Hi everyone, happy Thursday Joe for jazby's casewrakes.com coming at you with jazby's 14 a box multi-year Basketball mixer a lot of great stuff in this mixer. Thanks everybody for getting in We sold out all those sell-out packs all the mini breaks are done here We're giving this away at the end, which would be pretty nice. He's been playing some great basketball PSA 10 as well So we'll randomize that to one person the break at the very very very end new dice roll If there's any points, we'll give that to spot two on that randomizer There's all the boxes involved in this and pretty good stuff and a big thank you to all of these folks right here If you bought a spot straight up, I appreciate it if you bought if you won spots congratulations. Thanks for giving that a shot They're all the teams right there. All right, let's roll it. Let's randomize it three times one Two and three your names first then the teams Jason down to Kim Soo Min All right three times for the teams. Good luck teams one two and three We've got golden state down to Cleveland. Can we Cleveland Cavaliers want to be a little tiny bit bigger? There we go All right Jason with the Warriors Russell with the Spurs Kurt with the Rockets Nick with the Suns at ice with the Kings Daniel with the Pelicans Matthew with the Trailblazers Kim Jin Hee with the Gris Kim Soo Min with the Nuggets Mike You have the Wizards Jin Hee with the Hornets Kevin with the Sixers Jeremy with the Celtics Zach Gwynne with the T Wolves Matthew Shear with the Raptors Chris with the Pistons Aaron with the Bulls Pablo You got my Lakers James with the Nets Courtney with the Mavs Chris with the Heat Courtney with the Pacers Jason with the Magic Colin with the Clippers Mike with the Knicks Matthew with the Bucks Chris Matthew with the Hawks Chris has the Bucks So why with the Jazz Greg with the Thunder and Kim Soo Min with the Cleveland Cavaliers, there you go, so we Let's put this all on one page. Let's alphabetize by column B by team Let's box this up make this look really nice for the shipping team I'm gonna pause the video when we come back We're gonna see if there's any trades and they'll have the break itself stick around Kings up for trade So we want the Kings. All right be right back All right, welcome back everybody there was a trade as you can see on your screen right there So Tice is out of the King's spot and is with the Clippers put T next to there for trade In case you're following along see who draws first blood in the trade and Colin gets his Kings. He's a Kings fan So there you go. We'll put T right there and so now we can go ahead and Print and rip good luck everybody and like I said we are booked up for a couple hours or so so you can go hang out with With our personal breaks channel at Jasper's breaks on Instagram follow us there and follow us on Twitter at Jasper's breaks both spots And remember at the end if you end up having a bad break or you're not happy with the team that you have Don't worry remember. You got a chance at that Michael Porter, Jr. Jemmen 10 from prison basketball. He has been crushing Crushing it slide these over here. Actually, you know what let's get this college one out of the way here Let's start off with the bang 2019-2020 Contenders draft picks basketball. I'm actually gonna pull up a checklist really quick because it might be some players that Just not familiar with I mean, hopefully it's just these three guys would be half the autographs would be nice I don't think it works that way, but one one can hope Contender alphabetical Stand by folks hang tight. Let's get this checklist up So this is I'll drop it in the chat, too. We use group rate checklist calm We're not affiliated with them, but it's they do a good job Shout out to them. Thanks for doing God's work making up making those checklists. There it is Thanks, everybody Thursday the third thanks for spending a bit of your Thursday with us as much appreciated Let's go. I think we got a prism and some mosaic a mosaic box We might close out with these three boxes right here Maybe look for a Yanis in here some 1617 right here some 1718 prism right there Maybe that there's a blaster box back there Maybe we'll close with that look for some Zion's or John Moran's at the end. All right. Good luck, everybody And we got Jalen McDaniels Is your autograph? Forget we're Jalen McDaniel. I have his name sounds familiar. I don't know where Jalen McDaniels ended up going He went to the Hornets Charlotte with that one that's gonna go to Ginhee Charlotte with that one And we'll do a recap at the end too. So fear not You're like, oh man, I don't want to watch this hour-long video. You don't have to there'll be a recap at the very end We got a mere coffee a golden gopher Mere coffee is a clipper. So Tice drawing first blood in that trade I don't recall if he's been getting any action, but an autograph nonetheless And we got a DeMar de Rosen season ticket card eight out of 23 cracked ice That'll be for Russell and the Spurs and we got DeQuand Jeffries a Golden hurricane Quan Jeffries is a Orlando is and Orlando magic a Orlando magic He's with the magic and that'll be for Jason Jason F with Orlando Save save a Yanis to We got Marvin Bagley for the Kings 83 out of 99 This guy just would stay healthy be pretty great calling with that one We got Kai Bowman Spurs, maybe no Warriors So that'll be for Golden State Jason We've got Spurs, Utah right now Spurs Jazz at the moment spur jazz up 61 50 and Spurs are a half game out of a half game out of a The possible play in spot now. I don't think it's gonna happen. I think the Sun's one today The Grizzlies one today That might have effectively cancel their chances. I don't I'm not really sure how that math works We got Goga Batatze International ticket autograph for the Pacers Courtney Goga has been quietly quietly raising his stock It'll go by whatever the Whatever it is on the team card On the card itself whatever teams on the card Imagine if we had to do updates on all the teams and we're doing like 16 17 product like we the shipping team would hate us Jordan pool For the Warriors. All right, let's move into some 2017-18. In fact, we're gonna do this mega box right here 17-18 mega. All right. Good luck. Got BAM out of bio. That's pretty nice and John Collins is strong too They may creep up into well BAM out of bio this is really solid for John Collins may creep up into that Kind of upper tier rookie BAM out of bio is nice for the heat It's a Lonzo hollow Ones are still carrying some good value It's Lakers edition for Pablo and a nice Donovan Mitchell rated rookie Tri-color here Do I loo with the Utah Jazz nice Jason Tatum rated rookie. That's nice too Celtics that'll be for Jeremy Port won that spot One of our randoms.  So Lonzo Tatum out of bio Mitchell so on and so forth Calvin Johnson five or six from the field 14 points already dust off your Calvin Johnson rookie cards we're at a point folks where Where I mean especially for basketball any of the rookies even if you even if they haven't played a lick just top-loaded Protect it make sure dust doesn't get in there. Just throw into the back of your drawer put in a shoebox something like that Never know Like who was thinking Kelvin Johnson prism silvers were gonna go for you know 50 to 100 dollars this week Some of you who may ended who may have ended up with the spurs may have some Kelvin Johnson sitting in your collection from from our breaks On jasky's case breaks calm All right Jarell Taylor knew but Cuz family 99% of the people I don't think had Kelvin Johnson on their list slide this over here Let's get this stack going right here Good luck everyone. I Mean even this guy could have a revival. You never know Well LeBron James Cavs edition even these non-rookie year LeBron James cards have carried some extra weight Which has been awesome Kings Colin Sullivan will get the rookies deer and Fox Lonzo Lakers Pablo Nice Jason Tatum right here rated rookie. It's another one for Jeremy Port and the seas Save those yannises to the other only vet commons that end up shipping Yannis second-year Luca Donchiches LeBron James Yannis onto the combo that'll be for Chris and the Bucks Steven Adams Thought I was flipped around thought I was gonna be an autograph. It's not but it's out of 199 Greg There you go Steven M. The Thunder in those seeding games. They've been playing some good basketball They could be a threat You know in the in the playoffs, which are gonna be starting soon. I'm excited. We've got that's not a graph We got Zarydunas Ilgowskis Significant signatures for the Cavs Zoom in with that there you go Jimmy Butler singing Hull notes as Lori marketing for the Bulls Aaron And obviously all the rated rookie cards will ship There's JR Smith Press proof to 199 for the Cavs Summon I can't wait for as a Lakers fan. He's on the Lakers now. I can't wait for playoff JR Smith What wackiness will he do? Will he You know will he go off and go go six for six from three or will he forget how much time's on the clock? You don't know You don't know with JR Smith Is he gonna hit a clutch bucket?  I don't know another Donovan Mitchell For Dwight and Rudy go bear Jersey Kings Relic for the Jan another one for Dwight Chris Gerard with the The heat and the bam out of IOS West Matthews press proof 299 gray border on this one Mavs that'll be for Courtney Markel folds. I don't know. Why are you sleeping this guy? Kevin Mendoza's thinking I don't know former number one overall pick Does he have does he bloom late like like Victor Oladipo or something like that? Maybe possibly Might as well hold on to that see what happens. Maybe the change of scenery in Orlando will help him long off season to To to kind of get his get back on track All right, we're still in 1718. I think this might be our last 1718 box. I Think we got one more back here at 1718 Let's just do this one for now. All right, so Donner's optic of basketball. Good luck everybody In NBA action the Celtics fell to the Wizards Wizards 196 90 How did our how did our boy Rui Hachimura do did he play? Let's see a little quick rookie report here Hachimura DNP right quad what about oh Thomas Brian When 2026 points nine rebounds a couple assists and four steals. Yeah, you're right Jarell With a plus minus of 11 playing a little defense, too Let's cross 30 minutes 10 for 15 from the floor two for four from three Almost had a double-double. All right Do we do we What about anyone on the Celtics now nothing too exciting for the rookies on the Celtics? Ojula had 13 points. I think he's might be 1718. What else now? I could my Lakers got dusted by the Kings I think either team really had anything going on there and nothing really to play for Pelicans easily Dan In the Lakers game dust off your Deon waiters cards 19 points for the Lakers today LeBron only played 15 minutes LeBron had 17 points in 15 minutes who said he was old How did our boy town horn Tucker do town horn Tucker played 34 minutes 14 points for rebounds three assists three steals Not bad Queen cook 12 points J. R. Smith 11 points Did Costas onto decumpo Yanis's brother got 11 minutes in the game today. Wow seven points For costas two rebounds one assist a plus minus of plus 14 playing a little D as well Three for three a perfect three for three from the field one for two from the free throw line There you go There you go Yanis to the Lakers confirmed Or is Alec Peters signature series for the Sun's Nick Stober? I'd be shocked if if THT played over JR or Deon waiters I think THT has earned some playoff minutes for sure, but Over Silver stuff Overseason veterans. I don't know. I don't think so but I think the Lakers do like that they have that option on the bench in case of You know case of injury basically warriors Jason Tall Kings Pete Marovitch Who's my save one of those nice Jason Tatum right there and silver dame Battleship That's an insert for Yanis Silver dame trailblaze. I don't know if these or it's hollow. I guess an optic.  