Very often we get complaints about throughput, which seems to be not high enough for some STM32 users. In this chapter we will show conditions affecting USB devices throughput. In this case our focus will be on USB CDC device example. Then our results with various scenarios in setting will be presented and at the end of this part a few tips how to optimize more project with STM32 for reaching higher speed can be done. Probably the most crucial fact for USB devices throughput is that USB communication is driven by host. So even if device is able to response on each host request immediately throughput may be low due to host low frequency of in and or out packet sendings. This is very typical problem with commonly used virtual comport application used on PCs. These applications are not expecting so high throughput from the devices as they are on emulation of the obsolete communication devices with lower speed. So the bandwidth achieved with this program is usually limited. Another problem may come with CDC endpoint standard behavior. Bandwidth is not guaranteed for this kind of endpoints and share among all devices on the root hub. So you cannot expect same results with device connected on directly to root hub or if you connect as one of multiple devices to some of hub downstairs in the USB bus topology. Topology of USB connection inside your PC where your device is already connected can be displayed using device manager or some dedicated application as you can see on the slide. On the pictures on the slide on the left side you can see a device connected to root hub to hub but on the right side the device is connected already directly. So using this way you can explore the USB ports on your PC and choose the one which is connected to most directly for the usage and also try to skip other connection on the same root hub. If you go back to problematic of USB host and frequency of sending the request on the device now is the question how do actually get the information about the throughput and where is the limitation if it's from the host or from the design. Probably the simplest method to verify this is using USB analyzer. So as you can see on the picture if you can observe in communication trace NUX from the device speed is limited by the device because it's not able to answer on all the requests from the host side on time and need to not acknowledge time to time some of the requests which need to be repeated and that is limiting bandwidth at the end. On other hand if there are no NUX in the communication device is able to all the requests from the host on time so it's very probable that even more requests from the host side could be proceed by the device in the same time so now the button like is created by the host. As already mentioned virtual comport applications are not so convenient for throughput test as performance during testing with such software was never good on our side. We mostly use for this test the command line application so the direct console which is showing better performance mostly on the linux so here you can see the commands to work with the comport using either Windows or linux and the commands are for both directions so either in or out. To get directed to the results we were able to test in our testing project generated by Cubemix without no additional changes in libraries so both driver and middleware layer were not changed. For the test we used STM32F723 which already embeds USB high speed phi so it's the easiest solution for native USB high speed and we use CDC device application for the test. All of these tests were performed on the same PC with USB 3.0 as host root hub. In first test device speed in in direction so the data are sent to from the device to the host with default Windows 10 driver and common line application four megabytes per second were measured. With application created in the C sharp and various time driver on the PC side and in the microcontroller side like in previous case more than double was achieved so 8.75 megabyte per second so here we can see that even the command line is not so efficient but it's still easier than writing a dedicated application for testing in the C sharp. Using lip USB which already was mentioned and again the command line interface 6 megabyte per second was reached so that's again 15 percent increased to the measurement with the original Windows 10 USB C sharp. That's his driver but very best results were observed when Linux was used instead of Windows 10. Here the bandwidth using the command line commands mentioned in the previous slide raised up to 12.5 megabytes per second. So as was mentioned previous measurement were done without additional change in USB libraries or drivers and for more experienced users we have here some tips where to focus to improve the driver or the middleware for tuning the throughput inside of the STM32 project. Very first place where to go is the configuration of the FIFOs. So bigger the FIFO of the use endpoints the more space for the data and more time for the application to proceed so increase the FIFO space but the user still should be aware that the setting of the FIFO in the both registers and also our libraries is in terms of words instead of the blights which maybe could be expected so it's important to be careful to do not allocate the buffer outside of the FIFO or the bucket buffer. Second thing we need to focus is the interrupt so as the USB functionality is completely handled by interrupts inside the STM32 USB libraries we should prevent other long interrupts and high parity interrupts to get the most processing time to the application and the last and probably most demanding steps would be to improve the USB library by yourself so there is the possibility to change the library in the way how it's handling and checking the flags to reach exactly your application for example you can observe that the flag and function which is used most in your application is not the first which is checked in the interrupt so the order may be changed but this here you should be very careful to do not put there any miss functionality any bug and probably the performance gain which you can get using these changes won't be so crucial


Jerry of the circus he said he was getting better and could walk without a cane now is he going to get out of the hospital soon oh he said it would be a little over a month before the doctors would let him leave but he wants to get out now he said he feels fine well if he feels good why they let him go oh I don't know but I guess they want to make sure he's really all well didn't you tell me that he was going to join up with the circus when he leaves the hospital he might mr. Randall promised him a job if he wants it he used to be with the circus didn't he uh-huh but before that he was a big game hunter he got all sorts of wild animals for circuses and zoos really sure and he was one of the best in the world that's a real dangerous business all bad sure is well here we are now let's see if we can find whitey there he is way and back in the tent see him oh yeah whitey oh whitey hello Jerry you're busy no come on in come on Tom I'm coming what's on your mind Jerry something special I thought I'd take Splendor out back and give him a ride maybe try a few tricks with him okay Jerry go get him hey Jerry do you think I can ride him too oh I don't know that's why should I sure why not whitey Tommy or wants to ask you something we're what is on the fireway well do you think that has I'm well I was wondering could I ride Splendor to just a little well now I don't know Tom ever done any riding oh sure I can ride all right I used to ride a lot Splendor's a little spirited you know he gets pretty time I know it if you think you can stay up in the saddle go ahead and give it a try well gee thanks whitey it's okay my boy but be careful you don't want any accidents around here come on Splendor come on I got the saddle Jordan no but I've got his bridle you get the saddle Tom there it is hanging over there okay I'll bring it we'll go right off the back way here come on Splendor you're going to get a little exercise I think you can manage by yourself oh sure be sure you get that settle on good and tight pull that since strappers tight as you can okay I will come on Tom yeah I'm right behind you if I'm not here when you get through riding and you'll be here side Splendor up all right whitey hey Jerry well put it down that will settle him up right here well Splendor easy now you put the bottle on first don't you know here you hold on to this holder okay now come on Splendor open up come on will there Splendor will he'll take the bit all right come on now that's it that was easy yeah it's getting easier every time but he sure didn't like that bit in his mouth the first time throw that blanket over now we're coming up there you are now the saddle there we are oh he sure is anxious to go well there boy well take it easy now hey look at your hand did you hurt it no why it's all red all that that's nothing I got that when I was picking the strawberries we had for breakfast boy they sure were good weren't they I'll say the first ones we had this year and were they swell nice and sweet well he's ready now I got the since strap couldn't hide you want to write him first can I sure go ahead John okay easy now Splendor I guess he knows he's gonna have a new rider that's right up you go pull out die Tom oh I'm going yeah whitey Splendor's not running away with him is he no he's all right whitey I got a scare for a minute I just happen to look out when Splendor took off with Tom and well for all the world it looked like that cold was out of control no a dump can handle him swell yeah see now he's heading him back this way boy sure knows what it's all about when it comes to riding good for you Tom you're a keen rider you're doing all right son Splendor told you he was spirited the fastest little pony I ever saw yes sir there's no doubt about that did you ever try to jump with him Jerry no not yeah why oh he wants to jump how can you tell did you notice when I got him way out there by the fence no I wasn't looking just well he sure wanted to go over the fence all right I could tell he put on a lot of extra speed and wanted a head right for it did you ever ride a jumper Tom yes sir I think you'd like to try Splendor oh I sure would well if he wants to make that fancy go ahead and give it a try but be careful okay come on sweater that boy is a natural born rider he handles that cold like you've been riding him for years he look at look at what he he's getting to the fans he's over he did it good for him boy that was your pretty yes it was Jerry his first jump and just as pretty as ever I've seen had a lot of room to spare to here it comes back now doesn't look at him oh Splendor seems that just sort of since he's got a good rider up in the saddle bring him over here Tom how was that for a high job oh that was fine Tom really fine they use your can ride Tom and better get down now and let's let the rest of you men we don't want to overdo a good thing you know okay well that's been here oh boy that was sure fun Splendor's a jumper all right isn't he I'll say he is he went over that just like a bird hey Tom yeah whitey did anybody ever tell you what a good rider you were no why well sir I think you're the best little horsemen in the world for your age I've never seen your equal I quit kidding me no I'm not Tom you're a natural born Jockey what why he says he means what I do let me take a look at those hands of yours here mm-hmm just what I thought what about his hands well they're bigger than yours aren't they yeah I guess they are they're almost as big as mine well what about it those big hands Tom and your strong arms have a lot to do you're being a good rider it takes good strong hands to handle a horse you got to be able to hang on to the reins hold your horse down then let him out when you want more speed I know what you mean I don't Splendor back till just before we got to the fence and then I let him go and game all the reins you want it's it exactly did you ever think of becoming a jockey oh well you'd be a good one yeah they're just about the right size how much do you weigh oh I think about 95 pounds just about right jockeys can't be too heavy can they no no they got to keep their weight down yet they have to be strong good and strong I'd sure like to ride a racehorse sometimes I'll bet it's exciting you bet it is can I write Splendor now he's all quieted down okay Jerry but you better not try any jumps with him if he starts with that fence back there you hold him down and turn him away from it okay I will I'll bet Splendor has right for the fence again he sure likes to jump yeah it kind of looks that way oh Splendor oh now can you get up on them all right there Jerry now you'll be careful Jerry tells me you're leaving us when we get the Fordham tomorrow is that right then yes sir I'm gonna try to find my father when I get there that's