What's up internet? Revisiting our budget build for 15K. If you didn't see that video, gumawa kami ng budget build for only 15,000 na gaming PC na maging performance. At that price range, the best value for money is second hand parts and we got a good deal with the 10th gen Intel na pasok sa budget. Ang problema na pili namin GPU, the 1650 low profile, hindi ka siya dito sa OEM case. So a lot of you guys are super technical and you have a lot of comments on how we can address that. And so I just wanted to comment on those comments. para sa windows mo. Maraming na kaming na tulungan. Dati, said hindi purchase ako. But now, I found a log of my life. Dati, ing-resin walang purchase ang life ko. But now, I'm a world-class Zumba instructor. So web developer ako and content creator for a YouTube channel. And ngayon, ganun pa rin ako, pero activated na yung windows ko. Kaya ako nag-hanap ka ng legit, mura, at original software. Check out cdkoffer.com. Check out cdkoffer.com. Check out cdkoffer.com. Sino bukantara namin everything we could do to keep this built-in tack. Kasi good price nga siya, good performance. One of the questions we got a lot was could we just change the case? Tagalin yung motherboard, ilipat sa ibang ITX case. Since yung issue lang naman, hindi ka siya yung GPU dito. Unfortunately, even the board is proprietary. If you notice here na back, the I-O ports don't line up with the traditional standard I-O ports. So hindi siya mag-line up sa classic na I-O shield. So we wouldn't be able to fit this into a standard motherboard. The front as well, all of these front buttons and front ports in a computer case, usually, they gain functionality because they're connected by wires from the case inside to the motherboard. Pero itong proprietary design, wala na mga wires, wala na mga headers. The front is literally mated or the motherboard and the front slot into place. So that also wouldn't work with a standard ITX case. So changing the case out of the question, next suggestion was riser cables. And yes, that is possible. One way to do it would be literally tanggalin yung tong front cover naka riser cable yung GPU. And the riser cable is just basically an extension cable between the GPU and the PCI-E slot. So the, let's say the GPU would be hanging here somehow, may stand ka dito or you could plop it on top of the other components. Gagan na naman siya. Yes, that's possible. Although, removing the top cover, finding a place to kind of position the GPU para hindi mayang, para hindi mahulog, all of that does seem a little junky even for a second hand 15K budget build. We did try to use a riser cable to position the GPU elsewhere. Pero hindi talaga kaya. Tumatama naman siya dun sa PCI-E 1X slot. So keeping it inside the case even with the riser cable natin option, keeping it outside of the case via riser cable. Yes, that's possible. Although that was a level of open type PC case that we wouldn't want for a budget build. Especially, we assume a 15K build is somebody who's just starting out with computers and perhaps they might not be comfortable with everything exposed kasi literally tatang galin mo yung top cover to make room for the GPU which would be hanging somewhere on top. Another interesting work around would have been not to use a 1650 and instead use an RX 6400. And yes, it's true. Mapapasok mo naman yung RX 6400 dito. It will fit. So we avoid the problem that we had with the 1650 even the 1650 low profile. The problem with using an RX 6400 is that it is a PCI-E 4 card. And when it's running at that speed medyo kapantayin niya yung 1650. But the available PCI-E slot here is PCI-E 3 and it's PCI-E 3 X4. So not the full bandwidth available to an X16 slot. So even that is technically not a problem. You can go into your BIOS and tell the GPU to run at PCI-E 3 speeds. Ang problema an RX 6400 at PCI-E 4 settings versus a 1650 they're around equal. But if you step down the RX 6400 to PCI-E 3 X4 you lose substantial performance let's say roughly around 10-15% which you can see in the performance in games. Bababa talaga frame rates mo. So while we could fit in an RX 6400 the performance wouldn't be as good as a 1650. Pero tutoga if we had bought a 6400 instead of a 1650 it would work. Although we would take a performance hit. Is it worth it to keep a 1650 but just find a different computer because we did find a 8th gen and that will be our next video yung next sa series na to of our attempt at a budget build using second hand parts. And so we can't see how much of a performance hit we take if we downgrade the CPU because we found one that will actually fit a 1650. But the idea of using a 6400 especially if you haven't bought a GPU yet yes, is possible if you don't mind that performance hit. So it's kind of obvious by now that we're kind of hung up on the 1650 we really want to use it. We tried to go with a low profile maparamapasok, di talaga kaya. Our tech guys even tried to deshroud it or basically remove this top remove the fans which would leave the passive heat sink element. Kung mapasok ba natin basically you're just lopping off this part without the active heating element mapapasok ba natin sa PC. And it almost succeeded except for the fact that the DVI port gets in the way. They actually tried monkeying around to remove the DVI port but it's soldered on. Hindi lang pedeng basta-bastang tanggalin. And so that leaves us stuck again. Even without its fans the 1650 low profile is just too darn large to fit into here. So that's really interesting about using OEMs about delving into the second hand market and dami pang ibang considerations and it's really forces you to become creative in the kind of parts that you use in the kinds of parts that you can fit into something like this. May pros and cons of going with this second hand OEM approach definitely one of the cons as we've documented in this video and the last video the parts are not standardized so they don't mesh well sometimes with off-the-shelf components. But what you do get when everything works is a very good performer for its price. And that one we're gonna talk about in the next budget build video. Paminsan may nagtatanong kumikilala ba kami ng computer shop na trusted yung hindi kalolo ko hin? Actually meron kami. Full service PC store ang hardware sugar. Nagbabenta kami ng PC components. Nagbabenta rin kami ng fully assembled rigs. We clean computers. Kasama na rin yung excellent cable management namin and CPU cooler repasting sa cleaning. We also clean and repaste GPUs. Nasamakati yung physical store namin and you can also buy from our site www.hwsugar.ph na 100% palaging up-to-date yung inventory dun. Kung instock yung item sa amin available yun sa site. We also ship nationwide. Thanks for watching and maybe one of these days mankita tayo sa shop.


If you really want to have global health and youth, we need to compensate all these things. So it's not only about the food, it's not only about the skincare, it's not only about the lifestyle, it's doing right all the things in synergy at the same time. Inset Out Beauty means to me about being well informed about the right products to use on your skin, but also what to put into your body, what to consume that's really going to nourish you and allow you to glow from the inside out. Inset Out Beauty means to me that it is all interconnected. It means that we all need to take care of our sleep, of our nutrition, of movement, exercising and of course of our skincare.


