Coming up on DTNS, Apple might make a car after all, Microsoft Flight Simulator goes VR and electric rays that can power their own ocean floor maps. This is the Daily Tech News for Tuesday, December 22, 2020 in Los Angeles. I'm Tom Merritt. And from Studio Redwood, I'm Sarah Lane. And I'm the show's producer, Roger Chang. Ah, we're very excited to have Blair Bazderich, cohost of This Week in Science, back on the show. Blair, welcome back. Hello, thanks so much. Happy to be here. Yeah, it's good to have you. We were just talking about what we all eat for breakfast, among other things, on Good Day Internet. If you want that wider conversation, become a member at patreon.com.dtns. Let's start with a few tech things you should know. Xiaomi confirmed its Mi 11 and Mi 11 Pro phones will launch at an event in China on December 28. The phones will be the first to ship with Qualcomm's Snapdragon 888 processor. Microsoft is active in a couple of security efforts to tell you about. First of all, Microsoft, Google, Cisco, VMware, and the Internet Association all filed amicus briefs in Facebook's lawsuit against NSO Group over NSO exploiting a bug in WhatsApp. And Microsoft, McAfee, and 17 other organizations announced plans Monday to form a ransomware task force. The RTF will identify gaps in current solutions and work on a roadmap to address issues and develop a standardized framework to help the industry deal with ransomware attacks. Technology and media holding company IAC says it will spin off Vimeo into a standalone public company in the second quarter of 2021. Vimeo says it has 200 million users. Last month CEO Anjali Sood said, quote, In the last seven months we welcomed over 30 million new members, seen over 60 million new videos, created, uploaded, and powered millions of live events that went digital for the first time more than the prior three years combined. The Wall Street Journal reports that according to an unredacted draft of that lawsuit filed by Texas and other states against Google, Google and Facebook agreed to cooperate and assist each other in responding to any antitrust action regarding their advertising agreements. Google says that such agreements over antitrust threats are extremely common and that both companies still say there is nothing illegal about their advertising agreements. The Washington Post sources say Facebook offered to license its code and users' relationships to a competing social network in order to avoid an antitrust suit. However, investigators investigating Facebook declined that offer. Earlier this month, FTC led the filing of two suits against Facebook's acquisition of WhatsApp and Instagram. Alright, let's talk about that Apple car. Reuters sources, two sources say Apple plans to introduce an autonomous passenger vehicle for consumers. At 2024 at the earliest, though they might get delayed because 2020, the car would operate using a battery designed by Apple to reduce cost and increase range. Supposedly that battery design eliminates some of the internal structure. They called it a monocel in the Reuters story, holding battery materials together, leaving room for more active materials, so smaller and more energy efficient is the promise. And Apple may be using lithium-iron phosphate for its battery instead of lithium-ion. Lithium-ion is what you have in most of your phones out there. Lithium-iron phosphate is less likely to overheat. The sources said they expect Apple to rely on a manufacturing partner to build the car, and it may just end up with Apple integrating their system into a car made by a traditional automaker rather than an Apple branded car. Apple would supposedly use multiple LiDAR sensors, some of them from Apple and others sourced from partners. Apple's head of AI, John, John Andrea, took over the car project earlier this month. And again, these are just sources, but that project's been on and off according to sources over the year. It looks like it's back on again. Oh my gosh, the Apple car rumor has been, you know, what are we, five years strong now? Although we're talking about 2024, so there's certainly some work yet to be done. Like you mentioned, Tom, 2020 has stalled a lot of projects. So who knows, this could be, I don't know, farther in the future. But I like the idea that Apple might have a technology for a battery that it could license out to all sorts of manufacturers. It could be the Apple car or something that Apple has partnered with, you know, a GM or Ford or whoever, you know, any number of the large auto conglomerates around the world to be able to produce something that's a passenger vehicle that can rival what Waymo is doing, for example. But if they can get certain things right by using different materials, what we may see as the automotive industry and what goes into cars five to 10 years from now could be drastically different. Yeah, I think the key to remember here is that Apple probably hasn't exactly decided how they're going to go about this, whether they're going to say partner with Hyundai or somebody like that to make the car and it'll be a Hyundai Apple version. That I think is on one end of the spectrum. The other is that they contract with someone out there to build the car for them. Some automakers already do that. Some automakers don't even build their own cars already. So it gets complicated when you look at the supply chain. But the fact of the matter is, if this ends up panning out in 2024, maybe 2025, you'll be able to go somewhere and buy a car that has got Apple technology in it. Blair, does that sound like the kind of autonomous car you would want to get yourself? I mean, I'm laughing over here because when I think about my iPhone, I think about how an update might get pushed onto my phone and then it suddenly doesn't work for a day until the new update comes out. And I get very concerned about an interface on a car deciding not to break because the latest update was buggy. Yeah, thankfully, we have a lot of experience with firmware on cars. Not that there haven't been problems over the years, but thankfully, usually cars don't break down for a day because of firmware updates. And hopefully Apple would be able to carry that on. This feels like something that'll be cache, right? Like somebody will be like, ooh, I want the Apple car because it's got the whole Apple system in it. The other side could be that Apple becomes, like Apple and Tesla would be my guess, would be the first to offer autonomous cars to people for them to buy themselves. From the Reuters story anyway, it doesn't sound like they're really targeting fleets or industrial uses, but consumer uses. Yeah, that was the thing about the idea of Apple having a manufacturing partner when you think of the iPhone, right? For one of the best examples, it's like, well, different companies put those iPhones together physically, but it's an Apple product. Would Apple be able to move that much merch on something that would clearly be a designer car? I mean, Tesla's, even the cheapest models of Tesla's are mostly outside of anybody's budget. And I can't imagine Apple would be going into the low-cost car market. It doesn't sell like a bargain car. No. It doesn't. And it's, I mean, listen, I want to see this thing. Trust me, I wanted to exist. But how many would you really be selling? Yeah. Well, in the Reuters article, they say they have to make at least 100,000 a year to start with. And Tesla took a long time to get to profitability. That's why it sounds like maybe they're better off partnering with an automaker that can already kind of move them into the supply chain. Yeah. This is one we won't know anything for sure for years. And it may even evaporate and go away again. But it's interesting to think about. Indeed it is. Also interesting to think about is flying and VR. Microsoft Flight Simulator is officially in VR. A Sobo Studio, that's the company that developed the 2020 version of the app, has now added support for open XR compatible headsets, which include Windows Mixed Reality headsets. That includes the HP Reverb G2, also the Valve Index, HCC Vive, the Oculus Rift, and even the Oculus Quest, if you're using a link cable to connect to a PC. Microsoft originally said that the VR feature would first arrive on the HP Reverb G2, so that's not untrue, but there are other platforms also available now. The beta test, which ran in October, does require a computer with an i5-8400 or Ryzen 5 1500X paired with a GTX 1080. These are not things I have. So I can't try this yet, but I really want to.  It's a whole community, and people are very enthusiastic. And I know just from looking at them, and it makes sense to be that the idea of doing this in VR is kind of ideal. There's a whole hobbyist community built around making your cockpit to have your Flight Simulator in. But even then, you have to design it on one particular plane or try to kind of design it towards a few of them. And it's not going to change. Whereas in VR, you're put inside the cockpit of whatever plane you're flying in VR. And it adapts so you really do feel like you're flying. Yeah. You know, you look left, right? You're in the cockpit. Again, this is not something that I've tried personally and no idea how to fly a plane. I was always horrible at Flight Simulator. All I did was crash. Blair, I don't know if you ever played any iterations of this app growing up. But what do you make of VR and flying planes? I think it would be so fun because it's something I would never have the guts to do in real life. I also wonder with all these kind of very specialized rigs for VR, if this is going to be the resurgence of arcades in our culture, you know, kind of after COVID, obviously. But I could totally see VR arcades where you'd have all these special setups for different opportunities that I would not buy VR equipment. That seems like way too heavy of an investment. But I would love to go pay for a session at a VR studio. Yeah. There were a few of those getting off the ground. And the question was, would things like the Quest making it cheaper to do VR at home mean you wouldn't want to go to a place and then of course we all locked down and nobody could go any places. So it'll be interesting to see if that bounces back. Hey, we mentioned some of the technology the US Stimulus Bill will fund on Monday's show, but also included in the 5,593 page bill, if that explains why we didn't get to everything yesterday, are some penalties against copyright infringement that you might want to be aware of. For instance, one provision of the bill makes it a felony, not a misdemeanor, a felony to quote willfully and for purposes of commercial advantage or private financial gain offer or provide to the public a digital transmission service of unauthorized media and it carries penalties of up to three