So Jeremy Port with the Celtics There's a deer and fox here and someone was saying who is it David? Maybe? Yeah, David was saying John Wall when if he's back in healthy that wizard team could be really interesting All right, Bradley Beale Hatchimura John Wall Thomas Brian they've got some pieces In a week or east Kings. That's for Colin Collins a King's fan. He's happy about the Kings win today Terence Ferguson 85. Okay, see playing some good hoop Wow Dan thinks that Dan was just saying a J.R. Smith you could see him one-upping himself from the previous Playoff time when he forgot what clock at what time it was and Shoot a bucket in the other basket. Wow Warrens and playing some good hoops. It's not rookie TJ Warren, but wish it was Alright, let's get into some more optic. This is 2018-19 edition grizzles saying But Kelvin was 16 in 19 minutes. No, it says no rebounds though in that time He needs to he needs to get that ball back needs to give up the ball Pass it shoot it someone's got a score Grizzlies keeping their playoff hopes alive. I think 119 They beat the Bucks 119 to 106. I think Yanis only played The honest played zero minutes because he was suspended. That's right.  I would have won a million dollars People would put it put the house on that if I said both brothers would have more minutes than Yanis. I said that last week Pre-suspension So thonus so costus got got some minutes with the Lakers Thonus onto to combo with he was on the Bucks got 18 minutes a couple points a rebound couple assists a steal Plus 12 On the plus minus For what it's worth Frank Mason the third is on the Bucks had 18 points today Jonas Valenchunas at 26 for the for the Grizz John right only at 12, but Dylan Brooks had 31 points a day John right at 13. He John right at triple double though. Yeah It's kind of weird 12 point triple double 12 points 13 rebound ten assists. There you go. Good for him Oh for four from three Oh for four from three. All right, here we go and Watching live Utah Jazz up 80 to 66 on the Spurs I don't think the Spurs have even had the lead and maybe early in the game Spurs had the lead, but Not since it's hardens to 85 Zach you watch your my videos all the way from Canada. Well, welcome The Andre Aten playing some good hoop ladies and gentlemen just across the the regular season in general His his value creeping up a little bit Sons could be a dangerous team next year Nick Stober with Phoenix More bomb up for the magic that'll be for Jason Trey young rated rookie. All right. This is 18 19. I was like Trey young That's goes to Matthew Shearer nice Trey young rated rookie. I feel like this team underachieved this year Well, I mean, what do you think am I wrong? Maybe I? Don't know. I think I think they underachieved this year. I think Trey young did fine but I Mean they have Kevin Huter Deandre Hunter John Collins Right cam reddish, you know, you're telling me that that that team only won 20 games. You kidding me? But apparently they have 42 million dollars in cap space. Oh, yeah, definitely still young Absolutely, I just I just thought they'd win more than 20 games Jared Jackson jr. Getting injured really interrupted the Grizzlies. No, yeah, I absolutely give them a chance. I just thought I Don't know. I thought they'd win like 30 or 40 games You know, I mean, I didn't think they're gonna win the division or anything like that I realized that they're still evolving but But I was like, yeah, Deandre Hunter cam reddish and Trey young And Collins, you know, like I thought more than 20 games. That's for sure SGA people of SGA, you know, okay, see this one goes to LA. This will be for Tice. I like the Hawks so I think I think actually Maybe I maybe I kind of like there actually let's do this hangar box really quick 1819 dollars Maybe I actually like that the Hawks All right, the defense that's what it is probably. Yeah, the defense isn't there. Yes, it's Jeff um Maybe I like that they only want 20 games though because maybe they'll be undervalued by Vegas So we take the over on wins for Atlanta And they have 42 million dollars in cap space I just read this right now on the Hawks team page on ESPN.com from an article by Bobby Marks two days ago so They could add I don't know who the big Freedons are this season, but they could add a lot of players Anthony Davis is a free agent I think so can you imagine? AD on that team AD with Trey young. I don't even want to think that put that put that out of my head now There's few few guys we like here. All right, Mo Bamba for the magic Jason Fractor people like Mo Bamba. We talked about SGA. Okay. See looked great in the reseeding games Devante Graham PJ Washington bridges among others You know one of these guys is gonna end up being could end up being a perennial all-star Once they evolve a little bit. They've got a good young core now. We're in Zion year hoops basketball What was the wasn't there another game? Maybe that was it? Did we go through all the all today's games? Oh? No, we had Mavs sons. That was the game. We were watching early. That was the early game Suns up one beat the Mavs 128 to 102 Think Donchage only play at Donchage only played 13 minutes. I don't think poor zingas even played Mar Marjanovic 18 points today. Look at Donchage still had 18 points in 13 minutes.  No assist though. He had a steal Donchage could have had like a 20 point double-double in like 15 minutes if they gave him a few extra minutes crazy Boy Cameron Johnson, you'll see him in this year, right? He's a rookie Cameron Johnson 1920 15 points five rebounds three assists two steals a block Plus 12 on the plus minus if you're into that sort of thing in 30 minutes Some good basketball right there Michael Bridges miles bridges and we just had miles bridges who's on the horn Michael Bridges 14 points for total rebounds three assists a block plus 26 on the plus minus Devon Booker doing Devon Booker things plus 27 seven rebounds five assists a plus 30 on the plus minus Either there wasn't good shooting from the Mavs or Devon Booker playing a little D Cameron artists pain is on this team yet 15 points plus minus the plus 20 Dario Sarich had 16 points as well Good game by the Suns take this whole mat with me. I guess right. Let's roll Hachimura, there's Cameron Johnson. We're just talking about him high voltage Paul George for okc Cameron Johnson will go to the Suns next over Hachimura for Mike G and the Wizards and we've got Utah Watanabe Hot signatures for the Grizzlies that'll be for Ginhee with that one. There's Zion Jalen Brunson per oh, I thought these were numbered they're not at all Zion Williamson that'll be for the Pelicans.  He's I Dust off your town Horton Tucker's ladies and gentlemen. I'm half kidding but With the way With the way the hobby is going these days With the way the hobby is going these days if he gets to playoff minutes and he He scores a significant bucket or two Watch those values continue this to rise Town Horton Tucker has gone from a dollar card It's like a five dollar card It could be a ten dollar card By the end of the first round Brooke Lopez to 199 We're brawn former teammates. There's Keldon Johnson his stocks rising their sake with the boy a autograph He's like the only main guy only rookie or only young player on that team worth a lick Chris Gerard Gerard Steven Gerard sprother Chris Detroit pistons. There you go, man. There's Wiggins to 25. Good question. Jeff What do you like more? Are you talking about Jackson Hayes or Brandon Clark? Who has the better career? What position does Brandon Clark play forward? So same position, right? Jackson Hayes also also a forward slash possible center. It's a good question. I haven't thought about that I feel like Jackson Hayes seems more exciting to like watch But I don't know what kind of what kind of minutes he'll get With you know with Zion on the court and some other other players on the court I feel like Brandon Clark Might have some more chances You know, especially with this guy too, right? It's John Morant plays a little more in the I don't know I mean maybe team construction like in a vacuum. I would say Jackson Hayes But I feel like on their current teams I would why would lean Brandon Clark is that crazy. I don't know. What do you think Jeff? You think Hayes is better, but Clark has the higher ceiling Hayes is better now Clark has a higher ceiling in the future Well, I don't know all I do like that we're having this conversation because That this the 2019-20 draft class has been a lot better than we think you can go up and down the list and you can at least find Find some potential You know in a lot of players like last couple years We kind of know like listen, there's some guys that you just know aren't gonna Aren't gonna be you know in the league for the next few years You know what I mean, but this year like you have to hit the pause button on a lot of players because you know, it's like goga You know like They're doing interesting enough things where even Terence man, they're doing interesting enough things where you're like Okay, I might hold for just a little longer There's a good question David Downer saying who wins a chip first Zion or Ja Let and let's just say they're they're have they have to be on their current teams Pelicans and Grizzlies respectively Now Jarrell is all over Jackson Hayes, Hela energy defense first reminds them of a Of a Tyson Chandler type. That's pretty good. Where should we go now? Should let's get in to get that totally certified goings back to 17-18 Getting close to the end here David, what do you think you would take the Grizz roster next five years over the pelicans roster, would you I don't know I mean, I'm biased because there's like three former Lakers on the pelicans like I think Lonzo may not ever be a scorer, but I could see Lonzo getting like 15 assists a night with that team You know, I could see I could see Brandon Ingram Come on Brandon Ingram is like it's pretty incredible Josh Harkins you a little bit from the outside. They still got a retic out there. It's still got holiday out there. I mean, I Like that Pelican seem a lot Now if they could I don't know what do they have to do if they can maybe I don't know what their cap space situation is But if they can get a couple free agents here and there Hey come run with Zion Be part of the Zion experience some players might be interested in that There's TJ leaf 137 out of 249 for the Pacers I Would I would lean I would lean pelicans To potentially win a chip first before the Grizz if those are the only two teams are talking about here Adam saying if John and Zion ever team up That team would have no money. They'd have to pay those two guys and that's it There'll be no one else on that team till a Keenan at 299 Better future Ben Simmons or Siakam Siakam Ben Simmons, I don't think has reached his final form yet, but but Siakam right now clearly is is Playing some really great basketball nice rookie roll call bam out of bio for the Miami Heat Chris G. Yeah, Ricky saying Siakam for sure Jarella going back to the pelicans. Yeah, all the pelicans need a Couple shooters, it'll be over. I Agree and just that Zion energy. All right, let's do let's stay let's stay in 1718. Let's do that blaster box Thanks, Ben Simmons has a has a high ceiling though. I know the the injuries are a little weird the the the lack of outside shooting he may need to develop that but I'm not selling on Ben Simmons Not not yet, but right now Siakam's pretty pretty good. All right, so 1718 blaster box I think this is the last 1718 box for the people who have 1718 teams. Oh, actually, no, we'll do court Kings after that So two more chances for 1718 We got Edmund Sumner It's a 25 No, I didn't know I've got nothing on that guy. They're semi-osula relic for the Celtics Jeremy Kyle weaver Panini elements a must-buy you all feel like well if you're if you're asking me Anything on jazby's case breaks calm is a must-buy That's why we put up on the site. We certainly wouldn't put non-must-buys on the site Everything on the website a must-buy. There's siakam right there. I'm biased though. Yeah pistons Chris, let's have silver Lucanard.  