Hello and welcome to NewsClick. Tamil Nadu is one of the states which is going for polls in the current year. On April 6th, the state will go to vote to decide who is going to rule the state in the next five years. To discuss more about the elections and the propaganda which is on the grounds, we have with us Mr. Vijay Shankar, the editor of Frontline. Welcome, sir, for joining us. Thank you. Similarly, to the previous elections, Tamil Nadu is witnessing a multi-cornered contest this year. There is a front led by DMK, ADMK, the Kamal Hassan-led party is leading another alliance and the ADMK splinter group is also leading another alliance apart from the NAMM-Tamilore fetch. So what is your take on the division of votes and who may get the advantage of the division of votes? Before going into the, I want to tell you something about the past history, the recent history of Tamil Nadu elections. Though there have been attempts by another front, third front, actually what is happening is this two-party, two-front dominance has not diminished. Actually it's still going on, that's what I feel. It's still, in my opinion, two-race horrors because I have some reasons to say this because recently we published one article by Mr. Srinivasan which proved that at least in the last six elections he took statistics, electoral statistics. There actually, there were other players in the last three, four elections, not like friends, but like NAMM-Tamilore Katchee and other smaller parties were there. But what he has proved it in this election, electrically, statistically, he has proved something. Though there were other parties, actually the vote, about 75-80% of the vote went to either of these friends. So that situation has not changed as even now, that's what I feel. I think because the statistics prove that. And even on the ground, though there are other players like Kavla Vasan and another person who is in the news very often is NAMM-Tamilore Katchee Seemaan. Otherwise, there is no other significant player because most of them are with either of these friends, two friends, ADM-Kilette friend, DM-Kilette friend. So the vote split theory, it is yet to be proved. We have to see it. Opinion polls show that again, this two-party dominance continues. There was an opinion poll by Pudiyath Alemurain which showed that mainly other parties like Makkal Nidhimayam and NAMM-Tamilore Katchee are very insignificant. Makkal Nidhimayam will score some, make some advances in terms of votes in urban areas. NAMM-Tamilore Katchee is not even making that kind of advance. That is the opinion poll. So I don't think there is any major change as of now in the scene. I think these two friends will share it. But I think from the ground level reports we get, it is clear that it is remains two-front election. And very interestingly about in 170 to 80 constituencies, DM-K and ADM-K are facing each other. Definitely there is no place for another player in these two 176 seats. What is happening in Tamil Nadu is, 2016 assembly election when Jailalitha was alive, such a tall leader was alive, Karnatidhi was alive, then DM-K almost managed to come close to power. DM-K Congress Alliance won about 109 seats. But 109 seats. And the difference between ADM-K and DM-K front was just 1%. Even when Jailitha was alive. So now Jailitha is not alive. Another factor is, after that election, these Makkal Nala Kootani were left parties, VCK, Vidvalay Chirithayal Katchee and Makkal Mdm-K and Vijayagans party, they were all together. They tried to break this dominance. They were not with DM-K then. But 2019, they all came to DM-K. There was a grand coalition which swept the election. So that alliance remains intact even now. But in the case of ADM-K, as I explained to you, ADM-K would be after Jailitha, it is not as united as it was and AM-M-K is taking away some votes. What is happening in the South is because of AM-M-K factors. This Sashikala took a vote against, that is she will teach a lesson against Edapadi. She feels betrayed by Edapadi and in a way, O Peppaneer Sivam. So now she has a personal reason to ensure the defeat of the BJP. And AM-M-K in southern districts where the ADM-K is strong, they are very clear. They made it very open that they are going to defeat of the ADM-K. Though the leader speaks about DM-K, they are on the ground. I think they are going to damage ADM-K in a big way. And what has contributed to this factors, this one year reservation. One year reservation 10% within MBC has actually reduced the space for reservation for this denotified communities like Mukulathur and other sections. It has been reduced to 2% so they are all angry with the ruling party for having considered 10.5% of the 20% to 1 years. So their reaction will benefit AM-M-K because AM-M-K's base is against Mukulathur. So that again is going to cut into ADM-K votes in community-wise also, not only politically, but community-wise. And DM-B-K which was part of the DM-K alliance last year, DM-B-K also walked out feeling insulted by ADM-K. And they are also very clear that they are not going to come to power. So they are going to work against, they also vow to defeat the ADM-K this time. So defeating the ADM-K with a very powerful combination is working in the southern parts of the country. That is what is happening. And Kamala Hassan. Kamala Hassan has some support in the urban areas, urban upper class. So Kamala Hassan's target is mainly DM-K. So what in my opinion will happen is Kamala Hassan will take away again urban votes which would have gone otherwise to ADM-K to be the so-called educated sections, upper class, upper class sections. One ADM-K vote again will erode. So my opinion is, my point is DM-K alliance is intact. There are so many factors which is eroding the ADM-K's vote base. They may lose about 10 to 15 persons, which means it's a complete wipeout. And one more factor, ADM-K made a blunder was in the manifesto. They promised to name Madurai Airport after Muthuramalinga Thever. Muthuramalinga Thever is the leader of Thevers. And Thevers and Devendra Pulavelala, they don't, you know, the traditional rival with their riots in the 1990s. Devendra Pulavelala, that is, that is a section of the list. So that has created a lot of resentment. So all these Mukulathore votes again will consolidate behind the ADM-K, which will damage the ADM-K. That is the scene. One more factor is, this Dalit party called Puriathamilagam, they wanted Devendra Pulavelala. They wanted seven castes to be brought under Devendra Pulavelala. They don't want to be called Dalits. So that, so now they are known as, you can't call them Pallas. Pallas are the major group. So Devendra Pulavelala, when you say call them Devendra Pulavelala, the real Vellalas, they are resenting it. They are not equal. So this cast factor is going to damage the ADM-K in the south. That is the real scene. In urban areas, Mukul Nalakutani, that is the Mukul Nidhi Mayam, will also cut into ADM-K votes. So in my opinion, because of these factors, ADM-K is going to be the real noser. Another major player, or they claim to be so, is the Barja Jansha party. So after allying with the ADM-K, in the general election, they are again aligned with the same party and are contesting in 20 seats. So even though their record is digital, their sister organizations or the Sangh Parivar organizations have been doing different kind of ground works in different parts of the state. Do you think that this ground works benefit the BJP in these 20 seats? No, not at all. See this ground work, it is happening. Actually what they are doing is, they are approaching temples. They first go to temple, they build some kind of organization around temple, they mobilize people, but it is too late in the day. And basically the problem, BJP's problem is, it is completely, it has managed to alienate people of Tamil Nadu with a lot of things, which they did, the kind of rhetoric they have been, they are talking about Hindi, Sanskrit. And interestingly, BJP's candidate list is in English and Hindi in Tamil Nadu, which is ridiculous, people are very angry about it. So BJP, whatever they do on the ground may have some impact in the long run. BJP's real chance or any party, any non-derivident parties chance may emerge only after this election, because this election reminds a kind of competition between the new parties. But what is happening is, AIDMK after Jail Litha's demise, actually it is highly divided. One, this Chief Minister and Deputy, actually the Chief Minister and Deputy, they don't get along, it's very clear and their differences are in the open, they try to, somehow the BJP is trying to keep them united, it's very clear. And AMMK, Amma Makkal Munnetra Kadagam, which is founded by Sashi Kala's nephew, and that party is also reminds outside this set of AIDMK set up. So what is happening is AIDMK is slowly, is on the wing, slowly it's going to lose its base, it's already losing its base, they are so divided. So in this situation, definitely this two-party system, that is duopoly, they call it, two-party dominance, definitely after this election, most probably if, what I think is AMK is going to win the election hands down, after that AIDMK party is facing a big existential crisis, there is no charismatic leader. One common complaint against local people, even AIDMK cadre, somehow they don't find these leaders very attractive, both the leaders, EPS and OPS, and they are highly divided also, and they are divided in caste lines, caste lines also, because the OPS belongs to this Mukulathore community, which is dominant in the south, and they also built a party like that, they built their bases among their caste groups. He's from, EPS is from Kongubelala community, which is dominant in the western region. So what, BJP or any party, I'm not looking at BJP's chance in Tamil Nadu, and there is a possibility of some space opening up. This is the first time after 67 AMK came to power, they have been alternately ruling their state. First time after this election, in the last next five or ten years, there may be, really there is going to be a space for, there will be a space, definitely. Who is going to take it? And BJP with its powers hold over the power, hold over power at the centre, it has been trying to take this space, and they are keeping completely, AIDMK leaders are completely, they're not asserting themselves, they're not in a mood to do it, because they want to be in power, they wanted to be there. AIDMK remained united only because four years after JLITAs did, they had power, that is why they remained united. Otherwise AIDMK would have split long back, or would have not disappearing, slowly the erosion is happening. So even BJP will be growing at the expense of AIDMK. BJP, you know, they claim that their, BJP has present in almost all the villages. Not all the villages, but it is happening. What is happening is they are recruiting local youth, unemployed youth, and they build, they bring them around, and they put them in temple committees, and they are given all powers. You do whatever you want, is what is happening on the ground. Whatever you want, you do it on the ground. In the name of temple, they could go for collections. So if there is some action, we'll take it, because our state power is with us, central power is also with us. So with that, they are trying to infiltrate and see what has happened is over a period, even the Dravidian movement. This because the dilution of Dravidian ideology is stark in AIDMK. Because even Karnanidhi, Mr. Karnanidhi till his last movement, he hold held on to it. In DMT, even now you'll find a lot of people talking about ideology, social justice. In AIDMK, you don't find this letter rhetoric at all. Jelitha, she ensured that 69% preservation. But what is happening is, there is no real ideological basis for AIDMKs. They don't talk about it at all. At least DMT, there is some talk of all the social justice language, all these issues, they keep it alive. In the case of AIDMK, they are completely, there is an ideological vacuum, that will... So it is easy to attract AIDMK, at least the leaders. One likely scenario is with the power, BJP is going to be power and power for some more at the center, it's going to be power for some more years. So with that, they can actually attract a lot of AIDMK leaders at the state and district levels. But on the ground, we don't know what is going to happen because... See, after Jelitha's... Actually, if you look at AIDMK's history, MGR formed the party. After that, after his demise, Jelitha continued and there was a split in the party, later she took hold but Jelitha ensured that MGR is also forgotten. She used to invoke MGR's name only at the time of elections. Now what has happened to Jelitha after her demise, that name doesn't invoke any emotion. Even now people in villages talk about MGR. The loyalty to MGR is still there. But Jelitha doesn't command any loyalty. Jelitha's loyalty, whatever loyalty is out of fear and out of the fact that only she can win elections, even all these leaders, they were loyal to her only for one reason. This loyalty is very superficial. They knew that only she can win elections. Now in the last two years, you don't hear much about Jelitha. They don't talk about it. Some they always talk about Amma. But even now, what is happening is if you go to Chennai Marina, you can see hundreds of people visiting MGR Samadhi and Jelitha Samadhi. But I don't think this ADMT cadre on the ground, they don't have any kind of great attachment to the party any longer. So this space is going to be important. So BJP also I think has a long-term plan. So with their power at the center, they will try to capture or they'll try to bring in district level and local leaders with a lot of other methods, not only you know how BJP uses agencies against top people. And with the lower level, you have you'll offer anything to bring in leaders. So the real danger or whatever in my opinion is this middle level leadership and top level leadership without any comms they may go and join BJP. So then there is a possibility for BJP. In this election, I don't think there is any scope for any. They can't even win a single seat. That is the ground situation.