years, five if the person should have known that the work was being prepared for commercial or public performance and 10 years for multiple offenses. You're talking about 10 years in jail. Yes, penalties of up to three years in prison, five years in prison or 10 years in prison. It does not apply to users of a service and the public knowledge believes it won't impact streamers who include unlicensed works as part of their stream. So it's targeted at I'm taking the NFL network and rebroadcasting it on my own service. That's who they're going after. It's hard to tell from the language how it might be applied though, because, remember, copyright trolls will try to take what you wrote and doesn't matter why you wrote it and try to apply it to their own ends. The bill also includes the case act, which creates a copyright claims board to be part of the U.S. Copyright Office. That claims board would rule on copyright claims and order payments of up to $30,000. Now, it's being compared to a small claims court, but it's not a court. It'll consist of claims officers, not judges. If a claim is filed against you, you must respond in a specific manner outlined in the law within a limited period of time, or you will be stuck with whatever penalty has been decided on, and this will make it easier to file cases. Again, the reason they're saying they want this law is to help smaller artists fight back against copyright infringement, but it does seem to favor companies with large amounts of lawyers as defendants. In fact, the Electronic Frontier Foundation warns that this could increase copyright trolling, the practice of filing a lawsuit against someone without the resources to fight it, so that you can either stick them with a penalty or get them to settle out of court, because they can't afford to fight it. When this sort of got broken down in the first article that I read late last night, I was like, hmm, there are times when someone's like, hey, want to watch this movie? I have it, and I'm like, yeah, sure. I'm like, am I going to jail for three years, even if I can argue that I didn't really know what I was doing type thing? That does not sound like what's going on here. It's like you mentioned, Tom, if I am carrying an NFL stream willfully with the knowledge that I shouldn't do it and offering it to the public on some level and keeping money out of the hands of the original broadcaster, that is different. But it is ambiguous enough that I know a lot of people were kind of freaked out because there are lots of ways that people get information. The copyright stuff sounds like a mess, honestly. It sounds like just a total mess. And I mean, we on DTNS have copyright infringement issues on putting stuff up on YouTube every so often, and sometimes they make sense, sometimes they don't, and a lot of people have it worse. And this doesn't seem to lay out anything that makes more sense to me, and it seems like it will hurt smaller creators rather than help them. Well, yeah, there's so much going on here and to boil it down and make it simple. These were not necessary to be part of a stimulus bill. They were forced into the bill before people could have enough time to really read it and understand it. And when the major downside of this kind of thing is that, if even if you assume the best of intentions, let's just pretend they had the best of intentions. If you don't have enough eyes on it looking at how it could be misinterpreted and misused, it will be misinterpreted and misused. And this is going to be signed into law. This is going to be a lot. It does have a sunset provision, I think of 2023. So it may not stay law, but Tom Tillis, Senator Tom Tillis, who was behind these provisions, has a whole DMCA revision that he wants to push forward now as well. So we're going to get more of the same coming down the line.  This impacts you too. Yeah, it's very frustrating to hear about how this has all progressed and just how needless things have been tacked on when really just people needed a lifeline. That's all this is about. Yeah, the bill was supposed to be about helping people get through COVID, and get through COVID restrictions, recover from economic disruption. We'll float you a little money, but if you stream something, boy, you are really screwed. And I know that channel works. Are you trying to make money by streaming now in addition to your normal day job? Well, let me make it harder for you. Yeah, and again, it's all about the misinterpretation. I'm fine with like, oh, they want to stop these industrialized people who re-broadcast streams of television networks. Fine, I guess I don't have a problem with that. They want to make it easier for a small artist to be able to protect the copyright for their songs and movies and independence, doing something for small artists. I'm all for that. I'm not sure that these are properly secured and targeted and vetted to be able to do that. And I guess we'll find out, because the folks who want to take advantage of this kind of system will not wait around very long once it's signed into law to take advantage of it. Hey, what do you want to hear us talk about on the show? One way to let us know is our subreddit. Submit stories and vote on them at dailytechnewshow.reddit.com. Well, the ocean's a big place, a vast ocean floor. And as much as we've learned about the ocean, it's not as well mapped as we would like it to be. If you want to find resources like minerals or fuel or fish or other marine wildlife, we're going to need better maps. Blair, who can help us here? Oh, just electric rays and stingrays. So yeah, these rays, these cartilaginous fish related to sharks, they are benthic animals, which means that they move around on the very bottom of the ocean floor. And Ryken Center for Biosystems Dynamics Research in Japan have completed a feasibility study seeing if these little sharks or these little rays can actually be equipped with pingers and cameras to help us explore the bottom of the ocean. So they used pinger technology, which, you know, it's just it emits ultrasonic sound and when the sound is picked up by a receiver, the position of the receivers at the time is detected and then they can calculate a position. And so they use that along with cameras to figure out if this was even possible. If they could put these little backpacks on these rays, I'm sure they're not backpacks. They're much smaller than that, but I like to imagine that they're backpacks. And going to camp down into the ocean and do this job for them. Less invasive, as you might imagine, than bringing down ROVs, remotely operated vehicles or other technology, just send those animals on their merry way. So they started with a water tank. They used cameras in three places, front, side, and top so that they could verify the position of them while they were using these pingers. They allowed, they did 3D reconstruction of movements over time. And then based on that and the pingers, they found that it was very effective. So then they plopped them into the ocean. They found a part of the ocean that was pretty flat and was well plotted already so that they could verify it. And it was all about 60 to 120 feet deep. So still close enough that they could check in on everything periodically. And they found that the positions of the rays were very accurate. They were about 10 centimeters within the public map at most when they were recording all of the stuff. So all of these feasibility studies tell us that this is something that could be done. So the next test would be to throw them into a spot that is more varied, is less flat to double check the efficacy as it gets more complex. But one of the things that I absolutely love about this is they were actually able to use the electric rays to charge the pingers themselves because they're electric. That is insane. Yeah. So they have enough charge in the electric rays that they could charge a battery. And I get it that these are just the pinger that's getting charged, right? Not the camera. So it probably draws very little current. Totally. Yeah. And so that's just part of their, they have their own little electric current that runs through their body. Yeah. Yeah. So they can use that to power the device. But then also in the normal sting rays, they were able to use little batteries. Do you need an adapter when you want to use electric ray to power your pinger? Yeah. Yeah. So a pretty good adapter if you wanted to charge your phone off of an electric ray. But yeah, no, they were able to use the electric field effectively so that there wasn't much adaptation of the technology necessary. So they think that it could work for a pretty long-term monitoring because of that, because the rays charge it themselves as they go. So Blair, I know that, you know, a lot of the sort of mystery of the sea is because there is so much life so far down that, yes, like an ROV, you can get, you can get close, but it's pretty taxing. It's hard to do. And there's a lot of wildlife that doesn't come back to the surface in any way. So you got to go down to even know what's happening. Our electric rays and sting rays known to be able to be on the surface, get a camera attached, all good. And then go down more than, you know, any other, I don't know, somebody that lives in the sea. Yeah, yeah. So these guys aren't abyssal. So they're not going back down where like the blob fish lives.  So they're not going that deep, but they are going pretty far down several hundred feet down for sure. And so it's a good start. If there's other things that we could potentially do with this technology, if you kind of, if this proved really effective, you could start to adapt it to try to use it on marine mammals, things like sea lions or elephant seals that go miles and miles and miles below the ocean. If you could find a way to make it work for you for this use. But I think the thing that makes the rays so special here is that they really do just comb the bottom of the ocean. They are stuck flat to the bottom. And so the pings are really similar to the actual sea surface. So I think that's kind of the difference is they'd have to adapt it to be able to deal with the fact that an animal wasn't going flat against the sea bed. Right. Because that challenge is, is how do you bring a benthic animal? How do you bring an abyssal animal up to atmospheric pressure?  