Hi everyone, this is Chih-chou. Welcome back to my channel and welcome back to another comic book haul video.  It's basically a short box. Okay. And the guy went to his home depot and bought the box and threw all the comics in here. And what I mentioned in the past is that usually the way it works is I do try to look for really good deals on eBay, usually in January and February. That's a couple of months that I do actually actively look on eBay because it's after the holiday season. So a lot of people are tapped out and you can usually get pretty good deals and you can usually find people, collectors that are getting rid of some of their collection, maybe you're just cleaning or whatever it is. And I ended up finding someone that was getting rid of some of their books. And one thing we did during our previous live stream session is I showed you guys how I surf eBay and how I end up finding the books that I do find. One thing I believe I forgot to mention during that live session, Twitch session, where we went through eBay and I showed you that stuff, one thing I believe I forgot to do was to let you know that actually some of the searches, whenever you search on eBay, things pop up. But towards the top, there's a little button usually available that says follow this search. So I do have searches that I follow and just a handful of those I have emails being sent to me from eBay, showing me new listings and stuff. So all of the comic books that we got from this haul, I basically found because an email came to me from one of the searches that I follow from a complete lot. It's Quantum and Woody and I'm always looking for Quantum and Woody sets if I can get them what a good deal, especially the early, the first appearance, the first run which I've only been able to get one set that we did a comic book haul video on where I showed you, I got the complete set. And I did pick up another partial set as well. But basically I look for Quantum and Woody. That's one of the searches that I have because what I found out is people who have complete runs of Quantum and Woody, they're in general sometimes the ones I've gotten anyway, they're collectors and they tend to have collections, other collections I'm also interested in. So keep this in mind, if you're searching on eBay, when you follow a search, put a type of search that you want to follow where if it's a complete run or if it's a single issue, the person putting up that comic book or the company or the store that's putting up that comic book might also have collections, other collections that you might be interested in. So that's what we're going to look at right now. It's a collection or a bunch of lots that I was able to get my hands on from a search that came to me from one of the searches that I follow from Quantum and Woody and you can guess there's Quantum and Woody on here. And one other thing I want to let you know before we get in, before we crack open the box, what I'm also doing for this video is I'm also live streaming this shoot on Twitch. And one of the reasons I'm doing this is because I've had a lot of people ask me about some of the equipment that I end up using and some of the setup that I end up doing and stuff like this. I've mentioned to people what the sound system I'm using, which is extremely low budget. I ended up getting this La Palma for like $15 or something. I bought four of them just to make sure I have them for a while and I open box camera. We talked about that and stuff like this. But basically I've done a little upgrade with my hardware and my software to kick it up a notch with the live stream. So I'm doing a little test run. So one of the reasons I'm doing this live stream of this comic book haul video is to show you guys the type of setups that I'm using and the type of setup that I'm going to start using and sort of share with you as much as I can, as much as I learned really, as we sort of grow this to incorporate multiple angles into our videos, especially when we go into the kitchen, when we start doing some food videos, cooking videos and incorporate that with recipes and mathematics and especially do some mathematics videos with different angles and different cameras. And right now the best I can do with this setup right now is have, and I have a fairly big mic here for the live stream. If you do, I will be loading on the live stream after this video comes up, but you'll see the mic that I've bought and there's a couple of cameras on either side of me that are doing a live stream right now. Okay, so that's my sort of long-winded intro to this comic book haul video because I'm sort of doing a test run and sort of layering my work as I mentioned that I like to do, right? And there's a couple more people that have joined on the live stream. Three cameras, three cameras brother. So there's people commenting on live streams, so I'm just replying to the comments. So every now and then I'm going to check back to the live stream just to see if anybody has any questions regarding the comic book haul video, okay? So let's crack this baby open. I'm excited to get this. There's some stuff that I want to read here and some stuff I already have and I think they were amazing buys from me and I wanted to get more of them, right? So let's crack this open. I'm not going to lift up the box again. Let me show it to you. I think I showed this box in the live stream. I didn't show it since we started recording, but this is the box that we have, okay? And I'm going to crack this open. Apologies. I have to apologize to the live stream people because the mic that I have here right now is very sensitive. So I'm guessing the sound of me cracking this open is coming out pretty loud. Like a nice package. Very nice packaging. I just cracked open the top of this and the guy's done a fantastic job packaging this. He just had a comment on Twitch saying someone's watching from the UK Tink 10 mention that they like what I do. So thank you very much for being here. Fun to share, fun to share. Okay, I definitely have to show you guys how this guy has packaged his comic books because they're phenomenal packaging. This is the way I like the package just to come to me. Take a look at this. Now, if you're selling comic books and if you're getting comic books, this is the ideal way you want your stuff package, right? I bought a few different lots from the guy and he's bubble wrapped each package and cardboarded each package like another layer of cardboard within each package. Okay, fantastic job, fantastic job. And this was his private collection. He's just a, he's a collector that's selling some of those comics, right? Now, let's, you know, he hasn't listed which one these are. So I've opened up the eBay listings that I won. Okay. And as we cracked these open, right, beautiful packaging. Look at this. Fantastic. Full respect to anyone that treats comics like this, right? So because what they do to make sure that the stuff doesn't move around, it could be actually just here. Let's check it out. Sometimes people put tapes on the bags that the comic books are in on the middle so they hold on to the cardboard that's in there as well, right? So they don't even move around very much, which is a good idea, right? But it makes it a little bit harder to grab the books, right? You have to be a little bit careful when you're breaking the tapes, right? So join them, definitely lock them. Okay. We're gonna crack open the bubble wrap. Apologies about the sound. Easy peasy and he's got them wrapped on the side here too. So serious. Let me show you this. And this is, so this one is, I wrote down the comic book halls and who the artists and stuff are on this, right? Because most of these writers I know, half of the artists I know, I know that I've read stuff off, most of the writers I know basically. So basically, this one is Uncanny Avengers. Okay. And it's the full set of Reminders, Uncanny Avengers. Okay. Rick Reminder. And as you know, and the artist for this is John Cassidy. And as you know, I've mentioned Rick Reminder before. He does, I did a full-on recommendation for Tokyo Ghost and he does deadly class and he's done low, I believe. I believe I wrote down this. No, no, low is, does Rick Reminder do low? I don't know. But basically deadly class Tokyo Ghost and Black Science, right? And he's done a lot of work for image. Okay. He's done a lot of work for Marvel, a lot of work for DC as well. So this is number eight. And this was the full set of Rick Reminders, Uncanny Avengers. Okay. And these are supposed to be in mint condition. One thing I do need to open up though, I need to open up because I'm going to put a little, little spreadsheet together to listing the prices, basically the same way we did our first comic book haul video, right? So let me crack that open, the spreadsheet. And here's the spreadsheet, right? So I grabbed the Uncanny Avengers number, the Uncanny Avengers from 2012, right? The first volume of Uncanny Avengers from 2012. It contained issues number one to 25 plus the annual, all in near mint condition. Okay. So it's the full Uncanny Avengers for, for Reminder's run, the first run that came on 2012. And I bought them for $25. So they came out to, and that was $25 Canadian, by the way. So everything I bought was in Canadian prices. And in the spreadsheet, I've set up a little conversion column in US prices. So it came out basically if you, one Canadian dollar is worth 80 cents US, which it is right now. I bought the full run of Uncanny Avengers, Rick Reminders Uncanny Avengers for $20 US. Okay. And it includes the annuals. So 26 issues, which basically came out to 77 cents a pop, right? Near mint copies. And these look like to be amazing copies. Okay. I'm not going to crack any of them open right now. So that was number eight. This is number nine, right? And it's Rick Reminders. I don't know this, this run very, very well. Here's number 10. Here's number 11. Hopefully this is coming out on the Twitch livestream as well. Let's see. Is it coming out? Yeah, it's coming out. My little cameras are a little bit too big, I think. Hello, Chicho. I love chat. Rick Reminders does low. Yeah, that's right. Rick Reminders does low. I read the first volume of a little fantastic, beautiful artwork. Low is absolutely the artwork for low is mind boggling. It's fantastic, right? And I don't know this run, but I love Rick Reminders work. Deadly class is absolutely fantastic. I read some of the earlier issues, but I haven't read it for a long, long time. I'm still collecting it. Okay. So this isn't, you know, I bought all 25 and this is a smaller package. So here's these guys, right? And this goes from 13, 14, 15, 16, 17, 18, 19. All right. Here's number 19. Beautiful. Let's see if we can find. Here, let's put this guy. I'm going to put this guy here. Okay. Now that we know how to open these packages, let's open up another one. Hopefully we get the rest of the uncanny Avengers. Okay. I'm going to open up the bubble wrap first. Again, apologies for the sound. The trick is to take out the bubble wrap first. Easily done. And so to open up the top part, okay, I sort of lay this down so the comics, because in the bag, they loose, right? So they shift around the bag, which is okay thing because if they were tied up in the bag, I might end up cutting the comics. I don't want to do some weighing this down. So up here, when I cut with a knife, right, there's no danger of cutting the comic books. Okay. I'm going to cut the side. It won't crack anything open right now. I haven't bought