Hi, my name is Nick Chay and I've been a photographer for 6 years now. When it comes to photography, obviously there's important things like your camera or your tripod and SD cards. What it comes down to the most important thing is your camera, the body that you shoot on and ultimately the quality of the image that it produces. With Apple's new flagship device, the iPhone X and their dual camera and the improved photo mode and better video quality, we've gotten to the point where mobile photography has really taken a step up and is starting to be able to compete with professional level DSLRs. So I decided to run a few experiments. I took two portraits with similar background, similar editing and tried to get them as close as possible. Then I went to our student center and decided to ask a few questions. The first experiment was which photo they liked better regardless of what brand or which one it was. The second one, whether or not people could tell if it was an iPhone or DSLR. And the third experiment was if I changed the names and told people that the iPhone was actually DSLR and the DSLR was actually an iPhone, would people have the same opinions? Would that change anything and would that have any influence on their final decision? The iPhone X runs for about $1,000 compared to my $3,000 professional level camera. Aside from the price, there's more important things like portability, comfortability and just friendliness when you're using it. Obviously, it's way more convenient to have an iPhone in your pocket that works just as well as a professional level camera rather than having to carry around a huge camera body, another lens, SD cards, and a backpack to carry it all in. Obviously, there are a ton of variables and opinions on what makes a photo good and I can't control for all of them, but I hope this gives a better idea of whether or not an iPhone can truly replace a DSLR camera. So, the point of this experiment is to see whether the sacrifice of quality makes up for the convenience of portability. Let's get to it. iPhone X can replace a DSLR camera. So, I took two pictures, one with the iPhone, one with the DSLR. So, first off, let me see if you guys can figure out which one it is. Do you think this one's the camera? Do you think that one's the iPhone? Yeah, I think that one's iPhone. So, what about this one's iPhone? Is that the camera? That one. Why do you think that? I think that's the camera. I think that's the camera. That's the camera. Why is that? Because that one looks like a portrait. Okay, well actually, this is the camera, this is the iPhone. Oh, I nailed it. I told you. Okay, the phone's better. Do you think the phone's better? Yeah, it's better. Do you think this one looks better? Yeah. So, it's your photographer and videographer, so I'm going to ask you which one do you think is the better one? iPhone X. iPhone X. The same thing. This puts you closer to the building. No, iPhone X. Yeah. Is it right? No, this is the camera, this is the iPhone X. This is the camera, this is the iPhone X. And it's videographer yourself. Do you think that iPhone X can start with the iPhone X? I think it's just like kind of how you think it is more like the person behind the camera than the camera itself. It just has the quality to it. Okay, awesome. Thank you so much. See you later. Okay, wait. Let's do this one first. So, they have different opinions. He likes the second one, she likes the first one. Okay, what about this one? One, two, three. Okay, dude. Different again. So, do you think it's the iPhone? That's definitely a tough question. Which one's the professional one? I think the second one is the DSLR. I think the first one is the DSLR. This one's DSLR? I think the ones that I like are the DSLR. These are the iPhones. And then the second one's where the DSLR. Interesting. I don't think it completely replaces it. They have completely different DSLRs. This is the iPhone X. And this is the Sanitation DSLR form. So, which one do you like the most? The first one. This one? Yes. What if I told you that this is actually the iPhone and this is the DSLR? So, ultimately, it doesn't matter what the brand was. I think it's just the quality. So, if I told you that this was taken on camera, it's the quality. Awesome. The iPhones. The DSLRs. Can you tell the difference? For the one on the left, I can tell the difference.  Yeah. I think the one on the right was taken with the DSLR because the picture's kind of more vibrant and kind of the focus of it is more there as opposed to the one on the left. I think the one on the right was also the DSLR. Yeah. So, do you think we're at the point where iPhones can start replacing DSLRs? So, as a photographer, I don't have to carry around this huge $3,000 camera and I can literally just carry my phone. I would say the pictures are both very nice and for those who can't see the other option, people would be very content with having the iPhone 10 and a camera picture with that and put it into an editor. It will look just the same. Alright, so I'm going to tell you guys that this is the DSLR and this is the iPhone 10. Which one do you guys like better? I like the DSLR better. I agree. I like the DSLRs. Why is that? To me, it seems like the colors are a little more vibrant. Yeah, it's a little brighter. It's also a better picture of you. What if I told you guys that they're swamped? Oh, shoot. This is the iPhone 10, this is the DSLR, but you guys like the iPhone 10 better? Yeah, I think the reason that I like this one better is because the light is better on the left. Yeah, the colors, I don't know. So in certain situations, there's a lot of variables that play into factor what makes a good photograph, right? Yeah. Alright, so I just got back running a few tests and as you can see from the video, there were a lot of differing opinions. Obviously there are a ton of variables that make a photo appealing to certain people, but for the majority, I do think that people were very impressed with the quality that the iPhone 10 was able to produce, especially in the experiments where we ran and still were more drawn to the better quality image. So as a photographer, my conclusion is that it depends on the occasion. What I mean by that is it depends on the event, the amount of equipment needed, whether or not I'm going to be doing photo and video. And so obviously if I'm shooting a wedding or an event or a portfolio for someone that I'll be bringing my professional level DSLR, there's no way I could show up to a gig with an iPhone.  But if I'm going out with my friends, I don't want to lug around and then I backpack to carry out all my camera gear, then yes, the iPhone 10 is a perfect substitute for my DSLR. It has amazing portrait mode. If you throw it into Lightroom, do a few edits, then the images turn out really stellar to the point where people can't even tell the difference, especially if they're going on social media sharing sites like Instagram or Facebook. That different quality is so minimal that people aren't even going to be able to realize it. Overall, I was very impressed with the photos that came out of the iPhone 10. They were very comparable to the DSLRs, meaning that Apple, along with Samsung and all these other companies are really stepping up the game to bring mobile photography into the hands of professional photographers. For the everyday consumers, it's a huge advantage to have that portrait mode, especially if you just want to take cool pictures of your friends or subjects with that blurry background. But for the people who are more interested in having that more professional look and that higher quality, then obviously the DSLRs are going to be for you. I hope this video was informative and got my message across. I just want to say thank you to Kachina and all my classmates in writing seminar for making the past semester a great time. I learned a lot as a writer and hope to continue growing throughout the next four years. Thank you so much and I'll see you guys next time.