Yeah, that's definitely five steps down the line, but interesting. Yeah, yeah. But if people don't realize it, we have not mapped the ocean floor very well. I don't know what the exact comparison is, but what we know about the ocean is usually compared to what we know about the moon, which we've been to a couple of times. We're on top of the ocean all the time. We just don't go down to the bottom of it very often or in all the places that we would need to have a really good map. Yeah. Yeah, I think some people, depending, it depends how you measure it, of course, right? But I think some people like to say that we know more about our own solar system than we know about the ocean. Yeah, yeah. I mean, that's the kind of comparison you make, right? It's not like we know way more about land. We're so land-centric and most of our planet is water. Yeah. I think also it turns out we're a lot better at making things for zero gravity than extreme atmospheric pressure, which is what happens underwater. All right, real quickly, law enforcement agencies from the U.S., Germany, France, Switzerland and the Netherlands have seized the web domains and server infrastructure across five countries, VPN services, insorg.org, safe-inet.com and safe-inet.net. The three are believed to be operated by the same individual or group and widely advertised on both Russian and English-speaking forums. The U.S. Department of Justice in Europol claim the three companies' servers masked the identities of ransomware gangs, web skimmers, fissures and hackers involved in account takeovers. Europol said it plans to analyze the collected information and begin to take action against some of the services users.  These aren't VPNs that you're likely to have been subscribed to or using. They are very specifically used by people who want to engage in illicit activity and keep it hidden. They are so-called bulletproof VPNs that means the operators will change customers' IP addresses or move the servers so that when they're asked to cooperate with law enforcement, they don't cooperate with law enforcement. So there you go. That's going to make it harder for criminals to hide on the Internet. Not impossible. They'll find other places, but worth noting. Well, also worth noting, and boy did I have fun with this this morning. A team from Databots has used a recurrent neural network, or RNN, to create a YouTube stream featuring a base solo that never ends. CJ Carr and Jack Zukowski from Databots trained the RNN with two hours of base improvisation from YouTuber Adam Neely. There's a lot of base stuff online, you know, you might know them already. Then they limited the dataset to mainly faster licks because the RNN seemed to prefer faster tempos. Just had more fun with them. Databots previously experimented with an endless death metal stream. Now death metal is probably not something that I would stream while preparing for DTNS as I do every morning. But I did turn on the YouTube base, NeverEndy Base stream this morning, and I kept it on for like the whole morning. It's great. That's an AI, just in the core. Right. It reminded me very much of, I mean, it's actually a lot more talented than all the college guys I hung out with who tried to learn guitar. But it kind of reminds you of like that one person in the corner who's just kind of riffing, you know, they're just kind of doing their thing. It's all instrumental. I found it very soothing. I really like this. It reminds me of a NeverEnding Seinfeld episode for just going segway to segway to segway to segway. It's so Seinfeld-esque, yeah. Well, it's the deal with datasets. Yeah. So go check it out, folks. We'll have a link in the show notes so you can just search Infinite Base on YouTube. I think you'll find it. Oh, it's so good. Good work, robots. Getting there. All right. Let's check out the mailbag. Let's do it. So we talked to Alison Sheridan about Apple Fitness Plus, which she has tried out. She's the only one of us who had, at least as of yesterday. David from Brisbane, Australia says, when it comes to tracking metrics during fitness, whatever the brand we're using. So it doesn't have to be Apple, any brand. It seems to assume that because we can track it, it must be important. I referee soccer and I attend referee fitness training. I was shocked when the person next to me said his heart rate was 140 BPM. Mine was 190. We were doing the same exercise. So shocked that I went to the doctor to see if something was wrong with me. She ruled her eyes and said, I should stop looking at the metrics on my watch and that my max heart rate is my max heart rate. And if I'm sweating during exercise, I'm doing myself good and I'll be progressing in my fitness. She also pointed to several medications that I'm on. Those would increase my resting heart rate. David says, I love data and the insight it brings and maybe some of the things our devices are tracking are genuinely helpful. But I'm concerned that we're being led to pay too much attention to certain metrics because they're easy to track and feel scientific rather than because they're of any real benefit. I like to see independent research of these fitness programs before I put my money down. Yeah, the data is helpful if you know how to interpret it. The data itself doesn't mean anything on its own, right? That's the key with all of this stuff. And I get what David is saying where people push like, oh, look at all the data it collects, but that's not the end of the story. It's like, OK, now what do you do with that? Do you have a reliable way to interpret it? Is the service you're using to interpret it reliable? Dave, maybe you need to consult with your doctor as David wisely did here. So yeah, that's all important stuff. Yeah, its calorie counting is important too, right? But you can't just have 2,000 calories every day. Every person is different. Also, if you have 2,000 calories of cookies or if you have 2,000 calories of salad and eggs and bread, that's a very different count. But it's the same data. Yeah. I don't think that collecting the data is bad. I just think that David's making a very important... And I don't think David's saying that either. Just a very important point of that's not the end. It's not just I got the data. It's making sure you have a good, reliable way to interpret it. Indeed. Well, if you have feedback on anything that we talk about on any of our shows or stuff that you'd like us to talk about, or questions or comments or cat photos, feedback at dailytechnewshow.com is where to send those emails. You can also shout out our patrons along with us today. Master Grandmasters include Pat Sheeran, John Atwood, and Daniel Dorado. Thanks to all. Also, thanks to Blair Bazderich for being with us today. Blair, what's been going on in your world since we saw you last? Oh, you know, we're just about to count down to our 10 most important stories of 2020 on This Week in Science. That'll be next Wednesday. And then also, yes, I see Tom has his. We have This Week in Science, Blair's Animal Corner. Yeah. So I realize you probably didn't use your calendar this year at all, but let's hope that we can use them in 2021. And so there's fun animals. Here's a giraffe weevil for May. But yeah, those are available at twist.org. Just click on the picture of the horned frog on the homepage. You can buy one or 10. The white sturgeon I particularly like. I'm looking forward to June. I worked with a white sturgeon named Lenny. He was my inspiration for that one. Well, thank Lenny for me. Hey, folks, we have some stuff for you as well. If you stick around with us, we're offering Patreon loyalty rewards. Patreon has a new ongoing program. It doesn't end after three months anymore. You get something every three months. They have to be at the right level, but you can get a unique sticker, a mug, a t-shirt or a hoodie every three months, as long as you stay a patron. That's not all levels, but if you're at the level that has that, you'll get either a unique piece of art from Len Peralta with the DTNS7, your anniversary logo, or three months after that one with Roger in it, three months after that one with Sarah in it, after that one with me in it. You can get all the details about the levels and the art at patreon.com. We're live Monday through Friday at 4.30 p.m. Eastern 2130 UTC. And you can find out more at dailytechnewshow.com slash live. We are back tomorrow. Rollin' right into Christmas week with Scott Johnson. Talk to you then. This show is part of the Frog Pants Network. Get more at frogpants.com. Bob, I hope you have enjoyed this program.


Okay, so right so before we start today, I just want to make a quick announcement. So there's a PCMI cross program activity today it's a panel discussion on career arcs, which everyone is strongly encouraged to go to it so this panel discussion is at 1230 mountain time and the link is in the email sent out. So this does conflict with the usual time for the TA session and I believe the TA session today is likely to be moved. Probably just sometime after that, but maybe just confirm, maybe I can confirm at the end or or check discord for to confirm the time of the session so okay. Okay. Hello everyone and welcome back. We're going to have some fun today. Let's see here. We go. So, yeah, we're going to be talking about reduction theory of. There's a lot of adjectives. There's a lot of definite binary integral quadratic forms I think I got them all in there but let's just just say what we mean so let's recall. Let's recall the context we had last time. So, we're interested in quadratic forms over the integers, and we're looking at the smallest non trivial case which is two variables. A binary quadrant that's called a binary quadratic form integral is a polynomial of the form. f of x y is equal to a x squared plus b x y plus C y squared, or a b and c are integers. So it's a homogeneous degree to polynomial in two variables. And we're also going to denote this as left brackets a comma b comma c right brackets, but be careful. This is not consistent with the previous use of this notation where this stood for a diagonal form in three variables, a x squared plus b y squared plus C z squared. And b here stands for the cross term x y. And in general, not every such guy is going to be diagonalizable, but actually see this as a consequence of something we do today. So there's no use in the diagonal notation it does it's not general enough so this will be the new notation, they'll just put this caution symbol here to not consistent with, with previous. Okay. All right, and the most important quantity in the most important invariant you associate to this. Such a binary quadratic forms is discriminant, which is just the usual expression b squared minus four AC. And we will assume, we will be assuming that the corresponding if you take the same polynomial but view it as a real quadratic form. And then we'll be assuming that that's positive definite. And as I explained last time that's equivalent to some other properties of this form, for example, you know, f of the x y is greater than zero, unless of course, x y is equal to zero. It's not negative and in fact strictly positive unless you're evaluating at zero. And it's also equivalent to a numerical