what your father do I don't know but Mr. I mean Colonel Alger said that he thinks my father has a horse out of the county fair that he's gonna put in the big race take it easy Jerry don't run too fast I want but he sure wants to go he keeps heading for the fence all the time well head him off the other end of the line practice those turns with him all right yes look at that boy go he sure likes to ride what were you saying about the fair in Fordham Tom well I said the Colonel Alger thought my father was out at the fairgrounds he said he had a horse yeah you did say that well it won't be hard to find him then by the way I understand that's gonna be some race out at the fair the big one I mean they're sending race horse from all over the state for that big handicap is that so yeah that's what I hear well then I guess my father won't have a chance why what do you mean Tom well Colonel Alger said everyone was making fun of dad because his horse used to be a fire horse and they said he didn't show any speed at all well I wouldn't put any stock in here say your daddy most likely knows what he's doing look at splendor take those turns first one way and then the other yep he's learning fast say I meant to ask you how Charlie feeling is he getting over those measles all right well sure he's getting better all the time and the doctor is sure having a hard time keeping him in bed well I guess it's pretty tough for a fellow as active as Charlie Alger to have to stay in a bed he likes to be up and doing things did you know the ruseff troop is going out to the fair in Fordham mm-hmm yes they're going to put on some stunts and trick riding is a sort of an advertisement for the circus here he comes and boy look at them gallop how is that for speed all right Jerry all right splendor is taking those commands nice now bet he is I think he likes to take commands and make turns and everything well he sure seems to sure horses like to learn next time Jerry comes over here to ride splendor you come along Tom and oh that's right you're leaving the outfit tomorrow I forgot what were you going to say why do well I thought I'd like to have Tom do some more jumping with Blinder well can I jump with him well now I don't know Jerry trick to it and Tom seems to know that trick you might be able to learn but well I wouldn't want you to take a fall while Splendor's making a high jump hey there's Betsy oh hi ya Betsy I'm looking for you Jerry and for you to hide what's up while you're all excited might be serious what might be serious well dr. Williams is very much afraid that the whole circle of circuses taken down with the measles you don't know that bad that looks that way the doctor thinks that someone caught the measles from Colonel Algern and now it's spreading all around come on we've got a hurry hurry to where for what I'm so excited I forgot you see we all have to go right over to the hospital top and get in line dr. Williams is examining everybody in the circus what yes one of the car boys broke out with red blotches all over his face and then mr. Murray turned up with the same kind of blotches on his arms and well the doctor thinks maybe everybody is breaking out with a measles and he wants all of you to come right over now for an examination Tom and I don't have to go we've had him when we were little and we can't get him again I don't know about that Jerry but I was told to get you and have you require for examination too well I guess Doc knows what he's doing we better get along and sure would be awful if the whole outfit took down with the pesky measles and had to be quarantined hey look whitey look at your arms you've got funny little red marks what say this is no joke come on let's get right over to doc Williams and find out what this is all about


Today we're going to revisit how to find the best route when you don't want to drive down sharp grades with a demonstration and a big shout-out. If my big solar setup still works, if I need a generator, and if someone wanted to do this full-time RV thing, where do they even start? Happy Sunday birdwatchers! It's Robin with Creativity RV, and this isn't all new. Sunday morning view cue where I answer all the best viewer questions about my full-time RV life, and we're going to get right into it today with a demonstration and a question about driving on steep grades. You might remember that I talked about this in my last video because when you're pulling a lot of weight behind you, driving down some steep hills can be scary. It was for me when it first started, so I told you what I use, but I got another comment in the last view cue from Rocky5PR, and I want to give this person a big shout-out because I checked out this website that he recommends, and I am a huge instant fan. He said, with sort of surprise, you didn't mention flattestroute.com. Rocky, I went and checked out this website, and you are right. I was amazed at how much information this website gave me, so I'm going to hop over to my computer, and I'm actually going to give you guys a demonstration on a route where this website showed me the best way to go, exactly what was coming up, how much time it was going to take me, and more. This is flattestroute.com. You can see here I put in a trip from Susanville, California to Klamath Falls, Oregon, and the route that came up was the same one that came up on Google Maps and in my car, but here's the difference. You can zoom in, and you can see that the map is color-coded. Green is easy, yellow is a little bit steeper of a grade, and you can go through the map and just put your mouse over these different areas, and you can see exactly what you're getting into for a grade going up or down, and then this graph on the right shows you the big ups and downs, and you can actually take your mouse along that area, and it takes you along the map and shows you what that looks like and what area you're in. You can see that they give you a steepness score, the higher the score, the more you're looking at for your big ups and big downs, and you can see that, you know, blue and red and black are a little bit hairy. Now, what I really liked is that you just don't have to look at it yourself. They tell you that there might be a better route. In my case, it said through Bernie, California, with a link, so I click that and went over to that map, which you can see here, same thing. You can go through it and go up and down and see where on the map you're going to be looking at the steepest grades. It also tells you how much longer it's going to take you, what the miles are going to be, and then they extrapolate the information between the two routes, and they say, look, it looks like the second one is better, and it has a better steepness score than the first one you looked at, so you don't need to compare routes anymore.  It was right here. You know, we went down this road, and I said to Doug, it's not that bad, but this is one of those roads when you're carrying a lot of weight, where you can really build up some steam, and it wasn't marked great. There was a really quick hairpin turn, and on our way back up that direction, we actually got behind an accident where a lady was coming down that hill too fast, and four people were flighted out to different hospitals because she flipped the car several times. Hope everybody's okay. Now, I couldn't find that flattestrout.com has an app, but a workaround I have for that is I just pull the website up on my phone. I have an Android's, if you have an Apple, this might be a little bit different, and I go up to the right-hand corner, hit those three little dots. I say, add to home screen, and then I pull that little icon that they give me into the same folder where I keep all of my other camping, weather, and routing apps. So it's right there. That takes me to the next question, which is on my solar setup, and if I can get by without a generator. The first question comes from Granny Annie's Traveling Bucket List. Question, have you experienced a solar setup on your RV? If so, what did you use for a mainframe to attach the solar setup tilting arms and brackets to? And then Margo asked, I'm interested in how the solar system you put in is working. Are you still able to live without a generator? Okay, so yes, Granny, I have had solar installed. I've had it installed on all three of my rigs. On my first two, I couldn't fit a lot of panels up on the roof, so I had about two, three, 400 watts between the two and two regular six volt batteries. But on this rig, there was a big space for us to have a generator put in, but it did not come in the price of the rig, and it was almost as much as solar would be. So I decided to put that money into solar. I had 1280 watts put on the roof and six lithium batteries, which has been rad, because if you don't know the difference in the batteries, the little six volts, you cannot run down below 50%. It's very bad for them, but the lithium ones, you can, and they just hold more power and they're lighter. Now I thought putting that much solar on mint that we would never need to use a generator again, mostly true, but not always. Here's what I found. That solar powers us up to 100% on six lithium batteries when the sun is high. But in the winter, the sun is too low because I don't have a tilt system, which I'll talk about in a second. We weren't getting quite as much power, so we had to conserve. But luckily we have a jackery, which is a little portable power station that we charge with 400 watt portable panels, which are tiltable.  Now I asked for solar panels that tilt when I had my install from Northern Arizona wind and sun, and ultimately I did not have the tilt put on for three reasons. First of all, they wouldn't do it because they didn't want the liability and they told me some horror stories about panels and systems that flew off on the road. Second, when those panels are tilted on the roof, they're more likely to fly off in high wind and I camp in some pretty high winds and I didn't want to worry about that all the time. When you put them on a mounting bracket so that they can lift up on the roof, it can add four, six, eight inches depending on your system. This rig is almost 14 feet tall. There are some bridges that I barely get under already and honestly, I just didn't want to worry about it. And like I said, we planned not to have a generator, but you might remember that I went back to Colorado in March, which is really snowy to mooch dock on a friend's property, and the solar was not going to get it done and I did not have a hookup there. So I went ahead and purchased a generator that sits outside my rig. It is not an onboard generator. Now we just keep that in the back of the pickup truck underneath a cover. I thought we really wouldn't use it that much, but when we got down to Arizona and California this last year and we were staying in an LTVA, that low winter sun was not getting it done. We got there in October and it was blazing hot. It was in the 90s. Every couple days in the afternoon, we would run our air conditioner for an hour or two up in the bedroom and just chill out up there, which we couldn't have done without the generator. Near to the summer when the sun is high, it powers everything just fine.  And here's one last minute question that we just added on from LK. She, he said, I have never had the desire to travel. I wondered if we start this full-time thing, where would we start? Well that's just a loaded question. Are you in an RV now and you just didn't want to travel in it? Or are you in a sticks and bricks? Are you solo? Are you in a couple? There's so much to go over. But I would say this, if you want to travel full-time and you want to stay in RV parks, that is a real struggle right now. Know that you have to reserve things way in advance and then call to confirm. I'm going to tell you more about this soon. If you want a boondock though, like I do, on BLM land or national forest or at overnight spots like a harvest host, have an arsenal of camping options at the ready. You always want to have a plan A, B, maybe a C. Go back in my playlist and look for five camping options in five miles or never get stranded. There are resources that help you do all of this. Start out in an area that you know or you know where there are several camping spots. Get your feet under you. Get comfortable going down those roads and setting up camp, hooking and unhooking if that's the kind of rig that you have. Going and finding the water or the propane or the gas or the groceries and then maybe circle out from there. And spoiler alert, in about three weeks we have a new series coming out where we're actually going to show you how we do all of this while we go down the road. So look out for that. If you have a question for me that you'd like me to answer, put that in the comments below too. I hope you're all doing well out there and staying happy and healthy. I'll see you next Sunday. Until then, everybody have happy travels and be free.