You never know You're gonna hear me say that a lot you never know especially with hoops the way basketball is going Sometimes these box hoppers can be autographed. Why is Harris was on the pistons? Oh my why am I blanking on the Tobias Harris and Harris pistons edition Halkusma level three and Yanis onto the compo five out of 25 nice bucks That will be for Chris G whose Go to Pablo in the Lakers think level four is the shortest printed one in these court kings I like the street art background right there. Remember remember how it's called street art now You should just be graffiti now not anymore. It's art. I Like that. All right. See it Oh gee, I know Obi Raptors Matthew Shearer with the fresh paint autograph and Jimmy Butler Performance art and that's bulls edition going to James 183 out of 299 Box topper and make the season for me box topper. Yes. It's James Harden Rockets There you go. That's smart right there. Nice There he is Houston that goes to Kurt. All right, let's go 1819 Donner's basketball blaster edition OG's auto is awful. Yeah, it's not so good. It'd be something Anna Obi needs to work on in the offseason Oh, yeah, no, he's like Oh gee, what do you what are you gonna do in the offseason to to improve your game? Take it to the next level. Well We'll work on my footwork Lift some more weights work on that auto that autograph Those are the key things I want to work on the offseason If I see that if I ever see that quote, I would make me so happy And I'd like to announce now that I'm going to Jaspie's autograph camp in Hermosa Beach, California We'll help you. Oh, yeah, I know Obi guest speaker Ozzie Smith will be there Randy Johnson will be there Mariana Rivera will be a guest speaker at Jaspie's autograph camp Yeah, can't Kam Reddish to Amber. Yeah We'll invite Kam to Jaspie's autograph camp bag. We we've got a Jalen Brown I think Adam was saying earlier that Jalen Brown doesn't get the respect he deserves. There is He's Jalen Brown for the Celtics Jeremy There's Michael Bridges. He's been playing well rookie jerseys and there's Michael poor junior rated rookie Denver Nuggets Kim zoom in with this one rated rookie nice Does 2020 Panini elements legit come out tomorrow? Well, it says Friday in the item description Kyle weaver We certainly wouldn't lie to you Why would we put that if it was not legit coming out tomorrow? That would be crazy. There's Trey young right there And TJ Warren some playing some good basketball Lonnie Walker Troy Brown Junior and my Conley exact. That's weird. I feel like someone else mentioned that today earlier I'm not sure Shoot us. Uh, shoot us an email on that click that support button and shoot us an email. We'll look into it TJ warned you're not not a rookie, but it's a different parallel right here and Trey young a Matthew cheer up. I don't know what other advice I have for you Unfortunately his axe. Sorry try maybe restripe. I guess if you can't find your email, you can't reset your password It's kind of odd All right. Let's do this 16 17 mosaic basketball Spurs, I think are keeping it close Utah's up 105 100 now, although I think spurs have officially been eliminated with With the winds today by the Grizzlies and the Suns. Yeah, I think they're eliminated officially so right now I think the trailblazers Need to win What's here? So the Grizzlies with the wind have clinched a playoff berth play in berth They're currently in the eighth spot, but the trailblazers haven't played yet So if the trailblazers win Then they jump into the eighth spot. I think and then Grizzlies will still be in the ninth spot So Grizzlies are in If blazers are when they're in blazers lose. I think Phoenix is in Yeah, I think Phoenix is in all right, so that Portland game gets to be really important this game has has no importance There's Pascal Siakam rookie members 16 17 rookie Pascal Siakam Speaking of Siakam There's Jamal Murray. I like his game too so rookie Jamal Murray for Kim zoom in in the Nuggets and Matthew Shira rookie Pascal Siakam out of what they called it prison mosaic back then now They just call it mosaic another Jamal. You just saw one right there Pascal Siakam rookie and there's been many other big hits like that this mixer We'll do a recap in a little bit another Jamal Murray Pursume in those aren't numbered. Oh, I think all of these will ship anyway And but do we like buddy healed buddy healed Pelicans edition going to Daniel W Does he still have some upside with the Kings? alright Let's do this We're wondering I'm saving Crusade for last this is 2013 14 Crusade Yeah, Chet Lincoln you'll miss it. Yeah, buddy healed was on the Pelicans for like half a second half a season I think and they traded him that for his first season went to the Kings For cousins I think for boogie cousins among others. There's dame. We'll see him tonight To 99 there you go Dan Tappy. He's he likes the buddy healed. There's Luca Donchich second-year Luca Donchich for Courtney in the maz 21 out of 49 Carson Edwards is your die-cut relic heiress of the throne Jeremy Port We got Trey young autograph three out of 25 The second-year Trey young, but it's an on-card auto one of Commemorating his 16 assist performance Lana Hawks Matthew Shearer and then coming out of college. I think people are like Trey young just a shooter But no, I think everyone he surprised the entire league. I think with just how How well rounded? He ended up being all right final box coming up. There it is Crusade 13 14. Let's go Yanis. Let's go rookie Yanis Spud web autograph for the Hawks Matthew Shearer Vince Carter Mavs edition Kelly Olanik and Yao Ming Tony Parker to 249 for the spurs that will be for Russell. There's Anthony Bennett. There's Dirk Nowitzki Little extra chalk on there When you end up seeing these cards we got Gordon Hayward two-color relic jazz edition For do I loo there's Rudy Gay to 20 out of 349 Is anything event the biggest bust in sports history? Might be Because it he didn't have injuries, right? Because I don't consider players who have unlucky injuries. I Don't count those as busts. You can that's unfair, but I think he wasn't injured I think he just he just washed out, right? Pete Marovitch to 16 out of 249 Jamarcus Russell for my Raiders might might be a bigger bus. I think Derek Fisher Mario Chalmers JR Smith next edition to 249 Xavier Henry John Lucas Yeah, well, it might Russell might be the worst traffic the Raiders have ever made since James might be the worst ever in history See Johnny football though Damien. I think that was many years after because when Jamarcus Russell was signed Remember there was no rookie Scale I think rookie scale for contracts Jamarcus Russell held out for like a zillion dollar contract or something like that and Like the teams were like hamstrung you had to sign them eventually there's that 34 a nickel of a tomb and No, Yannis in this. Oh, well we tried and it's Dr. J though so not only did the Raiders sign Jamarcus Russell for a hundred million dollars plus Most mom sir a lot of it was guaranteed to and ended up being terrible All right, Joe for jasper's case breaks calm if you had a terrible break guess what you still have a shot at this You're like Joe. Yeah, that was not a good one for me, buddy. But guess what? Just lucky just a little luck on the randomizer get that PSA 10 Michael poor junior from 2018-19 Panini prism So Don't fret just yet. It's not over James Not over yet Let's flip back to the screen right here. We got Matthew shear and the Hawks down to Mike G and the Wizards Get all those names into a blank. No points here, too. That's good. There's all the names right there Let's roll it. Let's randomize it name on top after two snake eyes Name on top after two. Good luck one easy for me to All right, big. Thanks everybody right here for getting into the action appreciate it. Sorry James Two down to 30. Thanks everyone for getting in name on top is gonna be Greg Greg B I don't know if I called your name too often in this break Greg. You had a couple S. He had the thunder No, you don't even have those S. G. A's. I was clippers. I Don't know if you really had anything in this break Greg, but you know what you do have You have this so there you go Congrats to you. Thanks everyone else for getting in really appreciate it We always do these kind of crazy mixer. So look out for more on jazzy's case breaks calm I'm Joe and I'll break more with you next time. Bye. Bye


Sri Lanka is one of most beautiful country in the world. With the paradise for the sun-lovers, Sri Lanka has more than 240 kilometers of coastline with beautiful, luxurious beaches. It offers swimming, windsurfing, sailing, canoeing, speedboat racing, diving and viewing the coral of Kandakuliar in Karaitivu. Today Sri Lanka is currently experiencing its worst economic crisis in history. With long lines for fuel, cooking gas, essentials in short supply and long hours of power cuts the public has been suffering for weeks. The Sri Lanka government has said it is seeking a bailout from the International Monetary Fund while asking for fresh loans from India and China. On the surface the cause for Sri Lanka's current economic crisis can be attributed to economic mismanagement by successive governments and a string of ill-advised decisions, such as banning the use of chemical fertilizer and deep tax cuts promised by President Rajapaksha during his 2019 election campaign months before the COVID-19 pandemic decimated. Sri Lanka's major tourism sector and also stopped foreign remittances. What's behind Sri Lanka's economic crisis? Sri Lanka is facing one of its worst economic crises in recent decades. With skyrocketing inflation, weak government finances and COVID wreaking havoc on the economy. Skyrocketing inflation, weak government finances, ill-time tax cuts and the COVID-19 pandemic, which hurt the important revenue-generating tourism industry and foreign remittances, have wreaked havoc on the Sri Lanka economy over the past several months. Prices of food items, for instance, shot up by as much as 25% in the last month alone. Shortage of food and fuel Meanwhile, the nation's foreign currency reserves plummeted by about 70% since January 2020 to around $2.3 billion, 2.1 billion euros, by February, even as it faces debt payments of about $4 billion through the rest of the year. Sri Lanka's current reserves are only enough to pay for about a month's worth of goods imports. A shortage of foreign currency has meant that the country has been struggling to import and pay for essential commodities like fuel, food and medicines. These challenges has led to cuts in electricity generation with only four hours of power a day and long queues outside fuel stations. Even the newspaper and printing industries have been hit by a severe shortage of printing material, forcing cuts in publications and school examination postponements. Tax cuts and pressure on public finances The economic emergency poses a significant challenge for President Gotabaya Rajapaksa, who came to power in 2019 promising rapid economic growth. During his presidential campaign, Rajapaksa promised to cut the 15% value added tax by nearly half and abolish some other taxes as a way to boost consumption and growth. The tax cuts led to a loss of billions of rupees in tax revenues, putting further pressure on the public finances of the already heavily indebted economy. Then came Covid, which dealt a huge blow to the tourism sector, which accounts for over 12% of the nation's total economic output. Sri Lanka's public debt, which was already on an unsustainable path before the pandemic, is estimated to have risen from 94% in 2019 to 119% of GDP in 2021. India, China and IMF to the rescue, to address the economic problems, Rajapaksa's government has restricted imports of several items which have been declared non-essential. It has also approached India and China for assistance. It's reported on Monday that Colombo has sought an additional credit line of $1 billion from India to import essential items, after Sri Lanka Finance Minister Basil Rajapaksa signed a $1 billion credit line with New Delhi earlier this month. In addition to the credit lines, India extended a $400 million currency swap and a $500 million credit line for fuel purchases to Sri Lanka earlier this year. Meanwhile, Sri Lanka has asked China to restructure its debt repayments to help navigate the financial crisis. The country is also in talks with China for a further $2.5 billion in credit support.


