leave. So it was kind of just a really crazy experience.  And they just you could just feel their like heartfelt emotions for us.  That was we didn't even know that was a part of our pitch. They just their shark tank had it and they added it. So it was kind of wild, really hearing our dad's voice like he was in the room. No, it really was it was booming.  And the emotions just hit because we were not expecting that to like that wasn't a part of us like practicing or anything. So when they were playing that, it was pretty pretty crazy. Yeah, I can imagine that being unbelievably emotional, especially I imagine it's loud. Oh, my goodness, it's the last thing you're expecting. I think that that part of it was honestly like the most like shocking to really feel like our dad was in the room with us and a part of the pitch, which was cool. And because it's a literal video that like he sent that in for his audition. Yeah. So yeah, that was crazy. Yeah. And it was Father's Day weekend also that we were. So it was a lot of emotions, but that's the one thing Mindy mentioned. She's like, it just felt like the universe stopped in some way where it was like the Father's Day thing got to, you know, it was like all the elements were there. Yeah.  Just getting her email the night before his one month anniversary just felt like a sign just because like in the beginning, especially to like anniversary just hit you different. And when it's like a month or like how how did this happen? But when you got when we received an email like that, it kind of gave you like hope and they're still here. Things are happening and we're going to be OK. So that's all because of Mindy. So we love Mindy way more than she could ever love us. She really is the best. I think what I love about her is I don't think she knows how how much of the best she is. Like, I'm like, you're like the coolest person. I don't think she is for her. She's just like having fun and doing her job. And, you know, she just likes it. But I'm like, no, I think you're a legend. And she's like, am I, you know, it's the one that's changing lives. It's her. Yeah, totally. How long after year you guys were on? Did it air? Did the show actually air? We filmed in June and it aired in October.  I know you don't always know if it's going to air or not. But there's maybe some window. And so what happens after after you guys there? So we sold out of all our inventory, right? Yeah, when the episode aired, we sold out of all our inventory. And right before episode aired, Damon connected us with William Sonoma. His team was working really hard on that connection. So before it aired, we started conversations with them about the possibility of a licensing deal. So when you're doing that, too, you can't go and have your own boards produced. So we were just taking what our inventory was from when our dad was alive. Seeing what the response was, you also kind of get your investment from the sharks right before the show air. So it's not something that like you now have right after your pitch. And there's all this diligence. They're making sure your business is legit. But yeah, William Sonoma, we are so grateful to be able to have that opportunity with them. My dad also brought his cutting board in the past to William Sonoma's store to see how he could get it in. William Sonoma connected us with their manufacturer Epicurean and how our cutting boards made in America. And it definitely took the pressure off me, Krishnan, here a little bit, having that support from William Sonoma because just after continuing my mom's Pilates studio, I saw how much like I myself put myself in her shoes where I like left college and I'm so grateful that I did that. But to be able to do something to continue our parents legacy and not have to fully make Kira like really take out leave high school or college to focus on this, put that pressure on my brother. Like everyone can follow their own dreams while we continue our dad. On your own. Yeah, it is really cool. And how did you guys divvy up roles? Like who did what? Or was it just like constant chaos?  I feel like we all kind of just whatever our strength with that. But I like Kira be a cheerleader in high school. Like I didn't want to put that pressure on my siblings or whatever I could do. Like they're we're all so supportive of each other and whatever they want to be a part of their part of. But I know how hard it is to 14. Yeah, I'm so young. I would like to do some social media. Yeah, yeah, that was mine. We'd be online just cutting up some things, talking to all the people joining. And then I'd be like, don't say that.  I feel like it's just been something that keeps us close. And we have something great to work towards work for and to be able to continue our debts like to see within is pretty beautiful. And what happened in the local community? Anything that once you guys were on TV, did they just like rally behind you guys in a real way? Like what kind of what did you just see locally? I'd have to say that our whole town really rallied behind us for years before that with like our mom fighting cancer. We had fundraisers for her in the whole town really, like so much love and support. Our town's very close. And when Shark Tank aired, we actually had our viewing party at the middle school and high school. And we added a whole fundraiser for someone who was fighting breast cancer at the time where proceeds from that event went to her. So it's pretty special to be able to continue doing good things in all different ways. That's amazing. And where is the company at today? What is Cupboard Pro today? So today, Cupboard Pro is one of William Sonoma's number one selling cutting boards. We still have the licensing deal with them. We recreated our dad's cookbook and they sell our dad's cookbook as well. And we have all new accessories and different things in the works now to hopefully be out next year. That's amazing. And what are you guys all working on personally? Are you still at the Pilates studio, Dan's studio? What are you working on? So I sold my own Pilates studio in January of 2020. And I actually graduated from FIT with an interior design degree in 20 December of 2018. So it was a couple of months after Shark Tank aired. I ended up going back to school. So I started my own little interior design business and I have different clients working on different things. I got married in September and I'm actually pregnant with my first child. So. Oh, wow, congrats. Thank you. That's amazing. That's a lot of good things. Yeah. How about you? Thank you. Mine's not as exciting as Kaley's. I'm just still a student at the College of Charleston. Very nice. Cleaning business. Charleston's a great place. That's a great place to go to school, I think. Probably one of the one of the most perfect places to go to school. Yeah, I love visiting her. So it works out. Yeah, that's so perfect. And you're studying business and is the hope to keep running the company? Is the hope to start your own business? Do you have because you've seen it? You've seen it so many different ways, right? And so you must have the bug. I mean, definitely keeping the Cupboard Pro alive always and doing whatever we can. But I do want to one day have my own thing because that would just be fun. And it's actually definitely coming like from whatever your passion is. And what would it be? Do you have an idea of what you would start? I love the whole fitness industry and like mindset stuff, all that. I don't really know exactly what it would be like. And what's your brother doing? What is he up to now? Our brother works out of that in our hometown. So he's there full time and loves working with animals. So that's a tough job. Good for him. Yeah. We interviewed a vet company that's like it's called Modern Animal and they're trying to go big. And I had no idea what vets actually deal with on a day to day basis. And it's it's pretty brutal.  