You can't let that water bottle matter. How hard you try. One, two, three, wait. Watch what happens if I hand you the bottle. You've become completely stuck to the couch. What? Stop grabbing onto me. Let me grab you. See it flat on the ground for me. This will help. That's actually crazy. I am kind of scared. Aren't you? Is it too late to back out now or? I think it has to do with it. Nice to meet you guys. Andrew. I'm just going to be honest. I don't think any of us have ever really seen hypnosis. And I feel like 90% of our followers probably have no idea. So could you explain it to us? So a lot of people tend to have the idea that they've never seen hypnosis actually live or in person. Be creative, believe strong enough that it becomes your actual experience. Has it ever like not worked? Or has it like worked? So the only times it doesn't work are when people don't want it to work. So I guess there's only one way to find out. If something happens, I don't think we should go first. I don't even look at me. It's two against one. Go, go first. Here we go. So what we'll do is I'll take your hand if you all have your focus right in the center of this hand. And in a moment, when I say that word sleep, that feeling of relaxation will spread from your arm to the rest of your body. So close down now, let yourself sleep. Focus on the sound of my voice all the way. Absolutely incredible. Lift yourself. One, two, three. Awake. Eyes open. Good. Hey, come on. Pretty good. Close all the way down. You need to be next time that you open your eyes, you're going to find that your own first name is suddenly just vanished from your mind. I know we did like the intro at the beginning. And again, your name was David. So if we just give your name back, and now you know what it was. Andrew. There we go. But just like we could give it back, we could take it away. Imagine I reach him and I steal your name. So now it's gone again. Your name was Stephen. Stephen. Do you remember when you said that? I just remember saying Steve. I just don't know why. Who's Stephen? Is that like your best friend or something? Why didn't you say Alan or Alec? So deep breaths in for me, breathe out, and return to sleep. All the way down in the basic state of relaxation. However, next time you open your eyes, you're going to find that you've become completely stuck to the couch. You cannot get up. It's like the world's strongest superglues underneath. One, two, three, awake. Good. Right. But notice what happens. If I ask you to like to try to stand up, notice what happens. Wait, what? He's going to be like that. He's like that. Ow! One, when you reach number one, he'll be unstuck. Three, two, one. He's like crazy. All right. I'm going to use this. I think it looks right here for me. Sleep all the way down in the basic state of relaxation.  You can't lift that water bottle no matter how hard you try. One, two, three, awake. Are you good? Yes. All right. Now, this is a normal water bottle. But watch, notice what happens if you try to pick it up. But this is so heavy, you can't even hold it. Watch what happens if I hand you the bottle. It's so weird seeing him like this. All right, watch. You've found your hands and become stuck to the bottle now. Give it to me. Hey, give it to me. Give it to me. Do you remember anything? But at the time when you're doing it, you're just like, I don't remember. I think when you're in a dream. Yeah, exactly. It's almost like a dream. You know how you remember how you can explain how? That's crazy. All right, guys. Good luck, guys. Center twinkle. Center twinkle. Good luck, guys. Really good luck. Be safe, guys. Bye, Andrew. I feel like you should go first. No, you should go first. No, you go first. Because I'm older. I'll tell you what. We're going both at the same time. Both of you take your two hands for me like this. Bring them together and lock all of your fingers. As your eyes just focus and fixate on that spot, let your mind just become completely focused. And as you let yourself now sleep, all the way down that amazing state of relaxation focus. That's right. All the way down. Even be one, two, three. Wake up. Doing good sitting up for me. Sound good. All right. Why was I twitching? That feels pretty relaxing. Pretty good. Yeah, no, no. I actually got like tired and I was in my hands really. I don't even know what just happened. Deep breath in. Breathe out. Sleep. All the way back down. Focus inside my voice. All the way down. That's right. All the way down. Next time you open your eyes, you're going to find that the both of you have just completely forgotten your own first name. Absolutely amazing. One, two, three. Oh, wait. I'm good. Sit down straight and tall. So I know we did the intro. So people know I'm Zach Vincent. I'm the hidden fist. And you were... My name is Jeff. And you were... Who's this? You know his name, right? What? Wait, what?  Alan? Right? Yeah, you know his name. What was your name? I just said it. I just said it. Wait, I'm so confused. Wait, what? No, it's no, no, no.  It's kind of like... Wait, what is my name? Alex? Shane, what's my name? Alex, right? Wait, can we take a quick pause?  To sleep. All the way down. To sleep. Eyes closed. All the way down.  Next time you open your eyes. One, two, three. Oh, wait. Good. First time I've either of you have done any hypnosis before, right? What? Wait, what? I was like, it's just hypnosis. It's not my thing. Say it one more time. It's just hypnosis. So, both of you, deep breath in. Breathe out. Eyes closed. All the way down to the basics. Every relaxation, focus. Next time you open your eyes, you're going to find that you've become completely stuck to the couch.    One, two, three. Wait. Good. Notice what happens if I ask you to try to stand up. Wait, what? Wait, wait. What? Stop grabbing onto me. Let me grab you on. Stop. Stop. No, no, no, no. Wait. Let me try to get you up first. When I can't. Watch your feet suffer. Watch this. Feet. Put your feet flat on the ground for me. This one. When I snap my fingers, you're no longer stuck to the couch, but your feet are stuck to the floor. Watch. Now you can get up easily. You can stand up easily. But now watch what happens if you try to take a step forward. You notice that you can't. You're stuck. Oh, there we go. Don't fall over. Watch. Again, I snap my fingers, your feet are unstuck, and now you can take a step easily. It's like I forgot how to walk. Wait, what? Why am I walking on my side? I'm walking on my side. Actually, this will be kind of interesting. What we'll do is actually, both of you put your right finger on your right hand and then it catches. Eyes closed. All the way down. Next time that you open your eyes, any time that I clap my hands, I become completely invisible.  One, two, three. Wait. I'm good. All right. Good. What were you doing? Wait, are you okay? You stare at something for too long. You don't see it anymore. Wait, what? Okay, wait. So, wait, can you show me the footage? I want to know what actually is. No, no. Show me the footage. Watch.  Right? The bottle? Yeah. Tell me to drink it. It told you. It was going like, drink me. Drink me, Alex. Still talking to you. What else is it saying? Go grab another bottle of water to clear up your skin. Drink all it's friends. Keep focusing. Follow it. All the way down those eyes. Close. Next time it's you open your eyes, any time I stomp my foot and make that sound, I actually turn into big foot. Two, three, awake. Good. All right. Watch. All that I really have to do is this. Watch. What does he look like Chewbacca? What I'm talking about, right? Alex, are you okay? All right. Both of you focus right here for me. Deep breath in. Eyes closed. Let your shelves sleep. All the way down. The next time that you open your eyes, you're not going to recognize the person sitting next to you. One, two, three, awake. Why are we sitting so close to each other? Dude, I mean... Wait, why are we? Well, who were you here with? Who was it? Why was he here the whole time? Because that's how we... Because we were already... Did it work? Yes. Like, I was laughing so much. It did. No, it did not work, right? So we're not, at least I'm not exactly sure like what 100% like happened inside the video. I feel like... This has been a weird morning. Yeah, like a very, very weird experience. It feels like we weren't filming, but... Yeah, so like I'm not really going to know like what exactly happened until like I watch over the footage. So that's going to be it for this week's video. Hope you guys enjoyed. We say this every time, but that was actually the craziest and weirdest thing I've ever seen. So yeah, if you guys enjoyed the video, be sure to give this video a like. You guys should comment down below what your favorite part of the video was. Okay, at this point I'm just using it for like my own good, but maybe if you guys want to see who my crush is and hypnotize her to fall in love with me, because I could use that. I could use some of that right now. But anyways, we wanted to give a huge thank you to Zach. We'll be linking this channel down in the description. We want to see more really cool hypnosis videos. Thank you so much. Thanks for having me guys. Yeah, thank you so much. Bye bye. But yeah, thank you guys so much for watching once again and we'll see you guys next time.


Hello and welcome to this introduction to the STM32L5 series training session. It describes the feature sets available in the STM32L5 microcontroller series. The main challenge faced by our customers is to protect from hackers at logical and board level attack and to add more security. The second concern is related to the low power consumption in order to reduce the battery size and increase the lifetime. In the end, the target is to define the right microcontroller which fits the application requirements the best in terms of integration, size and performance. The STM32L5 microcontroller is the first STM32MCU based on the Cortex M33 core. It provides more security with trust zone and ST security implementation. It offers lower power consumption, reusing the STM32 best-in-class ultra-low power technology and adds innovation. Thanks to a large portfolio, high memory size and better performance, it can fit into a large number of applications. In addition to trust zone, it shows a full set of security. Besides a symmetric keys accelerator, a true number generator and a certified crypto library, it provides a private key accelerator or PKA for asymmetric cryptography. In terms of memory and IP protection, it offers an active temperature and voltage tampering. Two new features are also reinforcing the security level and help the design of a route of trust. HTTP or Hyde Protect, which provides higher protection by a hardware secure boot with secure keys and a unique boot entry, which defines the address to boot. The second main message is about power consumption. The STM32L5 series reuses the STM32L4 and L4 Plus technology, achieving best-in-class power consumption. It integrates an optional SMPS or DC to DC buck voltage regulator which can be enabled or disabled on the fly to avoid external noise for external RF or data acquisition. This excellence in power consumption is proven by the EEMBC benchmark scores, 370 ULP marks with the core profile benchmark and 54 ULP marks with the peripheral profile benchmark. The STM32L5 series offers best-in-class power consumption figures with a full flexibility. The user can select the mode depending on the wake-up time, the state of attention, and the number of sources required. The STM32L5 series provides more performance thanks to its new core. The Cortex M33 core gives 20% more performance running at the same frequency and so achieving 1.5 D MIPS per MHz leading to 165 D MIPS or 442 core mark. Furthermore, it supports a new version of the STR accelerator. In addition to the internal flash memory, this version also supports external memory thanks to its 8 kilobytes instruction cache which enables a higher performance as well as a better power efficiency. Here is the block diagram of an STM32L5 microcontroller. This product embeds a long list of digital and analog peripherals as well as a large memory size, up to 512 kilobytes of flash dual bank and 256 kilobytes of SRAM. In addition to a USB crystal-less port, it provides an extended connectivity with a USB Type-C port with power delivery support and CAN-FD. The STM32L5 series is available in 7 packages with several options with or without the crypto accelerator, 512 or 256 kilobytes of flash memory and with or without the SMPS buck converter, so a large portfolio.