To add an image to your project, tap on the image icon. This will open up the images tab. From here, you can choose an image from your photo library, or find one in our free stock libraries, Unsplash or Pixabay. Tap on the image you want, and it will add it to your project. From here, you can edit your photo using the image editing tools in the bottom menu. You can add a filter, adjust things like the saturation, contrast or exposure, you can also crop your image into a square, a circle or custom. To resize your image, simply use two fingers to pinch and adjust. You can also move your image around in your canvas. When you've completed an action and you want to move on, tap the tick. If you'd like to change the background color, tap on it, tap edit, tap color and choose a color. To add graphics and illustrations to your photo, tap the graphic icon. This will bring up the graphics library. Search for something you'd like, then tap on it to open it in your project. If you'd like to move the graphic behind the photo, tap on the layers icon. This will bring up your layers view. From here, you can move layers around and lock them. Now your graphic is behind the photo, but it's white, so we need to change the color. Simply tap on color, you can use the eyedropper or choose a color from the color wheel. Then pinch with two fingers to resize. When you're ready to save your design, tap the export button in the top right corner. From here, you can tap the yellow save button to save it to your library. And that's all it takes to turn your photos into stunning design.


Hey there, chameleons! It's Camille, and today I'm doing an ombre wig application and review video, so I hope you enjoy it! So to get started, I've French braided my hair. My hair is originally straight before this, and then I braided it very tightly to my head. I actually filmed a tutorial for this, and it'll be up on my channel soon, so don't forget to subscribe if you want to see that.  Now I'm using this wig. It's by Premier Lace Wigs. It's the Rihanna Style Ombre Medium Brown. It's linked in the description box down below, as well as all of the details about it. So it came in this plastic casing to keep it all together. There were instructions, there were a set of eyelashes, a necklace that you even threw in there, and two wig caps, which are awesome, because you need wig caps.  So you just want to remove that and let it stick to your arm with all that static electricity.  I love the curls so far. You can always change them, but the curls came pretty nicely. So first you want to take off the net part. This is just extra lace that's on any new wig, and it's just so you can sort of fix the hairline however you want it. And that's how you know it's a completely new wig. The inside of this one has clips on it, unlike the last one, which is awesome because it will actually stay on your head better without having to DIY clips onto it or bobby pin. And these are just really great for going around that wig cap. It also has an adjustable strap so it will fit people who have smaller heads, larger heads. You can adjust it to whatever your size is, and it's just very easy. So first I'm going to take out one of the bald caps, and it looks tiny, but it will fit over my whole head. You're also going to need some bobby pins to pin your braids, and I'm taking off the hair scrunchies because they're thick and you don't need that.  Doing the same thing on the other side, remove the hair scrunchie to get rid of any of the bulk, and then wrap it and tuck it underneath the other braid. It will be a lot less hair if you have short hair, but you can still do this with long hair. Just kind of make your hair as flat as possible. Now you have these little horns, but don't worry about that. That will all flatten under the bald cap, which looks like this.  So it looks more like your scalp when you have the actual wig on. Don't forget to tuck down any strays, and now you're looking like a straight up gangster girl. Ooh, yeah, rocking that, shaping my head soon because I look good.  Here's all that extra lace that you don't need, and I'm pretty much just going to cut it all off, keeping the same hairline, just because the hairline was fine for me, but if for some reason you want to change that, that's all up to you girl. But I'm just getting scissors and cutting as close to that hairline as possible, so there's no extra lace. Now I'm going to adjust the strap based on my head size, and just hook it right in there. Trying to do it without looking does not work, note to self.  You want to readjust it and kind of look for those clips because you want to use those around the bald cap, so you're going to find it, pull the clip forwards, tuck it under the bald cap, and tuck that into your actual hair, and then clamp down. So you just want to go right outside, tuck under, and then invert it so then it will stay attached to your head, so you can do flips and tricks, which you probably actually shouldn't with the wig, but you know. Anyways, now you can pull down some of your baby hairs, so it looks more natural, just kind of adjust those however you would do your babies, and you can pull some out the comb, you can weigh it down with some gel, whatever your thing is. Now you want to go in with some concealer foundation and a flat brush, and I'm using this to disguise the part, so it will look a lot more like skin once you start painting it on, covering some of that lace, and blending it in with your actual part. To get it looking even more natural, you want to go in with some tweezers and just pull out some of the hairs that are around the part, so it's not so perfect, because as Hannah Montana says, nobody's perfect, and even though she got away with wearing a wig and people thinking it was real, you won't if you don't adjust the part. No one has a perfectly straight, exactly thin part like this, so just roughen it up a little bit, but just don't over-tweez. I'm just gonna take out those extra strands that I just tweezed off, and get rid of it. Now you're pretty much done! Just adjust your wig, curl it, straighten it, cut it, whatever you want to do to it. I liked how it was curled. This is kind of scary for me, because now that I see myself with brown hair again, I kind of want it! Paradise tutorial coming soon! Maybe, probably not. I don't know! I hope you guys liked this video, I hope it was helpful or entertaining. Don't forget to subscribe to my channel for more videos like this. I have all my social media linked in the description box down below, including my Facebook page, which I think you should all like, so you can get updates on everything, including my new blog posts. I'm very active on Twitter and Instagram. Pretty much I just want to talk to you guys all day, so don't forget to check that out. But thank you Chameleons so much for watching! I wish you all a happy, healthy life, and I will see you in the next video. Kisses!


Hello, everyone. I'm Parvani Kripati and I'm here to share with you all my first experience with open source. So last year I got selected for Google Summer of Code program and in that program my aim was to develop a RISC-5 processer model for the RTC community. So let me first introduce you all with the RTC community. It is an open source organization developed by the students and professors of University of Campinas, Brazil. So the aim of this platform is to enable people, researchers and students to develop their own processer models for developing their own prototypes and test their performance on this platform. Going on to the RISC-5 architecture which I implemented, this is an only open source instruction site architecture designed at University of Berkeley and it is very advantages in the terms of when it comes to speed, when it comes to design area and the cost. Now why RISC-5? So the main advantage of RISC-5 is that it is the only open source instruction site architecture. This allows you to customize it, to develop it and according to your needs and then add features accordingly. The base model of this, the base model is the instruction site, is the integer base model which is a very basic simple model and it just has around 40 instructions. So many designers say that when compared to other commercial architectures, the CPU is designed using this RISC-5 architecture is very fast and consumes very less area. One of the reasons is that the position of the most significant bit is actually fixed and this speeds up the sign extension process, thus making the whole system a lot faster. Now for what I contributed during this project was that the ArchC community have this multi-processor system on chip design platform on which you can design your own MPSOCs and test its performance. So by adding the RISC-5 processor model to this MPSOC bench, I was able to improve its analyzing ability and also the verification ability and you one can add this RISC-5 processor model to your own design and then test its performance and check whether this design is optimal or not. So I started off by implementing the basic integer instruction set and then further extensions for single precision floating point and double precision floating point were added. The major challenge that I faced during this project was that back then the RISC-5 architecture ISA toolchain only supported 64-bit format whereas this ArchC platform only supports 32-bit format. So with the help of Dario, another member of the ArchC community, I was able to develop the 32-bit toolchain and this obviously managed to complete my project. So this was the brief overview of what I did and if anyone of you is interested in working in the field of computer architecture, you guys can come and talk to me during the coffee or lunch break or if you guys have any questions right now, you can ask me. Thank you.