and welcome to a video that could've been an email. The only morning show where we answer the universe's greatest questions, submitted by you. I'm Simmons. And I'm Gryth. Today's question comes from Jane Donut, who asks, do you guys ever leave Red Base? What? Are we even allowed to leave Red Base? That actually reminds me of an article I scrolled past yesterday that said, more people than ever are choosing to spend over 90% of their time indoors. Huh, no wonder doctors are declaring the next big pandemic as just vitamin D deficiency. As we move into this interior utopia, many say we've gone too far. But have we really? Does anyone actually like it outside, like actually enjoy it? I've heard outside is quite lovely this time of year. That's just Hollywood propaganda.  It's either too hot and your skin starts leaking or it's too cold and it hurts to use your hands. Inside is like the Goldilocks of locations, but unlike in Goldilocks, real bears live outside.  What do you even do out there anyway? Go for a walk? Why? Sports? They got boring decades ago. Go to a barbecue. Last I checked, we can cook inside. That's why they're called Kitch Inns. So clearly this is a hot button issue for some reason. So let's throw it over to our in the field correspondent, Sarge, who's outside right now to see what it's like. Sarge? Uh, Sarge. Sarge, I'm here with- Sarge?   Can't you see I'm trying to do some reporting? I think there's some kind of delay, Sarge. I think there's a delay? Yeah, so maybe just- Well, I was gonna say- Well, why don't you- Hey, Griff, I stand over here. You know what? I'll just use my superior five-star intellect to anticipate what you're going to ask and respond ahead of time. Go ahead. So, uh, what's it like out there, Sarge? Yes. This isn't working. It isn't working, Simmons. I told you this would happen if you had half these on a morning show with the old Agent Orange soda over there.  Just talk about the outdoors. But back to the great outdoors. It's great! There's nothing quite like cracking open a cold one around a warm fire after a long day of camping and waiting for Griff to come around the corner to get him when he least expects it. I also enjoy running after Griff, bicycling over Griff and hunting Griff. When the weather's nice, I also enjoy playing golf with Griff's. BEEP Huh. Looks like we lost the feet there. Weird. Must be the weather. I think you're right, Simmons.  So, there you have it. No one likes it outside. Yeah, and take it from us. Two dudes who never left the safety of their own full-body armor.  And that's all the time we've got for this episode. It's been 20 years. Have a great morning and an even better day.


Γεια σας και ευχαριστούμε, είναι Vicky εδώ, με την νέα κέντρα νέα. Θα παίρνω με πολλές τεκνικές και κολάσεις και θα κάνω με τον κέντρο μου, αυτό είναι ένα καστωμένιο που χρησιμοποιηθήσαμε με νέα κολλοπίδια. Παίρνω πώς έκανε αυτή χρόνια και το σημαίνω ότι είναι 8 x 8. Λοιπόν θα παίρνω ένα από τα παίρνια και θα παίρνω με τα σπράσια μου. Αυτά είναι σπράσια σπράσια από διελούσια και έχουν πολλές τεκνικές στον κέντρο, που θα σημαίνει στην κομμάτι της κομμάτρας. Λοιπόν, πριν ξεκινάτε να δοκιμάσεις με τον κομμάτι, δημιουργήστε ότι θα σημαίνετε πολύ καλύτερα. Τα τεκνικές που χρησιμοποιηθήσαμε είναι σπράσια και κομμάτρα. Θα σημαίνω όλα τα παίρνια μου πρώτα με νέα. Αυτό θα σημαίνει να σημαίνει αυτά τα παίρνια να μπράσεις εύκολα και δεν θα πρέπει να σημαίνει πολλές τεκνικές. Θα σημαίνω μόνο με κάθε παίρνια. Τώρα πάντα θα σημαίνω με νέα. Θα θέλω να έχω την παίρνια να είναι πολύ λύτη και δεν να σημαίνω. Δηλαδή δεν είχα σημαίνει πολλές τεκνικές. Λοιπόν μπορείς να δοκιμάσεις την παίρνια σου για να σημαίνεις πρώτα. Είμαι ένα very impatient crafter, οπότε είμαι χρησιμοποιημένος στον κέντρο. Να δώσω ένα σύστημα. Τώρα θα σημαίνω με το παίρνιο μου με ένα στένσυλο. Είμαι χρησιμοποιημένος με ένα στένσυλο από το crafter's workshop. Αυτό είναι ο Γραφίτι Κυσής. Είναι ένα από τους στένσυλες που μπορείς να χρησιμοποιήσεις πριν και πριν σε κάθε πρόγεκτο. Έχει ένα πολύ γενικό σχέδιο στις κύρκλους και εξίσεις και λίγο χρήμα. Έτσι έχω σημαίνει ένα λίγο από το στένσυλο. Πριν το βάζεις στον κέντρο και σχέδιας το σπράστο, αφήσεις ότι θα σχέδεις το νόζυλο. Αυτό είναι το μάικα δεν θα σχέδεις το νόζυλο και η επόμενη φορά θα χρησιμοποιήσεις το σπράστο. Θα θα είναι καλύτερο να πω. Τώρα θα σημαίνω με το παίρνιο μου και το same στένσυλο. Τώρα θα σημαίνω με το παίρνιο μου με ένα στένσυλο. Ξεκουσάω σε δυο άλλες περιοχές. Και υπάρχει ένα lot of water on my stencil. Ξεκουσάω να αφήσω μερικές σπλαττες για το μάικα. Θα θα σημαίνω λίγο για τα λόγια για να αντιμετωπίσω με το λόγωμα που είναι στον κέντρο μου. Παρακολουθώ ότι αυτά τα λόγια αφήσουν ως λόγια. Και τώρα μπορώ να βρεις ένα κομμάτι στήνυο στον κέντρο να αφήσω όλα τα λόγια. Μπορείς να δεις ότι έχω κάποια σύμφωση. Τώρα θα χρειαστεί να αφήσω μερικές περιοχές. Αυτό είναι γιατί θα σημαίνω με το παίρνιο μου. Θα κάνω το ίδιο πρόεδρο όντως είμαι ευχαρισμένη με το πρόβλημα μου. Αυτή είναι μια καλή τεκνική για να δημιουργήσει τη βασικασία στον πρόβλημα σου, μετά να αφήσω μερικές περιοχές. Και παρακολουθεί στήνυο στήνυο στήνυο, όμως υπάρχουν πολλές πράγματα στον κέντρο μου. Και εδώ είναι ένα κλώσ-α-πλούκ για το πρόβλημα μου. Τώρα θα παρακολουθήσω μερικές λόγια σήμερα. Και θα παρακολουθήσω μερικές λόγια στήνυο. This is by Studio Light and it is designed by my friend Marlene. She has a whole collection of products for Argenalink and she just came up with a new collection, which I'm going to show you in an upcoming video. So I went through all those collage papers and I was looking for something very colorful like this one that I chose. And I'm cutting out strips of this collage paper. And I will use those strips to create a border all around my background. To stick my strips of paper I am using my matmedume. And as you can see I haven't been knit at all. I am not measuring anything and certainly they are not completely straight. So I'm just going to stick them all around. It is a great way to create a border which is very colorful. And I will be working on the border more as we go on. And you will see how I'm going to add shadowing and highlighting and little doodling to bring everything together. Now I'm also going to cover up completely all those holes. I don't care at this stage how the holes are going to be covered. I can always put them back with my punch. Now from the same collage paper I'm going to cut out even more pieces which I'm going to stick on top of my page. Looking as if they are heels at the background. Again I'm using my matmedume to stick everything down. We can go over the collage paper or only at the bottom if you like. And I like that they are so thin that they become a part of the page. As I said before today I'm going to play with collage. So I'm going to use some die cuts from this block. I'm going to show you where I got the actual idea. This is a little design with a girl and the heels at the back. So that's where I got the idea for today's project. Now I'm going to look through those two books and try to decide what I want to stick on top of my page. You will see that I'm going to grab the little tree. I'm also going to grab the girl. And I'm also going to look for flowers as well as some clouds to complete my scene.  They are all designed by my friend Marlene, by Art by Marlene. And they are by Studio Light. You will find all the links that you need down below in the description area. Just like always. When you take the designs out of the page you will find that they come with a white border all around. You can embrace this border. It's going to help all those die cuts to stand against the background. But I decided that I don't need that white border so I'm using my scissors to cut it out. Now I want to add some shadows around my border. So for that I will use my scribble sticks. I'm actually using the marine color. This is the marine stick. And I will go all around the border and then dilute it with water. It's really easy to go over different surfaces with those scribble sticks. And they are water soluble. Now you can see that my border is not even at all up at the top. I am going to fix that later on. I will decide that I don't like it as it is. But for now I'm working with what I have. Now as I'm going over that color with my water brush you can see that the water dilutes the paint. And it kind of helps to have a separated area such as the borders from the heels and the sky. It kind of separates all those different elements. Now I'm happy with how my background is looking at the moment. I played around a little bit to decide where all the elements are going to go. And now I'm going to place them down. For sticking them down again I'm going to use my matte medium. I find collage is a really fun technique to create our journal pages. It is very relaxing for me as I play along with all those different cutouts to try to decide where I want them to go and how to put everything together. And I absolutely love Marlene's style so these are perfect for my book. I love all those happy colors and the lovely designs. But of course remember that you can use magazine cutouts to create beautiful layouts. I'm also going to stick down the clouds and here is the finished piece. Now I'm going to add some details with my acrylic markers. I'm going to use different brands here with different nibs. I have some by Liquitex, others by Pebeo. You will find links to different products that I'm using down below as always. But I really have a big collection of acrylic markers. It depends on the nibs that I find every time or the colors that I like. So I am going to use the green ones here to extend the stems of my flowers. I'm using a green deco marker here and I'm also going to add some black lines. This is a black pen by Ranger. You will find that if you use this pen it writes over everything. It's the perfect pen for mixed media but it takes quite a few minutes to dry. So make sure that if you use it you don't smudge it. Now as you can see I'm going all around the border to add black lines to help all the elements separate from each other. And this is where I decided that I don't like the top border which is really easy to fix. I went ahead and cut out a few more strips from my collage paper and I'm going to stick them down with my matte medium again. And of course once this is dry I will go ahead and do the scribble stick and the black line to have a nice and neat finish at the top of my border. You can call this page done of course. I can go ahead and add little details forever. So I'm using my black pen to add some doodling around the border. I'm also going to use my acrylic markers in different areas. This is going to highlight the colors, make them more vibrant. So I'm using oranges and reds. Just adding color here and there.  For doing so I'm using an acrylic marker by Liquitex. This is a Liquitex paint marker and the color I'm using here is titanium white. And if you are wondering about the nib they come in different nibs these markers. I believe the one I'm using here is the 2mm chisel nib. So I'm adding highlights pretty much everywhere. On the apples, on the trees, the clouds, the flowers, the flowers on top of the girl's head and also drawing some lines all around. Just adding touches here and there. And finally it's time to add my quote. For this page I decided to go with the one that says BU the world will adjust. It comes from a booklet by Dilusions. And this booklet is actually full of different designs and black and white, doodling and lots of quotes that you can use for collage art journaling.  So all I need to do now is to put this page back on my disk bound journal. Of course I don't have the holes anymore. So I had to use my happy planner punch. And I think this planner punch is out of stock. I'm not sure I will make sure to link down below and to other options as well. I'm really happy with how my page came out. So bright and happy. I had so much fun creating it. Here are some close up photos on today's project. I hope that you had fun and you got inspired to play with cutouts and collage. Thank you all so much for watching. Don't forget to subscribe if you haven't done so already. And I'll see you next time.