It's like a dying industry and everyone's trying to solve it. So good for him kudos to him. Well, look, I appreciate you guys coming on the podcast and at least sharing the follow of any tips or people who are watching, listening on either what to do before the before Shark Tank, what to do after. I love the tip that you gave, which is you you sort of wrote a thank you note, which is I've never heard that. So that's amazing that you just it's within your personality to do that. But are there any other things that you guys just did, whether it's superstitious or just like lucky or just things you thought of doing prior to the show? Before I went on for about two months straight, I watched every possible Shark Tank episode that I could to know what kind of question could be asked, how it could be asked. It was kind of like film study, I feel like, just really putting myself like mentally in that position and how I would respond. So it worked out, I guess, because that practice really helped. And to really know everything about your business, the honest answers and to believe in yourself and your company, because I feel like if you don't believe in your company, then no one else is going to. Yeah, that's absolutely right. Are you guys still in touch with any of the sharks? Yes, we actually just filmed an update in November this past November with and Laurie and Matt were there. And there's a lot of exciting things you said coming from Williams and Sonoma, some new products coming soonish this year, next year. We're probably aiming for a 2024 launch date with all of that. But we're still continuing with our charting deal. 20 percent of our proceeds go back to a 9-11 related charity to support families like ours that have gone through or going through a similar thing. So we continue to be able to donate 20 percent with our sales, which is really awesome as well. Yeah, that's amazing. Are you guys are you guys personally involved in that charity as well? It's a charity that my dad really supported. So they helped us a lot. It's called the FDNY Foundation and just an example of their support. When my dad was sick, they would drive him to all of his doctor's appointments from Long Island to New York City at Sloan, which was really helpful because not having not having our mom there for that was tough. So, yeah, they definitely took a lot of pressure off. A lot of different aspects. You guys have been through so much.  What are you thinking about? What's going on? Are you thinking like I think every time I I'm not a dad yet. But when I think about becoming one, I always think there's going to be so many things I'm going to want to create to make other dads like because that's how I'm wired where I'm like all of a sudden you get you get to see what's broken because it's like a little child running around breaking everything. And then I'm like, I'm probably going to invent 14 products when I have a child. I know I get like that, too. I don't know. I really just love the way that our parents raised us. It was always family first, just treating each other with kindness, love, respect. She already went through the hard part because she adopted me when I was a teenager. Yes. So yeah, this is my sister child. The baby might be easier. Adopting a little teenage girl is a lot different. A little newborn baby, I feel like. Yeah, that's that's a lot. That's a lot to think you're 10 years apart. So when I was 10, here was like my little doll. And I did everything with her even going to college. Like I was home all the time.  So I never actually like left the home and went away. I was commuting when my mom was really sick. And then just after she passed, my dad was a New York City fireman. So I would be the one getting Kira ready at like eight, nine years old after my mom passed away and making her breakfast, bring her to school. So I've had all the different different experiences with Kira. Well, you guys are still tight knit, which is amazing to see. And you guys can still laugh together and run a business together, which is pretty amazing also and very rare. And so it's super special to see you guys in this light. And and kudos to you to keep your father's dream going. And just tell everyone where they can support you guys, whether it's via follows or just purchase the product, obviously, Williams and Sonoma, but tell everyone where they can support. Sure. William Sonoma, you can buy the cutting board and cookbook with the firehouse chef online or in store. And we have our Instagram or you can follow us. We're definitely more active, I would say, on our personal Instagrams, mine's at Keeley Young, mine's at Kira Young. And then no, hers is young Kira. Oh, what? Sorry. Yeah. Got it at young Kira and our Erase that. Our business Instagram is the Cupboard Pro. Yeah. All right, guys. Well, look, thanks so much for coming on the podcast. Thanks for sharing your story. I had to talk to you guys after Mindy said just such amazing things about her experience with you guys and you guys on the show. And so a big thank you. Thank you. I appreciate that. Thank you so much. Thank you so much for the support and making it to the end of the episode. If you haven't already, please review and show the episode with your friends. If you never want to miss a beat on all things entrepreneurship, make sure to follow us on socials for daily content. See you next Tuesday for another great episode.


Hashtag kidding! Who's van is it? My mother-in-law! An amazing woman, alright. She's so sweet because my mom is in town. On Tuesday night she gets the date night off as well. So she goes to like the library and a lot of mine is. We... She doesn't want to drive stick so she takes our car. And so then we get... Now we're riding in style. Riding in style. Um, Han why don't you pick up that letter on the floor and open it up? Okay. Let's just... Oh my my. Oh my my. This is a great one. Hey, hey, hey. There it goes, my husband. Is it the right amount? Yes it is. Oh, oh! Bam! Safe numbers. 22 years folks right here. I did win a little prize money last year in 2017 but this check is from the 10k, six miles, six miles, and veiled from two weeks ago. It arrived in the mail. Thank you GoPro for putting on a great race and providing some prize money for the runners that work hard, work hard, work hard, and put in the blood, sweat, and tears to try and uh well, frankly, provide for the family, buy some running shoes. This check allows me to do a lot. Buy vitamins, buy running shoes, buy a rec center pass so I can lift more and get stronger for the 100 mile race coming up. Nice. Bottom line is I'm grateful and... But folks, 22 years. This just doesn't happen with a snap of a finger and you gotta work hard and you gotta stay consistent and keep believing and I think I already mentioned this two weeks ago, but the top three runners first is 34 years old, I was second, 33 years old, and third place was 37 years old. So keep dreaming if you're in your 20s or in your teens and you're like not winning races yet, don't worry. Keep fighting and you'll get there someday in your 30s because I'm gonna win right now. Come on. Most people's hobbies or passion cost money. It costs the family money to like invest in the equipment or like if you're a skier or something like that, but my husband's hobby brings home the bacon. And he didn't mention this but he also is giving part of that winning. It's a substantial part of that winning to me so I can go have a spa day. I appreciate that. Spa day for the true love. Our ritual stop at Sweet Cow. We got a little sweet tooth tonight, a little sweet cow ice cream. Not Andes. Whoa, whoa. I know. Haley, if you're out there watching, you're gonna have to make it to Andes at some point. It's, it's, it's superb, but every now and then we change course just a little bit, just deviate a little bit. Oh man, okay snarfs. Let me tell you about snarfs. It started in Boulder in the shack, a tiny little place. It is the best sub shop in Denver. No doubt in Colorado. No doubt. Snarfs, S-N-A-R-F and it's, I could say it's because of the fresh ingredients. I could say that, but the real reason is the bread. It's all about the bread. The bread is fresh baked, but how they bake it, I do not know. We just love snarfs. We love it. We love it. So in it's a way to save a little money because you don't have to tip. All right, waiting for sweet cow as people stare at me talking to a camera inside a minivan. I have to sit outside because we don't know the code. Okay, why don't we be close? Fun, I love that idea. I love that idea, honey. Hang on. I'm still just eating whipped cream. Yeah. Yeah. Oh my. Oh my. There it is. I mean, I don't think it trumps Andes, but it's a real close second.