So thank you very much and thank you Maxime for existing so Yeah Important words. I was the thesis supervisor of Maxime I was just in Moscow a couple of weeks ago and met a certain person to my mention she mentioned Konsevich and I said yes, I know I'm in fact I was his thesis supervisor ha ha she was telling every minute this guy thinks he was the advisor of Maxime Konsevich So it's true that for instance, we have another man in the front row who much more reasonably considers that he was the Advisor, but he was it. He was only the teacher. He only taught Maxime mathematics But Maxime's PhD was not obtained in Moscow But in bond and I was his thesis advisor and I even Suggested the problem not that I understood it Even the problem let alone the answer But this was in 1991 when Maxime had just arrived fairly we I think 1991 He had arrived fairly recently in bond. I had met him before in Moscow in 87 So he was already a lot older than when you first knew him But he was still extremely young and of course I was completely overwhelmed and read if my memory is right We invited you to bond once before and then you came later as the second visit at the second time not as a visitor and Well, professor Hertzberg said to him, you know, if you don't have a PhD we can only pay you like a student So you need a PhD. He didn't have I mean not the doctor. He didn't have even the candidate He didn't have a PhD so he needed a PhD and so he needed a supervisor So I was assigned and then the abides talking happened and I tear gave the as always the opening talk No, no, it wasn't before No, it was 90. That was 90. Yeah, I wasn't sure which year and So I tear gave as always the opening talk and he explained Witten's conjecture and at the end I brought you know, it's a long row. I brought Maxime down to introduce to a tier who had been my first thesis supervisor Said Mike, I would like to introduce you to my to my student Maxime who will probably be I imagine will be able to prove The conjecture you just told us and so I tear looked at me like he's completely mad.  I mean, we're not talking high school mathematics here And so he looked at me like, you know, I've known you a long time job, but now you've really gone Over the tent and then I think two days later There's the famous boat trip on the Rhine part of the Arbide Stuggan and the entire my memory at least the entire three Hours going and the entire hour and a half coming back because you know, it's downhill it goes much faster I tear and and Maxime were together discussing and Maxime showed the proof. Well, it wasn't complete I think it was not completely finished till the end of the week But the idea of the proof in two days after hearing of the conjecture he had Solved it essentially and of course it was considered sufficient in bond for a PhD and It was considered sufficient by the IMU for a Fields Medal and so Strangely enough It's a it's a long thesis. Yeah, I mean But it was yeah 25 pages He hadn't yet read Kacna napisad, yeah, I'm at my teacher's to pa English key So he didn't know how to make it shorter now. He would write it in three pages and nobody would understand So I've been very disappointed since then sorry Well, that's another good if I start telling about the qualifying exam. I'll never get my talk But so I'll just say that I've been very disappointed. I haven't had a single student since who was that good? And It's not going to happen Okay, so he's also a wonderful friend and a wonderful person and I'm not going to say anything more, but the thoughts are there So I'm supposed to give a talk which means I have to take Roll up my sleeves so the title of my talk had three ingredients One of them I understand more or less which is partitions and One is quasi-multi-reforms, which I understand very well and One is Siegel Beach constants, which I don't really understand at all But since that's the part that led to this work The only part that's specifically connected with something Maxim is done I decided to have to include it But luckily we had a beautiful explanation yesterday by Anton Zorich is Still there and so I can skip trying to explain things that I only have understand So since that was explained I won't say anything But I won't say something first about the first two ingredients and then how it all comes together So the last part this is essentially going to be about well first of all Some old work, but in particular something I did many years ago with Masanova Kaneko and then a wonderful generalization which my main point is to Popularize because maybe some of you don't know it and also that's a beautiful theorem of Bloch and Okunkov Of a few years ago, and I found a very very simple proof So I'll be able to essentially give you a complete proof of their theorem and this part is joint work well with Dabe Chen and Martin Müller Except that I actually have I'm not even sure I've ever met Chen the joint work is Chen Müller and Müller and me There's a part on Siegel Beach and a part on modern reforms and you need them both But they're disjoint and the common link is Müller and just for people who were at the talk yesterday and saw the very unusual Spelling of Muller. This is the standard one the one that we saw on the slide yesterday is based I think on the French champagne Mouet and Chandon, but it's it's just it's too much Okay so Okay, so I want to start with partitions because they're certainly the easiest whoops this thing doesn't go up as high as one expects So a partition is a partition of a number and the notation will be lambda is in N And I'll use P the script P for the set of all partitions So I won't specify the end and you could put Pn and the usual rate wave writing and partition Would be lambda one lambda two and you can either stop at the last positive one or you can just take a sequence of integers Decreasing to zero and only find out the many and then to say that it's a partition of N means that the sum of the lambda I is N So that's what a partition is and of course when you do You know what I'm always Saying but people always laugh that I always say that the only person who really lives by this precept apart from me is Maxime whenever you see numbers in life a sequence of numbers and it happens Then you should always make the generating function and take it seriously and it almost always helps you sometimes it doesn't So here if you take P of N to be the number of partitions Of a given number N then of course what you should do is make the partition function There's the generating function and from zero to infinity P of N q to the N And so this was noticed already very long ago by Euler that this is equal to Well, I'll have a notation that I'll use sometimes. It's convenient Just so I don't have to write so much the infinite product 1 minus q times 1 minus q squared and so on Which is called the infinite q pochamber symbol sometimes This is just for convenience that I write that and Euler It's very trivial of course to prove that now Euler notice this fact But Euler was then interested in computing the partitions Let's say up to 50 and so to do that he said well if I can compute the Generating function of the reciprocal so q infinity itself this product then that should be easier And then I can always invert a power series and so he found well this famous That this is the sum where every coefficient of q to the n is zero most of them are zero and otherwise plus or minus 1 So this is the sum Minus 1 to the n q to the 3n squared plus n over 2 and then if you know from high school that when you see a quadratic of Function you should complete it to a square to a perfect square then you immediately see that actually the natural function Is the product of this thing with q to the 124th because then you've completed the square So it now becomes 6n plus 1 squared About multiple forms you know that if you have Q to a perfect square and you sum over a lattice Maybe with some congruence conditions like n is congruent to either 1 or 7 multiple or 12 And you take the alternating sign this thing is a modular form And of course it has a name It's called Eta and so this is a modular form in this case of weight a half and it's the dedicated Eta function and Again, I assume that people do know what a modular form is although very briefly remind you So you take a variable tau which should now be in the upper half plane This is meant to be a German age So the upper half plane is the complex numbers with positive imaginary part and then one always writes so this I won't repeat Q is e to the 2 pi tau and therefore if tau is in the upper half plane q is in the open unit disk and then to say that it's a modular form in general means that If you take any matrix a b cd and sl 2z then well here there's a 24th root of unity So there's a little factor But you have cd plus d to a certain power called the weight which here is a half times 8 of town This is what what Dedekind discovered so in other words if we start with partitions We immediately see that we're led to at least classical modular forms which have this transformation law so now Let me introduce the second ingredient of my title before putting them together at all So that's quasi modular forms. Maybe I'll go here So modular forms if I take m star of SL 2z so all modular forms on SL 2z I won't write the group each time to m star then it's very well known That there are Eisenstein series e4 e6 e8 and so on which are modular forms of that weight and It's very convenient to introduce Short-hand notations due to Ramanjan actually there's also a an e2, but it's not a modular forms It's not in this ring. So we have Eisenstein series pq and R. I won't define the general one But I can define these three since it's so easy p is 1 minus the sum from 1 to infinity n times q to the n over 1 minus q to the n the q is 1 plus 240 times the same sum with n cubed this is the same and R is 1 minus 504 Times the same sum with n5 and then the remake the remarkable thing is that these functions q and r are Multidiforms of weight 4 and 6 so they transform exactly like this now There's no funny factor just with c tau plus d to the 4 and c tau plus d to the 6th And of course It's very well known that you have this relation and the 8th that I had before there are two relations With eta and this one of which is that the 24th power of eta is q q minus r squared Well divided by 1728 and also that the derivative of Eta divided by 8 so the derivative is just the derivative But it's very convenient to divide by 2 pi i that we don't keep getting formulas full of 2 pi i I want to work over q and so this in terms of the q and it's just q d by dq and then you see very easily that the Derivative d eta by eta is exactly 124th times e2 which I'm calling p so you already have a connection between what I told you here Which is that this dedicated eta function which is a multidiform is related to the multidiforms q and r but there is a bigger ring called Quasi-multidiforms the word was introduced in fact in this paper of Canneco and myself that I mentioned about 10 years ago Quasi-multidiforms you can define it for a group of I Won't bother with but the answer here is it happens to be a free algebra on three generators, which are exactly this pq and r So since that's not my main object. Let's just take this as a definition. It's certainly included because you check Thank you I for me I want to work over q all the time just to keep life simple Thank you. So now you have something very nice. We just had this derivative and you see that the derivative of eta Over well 8 isn't in this ring, but 8 of the 24th isn't that would just multiply this logarithmic drift by 24 so you see that the logarithm the the derivative d of of 8 of the 24th is in this ring but in fact D acts on this ring and Let me tell you so So there are the famous formulas very easy to prove that Ramanujan found and just for completeness I'll write them on the board So the derivative of p is p squared