So it is a pleasure to introduce again Philip, who will continue with his talk on torsars of affine cuffs. Okay, so, so today I have, I have, thank you for the introduction and today I have two good news first I have the real picture of the place okay as you can see, okay so thanks to Dana. So the, if you have not been which is my case, it looks like that. Okay. Secondly, I'm able to to answer your marks questions. So it's very rare, which I should say. And, and, and the result is, is as follow is that the tension bundle. Okay. Okay, so it does not. I check everything, but, of course, just to increase the suspense. Yeah, yeah, no, no, no, of course, I mean this is like the open round because the repetition is always perfect. And, okay, so, so, so this is this fact so the, why does not work. I'm sorry, I don't know what what happened about, I will share screen with that thing just to answer your mark question so I guess this thing works. So, so, so let me the theorem. So, here it's works. This is what which is not really a theorem. Okay, so, so on the two complex fear, the tangent bundle is three. Okay. And, and then. So it's, it's very related of the of the of the, I mean, the example of, of yesterday was really related with with. it related with a field of real numbers, okay? And how far as I know, you cannot trace an elementary proof of this statement in the literature, okay? So the proof I give is just a consequence of a huge theorem which is called Merti-Swan theorem which says that direct stably pre-projective modules over some x, so this is affine smooth surface C, half free, okay? And of course the tangent bundle is stably free because if you add the normal bundle, you get a trivial bundle, okay? So this is Invention Espeper in 1966. And of course I was not very happy about that. I mean, I was happy to transfer a mark but I was not happy to just apply this big hammer. And my colleague from Dijon, Adrian Duboulot, asked two proofs, two proofs, but none of them are really simple, okay? So I will not reproduce, it would take a lot of time, okay? But it seems that there is no elementary or very easy proof or exercise proof, I will say, of this theorem, okay? So that's one thing. It's a strong mark for the question, okay? And let me erase that. And just since I'm there, okay, just to say a few words what we have done yesterday. And so yesterday we introduced flat topology. And the main story was that we relate for G over R wing affine group scheme. We relate the H1, RG. So this is the torso size with the H1 here. So I can put FPPF time to time if it's not, if there's nothing written, it means this is FPPF, okay? So and if it's affine, so always we have a map like that, okay? Which is given by trivialization. And if it's affine, we can go here by this sum, okay? So this is what we have done essentially yesterday with example of GLN and so on. And so the descent is a key tool, okay, for dealing with these things. Okay, so now I come to my Philippe while you're doing that. You hinted there's a problem with the scent over arbitrary schemes, but what about sufficiently nice schemes like quasi-projective schemes? Yes, yes, yes. No, no, no, no, I did not say that quasi-projective is when you do Galois descent, okay? If you want, I think a robust statement, what you need is, you know, to work with the proge, okay? And to use the, you need, if you twist with a group scheme, we need ample invertible shift, which is G equivalent, okay? So it's, for example, it's what happens when you do, when you deal with borrowers, save riskings and things like that. Right, I see. But so in my world, it's usually there. Yes, yes, yes. Okay, so now this is crazy. I updated my, so let us see what we see here. Oh, I'm very sorry. I mean, I changed and, okay, well, I will start again this thing. Sorry. Okay, so now it seems to work. Okay, so let me do first some examples. I have not enough to finish yesterday. Okay, so the vector group scheme. So what I'm saying is that for a vector group scheme, the, the chronology is always zero. Okay, so it's my claim. So it looks a bit analogous with what you know about the chronology of current shifts, which are, which are, which vanishes over affine schemes. Okay, and so what, and, and the reason is again, this is a atmosphere complex. Okay. And this atmosphere complex, if we take a flat cover S over R, okay, to say that this complex is exact at this element, okay, says exactly that a co-cycle here is a columbary. So to be a co-cycle is to go to zero here, but then to be a comodary, this is to come from here. So then this chess comodology is zero. And, and, and, and of course, there are no torsos. Okay, so this is good to know. Okay, so there are, and once again, you could, you could think about this result as a negative result, but no, it's a positive result. Okay, it's good to know that certain obstruction vanish. Okay, another important case of, of groups are the so-called finite constants, group schemes. Okay, so you, so to distinguish the thing, I did not buy gamma, an abstract group, okay, and gamma R, this is the associated group scheme. Okay, so in term of groups, this is of scheme, this is G, and you take just gamma copies of spec R, okay, and, and you can, you can multiply them by, by, by the group flow on gamma. And the coordinate ring is, is R gamma, so I put parenthesis to, to avoid any confusion with fixed flow. Okay, and then in this case, since we are building in the, in the affine setting of the spec R, what is a gamma torsos? So we have the, the, the right notion in algebra, it's called Galois algebra. Okay, so it's a special case of et al algebra if you know, and this is for algebra, the place for algebra is essentially the job of, of Galois extension for fees. Okay, except that we are, we, it can, I mean, it can split in, in, in, in, in different, in two pieces. Okay, so, and, so Gal, finite extension is what is, is, is, is the, yes, is the most famous example of such an occurrence. And as for GLR, we, we, we, we have the case of groups which are groups of forms. So remember that many things we, we could deduce yesterday, including Hilbert Quotendick-90 was to interpret GLR as automorphism of the, of the, of the trivial vector bundle. Okay, so this yoga of forms were provided, you, you have descent. Okay, so if everything is affine, everything is nice. So one example, I discussed yesterday in, in the, in the question session was, was PGLN. So one other is the orthogonal group O2N with the hyperbolic quadratic form attached to RN. Okay. And so this is an exercise you, you could do for tonight. So a regular quadratic form of RN2Ns are locally exomorphic for the flat topology to the hyperbolic form. This is not totally trivial. If two is not invertible. Okay. And then descent theory gives you this dictionary. So namely an equivalence of category between group O2Ns of regular quadratic forms of RN. No, it's not an O2N torsos. It's not totally correct. So H1, this H1 does classify the quadratic forms. Okay. And one last example is the, the example of the, the, the similar group SN, okay, which is the group attached to, to, to, to the constant group SN. Okay. So, and SN occurs as the automorphism group of the, of the trivial algebra. Okay. So, so here we add a structure. Okay. So one way to think about SN is to see SN inside GLA. Okay. So and then the, the finite algebra of the grain are locally exomorphic for flat and I call you also for etal, for etal and flat and 2RD or for 2RN, sorry. And so the same yoga, the same yoga of descent shows that there is an equivalence of categories between category of SN torsos and that will find out the algebra of wrong head. Okay. So, and of course this is interesting to, to, to write this, to write maps. Okay. And, and the inverse functor. So, for my finite etal algebra, I can get an extra, sorry, is, is, is nothing but the Galois closure construction. Okay. For rings, and which was done by Seier in this fantastic paper of 1958 but which, which occurs in several places and the reference to, this is a recent interesting paper by Bargava and Satriano. So I recommend for people interested. But also again, everything comes from this kind of condition. The beginning comes from this kind of consideration. Okay. So I will go quite fast on functority issues, but as I said, yesterday, functority issues are really important when we play with, with, with, with more groups. Okay. And, and also, if you remember the, the CUMER seconds and the CUMER morphism, it's, it's clear that that was one of the reasons why etal or the risky topology was not enough. Okay. So I just consider a basic case on functority. If you like that, you can open the book by Giro, for example, and which is, which is quite hard to digest. But so there are many, many, many possible functorialities and, and this is, this is not always easy. Okay. And because the difficulty comes from the fact that we are dealing with non-commutative groups. Okay. So, so, so here we consider this basic thing. So we take a monomorphism of R group schemes. It means that for each string, GX, GS injects in Hs. And we say that an R scheme X, equipped with a map from H to X is a flat question of H by G. If for each, each R algebra S, the map XS to XS and use an adjective map, Hs mod GS to XS. So, but it is not what we want only that. We want the morphism to be couvant. We say that in French. We want to be surjective in the sound of flat sheets. Of course, I did not define flat sheets, but this is the theory which is beyond this definition. So for, for each ring S, and for each element X in XS, we want to find a flat cover S prime of S, such that when you extend X to S prime, it belongs to the image of this map of X prime of F over S prime. Okay. So, so, so we have, we have these two conditions. And if it exists, a flat, it's easy to show that a flat question is, is unique and even up to unique isomorphism. Okay, but be careful. It, it does not exist. I mean, there are many cases when it does not exist. Okay. And, but if it exists, and if furthermore G is a normal in H, so last before, what do you mean J normal for group skis mean GS is normal in HS for all S, then X carries a natural structure of a group ski. Okay. And then we can talk about, in this case, about an exact sequence of a group ski. Okay. And if you want to be really precise, we say for the flat topology. Okay. So if you want to replace flat by other content, the topology, you have to, to ask the, the ear in set to put flat, for example, you have to put the risk or okay. Okay. So what is now with flat questions is that if you have a flat question of H by J is defined automatically a G torso. Okay. So this map H, the G acts on the right on, on, on, on, on edge. And this map is a torso. Okay. So, so flat questions or questions are our way to, to, to, to provide a torso, not from the bottom, but from the top. And so my colleague, Reichstein in Vancouver, he really thinks about torso as, as, as, as questions. And then in that setting, we have an exact second of point and sets. Okay. So because, so it means that the, the pre-image of, of the, of the unit element comes from the, the previous map. Okay. So we have a map G air to H air to X air. And then here we have a very interesting map five, which is called the characteristic map, which goes to the H one of G. And then we can push to H one of H. And so I, I, I define yesterday this map here by co-cycle, but I, I said this is also interesting. And what is this characteristic map? This is when you have a X. So maybe this pen will work. Let's, let's see. I make a new try. So this is totally wrong though. Okay. If you have H of X, if you have a point. So yes, this is spec X. So yes, this is a G torso. You can just bait change this guy. Okay. This is F minus one X. And, and, and then so this is a very natural map. Okay. And this map, let me explain why this is the option because a point X, which is X here. Okay. To say that the torso is trivial. So namely to go to one here. Okay. This is that this guy has a point. Okay. So namely, this torso has a point. This is to have a section here. But if I have a section here, it means exactly that my point lift to H. Okay. So this exact sequence is easy to establish. So this point one is a bit other. Okay. But this is in the north. Okay. So you mentioned, you mentioned this problem about saying that this flat quotient doesn't always exist. Yes. And if your base scheme is just a field, and G is a linear algebraic group, does this always exist? Okay. Over a field, it always exists. Yes. Okay. And I see. And then the usual G mod H, the usual H mod G is then a flat quotient. Yes, exactly. Okay. So for over over a dodecagon, okay, the criterion, this is an entire amount of Renault result over a dodecagon wing. This is the condition G is closed in H. So over a dodecagon, but but but over say a base of the motion to you have, for example, correct counter example in this Renault thesis, I mean, this lecture not 190. Okay. Great. Great. Thanks. Okay. No, no, no, no, we don't, we don't, we cannot do what we want. Okay. So no. Okay.  