Hey guys, Thundie here and you guys have asked for a speaker test on the Galaxy S21 Ultra. So I'm going to give it to you, but not just the Ultra, the S21, I'm going to include also the Galaxy Note 20 Ultra, the Galaxy S20 Ultra and the iPhone 12 Pro Max for reference. Now first off, the S21 Ultra and the S21 are brand new from Samsung and of course they have stereo speakers. Actually everything here has stereo speakers in terms of all the devices we've been looking at in this video. But the Galaxy devices from the S21 Ultra all the way to the S20 Ultra supports Dolby Atmos. So you've got Dolby Atmos sound and we had that all set up. What we're going to do now is play all your clips for each device. You take a listen and we'll find out. So let's go ahead and listen. Alright guys, that was pretty cool. They all sounded great including the iPhone 12 Pro Max, the S21, the S21 Ultra. It's a really nice vivid sound altogether. Now I can't give you a finalized pick on who has the best sound but I will tell you that they all sounded pretty good. I want to know what you think on which device has the best sound in speaker. Is it the brand new S21 Ultra or is it the S21 or is it even the iPhone 12 Pro Max? Leave your thoughts down below. Let's find out and I'll give you my own final decision with my full review coming next week. So this is Thundie E saying thank you and always enjoy your entertainment.


The peace and blessings of Allah be upon the servant of Imam Hussain, alaihi s-salam. Muslim Ibn Aqeel is the trustee of Ibn Abdullah al-Hussain, peace be upon him. Muslim Ibn Aqeel, Ibn Abu Ghalib is the cousin of Imam Hussain, peace be upon him. Ibn Aqeel has a very high status in the eyes of Ibn Abdullah al-Hussain, peace be upon him. Tonight we will end this program, this journey by speaking of Muslim Ibn Aqeel, peace be upon him. The companions of Ibn Abdullah al-Hussain, peace be upon him, may my soul be sacrificed for him. The companions, the Ansar of Hussain, peace be upon them, are companions like no other. I will narrate to you a tradition, Imam Ali Ibn Al-Hussain, al-Sajjad, peace be upon him. He says, my father Al-Hussain gathered his companions and began to speak to them. He said to them, I praise Allah, I praise Allah the most glorified be he. I thank you, I praise Allah, glorified be he and thank him for all happiness and all misery. He said, oh Allah, I commend you that you have honored us with prophethood. Meaning that you have honored Bani Hashim by giving them prophethood to Muhammad, peace be upon him. And taught us the Qur'an and made us learn jurists, you taught us the ilm.  I do not know anyone more loyal than you my companions. You see, Muslim Ibn Aqeel has this derajah because he is counted amongst the Ansar of Ibn Abdullah al-Hussain. He is counted amongst those companions whom there are none like in this world until the day of judgment there will be none like the Ansar of Ibn Abdullah al-Hussain, peace be upon them. Muslim Ibn Aqeel was given a title even greater than being the cousin of Ibn Abdullah al-Hussain. Muslim Ibn Aqeel was given the title of Safira al-Hussain, the one who represents Ibn Abdullah al-Hussain. The one who is the trustee of Ibn Abdullah al-Hussain.  The prophet, peace be upon him, prophesized the killing of Muslim Ibn Aqeel. He said, when Ali Ibn Abi Talib asked him, he said, O Rasulullah, do you love Aqeel? He said, I love Aqeel. And I love Aqeel for two reasons. I love Aqeel because of Abu Talib and I love Aqeel because his son. Muslim will be killed for your son. Muslim Ibn Aqeel will be killed for the cousin of Ibn Abdullah al-Hussain, peace be upon him. Muslim story is a story that we hear time and time again. It's a story that is very lengthy and long. And we will try our best to summarize what we can from this story. After the demise of Muawiyah, the Khufins sent letters towards Imam al-Hussain, peace be upon him. It is narrated that they sent 12,000 letters. These 12,000 letters were sent to Imam al-Hussain, peace be upon him, telling him, O Abu Abdullah, it is time for you to come to us, to Khufa. We are lost. We need an Imam. And we do not know what to do. The minions, Omar ibn Sa'ad and his gang, Omar ibn Sa'ad and his posse, they heard of this news. And they found out that the Wali over Khufa and Na'man did not have much power to get rid of Muslim. So they decided themselves to write to Yazid. They told Yazid, Yazid, the Khufins have wrote letters to Imam al-Hussain, peace be upon him, asking him to come to Khufa. What will you do now? He said, we will take Ubaidullah ibn Ziyad, who was the Wali of Basra, and Ubaidullah ibn Ziyad will take his brother, Uthman, as the Wali of Basra, and Ubaidullah will come to Khufa. This is the story behind in the background. Now what happens now? What happens is that Imam al-Hussain, peace be upon him, sent Muslim ibn Aqeen with his letter. He said, furthermore, after accepting the letters of the Khufins, he said, furthermore, for you, the last of your letters arrived to me. The last of your letters have arrived to me from Hani and Saeed, for I have understood your request from your letters and what you said. We have no Imam. We have no Imam in our city and we want you to come, for maybe Allah will unite us under your guidance, and you can guide us as well. I have sent to you my brother, listen, I have sent to you my brother, the son of my uncle and my trustee whom I trust from my pajni, Muslim ibn Aqeen, to report to you about your affairs. To report to you about the affairs of the people of Khufa. If this report agrees to what you have written, I will soon be with you in Khufa. You must be clear to the fact that the Imam is only one who follows the book of Allah, Subhanahu wa Ta'ala, and serves Allah in all matters in affairs and justice, honesty and truth. This was the Rasala, the Kitab that Abi Abdullah al-Hussain, peace be upon him, sent with Muslim ibn Aqeen to go to Khufa. Muslim now arrives in Khufa. The Shi'a in Khufa, the Muslim in Khufa are happy for the coming of Muslim. It is narrated 30,000 to approximately 80,000 Khufans swore an allegiance to Imam al-Hussain, peace be upon him, through Muslim ibn Aqeen. Of these preachers and of these ulama and of these faqaa in Khufa, were Abbas ibn Abi Habib al-Shakiri and Habib ibn Muzaher al-Asdi. And the number adds up to 80,000. 30,000 to 80,000 people. Remember this number, this number is very important. 80 to 30,000 when he arrived in Khufa, swore an oath of allegiance to be with Imam al-Hussain. It continues. And like I said, the cohorts of Yazid did not like the fact that al-Naman was very weak as the Wali of Khufa. Ibn Ziyad began his way towards Khufa. Ibn Ziyad was very cunning. Ibn Ziyad, what did he do? He disguised himself in a green amama and he wore clothes, fully covering himself. And then he entered with a small group of people. He entered with a small group of people to try to deceive the Khufans. Everybody thought that it was Imam al-Hussain, alaihi salam. To the point where the Khufans would actually come and kiss his feet and kiss his forehead, he was very deceiving. And it was not until he got towards where? To the castle. To the castle behind Masjid al-Kufa, Qasr al-Imara. He also tricked Ibn Abashir. He forced his way inside and he said, Oh people, today I am your Wali over Khufa. He gained power and he began slowly to deceive the people of Khufa by money, by tricks, by sending the mothers and the daughters and the husbands and the brothers to whom he was going to revolt with Muslim. Why? To cause a problem. This was an issue. This was an issue because the Khufans' hearts weren't that very strong. To the point where it is narrated that he brought one of his slaves to act as a Shi'i to come to this Masjid over here and to listen to the ears. Who gave allegiance to Muslim? So they can quickly and deceivably become a spy inside. They gained the trust and Ibn Ziyad gave them 3,000 dirhams. He gave them 3,000 dirhams and he told them, Take this money with you. Take this money with you. And then if you see somebody in the Masjid, make sure you try to deceive him. Tell him, I really want to see Muslim Naqeel. I want to see Muslim. Tell him, in fact this man came, the slave of Ibn Ziyad and he found somebody in this Masjid, in Masjid Al-Kufa and he told them, I yearn to see Muslim Naqeel and I love the son of Fatima and I want to aid him after taking an oath of trust from him and a trustee of him. After taking an oath of trust from him and they began to trust him, what happened? They trusted him, they showed him Muslim and when they showed him Muslim right away Ibn Ziyad knew of the story of Hani Ibn A'ruwa and he knew that Hani Ibn A'ruwa was not sick as he claimed to be sick because Hani Ibn A'ruwa, when Muslim first came to Kufa he was in the house of Al-Muqtar and after that he took him to the house of Hani because the situation got very difficult in Kufa. Ibn Ziyad was very cunning, like I said he was very cunning to try to distract the people.