minus q over 12 the derivative of q is Pq minus r over 3 and the derivative of r is Pr Minus q squared over 2 so since it's a derivation. It's enough to do the generators So it acts on the ring and the last comment I want to make in this generality about quasi-multi-reforms is that we have three operators So D sends f to as I said qd by dq of f and this will be in mk plus in mk plus 2 if Let's say that I've an f in mk tilde so quasi-multi-reform weight k the derivative will be a quasi-multi-reform of weight K plus 2 then I've the weight operator, which is the Hamiltonian if you prefer it just multiplies A form by its weight so of course it doesn't change the weight But I've a third operator This is meant to be a German D for those of you who like the Zutelin script and this is Well, it's the derivative 12 times the derivative with respect to e2 of f so this will be a quasi-multi-form of weight k minus 2 and The derivative means if you in this definition, there's no problem since I haven't defined f In an intrinsic way is a function of tau you can do it by transformation properly, but simply as a polynomial in modular forms In a polynomial in p with coefficients that are modular if you do that you just differentiated in so the d by dp I've been using so the reason I want these three operators. Well, they'll come later So trivially the commutator of W and D is 2d because W is the weight and D raises the weight by 2 and for the same reason the commutator with D is minus 2d and then finally the commutator of D and D is W and so this is an SL2 which is acting on quasi-multi-form So that's kind of and the primitive part the kernel of German D is exactly the mutli-form So that's how to see mutli-forms in quasi-multi-form So we'll see explicit examples in a minute. So now I want to say what this has to do Well how these things first came together at least for me So the notion of quasi-multi-form it came from the physicists Well several also wrote but the main person was Degraff who had the idea many years ago of looking at mirror symmetry in Dimension one so mirror symmetry which heard of many lectures here you map curves holomorphic curves complex curves into an algebraic variety But you fix the homology class of that that the fundamental has is mapped onto and you count them in some suitable way and In string theory and in other parts one especially wants to do this for Calabria's so you want to count the curve sitting in the Calabria Oh, and so Calabria in dimension one mirror symmetry in dimension one Would Would be that I'm well you can't really speak of putting curves inside a curve if there's not room But you can still map a curve of genus G to a collab to a curve of which is a Calabria, but that means that it's A torus it's genus one because that's a one-dimensional Calabria And now saying that you fix where the fundamental class goes means that you fix the degree So the degree is some fixed number n and so you can ask the question How many coverings are there? Let's say top logical because then you can pull back the complex structure How many coverings are there? appropriately counted so we have sort of n of Well, I won't give it a name because I'm going to introduce an invitation a second It'll be confusing now. We want to count how many Coverings And it's usually you count things up to isomorphism when you count with the multiplicity Which is one of the number of automorphisms I don't want to explain it exactly but this will be certainly a rational number It's it'll be a finite some if you do it right I'll come in a second But I want to specify the ramification and the ramification is this I assume that it's generic ramification So that means that any critical if I have a critical value down here Then there's a unique critical point above it So at any point the pre-image is either n sheets or it's n minus one and exactly two cross Okay, so that's the That's the nature of the ramification and so we want to count this number and we want to make a generating function So let me define fg Of q to be the sum over all such well I take n from one to infinity of course n can't be one if the genus is not one because you can't have Map degree one so the first coefficient will be zero It's this number number of coverings with all the things I said Generic branching and countered with one of all the morphisms and then again You make a generating function q to the degree of the covering and the prediction of in predictor digrav With that this one was supposed to be a multiple form a quasi multiple form and in fact of Of weight 16 minus 6 and there are two parts one you have to find a way to compute it What those numbers are you don't want to you know go to your computer and say please draw pictures of Riemann surfaces, so you need a combinatorial description that comes from group theory and he gave and then he had also a proof of this proof in quotes using ideas from Sort of quantum field theory and path integrals, but it wasn't a proof in any way. This is Certainly a theorem and an explicit concrete proof was given in this very short paper by Kaneko myself Which appeared in the same volume as daycrafts paper where he explains the the whole thing So this was I don't know 10 or 12 years ago So this is the theorem. Let me give you an example just so that you see something concrete So G equals one this and actually this is only of G is at least two for G equals one There's a slight exception because of your you know Taurus is an infinite group of Voldemort's by translation and it somehow messes up the counting. So for instance F2 explicitly is 10 p cubed it has to be of weight six because G is six a G is two and in weight six we have only three a base of three elements p cube pq and I Have no idea why I wrote this this way five p cube minus three pq plus two r divided by five thousand It's not five thousand, but it's fifty one thousand eight hundred and forty up to a constant There's a this the second line has a constant term the line before it doesn't have a constant I've no idea what you mean There aren't the q to the zero Well, okay, the the coefficient the coefficient of q to the zero is Equal to the coefficient of q to the one and they're both equal to zero So you can start with zero one or two those coefficients are zero It certainly does oh Thank you. That's all I was trying to say. Thank you very much. We have somebody in the audience who can subtract five minus three Plus two. Thank you very much. I thought you were complaining about the first slide The second line it's much better What's more you're even right and then you know with what I had there would have been horrible denominators But this thing actually is integer coefficients, which wouldn't have been true at all for the other But that's an accident in no other genus do you get integers? You always get a bounded denominator Okay, so that's the the theorem and This the theorem of well as a formal theorem of kanei koni with the formal proof But the it was the discovery of daycraft and also if the physicist rod as well, I should maybe also written Now comes the amazing General but let me say the combinatorics as I said there are two steps. So two steps one Give a formula and a computable formula that you can put on the computer so to speak for FG And then which the daycraft had already done and the other is then show the quality modularity So let me tell you the first part first and I go back to partitions If you have actually for any finite group, but let's now think of SN So this is meant to be a German S as you all know Irreducible representation of SN correspond to young diagrams and therefore also to so I remind you that any partition For instance the partition 3 plus 1 can be represented pictorially by a young diagram or young tableau or whatever it's called so you've N squares which On the upper left hand corner the way that I do it if the lower right hand quadrant So that corresponds to unique irreducible representation and now let me define an Invariant new T of lambda you could do this where T is any currency class in SN You could do it for any finite group. It's not it's proportional to if you take the pie If you have any element of the group For any finite group if you have an irreducible Representation and a conjugacy class you of course have the character on the well-known character table by evaluating well by taking the trace Pi of T or pi or whatever the element is, but you have something else which is the action You take all elements So this is really the context class, but I didn't write you take all elements in the symmetric group But you're in the context class of here It's the trans the transposition so T is the transposition you interchange two of the two numbers one and two The currency class consists of all transpositions of I and J and you see that that's relevant for our problem Because here if I do monogamy I number these points and then when you go around and come back You've interchanged two for the monogamy of the of the singularity you went around so it's reasonable that this plays a role But now that's why we need T But for any element of the group you take for any consequence the class of the group you take the sum and Then these some of the corresponding matrices is a scalar by Shor's lemma because it's irreducible Representation it's a number and so this is as an integer and it's actually this is a very trivial formula That these two numbers are proportional you have to multiply it's true for any group and any come to see class So if I've a finite group of come to class in irreducible representation Then chi p of the come to see class is equal. Well, I'll do it the other way new See the way I'm writing it of pi is equal to chi p of side of multiplied by the size of the come to the class And divided by chi p of one which is the dimension of pi to completely trivial Just take the trace of this element the trace of a scalar is the scalar multiplied by the dimension of the space But the trace of a sum is the sum of the trace that they all have the same trace You just get the trace multiplied by the size of the come to the class But these these are integers and these are inches, but the ones I want our new C of p So the answer is let me give you a little picture. So if I take n equals 4 Then you have the following diagrams and The value this is the picture of new and then the value of this sorry, this is the picture lambda Which corresponds to a pi and then the new t of lambda will be 6 to 0 minus 2 and minus 6 it's it's easy to see that it'll change sign When you flip the young diagram, which means the contra gradient representation So now we associate to this now a polynomial h4 of u which will in this particular case be U to the sixth plus u squared plus 1 plus u to the minus 2 plus u to the minus 6 h n of u is just the sum over all partitions of n of U to the power of this funny invariant you what am I calling it new t of lambda? Okay, so that's That's that And then what you show is the following if you take Hn now, I hope I can do this You some over all partitions which again you shouldn't relieve and ever put the end you just some right from the beginning I could have just said you take U to the power new t of lambda times Q to the size of The partitions so if you take this sum then this is a power series in Q of course But also an x and this will exactly correspond the coefficient of Q to the n times X to the k or x the k maybe over k factorial if I remember correctly will exactly correspond to coverings n-fold coverings generically branched as I said n-fold coverings Of t2 with k branch points downstairs Now if I have a Riemann surface of genus g then the and here's a genus, you know Taurus Then you if you find trivially that because the other characteristic downstairs is zero that in the Riemann-Horwitz formula n doesn't play any rule and the k the number of so I've k branch points Then k will be independent of n. It'll simply be 