So even. Okay. So it seems that I can continue. And, okay, so we we can be even more precise about about the image of this map. So today, so what is harder is to say when a torso here comes from here. And then if it is a fine or if it be when they send when can produce a category of G torsors of the spec A is a current to category of couple fx, where f is a H torso. Okay, which which is actually the the torso push by by by this map from G to H and some x in this twist of this x by f. Okay. Well, I will use that later. Okay. But but then so so we have criteria ones to decide about the about the image of this map and remember yesterday, it's exactly what what what I wanted. And and of course, these things are if in the case x is a group, it takes a better form. So namely, this trivialization things is this fx is enough is an x torso. Okay. So so what I'm saying here is that a torso if if x is a group, okay, a torso over G for G, this is nothing but a torso over H with the trivialization of the underlying torso over X. Okay, so maybe maybe it will be clear here with example. Okay, so for example, with the Swansea's correspondence, an example that the category of SLR torso is equivalent to the category of pair m theta, where m is a locally free air module of front air and theta trivialization of the determinant.  And it's, it's, it's very convenient to think like that. Okay. Because SLR is not as as as on the ball as object for this purpose. But this is right. This is the best way to see. Okay. And also, we have the cumular exact seconds. Okay, mu D GM D. So, so GM, this is an example of cautions. Okay. So, so it's easy to check that locally for the ethyl topology, for the flat topology any element is a deep power. Okay. And similarly, the category of mutator source is given to the category of pair m theta where m is an invertible model and theta trivialization. So if you remember the lecture of yesterday is exactly the kind of object I consider it. And that was my interpretation of, of this h one air mu D. I did not have h one mu D, but no, we have it. Okay. So it was a question on about ethyl covers. Okay. So ethyl covers are, are flat covers, but they are nicer in the sense that the map is smooth. Okay. So smooth morphism relative to motion zero. If you don't know the definition, so there are many alternative definition. So for example, you can add that S is a flat air module. Flat and finite presented air module. I forgot the finite presented. So just for each air field F, S tensor RF is an ethyl F algebra. So what is an ethyl algebra? So this is finite geometric variation of algebra or few prefixes. This is a product, a direct finite direct product of finite separable field extension of F. Okay. And, so the standard example of, of, of ethyl, this is a localization. Okay. You just invert an element F. The cumulomorphism is interesting. So the, because it's ethyl, if you only if actually the, the, the exponent D is invertible in the ring. Okay. Okay. So here this is a, and more generally, if D is invertible in air and you have given a unit in air, then when you extract the, the, the D roots of air, so that's the, the, the right way to do it. Okay. To write this, this caution ring. This is a finite ethyl algebra. Okay. So, and, and you cannot do that. Okay. If D is not invertible.  Okay. And, and you are forced to, to work with, with, with the flat city. Okay. So I pass very quickly on that. I will not sketch just to, I mentioned it yesterday in one of the questions. So if G is alpha and smooth, then the, the torso for ethyl are the same things that the torso for HPPF. So in the smooth situation, you can always trivialize by ethyl morphism. Okay. I, I, I go very fast. But I already said that yesterday. So more important for us today, this is what, what is the relation with Galois Comergy. So we have seen Galois Comergy in, in the other lectures by Danny. Okay. And if we are given a Galois algebra S, so that's example of, of flat, flat, flat torso of group, of, of constant group gamma air. So then you write the, the, the action isomorphism. And if you look at in terms of algebra, as this is an isomorphism between S tensor, S tensor, S tensor, S tensor, S tensor, S tensor R. And this is the group algebra and gamma, but these things can be written as gamma. And then the natural question is, how can we write co-cycle in that setting? Okay. So our co-cycle, this is, as you remember, this is an element in G of S tensor S over R, which are to find a certain relation. But I can't write this G times G, S tensor R as G S gamma. okay so so then a co-cycle is the data of some element z gamma and where z gamma is in g of s okay but so the answer is is z okay it's enough to write what are these p1 lower upper star and so on okay so we we need to introduce the action of gamma so it acts on the left on s so it acts on as well on the left on gs and actually on xx for any scheme x okay so typically if if if s is c and and and r is a real number this is just the the the conjugation action and and so the condition which is here so this relation reads as follows so an element like that is a co-cycle if only if we have this relation this co-cycle relation z sigma tau is z sigma sigma z tau for all sigma tau in gamma so the nice thing is that it was it was settled by ser I think and and then nobody tried to use the opposite convention okay so so it seems there is a single Galois Comology convention for dealing with co-cycles okay so so then what I'm saying is that when you deal with with Galois algebra and typically with fields and Galois field extension to to to use co-cycles this this refers exactly in this what you find in ser's book commercial was it okay and so I'm sorry can you yes can you go can you go back one slide can you remind me of the notation I don't remember seeing this what's this s upper parentheses gamma I mean gamma times oh okay just the maps from the set gamma exactly exactly oh okay that convention because if I put gamma you know we could you could think it's about invariance yeah yeah no no I just wasn't sure what it meant great now I know thanks okay and everybody with you good thank you okay so so so this z one we defined yesterday then is that nothing but a set of Galois co-cycle okay and when you take homologies we get a set of none of a young goal Galois Comology okay and the common the co-relation reads like that so two co-cycle z that prime are common August if that gamma is g minus one z prime gamma sigma g for some g in gs and similarly as yesterday the g you are here is not unique and is not there is no reason to find a better than you have a large choice and this part of of the theory okay and the simplest case is g sigma is the gamma right exactly yeah sigma is gamma yeah okay okay I I do not try my to use my pain for that okay so interesting case with g is constant okay so I I have gamma already no I have this state this constant with with associated group theta so what what is the set of co-cycle that so here you have to be a bit careful that the home of air groups from gamma s to theta s okay so I have two finite constant group scheme and the h1 you mod out by theta of s okay and the case you know is is when s is connected so in this case theta what you get are really home of abstract group from gamma to theta and the h1 are home from example gamma theta mod theta okay so this is exactly what you you you do when when you make your what you read okay so be careful the rings are in general are not connected okay so so so the right formula is that okay so I will use exactly that later okay more and as I said yesterday you can read the notes so if you you you specify what means this sounds first fully flat this sounds in that special case what you get is Galois descent so namely that the category of air air module is equivalent to the category of of modules over s equipped with a semi-linear action okay but also I mean on the same level if you take say purely infinitesimal things what you get you know is connections and things like that okay so so so it's it's one of the way to promote the the the first fully flat this something okay but I will not use Galois descent today okay I just will use co-cycle description because at some point you when you need to work with with torsos you have no other chance that no other ways no other methods than to to to compute and to work with co-cycles okay and a related issue is is is our isotrivial torsos okay so so there that's a very important class of torsos so we say that a torsos under an air group scheme g so no I'm I'm back in the general framework is isotrivial so it's this isotriviality condition is is a is quite present in in in grotendic hg s3 okay so it applies to all to all sorts of their notion if it is split by a galois finite et alcove okay so by galois I do not mean connected okay so and and so this is a subclass of torsos okay which can be explicit by balouac homologic computation okay and and then it means in the in the for concrete problems okay if you have a torsos you want to try to understand it the the first question is to ask whether it's a it's a isotrivial or not and so there is a result by peonsola and myself okay so if you take the rhythm of Laurent polynomials over a field okay so this is a result that all torsos over a reductive group scheme or reductive group over a field isotrivial okay so this is not I mean it's not trivial it was for gabber but it for us it was not okay but okay and a special case of the story are the so-called loop torsos and so loop torsos are torsos coming from which have even a much better shape okay so loop torsos you put conditions with with values for the co-cycle so loop torsos are really related with representation of algebraic groups okay so but they will not occur a bit today I will tell you later on an example okay okay and so why again I killed my file I don't know why this program is so is so bad but it's a this is a this is what we have in Lyon okay well no okay I recuperate my file I'm happy I come back to to to the dodikin ring case okay because the the main topic is about our fine curves and our fine curves are our special nice case of dodikin rings so I took the formalism of yesterday for SLN but I will I will apply it to to to other groups okay so we are given a dodikin ring with fraction of capital K and we take an element f of r and we look at the complement of spec r f of the localization is spec r so this is a finite set of points you know and I see that finite set of maximal ideal p1 pc and and I have notation with hat for completion and and capital K hat for completion of the fee and if we if we do the exactly if you repeat verbatim what I have done for for gln yesterday if we take a torso under g which is trivial over this open uh substance spec rf and which is low