from this seller before, but I talked with them. This is nice. I'm going to show you this. This was I was very happy to get because I don't have this one either. But with this seller, basically, I was tracking the, I got the email notice for Quantum of Woody because it's a search that I follow on eBay, right? And I checked out the rest of the stuff with the rest of the stuff he was selling and I followed the Quantum of Woody for one cycle so that I didn't bid on or in the first round. Okay. It I let it go because I was tapped out. I already bought the cameras and the mic and the computer. So my comic book budget was completely tapped out, right? So I had to let the cycle go once. I let it go twice and the email reminder came to me on the third go and I had enough funds where I could buy the Quantum of Woody and the guy was local so the shipping wasn't going to be very much. He was in BC, right? But I checked out the rest of the books that he had, right? And I liked what I saw. So I made them. I bought, I won a couple of them and they were auctioned. They weren't bought now. So no one else was bidding on them. Fantastic for me, right? And I won a couple and I contacted him and he had some other comics on there that I had already gone through the cycle. One of them was the Quantum of Woody, right? So because the other the other lots were coming up before the Quantum of Woody. So I contacted him. I won a couple of those ones and I mentioned about the next comic books coming up, Quantum of Woody and the Avengers. And what he ended up doing is taking the listings off, putting them together as one package and putting the original starting price what he had, which was $25 for the Avengers and the Quantum of Woody lot was $20. So $45 I ended up buying the other two sets that I wanted, right? Now, this is another set that I ended up getting. Sorry about the long-winded little, what do you call it, sort of rundown of what it is that I ended up getting, but we ended up exchanging a few different emails just because I really wanted to make sure I got the sets that I wanted to get and he was totally cool with it. This is the other set that I bought. This is the other lot that I bought and I bought this for $30, okay? And it's Ed Brubaker's Daredevil run. And it's Daredevil number 82 to 119 plus number 500 plus an annual and it's the complete Ed Brubaker Daredevil run. So here's number 93 and I don't have this either. I've been looking for this and I bought it for $30 Canadian. Let me tell you what that is in US. So Daredevil, so in US dollars it'd be $24 and it's 39 books. So basically ended up costing me 62 cents a pop, bagged and boarded. Ed Brubaker's Daredevil run, right? Fantastic. Number 92. Ed Brubaker's huge. I've read some of Ed Brubaker's work, Velvet's Killer Be Killed, like fantastic series, right? I'm a little bit behind the Killer Be Killed, but Killer Be Killed, a self-published book, he's doing on Image and it's supposed to be there, you know, it's being optioned for TV series or something. I'm not sure. A lot of comic books are being optioned. Nothing says that they're going to come. Number 89. They're ever going to be made. Number 88. Look at this. The secret life of Foggy Nelson. Ooh, Foggy. What have you been up to? Apologies about the glare, but it's really cloudy outside, rainy outside. Great time to go through a comic book haul video. Look at this one. 86. And who does the artwork for this? Oh, I didn't write down the artwork because I'm assuming it's not one person. Does it say on here? It should. Brubaker Lark Guadino. Let's see. Take a look. See if you can see the names, right? It could be the same person doing the artwork. We'll check it, right? Okay. So basically 39 issues of Daredevil for $24.62 a pop. And this was, actually no, this wasn't the lowest price I paid for it, for one of the lots. I got a couple of smaller lots for a little bit better price. And this is 82 and it starts off with this one. Okay. Brubaker's run. So fantastic. The whole thing's not here. We'll definitely check to make sure everything is here. So let's put this guy on the ground here. Let's do, we'll keep the Avengers there. Let's grab another one. Let's grab this one. I hope everything is going well on the live Twitch. There's a place where I can find out how many people are watching, but I hope it's all going well. There's a few comments there right now saying hello, hello, hello, hello guys. So let's open up the bubble wrap again. Let's make sure the sound is still recording. Just have to open this. My bloodshot box cutter thing. This thing, I've used a lot. I thought it was pretty hokey pokey initially to a certain degree, but it's a pretty good package opener for comic book calls anyway, which is fantastic, which is what we want. This is the rest of the Daredevil run, right? Beautiful covers. Look at this. I haven't read any of this yet. Look at this one. Beautiful. Yeah, see the artist has changed now. It's Ajah. The ink or the letter I guess is Jordanio. I don't know the name, but I'm not sure if this is going to zoom in. Maybe you can see it. So the artist has changed and then Lark is back on this one. Fantastic. And one way to, great way to ship comic books is to get them to face each other. See if you can, here I'll show you. See the way he's shipping it? So this one, and then a comic facing it, and then another comic facing the other way, so you have the covers facing each other. And I believe that's supposed to protect it more. And the artist for this one is Man. It's a beautiful car. I don't know what this is, this character. At some point I'll do a run on this marathon on this and read it. If you know who the character is, let me know. I don't know what this is. Daredevil's got a gun, a machine gun at that. Look at that. So we assume it's Daredevil. We don't know. And here's another artist that's on there, Azateka. So the consistent creator for this is Brubaker. So I wonder if that's 39 books? I don't think so. We'll crack everything open and then what we'll do, we'll crack everything open and then do the count of the comic books to make sure everything's there. Lady Bullseye. That's Lady Bullseye. When did Lady Bullseye, someone in the live streams mentioning who that character was? This one. Lady Bullseye. No Lady Bullseye. This one. This is Lady Bullseye. Very cool. Should have known with the Bullseye in the forehand, right? So is this character new or I'm assuming it must have appeared a while ago or is it at Brubaker's creation? I actually like this live streaming the comic book haul videos because if I don't know something, I can ask you guys and you'll give me the answer, which is fantastic. The beauty of sharing, I guess, right? The learning. The learning is an important part of sharing. For me anyway, we're too close to the comic book, but I gotta be careful. Get too excited. This is the rest of the uncanny Avengers, right? Take a look. And these books look like to be like mint condition. All the description that he put on the sale items basically, here, let me read you the, I haven't shown you that one yet. Here's a daredevil one. This is basically the same style of description he would put on. He says, this auction is for issues number 82 to 119, 500 and annual number one, which covers Ed Brubaker's complete run on daredevil. The issues are an approximate near mint shape, right? And they are. That's what they look like anyway. And then he's got the daredevil. So take a look. He's got both daredevil and uncanny Avengers in this. So let's see. Uncanny Avengers. Let's do the count on the uncanny Avengers. I don't think it's 25. Is it going to be 25? No, it's not. It won't be. But here's the rest of the uncanny Avengers. There's another seven here. Reminders uncanny Avengers. Very nice. Here's Havoc. Havoc and Scarlet Witch. Look at this. I'm guessing this is the full uncanny Avengers team, which is Rogue, Havoc, Scarlet Witch, Captain America, Thor, maybe. And uncanny Avengers has had three volumes, I believe. The second volume would cancel real quick. I think it got canceled after five issues. And there's a third uncanny Avengers. And here's daredevil. The covers are daredevil of this runner. Beautiful. Absolutely beautiful. And I've looked on eBay for Ed Breaker's run for daredevil. It goes for a lot higher than what I paid, which was, what did I pay for this? Daredevil. $24 US for 39 issues of Ed Breaker's daredevil. Wow. Here's issue 500. It's thick. It's double-sized. Oh, here's Lady Bolzai again. Check it out. Taking on the kingpin. What? Hopefully it's coming out for the live stream as well. If it's not, I'm going to hold it down a little bit. Fantastic. Hey, there's multiple stream options. Sweet. That's good. I guess it's related to the different angles on the camera, right? Love the multi-camera. Love the multi-camera. Got a few ideas about the multi-camera. How's the sound, guys, for the live stream? Is it coming out okay? It's not too loud? It's good. Okay. Thanks, Dante. Appreciate it. Oh, here's more daredevil. There's a lot of daredevil. 39 issues of daredevil. Sweet. Bunker down. Oh, look at the 100th issue. That's a double-sized as well. I'll show you the daredevil covers, but it is what it is, right? It is what it is. Seriously. Bagged and boarded. Near mint condition. Came out to 62 cents a pop, US. January and February is a good time to buy books on eBay. People are dumping. Some people are getting rid of their collections anyway. Look at the issue 100. Thick. Oh, you got iron fist and Luke Cage and Spider-Man in here. Take a look. Sweet. Nice covers. Beautiful covers. I'll have to check out the interior pages. Who's this? Can't tell who that is on top. Awesome. Awesome. Look at this cover. It must be a variant cover or something. Oh, Matt, I love you so. Why must you slip further away every night? Not even they could see their their blind love. Look at this. That's totally different than all the other covers. I don't know if that's the regular cover or variant cover. That's daredevil number 94. Is that supposed to be in the lots? Yeah, it's supposed to be in there at Brubaker. It doesn't mention it, but it's part of the run. Cool. Very cool. I'm guessing that's almost, that must be all the daredevils. Let's take a look at these guys. Let's take off the wrap. Out of all the comic books I've bought on eBay and I've bought a lot, I've been buying for a long time, right? Only a few people wrap their books like this, which is very unfortunate, very unfortunate. More people should be wrapping their books like this. They be guaranteed repeat customers as well. Repeat buyers. I would buy that from people selling my books for sure. Now, I'm going to show you something that I've shown you before. Okay. And as far as I'm concerned, this was one of the most important comic book minis that came out. I believe it was 2007. I don't think it was 2006. It might have started in 2006, but this is black by black mask comics. Okay. And I've shown you the set before. I have this set. I have multiple copies of a couple of the issues. Here's number one. The concept is what if only black people had superpowers, right? So this is issue number one. And this is very relevant. Hands up, right? And the creators of this are Kawanjo. I can't pronounce the names. I'm sorry. On the front here, it says Eagle Randolph, Lit, Riggs, Stern, Sharp, Tim Smith 3 is one of the creators. Ashley A Woods, Varian Covers. I believe Tim Smith 3 is. Yeah, Tim Smith. And the writer for this is Kawanjo Osojito. And Jamal