Okay, so this is the last lecture. So because this definition of mixed elliptic motive, universal mixed elliptic motive is long, I will repeat it very quickly. So recall. So the first thing is a universal mixed elliptic motive. It has various pieces, and I'll say over M1, 1. Once at one time I'll need it over M1, vector 1, and I'll have some notation there, but MEM will denote motives over this guy. What does it consist of? First of all, the first thing is a V in MTM, and by that I mean MTMZ. So a mixed-tape motive. And it's weight filtration. So the whole story is a little confusing because there are two weight filtrations. The weight filtration in MTM is denoted by M dot, not the usual W dot. And there's another filtration, W dot, which is also weight filtration. And it's a filtration by sub-objects in MTM. And then the second piece of data is a representation row from SL2Z, which you can naturally identify with pi 1 M1, 1 with base point DDQ into the automorphisms of V, Betty. You take the Betty realization of this guy. And that preserve the filtration W dot. They don't, it doesn't preserve the actual weight filtration of V, Betty. And then such that each GRWM of V, Betty is a sum of symmetric powers of H. So Heway just talking about a representation, the irreducible representations of SL2, the algebraic group are just the symmetric powers of the fundamental representation. And N will be congruent to M mod 2. But you don't, this will follow anyway from the other conditions. And note that this implies here, for example, that T, which is 1, 1, 0, 1 acts unipotently because you have the filtration W. And each graded quotient is one of these and Tx unipotently on each graded quotient. So we're going to set N equal to log T. And you can add this as a requirement here or later, M dot is equal to the relative weight filtration of N, this nilpotent endomorphism. So this is nilpotent acting on V, Betty, W dot. So if you have a filtered vector space and nilpotent endomorphism, you may have a relative weight filtration. And here you definitely do. It's part of the requirement. And so this also leads you to a filtered local system, VW dot, over M1, 1 analytic. I guess here I should have put A. All right, so we want this to be part of a variation of mixed hard structure. And I should point out here that the fiber over DDQ, in the sense I mentioned before, is naturally V Betty, right? To define this, you would have to take Deline's canonical extension of this and then put a Q structure on the fiber over the cusp. And it would be exactly this naturally identified with this rational vector space. And then the next part of it, maybe I'll put it over here, is three. You've got a Q vector bundle. So this is the Dirom story, V over M1, 1 bar over Q. So that's the same with the connection, with a flat connection. So NABLA will take V into V tensor omega 1 M bar log P, where P is the cusp. So it's got logarithmic singularity. And also you want a hodge filtration, F dot, of this guy here. And you want that NABLA of Fp of V is contained in Fp minus 1 of Griffith's Traddon's Facility. And a weight filtration. So all of these are defined over Q. These are by flat sub bundles. And you want the fiber over the cusp P to be naturally identified with V. You fix an identification with the Dirom realization of your mixtape motive. And then the next thing is you have an isomorphism. If you take V tensor C, it will be isomorphic to the canonical extension of V tensor OM1. So this is a flat vector bundle. And it should be isomorphic to this guy here. And all these isomorphisms should be compatible with the isomorphism between V-beddy. This should respect the isomorphism with V-beddy tensor C as isomorphic to V-derom tensor C. And then together, so this V and V is an admissible variation of mixtod structure. So we've got this admissible variation. And the last part is the allatic etal part is that, so let me just draw a diagram. So we have SL2Z. This is isomorphic to pi 1 of M11 analytic over our base point. We have a representation here into ought V-beddy, but I'm going to tension it with QL. So I just think of it as acting on this larger vector space. This guy here maps into pi 1 of M11 over Q-bar ddQ. And this map here is just profinite completion. And then here you'll get ought of the allatic realization of V. And these guys here are isomorphic via the comparison isomorphism. And so you're getting a representation here. And I'll call this rho hat. And the last condition is that for each L, maybe I'll call it rho hat L. For each L, rho hat L is GQ-equivariant. GQ acts on the allatic realization here. And it also acts on this geometric fundamental group of M11. And that's the whole definition. So it's a mouthful. That's why I repeated it. And so the other thing I want to review very quickly, I'll say it in just two words, is relative completion. So this is the second thing to recall. And here I'll just do it for SL2Z. So C is equal to the category of finite dimensional representations of SL2Z over Q. So representations on the finite dimensional vector spaces that admit a filtration. It can be any filtration that has this property. I'll call it W dot such that the action on each graded quotient of this representation is actually a representation of the algebraic group SL2 and then SL2Z. So this notation is meant to mean that the action of SL2Z on each graded quotient factors through a rational representation of the algebraic group. It's actually a polynomial representation. It's the same as the sum of this at h. Sorry? Which is the same as the assumption answer. So this is just abstract. I just correspond to your assumptions. It's a sum of this. That's right. These local systems will be in that representation will be in here. So rho is in here. But we can just consider all these representations anyway. And so in other words, each of these here is just a sum of things SMIH, right? And so this is a Tanakian category. So the relative completion of SL2Z is the Tanakian fundamental group of this category with here. This is just the underlying Q vector space. So and what does this guy look like? It is an extension. Maybe I'll put here. We have 1 into url, into grl, into SL, I'll call it SLH. And we have SL2Z here. And I'll be a bit sloppy. I should write Q rational points here. Here's the standard inclusion. And there'll be this lift here. This lift is the risky dense. And this guy here is pro-unipotent. And I can't remember exactly how much I told you about this last time. I'll say a few more words in a minute in a little while. But excuse me. The filtration is not this. In the category, the filtration exists, but it's not part of the category. No. It's just representations that admit a filtration. But when you are aware of the morphics from one representation, and nothing else, no, the filtration is. For example, I could, if I had a graded quotient isomorphic to this, I could refine the filtration to split each of these into a different graded quotient. The identity mapping and its inverse should be mapped in the category. So we'll compute what this is. It's going to boil down to computing what this group is here in a little while. And then the goal is that we want. So our goal is to compute pi 1 of MEM with respect to some useful base point, usually Betty or Durham. So let me just remind you of some things we said last time. The first is that we have pi 1, what I call geometric. The geometric part of pi 1 of MEM. And then I map into pi 1 MEM. And then I'll map into pi 1 of mixtape motives. So if I take a representation of that, it's a mixtape motive. When I pull back here, it's a geometrically constant mixed elliptic motive. There's a section here coming from the base point because we know the fiber over the base point is a mixtape motive. So if you take a mixed elliptic motive, and you forget everything, but the initial V, you've got an object of this category, and that's giving you a section. Also, I'll be sloppy here just to save time. If I put QL here, just to give you an idea of what might be going on here, I could put pi 1 of M11 over, say, Q bar, pi 1 of M11 over Q, and then GQ here. So we've got the standard exact sequence. There's going to be a map here. This is going to correspond to looking at the monodromy. Basically, this representation here, there's a natural map here, and we'll get an induced map here. So this exact sequence really does look like the usual exact sequence of the geometric fundamental group, the arithmetic fundamental group, and the Galois group. All right, so that's the end of the rea. And I think you'll be a musician. Yeah. So that's the end of the review. So now the little test. So let me do the density argument because I was trying to explain that in a clumsy way last time, and it wasn't complete. And also, I have to do this stuff anyway. So I'm going to start with a theorem. And it's a special case of a general result, which I don't want to state. And the general results, actually, with something I've been trying to write up for years with various people. I'm now going to write out with Greg Perlstein. I'm going to be with Matsumoto and Teresoma. But I discovered that when you try to write a paper with four authors, nothing much gets written except. But anyway, and so this version is written up. I put a paper on the web the other day called Hodge-Durham Theory of Modular Groups. And it's proved in there, or it's sketched in there. So the theorem says this. For each base point, and I should point out here, the base point we most want to use is ddq of m11 analytic, the coordinate ring of g rel. So implicit in this is that I'm identifying sl2z with the fundamental group of m11 analytic with that particular base point. In the case, the way I've set things up, sl2z is naturally isomorphic to pi1 of this with this base point. And this is the one I'm going to be using. So you might as well just think of this base point. It has, well, is a hop-algebra in the category of end mixed-hodge structures. So that by this, I mean it's a direct limit of mixed-hodge structures. And the coproduct and the product and the antipode all preserve hodge and weight filtrations, queue structure, and so on. So for all finite base points, I prove this in a much older paper. And the second part is, so let's let mHS m11h, this is going to be equal to the category admissible variations of mixed-hodge structure over m11, whose weight-graded quotients are sums of, sums as variations of, as polarized variations of hodge structure, sums of symmetric powers of h tensed with a, where this is a constant hodge structure. So note I'm allowing