criterion involving D. So, well the discriminant of f should be less than zero. And, well, to separate between the positive definite and the negative definite cases you can say that a and C, or, or definitely just one of them has to be positive. So both a and C are values, right, on one comma zero and zero comma one respectively so they have to be positive. And that actually is enough to guarantee that your positive definite. Once you have this condition as well, because you're either positive definite or negative definite right and that shows. Yep. Okay. So from now on. This form means such a guy. So it's a positive definite integral binary quadratic form. It's an app like this satisfying these equivalent conditions. I'm going to say form for that. And you'll have to remember that it's positive definite, for example. And we talked about the notion of equivalence of forms that we're going to be using. So, so also recall, we say that F is equivalent to F was strictly equivalent. Sorry, F is strictly equivalent to G. If and only if they're related by an invertible change of variables with determinant equal to one. So in general, invertible matrix over the integers will either have to turn one or minus one and we want to only look at those were making a finer equivalence. We're probably determined to be equal to one so concretely, this means that you can write that if you take F of like alpha x plus beta y gamma x plus delta y. Then this gives you G of x y or some alpha beta gamma delta in SL to Z. So the group of integer two by two matrices with determinant one. Okay. Now. So these are going to be our favorite characters from now on. So, well, also, but I just accidentally hit my camera. Okay. So also recall. That's if F is strictly equivalent to G, then, in particular, the discriminants are equal to each other. That was something we also saw last time. However, let me remark. There are many so that so well, well this is not invertible. So this is not a if and only a so discriminants being equal is also a coarser equivalence relation than being strictly isomorphic. And in fact, you can put a whole bunch of different rather natural equivalence relations sitting in between the strict equivalence and the equality of discriminants so let me just write a chain of implications here. So F is strictly equivalent to G implies F is isomorphic to G, and this I mean in the sense of, you know, using GL to Z instead of SL to Z so arbitrary invertible linear change of coordinates. This in turn implies that F is isomorphic to G over the chaotic integers for all primes P. Which in turn implies that's F and G or isomorphic over the rational numbers. Which in turn, I guess implies that that already implies that the discriminant F is equal to the discriminant G. And in general, all of these implications are strict. Well, maybe there's one implication that's not quite obvious I guess this one. So why if they're isomorphic over the chaotic integers for all PR the isomorphic over the rationales. Can anyone give me the reason. I think you can use. Can use what I'm sorry. Yes, exactly. So if you're isomorphic over ZP then you're certainly isomorphic over QP. But on the other hand, oh, G, sorry, since F and G are both positive definite they're also isomorphic over our, because there's only up to isomorphism there's only one positive definite quadratic form over R with, you know, plus plus in terms of solicitors law of inertia. So in Kowski, this implication holds. And, yeah. Actually wait does this implication or I think I'm going to end up confusing myself here. This only implies that these are equal up to squares. Oh geez I might have confused myself here. Yeah, let me let me put just I don't want to have to, I don't want to get something wrong so let me put this here instead. I might have been mistaken argument there. Anyway, in general, all of these implications are strict. And you saw some examples of strictness of implications in your problem set. From last time. Now, we're going to be interested in. Oops, be studying the set XD. So, for, you know, for D a possible discriminant so D a negative integer, the set XD, which is the set of equivalent strict equivalence classes of forms of discriminant B. So we're going to be taking this course invariant the discriminant and asking to classify all of all possible forms of a given discriminant. And it turns out that this is sort of the best question to ask in this two variable case. I want to mention that outside of the two variable case, the classification by discriminant is what much too course to be kind of a useful way of classifying forms. And in that case it's this one that is the appropriate general notion. And this is called this in this case we say that F and G are in the same genius. And the thing you want to do in general is ask to classify equivalence classes of forms of the same genius. And that gives results somewhat analogous to the ones that we're going to be talking about here in the two variable case. However, in the two variable case because of special phenomena that appear which will explain. It's actually actually get a better result by classifying forms with the same discriminant rather than forms of the same genius so slightly coarser. And in fact you can understand forms of the same genius in terms of the results will describe about forms of the same discriminant. So, in generality you'd want to look at classify forms of the same genius but we're going to look at forms of the same discriminant because that's the nicest thing in this special case we happen to be considering. So that's that's just a general remark to orient you a little bit in the general theory. Right, so we're interested in studying a set let me check my notes and make sure I didn't. Right. So what's the first question you can ask about a set. Is it empty or not. So, for which D is XD non empty. In other words, given a negative integer D when does there exist a form of discriminant D. So, well, note that by definition D is equal to be squared minus for AC. So it's congruent to be squared mod four. In other words, D has to be a square mod four, which means that D has to be D has to be congruent to zero or one mod four. And conversely, if D is congruent to zero one one four it's actually not difficult to explicitly write down a form of discriminant D. So, if D is congruent to zero mod four. Then the form F equals X squared plus D divided by four Y squared. As discriminant F equal to D. And if D is congruent one one four, then F equals X squared plus X Y plus one minus D over four shows remember that he is negative yeah. I didn't remember D was negative. That's better. Plus one minus D over four Y squared has discriminant D. Sorry guys, my pen is failing me here. Has D equals D. So the answer to this first question of when the set is empty is the set is empty is non empty if and only if D is congruent to zero one four. So, so called, I mean these sort of obvious forms you write down actually do play a, they're not just random examples actually play a special role in the story so these specific forms are called the principle form of discriminant D. But or you might more generally call a principle form anything in the city, anything which is strictly equivalent to one of these two forms but we'll see that these two also have a special extra property which characterize them in their equivalence class. And the principle forms do play a special role in theory so it's not just an example, you can write down. Okay. Now I want to give you a first reason why this question is a natural question. So, why it's natural the group forms by discriminant. So, here's one answer, they're actually use several possible answers but here's one. And it has to do with this question we're interested in of which primes are, which primes for example are represented by giving quadratic form. Here's a theorem. So let D be negative and D congruent to zero or one or one mod four. So D is a possible discriminant in other words, let P be an odd prime, then P can be represented by some form of discriminant D. So, if and only if some condition you can actually check. So if and only if D is a square one. Now you can't say D is a square one P you can necessarily say that he is represented by some, I mean you can fix the form beforehand you can't necessarily a form of discriminant D before and you can necessarily say that he has to be represented by that form, but he will be represented by some form that has the correct discriminant. So here you sort of have to take all forms of discriminant D into consideration a priori to get the, the good result about primes being represented. And the proof is not too bad so let's actually give it. So first let's give the forward direction. So let's assume that P is equal to a x squared plus b x y plus C y squared. Well, P has odd petic valuation in fact petic valuation one it follows that it can't be the case that both x and y or congruent to zero mod P. Because if both of them were divisible by P then you'd be able to have you'd have a p squared on the right hand side but you only have a P on the left hand side. So that's pretty simple but let me reinterpret this. Let's, this means that if we reduce these numbers mod P. This is non zero. Right. So in other words it's a non zero vector in the two dimensional vector space over this finite field. So, any non zero vector over a field can be extended to a basis. So we'll extend this guy to a basis of the two dimensional vector space in Fp, right some two, and then calculate the discriminant in this basis, or the discriminant mod P I should say. So then, what's the matrix going to look like. Well, our basis vector annihilates the quadratic form because P is equal to our quadratic form. And therefore, when you reduce mod P zero is equal to our quadratic form so we're going to look something we're going to look like zero BB and then some C right. And then the determinant is going to be B squared. So we'll get B squared is congruent to D mod P, which is exactly what we wanted to show. So that means that D is a square mod P. Okay, so here I'm using that the determinant is the, I mean the discriminant is minus the determinant of the matrix associated with the quadratic form. Okay, so now let's do the other direction. Well, it's actually pretty simple. I mean, if D is, if D is a square mod P. So then we can write D is a D is equal to B squared plus four times. Oh, sorry, sorry, so D is a square mod P. It's also a square mod four, because it's congruent to zero one one four. And then P is odd so four and pure co crime so that implies that D is a square mod four P. So I'm actually going to use that I'm going to write D equals B squared plus four P times a random number which I'll call minus C. Right. I'm allowed to do that. I'm not supposed