2g-2. That's because downstairs We have other characteristic zero so I could have used k instead of g But the point is these coverings may not be connected And so if that that will correspond to that now if you take the log of this then this will correspond to connected Coverings and so that's the generating function we want but of course taking the log won't change the quasi-modularity So actually in the proof one shows that the coefficients of this thing are quasi-modern forms And then when you take the log rather still quasi-modern forms the log for power series has coefficients in the same ring as the original power series So this is the way that you compute it. So I hope this is clear. So you write down this sum I wrote down explicitly how to define This so new t and an example for n equals 4 how it looks you take these polynomials Then the co you take the log of this power series and now the coefficient of q to the n The x to the k over k factorial or k is 2g-2 will be exactly the rational number. We're talking about So now comes the theorem of block and Okonkov So first let me introduce an abstract ring. This is not quite also the notations are not theirs so if you know their paper you still have to listen unfortunately because I've changed every single notation slightly these qk are related to their pk But shifted by one and divided by factorial, but these are more convenient that they don't have a q1 You'll see in a moment why they might have felt they didn't need a q1, but it's very very useful So let R be the abstract polynomial algebra in if the many variables where qk has weight k So of course the dimension of this space in a given weight k is simply the number of partitions of k and There's a nice fox base representation I won't I'm not going to talk about that at all, but it's implicit in what they but explicit But they didn't implicit in what I'll tell and now this will act on I mean if I have an f in R and a partition of some number then I can evaluate f of lambda So what I'm saying is that there exists specific functions q1 of lambda q2 of lambda q3 of lambda and it continues But the board you know the margin is too small to contain it So we have a series of specific functions and they're very explicit q1 of lambdas function You all very well. It's the function that assigns to every partition zero That's why block and a kunkoff didn't include q1 But I wanted my abstract ring the q2 of lambda is the size of the partition So the simplest imaginable invariant is that it doesn't work unless you correct it by subtracting a 24th and q3 of lambda is Exactly the new t of lambda that I told you Okay, so q3 is the function that we needed and so their generalization will say that I mean we knew I knew and in fact It's written in their paper that I suggested that such a form might be true But I'd forgotten and was trying to wonder how they could possibly have discovered this thing But anyway, there's I knew that there were higher functions But I saw no way to prove the cause of modularity So the statement is going to be that not just if you do this procedure that I just did with If you multiply this out, then you'll find that in this series, of course since this is the sum What was it e to the u but I've replaced e by u by e to the x So this becomes e to the new t of lambda x and so the coefficient of lambda to the k is of course simply the sum over all partitions of The very special function u t to the n to the k Which will be an even integer later 2g minus 2 times sorry new t of lambda times q to the lambda Or yeah, it says sorry. It's it's it's essentially that So the the thing we're talking about is this and so they had the idea given that new t is just The third of a list of infinitely many interesting functions Why not take any polynomial in these special functions which I'll define in a minute And then their theorem is going to be that you'll always get a lot of reform But let me first introduce the notation I want So I define the q bracket of a function Let's say that I have any function on on the set of all partitions as I say I'll always work over q just because everything All the all the examples are rational So if you have any function arbitrary no conversions, then the q bracket is the q average You could I recall the q bracket But maybe q average is better and you do what you always do in statistical physics in Statistical physics you have a large ensembly you have a lot of states and you have an observable Which in this case is f so the states would be partitions and you've an observable And you want to know the expectation value same quantum theory So what will you get when you evaluate f of lambda? So what would you like to do is take f of lambda the sum of all partitions and divide by the sum of one But of course that doesn't make sense both sums are infinite. So what you do in statistical physics, of course You put in a factor In statistical physics it would be e to the minus one. I forget where the case. I think downstairs one Sorry the energy sorry It would be e to the minus one over Boltzmann's constant times the temperature And then you would take if the state is energy a to the well then you would wait with with that power So you typically have a constant q which sort of depends on temperature and you wait So here we'll have a constant q which is just called q but you can think of it as you know as q gets bigger and smaller It's it's getting hotter and colder and so this is the definition and then of course downstairs you do the same So this is the q average But of course what it is is a q series. I mean a power series. There are a lot of queues That's a capital q for rational numbers. It's a power series in q So we associate to every function and it's a formal power series I don't care about convergence at all although actually all the ones I write down are convergent for q less than one Because they're in fact going to be modular. Okay, so this is the definition But now if you remember that this denominator, that's what I started with if you just sum over all partitions q to the Size of lambda. That's the same as p of nqn. So that's this a to so I could also write this thing as I take the sum over all lambda in P and now I simply take f of lambda q to the lambda and now I have to divide by this denominator But the denominator is the reciprocal of this I have to multiply by eta Which is of course a multiform But then a to remember was not quite right. There was a q to the 124 So I have to actually wait now with q to the lambda minus 24 You see that's nice because that lambda minus 24th is exactly our q2 So the result I was talking about before would be the special case q3 of k a q3 of lambda To some power which is 2g minus 2 let's say well I pretend you didn't have to take the log and then q to the power q2 of lambda, okay, so it's completely in terms of this ring and so now I can give the theorem of blog and Okonkov, which is I don't know 2003 if I remember correctly anyway about ten years old And it says that this cube bracket is a model or a quasi-modular form For every f in this ring which they call the ring of shifted symmetric polynomials so that's an absolutely wonderful theorem and But it turns out to also have a wonderfully simple proof when did I start when should I stop and so on Of course the remarks about Maxim don't count, right? That's Sorry Left or since the beginning. Aye. Aye. Aye. It's all about positive and negative things. Well then We know I really don't want to skip my sketch of the proof because it's really a beautiful proof very very very Simple, let me say briefly with these q3 of lambda r. So you start with the partition Let's take a very simple example and You are going to Let me say how I define qk of lambda Qk of lambda well you take what are called Frobenius coordinates. You take the longest diagonal In this young diagram. So in more complicated. It's the as far as you can go down the diagonal It's the here's index if you think of this as a researcher who has one article with three citations Maxim is more But if you have one article with three citations and two with one citation, that's all that you ever wrote then absolute value of lamb is the number you have citations and And here's index is you know the smallest I for which your I've papers I citations in the literature That would be this diagonal so in general if I do more complicated Thing you know then you would have You would have this diagonal and so that divides up your the young diagram into this piece this piece And then the pieces below so you'd have lengths here, which would be seven halves One half and then you would also have minus a half in this case and minus five halves So you would associate to see a finite set of This will be contained and I'll call it F for fermions, which is z-shifted by half Okay, and then this finite set well it'll have equally many it'll be a balanced fermionic set so This it's balanced in the sense that it's a finite subset It's equally many positive and negative things because that's the length of this diagonal and of course the sum So if C lambda if I take the sum of The absolute value of C is in the C lambda then that's obviously the total area of these young diagrams So that's the size for partition, but now the Q K is the sum of This shift and I think it's Q to the K minus one that's right So I have to take C to the K minus one, but I have to put a sign Sign of C times C to the K minus one so you see if K is two Then sign of C times C is just the absolute value of C So I get this but then you have to divide by K minus one factorial and you have to subtract a constant Which is the coefficient? The kth coefficient of x over two over cinch x over two or maybe sign So I've forgotten the sign so in particular if K is two you see this is one This is sign of C times C So it's the absolute value still get lambda and the thing you have to add is minus a 24th And so that's the the official definition of Q K of lambda, okay? So that's now of defined to you So at least you now know the statement of the theorem because that's a complete Definition but of course it looks very artificial and there's a much nicer way of doing it, which is also in their Paper this one doesn't work this one works The way that you do it is you're going to associate to lambda the set X lambda which will be simply you take the elements of your partition and you shift them all down That's why it's called shifted symmetric you shift The jth entry down by j minus a half Okay, and so this is going to be in a set that I'll call x fermionic Zero which is a set a subset of x fermionic. So these are Sets in in the fermions finite sets Sorry, not elements, but finite sets such that it's bounded above and co-bounded below So if you can think of this like D rocks C of electrons You start with the vacuum where all of the negative states minus a half minus three halves and so on are all occupied and Then you have a new state Which is not the vacuum state where some of the positive energy states so the energy here is a half integer And some of the positive states are occupied but only find at the many and and there are some holes on the left hand part and Balanced would be that it's balanced in the same sense that we just had that there are as many holes on the left So if I if I make one hole by take this point and move it here So you apply some operator like that then you'll get a balanced set and that's X of zero where there are as many Post-developments as negative holes and so in fact you can see very easily that XF lambda plus is C lambda plus and X lambda minus is all negative fermions minus C lambda minus so that's why you know, that's a correspondence. So now if you do this Oh, I was lucky and I need that dangerous implement Where is it? Okay So I'll pull this thing down Down so Okay, so now what I do is I associate to any X in particular X lambda I associate to X a Formal parses, but well, it's actually half power