Good morning, everybody. First of all, I would like to, as a veteran who was here at the creation in 2008, congratulate Thierry de Montréal and his extraordinary team to give us the chance to see each other again in person in this terrible crisis that we've had. It's wonderful to see each other again in person. The preceding panel focused on economic trends, although they touched about some power questions looking at the relationship between China and the world. This panel will focus on power relationships in present international politics. There's a lot of noise and there are many tumultuous events. We want to look beyond them and identify how various key actors and regents, US, Europe, Russia, China, how they are affected by the tectonic shifts in the geopolitical structure and how they shape them. And one of the typical examples of a lot of noise and tumultuous events is the withdrawal from Afghanistan. But what we have to ask ourselves is, what is the geopolitical meaning of the withdrawal? Who is to gain? Some argue the US lost, others argue the US gained from this withdrawal. Similarly, the AUKUS pact between Australia, the US, and the UK cost an enormous shock. Lots of consternation, disappointment, hurt feelings, breach of trust, all of that. But we have to look beyond it. Is there a new constellation of power, a new power balance emerging, between China and a group of Western states led by the US? Where does France place itself? Where does the European Union place itself in this? What role does the US accord to Europe? What do we have to think about the notion of European autonomy in this context? What does it mean? Is it real? Is it a genuine alternative? And looking at the somewhat inept way the US pulled out of Afghanistan and organized the AUKUS pact, how is the US, the Biden administration, going to balance this what the economist called ferocious complexity of reconciling on the one hand confrontation and conflict with China, with on the other hand the need for cooperation on global issues such as climate or health. And if we look at Russia, again, we have to go beyond the conflict, which is definitely the case, and at the moment, relations with the world, with the West are not really at their lowest point. We have to look beyond it and ask ourselves the geopolitical question of the future. And the Russians too, is it in the West's interest to have a Russia that remains a satellite of a powerful and rising China? Is it in Russia's interest? And as we sit here, the Trade Technology Council created between the European Union and the US to reset their economic relationship is actually meeting in Pennsylvania. Are the two blocks still the most important power block in the world in terms of economics? Are they going to reformulate and redo the orders and regulations of the liberal world order and do this at a time when the Biden administration has not discarded many of the protectionist and America first ideologies of the Trump administration? So these are open questions. And to conclude, my questions, two general questions. Is the emerging conflict with China going to structure and dominate the world in the same way as the old Cold War with a bipolar structure between the US and the Soviet Union once did? And in its combination of confrontation and interdependence and cooperation, on the other hand, how does this conflict differ from the preceding Cold War conflict? And how does this affect the likelihood of war, which, after all, the Cold War avoided, thank God? So with these questions, I turn to the panel, where we have wonderful expertise gathered here. I'm not going to introduce them with all their background because you can read that in the introduction. And I will start, and please, since our time is limited, confine yourself to six to seven minutes. I will hold up my watch when the time comes.


please Igor. Okay, thank you very much for inviting me to give this talk and even more thanks for arranging the whole seminar program all together. It keeps him up to a.m. every Thursday. But yeah, I enjoyed it. Thank you very much. So today, indeed, I will be talking about Vile Sums in all sizes, which is a joint work with Chenggao Chen and Bryce Kert. We did it while all three of us were here at UNSW and now we all left different locations. Okay, so let me start with basic definitions. Assume that we are given a vector of real numbers, which belongs to the unit cube of dimension D. And in fact, it's more convenient to think about this cube as a torus. So we reduced everything model 1. So T of D is this d-dimensional unit torus. And then our main object of study is the first following Vile Sums, which we denote as sub D, UN. It depends on three parameters. D is the degree of the polynomial in the exponent. Ui's are the coefficients. And n is the variable of summation. And E of x is the exponential function of x with a coefficient 2 pi i. So these are all unit vectors. And as the name suggests, they're named after Hermann Weierl, who introduced and instigated the sums and, more significantly, he foresaw the great value for mathematics. For more than 100 years ago. And just in case, if you forget the definition, instead of putting the title, and my name is the footer of each slide, I decided it would be more useful to put the definition. So hope you can see it. So this definition will be always with you. And there is another set, which will appear a little bit later. We'll talk about this. Okay, so when Hermann Weierl introduced the sums, he almost immediately found two very important applications, and actually that was the motivation to introduce the sums. In 1916, he proved that these sums can be used to show some uniformity of distribution results for fractional parts of values of real polynomials. And actually, even now, 100 years later, we don't really know how to deal with this question. Otherwise, how to show the fractional parts of polynomials are uniformly distributed. There's only one exception. There's a dynamical approach of Fürstenberg, which however still loses to the original approach. It gives quantitatively weaker results. And after this few years later, he also found another exciting complication, namely, he proved the first so-called subconvexity bound for the even-zeta function, which is this first non-trivial result towards the still-opened dintel of conjecture. And of course, since that time, lots of other applications were found, and here is certainly a very incomplete list. People obtained new bounds for the 0th region of this zeta s solved some other problems, like varying problems, and many other additive problems found bounds of short character sums and also L functions for highly composite modeling, investigated low-lying 0s of L functions, and so on and so forth. This is a very incomplete list, as I said, and there are actually very surprising complications, which I know at the first glance has nothing to do with number theory. And maybe later, I will describe some of them. For example, last year, Erdogan and Shakan found links between bounds of some very special wild sums and partial differential equations. A very unexpected link. Okay, now let me first explain what we know about wild sums, and then we talk about things which we do not know. And our knowledge can be kind of classified in the two classes, average values and individual bounds, and it starts with average values. So for average values, of course, you can trivial it, prove that the average squares and it's nothing but a partial identity, a completely trivial result. But if you make one step forward and decide to look at higher moments, for example, for the first moment, the equation becomes highly non-trivial. For example, there's no easy way to say anything meaningful against the first moment of the wild sums. And finding estimates of this type, integral of all possible vectors, u vectors of coefficients of polynomials over the unit cube td, for s2, free and so on, are usually known under one collective name, Wiener-Graduos mean value serum, but it's better to use plural because there are several modifications. So well, it depends on your point of view whether you want to