Ingle does the covers, I believe. No, sorry. The Kawan-Randolph does the covers. And he did some of the covers for We Are Robin as well. Some of the work for We Are Robin as well. I bought the black number one solely for the cover. Very interesting comic. Yeah. This cover for someone just mentioned that they bought the first issue just because of the cover, as did I, right? I was following them. I've been following Black Mask. I've been getting a fair bit of comics from them. I like the independence, right? Lion Forge, Black Mask, Action Labs, Valiant for sure, right? So I've been buying a lot, but I picked up two or three covers of this when it came out because I thought it was brilliant. And then the cover for number two came out. And when it came out, I bought five copies of this. And I bought a couple of sets since then on eBay. In my personal opinion, this is one of the most important comic book covers in comic book history. Most definitely for our time since the turn of the century. Okay. Someone's bowing out. We'll have to watch of it later as I am working in London. Thanks for the speech. My pleasure, brother. Hope you had a nice little break from work. But this is one of the most important covers in my opinion. And if you want to know how important this cover is, you need to check out Billie Holiday's Strange Fruit, the song, the lyrics, the story behind that song, how that song came to be. Nina Simone did a version of this, Strange Fruit. That's how I got exposed to it first. And then I found Billie Holiday afterwards, right? Nina Simone's version of Strange Fruit is blue. It blows me away. I think it's one of her most powerful songs. And from there, if you're still interested, and check into how Billie Holiday died, what the US government did to Billie Holiday, how they made sure she died. Okay. It's important. It should be taught in history classrooms across the United States, right? But there's also a documentary, I believe it's called Eyes on the Prize. It's a 12-hour, seven-hour documentary that is well worth watching. I mentioned it before. Okay. If you want to get the meaning of this cover, very important, very important. And I will continue to buy sets of black from Black Mask Studios, if I can get them at a fantastic price, at a good price. And this is an important cover as well. Very, very important cover. Take a look at this one. Cover for number three. I'll let you read the words. Let's see if you can get a nice read on that, right? Thug. Remember when they were called thugs? People were called thugs. Children were called thugs. Crazy. Stay woke. Beautiful, powerful, powerful. This is cover for number five, right? Tattooed on the fingers of the person holding the bars is stay woke, right? This is an American flag upside down implying distress, right? Powerful, powerful, powerful. I'm going to put these up on the wall. So this set, I ended up getting black number one to six in near mint condition from Black Mask Comics. I paid $15 Canadian, which is $12 US for six comics, and it cost me $2 a pop. And these comics, $2 a pop, were more than double the price, almost triple the price of any of the most expensive ones, right? So I paid the most for this set from Black Mask Studios, and I would gladly pay the same price again for this set. And I've read the first three issues of this, I haven't finished the whole thing yet. I have a lot of comic books that are a lot of sets that I still need to finish. And maybe during the summer we'll do a catch up session, right? Now, as I mentioned, the reason I was able to get my hands on these books was because a notice came to me regarding the Quantum of Woody comic book lot, right? So I bought a Quantum of Woody comic book lot, and it cost me $20 Canadian, which is $16 US, and it contained, let me pull it up, like a Quantum of Woody, Quantum of Woody. So it contained Quantum of Woody, the 2013 series, from issue number 0 to 12. It contained Quantum of Woody must die, number 1 to 4, and it contained Q2, the return of Quantum of Woody by priests, but the original creators of Quantum of Woody in the first volume, right? And this is the Quantum of Woody, the return of Quantum of Woody, okay? This is issue number 5, and this is by priests and bright, Mark Bright, I believe, is it Mark Bright? Yeah, Christopher Priest and Mark Bright, okay? And they're the original creators of Quantum of Woody. This is number 5. This is number 4. I haven't read this one. I read the series, the 12 Issues series from 2013. I haven't read this one. And Quantum of Woody must die. I don't think I've read it. I'll have to take a look at it to make sure I haven't read it or I've read it or whatnot, right? This is number 3. Here's number 2, and here's number 1. And a new Quantum of Woody series just started, I think it's two issues deep, right? So this is by priests and bright, the creators of Quantum of Woody, fantastic, right? And here's Quantum of Woody number 0. I've read this, it was fun, right? And this is from the 2013 series, okay? And this came out after Q2. Quantum of Woody must die, I believe, anyway. I hope so, anyway. Let's crack another one. So those ones, as far as comic book haul go, they cost me $16. And in that lot was 22 books. So it ended up costing me $0.73 a pop bag and boarded in near mint condition, from what was listed. I'll check them, but these comics look fantastic. Just pops out, beautifully packaged, man. And this guy only had on his feedback, on eBay, he only has like 62 feedback.  I didn't check to see what else he had sold, but I'm glad I got my hands on this. I bought everything he had listed when I started talking with him and making the offers. Prison industrial system, was that one of the things that was mentioned on Black. Prison industrial complex, yeah. Quantum of Woody, number one. Number two. This series is hilarious. Really, so funny. This was my first exposure to Quantum of Woody. I hadn't, and I haven't yet read the Priest run yet, the first series, the first volume that came out. But as soon as I started reading this, I could not believe how funny it was. It blew me away. Right? It was at par with the 2012 Archer and Armstrong series. And that's a fantastic series. These look like fantastic copies. This, like, the coats. This is so funny. So funny. Still cracks me up. Still cracks me up. I might give this another go. I don't remember it too much. And that was, when did it come out, 2013? I guess in three years, four years, I totally forgot what was going on at the end. Been reading a lot of comic books. This is from Reservoir and Docs, of course, right? I don't know if it's the variant cover or the regular cover, but this is from Reservoir and Docs. I saw Reservoir and Docs at university in a little small theater, blew people away when we saw it. Quantum of Woody, number 12, and this is number five, out of order. All new arc, must read valiant. That's the Quantum of Woody. You know, since we got the Quantum of Woody, let's do the count for this. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. Yeah. Okay. So that's there. And the number zero we had already. Filing up. Nice. Let me show you this one too. Now, this one I've heard about. I've been, you know, it's hit my radar a little bit, but I haven't, I don't know too much about it. But I've heard great things and the covers look for you. Beautiful. Let me show you this. I don't know anyone that's read this, so I can't comment on it. And I haven't looked up the reviews because I usually don't look up the reviews. This is, is the artist's name there? Yeah. So Jim Zubowski. So this is Wayward. Okay. I bought the Wayward set. Seven issues, right? And I ended up buying this Wayward, number one to seven, five dollars Canadian, four dollars US, right? Seven books, ended up costing 57 cents a pop. Neumann condition, back then boarded, right? And the writer for this is Jim Zub, Zabowski, I believe that's his name, or Zbavvich, okay? So Jim Zub, right? And Stephen Cummings, okay? And Jim Zub, I've, I made a note here. I didn't, he didn't, I remember seeing his name for some reason I was connecting him with invaders is him. I guess it's because of his name being Z starting. But Jim Zub, he did Samurai Jack as well. Samurai Jack I was reading. So that's probably the reason I recognize his name. But Wayward set, number one to seven, Jim Zub Wayward. Yeah, Wayward is a great series. Awesome. Someone on the live streams mentioning that Wayward is a great series. Looking forward to reading this, man. Looking forward to reading this. And I can't believe, like, just the price I was getting these, like, and the covers for these look beautiful. The cover for number three is absolutely fantastic. Does the series end? I don't know if the series, I think it's a continuing series from, because I've seen it recently, I believe on the shelves or in the last few months anyway. So this is supposed to be a continuing series from what I understand. But this is a beautiful cover. Look at this. Absolutely fantastic. And this is, I believe, being looked into, to create a animated series, TV series for this, right? That's one reason it hit my radar. It's ongoing, someone's mentioning. Fantastic. I'll definitely give these a, give these a read. And if I like it, I'm going to track down the rest. I think it must be into the 20s at least, number wise. And this is Image Comics, right? Like, there's a few, few different publishers that are just killing, killing the comic book industry. DC for sure is one of them, with the rebirth, they're doing a fantastic job. But Image, Valiant, Black Mask, right? Action Labs, there are some amazing comic books being put out there, right? So wayward, number one to seven. Nice buy, nice buy. Very happy to have this. I always like getting sets on the cheap as well, for sure. Any sets on the cheap, but sets that I haven't read, that are still ongoing, the early issues that I might delve into, right? Now, here's another set that I picked up, is, let me put them in order, Men of Wrath, number one to five by Jason Aaron. He's the writer, okay. And who's the artist for this, does it say? Oh yeah, Jason Aaron and Ron Garni, okay. And I ended up buying this for five dollars, five issues, five dollars Canadian, so four dollars US, ended up costing 80 cents a pop, right? And this is a mini-series, this is completed. Issue number one, let's just put it like this. Here's issue number two. And Jason Aaron wrote Southern Bastards, if I'm not mistaken. I believe so. Southern Bastards, Southern Bastards, must be, here's issue number three. I've seen some kind of cover like this on Daredevil before, with the stained glass in the background with bullseye and stuff. Number four, and number five. Nice, nice, nice. Fun buys, excellent buys, excellent buys. We've got two more to open up. We're not cutting into any comics. Did Cheecho ever heard of the Hedge Knight Game of Thrones comic books? I've got the Game of Thrones comic books. I picked up the first couple of issues. Might not be what you're into with your damn good series. Hedge Knight Game of Thrones comics. I've got a few copies issue number one, the variants as well of the Game of Thrones comic books. I didn't read them, I just bought them because I like the series, the Netflix series. Here we go. Let me put these guys in order. They're all facing each other. Oh, sweet. That's them. That's them. Okay, sweet. So it's a good read, the Game of Thrones comics. Ah, maybe I should start reading it. I love the series. And the covers, I just wanted to get the