series just t to the X So this will be so in my example. It would be t to the now of Covered up the example also for you. So dt to the Where's it gone? It was seven halves I think was the biggest t to the seventh half plus t to the one half plus now t But now it's not the the things that you see but remember it was the holes in this case So in this particular case it would be well It's the lambda j minus a half and in fact the next one would be t to the minus Three halves plus t to the minus five halves and so on so in our example And you see that after a while since you've everything this whole thing will be it'll have a la romp part Shifted by t to the one half plus and then the in detail after a certain point will be the same as it would be for the vacuum Which is this so although this is an infinite sum it becomes a rational function And so then I can define now capital W x let's say z is W little x of e to the z and once I put in this rational function that makes sense and this will be Exactly the sum k from zero to infinity q k of lambda Well q k of x, but then q k of lambda Will be q k of the associated x lambda in the way that I just defined and then here It's simply z to the k minus one you see that the Q zero for any set x will always be one because there it's always a finite set plus this thing Which in the new language is z over two over sims z over two and of course that has a simple pole with rescue one So the the one over z term will always be one So for any such bounded set, but q one of lambda q one Of x if you do the calculation will be the excess of the number of positive Occupied things minus negative unoccupied so q one of x equals zero if and only if X is in this set for amionics So as I said, they didn't use the q one because on the partitions which correspond to the balanced ones You don't see it, but it's actually there Okay, and now I'm going to give you a Nice theorem which and I'll try to sketch the proof because it's so Once you've set everything up It's it's literally one line and then that theorem very very easily like in two or three lines that I might skip implies the on the block of kunkhoff theorem, so that's The next thing I want to show you but I want to get the formulas right so I want to first define the derivation From this ring to itself, which is the derivation that sends since it's a derivation. I'll have to say what it does on generators It sends qk to the previous one and of course q zero. I should say it's one It's not one of the generators and obviously q one goes to zero Okay, so that's the generator and I also need the classical theta series Which depends on q which is the sum over all fermions which remember just half integers of minus one to the integer part of new Times uh e to the power new z times q to the power new square root over two So this is the Jacobi theta function, okay, and it's very very easy to see It's well known. I'm Jacobi found that the this is an odd function. That's trivial and Because it's an odd function. It has a power series with odd powers And so I can pull out the derivative, but that's what Jacobi found by his triple protoform That's the cube of eta is the constant the leading term and then it'll be one plus h two Z squared plus h four z to the fourth It's a power series expansion and each hk I could write down a couple is Quasi-modular form of weight k That's the only thing I'll actually need about multiple forms is that this and this is a one-line proof Maybe since it's one line. I'll write down the line and then you have it Well, you and you have to prove the proof which is another one line But the the thing that you do in one line is that 4n times n plus 1 times hn if n is at least 2 Is 8 times the derivative of hn minus 2 Plus p remember our p the special cost model form times hn minus 2 That's a very very simple formula to prove and you start with h 0 is 1 and so since I already told you the Differentiation which raises the weight by 2 and preserves quasi-modularity these h's are completely explicit quasi-modular forms So those are my two ingredients is this derivation and the theta series and now the theorem is the following That for all f which are in the sub ring of q Which is actually the sub ring they used where now I don't include our one so the sub ring of Sorry, it's a polynomial So this is the sub ring of r where q 1 does not appear Then you have the following statement That theta you see this is a power series And so you can apply a power series to a derivation and apply it to an element and that makes sense because since this Derivation reduces the degree by one on any polynomial large powers of d will kill it And so I can evaluate a whole power series and the claim is That the cube bracket is simply zero So now if you think what it means the cube bracket I can divide by 8 of tau cubed and by z if I want not by z So this tells me that if f is d of g Where g is in the sub ring Q 2 and so on then if I take this expansion then I'll find that zero is equal to the cube bracket of f plus h 2 times the cube bracket of d squared f plus dot dot dot and And so by induction since this thing is lower weight I've already shown that it's quality modeler h 2 is quality modeler So by induction this gives the quality modernity and the only slight problems They're not every f is in the image But it's very easy to show that every f of positive weight is the sum of d of something in this sub ring plus something Which is divisible by q 1 and for that the cube bracket is zero So in one line once you have this theorem you you find an inductive formula It's completely programmable. You can make a table that you can actually not only prove but also evaluate the cube brackets You prove the multilayer didn't evaluate the cube brackets as Polynomes and pq and r directly you never have to write out their q expansions as we did for the first two years for Computations with huge sums over over partitions And then we could only get 40 terms of the series because there are a lot of partitions when you go beyond 40 or 50 But now it's completely algebraic So let me sketch the proof of this theorem and then I want to say a little bit about the Siegel Beach part, but I'll just Okay, so the the idea is this if you have a set x in Xf and you also have a number in f which remember was a half integer then I can simply translate x You just shove it let's say this is three halves you translate by three halves And this will be this xz which is defined exactly the same way as xf was which is probably disappeared So these are subsets now the integers which are bounded above and co bounded below in other words their complement is bounded below So you have this rather trivial change and so now I claim Well every element x and z so here you have an x which is a subset of z Which is bounded above unbounded below then it has a shift the number of positive things I mean the it turns out that there's a unique new that you can shift it by So under this map, actually I want to write it the other way I think I can say given any fermionic number new so half integer and any partition Lambda then I simply map new lambda to the set x lambda that we had before which was one of these things for the Fermions and then you shift it by new and that no longer will satisfy the boundedness condition But given this x then q1 of x. It's very easy to see is going to be exactly new And therefore if you give me an x in this set you evaluate q1 And that tells you what new you need and that tells you which new to subtract and then when you subtract it You get a bounce fermionic set and you interpret that as a young diagram And so you have this very nice projection Which actually is certainly known and is used in other forms in particular in connection with the free fermion free fermionic field But I don't want to talk about that and it plays no role. So now under this correspondence, you have a very simple lemma So the lemma is simply this if I have again an x in this set of Subsets bounded above and co-bounded below of xz and here I've a pair consisting of a half integer and a partition Then and if I have any f in in the block of kunkhoff ring Then if I want to evaluate f of x, it's very easy I take e to the power this new this half integer times my derivation and apply to f That remember was a well-defined thing and then I pull it well It's that for lambda, but that means on x lambda. Well, sorry the way I've done it just on lambda Okay, let me prove this because it's really completely stupid since x is equal to new plus x lambda That means that w small w x of t Which remember was the sum over all x in my set bounded above and co-bounded below of t to the x Well, you've just shifted so this is obviously just t to the w times the corresponding thing for the partition That's completely obvious So that means that when I go to the capital W, which was the generating function remember qk of lambdas qk in this case of x z to the k minus 1 This was the same thing where you change T to e to the z. Well, then this will be e to the new z times W well times the same thing Qk of lambda z to the k minus 1 so now if you compare then you see that qk of lambda a Qk of x will be qk of the partition I won't write the lambda plus nu times qk minus 1 plus nu squared over 2 times qk minus 2 and so on but then remember our derivation d sends qk to qk minus 1 qk minus 1 qk minus 2 so this is that's this of in the case of qk But then if you have a derivation then the exponential of it is an automorphism and the qk are ring generators So you get this formula for free and in particular you have two Special cases that already well one of them already mentioned q1 of x is the nu and q2 of x is the q2 of the partition Which remember was the size of the partition minus the 24th There's no contribution from q1 because it's 0 for partitions But then there's a contribution from q0 so nu squared over 2 so you have these two formulas And now let me prove the theorem which is above So now proof of the theorem so I'm going to write I Want to take the q bracket of e of theta of this derivation applied to F F remember within the subring not involving q1 I want the q bracket of that and I might as well Divide it by 8 of tell because I want to show it 0 so it's equally 0 if I divide by 8 of tell so by what I've already told you that's the sum of all over all partitions of q to the power the size of the lambda minus the partition But now I have to insert theta of Delta right so therefore I have to put here theta of Delta which by the definition of the theta series which is still there is the sum So this is all partitions. This is all fermionic numbers half integers. I've minus one to the integer part of nu times e To the nu times the derivation because I'm involving my thing here That's the next term then I've e to the nu squared over q to the nu squared over 2 so that comes here Okay, and then I have to take this and apply to F At lambda Right, that's that's the recipe. That's what the left-hand side is but now by the lemma The lemma said that there was a bijection between pairs consisting of a partition and a half integer shift and Under this lemma. Well, this thing is just q2 of x because that was the the second formula written here The size of lamb minus the 24th plus nu squared over 2 similarly Eat the new Delta f of lambda. Well, that's just f of x f f of x and Finally minus one to the integer part of lambda is minus one to the power integer part of q1 of x because q1 of x is lambda So now I've written in terms of lambda and I promised you a one-line proof And I'm almost at the end of the line, but it's still going to work This is zero and the reason that that zero is completely obvious. You have an Involution an Involution which is really stupid you map x to x star which is the stupidest thing you can think of if You're set which is now a subset of z if it contains zero you take zero out and if it doesn't contain zero You put zero in That's