trivial at the completion you look at the different at the differences between uh the trivialization you have here and you have here so that provides the trivializations uh elements of of of these things and to to get a well-defined map you have to model by the possible choice and and so on the left this with this is with g by g mod f and on the right each uh each uh group here is modelled by g of the completion okay so this is harder way to to to to to to estimate torso and as for gln and as for any other group scheme this map is injective okay so this is a variant that the co-cycle determines the torso okay and uh you you can also also the whether this injective map is is is is uh is uh subjective uh so this is kind of this is a patching problem so I do need I do not need it today so I'm not discussing things that but uh but uh uh in many cases uh this kernel or here so and which is the way to express once again torso which are killed by this rf and all this localization uh are described by by by this double coset and and the corollary uh we have if if this double coset is trivial okay but then by this set is is trivial okay then there are no torso slides and it it happens in one case is when this grf is done here okay because this thing is open okay so if you take grf times this product of of of of of of of of groups here it will be discreet and dance so uh sorry now it will be open and dance and then it will be everything okay and as we have seen yesterday so sln was a case where where it's true okay so let us see other groups okay so I just uh brought brought again uh what uh what uh the previous statement okay uh and so so then I need to introduce some some terminology so uh a nice case when it works it like sln this is the case when g is a semi-simple simply connected split split is really important here air group scheme then in this case uh we can even get rid of these things I will explain why then any torso over air which is trivial over sublocalization rf is trivial okay so so I I have to say what is a semi-simple simply connected split split simply connected means here is the sense of semi-simple algebraic groups okay so this is a springer book or semi-simple group schemes okay so uh it is not is not exactly related with with uh simply connected in algebraic germany okay so be careful so maybe the the way to to to to to say when a group is simply connected this is to to look at the the case of the field of complex number where this notion okay is coincide with topological one okay so for example sln of of c is simply connected topological space uh for the simplistic group this is the same thing for the spin group this is the same thing but as you know if you if I take uh special or tonal groups uh if it is it is not simply connected okay and uh and then uh let me just explain why we can get rid of this simplification is because we have a local result by Brouillatis okay in this setting so g semi-simple simply connected split telling us that the h1 of the of this locality injects in the h1 of uh this uh of its fraction fee okay so so maybe somebody will talk about this cot and excess conjecture uh so which is a general uh statement or conjecture when this this kind of statement occurs okay and then of course if you take an element here which is trivial over rf it it will be trivial over capital k and then it will be trivial over the completion and that because of the statement will be trivial so that's the reason why you can get rid yes the corollary doesn't have a statement what's the statement no the statement says that the kernel is trivial oh but the first corollary oh no it has to be one yes oh yeah you're right you're right okay great no great it's not okay so you put one okay it was better on the previous pages but the repetition was not good okay that's here i'm here i'm i'm i'm guilty this is not the technique okay okay so uh so so this is nice okay because uh uh if you push to the limit okay it will tell you uh that many torsos are actually trivial okay uh but so so like that uh we get that uh uh h1 the risk of sln is one okay because if you are trivial the risk he uh it has to trivialize over some localization okay and but sln of course uh is not so surprising uh because i you know i can play with uh with with uh project projective modules of of dedicating okay but take e8 okay the garibaldis favorite group okay so this is a this is a big group okay so when i say e8 this is a split e8 okay so this is the most complicated uh or the most exceptional algebraic group and uh and then uh the the trick to to to to approximate the the fact that elements of this group are products of one parameter element is true okay and then we have a result for group of type e8 okay and then the technique makes sense okay but you can you if you have you you can take your favorite group okay uh well so that's why this is the same statement okay so now i can i can't really with with the topic of of the lectures so i find curves over fields so i start with uh i find curves over an algebraic lawsuit and here we have a we have a clear term okay if we take a semi-simple algebraic k group because semi-simple this is sln s o m spin pgln okay uh where k is an algebraic closed field you take a smooth connector affine curve affine curve okay so all the all the walls are important here then the h1 is trivial and uh so it it will be uh an application of uh there are ingredients okay in that statement so the main gradient is is an important result due to robert steinberg we state that when you go to the function field k of c then is vanished so that's a result of non-nabean galois homology so we which was conjectured by ser so it's called ser's conjecture one okay so i this is a very nice statement to prove if you want to on to learn okay so that's a really a good way to learn galois non-nabean galois homology uh let me just illustrate what happened on us on the case of pgln okay so as i said yesterday uh h h1 of pgln classifies this uh over a field that classifies the central simple algebras okay so we have this dictionary and to to say that h1 kc of pgln is trivial and this is to say that all central simple algebras over kc are matrix algebras okay you have no division algebra at all and uh if you open a book on a on a on broer groups and so on uh the the way to prove this statement is the center okay so center rams tells you that uh if if you have a hyper an hyper surface of with in pn with n variables for such a field then it has no zero it it has a non-trivial zero and we apply uh this same syndrome to the reduced norms of the algebra okay so uh anyway center rami is part is used by by by standard okay but that gives you the flavor of of of this vanishing and this vanishing is something which which is only uh over fields okay well no uh i need a second aggression for later which is about the picard group because the picard group this is h1c gm that classifies invertible invertible bundles and and for a curve uh over uh a very close field uh this group is divisible divisible and this follow of the structure of of this pixie of the picard of the compactification so let me explain that this divisibility argument so here uh maybe i did not take take in enough precaution with uh with uh with a positive characteristic okay so if if i'm wrong please tell me so what is the structure of of this picard uh we have a degree map okay so we have when you have a bundle you have the degree which goes to z okay and uh and the kernel is called the the there are the the the key point of the so-called jacobian varieties okay which is an abeyan variety and then what happens if you if you if you remove uh s-point from your curves the the picard group of the compactification maps on to the picard group of c okay the degree disappears because you have a split here uh you have a splitting here which is given by uh just the class of a point now you take the x1 and then uh the the map from the jacobian to pixie is on to but see this is abeyan variety this is this is divisible and pixie is divisible uh and uh let me sketch the proof of the theorem of this vanishing okay so the uh so we want to show that h1cg is one for g submissive well so the the simple the the easy case for from what we have done is when g is simply connected so we are given a class gamma in h1cg and according to stengbert theorem okay you you can find a map f in kc okay so now because everything which is true at the level at the function field level is true at at at some localization okay so hope to localize the curve uh enough that class gamma is key okay but but then the above color we tell us that h1cg because this condition is already enough to ensure that this is trivial okay so for so for the simply connected case i do not need this consideration of picard more so no i will i will prove the other case so it's a big technical okay so if uh maybe you will not understand the first time okay but there are notes and other references okay uh but i i would like to do it because that's that's a fundamental step oh for the general case what we do we we try to to to use the the case we have okay so the simply connected case and then there is a notion of simply connected covering for algebraic groups okay so for example the simply connected cover of pj and sln for so and this is spin n and so and since my i work with uh with uh over an algebraic algebraic close field i i have a max i have maximal split right inside my group okay so for example for sln this is a diagonal torus okay and so i pick a maximal torus on gsc and i can push it so here there is a quotient so mark this is a exactly a flat question and i get a max a maximal torus of g and and then there is notion of boral subgroups so if you want uh upper triangular matrices for sln case okay and what i claim is that the map from h1cb to h1cg is on two so it means that any torsors g torsor admit a reduction to the boray okay so if you prefer i admit a kind of of trigonal realization and here i use uh uh more more more techniques okay so i i use this variety g g mod b which is a which which is called the variety of boral subgroup of g i can tweak this by my torsor so i'm given a torsor uh c torsor dot g so and this thing is is well defined okay it's it's it's really a scheme okay and actually it has a very nice improvisation this is a scheme of boral subgroup of of the twisted group scheme e g okay so uh so we understand really why the what object is this why and no uh i apply steinberg but steinberg tells me that uh that my torsor is trivial over over over the generic point of the curve so in particular uh this y will have uh a point over this field kc yes but the nice thing with g mod b okay uh this variety of complex flags is is to be a projective variety and this y is also projective over c okay so this is part of the of the construction okay and then this y being projective uh i can apply the value of the criterion of preparedness so c uh is is dimension one so so my my rational section here extends to to to to c and then uh it follows that my this scheme y as a c point and this is to say that this twisted guy carries a boral subgroup scheme okay or according to the the function to briefly i discussed okay the fact this yc as a point tells us exactly that the torsor e and makes a reduction to b okay so this is the main step of the of the proof okay and so this is a classical argument i was okay but and no i have to to to to fight a bit with this b so b thinks about the group of upper triangular matrices so this is a semi semi semi direct product of uh strictly upper uh triangular matrices by the torus and this this t uh acts on u and it acts now we can filtrate this u okay but by subgroups such as the questions are isomorphic to to vector groups and no what happened is that i used what i said before that the the the h1 for for vector groups is is trivial so because the curve c is affine and and a devisage argument okay so if you plug everything