to give me what I want, because, well, this means that D is the discriminant of the form F where F is equal to P x squared plus B x y plus C y squared. So it's sort of, you sort of just do it. Just do it by hand you can easily solve the equation discriminant. Well, anyway. Okay, so that was that theorem. So that says that yeah prime is represented by some form of discriminant D if and only if D is a square mod P. Let me make a remark. I didn't understand the other indication. The second one or the first one. So our assumption was the D is a square mod P. However, we also have the hypothesis that D is convert to zero or one mod four which means that D is a square mod four. The hypothesis that P is odd. In other words, P is relatively prime to four. So by Chinese remainder theorem it follows the D is a square mod four P. Are we so far. Yeah. So let me just write what that means. It means we can find a B such that B squared is congruent to D mod four P. So D is equal to B squared plus some multiple of four P which I'll call, which I'm calling minus C. And that equation just says that D is the discriminant of this form. Yeah, so it's very, it's very elementary. I'm not doing anything fancy here. So let me make a remark. So this condition we had that D P equals plus one can be rewritten as a congruence condition, modulo D. A congruence condition on P modulo D by quadratic reciprocity. So let me pick up into cases of, according as to D is zero one one four but in the end you find that indeed, in all cases, you get a congruence condition what D so there was an exercise, sort of similar to this a couple of problems that's back. So let me give it just us just a very simple example so. So D was minus four, which is the discriminant of this x squared plus y squared. So D is equal to four on P is equal to plus one if and only if P is congruent to one month for. So it's a congruence condition mod minus D, which is the same thing as one. So, you know, so because of this observation, this condition is actually it's very easy to check it for a given prime. It's just a congruence condition of the discriminant. Okay. Now let me explain a corollary of this theorem. So, if it just so happens that the size of this set of equivalence classes of forms of discriminant D is equal to one. Then if F is any form of discriminant D, we have that P is represented by F, if and only if well yeah just the same thing. Again this nice congruence condition on P. Well, you have any form of discriminant so we know that this condition in general is equivalent to saying you're represented by some form of discriminant D. But when the size of this set is equal to one that means any two forms of discriminant D are equivalent to each other and the equivalent forms always represent the same set of integers. Because you can get back and forth by just do linear transformations in the variables. So, if it's represented by some form it has to be represented by F. Yes, there is. In this argument that we just gave in the case where XD has just one element in it. Well one class basically of strong equivalence. And we don't we didn't use anywhere the fact that the equivalence is strong right because the fact that they represent the same elements is from regular isometry. That's right. So you could say something a little more refined here. Absolutely. So that is, it's not the strongest possible statement of this form and that's a very good remark. Yes. And so. Yeah. Okay. But anyway, this is going to be sort of our general strategy. So, so now, what's the main theorem about this set XD. So, which we're going to prove today. So for all D less than zero, well, I might as well add the common zero one month for all the theorems true without that. XD is finite. So may not always have just one element. But it is always finite. And it's cardinality is denoted HD. And it's called the class number. But actually the theorem is the term that we're going to prove is actually better than just this abstract statement that you have a finite set. In fact, in fact, given D, one can explicitly produce by a very efficient algorithm. So you can actually make this completely explicit and algorithmic. So given D, you can actually figure out what the class number, you can calculate what the class number is and find representatives. So you can actually make this completely explicit and algorithmic. So in D, you can actually figure out what the class number you can calculate what the class number is and find representatives for all equivalence classes of forms of a given discriminant. So before giving the formal proof, let me say what the basic idea is. So basic idea is to, well, try to make a and B as small as possible. So you maybe you start with a random form ABC, right, and then you apply you want to apply changes of variables in order to make aim a or smaller or be smaller. And then either one you win because eventually it'll get down to something really small and then they're there only be finally many possibilities. So that's the basic idea. And to do it, we don't have to use. So priori use arbitrary elements of SL to Z. In fact, you only use two specific kinds of elements of SL to Z. So you use two kinds of change of variables. So the first kind is where you want to sort of swap X and Y, but that actually has the term that minus one so you have to swap them and change the sign of the other. So this corresponds to some matrix like I guess zero one minus one zero. And what does this do on the level of forms. Well, you can just plug this in so plug wine minus accident here, take f of y minus X and rewrite it in standard form and you'll see that what happens is you get ABC. You can see that you can transform this into C minus be a. In other words, you're allowed to switch A and C as long as you change the sign of the. Okay. So that's the one kind of variable change. And the second kind of variable change is, so let me make sure I get this right so you can send X to X plus K times why. And you can just keep y fixed. So it's kind of a sheer. And this corresponds to, I might get this slightly wrong, but I guess it's this one. I could, you could make it for trying to use the upper triangle instead. So the guy here is going to be an integer. But actually this is just the, you only need this one because this is just the cake power of this element here in the group SL to Z. I'm sorry to ask another question but for the first kind of changes, should it not be the first column zero one and then minus one zero because X goes to why and why goes to minus X. Yes, indeed. Thank you. Although, there is a question of whether it's the matrix or it's inverse the kind of, I love what I mean by this really. You see what I'm saying.  Right. So, if the notation he's using in for a certain matrix is correct, then the first matrix is correct. Yeah, I think it's consistent whereas before, there is this comment that wasn't being consistent so that's, yeah, so that's good. Anyway, in a case we're not going to think about the major season so it's just going to be about how a B and C change that's all we care about so let me write down how a B and C change here. So a B C. You can plug it in and check for yourself.  It goes to a, and then B plus to a K, and then whatever the third one is it's not going to be important for us. And third one was actually determined by the other two right because we know what the discriminant is. So we can always solve for the other one so we don't really need to keep track of C and we can just focus on a and B that's also part of the general strategy. Right. So what is this move do it keeps a the same, but it replaces be by a multiple of two a. Okay, now let me state the most precise form of the theorem explaining exactly how to find this set of representatives, and this is called reduction theories this is the very topic of this lecture. So, the definition that you use is this. So a form. C is reduced. If. So, if. So, I am C are always positive right, we said that the B can have candy negative it's possible for B to be negative. But but now we're going to call the form reduced if the absolute value of B is less than or equal to a. A is less than or equal to C. So we're trying to make B and a small and this is the precise way in which we'll be able to do it. And more over, if we're in a boundary case. So if either a equals B. Sorry, a equals absolute value of B or a equals absolute value of C then there's actually going to be some freedom and we want to fix that freedom. So we're trying to make sure that B should be non negative. So we just have to handle the boundary cases. We didn't throw this in there would be two in equivalent forms. Two forms to reduce forms which are equivalent if we're not going to want to have so we have to choose one of them and we do it by fixing the sign of the end if you're in a boundary case. Okay. So, that's the definition of what it means for a form to be reduced and now the most precise form of the theorem is the following and I guess this is Gauss's. He actually, I mean he, he did everything we're going to be talking about not not everything we did a lot of the stuff we'll be talking about and more. In his book, this position is arithmetic thing. Actually, there's a fun story about that book. So maybe I should plug this book. So it's definitely the most famous book in all of number theory, one of the most famous books in all of mathematics. And it relates to the subject of our, our lectures in more than one way. So first of all, he discusses a lot of this theory of binary quadratic forms integral binary quadratic forms that we'll be talking about here in the second half. But he also was gave the first proof of quadratic reciprocity. And we also gave a proof of quadratic reciprocity in the form of Hilbert's product formula. What I said was due to Tate, but then here's a funny story which Tate likes to tell that he came up with this calculation of K to a few. And then, once he had done it he realized that he'd seen that argument before somewhere, essentially. And it was in the various for first proof, the Gauss ever gave the law of quadratic reciprocity which is one of the first things he does in this book this position is arithmetic thing. And if you read this book and look at Gauss's proof of quadratic reciprocity the first one that he has in there, then it is, in essence the same as the proof of his calculation of K to a few and the deduction of a bit reciprocity from that so it could be fun for you guys to. I well this book you can't go wrong reading this book it's just got so much amazing stuff in it. But just, that's just a little plug for one of the most famous books in our subjects history. Anyway, I digress. So, every form is a is a strictly equivalent to a unique form, a unique reduced form. So, in other