treat it as one serum or as a series of serums. And I will be abbreviating this vmvt and it solves the problem with the sexta s that will be compressed in t. So why do we call it Wiener-Graduos mean value serum? It's because Wiener-Graduos in 1935 obtained its first non-trivial bounds on this quantity, j sub ds of n, and actually with the right saving. And we talk about this a little later, what is the right saving? How can you estimate these moments? So the saving was right, but he achieved this right saving for the values of s slightly higher than one would expect. And we know that it was kind of right expectation. So he went a little bit far up with the values of s for which it has its bound. And he also linked this average bounds to individual bounds. So he obtained first non-trivial results within his method, which for large values of d were much stronger than the result of Hermann Weierl and some other researches. Well, so he kind of opened this direction and after works of many other people, most significantly, Wiener-Graduos, Korobov, Kratzuba, Ford, Rohn, Wuli, and I'm sure many other people. So 80 years later and several dozen papers later, we have the following, which is the optimal Wiener-Graduos mean value serum. So what we know now is the following estimate. So for any s starting from 2, this integral, the average s moments of while sums can be estimated by a sum of two terms, n to the s, and n to the 2s minus d times t plus 1 over 2. And I ignore a little over in the exponent. I will never mention them, even of course they will present in essentially all estimates I write today. So this result, which is optimal, and I will explain in a second why I call it optimal, is due to Tera Wuli 4d equals 3. It was appears 2016. Then at about the same time, Wolfgang de Meta and Goetz establishes 4d greater equal 4. And actually this same method didn't seem to work for d equals 3, as far as I understand. And a few years later, Wuli made another step and extended this estimate to more general exponential sums with linear combinations, not just powers, n, n squared, n to the d, but with linear combinations of d arbitrary polynomials. And later it will be important for us. So we'll come back to this. So now let's discuss why this believes that this bound is optimal. And it also kind of will help to understand how one can deal with the sums and how they behave. So I want to consider a generalization of my integral j d ds. I want to introduce another parameter, vector lambda. So it takes the same moment as before, the s moment of while sums, and twisted with an exponential function of the scattered product of lambda and u. So there is a linear function here in the exponent. Okay. So what I want to notice? I want to notice the following identity. If we take the sum, write them to s power, and open up, I will get linear combinations of the power sums, because instead of writing the absolute value of s squared, I write it as s times the conjugate s. Conjugation just changes the sign to the exponent, and I do it s times, so I have two s terms, two from which square of s, and they appear in this shape. And of course, this linear function of lambda will be combined in the same exponent, so we have this expression. It looks very clattered, but actually they have very simple structure, which we'll see on the next slide. And what I want to do next is to change the order of integration, to change the order here, and integration will move inside, and the product will go outside. So for hj from 1 to d, I have this single integral over uj. And now it's time to recall the orthogonality relation, then when you integrate this function between 0 and 1 for some integer w, you have two possible answers. It's one when w is 0, and it's 0 otherwise. So you will see that these integrals, they become characteristic functions of this exponent being 0 or not. So if you put everything together, you see that this integral, this twisted moment, is nothing but the number of solutions to this system of equations. You have d equations, and in each equation you have s powers of n, of n on the left-hand side, and another s powers on the right-hand side, plus this shift lambda j. Okay, now, so this number of solutions to the system of equations. Now let's come back to the previous slide and look at this. And I claim that it's obvious that when lambda is 0, when lambda is a 0 vector, which corresponds to our initial definition, this integral is larger than all other integrals, because it's what happens.  So applying the absolute value, I can only increase the value. So lambda equals 0 gives me the largest possible value. So we have this. And of course, when lambda is 0, when it's missing, I obviously have n to the s solutions. I take n1 equals to ns plus 1 and so on, and I don't even want to worry about taking permutations. I don't care about constants. So this trivial solution already gives me n to the s. So n to the s should be present in any up about, because this value is at least n to the s. And it was the first term in the optimal binogratzous mean value theorem. So we are done with n to the s. Let's look at the second term, which was more involved. And here, I want to combine three very simple statements. The one which we just used, then when lambda is 0, the value of this function is bigger than all other values. Great equal. We don't know whether it's larger or not. Now, if you sum over all lambdas, if you look at the system and put solutions for all possible lambdas, which means you have no restrictions at all, whatever lambda comes, you call it lambda j. So sum of all lambdas without any restriction is the same as the total number of variables we have, which is n to the s. And another thing for this thing to be nonzero, lambda j should be at most this, because lambda is given by this expression. And each term is at most n to the j. And we have two s terms here, but they're positive and negative. So by absolute value we have this. And we are done now, because we have that many possible values on which this function is supported, values of lambda. For each of these values is at most this. So the quantity we have after j sub d s n without any lambda at all times this number of possibilities for lambdas is greater than the sum of all j's of all lambdas. But we know what's the answer here, it's n to the s. So if you put these things together, you see that this term must also be present in any upper bound. This means that the upper bound, which I gave you a few slides ago, which contain both terms is optimal. Okay. So now that was about average bounds, what's known about individual bounds. And here I wish I could tell you something that the result is, that at least some of the, one of the results is optimal, I can't, our knowledge is very, very sparse here. So what, what do we actually know? So if you use the Vinaigradoff's method, plus the optimal form of the Vinaigradoff's mean values theorem, then you have the following result. Assume you have this vector u, the vector of coefficients. And also assume that for some nu between 2 and d, the linear term is no good for us. You have to know something for, for non-linear terms, you know, your polynomial. It's a member that you, you is a vector of coefficients of the polynomial, like in this definition in the photo.  