more than just take things here. If you restrict it to just qn, this result is not true. Those are called Eisenstein variations. OK, so then the fiber at the base point induces an equivalence of categories. It's going to take mHS into what I'll call hrep grl. And what do I mean by this? So this is hodge representations. And so what do I mean by this? A hodge representation would be a v equals a mixed-hodge structure plus an action v into, well, if we write it down as an algebraic group, right? So if v is a mixed-hodge structure, then it's a mixed-hodge structure, deep statement. And if v is a representation of this group, you have this map here. And now this guy we know has a mixed-hodge structure. That's what part one says. And so we require that this be amorphism. So this is amorphism. You could also say, if the group's connected as it is here, if you took the Lie algebra of the group, that's what it means for this to be a hodge representation. So implicit in this is somehow that when you take the fiber at the base point, you get a mixed-hodge structure that the monodrami representation's amorphism. And conversely, if you have a mixed-hodge structure and a hodge representation, you can construct the variation. And the variation will be unique by the theorem of the fixed part. And so there's a version of this I proved with Steve Zucker almost 30 years ago. Well, he probably proved it 30 years ago. It was published in 87. That's the unipotent case. But this is the general case. And so this is useful to us. And so let me restate this in another way. So the corollary is that pi 1 of mhs m11h is isomorphic to pi 1 of mixed-hodge structures, semi-direct product G rel. So why does this make sense? So this is a Tanakian category. This is the fundamental group of this Tanakian category. This group has a mixed-hodge structure, which is the same as saying this group acts on it. So I can form the semi-direct product. And if you think about it, a hodge representation of this group is the same as a representation of this group. Representations of this, representations of this whole group equals hodge representations of G rel. It's a similar situation to this. I mean, we've got a splitting here coming from DDQ. What's a representation of this group? You can think of it as a GQ-equivariant representation of this group on a Galois module. Just exactly the analogous statement. All right, so now here's an observation is that we can take MEM, and we can take it to M. And we just take a mixed-hodge. We take a mixed-elliptic motive. It's just going to go to the underlying variation of mixed-hodge structure. And now, if we use the fact that the hodge realization of a mixed-hodge motive is fully faithful, you can see this guy here is fully faithful. So another corollary of this, or I've written down as a theorem in the notes, we'll put it on the next one. So theorem is that G rel, the natural representation I constructed last time from G rel into pi 1 geometric of mixed-elliptic motives, is subjective. And this is good because we're going to be able to compute what this guy is. And that's going to bound the size of this. So you remember that somewhere on those boards here, we have pi 1 of MEM as an extension of pi 1 of mixed-hate motives, which we know by pi 1 of MEM. And so let's do the theorem proof. This statement here, this statement right here, says that star is equivalent to saying that pi 1 MHS, semi-direct product G rel, subjects onto pi 1 of MEM is subjective. And now, and then you just look at the diagram here. You'd have G rel pi 1 MHS. The idea is pretty straightforward here. And this is pi 1 of MTM. And in both cases, you have a splitting given by DDQ. You can check everything fits together. This map is just induced by the hodge realization of MTM. This is subjective. This is subjective. And you can, this guy's subjective. Another corollary, this may be obvious from Tanakian considerations, but we get it for free here, is, well, I'm going to explain, this guy cannot be an isomorphism. So I'll just give you a little preview. The unipotent radical of this is generated by Eisenstein series and cuss forms. The unipotent radical of this is generated only by Eisenstein series. And so the kernel is generated by cuss forms, and they're going to impose relations down here. And here, this is huge, but you might ask, what's the smallest category of mixed hodge structures you can put here? And Francis has something that he's pushing, is what he calls mixed modular motives, basically the hodge structures you got out of modular forms. And so it's going to be bigger than just pi 1 of MTM, because it's got to be big enough to contain the hodge structures that correspond to cuss forms. So a corollary, I'll just put it over here, is that the kernel of pi 1 M E M, there's a map to GL of H. And remember, this is the fundamental group of the category of semi-simple mixed elliptic motives. So anyway, the statement is this is pro-unipotent, but that should just follow because the kernel is what controls extensions of these, and you can only extend by stuff of lower weight. But anyway, it certainly implies this result. So we're going to denote the kernel. So the kernel I'll denote by U M E M, I think. So we've got an extension 1 into U M E M into pi 1 M E M into GLH. And again, you can split this guy apart into its geometric part. You've got U M E M will map to what I call k. This is the Lie algebra. Let me put Lie algebras here. So they'll be the geometric. So I can break it into the part coming from mixed-tate motives. So this guy here is essentially the free Lie algebra on Z3, Z5, Z7, and so on. And this guy here is the geometric part. We're going to see that this guy up to the action of SL2 is generated by Eisenstein series. That's our next goal. So let's understand. So start up here. You've got 1 into U rel, and it maps into G rel, maps into SLH into 1. And this guy's pro-unipotent. And now Levy's theorem says we can choose a section here. So Levy says we can do this. So this says our G rel is isomorphic to SLH, semi-direct product U rel. Not only that, I should say this is unique up to conjugation by an element here. So it's unique up to conjugation by U rel. Sorry? You don't have a canonical one. No, not that I know of. I mean, you can write down. You can prove there are natural choices, but not canonical choice as far as I know. OK, so once you put this in here, this group acts on this. So it says that U rel, that's the Lie algebra of the kernel, is acted on by SLH. So it's a pro-nipotent Lie algebra in the category of SL2 representations. So this is pro-nipotent Lie algebra in rep of SLH, I should say, sort of completed representations. So now we want to write down a presentation of this. And I should also point out this guy is exactly U rel. So knowing this is exactly the same as knowing that. You lose no information by going to the unipotent radical. So the next thing to do is you've got U rel, and it maps into H1 of U rel. This is subjective. And it's also SL2H-equivariant. So you can choose an SL2-equivariant section. So this is SL2 or SLH-equivariant section. You just choose one. So this will give you a suggestion from the Lie algebra generated by H1 of U rel. Well, it'll map to U rel. So the section induces this. And I have to complete, and this will be subjective. So this is the first step towards writing down a presentation. In fact, it will be minimal because you can't make the generating set any smaller here because it's an isomorphism on H1. And now what about, so let's let the kernel be R. So this is some ideal. So this is contained inside L bracket L because it's a maps an isomorphism on H1. And L is the freely algebra. Saving chalk. So HOPF, there's the standard argument for H2 of a group. There's a formula due to HOPF. It's easy to apply equally to Lie algebras. And if you use this condition here, you see that H2 of U rel is isomorphic to R mod R L. And so if you think about, this is all relations, modulo consequences of all relations. So it's essentially a minimal set of relations. So I want to stress this sort of argument gives you, by the way, I should say this HOPF argument, all it does is just telling you that the calm of a freely algebras trivial and degrees greater than 1. So what you can do here is you've got a map of R onto this. Choose an SL2, SLH section. And now a little exercise is you can see the image. The image of this section will be a minimal set of generators of this ideal. So what you will see is that U rel is isomorphic to the freely algebra on H1 of U rel completed. Modulo, let me, if I call this section Pc, you can look at the image of Pc closure. And the image of this guy is isomorphic to H2U, because the image here exactly has to inject. It's the ideal generated by image of Pc. So the relation between H1 being generators and H2 being relations is much stricter for pro, it's perfectly strict for pronopodently algebras. So the reason I wrote that up is because the game is we're trying to find a presentation of the unipotent radical of pi1 of MEM. You don't need any assumption about the H3, et cetera. To stasis, you have no assumption about the ideal homology book. No. By the way, this discussion works equally well. I just did it here for U rel, but it works well for any pronopodently algebra in the category of representations of a reductive group. All right, so here's a theorem, and it's by this person called easy, is that H1 of U rel tensored with SNH. And if you take the SLH part of this, invariant part, this is isomorphic to H1 of SL2Z. And I should point out here that this vanishes when N is odd. That's standard. The whole point is that minus the identity is central in here. You can see it with modular forms, but for group homology, there's an argument called center kills. And minus 1x, non-trivially on these, so it has to kill those when N is odd. And so this is good, because we want to find a presentation here. Well, we just have to understand these homology groups of SL2Z, which is going to be Eichler-Chimura. And now we need to at least bound H2. So 2 is that H2 U rel. So this is telling us about the SN isotypical part of this guy here. It's a representation of SL2. It's telling us it's this. And this guy is contained in H2 SL2Z SNH. But this group here is trivial, because this is because SL2Z is virtually free. So a group has virtually a property if it has a finite index subgroup with this property. SL2Z has a finite index subgroup, which is free. Just take any modular curve that corresponds to a torsion-free group. It's an affine curve, so its fundamental group is free. It's perfectly standard. And then a little spectral sequence argument shows that as long as this group here is sort of divisible, the cohomology vanishes. So what's this telling us? It's telling us that H2 is 0, so there are no relations. So it's telling us the unipotent radical is free. Corollary, U rel is free. So U rel, it's isomorphic, but not naturally isomorphic to dual tent completed. And you have to complete, you have to look at all the finite dimensional nilpotent quotients and take their inverse limit. And I should point out here, this here. So yeah, I'm going to point out something here, which maybe I don't have to do it, but I will. Because I think it makes everything clearer, observations and notation. So remember, H is the fiber of Deline's canonical extension over the vector ddz. And so H, the first one is that H has a natural basis. So H we know is isomorphic to Q plus Q of minus 1. That's a calculation we did in an early lecture. And I can write out either the Dirac, or write out both the Dirac version and the Betty version. So the Betty version, this is Q of A plus Q of W. And in the Betty version, it's Q of A plus Q of B. And the relation between