a pretty silly Involution, but if you think of this Involution then our w x of t which was the sum T to the x well you just add or subtract it So when you do this you're going to simply add a subtract one And so if you remember that the coefficient of z to the k minus one and this is qk says qk of x star Will be equal to qk of x for all k different from one But q1 of x star will differ from q1 of x by plus or minus one Well, it'll differ by one in one direction the other So if you just apply the Involution this sum is zero because the terms just cancel in pairs every time you have x and You go to x star f of x only involves q2 q3, but those things don't change So this is invariant q2 is also at least q2 so it's also invariant But this q1 goes up by one so minus one to the integer part of q1 goes up by by a half Although it's hard to change the sign and so everything cancels in pairs So you now get you know the simple one-line proof of a theorem that immediately applies block of kunkhoff now my time is I hope still positive Very small Can I apply a scaling of a minute? Okay? I wanted to talk briefly about this Eagle Beach constant thing So maybe I'll just say then in words how it works That's what started this whole thing and it's a very very complicated investigation with Mulder It's been going on for three years using this block of kunkhoff theorem So basically the the the numbers that I'm not going to then repeat the forms We had them all in In selection I spent all three in the morning last night with my colleague on the phone Well, that was only half an hour But then myself Preparing that I would be able to explain it correctly and I won't need to but we had a famous a formula from the paper of Well Maxim and collaborators. I'll remind you it was a formula that said that if you're looking at the stratum 1111 of flat surfaces or any m1 up to mn the stratum of flat surfaces So a Riemann surface with the differential form with zeros of order m1 up to mn Then there's a Lyapunov story and some of these positive Lyapunov exponents Maxim had somehow I think it was mostly him or all him I'm not at all in this field had seen that this has everything to do with intersection numbers and multilize spaces That's one of the things that goes in but here. There's an explicit formula Which I remember correctly is like this and then plus well, there was a factor pi squared over three But I want to renormalize it so the thing called C area of the well If the particular surface you're looking at and see areas the thing that Anton explained in his talk when you have these cylinders Then you take you you you count them you count the number of G desicc's up to a certain length But you count them according to the weighting them by the area of the cylinder which which they stay so the original zeal beach console It was le each you didn't put in that area factor But it turns out this had much better properties and then you do something with the asymptotics of that But the final upshot when you take the average in some sense or the generic point of the stratum is that your relation between one thing You want this number which is very simple and for the principal stratum, which is the only one Well, no we have results in principle for all stratum, but we've worked out in great detail for one one one This is just a simple number and this is the note This is another number you want and because of the relation with modular spaces These things are related to the volumes of modular spaces or various strata in this space of flat surfaces so They can be expressed that was done well in particular by the work of Chen and Miller You can express that all in these blochokunkov terms and therefore show that the generating functions of various counting functions Like this so you have to go to urban space and look at the ramification like I did in the special case of coverings of Taurus by well It's still coverings of a Taurus, but with more generic ramification You can translate the problem of evaluating these numbers into a problem of counting of averaging functions of repartitions of these q brackets Which we know are q-multiler and in particular these numbers like c-area Which the prediction was that they're always rational because there's supposed to be intersections in the in some multilize space And in many case it was known I mean that's Eskimaeser and Zorich and so on there are formulas But it's very very hard to get your hands on either the volumes or the areas or if they also Chen and Miller found that you should also do other counts. So here we're counting the cylinders by the By the width times the height of the cylinder Because it's the area, but it turns out. It's also a good idea to put powers I forget exactly the formula so there are way to things so they make higher Zegel beach constant cp which are also interesting and c-1 would be the c-area But the others are perfectly good in variance also and it turns out that the c-area are exactly given by the limiting behavior as towel tends to zero of the quasi-modular form associated to a very particular element of the blogger kunkhoff ring and so you can answer all of these questions completely if you can calculate Some rather difficult generating functions in this blogger kunkhoff thing really calculate these things So we have many many results on that and I was going to say something about them, but maybe it's just Hopeless maybe I'll just mention one thing because that's at least still on the the level of partitions so purely combinatorial So what Chen and Miller found is that to compute this cp for any p including p equals minus one So the area case but also these weighted here You have to normalize the whole area of the surface to be one What they found is that those cp Correspond in some way to tp and then q3 q3 so I already told you for the generic ramification If you have a bunch of ramification points K of them, but they're all generic then that corresponds to the q3 But now if you multiply this thing k times would actually have to take the so-called cumulants It's an alternating sum by inclusion exclusion then this thing corresponds to cp and tp is a function of So this is the q after remember the q average is defined for every function It's only quasi multilayer when when you're in this special ring, but it's defined So they found that the cps were in terms of tp and tp is defined as full as it's one over n factor I'm just a partition of n But then it's a huge sum of all the other partitions of n of the size of the comms t class times Times the square of the character times sp where sp of a thing is very simple It's just the moment to the sum lambda j to the p. So you have to take this kind of transform Using the character table of the original thing into this I think the form is right there. Maybe the two is wrong, but then they so he asked me So this is you know lambda one to the p up to the last one to the p And then they did a lot of calculation you get completely weird numbers and they asked me to see if I could recognize them And so you start if you do computer calculation the first thing you'd find is that when you do this n Then TV especially one t minus one t minus one of n is a very beautiful number It's the nth partial sum of z of two and so that generalize that tp of lambda So this is the theorem that I found on the computer and Miller then proved It's not that hard to prove it was somehow Took a while for us to understand that that was the statement that tp is just the sum over all elements of the young diagram of the hook length of S to the power p minus one so the hook length of a thing is what you would think you have a thing and you go as far as you can To the top and the bottom and you add that up that's called the hook and the number of squares is the hook length So for this very simple diagram the hook lengths are one two three up to n And so if p is minus one if p is minus one then you get the sum one over n squared So this is a non-trivial thing, but then it turned out that T So another theorem less obvious although also not very deep is that if p is greater than or equal to one So non-minus one and odd Then these tps belong to the block of kunkhoff ring. They are polynomials So this tp is just a function you can evaluate it for any real number of p But if p is a positive strictly positive odd number then that function is a shifted symmetric polynomial and therefore it's Bracket and what's more its bracket when you multiply by any thing in the ring is Is a quasi-multi-reform so if f is in the ring and P is positive and odd Then by block of kunkhoff plus this fact tp times f q will be in the ring of quasi-multi-reforms And then what we found with a huge amount of computer calculation by looking at hundreds of examples into very very complicated Formula we slept and completely proved although we've proved many cases that this is the sum of certain differential operators row ij Which are very comfortable the first if i or j is less than five we got explicit formulas by the computer And then we finally guessed the general thing you have certain differential operators on the ring f and then you apply those Universal operators and here you take the i-th derivative of the Eisenstein series will essentially the e before we normalized of weight j plus p plus 1 So this was this is still conjectural But we've proved in very very many cases and the form of this is such that if it's true for positive p It's also true for p equals minus one with a very small Modification and using that you get formulas the exact formulas for the volume is followed these strata and asymptotic formulas and Also for the area things of the seagull beach counting But some of the results are now unconditional But the most general in particular for the area counting is still conditional on proving this thing Which we've only proved for you know a lot of special cases, but not in general. So there's this very very interesting algebraic and combinatorial structure on on this ring with these tps and That's you know very amazingly even though they're defined well in two different ways completely differently But they turn out to be polynomials in the qk and therefore to act on the ring And so there's a very rich algebraic and combinatorial structure so As said in the introduction, I'm somebody who likes formulas with no content and to Twist a quote that we heard from Maxim when he was a young man. I'm a specialist in special problems Okay, so thank you and thank you Maxim Some No, but when you take the average or something I mean for the generic no But but but no, but I said that that it's not but because But that you have an expectation that it's connected because of multi-spaces, but here well, this is certainly rational Nobody can take that away from me In fact, it's n over 8 in our case and this thing is rational because we have a formula for it Well, that was already done by Esken Mazer and Zorich But we have actually I was going to write down the generating function we find for the volumes It's a male. I'll just say it in words take the Horvitz zeta function at three halves, but shifted by half So you take one of well or even at one half So you take one over the square root of n well slightly divergent summed over n is a half three halves five halves Sorry shifted. So one over the square root of x plus square root of x plus a half x plus a half x plus three halves x plus five halves expand that as a power series in x It has an asymptotic expansion. It starts square root of x So call that why so why is a power series starting with square root of x? And now you invert that then x is a Laurent series starting y squared and the coefficients of that inverse power series of the shift And torpid zeta function are essentially the volume of the corresponding strata That's the the formula and for that you can get asymptotic formulas Not that easily, but we can I was going to write them down, but I've run out of time