words, yeah, so you make a list of all the reduced forms. And then you've got up to strict equivalence you've got exactly all of the forms. So no no to reduced forms are strictly equivalent and every form is strictly equivalent to some reduced form. So, yes, there is. And if we were to remove the condition of strict equivalence and replace it by equivalence would we lose uniqueness. Yes. I think thanks. Yeah, a lot of things become less nice. Okay, so let's give the proof. It's actually not too hard. So let's first do existence. So suppose given up. So existence of a reduced form in the. So, among all of the possible ABCs which are strictly equivalent to F choose the one for which. A is minimal. So it is a positive integer. So, you know, it's, it's got a lower bound right so just take the least least lower bound, but the greatest lower, I don't know, the minimal. Yeah. I can't find more ways to say it. Choose one for which is minimal among all possible options. So then the first claim is that. That's right. See, it has to be greater than or equal to a. So otherwise, you use the first move. And you can switch a and see right you change the sign of evil we don't care about that at this point right now we're making sure a is less than or equal to see. So that was one of the conditions in being reduced a is less than or equal to see. So that's good. So we can use the. So now, then use the first, then they use the second move to make sure that absolute value of B is less than or equal to a. And in fact you can do better you can do minus a strictly be strictly greater than minus a and less than or equal to a. Well, recall that the second move. It doesn't change a, but it moves be by multiples of two times a. So by shifting over using this you can always make it land between minus a right that's an interval of length to a. And you can choose which edge case you want to keep and we want to keep the one where B is equal to a and not the one where B is equal to minus a. Okay, so we've already satisfied the first condition in the definition of reduce so we just need to make sure the boundary cases that we can have the boundary cases be okay. So if a is equal to see them, I'm getting confused about these case distinctions. Okay. All right, so, so that if these, okay, minus B is equal to a. No, wait, wait, sorry. Actually, if absolute, yeah, absolute value of B equals a means that either a is minus B or a is B, a can't be minus B because we explicitly ruled that out by our choice here. If a is equal to be. If a is bigger than zero B is also bigger than zero so we're okay. So the only thing we need to worry about is whether is if a is equal to see, then we need to be able to change the sign of B, but we can do that using the first new again, which switches and see that doesn't do anything because and see or equal, but changes the sign of B. Yeah, well I won't write all that down, but you can arrange this by just using the first move again. The only case to consider is a will see and then you just use the first move to change the song. So that's it. Well for existence. Okay, and actually that's all we're going to need. But so I'll go quickly through a sketch for the uniqueness. So there were two right so F were strictly isomorphic to G and G prime, and these are both reduced. Well, then G is strictly isomorphic to G prime. So it's enough to show that no to reduced forms can be strictly equivalent. Right. So the key claim for this, which is actually an exercise on your problem set is that if F is a reduced form. And it's equal to a comma B comma C, then a is the least positive value of F. So, since strictly equivalent forms always represent the same set of values. This means that if you have, you know, two strictly equivalent forms then at least the a's have to be equal right. And then you have to do a little more argument to get the things themselves have to be equal you have to see that if you have two strict forms of the same a, then any variables which goes from one to the other has to be a move of type two. And then you can see that when they're reduced this K actually has to be zero because otherwise you'd be moving the out of the window that you're looking at. So, the main claim is this and you need a little extra thing about what possible moves there are when the a's are the same but it kind of falls out from how you prove this anyway. So if we go too much into this uniqueness. The existence is kind of enough for the applications we're going to go into anyway. Okay, any questions about that group. So let me make a let me make a remark. Well as I presented it it's not quite an algorithm right so how to give an F, how to find F prime equivalent F with F prime reduced. You can explicitly find the change of variables. So there's an algorithm, and it's very simple it's a so the algorithm says you can either make a smaller by a move of type one or be smaller by a move of type two. And you can add you can so in practice you can always just do it you try to make a and be smaller and apply moves of type one and two and you'll succeed. So you have to alternate moves of type one and moves of type two. So you have to go back and forth between them. And that's actually kind of it's a kind of it first it seems kind of remarkable the only need to use these matrices, essentially, to make any two forms equivalent. But actually, if you look closer at what's going on here you can use this kind of stuff to see that these matrices generate SL to Z. So what are the coincidence that we only needed to use those two kinds of chains of variables and in fact you can kind of read that off from the argument, plus a little calculation of what the automorphisms are reduced for which is also on the problem set. So, yeah, so it's you know a good way to prove facts about groups like this fact is to make the group act on something interesting, and then you learn things about the groups this is an example. So you have SL to Z on the quadratic form so you can learn that this fact for these two things generate SL to Z. That's those are the only things you need to make a form reduced. So it's not quite obvious that that implies this fact but it's, it's on the problem set. Right. What else. Now let's talk about the main corollary, which is the claim we had so XD is finite. And that, well, so we need to see that the set of reduced f of discriminant D is finite, because every form is equivalent for a reduced one so that set is fine at the next day is fine. Note that let me do a little.  Now because the form is reduced. So I find that a is less than or equal to C. So this is greater than or equal to for a square so C is greater than or equal to a right. So this is greater than or equal to for a squared minus B squared but absolute value of B is less than or equal to a. So I can also turn that any quality around and make this an a squared as well. Because it's reduced. And that's equal to three a square. So this gives you a bound on a in terms of the discriminant so a is less than or equal to the square root of a negative D over three. But it feels weird to write square root of negative you and negative is positive so I'm writing absolute value of D instead. I mean, negative D when D is negative. Anyway, you know what I mean. I hope. Right. So we have a bound on a. So it's also important then, by definition of reduced again B is bounded by a so absolute value of B is also less than square root of D over three and recall that as positive yeah. So that means that both a and B. So the possibilities for a B are finite. And then but D equals B squared minus four AC then fixes C. In terms of a and B. So here's a corollary. Well, just sort of reading things. So if F one dot the dot F H is the list of primitive forms of discriminant D, then well, H is equal to HD. And every forms of discriminant D is equivalent to a unique fi. So this is how you calculate class numbers. You just look at all the primitive forms of discriminant D there's going to be only finally many of them look at all the possibilities for a and B and so on. And then that's your list and then the number of them is the class number that's it simple as that. So let's do an example, our favorite example D equals minus four. Yes, there is. What do you mean by a primitive form. I cannot recall that them. Wait, what's the word I used. And I say, you mean reduced reduced. Sorry, primitive is the wrong word to primitive means something else. Ah, sorry guys. I didn't know that earlier. No, yes, I'm not reduced. Thank you so much for catching that primitive actually means something entirely different. So, yeah, just. Okay, so D equals minus four. So let's put this into practice so we saw that. So we know that zero is less than a is less than square root of four thirds right less than less than or equal to but it's an integer so whatever. So that that means that a has to be equal to one. And then absolute value of B is less than or equal to a and then if that means that B is either minus one, zero or one. But actually, since we ruled out, I mean, the border case condition gets rid of the minus one unless that's not going to be so relevant. On the other hand, we have B squared minus four AC. So core times one times C is equal to minus four. And that tells you that B squared is converted to zero for which and the only thing is either zero or one that tells you that he has to be equal to zero. So C are uniquely determined and therefore so is C that tells you there's a unique reduced form of discriminant minus four, namely, x squared plus y squared. A corollary of this is an odd prime P is of the form can be written as x squared plus y squared, if and only if P is congruent to one one four. So we did it we proved one of our desired results. And it fell out of the sky I guess but the proof is that we just saw that h minus four is equal to one. So there's only one equivalence class of forms of discriminant minus four on the other hand this was the condition for Peter we represented by some form of discriminant minus four. So, that's good news. I'm going to make another remark which is just a small remark that, you know, we worked on the standard examples of forms of discriminant D so this x squared minus D over four y squared, and x squared plus x y plus one minus D over four y squared, these are reduced. And they are indeed very small, so you can believe it but also you can just check that that is true. So, that's another justified that's so they are the reduced forms representing that equivalence class of principle forms. So that was good. And let me make one final remark. The class number being one is rare. In fact, there are only nine values of D for which it holds.  