Of course, there are infinity-mediate approximations of this type for any irrational number u sub nu. Well, then for this value of n here, and this value of q, and this value of nu, you have the following bounds. S doesn't exceed n, which is a trivial upper bound, because you have n roots of unity. So this is trivial upper bound. And now let's, you have some savings, because as you see, all these exponents are negative, so they give us no trouble. And there is one more term here, which can potentially be, be a trouble maker, because q comes in a positive power. But we talk about this in a, in a few moments. So here if you have q to the minus one, and you have to rise it to this power, which of course reduces your saving. And to the minus one, again, you have to rise to this power. And here's the only restriction is that q must not be very large compared to n to the nu. And right now we don't really have any other plausible approach to have, to have better bounds. So at some point we need some new ideas. And let me just tell you that the formulation looks very involved, but it's very easy to see that any, any bound of this type must depend on the finite time properties of the coefficients u1, ud. For example, if all of them are integer numbers, then of course you have no consolation. You have integer numbers in the exponent, so all those exponential functions take really one, all vectors are aligned in the same direction, so you just have n. So you have to know something about your i's, how much they're different from, from integer numbers, which means that you have to have some information of this type. Okay, so this would function at everything I can say about individual numbers. So in summary, what do we have? We have almost complete knowledge of average values of the world. Yes. Philip, you wanted to ask a question? Yeah, I had a question regarding the slide above. So the, the condition here is not symmetric in, in new or in the ordering of the Yes. Yes, sure. You would like to know this for the highest coefficient. Sometimes you don't. Of course you gain more when new it's large. So if you have this information for the, for the leading coefficient, it's better. Sometimes you don't. Okay. So this is why kind of the bounding proofs is, is your knowledge. And yes, it's not symmetric with respect, with respect to Q and n. Yeah, but this, this, this, what this method gives. Okay, so what do you know, you know, essentially everything you would like to about average values and know something that very little about individual behaviors of, of, while sums. So what I'm going to do next, I will move to the part with some new results and outline some recent results, which tell us something about the distribution of the while sums. Namely, if you look at while sums of, with scribe size and as my title suggests, they will be small, large or close to the average values. After this, we consider kind of an intermediate scenario, which interpolates between these two cases. We consider some kind of equations, which combine average and individual results. And then I show how this two different directions come together. And allow us to say something interesting about the while sums. So this is the plan. So let's talk about the distribution of values of the while sums. So I recall our main tool, the mean value theorem. So this moment doesn't exceed n to the s coming from the diagonal, diagonal solutions. And this term, which gives you the right saving. And because this is the right saving. There are immediate implications of this. So it tells us that when s runs between one and d times d plus one over two, exactly this expression. Then the first term dominates. And instead of n to the s, we should do the trivial bound. It's n to the, sorry, n to the two s, which comes for free without doing anything because it's a trivial bound on this sum. You have n to the s, which means that on average, this sum is n to the one half, which is natural to expect because you have capital N vectors and you treat them as random vectors. So you expect something with the square root constellation. It's not surprising that you have n to the one half. Now, let's take the largest possible value, s, before the second term takes over, which is this, as I said, this last term here and they take this. And it tells us that if you ask, what's the measure of the set of coefficients you, for which the corresponding sum is bigger than n to the one half plus epsilon. So you add a small epsilon to the exponent, a fixed positive number. And for this, you have this bound. So it's true for a set of very small measure, n to the more or less d squared epsilon. So you know that the number of sum, the measure of the set of large sums is small for each concrete value of capital N. Okay. Now, you can do actually something more interesting. You can interchange the roles of n and u. And it's not immediately obvious. So what is true if you apply the mention of Radimacher theorem is formulated here for almost all vectors of the coefficients u. And for all values and for all sums, you still have the same bound. So you have a universal set, a small universal set of bad values of u destroyed. And then for all sums of all lengths, you have the square root bound. It's not a very difficult proof, but it will take some efforts to prove. Okay. So now we are ready to ask our questions. So what about typical values? You know that on average, all these moments suggest that s, the sums s, the viral sums should be about n to the one-half. But typical values are not necessarily average values. And one of the equations, for example, is here. Is it true that for almost all u, for almost all coefficients, for infinitely many, or maybe even for all n, these sums are at least as large as n to the one-half. The fact that the average values of this order of magnitude doesn't allow us to make this claim. Now, what's about extreme values? For example, what can we say about the set of u, the set of coefficients, for which were infinitely many n, the sums are very small, less than n to the delta. Is it a massive set? Does it exist at all? Maybe it's true for all vectors u we do not know, at least not yet. The same thing you can ask about large values of s. What can we say about sets of u in the unit terms, in the d-dimensional terms, such that for infinitely many n, these sums are very large occasionally. So all these equations are already open. And right now, we are not even able to rule out such ridiculous statements as this. Say, for almost all u in the d-dimensional terms, for all n, all the sums of this size n to the little over one, this should be very, very wrong. Nevertheless, I have no idea how to prove that it's wrong. So as far as I know, it's still a possibility, even if I would probably bet my annual salary against it. Okay, so let's try to make some quantitative statements. And now I want to introduce a set E, which will be with us essentially all my talk. And it's also presented here. So the set E, it's a set of exceptional sums, which is defined as a set of u. The vectors define our polynomial in the exponent for which real sums, which correspond to this u, at least n to the alpha for infinitely many n. So these are polynomials for which they are large sums. And I define c to of d as the infimum of all alphas for which this set of is of measure zero. So when alpha is bigger than this number, then for almost all u, sums are less than n to the alpha. So the exception of set is small, if alpha is bigger than c to d. So what I said before about when Schroff-Demacher theorem allows us to claim that c to of d is at most one-half. Because for almost all sums, we have the square root bounds with little o, but because it's inf, it makes no difference. And it's certainly a very natural to make the following conjecture, which Chang'eau, Chen and I kind of put officially in our paper, but I'm sure many people would guess the same that this is just an inequality instead of inequality. That c to of d is one-half. That everything switches when you pass the threshold one-half. But almost all sums should be at least n to the one-half and at most n to the one-half in Chen'eau. But as I just mentioned in the previous slide, right now we are not even able to show that c to of d is positive. This ridiculous statement on the last slide would mean c to of d is zero, which of course is not, is not likely to be true. And you know nothing about this except for the case d equals two, where we have some extra tools and extra possibilities. And then we know that the conjecture is actually true. So when d is two for quadratic sums, we know the c to of two is one-half. Okay, so we know very little about the set, and we cannot say much about c to d, but we can say something if you measure the size of the sets, not via the Lebesgue measure. This tool is too powerful for us to handle. But why household of dimension? Which is kind of more refined tool, and it allows us to make some non-trial statements. And we will be interested