comic books when they came out, right? Just having my collection. And let me show you these guys here. So this is the Quantum of Mori Masudae. Okay, the rest of the Quantum of Mori set, right? Number one. Here's number two. I don't think I've read this one. It's not ringing a bell. I've bought it, I've picked it up, but I don't think I read it. Here's number four. Nice. So the Quantum of Mori stuff is there. We've got all of them. And here's the rest of the Uncanny Avengers. Nice. Here's issue one, right? So there is one, two, three, four, another eight of them here. Okay. And here's number one. If you're looking for the, because it was a few different sets, I couldn't, that's issue number one for the Uncanny Avengers, the first volume with Ramin there. And it goes all the way up to issue number two. I mean, T5. But we'll do a final count. Okay, let's just crack this open. Which one is this one? Where is this one? For sure, after cracking these open, one thing I'm going to do, I'm going to just pick two books randomly, two or three books randomly, and open them up to make sure they're near mint condition, because everything you listed was supposed to be near mint condition. And then I'm going to get on eBay and leave this guy phenomenal feedback, phenomenal feedback, right? Fantastic job. What a nice guy. I don't know this one. So totally forgot about this one. This is, and I know the writer for this. I've read some books from him, right? He's written a lot of, the writer for this has written a lot of Deadpool. And he's done something independent that I was recently reading. I can't remember. But the writer for this is Colin Bunn, right? And it's the fearless defenders. Okay. And it's the full set. There should be, there should be 12 of them. Let's check it out. So fearless defenders, number one to 12, Colin Bunn. And the artist for this, I wrote it down here, is Will Sinley. Okay. And this lot, I bought ridiculously cheap. I bought $5 Canadian, $4 US, right? So 12 comics, bagged and boarded for $4, 33 cents a pop, right? Here's number 12. I don't know how good this is. Colin is, I've read some of this stuff before. Nothing that has, that I can recall that has blown me away, right? There's so many people writing Deadpool. Can't remember who wrote which Deadpool, right? That's one of the problems of a character being written by so many people and so short of a period. And this is about, I believe it's a defender's theme that's all female, female cast. I didn't know about it. I hadn't heard about it. Do we have a date when this first came out? The date's not on here and I didn't write it down. Sorry. Apologies. Not giving you all the info of the comic books. That comic book calls about finding treasures and finding things you may like and stuff, doing a little search as well, right? So that's a good thing. This is number five. Put these guys down so they don't fall on me. Here's number four. Oh, sorry, this is number, oh, there's a variant cover. AU. Number four AU. I guess it's an alternate variant cover. He hadn't listed that on that. Maybe he had. I heard the description. $4 for, I guess, 13 comics. Nice. The fearless defenders. Number three. Fearless defenders. Number two. And fearless defenders. Number one. And those are, that's the comic book haul, right? Let's do a count because these ones, they came in different packages. Make sure we got them all. This is the uncanny avengers. It should be 26 books, right? And it looks to be 26 as far as the, this many goes, right? So let's count it. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27. We've got 27. I'm not sure why we have 27, but like the fearless defenders, they might have been an alternate variant that he included in here. Okay, I'll check it later. Maybe put them in order before I put them in a box. At least these guys will be in order, right? And let's take out the daredevil. Let's see how many we got. Daredevils. We need the Ed Brubaker daredevil. There should be 39 issues here. Okay. There's a couple of comments on the last year. So I'm just going to read those. I wish I remember didn't leave Marvel, but he's killing it on image. Yeah, I agree with you. You know what? I'm glad, I'm personally glad that he left Marvel because I wasn't reading Marvel, really, just randoms. So I'm glad he went to Image on his creating zone stuff. Deadly class. Tokyo goes was absolutely fantastic, right? And 72 Eternity. That's right. You're mentioning 72 Eternity. I've been picking up 72 Eternity, but I haven't been reading it. And 72 Eternity is supposed to be phenomenal, right? Issue number one is so hard, so expensive to find. There's two different covers. The one with like a Star Wars pose that they have is very expensive to find. It's crazy. Are you talking about Harrow County? That's right. Harrow County is Colin Bunn. Thank you very much for that. I've read some Harrow County and it was very good. Very good. I just wasn't into that type of story, so I didn't continue it. But that's the one that's where he sticks in my mind. That's right. Thank you very much for that. I was like, wait a second, he's done something that I've read that I liked. As far as the daredevil goes, we need to have 39 of these. I believe 39, 39, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 29. Let me put these back so they don't fall out of my hand. 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41. They must have included a couple of Americans in this. So there's 41. If I had my spreadsheet, I just took an image of it. If I had the spreadsheet open right now, active right now, I change the numbers and see the prices per unit drop a little bit. Fun stuff. So in total, basically we bought 117 books. It came out to $105 Canadian, $84 US. So 117 books for $84 US came out to 72 cents a pop. The shipping was expensive. The shipping cost me $36.91 Canadian, which came out to $29.53 US. So $30 US. So total came out, shipping included $142 Canadian, which comes out to $113.50 US for 117 books, which comes out to $0.97 a pop back then boarded near mint condition. Delivered. Guaranteed delivered in a nice package. Less than $1 a pop. So it's like a dollar bin buy. But very good condition. Fantastic runs. In dollar bins, you usually don't find complete runs of things. You find randoms. Great buy to start the 2018 year. Very, very happy to get this. I'm very happy to do a comic book haul video. It's been a while, it's been a while. And I was itching to do it, but I definitely hit my budget regarding stuff that I was getting for this work that I'm doing with the equipment, camera, mic, and definitely the laptop. That put a serious dent in my yearly budget. And we definitely need to have a budget when we're living our lives and stuff. And this thing, these books, I'm very happy to have in my collection, ended up getting a great deal. And I'm going to do some readings. I'm going to do some nice readings. I might decrease my poll list a little bit until I catch up with some of these reads and then start picking up my poll list at my local comic shop again. But I definitely have stuff that I will continue to buy always, always. So far anyway, never say always, I guess with comic books, but I will continue to buy for the foreseeable future some image titles for sure. All the valiance, black mask for sure, action labs, a lion for some of the stuff that they're doing and whatnot, right? Lots of great independent books out there. Lots of fantastic books coming out by the big two as well. Okay, that's it for now guys. I hope you enjoyed the video. We have a lot of things planned for 2018. I hope it will be a nice ride. I hope you'll enjoy. And I guess I'll definitely see you guys in the next video. Bye for now.


Okay, thank you very much to the organizers for giving me the opportunity to speak and I want to apologize to the audience for being a person of the last century and not giving a slide talk or whatever, more fancy talk. So I hope you can read it. And so the topic is about the ranked honest Thomas theory of a certain Calabria 3-fold and in some sense it's a degenerate Calabria 3-fold. So in some sense ideally you would want to study Calabria 3-folds which are strict, so which have H1 of your Calabria 3-fold should be 0. And you obviously see that because of this elliptic curve that's not true here. Okay, so this is some sort of, you can say some sort of degenerate case. Of what we should expect in the true sense. So the reason to study this anyway is in some sense that from this degenerate case you might still learn something for the general case and then the second thing is also maybe just the personal motivation that is kind of, in this kind of degenerate case you actually can really give complete explicit formulas and that's kind of very satisfying in this field. At least for myself, okay. And so the second thing I want to say is that the talk will be slightly technical but I thought it fits well because it kind of uses a lot of the techniques that also will be discussed in Richard Thomas' talk. So I hope that is also useful in some sense. And yeah, if there are any questions just start. So I want to just recall a little bit, recall this, what is the standard setting? Okay, so if you have a smooth trajectory 3-fold and we'll just do the really the really classical case of curve counting, Donaldson-Thomas invariance. So we look at what is called this Hilbert Schemes of Curves. Okay, so this parameterizes closed one-dimensional sub-schemes of this x and we'll ask two conditions. So the class of this one-dimensional sub-schemes should be this curve class better and then the structure sheaf, sorry, and then we also put a condition on the structure sheaf and we also fix that the third term character of the structure sheaf should be n. Okay, and so maybe as a warning to, I don't know who follows usually this talk, this is a little bit different convention but yeah, don't worry about it. It's equivalent to just fixing the holomorphic Allah characteristic of the structure sheaf. Okay, and so another way to view this, I mean if since this is a school, it should just say it is the same as the modispace of Kizika, semi-stable sheaves with respect to some polarization of this with Schoen character specified by this data. And so then we learned in Richard's talks that there's a perfect obstruction theory and let me just write it out one more time. Okay, so there's this, you look at this universal sheaf which I call always E in this, in this case. Okay, so you have this universal sheaf on this product of Hilbert schemes times this thing and you look at the map to the Hilbert scheme and then you look at this relative form between the sheaf in itself, you push it forward to the Hilbert scheme, then you still shift it and then you take this trace-free part, then to really get this map to the co-tangent complex of your, of your, of to do a Hilbert scheme you just still have to dualize. So in particular this obstruction sheaf, I mean forget about all of this, the obstruction sheaf of this obstruction theory at the point corresponding to this E would be just this x2 of this sheaf and then the deformation space, so to trace this x2 and then the deformation space would be the traces x1. Okay, and so in particular from this data you, you get this, you get this virtual fundamental class, let me just put it here. So from this data you get this virtual fundamental class of this Hilbert scheme. So I drop this, this data here, so this will be char class in the virtual dimension of your Hilbert scheme and