if you put everything together in in a careful way we can one can show that this map from h1 ct to h1 cb a objective and actually this thing the fact this map is objective has nothing to do with with c okay it's true for any affine scheme okay so so here uh you there is a big difference between affine scheme at this stage and uh general schemes what we kind of a compression phenomenon so so no we are aware this map is objective and and what it remains to do is just to complete the the proof by by by summing up the pieces by putting the pieces together so for that we put a big diagram okay so remember i i had a torsor ish here in h1 cg i know it comes from b but to come from b this is the same thing to come from t right because this diagram commutes and i write the the same maps vertical map as the level of the simply connected cover the interesting thing is that uh here i already know this is this vanish but this map from tsc to t is a nitrogen okay so for example it it has it means it has finite kernel and and when you the fact is an energy if you put it together without the fact the peaks the peak arc group is a divisional group it follows that this map here is on two okay and uh and then it means that the the torsor here from h1 cg come from h1 ct but it comes from here but then since this diagram commutes it it it goes by that guy and then it's distributed okay so we conclude that h1 cg is one and uh we we saw not work okay we can do the reductive case for example glm okay so uh if you have a reductive group so what you you can model by his deraille group with dg which is semi-simple and so the quotient is called the corradical torus and what can show that this map from h1 cg to h1 cs is objective okay and so this is a generalization of the bijection of h1 cg there h1 cg ms peak peak c so the fact that vector bundle of a c are classified just by the determinant okay of what we have seen yesterday okay and uh and now far as i know i mean there is no way to to to prove uh in an elementary way i mean without without the things i mean stengberg and so on uh this result okay so i'm on time so maybe there are questions at this stage no questions so can i ask a question here i just yes please so you you want g to be split right is it because you want to have a split torus inside it is that you know the no no no i i want i i i took in that lecture okay i i i'm taking the essentially the best relevant case okay okay so we can push uh uh we can push everything okay with more technicalities uh i i can take uh i i can take a group scheme which is semi-simple over c i don't need to take a constant one okay and and then the the generic fiber will not be split but will be so-called quasi-split quasi-splits means to have still a borel subgroup okay so the really important thing is not to have split the real the real important thing no no no i have to be a bit careful with what i'm saying uh no i i think i think it's i did not done but uh i think what i'm saying works for for an arbitrary semi-simple group scheme over over c non-necessary constant and then non-necessary split also the whole argument goes through even if g is not split i think i think i think you need quasi-split though right because you are looking at the group is going to be quasi-split the group is going to be quasi-split on at least on the generic fiber okay so this is what is really relevant in this argument so for the first part the first part is okay okay it's clear it works in an higher generality here for higher generality what you need is a is a borel so we will have a borel because uh staying one of the consequence of of of steng verstigram is that uh submissive in groups semi-simple groups over k of c are all quasi-split and and the same argument i explained tells you that if you have a a borel subgroup of of of your group extended to the to k of c this borel subgroup extends in a unique way over over over the curve okay so this is what is very nice in dimension one and we have to be a bit careful with the with tori in this case and uh but my guess is that uh the the the h1 of this tori uh are are still uh divisible so that's maybe you can main me and i have time from tomorrow i i i can try to figure out that or to find a reference okay thank you but this is this is the result which is used by i mean by the people i mean uh with g constant okay mostly okay so now uh i i will uh i will go to uh to non-angry close field and over non-angry close fields there are no reasons okay to to have borel subgroups and to place this kind of of production to tori and so on this kind of diagonalization of them okay so uh let me say that uh historically speaking so the the the first case which was uh really uh done was about about the case of finite fields okay so by harder again okay and here finite fields this is a the first example of of non-angry close field i mean the azios one because the galva group is is that is nice and if you want to understand the the torso in this case this is this is this is very complicated okay so this is the work of of gunter harder in the 70s okay so i will not discuss about uh about i mean it would have been an option to discuss about finite fields in general groups or general curves is not i i will discuss uh because that's an homotopic appearance okay i will discuss the the case of the affine line okay so for the affine line we have a complete uh understanding of what's going on so this is a result by ragonathan and ramanathan in in the 80s so we are given a reductive k group g over a field k now which is not uh very close then we have a basic shot between the galva combo g of g h1 k g and the the the h1 k t of g so of the affine line uh the kernel so i mean the torso which are trivialized over k s of t where k s stands for uh separable closer okay so this is the the and uh so uh if the field k is is perfect or if the characteristics of p is good for g okay so there are many definitions of good very good and so for for prime p compare with red group so i i i do not enter in the in in the text but what happened we know that uh this h1 here is one okay so so because if k is perfect this is just the the t of m above okay and if p is good it is something work and so in this case so it means that uh h1 k t of g so torso over affine line are exactly the torso over uh the base field okay and of course for many groups we know uh what what what are the torso on the base field and what is happened so the terminologies that we say that the torso are constant so the torso is coming from the base field as the so-called constant so here as you can see this is a this is a perfect homotopy statement okay and uh but we have to pay attention with the small characteristics so there are few exotic cases okay when uh uh it does not hold okay so uh the easiest to understand this is p g l p uh not over k t over g is p g l p over okay when you take a base field k imperfect of characteristic p okay so so it consists to to to write on fancy division algebra and uh these things tell you uh uh so what happened for for zen mode p because you know if you take g g zen mode zen mode pz is not reductive but still you can consider the the torso for the group z mod p so this is arstein schuyol theory and you as you know the affine line is not simply connected in positive characters okay so uh so homotopy uh is nice in character in zero in this setting okay and also if p is large enough okay so uh my goal is to is to is to prove this theorem in the last 20 minutes okay uh at least to schedule and uh and and and also again i mean this is a very important statement okay because you you you cannot play with more variables if you are not aware of of of these results for example okay okay so i simplify my life i i suppose uh that my field case of characteristics zero okay so in that case the statement is just to say that all g torso are constants and they i should say there are seven there are two three or four proofs of this theorem okay but all of them use the affine the projective line okay so i did not define torso of the projective lines but you can easily imagine what what they are okay so you you cut your projective line in two affine line and you just glue the things okay so so the theory for projective line is not more complicated that for for for and for projective scheme or for arbitrary scheme that's the theory for affine scheme okay now so what is this cotendic hardware theorem states instead the following so uh so let me just say cotendic was when the field is algebraic closed hardware this is a general case so g as before is a reductive to the group and we consider a maximal case split torus okay so by by border at least we know that all of them are conjugated so so this is a good notion and we have this finite uh constant group this constant the group you you model the normalizer of s by its centralizing and there's a theorem state that you can look at the risky torso coming from s okay i will i will speak later about the vague group you you can push them to to to h1 p1 okay so here uh what you have essentially are are variant of the of border right and and the statement says that they cover all torsors which are trivial on the zero fiber okay and if you want to be direction you have to model by by by this vague group okay so so this theorem tells you that we have a very good understanding of of of what are the the the the g torsor of the p1 and so let me say uh there is one proof of this theorem okay by peter oscillivane which is a quite amazing only by category theory okay um so uh okay so we we we have we have this result and so for us the important thing is that if we have a g torsor over kt which is trivial at t is zero and this this this torso extends to p1 okay since the off bundle is trivial 100 when reduced to a1 so i mean just that p car of p1 the mark from p car to p1 to p car to a1 this is zero then if you manage to extend your g torsor to the the the projective line then it is trivial okay okay so so but what is not so clear is why we why can we extend okay and uh because in characteristic p this is exactly what happened so we cannot extend to p1 for for the exotic guys can i ask a question quick question here so you said there is this is a maximal case split torsor of g um so does it necessarily have a maximal case split torsor are you assuming gay to be a frame no key is not a very close i for me this s can be zero okay it can be one if the group is anisotropic this group is maximal among the case split Torah is what i mean is that clear so so you're assuming that it contains a copy of gm um no no no no no no i'm not assuming it contains a copy of gm uh so that's the problem between french and english uh tor maximal the player no no this is correct i think in english it's okay too maximal could just be the identity okay okay no no because because that's a classical you know uh i had these problems several times okay but this no i understand there's a difference between case split maximal torus and a maximal case split torus there's that's the difference you're talking about yeah so that means it doesn't necessarily have to have a gm right it's it doesn't have to be isotropic yes and actually the anisotropic case this is the harvest case so g can be g can be anisotropic exactly exactly okay and this is this is the harvest case so s can be just no parabolic no borel nothing okay yeah so s can be just trivial exactly okay okay so in this case so in this case the statement is anisotropic case it is just this kernel vanishes okay it even the statement is even more radical i see okay okay thank you okay so this is clarifying and this is not a problem of english so good so so we have to explain why we can extend to the projective line so we are given a g-torso x over k ot okay and there is a trick which calls the torsion trick which has an up to change g okay we can assume that the torso is trivial and the fiber t is there okay okay so so the original method