So this isn't there. This is the theorem of a beginner, who was an amateur mathematician and when he presented his result, nobody could follow what he was talking about and they thought his proof must be wrong. I think 20 years for someone and people came up with another proof for a similar proof and then they finally went back and realized oh wait this finger guy who's actually making sense we just couldn't understand him because he wasn't really speaking our language. He proved this amazing theorem. It was an open question since Gauss. You know, what were the Gauss conjecture there only finally many values, which ones that they are exactly. Yeah. It's pinpointed by Hager and it led to his technique also led to a huge amount of advanced mathematics and important stuff in number theory. So, that's a pretty cool story. All right. So that was what I wanted to talk about for today. And tomorrow we're going to switch gears slightly because we're going to be looking at the same subject from a new perspective. So this perspective of quadratic fields and rings of integers and quadratic fields will have to do a little bit of an introduction to that and then we'll be able to make the connection between the two things, binary quadratic forms and quadratic fields. Yeah, something to do with the fact that they both have the word quadratic in them. Okay. Any questions. I have. Sorry, is this following the following. What's it called Cox's prime to the form x, y squared, or is this just a common argument. I put that book in the list because I know that book has a huge amount of stuff. I put that book. I actually haven't ever opened that book. So, I mean, it's about the very subject we're talking about so I assume everything I'm saying is in there. Like, I remember, like the same argument center so. Okay. I mean, but I have read Gauss I have heard this because you want to share with medicaid so that might be where I'm getting this from and Freddie is confirming that it is a good book. Sorry, can I just make a quick announcement before we can request, but so I think the plan is for the TA session to be right after the cost program activity. So 130 PM mountain time so. I might not be able to make that. But the TAs are so great that I feel superfluous every time I show up. So, if you look at the equals, so we are looking at the different binary quadratic forms, non-equal and binary quadratic forms up to action of SL2Z. That will be a bigger. Now if I look up to GL2Z, that will be as that will still be fine. In fact, the smaller than the class. Yeah, this the court, the set XD would get smaller if you use that. Yeah, why is that not given primary importance why is this class number given a separate name etc why is that not so important. Because if I want for example just to give an example if I wanted to state the analog of this so I said every form is equivalent to exactly one reduced form. That's only true with SL2Z. I would like to say for GL2Z you'd have to, I mean it's not a big deal to make them modifications. So maybe that's not the real reason okay the real reason will come later in that there's a beautiful formula in fact for HD called Dirichlet's class number formula. And it doesn't look as pretty. Right, the formula for the analogous thing with GL2Z. So there's just, it just turns out that when you work through the theory that if you understand the SL2 equivalence theory first then it's it you then you can it is possible to deduce what's going on with the other one, but you can't go to SL2 equivalence and the result for the GL2 equivalence just doesn't look as pretty as the result for SL2 equivalence so it's definitely smart to do SL2 equivalence, I guess for those two reasons. One of what's prettier and two you can deduce the other one from this but not backwards. Yeah. That's that's what I would say at least. There was also a lot of theories you had your hand up at some point. I wanted to ask the same thing that Sundara did. So my question is covered. I can ask another question. Please just so will so this result of what's it, what's his name, Hagnar if I'm not mistaken. I can find where I. Oh, there it is. Yeah, Hagnar. So will this result follow from the class number formula that we're going to derive. It's that class number formula is really amazing. I mean, it's hard to use. I mean, first of all, I don't think I. I don't know any argument why a priori argument for why the value of the formula is even a, you know, it's an integer when you look at the formula but that it should be a positive integer is actually not so easy to prove. And that the only way I know how to prove it is by proving the class number formula and so it's saying it counts something. And it's actually not so easy to make a direct analysis of the class number formula so that's not how it goes. The argument is much more interesting than that. Yeah. Thank you. It uses, I mean, at least in the modern understanding of what Hagnar was doing uses some sort of advanced topics in arithmetic geometry. For example, so in these rings of integers and imaginary quadratic fields they also play a special role in the theory of elliptic curves. Because they can act on elliptic curves and the morphisms ring of elliptic curves can be rings of integers and imaginary quadratic fields. And that's about all they can be. If they're not busy. So there are some special elliptic curves and this gives special points in a modular elliptic curves and these points are now called Hagnar points and it's by studying the arithmetic geometry of these modular spaces that you eventually whittle down the list of possibilities with class and the ones given by Hagnar's list. Thank you. It's a completely different kind of argument from the class number formula. I guess they're called CM points and then Hagnar points or if you map that modelized space to something more explicit than the image of them. CM as in complex multiplication. Exactly. I think I have even even one more question. I kind of come up as I'm looking at my notes questions are great everyone's questions are great. So please just feel free. So we saw in this in today's lecture that we get this nice generating set for SL to see which it's just two matrices actually, given by this action of SL to see on on the forums that we were considering today. So it's generally a good strategy for some groups, if you want to find a generating set that's nice have them act on some set and look at that action. Do you have any other example in mind of like a group action that gives me information about the group itself. There are too many to name. Let's see. This is kind of the classic example or you can make SL to the act on all sorts of related objects the upper half playing a tree can write down. But yeah, I. Oh boy, there's so many. Okay, I mean, there's a kind of generalization of SL to the SL to Z generalizes to a concept called an arithmetic group, which is more or less something like the integer points of some matrix group. So, it could be SL and Z G L and Z or an orthogonal group over Z. These are all arithmetic groups. And for every arithmetic group you can write down a really nice topological space. In fact, Romani manifold on which, on which the group acts, and the question and finding fundamental domains for this action. So that kind of like a notion of a reduced point in this topological space. So, you know, finding a representative for each orbit, oftentimes gives you by techniques of so called geometric group theory gives you an understanding of the group, for example, generators and relations or co homology calculations or things like this. And yeah so one of the best ways to understand arithmetic groups is through this action on this. So called associated symmetric space. So that's a big class of examples but I mean they're really just all over the place if you have any group, one of the main techniques in studying it is just devising some some nice topological space on which it acts. Yeah, what are some other good examples. Are there any popping to your mind. Well, I guess a lot of, I mean maybe maybe right now I guess many groups are sort of defined through like they act on or sort of automorphisms of something. But I mean I guess like. Yeah, I don't know. I think that I think there are like lots of examples of I guess things like mapping class groups and all these things are sort of studied by making them act on some, some object. But I guess maybe very concretely. Yeah, sorry maybe I don't have a better answer right now but I guess in general, I think one studies, one typically studies groups by in terms of their actions on. I guess on a more elementary level isn't there some proof of the silo theorems by making G acting on like subsets. Yeah, I don't know. Yeah, it goes but there's something that's act on some sense of giving cardinality yeah so that's I guess that's a minor example of this kind of phenomenon. Yeah, you have your hand up again. Or even because it just came to me like as a probably very elementary example, maybe the first example we see, could it be Cayley's theorem that even if you have a group acting on itself. You get Cayley's theorem and you can view it as a, well as a subgroup of some SN and maybe you can do some computations more easily if you represent it as a subgroup of a symmetric group. I don't know. I don't really see the action of a group on itself by translation I'm not sure that maybe maybe it's just because it's so basic that I don't notice it being used but I was going to say, sorry, maybe one comment is like if you look at this, I guess there's this book by Sarah trees and thoughts of sort of fun examples of like the compositions of groups and so forth that you get there. When you look at the group apps on like a tree or a graph or something and so. Right. Yeah, actually, um, you show that groups are free in many cases, like if you're a portion free subgroup of I guess I'll, and yeah, anyway, sorry. Yeah.  