here the virtual dimension is just I mean just first sharing class integrated over this bit. Okay, so if x is Calabria then you can define the dt numbers of your space just by integrating, taking the degree of this virtual fundamental class, right, then Calabria means c1 when this first sharing class vanishes and then just integrate. And really the challenge I would say a challenge in the field is to compute explicitly, computed explicitly, explicitly for a strict Calabria, for a strict, strict, strict compact Calabria 3 and here strict is this condition I put here. Okay, so for example, a Quintic 3-fold compute this number. And so yeah, so let's consider this degenerate case that I discussed. So, and so now I mean that's kind of very unfortunate notation wise but I will not call my elliptic curve E so, but in some sense I have to call it E, so I will call it like this double E. Okay, so I hope that doesn't confuse anybody. Okay, so I think it should be clear whether what is the sheaf and what should be the space but if not then just interrupt. And so on this thing here, if you just use this naive definition I just wrote up that it's just 0. Okay, and the reason for that is I mean there are many ways to formulate it. One reason is that there's a co-section. Okay, so there exists the co-section. Exists the globus co-section co-section of the obstruction space, so of this obstruction space, sheaf. So there's a map sigma from the obstruction sheaf, so I should say hip to space to, so to the trivial sheaf of the hibiscus and we ask it to be surjective. And so how do you define this global co-section? How can I get this board down? Okay, so that's kind of part of this general this co-section stuff I think will be also discussed in this, in all these lectures we kind of play a role. So I want to prove this vanishing by constructing a co-section. In some sense there are two ways to this and let me just sketch them. And the second one is kind of closer to what we will do later and it's kind of maybe more clear what happened. So the first one is really goes back I think to Bufreitz-Flenner, the semi-regularity map. And then I guess there's also some work by Maulik, Pander, Pander and Thomas. And then I guess even farther back you can trace this back to Glock and so on. And so the idea is the follows and the idea is to well what do we have to do? We have to find a map from this obstruction space to a trivial sheaf. And so let's look at this guy here. So this is, yes? Why the existence of global co-section is important? Why do you need this? Yeah, so because it will prove that all the Donaldson-Thomas invariance are zero. And at the same time we'll give you a way to fix that. So actually define your non-zero counts. Okay, so I will just ask you to wait a minute and then I will kind of state this conclusion. So the first thing is to find this co-section. And in some sense this co-section should be some sort of trivial piece in this obstruction sheaf that doesn't play any role for for deformations in obstructions. Okay anyway, so how does it go? So well let's look at a sheaf E. So let's say look at a fixed sheaf with this obstruction space. And so how do you find the co-section of this guy? Right, so how do you, and so the idea is to use a certain gadget that also already was introduced. So I take it, I can say that I can use it. So there was this a tier class and so you just apply this a tier class here to the x3 and then you take the trace. So here's that tier class and then take the trace and you land in a certain cohomology group h3 of x comma omega x. And so now you have to do this community composition in your head because what is this space here? So x is this product. So this will be h2 of isomorphic to h2 of the k3 surface, structure shift of the k3 surface plus h11 of this elliptic curve. Okay, and so here there's a, and so this is in particular, it means unique, there's a unique dual of symtactic form and then the point class. Okay, so that's really just a one dimensional space. So I didn't tell you what is, I mean, I don't know, we didn't discuss much what is this tier class. Just a way to, a certain natural class assisted in this homology group and assisted to do it, you can write down this map. And so the way to show this is really subjective. I mean I said we really want to prove this rejection. The way to do this is some sort of calculation that I want to not discuss too much, but one way to do it is really construct an element in here that you can show goes to zero here. And so the way to do this is you look at, you pick, you pick an h1 homology class in this, in a certain group and you take it in here. And then you can calculate, okay, calculate the image under this map and I will not discuss how to do it, but the image will be this certain thing. And so then you can show that if you take this v correctly, this will be non-zero. So what is the, sorry, it's a bit fast, so I didn't notice it. So this is, w e is the, what is the w e there? Ah sorry, so this is the point class. I mean I want to take a generator here which is the class of the point on the elliptic curve. And I want, I'm saying if you take the class of point of elliptic curve and take the dual of the complex conjugate of your subtractive form, multiply them together, this gives you a generator of this, of the space and so the generator of the space. And this notation, this v in that product, sorry, very left. Yes please. I'm saying you pick any v here and then you take the interior product with this ATIA class. I mean all of this is not important. The way, I mean I feel this is the most natural way to understand this course section because somehow if you look at the model space of K3 surfaces, there's a certain divisor, so this model of K3, there's a certain divisor with this class beta that we started with here.  So here's the beta and the d. And so there's a certain divisor in this model space where this class beta is algebraic. And so here I ask this beta to be, in fact, if beta to be algebraic. So I should say 1, 1 of s comma c under funcratuality. Okay, so I'm kind of messed this up, the order up of these things. So ATIA class is kind of a three form, is it? This one or what? Anyway, so sorry, so essentially what this says is that, so I have your point, so here is kind of the divisor where this class beta is algebraic and somehow you have this point here of this K3 surface. If you take an infinitesimal deformation of this K3 surface which is given by this vector v, then somehow the obstruction to deforming your sheaf along with it is exactly given, I mean, sorry. So then if you want to try to deform this sheaf along, there's a certain obstruction which is coming from the fact that this churn 2 of this sheaf should stay algebraic. And the obstruction to that is really given by this group. So in this way you can prove this is not true. Okay, maybe I did kind of a very bad job explaining this. So let me give you the second way to construct this co-section which is actually much more easy to understand. Okay, so the second one is using the action of the elliptic curve on your model space. Okay, so we have our model space here. And so in particular if you have a subscheme you can just translate it by this action of the elliptic curve. And so in particular this gives you a map from the tangent space into the first x group of your sheaf. So now you can just, I say, you can just look at this in a sheaf way. So, I mean, look at this in a relative sense. So you get the sky, get a really sub-bundle of this sky. And now by seaduality, so we have, so this is isomorphic to the dual one. And so now by seaduality, this produces you this co-section. Seaduality produces you this, or by dualizing you get this co-section. So omega e. And so because you have a sub-bundle you get a co-section. Does that make sense? Maybe that's, that's clear. And so, so why do we do this? So then once you have that, and there's this result of Kim Lee that has the two things. The first thing is that there exists, so the first thing is that this virtual class that we defined with the standard obstruction theory is just zero. And then this is kind of the first statement. And the second statement is that there exists a way to fix this. This is called this reduced virtual class. So this guy here will live in the usual virtual dimension plus one of your models. Okay, so the virtual dimension in our case was zero. So it lives in a, it's now a one-dimensional class. And so, so this class is then really non-zero often.  So, so the definition here is the following. So you're going to now define the t-invariance of the space. So this x cross e, then you just integrate over this class and you integrate a certain insertion. So, so a certain commodity class, natural commodity class on it. And the way to do this, again, is using this, using the universal, universal, let's say universal sub-scheme in this thing. So there's two maps to x and to the hibiscus. And what you do is, so you have this two projection here.  And then you define this insertion tau zero of gamma, which is just given by pulling back this tx by, to this thing, taking, taking a cut product with the churned, churned two of the structure sheath. And then pushing it forward to, to, to the, to the hibiscus. And so here what we do is we just take tau zero. And in some sense it doesn't matter really what you take. The, the most convenient thing is to take, you pick a class beta dual in the h2 of your k3 surface, such that the pairing with this given curve class beta is zero. And then you also tensor it with this, this class that I defined up before this, this point class here. Okay. And so in the essential way, I mean before in the usual carburetor three-fold, somehow there was just a single number that govern everything. In this case here, essentially this is the single number that governs all the curve counting in this, for this hibiscus. All right. So the cross-section is rare in this construction? Cross-section comes in here through this reduced virtual class. So, so kind of with the cross-section I know that this vanishes. And then Kim Lee also tell you a way to fix that. Okay. These two cross-sections are the same? Yeah, they are the same. And the construction depends on the choice of the cross-section. This reduced? Yeah, up here yes, yes. I mean up, well, yeah, so in some sense, so the reduced virtual class only depends on the k-theory class of your, of your perfect abstract theory. And somehow it doesn't really matter which kind of trivial piece you subtract from this course. So in some sense, it doesn't matter which one you take. But yeah, it's good to fix one. I mean, the reason we need one is to have this first statement. Okay, so what's the theorem? And so, so basically there are two parts and kind of the first part is, I guess, more exciting one, but the older one, and let me just still state it to, for complete list, these are old results together with Aaron Pixon and then Juniang Shen. And it says, if you take better to be primitive, so better is our class on the k-3 surface. So, so that will be always primitive. So let me just write the picard group here. So, so this is primitive of square, of square 2h minus 2. Then this dt invariant is easily computed as follows, just given as a certain coefficient. So there's a sign here which doesn't play a role. And then there's a certain coefficient of this egusa cusp form, one of this egusa cusp form. Okay, so, so you take this one over this egusa cusp form and then you subtract, you extract a certain coefficient here, free coefficient of this egusa cusp form. So, you take h minus 1, q tilde, d minus 1 and then p to the n. Okay. And so here the q and the q tilde are the free coefficient, I mean, as I said, it to tau and tau tilde. So q is e to the 2 pi i tau and then the same for the tilde and then p is the over z. And this guy here, this fellow is this, this egusa cusp form. So there's a certain Zege model form of genus 2, it's a cusp form. Yeah, some sense as nice as this discriminant model from delta of tau just now in genus 2. So if you, if you have any model form textbook, somehow you can just open it and you find this guy. And so now today I just want to explain the second part to this theme. So this is maybe, and this just tells you what to do in a, in a general case. So in a general case, you have the certain multiple karma formula and it stated as follows. So that essentially you can, so here this better was primitive. And so we can always reduce to this primitive case. Okay. So the, the way to do this is you, you sum up all the, yeah, means it, so, so better is the element in this lattice. And so it's primitive if you can't divide it by any positive integer bigger than one in this, in this lattice. And so, so the primitive case is somehow easier because, yeah, he curves out, you can control them more. But it turns out in this K3 surface case, the imprimitive one is in fact controlled by the primitive. Okay. So the way this works is you sum over the, all the devices of the GCD of this beta and this n. And then you have a certain sign here which is given by this n plus n over K. And then we have a one over K here.  