Welcome to Spotlight. I'm Adam Navas and I'm Katie Blake. Spotlight uses a special English method of broadcasting. It is easier for people to understand no matter where in the world they live. Imagine a long dirt road. The country around it is dry. Small brown plants grow from the dry ground.  He is very hurt. Thieves have taken his money. They have beaten him very badly. Without help he will die. Soon two people from his country walk by. But they do not stop to help. Both believe they are too busy. Then a third man passes by. He is not from the man's country. Instead he is from a foreign country. In fact these two countries are at war. But instead of passing by this third man stops. The third man shows kindness. He helps the injured man onto his donkey. He takes him into his home. He cares for the injured man and nurses him back to health. It does not matter to this third man if the injured man is from a different country. He sees someone who is in trouble and he helps. This is a story from the Christian Bible. It is called the story of the good Samaritan. It tells Christian believers to be kind even to their enemies. Most religions have stories like this. Most religions tell us to be kind to each other. These may seem like commands given with little reason. But did you know there are also scientific reasons why a person should be kind? Kind people are happier. And they may even live longer. Today's spotlight is on the science of being kind. Most people know that being kind feels good. Helping someone in need makes us feel positive. And giving a gift to a friend feels as good as receiving one. But for most of human history no one could say why. Giacomo Rizzalotti is an Italian scientist. He studies how the brain works. In the 1980s he and his colleagues performed an experiment that showed a lot about kindness. Rizzalotti used electrical scans to examine the brains of monkeys. When a monkey did something part of the brain would show activity. Then Rizzalotti discovered something amazing. Sometimes the monkeys would watch each other. When watching their brains were also active. But they were active in the same way the other monkeys were. When one monkey reached out his hand the other monkey's brain looked as if it were doing the same. It was not reaching out. But the watching monkey's brain experienced the event as if it were. Only special parts of the brain acted this way. Rizzalotti called them mirror neurons. Soon he and his colleagues discovered the same thing in humans. Mirror neurons are very important to the science of kindness. Mirror neurons are the part of our brain that lets us feel what other people feel. They allow us to empathize. V.S. Ramchandran is a neuroscientist. He studies mirror neurons. He spoke to Greater Good magazine about why he thinks mirror neurons are important for empathy. These mirror neurons are involved in empathy for pain. To really empathize with your pain I need to experience it myself. That is what the mirror neurons are doing. They help me to feel your pain. I experience the same pain as if someone were to hurt me directly. That is the basis for all empathy. Empathy is the first step toward kindness. It lets us know when people are sad or hurt. We know mirror neurons help us feel other people's pain. But mirror neurons also allow us to feel when other people are happy. When we see others smile, we often smile. In being kind to others, we feel their happiness. The happiness we get from being kind also may be good for our health. When we are kind, our brain releases chemicals called neurotransmitters. Some neurotransmitters make us frightened or angry. But the neurotransmitters released in kindness have positive effects. They make us feel good about our relationships and more confident. But they also help our bodies. Oxytocin is one neurotransmitter released during a kind act. Oxytocin prevents stress, but it also lowers blood pressure. Marcy Hall is a psychiatrist at University Hospitals in Ohio in the United States. She spoke to the Healthy at UH podcast about the benefits of these neurotransmitters. Kindness actually protects your heart. It improves your body's disease fighting system. It improves your energy. It reduces aches and pains. It can lead to a longer life. In the process, you also feel better about yourself. Kindness and empathy are also very important to human society. They keep our relationships healthy. The good feelings we get from kindness make us want to help other people. And they encourage different groups of people to work together. Without empathy, our communities would be very different. They might not exist at all. One well-known theory is described as survival of the fittest. This idea is based on a theory by Charles Darwin. This theory says that the strongest and best people are the ones who will survive and continue. But Helen Reese disagrees with this idea. Reese is the director of a program that teaches people how to use and improve empathy. She wrote a paper called The Science of Empathy. In it, she explained why people have empathy. Why is the human brain designed this way? If life were only the result of survival of the fittest, we would only fight people. We would not care about responding to suffering. Our ability to see others' suffering allows us to feel and understand their pain. The personal distress we feel often influences us to show kindness. The survival of humans depends on helping others. Being kind reduces our own distress. Kindness is not just something nice you do for someone. Kindness is part of what makes us human. Without it, we would not be able to work together. We would not want to. Henry James was a famous English and American author. He wrote many books that explored the human experience. In 1902, James spoke to a young family member. James told the young man these wise words. Three things in human life are important. The first is to be kind. The second is to be kind. And the third is to be kind. Do you think kindness is important in your life? Tell us about one time you were kind. How did it feel? You can email us at contact at spotlightenglish.com. You can also find us on YouTube, Facebook, Instagram, and Twitter. The writer of this program was Dan Christman. The producer was Michio Ozaki. The voices you heard were from the United Kingdom and the United States. All quotes were adapted for this program and voiced by Spotlight. This program is called Why Being Kind Can Make You Healthy. Visit our website at www.spotlightenglish.com to download our free official app for Android and Apple devices. We hope you can join us again for the next Spotlight program. Goodbye.


I'd like to invite to the stage Hannah James from the ANU College of Physical and Mathematical Sciences. The title of Hannah's presentation tonight is You Are Where You Eat. You've heard the phrase, you are what you eat. But I don't care about what you eat. What I'm interested in is where you eat. That is where people of the past ate because that tells us where they lived. Understanding where people lived and where they migrated is central to archaeology. And in my research, I'm looking at human remains from around the world and amazingly using teeth. I can identify where people ate and lived as a child. It all starts with the chemical elements, oxygen and strontium, and in particular the isotopes of these elements. Isotopes are just the different versions of an element which differ only by their weight. And what's cool about oxygen and strontium is different rainfall patterns and bedrock geology in a region leads to different amounts of each isotope. So regions have their own unique elemental signatures. The plants, animals and people in that region adopt this elemental signature from their food and water. So by comparing the value in a body tissue to a regional signature, we can see if someone is a local. My body tissue of choice is teeth. Teeth form in childhood and then never change. So they've locked in the signature of what you ate and where you lived as a child. If I were to jump down to the audience now and sample your tooth, take it back to the lab and analyse it. We could see if you're a Canberra local or if you moved here after childhood. If you moved during childhood, we may even be able to see when by comparing your front teeth around to your wisdom teeth which form at different times. This technique has forensic applications but I'm an archaeologist and I prefer to work with more ancient dead people. But measuring the teeth alone isn't enough. To understand migration, we need to be able to compare teeth to regional signatures which aren't always available. This is where my work comes in. I'm building maps of these regional elemental signatures with samples of rocks, soils and plants. Using chemistry and filters, we just separate out the elements we're interested in and then analyse and map them. I can then compare teeth from archaeological sites to these maps to create a picture of the movement of individuals and the makeup of communities in the past. I'm working on several case studies in the Pacific to explore the first migration of people into a previously uninhabited area. In France, to trace the seasonal migration of people about the landscape. And in the Caribbean, we're looking at a slave cemetery to understand the different generations of people within that population. By combining this new information with more traditional archaeology, we can build up a better picture of our own human past. So in the end, you aren't just what to eat, but more interestingly, you are where you eat.