You can also calculate co homology and all sorts of stuff. And this technique of having an act on a tree in this case this is a just the one dimensional case of a sort of geometric theory technique of having groups act on a lot nice topological spaces and works for certain low groups such as SL to see And it will be finitely presented SL to see so finally many relations to. So I mean, I give the presentation in the problem set I'm not asking you to prove it I'm just telling you for your information. But if you, well, maybe I'll just say it in words so I mean they're, I think the nicest presentation is, you know, they're, they're, they're two special reduced forms. One is x squared plus y squared and one is x squared plus x y plus y squared and, and what's special about them is they have more automorphism and all the other forms. The first one has automorphism group cyclic of order for the second one has automorphism group cyclic of order six. That was actually already on your previous problem set that particular example. And all the other ones have automorphisms or order to automorphism. And but it's always the same order to automorphism. It's just, you know, minus one zero zero minus one that's an automorphism of every quadratic form just change the sign of X and Y. So I thought that SL2Z is just the so called amalgamated product of cyclic group of order for with cyclic group of order six amalgamated over this common subgroup they have, which is the cyclic group of order two. So, and, and from this amount, I mean amalgamated product is some, you know, push out in the category of groups but what it means concretely is that SL2Z is generated by an element to elements subject to the relations that the first one has order four or the first one has fourth power equal to one the second one has sixth power equal to one, and the second power of the first one is equal to third power of the second one. So that's a presentation of SL2Z by generators and relations. And it's kind of fun that you get this big, you know, almost torsion free group generated by two small cyclic subgroups right. There's non commutative groups that you start with something of order four and something of order six they can generate like an infinite cyclic thing when you combine them in the right way. I mean the group won't I mean it'll contain an infinite cyclic subgroup right this one one zero one, for example, as the magic of non commutative words. So SL2Z you said it's amalgamated product of cyclic group of order six and cyclic group of order four over a common subgroup of order two, if I heard you correctly. Thank you. So you can see Sarah's book for the for a proof of that. There should be some way to do it using the action on binary quadratic forms of all but it's much more vivid to use this geometric language. I guess concrete also more generally like all these so and they're all like finally presented groups, and I don't know if that's something that's sort of obvious to see directly but I guess you see it by like making it act on something that's just going to be fine. Even if you don't have an explicit presentation. I guess for SL2 you can probably sort of do it by hand, right, but for SLN and around. Yeah, I mean it right because to do it by hand you have to solve this problem of giving a fundamental domain. I mean that is there the reduction theory has been generalized to more variables but it's some to make it explicit is not so easy. And when we say that SLNs are finally presented like SLN of Z or SLNs of SLN of Z. Yeah, well, you can generalize a bit, but yeah. Definitely. So, just now you've completed a review before the end of the lecture that study of binary quadratic forms has some relation with what extension of Q. So, is there. And I'm not similar to this for a binary quadratic forms and QE extension so on. No, this is so special. It's this is so special. There's also not an analog over a ring different from Z. I mean, really, if you if you prefer to set up slightly this thing just breaks it's a really special phenomenon. Yeah. I mean, there based on the question that the fact that you get a degree to extension as Dustin had kind of hinted at is more related to the fact that you're working with degree to forms, not that they're binary. Does that make sense. I mean, I don't know. I think you need both of them to be equal to, I think you need both that it's binary and that is degree to, I mean, yeah. That's, that's right but I don't. Well that could be a degree six extension right because it's not, I mean degree to extensions are always galore, but does that matter. I don't know if that matters. Sorry, ignore me. There's it's just really, it only exists in this case so it's kind of I don't think you can point to any one specific aspect of it. Right. So this is just a special phenomenon for the integers and binary quadratic forms. Yeah, I think this, I think this phenomenon extends at least a bit to binary cubic forms, where there's like some correspondence between binary cubic forms and then like cubic rings like rings that are degree three over the integers. Let's just figure out which of these cubic rings are actually like domains, and then look at the like corresponding fraction fields. I think it's not. Maybe it's not like quite as nice as this binary quadratic case, I think you can try and do something similar in this binary cubic. I didn't know about this so what is. What is this. Where is this done. So I don't know like, like the return from same thing but I think in like, I give us people. And I think in this like Davenport Halbron stuff where they're like counting like cubic number fields. And they're like the questions like how many like cubic number fields are there discriminates like bounded. I think the way they get, they get at this is by like comparing with counting binary cubic forms. Yeah, so not at all obvious to me why these would be related okay thank you for the okay I guess this is also what you were saying Freddie more or less. This is new to me is all I can say. Thanks for the information guys. Weird okay. Yeah, I think the correspondence is like, not so straightforward. Jesus. Yeah. That's interesting. I guess weird that. I mean, like in the binary in this case like it really is you're using the fact that you have two variables and the degree is to in some sense I mean it. So I guess there's some reason why you can get rid of one of the variables in the cubic number field case. Oh, you don't let it vary or something you don't let the coefficient vary. Like the way you do it in the quadratic this case is you use the norm function which gives you a quadratic form right in this case. What's going on. Yes, I think in the chat of left area is linked to paper that talks to this in that second section that maybe. Okay. Yeah, so this is giving asymptotic results. I guess it's probably not as exact correspondence. What you're saying. Yeah, very similar to the correspondence is like less with like the number fields themselves and more with just like rings that happen to be like rank three over Z. Right, and then you have to like, yeah, figure out you know which of these are domains and what happens in the in this case as well as just that almost all of them are domains and then I mean like there's discriminant zero or something the only thing you need to worry about. So I can see this. Okay, so they interesting. There is an actual bijection between the set of GL to Z equivalent classes of integral binary quadratic forms and the set of isomorphism classes of cubic rings. Wow okay it's some weird thing it's not the norm anymore. Okay weird okay whoa. Okay remarkable. All right. I stand corrected, but I don't quite see. It doesn't seem like the same construction to my eyes, but it does. Yeah. Okay, any more questions. I think I think I have one maybe can to verify something so when you give me that example of of that arithmetic group action on a on a remanian manifold that remanian manifold is called the associated symmetric space, you said. Yes, that's right. It's nothing. I mean it's nothing obscure I mean the, you just take the same matrix group and take its real valued points. You get some lead group and then you mod out by its so called maximal compact subgroup which you can make explicit I mean for SL and it's s on. You get some contractable space which is nice on which your group acts and. Well, it acts almost freely if you pass to a finite index subgroup and be free and then it doesn't the big, the big subtlety comes because it doesn't act co compactly. And that, if you get when it does act co compactly, you're very happy but oftentimes it doesn't for example for SLM, and then you have to work to like compactify things and like all sorts of stuff. The modular curves themselves give examples of this with SL to z acting on the upper, SL to z acting on the upper half. I mean it's like the, the first example of this. Yeah. Thank you. You're welcome. Okay, so I, as I said I don't think I can come to the TA sessions today but I'll see you guys tomorrow at the lecture. Bye.


The Book of the Courtyre by Count Baldazar Castiglione Read by Nicholas Bolton for Naxos audiobooks To the Reverend and illustrious Lord Dom Miguel de Silva, Bishop of Visor When my Lord Guido Baldo de Montefeltro, Duke of Urbino, passed from this life, I, together with several other Cavaliers who had served him, remained in the service of Duke Francesca Maria de la Riverae, his heir and successor in the state, and as the recollection of Duke Guido's character was fresh in my mind, and the delight I had during those years in the kind companionship of the notable persons who at that time frequented the court of Urbino, I was moved by their memory to write these books of the Courtyre, which I did in a few days, purposing in time to correct those errors that arose from the wish to pay this debt speedily. But for many years past, Fortune has burdened me with toil so constant that I never could find leisure to make the book, such as would content even my poor judgment. Now being in Spain and learning from Italy that my Lady Vittoria della Colonna, Marchones of Pescara, to whom I gave a copy of the book, had against her word caused a large part of it to be transcribed, I could not but feel some annoyance, fearing the many inconveniences that may befall in such cases. Still, I relied upon the wit and good sense of this Lady, whose character I have always held in veneration as a thing divine, to prevent any mischief coming to me from having obeyed her wishes. Finally I was informed that this part of the book was in the hands of many people at Naples, and as men are always eager for anything new, it seemed likely that someone might try to have it printed. Alarmed at this peril, then, I resolved to revise the book at once so far as I had time with intent to publish it, for I thought better to let it be seen imperfectly corrected by my own hand than grievously mutilated by the hand of others. And so to carry out this plan I began to read the book again, and touched at the very outset by the title I was saddened not a little and far more so as I went on by the thought that most of the personages introduced in the discussion were already dead. For besides those mentioned in the poem of the last book, Mesa Alfonso Ariosto, to whom the work is dedicated, is also dead, a gracious youth, considerate of the highest breeding and apt in everything proper to a man who lives at court.  Mesa Bernardo, cardinal of Santa Maria in Portico, who for his keen and playful readiness of wit, was most delightful to all that knew him. He too is dead. Dead also is my lord Ottaviano Fregoso, a man very rare in our times. Magnanimous, devout, full of kindness, talent, good sense and... Sample complete. Ready to continue?