W and B is W is equal to 2 pi i inverse times B. And so, right. And we actually computed this. If we used a tangent vector, if instead I used lambda times ddq, where lambda was not plus or minus 1, I couldn't say this. Because this would be a non-trivial extension of this by that. And you wouldn't have a natural basis. But because it's split, you have a natural basis. And sorry, this should be a minus. I'll put minus W equals that. But it's a Dirac sum, not an extension. It's not an extension. It's a direct sum, it's saying. So the point I want to make is that the SL2 has a natural torus. And we're going to have a natural roots in all of this. So I mean, the representation theory is trivial, but we want to nail down a torus and so on. So and also, if you take N to be log T is equal to A, A, D, D, B. And the Dirac version is equal to A, D minus A. Yeah, it's a minus there, minus A, D, D, W. These are all followed from calculations we did earlier. And it lowers weights by 2, lowers m dot weights by 2. It's getting confusing because we have SL2 weights, m weights, and w weights. They're all related. And so what we're going to do is we're going to set E0 equal to this guy here, minus A, D, D, B. And so I've taken a symmetric algebra of the pitch. When you write it, it has a different zero perimeter. Yeah, because I want to know how to apply it to two elements of symmetric powers. So what does this tell us? It says there is a natural choice of torus. So we'll have A will have weight 1, B will have weight minus 1. It's natural to give A has got weight. I want it the other way around, I think. Sorry, A has weight minus 1 and B has weight plus 1. So this has lower m-hodge weight than that. And I want N, N, and N E0 has weight minus 2. Has SL2 weight. These are SL2 weights. And another notation I want that Francis doesn't like is that, so S, N, E is defined to be the SLH module with highest weight vector E. So we're going to need lots of different copies of SN. And I want to be able to label them and write down their elements. And so this is just going to be the span of the E0 to the J times E mod E0 to the N plus 1 times E equals 0. So let's look at a quick review of Eichler, Chamorro, Mann and Drinfeld. And I put Zucker's name in there as well, because I don't know if Zucker belongs in here. And I think of it that he does, because I use the way he does the Hodge theory for SL2. The Hodge theory has to be put in the framework of the whole Deline machine plus Mann and Drinfeld. So one is that H1 of M11 analytic, S2NH, which is the same as H1, SL2Z, S2NH, has a natural mixed-hide structure. And when N is 0, I'll leave it for you. H1 and H2 both vanish. So 2, I'll take N to be positive. And then you have an exact sequence, the cuspital cohomology, which is the same actually as the intersection homology of M11 bar with coefficients in S2NH. This maps to H1. And then it maps to Q. It's actually useful to think of this as S2NH mod E0 times S2NH. It's naturally isomorphic to this, twisted by minus 1. It's the co-invariant. And so I have colored chalk here for a reason. So this part here, this is a Hodge structure of type 2N plus 1, 0, and 0, 2N plus 1. And this part here is of type 2N plus 1, 2N plus 1. So it's really a copy of Q minus 2N minus 1. So yeah, a question is, who is the first person to say these should be the Hodge numbers of this and write it down and explain it? I mean, that's why I give Zooka credit here, I mean. So right, and so this part here, these are the cusp forms of cusp form weight 2N plus 2. Yeah, we have M weights, W weights, weights of cusp forms and SL2 weights. And these are the anti-holomorphic, I'll just say, their complex conjugates. And this part here is corresponds to the Eisenstein, G2N plus 2. And I'll say a little bit more about that in one second. Three is that this is Mann and Drinfeld. This extension is split. By over Q, I mean I'm viewing these as Q mixed Hodge structures and not Z mixed Hodge structures. So to direct sum, the middle guy is naturally a direct sum. And you split it using HECA correspondences. You were asking for the extension, and you said you see there's no extension. That's correct. So I guess I think Mann and proved it in weight 2 or something, and then I guess Drinfeld probably realized the same argument applies in general. And let me just say a word about how cusp forms determine or modular forms determine the cohomology class. So F is a modular form, say a holomorphic modular form of weight, say 2N plus 2. You can define omega F to be equal to, I like to think of it this way, it's F of tau times W to the 2N. So I define this guy to be a section of the bundle H. Remember, W at tau is equal to 2 pi i times dZ, which you think of as an element of H0 omega 1 of C mod lambda tau of the lattice associated with tau times dQ over Q. And if you recall, if this looks different from the usual picture, this is just minus 2 pi iB plus log Q times A, or if you like 2 pi i tau. So it's the usual formula. And this guy is an element of, it's a one form on the upper half plane with coefficients in S2NH and it's SL2Z invariant. But that's the same thing in this Orbefold world as a one form on M11 analytic with coefficients in S2N. Here I should have put, that's right here. So, and this gives you a class in F2N plus 1 H1M11 analytic. So that explains these things here. And the other classes are the complex conjugates of these. Yeah, why is it in this level of the Hodge filtration? This guy is in F1, so it's 2 Nth power is in F2N. And you're picking up, this guy's giving you an F1 here. OK, so I'm going to let B2N plus 2, this equals the normalized Hecker eigencust forms. I never know how to punctuate this of weight 2N plus 2. So we get a basis. Yes. And so what's a basis of H1?  And if I need to, I'll denote this by psi 2N plus 2. So that you get a natural basis of cohomology out of normalized cuss forms. But you need a cuss form in its conjugate, but you only need one Eisenstein series. And this guy spans the copy of this guy here. OK. All right, so maybe I'll write this down very quickly. And then I'll stop for a break. But let's try to put this together. So let's let EF double prime F in B2N plus 2 be a basis dual to the above, a basis of H1 dual to this one up here. So remember that H1 of U, recall that H lower one of U, rel, it's going to be isomorphic to the direct product of H1M11 analytic S2NH dual tensor S2NH. So these are just vector spaces. And this is where SL2 acts, right? And U rel is unnaturally freely generated by this guy here. So note that we have lots of copies of S2N. We have ones for a different thing. So basically, the trick is to find a basis of this. So yeah, and also we have that U rel is isomorphic to the freely algebra on H1 of U rel completed. So this is telling us that U rel is isomorphic as an SL2 representation, but not naturally. It's isomorphic to the freely algebra generated by the direct sum n greater than or equal to 1 S2N E2N plus 2. So that's saying you take a copy of S2N for the Eisenstein series. This is the guy dual to the Eisenstein series plus the sum over the eigencust forms. I'm just trying to give you some idea of how big and nasty this thing is. Well, it's not nasty. It's actually very beautiful. But for every cuss form, you get two copies of SN because the cuss form gives you a two-dimensional hard structure. And it's S2N EF double prime. And so think about what elements of this are. I mean, an element of this would be something like EF prime times E0 to the J, right? So the usefulness of the E0 is it generates all of these pieces, right? And so everything in here, so this is somehow, I'll be very sloppy, it's generated by E0 to the J times E2N plus 2s, E0 to the Js times EF primes, and E0 to the Js times EF double prime. And you have to take all ns and all fs and so on. Notate twists. Well, I'm going to put these guys in the natural place. So let me, yeah, so I've got my color here again. This part here is isomorphic to S2NH2N plus 1. And so this is going to have w-weight, this has w-weight minus 2N minus 2. That's why I put this 2N plus 2 there. It gives you, it's negative as the weight. And these pieces here, these have w-weight equals minus 1. Yeah, and this is actually, at first it looks bad, but it turns out to be very good. And in the notes, I won't put it on the board. I have a picture of the hodge types. So I, and all of this stuff I'm doing here is in this paper I recently put on the archive. The hodge-derarm theory, it's in detail. So let's take a break, a coffee break. And then the next goal is to see what this says about, so after the break, our goal is to understand basically U-M-E-M. We're not there yet. I mean, we being, say for example, Francis and myself, we're trying to understand what this guy is. And we can, we know how big the relations are, unless there's something very bizarre that violates the order of the universe. So we know, what do we know? So we have an exact sequence. So this is, this is the unipotent part of pi1 of M-T-M. I'm concentrating on the unipotent parts, or the pro-unipotent parts, because the reductive parts are all well understood. The reductive part here is G-M. The reductive part here is G-L-H. And the reductive part here is S-L-H. And so these are the unipotent parts of pi1 M-E-M, the geometric pi1 of M-E-M. And this is the geometric part of mixtape motives. And so we have a corresponding exact sequence of pro-unipotent the algebras. u, I think I might just call it u-g on, maps to u-m-e-m. And it will map to k. And remember, this is the free guy, Lz3, Z5, and so on. And, right.  And this is silly, because what's the proof? We have pi1 of mixtape motives. It maps into the automorphisms of pi1 of mixed elliptic motives by the base point ddq. And it preserves the homomorphism to G-L-H. So that's saying that pi1 of M-T-M acts on this, and it also acts on that. So therefore, they're both mixed T-Motives. So h1 of, right. But yeah. So this, a corollary is, let's put it this way, a corollary is that h1 of u-m-e-m and h1 of u-g-m are, you know, have pro-mixed hodge structures of T-type. All their graded quotients are of type Pp, right, because they're hodge realizations of mixed T-Motives. But they're also quotients, they, well, but u-g-m is a quotient of u-rel. Sorry about all these abbreviations. We have many kinds of weights, and we have many kinds of adjectives we apply to the algebras. So a pronilpotent, the algebras in some sense generated by its h1. So we need to understand the map on h1. So we've got u-rel maps onto u-g-m. Remember, to make it absolutely clear, this is the motivic guy here, and this is the relative completion of sl2. And now I meant to put h1 in here. So the first thing we want to understand is how big can this guy be? Well, we know this guy here is the direct sum from what I did over here up there, is that this is the direct sum n greater than or equal to 1, s2n, 2n plus 1. That's the Eisenstein pot. Plus it's got the Cuspital pot. I'm going to write it like this, vf-dual, tensor, s2n, h. And here, vf is equal to the span of, OK, I'm going to be slightly sloppy. These guys are really defined over a number field, a totally real number field, but I'm going to pretend they're defined over the reals, or the rational.  