Ok, donc, on va commencer à nouveau. Donc, j'ai introduit des structures au minimum, et j'ai étaté des résultats en haute théorie. Donc, le but d'aujourd'hui est d'offrir les deux premiers résultats, c'est-à-dire la définition des quotas arithmétiques et la définition des maps de période. Mais avant de faire ça, je veux fixer des détails qui n'ont pas été expliqués aujourd'hui. Alors, je vais commencer avec des remarques et des polarisations, parce qu'ils vont jouer un rôle important aujourd'hui. Donc, des remarques juste pour fixer les notations. Donc, suppose que vous avez une structure de la vz, la structure de la haute théorie de la haute, comme la dernière fois. Et puis, la définition de la polarisation. La polarisation de la vz est de la forme bilinare, on la appelle la formule qz, de la formule vz, de la formule vz, de la formule z, donc de la formule bilinare. Et de la formule minus 1 à la formule n, c'est simétrique. Donc, c'est simétrique si la formule n est même, et si la formule anti-symmétrique est haute, comme ça, vous avez besoin de deux propres, sur la forme de la formation associative. Et c'est cette forme de formation qui va jouer un rôle important dans ce qu'il se passe, comme si vous définissez l'âge de la forme vz, de la forme vz, qui est la forme associative. Donc, de la forme vz, de la forme vz, de la forme vz, de la forme vz, de la forme vz, de la forme vz. Donc, la forme vz est complexe par la forme bilinare. Et ce n'est pas la forme de formation que vous avez, c'est la forme de la forme vz, de la forme vz, de la forme vz. Donc, c'est la multiplication par la forme vz, de la forme vz, de la forme vz. Et donc, c'est cette âge qui va jouer un rôle important pour comprendre les sets de la forme vz. Donc, l'âge est appelée d'une norme originale associative à la polarization. Ok? Et donc, c'est la forme d'un mot, de la forme vz, de la forme vz, de la forme vz, de la forme vz, de la forme vz, de la forme vz. Et donc, on a une partie de l'âge qui se tient. 0 to integral part of n over 2 of li and then the primitive homologi n degree n minus 2i of xq, where this primitive homologi is defined by taking the kernel 1 degree higher. So, where the primitive homologi n degree j, you can even define it over z, is the kernel of lr minus j plus 1 from ej xz to h to r minus j plus 1 plus 2, sorry, of xz. So, this is defined only up to the middle. Yes, sure. So, in all this, you have to do up to the middle. Yeah, you have to say that some spaces are automatically 0. Not by the definition, but I fix it. I'm not writing it. You're right. OK. And so, how do you get your polarization? Well, the claim is that on each of this pj, there is a canonical polarization, more of a qjz. So, the intersection of product gives you a polarization on this primitive homologi. So, to alpha-veta, you associate the integral part. There is a sign, which is always a pain, j, j minus 1 over 2. And then, there is this integral over x of this. Then, the claim is that this is a polarization of these abstractions. So, yeah, I should have mentioned this decomposition is in the category of hard structures. So, this is a z polarization of the hard structure pj xz. And then, well, you construct a polarization on this by taking an appropriate rational polarization of taking a sum of those qjz on those spaces, twisted by a sign, and then you multiply by an integer to get an integral polarization. Maybe I do not want to write everything here. Just take sum. It's very classical, but I just wanted to remind you what happens. OK. And then, of course, now, you extend this definition of polarization to families. So, suppose now that you have a variation of a structure. So, let vz be nabla f with the notation as yesterday. So, this is your local system of a z. This is associated with the holomorphic bundle, your flat connection, and the odd filtration, be the variation of a structure on a smooth quasi-projective variety s of a c. Then, a polarization for this zvhs is a morphism of local system q vz to zs analytic such that, at each point, you have a polarization. This is a polarized z structure. So, now, I've recalled all these standard definitions in hot story. So, I also want to make some remarks about the pure domains and the pure maps that I defined yesterday. And then, we'll move to the proof of theorem one and two. So, some remarks. OK. So, this is a difficult question. Of course, geometrically, you can get torsion, but then, in some sense, you kill the torsion. It's not very important for what I'm saying, but it varies depending on the references. You can admit torsion or not. It's not very important for what I'm doing. So, yes, apparently, you can. Some remarks on pure domains and pure maps. So, suppose you start with a polarisable deviation of a structure. So, this means that there exists a polarization. I do not necessarily fix it. At some point, I will fix it when I need it. We know that for any s in sn, the structure is given by this morphism. So, I'm moving to the letter 5, because h now will be this hodge norm. Yesterday, I used notation h, but it was a mistake. So, I cut it now. So, we know that you can interpret the fact that the fiber is a hot structure just by associating this morphism from this deling torsion. So, this guy is just c star seen as a real algebraic group, and then you have this morphism here that gives you this hot decomposition. And then, as I explained yesterday, you can define the associated memphotate group, gs. So, this is a memphotate group of phi s. So, by definition, this is phi s of s. Then, you take zarski closure of q, and you know that this group detects all hot stances in this hot structure. So, this is a subgroup in gl of your fiber, and what is easy to show, as proven by Dolin, is that this association s gives gs is a local constant outside the omega set, which is exactly the argilocus that I defined yesterday. It appears, at the beginning of the theory, you just know that this is a omega set in sn. And then, at the end of the story, you know that this is a countable union of algebraic subvarities. So, this means that this gs is directly a local constant. So, you define g as being the generic memphotate group of your variation, and its cram equipped with... So, you fixed a hot generic point, and it gives you, at that point, you get that morphism. Okay? So, now, yesterday, I explained, what was I saying? Ah, yeah. So, the pure domain, one interesting way of defining your pure domain directly using this generic memphotate group is just to say that you define it as being the g of r, conjugacy class of phi. And because I want to work... So, in general, this is not necessarily connected, so I fixed a connected component. And you see that it's naturally mapped to the g of c, conjugacy class of phi. And this guy is exactly what I denoted d hat yesterday. Okay? And what you get also is that, now, where is the polarizability useful? So, the polarizability correspond to the fact that if you look at the action, the adjoint action of phi of i, then this is the carton involution on the adjoint group. So, this is where you use the polarizability. And then, it's easy to show that this d is really a g of r. If you want g adjoint r mod m, where m is a compact subgroup, and that, in fact, this natural map is an open immersion in a flag variety, which is your d hat, d check, which is, in fact, g of c mod some parabolic subgroup. Okay? So, the same construction as for shimoir variety here, but it gives you the same. Yeah, you're right. You can take the g of c conjugacy class of just the character phi of z1 in gc, but it does not change anything.  You map it to, well, you have phi s tilde, which is conjugate to your initial phi. And so, it descends to sn to this variety s gamma gm, which is your quotient by g of z. So, gamma basically is an arithmetic subgroup. So, you can see it here. Okay? So, this I explained yesterday, what I want to say a bit about is the transversality condition is actually an important role, not today, but next week. Ah, le visor first. So, what is this transversality condition? I remind you that if you have a variation of structures and you have this graphing condition that the flat connection applied to the filtration shifted by one. So, how do you see that on the period map? Well, you have this generic mm40 group, and let's look at it at a complexified level. And this induces a hot filtration on the l'algebra, right, because you just composite with the adjoint action. Okay? So, this morphism, which is, of course, in fact, defined over r, give you hot structure on the associated l'algebra. So, you get this hot filtration there. And so, now, if you want to interpret the tangent bundle to the check, then you see that this is a GC equivalent vector bundle with fiber, which is nothing else, of course, than the l'algebra GC, modulo, the l'algebra of the parabolic, which is, in fact, the F0 GC for this hot filtration. And so now, the hot structure is of weight zero on GC, so it has a degree everywhere. So, let me rather write this as being F minus infinity. So, this is the full space, if you want. And then, what is the horizontal tangent bundle? Well, you see that the filtration on GC induces the filtration on this guy, which is equivalent under the action of GC. This is a GC module, so. And so, you get the filtration on the tangent bundle, and the horizontal one is just the first piece.  