So, so for the Euler characteristic, you just divide by this, by this K. And then for the curve class, the D doesn't change. So the D doesn't change. And then here I want to really take better over K. But better over K is not primitive. So I didn't gain anything. So instead, I just assume it is primitive. Okay. So, so let me just say what is this prim here. So, so we are prim over better over K is a primitive effective curve class, curve class on some K3 surface as prime of same square of square given by this of square equal to the square of this, of this, of this class better over K. So you just take your better over K. You assume it's a primitive class of the same square on some other K3 surface. You calculate, take the corresponding DT round determined by maybe part A and then just, just plug it in. Does that make sense? Okay. So, so I'm saying here I take better tilde, which is, so I take, so take better tilde. So this is a algebraic class on some K3 surface. And let's say positive. So there's actually a curve in there, such that, such that the square of this class is the same as the square of this better. And then the prim of this better over K is, I want to define to be this better over K. So is that as prime any other K3 with that property or you really construct this start? Yeah, any other, any other, it doesn't depend by deformation invariance. So, somehow if you have a K3 surface with a primitive algebraic class of a given square then any two of these guys are deformation invariant. K cannot be 2 because beta square is 2h minus 2. I mean if, if K has to divide better here. Okay, so otherwise it doesn't work. Okay, so anyway, so this determines everything. Okay, so what is, for this, for this geometry in, in rank one? Okay, so maybe I, okay, so maybe I can make some remarks about this short history. So the first one is about, yes, about this problem. So, so maybe there are three remarks I can make. So the first one is if d is equal to 0, then, so d is the degree over the elliptic curve, right? So, so we have, okay, three surface elliptic curve here, then we have a s here. And so if d is 0, so d is the degree over the elliptic curve. So if you, then all your curves are somewhat vertical. So just live in, in this K3 fibers. So in this case, this, this statement just reduces to the K3 surface and, and then somehow there are like three results that one can state here. So if beta is irreducible, this goes back to kawai-yoshioka. And then if beta is primitive, there's this, I mean, this was then done by Moly, Panda, Panda, and Thomas. And then beta-imprimitive, this was done, then, by, done by Panda, Panda, and Thomas. And so let me just make also a remark for beta bigger than, so strictly bigger. Then just this part b, so, so part b, b for divisibility 2, was, is also, holds also, was done by follows, by work of Younger and Bay and Tim, Tim Bill on the gramophilm side, okay? So we don't know, for primitive beta, we unfortunately don't know any, we don't know the gramophilm-pt correspondence yet for the space, but on the gramophilm side you can independently prove this. And so then the final thing is the higher rank, you can ask for the higher rank and this is, yeah, still work in progress, but I think this also can be done eventually. Yeah, so, but on the gramophilm side you can just do product formula, and then this b is kind of a consequence of this, of this multiple-cover contractual forgate resource. So, so how much time do I have? I didn't want it. Okay, so for the last half an hour I will just try to explain the proof, and at least the key idea in the proof. And, and sometimes this follows closely somehow ideas that already been used here and also show up, will show up a lot in this, this week later. Okay, so let me just discuss the proof. So, proof only of part b. And so, so how do you prove such a thing? Okay, so, and so the ideas, well we have this s cross e, and you have to break something and what you break is always the elliptic curve here also. Okay, so, so the first thing is you degenerate the elliptic curve. Okay, so, so you have to smooth the elliptic curve and you degenerate it to a non-elliptic curve. And then you can, so this is just, you view this as the p1 where you glue this two points together. And then there's all this business about the degeneration formula. With the degeneration formula, let's say reduced to, then k3 surface times p1 relative to two fibres. And then with some other work, some more work. So, plus, then you can actually just reduce to the case where you serve a k3 surface times p1 relative to one fibre. Okay, so degenerate elliptic curve and reduce to, to this relative dt theory, theory of, of the surface time p1 relative to, to this, to the fibre was zero. So here s0 is five. And so I will not discuss this, but this is kind of standard. And don't worry about the relative dt, it will kind of disappear in a second. Okay, we'll, we'll just reduce the problem in absolute dt in the end. Okay, so this is the kind of the step one. Okay, and then the second step that you do is always you localize. Okay, so you look at this geometry, you look at this Hilbert scheme s, s cross p1, and then you take s0.  So you take gm. And yeah, so essentially there's three different fixed laws I, one has to consider in this, in this, in this relative Hilbert scheme. So this one I call the extreme one, I haven't told you what it is. This one is what is called pure rubber. And this one is some sort of mixed guy. And in some sense, for us, what will turn out is that the second and third one will not contribute at all. And that's, both of that is not hard to see. Okay, so let me first tell you what are these three components? So what is this A component? A will be the one that is kind of most important for us. Okay, so let me just write it in some more picture schematic ways. So this is a sub scheme for curves inside really surface times p1 of the form. So, so here you have the fiber over zero. That is your relative divisor. And so here's our p1. Here's, I mean, my picture's three dimensionals. Okay, so here's, so here's our k3 surface. Okay. And so, so these are kind of really the, what should I say, the fixed curve that really lives inside just the surface times p1. And so how does it have to look like? So your relative over zero, so, so nothing can happen over this fiber over zero. So here you have really just straight, straight lines, maybe thickened or something. And that meet this, meet this fiber over infinity, then at the fiber of infinity, you have all kinds of fixed, fixed curves here. Okay, so you can have some sort of fixed curve here, or maybe I should use the same color. Fixed curve, fixed points. Localization occurs at the infinity. Yeah, so if you have the scaling action on the p1, so the c star, we use the gm action or c star action on this p1, just scaling the kind of this parameter. And then you have two fixed points, zero and infinity. And somehow if you have a curve inside surface times p1, that is fixed under this action, then you can have some vertical components over zero, some vertical components over infinity, and then just kind of essentially straight lines between them. And the ones over zero, we cannot have in this, in this, in this component, because we have this kind of condition that we're relative to this.  So, so this is really the ideal sheaves in this geometry, such that if you restrict this i to surface times p1 without this fiber over infinity, and this is just pulled back from this, from the, from an ideal sheave on this, of ideal sheave of point on this, on this s, okay, so here, so let's say, so where eta is a zero-dimensional sub-scheme, sub-scheme of length d on this s. So just to short form notation, the modus phase is the Hibbel scheme of d points on the surface, and so eta is just the point. Okay, so if you take this ideal sheave of our thing, you restrict to, away from this fiber over infinity, where all the stuff happens, I mean the screen stuff happens, and you just get kind of this straight tubes here, which means it's just the, the corresponding sub-scheme is just a preimage of this, of this fixed zero-dimensional sub-scheme of the surface. Okay, so there's really a lower side that sits inside the Hibbel scheme of this surface times p1, okay, so in this absolute, absolute modus phase. And so yeah, let me also look at the other two, so b, what is this b? So let me also just draw a picture here, maybe c first, okay, and I draw this for the people who know it, this relative stuff, and so in the c you just, everything happens over this, over this, over this rubber component, and, okay, so here you have the original p1 here, and then here you have some sort of rubber component, so where you, it's like a p1 where you, where you identify curves when they differ by a gm scaling, and then you have some sort of stuff here, and then here you, on this side you just have straight lines, so this is this component c, and then component b is kind of a mixed, mixed, mixture of these two things, so you have some, some green stuff here, and then you have this, this straight lines, and then here you have some, some stuff. Anyway, so these two components doesn't matter, and somehow the whole action happens on this a here, on this part, and so this is the guy we have to understand, and try to, try to make sense. Are there any questions about anything? Yeah so, so c, in c I really want to say there's no, the only curves that are allowed in the surface times p1 is just, are just pullbacks, I mean the corresponding curve is just the pre-image of this zero-dimensional sub-schemes on the s, so it's just like p1 times the zero-dimensional sub-schemes, so there's no, yeah, there's no vertical components in any way. On the other end here I ask, I mean allow stuff here and then also stuff here. Okay so, so we have to