in the following regimes. Small real sums when alpha tends to zero. So we are interested in sets of u for which sums are bigger than m to the alpha for very small alpha, which should happen all the time of course, but we don't know how to prove it. We will be interested in very large sums, and this should be a very small set. And we are also interested in the regime when alpha tends to one-half, which should be a typical case. But I put a question mark, we still don't know this. We assume it's typical behavior, we don't know this, but let's call it typical still. So these are three kind of interesting regimes of alpha. So as I mentioned, we will be using household of dimension and here for definition, which I'm sure all of you know. So this is a low about the infimum of s, such that for any epsilon, there's a cover of our set a, a, which satisfies this condition. The sum of diameters of the sets ui, the power s is less than epsilon, where the diameter is just the largest distance between two elements of u. So what do we know about the sets e of alpha? Well, we actually know something. We cannot show that they, they are very massive with respect to their back measure. But for any alpha, we know that they are reasonably rich, meaning that if you take this set e of alpha d and intersect it with any small cube inside of you, d-dimensional torus, and then look at the intersection. This intersection is a positive part of dimension. So there are no holes in this set. That's everywhere, but it's more than just everywhere, everywhere. And there's enough kind of mass in this set. So there are many elements everywhere in the unit, in the d-dimensional unit torus. More specifically, we have the following low bound. I want to define this quantity, kappa sub d, which is defined formally by this expression. It's probably not so interesting what probably is important to know that this limit is three quarters, and if three divides d, you have just the exact value kappa of d is three over four d. So if you have to minimize this and maximize or renew, you obtain this. And with this definition, we have the following flow bounds. So for any small cube in the d-dimensional torus, when d is two, the dimension of this set, or the intersection of our set e alpha two, and this cube exceeds three half, the smallest out of three halves, and three minus one minus alpha. And we have a similar result, except with this slightly weaker coefficient kappa d for d greater or equal three. So everywhere, you have some positive value on the right hand side, which is probably the most interesting part here. And in fact, when alpha is less than one half, then the second term doesn't matter anymore in the bound simplifies like this. It becomes three halves, and this is about three or four d. And of course, for alpha less than one half, we expect that this set would be a positively back measure, probably of full measure. We don't know how to prove it. So our conjecture, which says that theta d is one half, it's exactly the threshold between small and large values of sums. Because if our conjecture is true and theta d is one half, then this set should be of of the measure d. And in fact, we believe that even stronger holds that all these small cubes should also give you an intersection of a full dimension d. But as I said, it's just a conjecture, and we don't really know how to approach it. So so far, we only have this, have this low bound. Well, that was a result which holds for any alpha. If you consider a shorter range, then recently together with price curve, we obtained a strong result. So in the range between zero and one half, strictly less than one half, we almost have something which resembles what we want to prove. This household of dimension is at least d minus one over two d. This term must not be here. But well, so far, this is the only thing we can prove. Unfortunately, this doesn't work for the critical value one half. We really would like to have it with alpha which can take the value of one half. We can't. This inequality is strict, but we can do it for a potentially slightly larger set. So what we want to introduce, we want to introduce yet another parameter. My definition of the set e alpha d says that this sum should be greater equivalent to the alpha. Hey, we need to introduce a constant c. We need to lower a little bit. And in this case, if you allow to introduce a constant c, which of course you'd have to choose less than one, you may need to choose less than one. Then we have this inequality, the same as here, even with alpha equals one half. So this is alpha, which is now one half. And the price to pay is to put an extra constant here, which we see equals one would give us what we really wanted. We cannot guarantee that c is one. We already know it exists. So potentially it could be less than one. And so the result is a little bit weaker than what we want. Okay, now upper bounds. Well, as I said, we certainly expect that this set is of dimension, of dimension d in this range. So we don't expect any upper bounds in this range. So we only want to talk about alpha, which is strictly greater than one half. Namely about sets of k-efficient of our polynomials for which while sums are occasionally very large, larger than the average value. So here we have the following estimate, which looks very convoluted. And there's no point of reading this. It contains some fancy function u, which depends on d and alpha. But what's interesting here is that if you forget about minimizing over k and just take a concrete value of k, say d minus one, then this upper bound tells you that this function u, which is an upper bound on the dimension, is d minus something. So it's strictly less than d. So even we expect that the set e alpha d is of how is the dimension d for alpha, which are between zero and one half, inclusive of both ends. It changes instantly when we bypass one half. The dimension is certainly less than d. And this is what we can prove. So it's a non-trivial bound. And now, of course, you can try to push it more and try to get better bounds. And indeed, we get kind of several other results. We have another bound, which is a very strange shape. But I presented because it will bring me to the second topic, which I am going to discuss. Namely, we have two parameters, k and d, two integer parameters. And the only restriction is that k is between one and d minus one. It should be strictly less than d. Then now we start with a bound, rather than with alpha. And we adjust alpha to the bound. So we claim that the dimension of the set e sub d alpha is at most k, provided that alpha exceeds this. And again, we are interested in non-trivial bounds on the dimension. So d minus one is what we are after. We don't want to take d and it's not allowed anyway. So the largest value of k is d minus one, which still makes sense here. So with k equals d minus one, you substitute here. You see that when alpha exceeds one half, certainly it must exceed one half, plus this function, which has a linear function of d in the numerator and a quadratic function of d in the denominator. So it decreases as one over d, more or less, when d is large. So if you exceed one half a little bit, the dimension already drops below d minus one. So it's just we just pass almost instantly, you have a drop in the dimension by one, at least. And this bound is obtained as a byproduct of some results on our next topic, projections of while sums. But before we leave, just a few more words about the set e alpha d. So what's the truth about this? What do we expect? Well, upper bounds and low bounds are very far apart, so they don't come close to each other. However, surprisingly enough, there is one regime where we are not so much off. So we essentially know not the truth, but kind of symmetries here. Namely, it's a regime when alpha comes to zero. So when alpha is of the shape one minus delta, but delta tends to zero. In this case, you know that the dimension is cornered between two linear functions of delta. So when delta decreases, this dimension behaves as a linear function of delta. And more precisely, we have these estimates. So instead of one