And then, the easy limit to check is that the Griffith transversality condition, which is a condition that nabla F dot is contained in F dot minus 1 tensor omega 1 S n is equivalent to the fact that when you look at the period map phi from S to S gamma GM then this map is horizontal meaning by this that phi S lower star of the tangent bundle to S is mapped to T H S gamma GM. So, just a word I define here the horizontal tangent bundle on this dual to the period domain. Of course, it restricts to the period domain. Here, this is GC invariant so on the period domain it becomes GR invariant and so it passes to the quotient and so it defines an horizontal tangent bundle on those varieties. F0 GC is the algebra of the corresponding parabolic. Let's write this one. This was the remark of the manual. Ok, so you get this and why is this very important? Well, for people familiar with Shimura variety and Hermitian symmetric space they know that from the complex point of view those manifolds have negative curvature. Ok? This is completely wrong for those spaces S gamma GM. The reason being that they have vertical fibers which are kind of flag varieties of smaller dimension. So they are positively curved. But this result tells you that you are in a very special direction of that space and in that direction basically, more or less speaking, you have all the negative curvature properties that you have for usual Shimura varieties and this will play a big role next week. In fact already today, but we won't see it in that form. Ok, so now I want to make a comment about what I said yesterday about the algebraicity of those spaces that they are almost never algebraic. So of course they are complex analytic because they are coefficient of complex analytic stuff by a proper discontinuous action. But so first a lemma which is to say that the horizontal tangent you can ask when is it true that the horizontal tangent bundle is the full tangent bundle, ok? And essentially this is true if only if D is Hermitian symmetric domain in which case I will say that the variation is of Shimura type. So S gamma GM will be a Shimura variety in particular this will be algebraic and we say that so this is for next week too V is of Shimura type and in some sense this is really exceptional this corresponds only to a billion motifs. Ok The corollary of this I will not write it but I mention it is that if you ask yourself which hot structures, polarized hot structures are geometric right? If you are in weight 1 it happens that this Shimura variety will always parameterize geometric things because you can always associate an obedient variety to your polarized weight 1 hot structure but in higher weight this condition of horizontality tells you that most polarized hot structures are not geometric So So the next remark is that we can now answer when is it true that S gamma GM is algebraic but when is S gamma GM algebraic So as I said M is a compact subgroup and in fact it's contained in a unique maximal compact so you can look at the projection of your D which is GMOD N associated GMOD K which is symmetric space where M is contained in K and K is maximal compact in G So this is a vibration just notice that in general this guy has no complex structure there is no reason why there should be but the fibres are complex of manifold so fibres are of the form K mod M so there are quotient of two compact groups and it happens that they are holomorphic homogenous varieties so essentially flag varieties then we'll say that D is classical this is the termination of griffes that I don't like too much because in some sense they are classical but they are also really the minority so D is classical if D is classical to this to this symmetric space is holomorphic or anti-holomorphic on to Hermitian symmetric space so basically D is classical it's a bit of an enrichment in the case where this map is an equality and D is already Hermitian symmetric yes exactly so this is what I'm saying D is classical in the very rare case this J mod K admite a complex structure and so it is Hermitian symmetric and more of this projection is holomorphic or anti-holomorphic exactly so if you want ok I don't want to confuse you so we can do this that's correct and you don't assume anything about irreducibility like if you have a product of many factors no when I will say classical this means that all the factors are classical in that sense it is equivalent to your definition you have all these decomposition theorem for symmetric spaces that you can decompose according to the factors of the group and I'm really asking that this is true for all factors and this is equivalent to this then the theorem is that basically if you are not in the chimera case or in some sense in this enrich chimera case then the quotient is never algebraic well to do it simple now let's assume for simplicity that G is simple so suppose and D non classical yes in the paper they assume it is simple but you can ok then S gamma GM is not algebraic so this result tells you that in the vast majority of variations of structures then you are not algebraic so maybe I will try to give rough idea how it works because I think this is of interest even though I will not use it at all but it's interesting if you have never seen it so I mean this is an entire paper and I want to spend much time but I will give a rough idea so so what is the proof so now suppose we are in this setting suppose D non classical and G joint simple then what happens is that you look at those fibres the fibres of P are homogeneous projective varieties and in this setting you show that you can deform them so when D D non classical so this is the deformation argument implies that one can deform these fibres to obtain compact homogeneous projective varieties compact let's say compact flag varieties in D not containing the fibres and not only this you can deform but in fact you really have a lot of them in fact there are enough of them to connect any two points by a chain like any two points D by a chain of such flags so you have your D mod M you pick any two points then you will find a chain a finite chain of such flag varieties connected to two points in particular this implies that S gamma GM is rationally connected because those flag varieties contain a lot of rational curves they are themselves rationally connected now basically the idea and this is where I am not technically precise at all but the idea is that if S gamma GM where algebraic then this rationally connectedness implies that the pi1 is trivial this is morally the idea but the pi1 is gamma which is infinite no this is always the case because this G mod M is always simply connected gamma is an arithmetic group in a non compact really group so it's infinite it's a lattice no then the variation is trivial it just means that the variation is trivial there is no variation of structure if the what do you mean compact compact quotient or where G mod M compact if G mod M is compact there is nothing I mean the mem forte group the generic mem forte group cannot conforme which is compact otherwise everything is trivial there is no variation yeah yeah I'm always considering those things coming from the variation otherwise I have to distinguish the case compact and non compact of course in the compact case you are just in the case of usual projective homogeneous varieties and in the non compact case you are in those partial open flag varieties notice in the paper they have something to get rid of of course I mean all those groups are linear so you have to get rid of pass to finite index to get rid of the torsion and so on but here I'm going fast ok so this is a rough idea of course there are a lot of inaccuracies in what I said and you have to be careful about the technical details but I think this is really the idea of the proof ok so these are the remarks that I wanted to make in Israeli that if you think in terms of variation of structures then you are really in a non-algebraic situation generically ok so let's go to what I mentioned yesterday so 3 definitions of arithmetic quotients of s gamma gm so I remind you the theorem 1 that I mentioned yesterday so due to take a in myself there are 2 parts 1st is that this s gamma gm has a natural structure of RL manifold ok so this means that you can I remind you this means that you can find a finite atlas of charts for this guy such that the charts are semi-algebraic and the change of coordinates are semi-algebraic and the naturality will be explained in the proof and second and while it's part of the naturality but is that this is factorial for morphism of such arithmetic quotients so any morphism s gamma prime g prime m prime to s gamma gm of arithmetic quotients is RL definable ok so basically what is a morphism of arithmetic quotient explained yesterday this means that it comes from a morphism of algebraic groups from g prime to g mapping gamma prime to gamma and mapping m prime to a conjugate of m by rational element ok and then this induces such a map ok so let's try now to give the idea of the proof of this so I will deal essentially with one first so the first remark so those questions s gamma gm is gamma back g mod m so g is g of R so this is a semi-algebraic set and then this is a classical result that if you take the quotient by a compact group of something semi-algebraic then it still has a semi-algebraic structure so as m is compact the semi-algebraic structure on g which is g of R gives rise to a unique semi-algebraic structure on g mod m making the natural projection g to g mod m semi-algebraic so I say this is classical it has been known for a long time by people working in group theory there is a purely algebraic proof there is also a proof from a