understand this extremar component, and somehow the techniques are essentially, this is well known in some sense, so and what's nice here is we are on the surface times, well let's say we're in surface times p1, but what really matters here is this, this part of away from zero, so really can just view our sheaf also just restricted to the surface times a1, where a1 is this, is p1 without the, without the zero point, okay, so I mean all the data of this curve is somehow totally determined by just restricting away from the zero point because there's just, and so in fact we can just work on the surface times a1, and so, so the way to do this is then the following observation, so this is kind of textbook edge write geometry, so if you have a current sheaf, current sheaf on surface times a1, this corresponds to a coherent os of t module on the surface, and then if you, and so, so the way this to do this, if you have a sheaf here on this three-fold, you can just push it forward, okay, and so this gives you then this os of t module on this thing, and then if you also ask it to be gm equivalent, this corresponds to, on this side to saying that this sheaf should be graded, okay, so grading with respect to this t is just the same as having this gm equivalent structure, so in particular if you have that you can, you can just write your, your push forward as kind of a direct sum of this, the sheaf, and then there's this feature that we look at current sheaf, so at some point this, this kind of direct sum has to stabilize, okay, so we have some e r minus 1, t r minus 1, but then we have e r, t to the r, and at this point it becomes stable, okay, so, so here's the point of, which I call point of stabilization, of stabilization, stabilization, okay, and so this multiplication by t here, I mean, gives you maps here, between the skies here, t, and you can show that, I mean, now, now it's kind of, that if you flat over this a1, so if you take our ideal sheaf, this will be always flat over a1 because it's torsion free, and so this will correspond to the fact that at t from this e i to e j is injective, okay, so really you have injections here if you start with a flat sheaf, flat over a1, injection between these two things, so in fact everything of here just corresponds to also finer filtrations, filtrations, yeah, on, on this, on this s, finer filtration of current sheafs, and so the way to go about this last is I'm often just sending this to this, this inclusion, okay, so, so this is kind of the, I mean, I guess standard standard two, okay, so how much time do I have now? Okay, so, okay, so now I should, somehow, so let me still say some words about this d equals to zero case, okay, so the d equals to zero case is this, this classical case, okay, and so in this case the, the Hilbert scheme, I'm sorry, this, this component a here that we had is just this nested Hilbert scheme, okay, and so the idea here, so the idea to solve this, determine our invariance in this case, so this is the idea of Pt, is to consider this, our obstruction theory, and so obstruction theory of our Hilbert scheme and restrict it to our fixed component and take the fixed part of this thing, and then observe that this has a second obstructed theory, so should have, so, should have, so, so maybe I put it here, construct a second, second cross-section for, for this fixed part of the obstruction, so you restrict it to this, to this component, and so from this you get that most components, most components finished, and so in some sense it's kind of very natural to expect such a thing in, in hindsight of course, and the reason is as follows, so if you, if you think back to the beginning of the talk, some of the obstruction theory by part one was kind of constructed by thinking about this class better and kind of trying to deform your k3 surface to a loci where this class better is no longer algebraic, so now in this fixed loci we have not just a single class better, but we have the churn characters of all of these sheaves, okay, so we have kind of r plus 1 sheaves here in this, in this fixed loci, and they have all kinds of churn characters, so in particular, I mean we already use this reduced class, so we know that somehow we cannot touch the churn 2 of this, of this sheave e, so this kind of global churn character is kind of, we have to keep algebraic, but we are free to kind of move all of these individual churn characters, I mean, I mean we have, we have no condition on this individual churn characters, so in particular if you have some, some guy here which has a churn 2, which is not linearly dependent to this, to this class better, we can just move our k3 surface to some place where this class is no longer algebraic, so in some sense if you look at this thing, you in some sense expect that you should get r plus 1 different cross sections, and just by this kind of intuitive thing that now to get a non-zero number you just don't have to protect the algebraic city of a single beta, but you have to kind of keep all of these churn characters algebraic under different measures, that's kind of in theory, I mean already this argument allows you to kill a lot of this term, I mean not kill a lot of these contributions, the issue is that what happens, what may happen is that churn character of these guys could be all kind of linearly dependent to this class better that we already have, so Sagan is not quite sufficient, and so instead what they do is they consider the sky here, I mean the structure sheath, so in this case d equals to 0, we have I mean over p1, so in case d equals to 0 really everything happens over this a1, so you have just the sheath here, okay, and so in that case they look at the sheaths, and then because of d equals to 0, at some point you hit this case where you stabilize, and then what comes afterward is all 0, okay, so you just get 0 at some point, and then they kind of do something amazing which is that they construct explicitly deformation deformation of this OZ, so of OZ over this over some field of dual numbers, this yields you a tangent vector, so from this you get a tangent vector conclusion in this x1 pi of ee, and then yeah so I should put a t here, and then again by this by the serduality argument we get the co-section, yeah it should be the torus weight, so I mean in some sense on surface times a1 this is Calabria, but equivalently it's not Calabria, so someone this is this t, so I just forget about it, and so this really very nice geometric construction is gives you additional tangent vector, additional deformation, and then it produces you this, so in our case we have to go to d positive, and then there are somehow two issues, the first one is you really have to work with the deformation theory of surface times p1, so you're not Calabria, and then the second issue you have to deal with is that your sheaves, I mean in our case we don't have the zeros, in our case the zeros still stay the structure sheave of our zero-dimensional sheave eta, so we have kind of this additional terms here floating around, and then you have this non-Calabria, and then so anyway so you have to kind of try to extend this argument to the surface times p1 case, so this is what I try to explain in the last 10 minutes. Is there any reason to take this, why do all numbers, it's why you non-obvious? I mean you have to I mean you have to, I want to construct a tangent vector, and the tangent vector in, so what's the tangent vector of a scheme, it's, I mean in the model space is in the Siebel scheme, it's just the sub-scheme over the x cross the dual numbers, anyway so this is kind of a very nice argument, second tangent vector, but we kind of do something more direct, I mean more stupid. How do you know it's a difficult section? Yeah so this is, I didn't mention this here, so I'm going to do this first, so they prove that this is a different course section if and only if you're, if your OZ here is uniformly stacked, so meaning that, so I should have said this, so this is different if and only if, so yeah so so this gives, gives the same course, same course section as before, if and only if all of these guys are the same, okay, so all of them are the same and then your curve really looks like a fixed curve with in the same structure, and if you, if you, if you have one of them different, then you get a new course section, then you get a vanishing by this Kim Lieben method, okay, so, okay so now, okay so there's kind of, I think two observations here, so let's do the general case, so, so let me just rewrite this line that I just erased, so this is our sheaf here, so now e would be our ideal sheaf in the surface times p1 or a1 for simplicity, okay, so it doesn't matter, so this would be, so we have this the non-trivial stuff, and then we have this stabilization guy, so t to the r, t to the stabilization point, then we stay, stay the same, and here we get this ideal sheaf of this it, and so the first idea is to, I mean, okay this looks kind of complicated, I mean, so you have kind of this non-trivial stuff here, nevertheless you can just view this, you can in some sense just cut it here in this, in this stabilization point, and just view it as a filtration, where you take e0 sitting inside e1, and so on, under er minus 1, then you have er, so er is this, this guy, then you just add a trivial piece, os, okay, so there's a natural unique map here, and this is just os, it's just c, right, this is the ideal sheaf of this, it's just c, and in fact there's a canonical map to os, in case you just view this as a, just add this kind of stupid term here in the, at the end, and so this tells you, also in this case here, this a and beta that I wrote, it's just the union of nested Hilbert schemes, okay, so, and so here the nesting, so we have this, this parameters, so n, I don't know if I should give you a definition, so this is, would be just the sub-schemes inside er, and then inside os, where this ei, ej's will be just ideal sheaf of points of length nj, and then you tensor with the o of minus a beta j, okay, and so then with this convention there's a certain constraint you have, so you have beta 0 plus beta r is equal to beta, and in our case the last one will be just the ideal sheaf of this eta, so this beta r will be just 0, and then you have a second constraint here, the sum over the nr, and well, my notation should be like the summation over ni plus 1 half beta i i square is equal to, I forgot, so n minus r times d or something, okay, and so, so once you have that, so that's kind of the first observation, you just, you don't need to construct a new kind of crazy model space, you just work with a nested habit scheme, and then step two is this co-section, okay, so co-section, and so what you prove is that the following, so now let's look at this, our guy here, so this is our fixed part of our obstruction theory, okay, restricted to this, restricted to this low side, now what you can prove is the following, this is just a cone of the following guy, so this is a direct sum r home of ei ei, mapping to the direct sum over r home of ei ei plus 1, and then you put a trace free part here, and then you have to get the sum and rights, so it matters here, so take the deformations of the first r's, then here these ones you go for, actually 0 to r minus 1, so if I put here r then you get just a rank, I mean the theory for respect to d equals to 0, but here we put r minus 1, because we did this cut, so and so here this 0 here, let me just define this, and so here this r home ei ei 0 is the