minus delta, switch to alpha again, so delta becomes one minus alpha. So the lower limit is greater equal three, and the upper limit here is at most just squared plus two d. So we know something at least in at least in this regime. But otherwise, we have no possible conjectures. We have no idea what one could expect for this function. So I will be very much welcome some input if someone has anything to say. Okay, now let's move to the next topic, projections of while sums. Again, we call that we have a complete knowledge of average values of the while sums. And we know something, but overall, very little about individual values. So the question is whether you can interpolate. Now, let me recall what we know for the individual values, something which I presented at the beginning of my talk. We know that if some coefficient with a nonlinear term is approximated by a rational fraction over q, up to one over q squared standard Dirichlet approximation, then we are in business, we have this value. So if you know something about at least one coefficient, we can ignore the rest. So if you control, say, ud, we don't care what happens to other coefficients, they could be integer numbers, and it's still a very, very interesting case. And actually, what we still have a non-trivial bound. So essentially, without doing anything, we have the following result for almost all ud and all other components of our vector u. We have this bound. It's because for almost all ud, we can choose q of this size, for example, which is a quasi-optimal choice here. Then these two terms become balanced, and the third term for this value of q gives us no trouble. It's dominated by the first two, which are equal in this case. So for almost all ud, you can always find approximations with a denominator of the size you control very well. So in this case, almost all ud, and almost all ud, and you don't care anymore what happens to other UIs. So for all of them, you have this bound. So this saving becomes 6, 1 minus 1 over d minus 1 in the exponent of n. So we have a result which holds for almost all ud and all other vectors. And now you can ask, what happens if you ask for family results of this type? So what do you want to say? You want to say that for almost all components of U on prescribed k positions, say the first, the second, and the tenth, and all components on the remaining d minus k positions. For all lengths of our sum n, you have a statement of this type, where x, x, x is whatever you can prove. The first bound you can do, depending on the positions you fix on k and maybe on something else. So it's kind of a prototype of the statements I want to prove. And this scenario was considered by several people. For example, by Flaminian Farni 2014. Then Trevor Woolley revised this topic and used a very different approach to obtain results of this type. And then finally, Chen Gao-Chen and myself last year also worked on this and we obtained several results in this direction which I will now present. But I want to reformulate it in a slightly convenient form. Instead of saying you want to fix, coordinate some positions, I will always be fixing them on the first k positions. But for this case, in this case, I need to permute the functions. And once I permute the functions, I can as well generalize it to a different situation. So I assume that I have d linearly independent with constants polynomials. Meaning that no linear combination of them is constant. It's not enough to say that it's nonzero. They need to avoid constant functions. And given a vector u, I define this function. I call it t not s, s is reserved for the classical, while t means that we have linear combinations of some polynomials. And of course, when psi i is t i, this is just a classical while sums. Now, I take my vector u and decompose it into two parts, part x of dimension k and part y of dimension d minus k. So my vector now has two components. And instead of writing t u n, I will write t x y. So we will separate two components of u. Okay. And now, following this work of Flaminion Favnii and Trevavuli, we are interested in bounds on the sums of the following four shape. They should hold for almost all x and all y. So some components, some components x are allowed to be kind of some of them because they're allowed to be discounted, some set of x, but we need it for all y's. Equivalently, we are interested in bounds on this quantity. So it should certainly have absolute value. And my apologies. We should certainly talk about the largest absolute value of the sums, maximized over all y's. And we want those bounds to be true for almost all values of x. So it's soup with respect to y and almost all with respect to x. So interpretation of this is the following. We take our sum t u n and project all the sums along y components. And we take the largest sum in each projection. And we want a bound which holds for almost all x coordinates. This is what we call a projection. We basically just take all sums with values of y, which is the same as projecting the sum. Okay. To formulate the results, we need to introduce another important parameter, which I call sigma with two parameters k and phi, of course. There is this vector function, functions we are given. So to sum of the degrees of polynomials in the y part, polynomials over which we maximize. And in this case, we'll improve the following estimate. Assume that k is between 1 and d minus 1. So this means that you at least have something in the y component, because k is the dimension of vector x. Then assuming that our set of functions as an integral of Ronsky, it's a necessary condition, which is easy to see. For almost all x, you have this bound. The largest radio of the sums, tips, phi, x, y, or n, doesn't exceed n to the one-half. Of course, one-half must be here, plus some losses, or what one would expect, what you notice for pure pile sums. And this gamma, which I call the gamma w in honor of Wuli, has this form. And last year, Shangar and I managed to get a little bit better. We reduce this exponent gamma by subtracting 1 and 2 from here, which you can see gives a larger value. Okay. So the bounds are non-trivial. If this exponent extra, extra exponent gamma is less than one-half, because we already have one-half, which means that sigma, the sum of degrees in the y part, should be at most this. And please know that if we, in the classical case, when you just permute these functions t, t squared, t sub d, it's always satisfied. In the more general case, well, it's a matter of luck. It depends how your functions look like. If they contain polynomials of huge degrees, then, of course, you have nothing, but this condition will be violated. So in the general case, we don't always have non-trivial results. And the problem still makes sense, and it would be nice to find a method to address it. We also have some specialized bounds, but I probably skip this. It's not so important. But I want to come back to the link I mentioned. Link between projections of some, something which we have discussed, and now bounds on the dimension of this set, e alpha d, set, which we defined here. To establish this link, we need to consider a new scenario, a slightly more general scenario, where we consider arbitrary projections. Before, we just discovered y coordinates. Now we want to do something else. So we can see the collection of all k-dimensional linear subspaces of i d, and pi sub, double sub v denotes orthogonal projection onto this linear space v. So we project not necessarily with respect to y coordinates, but we consider projections on other spaces. And very much as we did with e of alpha d, we consider this set, which defined exactly the same except of the while sum, I consider my more general sums with linear combinations of other polynomials, and the set of u for which the sums are large. So the equation we want to ask is, given this set of polynomials phi, for what alpha the measure, the big measure of this projection is zero, for all possible linear spaces. Well, what can we say from our previous result? From our previous result with this exponent, we know that if alpha is greater than this exponent, then the