minimality point of view in terms of definable quotients for a proper for a proper closed relation so I will not insist on this what is interesting is of course the other quotient because gamma now is infinite so it's not a stupid thing taking quotient by an infinite group it is forbidden for minimal reasons so this is the part which is not completely trivial so what is the idea the idea is that if x is a definable space so here definable in some minimal structure then the definable geometric quotient x mod r exist if r in x times x is a closed definable et tal equivalent solution so what do I mean by this quotient this means that this quotient will be y y is definable the fibres are exactly the r equivalence class and y as a quotient topology yes no no no no you want closed y as a quotient topology and here the important remark is that because you are assuming that this is definable et tal means finite et tal because you have r in x times x the first projections map this definable guy to this definable guy and has to be a definable map so the fibres have to be finite et tal is really finite et tal it is automatic because the fibres are a discrete set which is definable but this is what I explained yesterday is that I should have mentioned this but this definable setting the continuity is uniformly bounded but not constant it's constant and connected stuff but why you cannot have special points where so you have an et tal equivalent selection yes and like in a dubai geometry you can have an et tal equivalent selection yes and the et tal mode is of course finite et tal after stratified yes ok, when I say finite ok, you mean finite et tal after stratification this is what you mean yes, that's correct I completely agree I was not saying that ok so this is an easy exercise in fact so of course we cannot apply this result here because gamma is infinite so as gamma is infinite we cannot apply to the equivalence relation r gamma inside g mod m cross g mod m right where this equivalence relation given by x, y if as exist gamma gamma such that x is gamma y ok but the idea is to replace g mod m by a sufficiently nice finite et tal set and then I can do this so we look for a finite et tal set a semi algebraic finite et tal set f for gamma and then we apply to f and then you apply this equivalence solution that will be et tal ok so the idea is to construct such canonical fundamental set and such things has been existing forever namely since the 60s this is a theory of Ziegels sets so let me explain this because this is a crucial point in the paper so f will be a finite union of Ziegels sets exist so of course if s gamma gm is itself compact then there is nothing to do with just take any semi algebraic finite et tal set and it works but the problem are the casps so let me try to recall briefly the definition of those Ziegels sets and we really need it so I'm sorry to impose you this but this is really needed to prove that this is the way you prove that the map itself is definable you will use those Ziegels sets so it's a bit heavy but there is no choice so what is the intuition the intuition is that the casps are essentially parametrized by gamma conjugues classes of parbolics so now recall few facts so the casps of s gamma gm in one to one correspondence with the gamma conjugues classes of q parbolics p of g and the basic result of Borel is that this is a finite set so this is a finite set and I fix representatives that I will call p1 ps yeah proper as yesterday ok the way of going to infinity I'm not very precise here I'm just saying that canonically to that space you le shimura case it wouldn't be very good it would be Borelser yeah it would be Borelser I will explain ok so now I fix one of those parbolics so let p a parbolics subgroup and then basically you will associate to a pair of parbolics and a maximal compact subgroup d'un set in the group G and then a Ziegelset in G mod m will just be the projection in G mod m of a Ziegelset in G so how does it work well the first thing you do is that you write the canonical decomposition levy decomposition in a two reductive part and a unipotent part so this is a levy decomposition so recall that this guy is a unipotent radical of p so it's canonical this guy is a quotient p by np so this semi direct product is not canonical it's only canonical as a quotient but now I will use a compact subgroup so that at the level of our points I get a canonical lift so this is what I want to do in the next step but first I continue writing a bit of this so you write ok so you can write it as being ap np times np we are now ap is a connected component of split center of lp ok so now you get associated to this and the choice of a maximal compact so now if you have p and k a maximal compact then you get really the composition of G plus de points as being apk times npk times npk so this is your parabolic decomposition of G so I just have to explain what it is so this guy is a unique real levy of pr stable lifting lp and stable under the carton involution divided by k so it's a bit technical so now what is the Ziegels set associated with those data so the first remark is that this apk it is really the same thing as r larger than 0 to the power certain number the cardinal idea of the set of simple roots of ap acting on np the roots of the root system of root system phi apnp so it's a nice linear space and the notation will be that to a associate a to the power m minus alpha 1 a to the power minus alpha of r r is this cardinality so these are the simple roots ok so with those notation what is the Ziegels set so you want to describe a very particular set using this decomposition so a Ziegels set of G associated with pk is a guy sigma of the form u cross apk t so basically you want to look only how to go to infinity in the direction of this torus so times w inside this is the composition np apk and then w will be in mpkk ok and what are the properties u and w are relatively compact and semi algebraic open semi algebraic this makes sense in this uniponent group and in that group and apk t and apk t so t is a real number and you are just taking all the guys in apk such that a to the power alpha is larger than t ok so if you are in SL2 or I will go to the quotient so this was for G and now for G mod m so a Ziegels set is the image and G goes to G mod m of a Ziegels set of G associated with some power p and some k containing m so that you don't hear the action on the right is nice ok so this is the definition and of course you have the standard picture that you know for SL2 as the example if you take so SL2 ok so 2 so this is the usual power plane h so you are in this situation so now what is a Ziegels set well basically a Ziegels set by for instance you fix the real part between 0 and 1 and then you ask that y is larger than some t and with the right choice of normalization this t is really the t appearing here right there is only one root because you are in rank 1 and you are asking that you go to infinity so this will be of this type ok hmm so so what are the theorems that we use the first one is a result of Borel so it says that there exist finitely many Ziegels sets sigma i associated with p i so p i is this finite collection of gamma conjugacy glasses and some k i whose image covers s gamma gm and so you take the union of those finitely many and this gives you a fundamental set that you want to consider and then you take the etal quotient relation of those and the second point of Borel theorem is that this is fine you can do that because for any sigma 1 sigma 2 Ziegels sets then the number of gamma pushing one to the other is finite different from is finite you can make them bigger enough to cover everything sorry yes I said that I was not in the case where s gamma gm is compact because then everything is trivial ok ok so this is the first result so this means that you can do the procedure that I said because of this property the equivalence relation now which is defined is a finite et tale in the sense that I mentioned and so you can take this quotient and this gives you the structure now what about the naturality well this is given by or so 2 so this is in fact what is strange is that this is a recent result which was not written at the time where this theory was developed is that the image of any Ziegels set so let f from g prime to g morphism of group kind of a Q the image of any Ziegels set of g prime by morphism by this by f is contained so of course this is not necessarily a Ziegels set but this is contained in finitely many Ziegels sets in finitely many sets in finitely translate finitely many translates by element of g of q ok and ok and so basically the theorem which is here from 1 and 2 here in respective order so I am not giving the full details but I think at least this is believable I would just like to make a remark I was precise in this way because I wanted to say to have a nice statement in terms of definibility of R alge in fact for what follows you don't really need it it's enough to have definibility and there you could just appeal to the classical Borel-Ser theory in fact those Ziegels sets are the basic tool to construct the Borel-Ser compactification now Borel-Ser compactification is a real analytic manifold with corners and so the only thing that I am saying is that the real analytic structure on s gamma gm expanding it's R alge structure is just the one given on the complement of the boundary of the manifold with corners so as you see basically this result is essentially completely classical what is not completely trivial