to extend x to the projective line was to use bryat's theory okay so bryat is a parabolic group so-called parabolic group scheme it's not what i'm going and actually there were two variants of that i i i will not do use this beautiful theory which i will i will just do it by ants okay so but it is going to be a bit tricky okay and so the idea is to to to to do to to to to make a weaker statement okay so the weaker statement is to find an integer d such that if you best change k of t by k td no it's not it's t1 over d sorry so here i should maybe right so sorry sorry i here and everywhere i mean that okay i ramify one time and and gamma is x i guess no no what's gamma ah sorry gamma this is a class of x okay sorry oh okay okay so i have that and uh no and of course the one i have to to to make it uh treval time well so the fact we can extend to the projective line so this is by some patching trick okay this is a local statement and infinity so this is something you you you can look just at one just at the at the point at infinity which is k t minus one over d okay so so this is a local statement okay and so that's why the way it's to reach or okay so no i use a result of which which is a four-clock result okay but we we we brought that one channels up in rajstein so there exists a finite case of group s of g such that the map h1 fs h1 fg is on two for any k field f okay so what i'm saying is that when you play with co-cycles you can always the arrange the co-cycles to come to have value in some finite groups okay so i'm not saying this group is constant okay is is but it's finite and in particular uh if you look at this loran series with a variable t minus one at infinity this map is on two so so your co-cycle in as infinity you you you as as a nice shape okay and here you we know very well the gal by the gal by group which is there okay so this gal by group is inertia groups semi direct product with the gal by group of the of the best fit and as as you know now this inertia group uh is a projectively meet of this new way and so we are given a co-cycle the co-cycle in s okay and what we do we uh we we consider is restriction to the inertia group but on the inertia group there are no gal reactions so this it has to be a group of morphisms and a group of morphisms of of this big project limit factorize at some finite level i call it d okay for some d actually we'll see that later there is an uniform d which is which is uh just the cardinal of s but then uh it follows and if i i i extract the d root of t uh i i can kill this part of the inertia and then this guy belongs to the image uh of h1ks okay so now because when you ramified you you multiply by by by d on the inertia group so so we can really kill we had a problem and we kill it by this extension and then this gamma belongs to that and to to to this image okay well so as i said d we can take quite big to be the order of of this group and and then our reasoning uh shows the reason before up to replace t by this d root of t by this d root of t that gamma is one okay because it's uh it extends to p1 then it is constant but i have one the value at zero which is trivial so the value at zero will then this gamma is going to be one but actually the the key thing in that argument is that i have some freedom instead to work just with the the the d root of t i can take the the d root of at for any a in k star because uh there will be no difference and here comes the comes the trick okay so that that what what i got just by this uh in inertia uh consideration the trick consists to introduce a new indeterminate u and to extend the setting over this field f which is k of u okay so we we extend we still are everything okay and so i know i have h1f of t and of course if i apply what i have done to this field i i i take as parameter a my indeterminate okay but but then when i consider the fraction field okay of this thing is kt isomorphic to ktx because i can take this guy x okay as as indeterminate but this is very nice because when i when i extend uh yes this is not yes when i look at at the generic fiber of my class gamma okay it is h1 kt of g but it goes to zero to the trivial guy to kt of x okay but here uh this is this is this kernel is trivial okay because uh uh i have many kt points okay when i have a purely indeterminate extension over over enough and i feel i can i can just specialize the things just by writing co cycles okay so so then my my class gamma is rationally trivial and if it's rationally trivial i i can for sure i can extend it to p1 okay so so this is a trick which is permits to to to to avoid any uh brain activity in uh in uh in ragunathan's terrain so i think this is a simplest proof uh existing of of of this terrain okay so i'm sorry so the extension say rationally trivial so you can extend it it's because the the torsor information of always coming is coming from the completions at finitely many points and you just extend it to be trivially over infinity is at the point why exactly exactly okay great yes we extend trivially okay the the difficulty is is after us okay so uh we kill this so so this kind of argument of course you you can play them also in the setting of danie okay for example for this group that's the way i found that okay for with groups are easier because you have residues and so on you can you you have other means to detect whether a class is is is ramify or ramify here uh it's uh we we don't have so many tools so the we we need to go to the to these co-cycle things okay so what's next the the next case this is the pointer red affine line so a one was p one minus one point yes this is p one minus two points so this is g a and it's much more complicated than the affine that case okay so uh let let's let me simplify the statement to assuming characteristic zero so this is the result i shared with uh chernosov and tanzola but ten years ago so you take a reductive k group over a field k of characteristic zero then when you look at the map of our laurent polynomial in torsor of laurent polynomial in one variable what you see what you get is at the map on k w uh of this lower series g okay so so so the torsors are the same from this ring and for this field so maybe this is not as surprising if you think in terms of of of of galo theory and for those aware of of galo theory for rings and so they they they they are just same fundamental group and and so so so the the surjectivity is is is easy okay and is based exactly on the on on on on the statement i said with this finite group six okay so the projectivity you you just have to play with the case uh you reduce to a finite subgroup and and then uh co-cycle and this is easy the the hard part is the injectivity okay and uh the crucial step is to show the existence of a maximal torus for the relevant twisted group scheme okay so if you have a torsor a class of a torsor e here what what you want to do is to uh is to twist and to show the group scheme is uh is uh is uh is nice in the center it carries a maximal torus and then with that we can play one more time with finite subgroup and so on and uh and how how far as i know there is no there is there is one single process and this statement and this involves growth history and not buildings okay but twin buildings uh so we we need a uh uh this notion of twill buildings of of uh rona okay so uh yes well it's uh i guess this is a this is almost finished so let me just say that breuette history provides a description of this galois commercial set okay so this description is not nice uh for general group is nice for example if g is split and g is quasi split uh if if we know uh but often we know things here okay and so so then we we have a quite good on understanding of torsor of of of gm okay but we don't have uh if we i mean if how far as i know uh if we remove say three points okay so the next challenge should be to understand p1 minus three points and uh so overall very close field this is this is no problem uh we have we have seen it but over a general field uh it seems to be a very challenging question okay thank you very much for attention thank you uh so let's thank thank philip for for beautiful talk any questions to philip i have a question um so so you talked about torsos for the orthogonal group for even dimensions so for our dimension does wait wait one second i will stop the share i will take the uh i will take the yes the the basic uh yes please go on yes so i was asking about the torsos for orthogonal group of even dimensions you said it's it classifies um even dimensional quadratic forms right regular one yes are we from characteristic two i believe i'm not sure yes yes no no no this is too even in characteristic two oh okay but for all dimensions uh you know so do you have um does it do the same thing like classifies all dimensional quadratic forms okay so okay so uh uh uh no why is this pen it doesn't work yes it does it does the same thing but we we uh you have to be careful so i will draw with what i know uh you you mean uh uh i'm doing like that because my pen so you want to take n to be order yeah or yes but uh if n is order uh if if you are aware of two this is okay if two is invertible this works exactly the same okay but uh but if if uh if two is not invertible the the orthogonal group is not smooth i think and then uh it's it's uh it's much more complicated uh so i mean you i i'm writing uh things about that so i can answer you with more data separately okay what happened when two is not invertible okay but it's uh it's uh so that's a problem yes so so the good group is so q s s so q is is the the the special orthogonal group is is uh is okay uh in any dimension okay but not the orthogonal but for characteristic away from two then orthogonal still does the same thing is exactly that's what's exactly the same okay because because if two is invertible okay uh the so so-called non-singular quadratic forms are are regular also in even in odd dimension okay so so what is what is hard with this theory of quadratic forms when two is involved this is that for each author we we have one definition okay so there are several uh competing definition uh you know non-degenerate non-singular uh even time to time people say regular for one you know so so it's it's uh it's not something you you can discuss easily publicly i mean maybe we have to use a different apology instead of the flat one i mean the no no the one is good the flat one is good okay i mean for characteristic two i am saying like any other topology work no no no no no no no no no no no no no uh you mean if i if i work with two yeah if you work with two i mean if you work with with a special orthogonal okay uh then this is going to be smooth then you can work with et al okay but if you want to deal with h1 of of the orthogonal group uh i mean you it's well defined you can deal with that okay but then uh you you need to to deal with the flat things okay if you want to to to get all possible torsos okay because the first case this is you o1 okay and o1 this is mu2 as you know okay so mu2 if you look at at mu2 torsos for et alcobology you do not get much okay thank you um more questions for philip now but uh there is a uh you you can prepare for tonight and also you can send to to me i will be very happy to to to do my best to answer your question okay and and for this question of uh on uh yes on odd dimensional orthogonal group please send send me a mail i will answer it properly tomorrow i i have all the possible references in in in a paper in paper i'm i'm preparing with my colleague nair okay okay sure thank you any other questions for philip no questions at all so many of you in any case we can continue this at the discussion session if of course if philip will be there i will be i will be uh tonight tonight for me and and and learn for you yeah right so no then if if you get any questions later you will be you will have another chance to ask philip no but now let's thank philip again and