back measure of this projection, pi dk is zero, where pi dk is a specific projection, which we just discovered the last d-k components. Now we are after more general projections, not just pi dk, but projections on arbitrary linear spaces. And we have a series of results, slightly weaker bounds, which means our gammas, which appear since these results are larger, but we still have results, our method gives results for these arbitrary projections. And this allows us to address this issue, link projections of while sums to upper bound on the dimension, on the household of dimension as exceptional sets. And it's based on a classical result of Marstrad and Matilda, which is known as Marstrad-Matilda projection theorem. I don't want to give any formal definitions, just informally, it says the following. Assume that you have a set A, and you think about this set A is a set of our vectors U. And assume that almost all k-dimensional projections are massive. Massive means a full measure. Then the household of dimension of the set should be large. And as large as the dimension of the subspace which you can control. The largest of spaces you control, the larger bound you get. So it's exactly the link which allows us to use results on projections of while sums to obtain upper bounds on dimensions of the sets. Okay. Well, now I want to talk about something else about links between this direction and partial differential equations. I won't say much about partial differential equations, but at least I explain what sums appear in this area. And I think it was discovered by Erdogan and Shakan last year, who noticed that the previous equation with this very special set of polynomials, a polynomial psi of x, which could be anything, most commonly is x to the d, but not necessary. And this linear combination of phi of x and just x. So we have only two functions in your system. This function appears in the investigation of some partial differential equations. So what do you want? You want to answer a question of the previous type, namely, you want to know what's the smallest possible value of theta such that for any polynomial, phi of t of degree m, any coefficient tau, which appears here, for almost all x, these sums are less than n to the theta. So you have d equals 2, you have only two functions, k equals 1. One coefficient is, you maximize over one coefficient and take almost all values of the other coefficients. And you want to know what's the best bound you can get here. So it's exactly the previous scenario. And when Cengar and I looked at this, we saw that, well, it's a done deal. We just apply our result. Unfortunately, it didn't work. As I said, our previous results, they give some material bounds, but not always. It's never guaranteed. It depends on the luck on the structure of the exponent. Here, we didn't get lucky. So we didn't get nothing with the result itself, but the method works. So basically, we had to revise our method in this particular scenario. And then we obtained some results. And well, as I said, this function is related to partial differential equations. And here, there's two particular types where this question appears. And it was discovered by Erdogan and Schokhan in their very nice paper. Of course, this sums looks very strange. And nobody would probably look at them, but the existence is justified by these links. So what we know now, we know the following. So in their paper, they proved that you can take theta. Theta m is the best exponent, the best value of theta. To be the smallest out of these two quantities, this comes from the classical, while method from the Van der Kogh method. And this comes from Vinogradov, Burgain, Demeter. Goods. And using our method, we improved it a little bit. And we have kind of a very lengthy result, which gives specific exponents for small values of m. And then a formula, semi-explicit formula for larger values of m. But the upshot is, this function of m is about square root of m. So we ignore it. And this 0 and 1 also ignores this.  So it's twice larger or smaller, depending on the point of view, asymptotically for large values of m. So this is what our methods give. And just in the remaining few minutes, I want to outline very briefly ideas behind the groups. So how we obtained these results. None of these ideas is new, but we put it in a slightly different context. Well, the first idea, which was used by many people, starting from Vinogradov, is the continuity of while sums. So it says the following, you have two vectors of coefficients, u and v, and assume that they're close to each other. Because you work with continuous functions, of course, the sums will be close to each other as well. And you can make a quantitative statement of this type. Okay. So and this statement actually works as stated and has been used. But we noticed that this can be used only on the first step. And we introduced something which kind of was used recursively. And we invented something which we called self-improving government. So you use this principle as the first step, and you have a result. And you know that for almost all values of u, your sums are small. And please remember, they're small for all values of n. It's important. This is why changing the order between n and u was, in our results, was very, very important. So now for almost all values of u, the sums are small. Next time you estimate the difference between SU and SV, you use partial summation. And in partial summation, you obtain sums of this type, for which you already have a non-trivial result. So you improve your previous estimate, and you use recursively until you come to kind of fixed point of this recursion, which gives you a bound. It does give an optimal result, but it improves you, what one can get as a first step. So that was one of the things which allowed us to get new results. And the second idea is also a very old idea, which also goes back to Winogradov. In fact, appears even before his mean value theorem. Now it has a generic name, completion matrix. So what we want to say, we want to say that for almost all values of u, for almost all coefficients, this sums are small for all values of n. Again, it's important for us to have it in this order. We first throw away a bad cell, set of vectors u, and then we want this bounds for all values of n. This means that for each n, we have to control the size of exceptional sets. And we want this to be small. So if mu of n is this measure of the exceptional set for which the sums are not small, we need to show that this series converges. This is to say that for almost all n, we have an upper bound. But it will give you something this approach, but it's not so good because you have to deal with each n individually. You have to say if you want to deal with all sets up to capital M, you have to control M different sets. It's too much. This approach doesn't completely fail. It just gives a weaker result. So what we did used completion methods, namely instead of using these sums, completion methods allows us to control each of the sums for all n in this dyadic range with just one sum. And the sum certainly looks ugly and scary, but in fact it's harmless. All this kind of fudge which you have to add gives you no trouble. So what's the fudge? First of all, we have a summation between minus n and m, one over h. Well, we don't know how to end the harmonic series. Then in the exponent, we have the same linear combination plus one linear term. And linear terms usually change nothing in our argument. So this why the sums are not so difficult to estimate more or less is the same quality as for initial while sums. So instead of dealing with n sums, if you want to say make a statement from these measures up to capital M, you have to control only log m sets. You, for which this sums say supported on powers of two are small and it's much easier and this also allows us to get better results. And I think my time is up. So it's time to stop. And I would really appreciate questions or even more would appreciate answers. Thank you so much Igor. Please thank you so much for the wonderful talk. Please unmute your microphones as we can properly thank Igor.