This is the one form that represents the cohomology class corresponding to f. So this is a two-dimensional hodge structure. Sorry? No, no, it's of type 2n plus 1, 0, and 0, 2n plus 1. So, and this guy here is Tate. When I say it's Tate, I mean, at the, this is where it gets confusing. We've got to think of it at the base point, ddq. At ddq, the symmetric power of h here becomes mixed Tate. So, but this guy can never be mixed Tate. This guy will degenerate to a direct sum of copies of q of n or q of j for some j. This guy will never go away. You can never make these hodge numbers equal, right? So this part here has to be the kernel. This part here, and so this is a key thing to understand, this part here goes to 0. This is the kernel on h1. And we've got to be careful because h1 is not the canonical generators, but we have to use the fact that in the Land of Hodge theory or the Land of Motives, the functors grw and grm are both exact, all right? So what does this imply? So this implies that h1 of u, gom, the motivic guy here, mem, is a quotient of the direct sum n greater than or equal to 1 s2nh2n plus 1. So we know the generators there. So it's generated by Eisenstein series. You've got to correctly interpret these statements, but they're correct, suitably interpreted. So a corollary of this is that, well, I'll write down as a proposition, is in fact that h1 of u, mem, is equal to the direct sum n greater than or equal to 1 q of 2n plus 1 plus the direct sum n greater than or equal to 1 s2nh2n plus 1. So this is the Eisenstein part. So this is also the geometric part because this much is h1 of u, gom. And this part here is the part coming from k. This is h1 of k, the Lie algebra generated by the odd zeta values. So these are zeta values. So this is the mixtape motive story. So the Eisenstein series are the geometric part, and the zeta values are like the Galois part. Sorry? That gives the generators. They will give generators. They won't live canonically to generators. So the next thing I'll talk about discusses the Eisenstein quotient. Why do you know those Eisenstein elements can kill? Yes, sorry. I forgot to explain that. So proof, we only need to produce a mixta elliptic motive where all these guys are non-zero. And so proof, one is, well, the best one I know is that in the beginning I wrote down a mixta elliptic motive. So p, which is the mixta elliptic motive, which was defined to be the Lie algebra of pi 1 unipotent of e prime, elliptic curve minus the identity of over ddq, so essentially the first-order t-curve, with a suitable tangent vector as an object of mixta elliptic motives, or pro-elliptic mixta elliptic motives. And I should say over the thing with vector 1. And then there are various ways to see that all these classes are non-trivial here, but I have this paper called Notes on the Elliptic KZB Equation. So it comes out of this work of Levin and Rassenet that you see all these extensions are non-zero. So this implies, so you get a representation, say, of u geom into drp. And in a little while, I'll write down where the e2n plus 2 goes to something not equal to 0. You can write down the explicit derivation. You also get it out of, I haven't checked everything, or the elliptic logarithms. I'm sure everything is in the paper of Baylinson and Levin, everything. I need to check this, but I know the elliptic poly logarithms are motivating in this sense. So they realize the n-th elliptic poly logarithm restricted to the 0 section realizes the n-th one of these. All right, so what we have here is we have g rel, the le algebra of the relative completion. And this guy's definitely not a Tate hard structure. It's got a hard structure, but it's not Tate. Even at not Tate at, say, ddq, because you have these, in its pure quotients, you get the hard structures of cuss forms appearing. And now it's mapping into g of, well, say, maybe the geometric pi 1. It maps on to this one here. And this guy here is Tate. It's mixed Tate. It's a Tate hard structure. So you might ask yourself, what's the maximal Tate guy you would put here? And I want to call it the Eisenstein quotient. So what's the biggest quotient of this, where all the graded quotients are of the form s, n, h, sums of things like this, where I do not allow things like this tensor with vf. These guys are out. These guys are in. And see, when I go to the base point, ddq, h becomes Tate. So this guy will be mixed Tate at the base point. So that's what the Eisenstein quotient is. So g i's is equal to the maximal quotient of g rel in the category of groups with mixed hard structure whose weight graded quotients. I should say the weight graded quotients of its Lie algebra, but I'll be sloppy, are sums of s, n, h, r's. And it's also, you can check that it's pi 1 of the category of admissible variations of mixed hard structure over m11 with grw dot of v, a sum of s, n, h, r's. Meaning you can't have snh tensed with a constant hard structure like v of f. So this is a, well, it's interesting that this guy is not going to be the same as g rel because of these cuss forms. So we have factorization. We've got g rel maps to the Eisenstein quotient. It's called this because it involves, it's Duram theory only involves Eisenstein series. We're going to have the geometric pi 1 of mem. So we have this factorization. This is subjective. And now you might ask, what can you say about, first of all, what can you say about g ice and what can you say about this map? And I'll jump the gun a little bit and say there's quite a bit of evidence to suggest this is an isomorphism. So let's see if we can understand this. I'm going to jump ahead a little bit and jump back, maybe not. All right, so we want to understand this. So let's get some sort of lower bound on what it's like. And so let's say project is understand. So fact is that if I looked at pi 1 of mem, sorry, I didn't do this right. So if I look at g rel, and I'll call it 1 vector 1, so this equals the relative completion of pi 1 of m1 vector 1 ddq, well, appropriate base point here. Yeah, I know what I wanted to write. Well, I'll write this here. I'll call it 1 1 crossed with what I'll call ga of 1. Just a copy of ga, but the corresponding hod structure is of type minus 1 minus 1. And I also want pi 1 of mem 1 vector 1 is isomorphic to pi 1 of mem, the usual 1 cross ga of 1. So this has proved in that manuscript I've put on the archive that you get this product here. And you can also prove this. It's not hard. And the reason I introduced a base point is I want to consider the action on the unipot and fundamental group of the take curve that I wrote down over there. So we have an action, say, of g1 vector 1 into ought p, the appropriate base point. So this means li pi 1 unipot. And because of this statement up here, the map down to g1 1 actually is split. So we have a representation into here. And we can look at li algebras. And I should point out this guy here, remember, is it's a mixed elliptic motive. So everything here is take. So I'm taking, if you like, the take curve is a curve on the disk. And I either think of it in hodge theory, I think of it as being the fiber over the tangent vector ddq here, or in Galois theory, you think of it as, say, q bar q to the 1 over n. It gives me elliptic curve over this guy here. So this is a morphism of mixed hodge structures. Again, this follows from an old result. I also prove it in these elliptic kzb nodes. And so you get a map here from url, so I should say rel here, you get a map from url into the derivations on this p. So let me just call this p. And now I can take the weight graded of url into the weight graded here. And I end up in what's called dare0. So graded weight p is naturally isomorphic to the Freely algebra on h. So the weight filtration for this Lie algebra is its lower central series. Its h1 is canonically h, which gives us this thing here. So I'm looking at the Freely algebra on h. So this is canonical. And now, because Grw is an exact functor, you lose no information by going from here to the associated graded. This is a standard, hodge theory or standard motivic type argument. And now, we have the e to n plus 2s here. They go to certain derivations here. And they're easy to write down. I won't do it. But let me, so we're trying to understand if these e's satisfy any relations. And I should say, before I do this, this fact is through Grw dot u i's into here. And now, the e to n plus 2 is going to go to a derivation. And just to get the normalization right, yeah. So I'm going to define epsilon to n to be equal to the image of e to n times 2n minus 2 factorial divided by 2. I believe I have the correct normalization. So just a rescaling of these guys here. Thank you. So right, so this mapping class group G11, you can think of it as the mapping class group of a genus one curve with one boundary component. And so those mapping classes act trivially on the boundary. We're actually using a tangent vector, same thing. You're not allowed to turn the tangent vector. So it's hard to write down what that condition is here. But on the associated graded, you're looking at the derivations delta, such that delta of the bracket of the two generators is equal to 0. So that's what dare 0 is.  Yeah. And so H, remember, naturally has this basis A and B. So when I started doing this, which believe it or not was around 2007, some undergraduate named Aaron Pollack walked into my office. And he wanted a problem. And for reasons I'll explain in a little, so he's now a postdoc at Stanford. He got a PhD at Princeton. But he was just this young undergraduate. And there are reasons to believe, which I'll explain in a minute, that these epsilon should satisfy relations, one for each cusp form of each degree. So for example, the cusp form delta will give you quadratic, cubic, and so on relations between these derivations. Because we expected them, Matsumoto and I, to hold in the motivic case. We couldn't prove it. But they should hold here. And lo and behold, he found the relations. So let me write down the quadratic ones. So he says the sum j plus k equal to n jk positive cj epsilon 2j plus 2 epsilon 2k plus 2 equals 0 in their LH. If and only if there is a cusp form, now quadratic's absolute. I'm only doing quadratical. There is a cusp form f of weight 2n plus 2 such that rf plus of ab. So I'm thinking of this as the module of symbol. And I always think of these things as having meaning. And I'll write down the definition. But this is the even part of the module of symbol of f. It's the sum j plus k equals n. Here's j and k are greater than or equal to 0. So this doesn't have the highest order part. cj a to the 2j b to the 2n minus j. So what's the module of symbol? So rrf of ab is equal to the integral from 0 to i infinity. So the integral up the imaginary axis of omega f. And I'll remind you, if I write that out, that's equal to the integral from 0 to i infinity of f of tau. I'm actually leaving out a factor of 2 pi i, which doesn't matter, b minus tau a to the 2n detail. And I'll put 2 pi i to the way I normalize something. There'll be a power of 2 pi i here. But it doesn't matter because the multiple of a relation is a relation. And rf plus, so this is a polynomial in a and b. It's an element. And rf plus ab is equal to the terms where the degree of a and b are even. And let me give you a simple example. That's the eigen-coefficient model as well. Sorry? That's the failure to the normal. Yeah, but it's not obvious that no, because if you have, it doesn't matter. It just every relation corresponds to a cusp form. It doesn't have to be, right? So if I wrote down, every relation, quadratic relation is going to be a sum of relations. And you can find, whatever the coefficients are of the ones that correspond to eigenforms, you just write, express your f in that basis, and you. It's true. Sorry? Yeah, there's no, you don't need to specify any basis here. So for example, if you look at the weight 12 cusp form, what you get is e4 e10 minus 3 times epsilon 6 epsilon 8 is equal to 0. And I could write the weight 16 ones in the notes. But so this is just the first one. And now, so the question is, are these relations motivic? What do I mean by that? I mean, do they hold, but the corresponding relations hold between the e's? i.e. the motivic guys, the e2n plus 2s, hold in u, geom, you know, the mem. I should also mention, he found, in some truncation of the derivation algebra, he found relations of every possible degree for every cusp form. All right, so let me try to do this in maybe the next 20 minutes. So I want to, the answer is going to be yes. And I want to explain why. So. So there's still some explicit derivation in the reagent by elevation. Right, I could write it down.  