cone over r gamma of s os, r gamma over ei, sorry r home, and then the same thing also for this ei plus 1 thing, is the cone of the same guy here, mapping first to the same guy here, and then continuing with this natural inclusion with respect to ei, ei plus 1, okay, so here I can attribute this to, I mean Golampur, Seshmani and Yau, and also I guess Golampur Thomas, and then I mean it's really not, I'm epsilon square or something, okay, so now I don't have any time and more I guess, right, so I want to draw a final, I mean the final thing we have to do is just take the, understand this long exact sequence associated to this thing to find our cross-section, once we have that then we're done, okay, so let's just take homology of this, okay, so now we are hunting a second cross-section of this thing, so take a long exact sequence, and it's really kind of a beautiful nice thing to do because you learn a lot about this nested Hebert schemes by just taking this, stirring it, okay, so the trace free part, if you take the h0 of this term, you get home of ei ei, that's just c because the ei are ideal sheet, so they're stable, so this is just c, and then the trace free part just, the identity is killed out by this cone, so this first term is 0 and here you really get the Hormes model of this identity, so you get the first term here is the Horme, Horme of this ei, ei plus 1, Gauchner C times this natural inclusion, let me just write fi on inclusion from ei to ei plus 1, okay, so this is the first term, it goes to r minus 1, so this maps to the x1 here of this ei, 0 fix, okay, and so this term here really just describes you to, I mean, kind of tells you how to move, so the ei and ei plus 1 differ by a curve and some of this group tells you how to move that curve around, so this is this term, then the next term is, yeah, so we're here, then we're going here, and then the next term is here, so this is just a direct sum of this x i's, of this ei, ei, where i goes from 0 to r, and so this is also very natural because this just tells you how to, and when you have a deformation of the nested table scheme, it induces your deformation of each of the innovative members of the sheaves, so if you deform the whole thing, you also deform the individual ones, this map, and then the next one here is the x1 of this ei, ei plus 1, and then here you have the 0, and then you have er minus 1, okay, and what is this map here, this kind of interesting map, and let me just draw you one picture, so you have a deformation, so you have a guy here, let's say a guy, a tupper here, that gives you the deformation of the first sheave, I mean one sheave and another sheave, so let's say you're given a tupper for i, that just gives you collection of deformation of the individual sheaves, and then you have this natural inclusions here, and essentially this guy here just sends you to the corresponding condition that makes this diagram commute, so now I have to be careful, it's like fi, i for i minus this j, or something like that, the guy in the middle is not crest free, I'm just saying I'm just trying to describe you this map, so here, and so finally you end up with this obstruction space, okay, so now comes the punchline of the talk, so where's the co-section, and so this is really by trying to understand this sequence also, so let's just try to understand it, and so here we really took this r-gamma, and here I mean h1 of s or s is 0, so let's start at this first term here, so we have the sequence x1 of ei, ei plus 1, and so you go to this x1 of ei, ei plus 1 to the trace free part, and then you go around, you go to this h2, and then you continue, you have x2, and then you go to the x2 of this guy, which would be 0, and then 0, okay, and so now just, so the punchline comes in this thing to observe that this is just the home of ei plus 1 comma ei by said reality, do it, so in particular this is c if ei is equal to, if you are the same and 0 if they're different, okay, so if they're strictly contained in each other, so in particular if they're different, this guy is 0, and so this means this is subjective if they're different, okay, so if they're different it means you go to 0 here, they're subjective, so then you really have a map to this trivial piece, and so then what you see is that you get a map here to basically direct sum of all the i's where you're different, h2 or s, so these are trivial summands, then you show that this factors to it here, and then you win, okay, so the upshot is that in the end, so the corollary is that this guy here and better, so this, this is 0 if the number of under indices where these two things are different is bigger than 1, okay, so if then more than two steps in this filtration, then you get a 0, and so, yeah, so from this you, I mean you just do some calculation and then, and then, yeah, you get this, and you get, I mean it's kind of beautiful, you get from this geometric argument, you get the sign and you get this factor and then, yeah, I mean it's, yeah, okay, maybe I stop because I'm already over time. Questions? So you get this for the Iguzakas, we have this s3 symmetry, which can live to the derived category, yes, I guess the multiple curve you can express by the hacker operator, yes, you can do that, can you live that to the derived category? You mean this s3 action? No, I mean the hackers. Ah, to the derived category, I mean, no, because somehow, yeah, so this strictly does not come from a derived equivalence because somehow, yeah, I mean the invariants which are preserved, I mean, so for example, this, this, there's certain invariants you can write down for this, let's say beta 1 minus n1, you can write down invariants which are preserved under derived equivalences and which are different, so you can write down a certain number here, a certain number here, and they're different. If, for example, if this one is, if this has a divisor and this one doesn't, if the GCD of the second one is 1 and the GCD of this one is 2, then you can write down an invariant that tells it that there's no, I mean, which is a derived equivalence invariant, so you can never take one to the other. Okay, so, yeah, so derived equivalence will not help you prove the multiple-cover conjecture. Is it easy to describe this invariant? Yeah, it's easy to describe this invariant, sorry for not doing it, so, so the, yeah, so the way you do it is you look at this, look at the integral homology and you decompose it as like, homology of k3 surface times this class of a point on the elliptic curve plus, and then if you have a vector here, so this, I mean, if you have some guy here in the here, then you just take the wedge of this and you take the GCD of this wedge, and at least, okay, so that's, I mean, there's a slide, slide lie hidden here, so it's not, the auto group of autocorons of k3 times elliptic curve is not known. I think it's even not known for k3 surface, so, but under the natural ones, I mean, the ones coming from autocorons of the k3 surface and the ones from elliptic curve, you can prove that this quantity is just preserved under autocorons. And so, so the higher rank is that's, yeah, so the higher rank is not the torque, so, but yeah, I think this is also, I mean, I think that's like, yeah, I think three different approaches to it, so I think one of them would work. I have a question with multiplicity two, because Eugene was also wondering in chat. The fact that these hasted, hasted sheathness ideals come up just like in Richard's heading for the cotangent bottom of this, so is that just a consequence of the fact that you're using the same methods or is there a connection between the geometries? You mean that you get the same, I mean in you have s times e and he has the cotangent bottom of this?  So, high question is the fact that you can reduce to s times p1 is something that you haven't showed us and then it leads to the same things that. Exactly, exactly, yeah. And this reduction to s cross p1, I can refer to Pander Pander's lecture next week, so. So, then is it's no surprise that with the same thing should be solved? Yes, it's not surprising, yeah. I mean, what was surprising a little bit for me is that this issue with here can be solved too easily, but maybe this was just my stupidities. Could you explain briefly why components B and C doesn't contribute? Yeah, so here really the, if you calculate this, then the x2 really splits as a sum of two x, so for this component, ah, sorry, for this upstairs, really splits as of ee, I mean, maybe splits up to, maybe there's some factor, but let me make a small lie here, okay, so it's essentially the same as the obstruction spaces to these guys and maybe you have to worry that this, they match on both sides, where this is the adieu sheet for this guy. And so then you have a co-section here and you have a co-section here and they're independent, and so then you have again two co-sections, so this doesn't contribute. And here it's really dimension argument that somehow the reduced virtual dimension of this surface times p1 is, is 2d plus 1 and here this rubber just has a virtual class of dimension 2d, so there's some dimension argument that, that we don't need this, so this is actually, yeah, this is kind of, kind of lucky that this doesn't contribute. Any other questions? So you, you have this insertion of degree 1, how did you particularly choose the insertion? Ah, I chose this by tau zero, or you choose why, why I chose this one. Why do you expect much to cover for this insertion? I mean, you can choose, choose a different one and essentially you can re-express it in terms of this one. There's a different way to, you write everything and when you quotient out your model space by the elliptic curve and then, yeah. So essentially it doesn't matter which one you take. I mean, yeah, sometimes this is the, this is the, the natural one from several perspectives, but it's not, yeah. There's nothing particular about it. There's somehow a single number that you can attach to this model space and that's, that's, that's one. Well, to find if you're a reduced DTO fix, you make a choice of beta dual, such that beta, beta dual equal one. Yeah, and it doesn't depend on which one you take. Okay, it punishes when you reduce to the relative theory that beta dual, somehow. I mean, you have, I mean, I didn't talk about insertions, that's kind of also a different story. So you have to, you have to, yeah, deal with, yeah, deal with the insertions that you take. And here the surface times P1 relative to zero, it's not Kalabiao. So this one is not Kalabiao. It has the, this has reduced virtual dimension 2D plus one. In fact, you need a lot of, a lot of insertions here. And I just sweep them under the rock. Do you expect a similar story in the gromovitin side that you can reduce the non-primitive reduced gromovitin theory of K3 times E2 primitive in terms of primitive? Yeah, so I mean, on the gromovin side, you can just use the product formula and just reduce to gromovitin theory, reduce this question to gromovitin theory of K3 surface. And there's a conjecture for exactly that. That, I mean, any gromovin theory of, any gromovin environment of a K3 surface in primitive class can be reduced to primitive class. And that conjecture, I mean, I can refer to as Tim, to Tim, because he proved it in divisibility too. So, but in, in, in the general case, that's kind of an open question. So the expansion for a different point for the inverse. So that's related then to the high rank? Yeah, that's related to the high rank.