But they are explicitly defined. Yeah, they appear, you know, so if you look at the paper of Levin and Rassenet, they appear there. They appear in older work of Hirawaki, Nakamura, and they, in fact, I'll tell you how to find them.  I'm not good with formulas. If you look at dare0 LH, and you look at the part of, you look at graded weight minus 2n minus 2 of this thing here, it contains a unique copy of s2n. I think I put it in the right place. Yeah, it's a unique copy. You just take the highest weight factor. So I've written this up somewhere, but it's sitting in a drawer. They're just very natural, and they come up if you had to write down these derivations. All right, so I have to write, talk about cosmology and extensions. So we have forgetful functors. You've got M, E, M, and you've got, say, I'll call them Hodge M-E-M. And you've also got El-Addic, and they're both faithful. So for example, here you would look at the underlying variation of mixed hodge structure. Here you would look at the underlying least sheaf. And so what, so this story starts with Matsumoto. And exactly. And exact, yeah. So what we have is these induce subjective homomorphisms. Actually, on the El-Addic side, I have to explain why I get surjectivity, but I've got pi1 of M-E-M. And then I've got pi1 of Hodge. So I'm partly motivating what I'm doing here. And what surjects onto this is going to be pi1 of mixed hodge structures, semi-direct product with G-ICE. And on the other side, there's pi1 of El-Addic M-E-M. And the correct thing to put over here is the El-Addic weighted completion of pi1 of M11 over Z1 over L, DDQ. So this is where the story started, actually. Matsumoto and I, one day, sat down in spring of 2007 and computed this guy and got a strange answer that we didn't understand. And that was the basis. A couple of weeks later, I talked to Aaron Pollock. And he knew nothing about modular forms or freely algebras or SL2. But a week later, he had the first five relations. He's smart guy. So on both sides, so I'm going to talk generally in what I say will apply equally well on this side or this side. You get the same answer. And that is consistent with various conjectures and number theory about. But on both sides have an isomorphism on H1 of the geometric part, which is the part we're trying to understand. OK, so relations correspond to H2. So the H1 is X1. So on the hard side, what you get is that, and I don't have time to explain this, but it's an analogous thing on the Galois side that might be clear to. And I'm going to take f infinity. This means invariant unreal Frobenius, because everything here is defined over z, so it's defined over r, and that's an action. And this is going to be isomorphic to, and this direct sum, direct sum f in b2n plus 2 X1. Again, a lot of this is explained in this pre-print I put up. And the VFRs are OK here, because they're actually real hod structures, so I can write them down. These are these two-dimensional real hod structures. And I have to take f infinity. And so if you pull us apart, and I should say, this guy here is, oh, this is the hodge. This is in this one here. So this is going to be roughly in what I think I called mhsm11h, except it doesn't matter whether I restrict the graded quotients here or not. So anyway, what is this? This turns out to give you all the things that correspond to odd, there's a difference between odd and even, and here it's going to be r plus the sum of f in b2n plus 2 VFR plus, so this is a one-dimensional thing. It's corresponding to the plus part of the modular symbol. And this is for r even, and it's just equal to the direct sum of the f in b2n plus 2 Vf minus r. This is for r odd. So real Frobenius acts on this guy. It's two-dimensional. It has a plus 1 eigenspace and a minus 1 eigenspace. And they correspond to the plus and minus parts of the modular symbol. It's all written up here in Technicolor. And on the Galois side, it's a similar story. Let me just write the Galois guy here. This would be h2f, well, h2 of m11 over z, which you have to think about what this means, s2nhr. So it means if these are l-addict coefficients, unrammified outside l. And I'm told it's supposed to be crystalline at l. And then here, this would be h1f of, well, f, I'll just say, speck z of ql minus 2n minus 1. And this one here would be h1fgqvfr. So these are Selma groups. And you see this, if you look at the spectral sequence, maybe if I'd been a bit more honest, I would have put z1 over l or something here. If you look at the spectral sequence from this, from geometric, you know, cormology to this, you'll see this is the h2. So this is the Hodge analog. And they both work the same way. And so I'll make a little remark here is that these things shouldn't be too surprising because if you look at the realization maps from x1 mixed motives, whatever they are, or in Francis' favorite category, triple m, multiple, sorry, mixed modular motives, tensor r. If you look at these guys, and you went into the real x1 mixed hodge structures, rvfr, f infinity, this guy is supposed to be conjecturally an isomorphism. And you've got the eletic one. You've got x1mmqlvf, sorry, q. These guys are supposed to be, oops, too many things. These are conjecturally isomorphisms. So what should happen here is that you should be getting relations. This is telling us that this h2, there are all these different classes in h2 coming from cusp forms. And these should give you relations. And there's going to be two kinds of relations. And I'm only going to talk about one. These guys here should correspond to relations of the form. You can z2m plus 1, e2n plus 2. This should be another geometric quantity. And these relations here should be relations between, they should be geometric relations, meaning they're relations between the ease. So I'm not good at computing periods, but there's this guy, Francis Brown, and part of this special case, which says that the cup products, so maybe I need a whole blackboard to write this cup product down. It goes x1, mixed hard structure, m11, r, s2j, 2j plus 1. So this corresponds to a generator of the geometric generator of the unipotent radical. And now another guy, an x1, r, s2k. And then this guy is going to go into an x2, an x2, variations of mixed hard structure. Sorry, I'll just write the x2 by what I have over there. It's really an x1, mixed hard structures, r by vf of r. And these generators are Frobenius invariant, so they will land in the Frobenius invariant part. Whoops. Yeah, sorry. I screwed up here. Right, mixed hard structure. It's going to be an s2j. What I'm going to have to fix these nodes here. So it's x2, sorry? x1, x1, cannot go into x1. Yes, it does because of this guy over here. Right, but I've got to fix up what I've got here. It's s2j tends to s2k2n plus 2, so j plus k equals n. And now you can project this to an x2. It's the appropriate category here. An x2, first of all, of r by s2n minus 2r2n minus r plus 2, I think. Right, these are the various projections because the tensor product of these guys is not irreducible. It decomposes some of these guys here. And this guy here is, now you project onto some x1 mixed hard structure of r by vfr. And this is subjective. You've got to put the correct twist here. It's not vfr, it's vf, whatever. Subjective, every time this group is non-zero. And so what this tells you, I just try to summarize this quickly. Try to stop in one second. And I'm going to try to give a very brief, very concrete explanation of what's going on. I mean, this is all sort of abstract, homological argument. But the corollary, what's the whole point? The corollary pollux relations of all degrees, not just the quadratic ones. And the higher order ones, with a few exceptions, were all defined in some quotient. We didn't even know if they lifted relations in the derivation algebra. Lift to relations in u-g-o-m-e-m. Strictly speaking, I should say in the weight graded quotient of this. So all of these relations, and so they are motivic. And a remark is that if standard conjectures hold, so these are the ones that say these regulator maps, there are isomorphisms, then g-ice maps isomorphically onto pi-1-g-o-m-e-m. There's also the question of understanding the relations coming from Eisenstein series. It's been worked on. So and now I could stop here, but I would like to give you a very simple picture of why pollux first quadratic relation holds. I mean, all of this stuff is abstract, but there's a very concrete reason why it holds. And I'll stop right there. You have the one coming from the cusp form of weight 12. So what's going on? Why are we getting cusp forms when from, I'll just remind you, cusp forms were generators of in the relative completion. This was the free guy, but they become relations in u-ice. Why did this happen? So I'm going to draw a picture here. We have s2, so inside url we have, this is freely generated by a whole bunch of stuff. That includes s2 e4, s4 e6, s6 e8, s8 e10. And we also have v delta dual tensor s10. So this is really equal to s10 ef prime plus s10 ef double prime. So it's a really concrete thing. We've got a freely algebra, and among the generating stuff, we've got all of this stuff here. This has weight. What's the weight? The w weight of this is minus 4, minus 6, minus 8, minus 10. And this guy has weight minus 1. And if that's confusing, this has weight 10, and this has weight minus 11. And so in some sense, we want to work in the s10 isotypical part of the Lie algebra. So let's look here, and how can we get things in here? We could bracket this guy and that guy, and we're going to get something in the tensor product of these, but we can project on the highest weight. We're going to get an s10 with highest weight vector e4 e10. Plus, we're going to get an s10 with highest weight vector e6 e8. And while looking in this s10 part here, here it is, if you stare at this, you'll see you're getting an extension. This is a sort of mixed odd structure. We're getting some extension here, e, that maps to vf v delta dual, tensor s10, and the kernel is this stuff here. Now, it turns out everything here is isotypical, s10 isotypical, and you can get rid of the s10. One way to do it is to go to highest weight or lowest weight vectors. So you do that, and you get an extension of what weight is this in? This is in weight minus 14. So you're going to get something here, which is basically where you've got q times e4 e10 plus q times e6 e8.  And when you put the correct weights in and so on, note the difference in the weights here was 13. This gives you an element of x1 of q by v delta 12, two copies of it. Well, let me do it like this. You get one extension from this, and you get one extension from that. If you believe that everything here is motivic, the motivically, this group is only one dimensional. So those two extensions have to be proportional. And if you believe the extension is non-zero, if you take a homomorphism into another Lie algebra that sends this to 0, if this extension on some element here is non-trivial, it has to drag this guy to 0 with it. So basically, you're getting relations because of non-trivially extensions like this occurring between a cuss form which goes to 0 and it pulls other junk with it because it's sort of inextricably linked to it. Anyway, that's stopped there. So apologies for running over time.