is this but this is Martin's result alright so now a very important thing for me that I want to keep is that you see those Ziegels sets are in some sense nice from a group theoretic point of view but it's kind of complicated to make computations with them so the only thing that I want to really understand is Ziegels sets for SL and R so GL and R so this will be the following example let X be GL VR so V is my fiber of my variational structure and I can consider this symmetric space so this is a symmetric space of positive of positive definite quadratic forms on VR then I fix a basis a basis of VZ and C be larger than 0, a constant and then the definition is the following suppose that you have a point in this symmetric space so this is a quadratic form then I would say that it is reduced with respect to the basis E and the constant C so this is EC reduced if the following inequalities are satisfied so this is classical reduction theory you ask that when you evaluate your quadratic form on EIEI then in absolute values this is smaller than sorry EJ C B of EIEI for all IJ B B is that B of EIEI is smaller than C of B of EJ EJ for all I strictly smaller than J and C that product for I is equal to 1 to R of B of EIEI is smaller than C determinant B so the idea is that you are fixing in terms of a basis and a constant a lot of numerical conditions on a certain set of quadratic forms ok and then so this is the end of the definition then what is the claim qui est vraiment utile c'est que donc c'est le subset de X de B qui satisfait ces propriétés ok alors la première claim c'est que ce set est contenu dans un sigel de X et un sigel est contenu dans un JEC ok donc je n'ai pas vraiment compris le set en autres termes en termes de groupes de théorie mais ce que je sais c'est que purely en termes de formes de qualité ces numériques de condition me donnent une approximation et B qui sera très important c'est que si il y a un B de X qui est E'C' réduit pour des E' et des C' et E est une autre base E est une base pour laquelle condition C est satisfait pour des C oui, oui, oui oui, vous pouvez juste prendre un classique pour minimum un E est une base qui satisfait condition C pour un certain constat alors il existe un autre constat C' dépendant seulement de E seulement de E C E'C' comme que B est aussi E'C' donc ce part B est très important il vous dit que si vous savez que c'est E'C' puis en fait vous pouvez trouver une base si vous avez une base qui satisfait déjà une certaine condition puis à la suite de changer le constat vous pouvez imaginer que vous êtes E'C' donc ça va jouer un rôle important à la fin de la lecture ok donc je veux garder ça donc si je commence à érainer ça, j'arrête ok donc c'est ce que je voulais dire c'est la définabilité donc je veux aller à la définabilité de la map qui est la main partie de cette lecture donc c'est la main resultant de ce paper donc comme j'ai expliqué yesterday c'est reposé un théorème donc si vous commencez avec la variation qui est polarisable dvhs dvhs et S est smooth quasi-projectif c donc je vous remercie le statement si vous regardez la map c'est la map horizontal de la map de cette place classifiée puis c'est Rnx définable donc c'est précis donc j'expliquais qu'est-ce que si vous avez des variétés complexes et que c'est canoniquement définable dvhs ou dvhs ceci nous a juste prouvé que c'est définable dvhs donc la définabilité de Rnx avec respect à l'expansion de ces deux réels structures ici et ici donc c'est ce que je voulais dire donc ce qui est le sens de ceci le sens de ceci est que nous avons un final atlas de chartes ici, selon ces sets de ziggles je serai capable de trouver un final atlas de chartes ici pour que quand j'écoute cette map avec respect à ces deux atlas de chartes j'utilise seulement l'expansion réel de la map et les fonctions de réel de l'anétique respectées à l'impact c'est le sens de ceci donc peut-être je vais faire quelques commentaires que je ne vais pas dire le premier est que ce statement est facile dans le cas qui n'est jamais heureux en nature où ce gars est compact cette question peut être compact ce n'est pas le cas mais ok donc pourquoi est-ce facile ? alors dans le cas où ce gars est compact je claims que vous n'avez pas besoin d'une fonction d'expansion ce n'est pas possible dans notre âme alors comment est-ce l'argument ? c'est que vous commencez avec une chose très proche puis vous vous embellez en S-bar avec un déviseur normal en complémentation Borel Théorème vous dit que l'infinité sur ce déviseur normal est assez unipotent mais si ce gars est compact il n'y a pas de déviseur unipotent donc ça vous dit que le monorhomie est trivial mais ensuite la map de période va s'attendre à S-bar et ensuite vous allez juste obtenir une map complexe de S-bar pour ce compact et donc ceci est definable dans notre âme donc le plus grand de tout c'est que Borel Théorème sur le monorhomie unipotent a déjà obtenu une assumption et c'est bien connu pour une dynamique homogèneuse unipotent ou unipotent c'est beaucoup mieux qu'un élément toric donc c'était le premier remarque le deuxième remarque c'est que supposent vous d'assurer que vous êtes dans le cas d'un schéma de variété donc ce gars est algébrique et ce gars est algébrique et maintenant j'ai construit une map homomorphique qui est definable dans une structure homogèneuse entre deux stuffes algébriques alors je peux appliquer le théorème de Peter Zein et de Starchenko et il vous dit que cette map est automatiquement algébrique qui est un théorème bien connu donc Borel prouve que si vous avez une variété de structure avec un target ou un schéma de variété ou si vous avez d'autres maps homomorphiques d'une variété algébrique d'un schéma de variété c'est automatiquement algébrique ok torsion freinesse ce n'est pas un grand délire dans ce business parce que vous pouvez toujours aller finir dans un subgroup par prendre une étale couverture de votre manifold et vous utilisez l'algebraicité et vous utilisez aussi l'algebraicité non mais ok je suppose ici que je travaille avec ok vous êtes parfaitement bien négatifs donc je pense que si je te le met si je te le met ce n'est pas un grand délire ce n'est pas un grand délire de commencer pour puler la variété de structure ou est-ce qu'il y a un moyen d'formuler la condition que le type est un schéma de certaine type un schéma de gm un peu plus généralement que ce qui est de la variété de structure mais je n'ai pas l'experimentation mais c'est facile non mais la description que je donne si vous avez une variété de structure vous pouvez associer avec une map mais si vous avez une map pareille dans le sens où vous avez une map holomorphique qui est horizontale et localement lifted donc pour prendre soin de ce problème que vous avez mentionné cela correspond à une variété de structure vous appelez la gamme gm c'est la question de la choice de la gamme gm si vous n'avez pas d'autres choix pour lesquels c'est de la variété je pense que c'est assez général je veux dire si vous me donnez une classe intéressante donc ce qui est crucial c'est que la gamme est compact pour ce business pour construire la structure si la gamme n'est pas compact je ne sais rien excusez-moi une autre classe de subgroupes compact et une autre classe de gamme que ces gars sont complexes peut-être que je peux faire quelque chose mais je pense que je ne sais pas d'autres classes intéressantes pour entrer donc parce qu'il y a une question sur la case de compact si ce n'est pas correct oui donc on va voir la provenance ok, donc je vais tout d'abord essayer de décrire la stratégie de la provenance et ensuite aller aux détails techniques donc je vais me demander la dimension de S et donc comme je l'ai dit j'ai choisi une compactification donc une devise que je peux toujours faire et donc la picture de l'infinité c'est le delta donc c'est smooth et localement à un point ici j'ai un delta à la fin et l'ambient est parce que j'ai un déviseur normal c'est le delta star à la r times delta n-r c'est la picture de l'infinité dans les coordinates donc la première chose que nous faisons c'est que nous couvrons la barre de finitiellement ouvrir des sub-sets de cette forme et si vous voulez, vous pouvez ajouter un gros set de compact qui n'aura pas de rôle dans la définitive et donc vous voyez que ce que nous avons réellement réduit c'est que nous sommes réduits pour prouver que la map local j'arrête mon pi à chaque de ces donc j'ai basically un map pierre et j'assume qu'il y a des modes triviales je peux toujours supposer que r est equal à n donc j'ai un map pierre local comme ceci et nous voulons prouver que c'est un x definable r un x definable donc c'est vraiment une question locale donc comment nous faisons ça ? bien, j'utilise Borel theorem pour ceux qui ne sont pas familles je dois mentionner que Borel theorem une façon de prouver Borel theorem est exactement d'utiliser la property négative dans la direction horizontale ce que j'ai dit avant c'était vraiment relevant ici donc Borel vous dit que l'endromie de vz sur delta star à la n est quasi-unipotent pour la finite étale ce n'est pas un problème pour cette question de définibilité sans plus de généralité on peut s'assurer c'est unipotent et donc nous sommes dans la prochaine situation maintenant j'ai fait le même diagramme pour mon map pierre mais tout est local maintenant donc la couche universelle de ce produit de puncture c'est juste un peu de paix et puis j'ai ce map homomorphique horizontal de Twilda et ici c'est l'uniformisation donc le fait crucial est que dans la restriction de Ziegelset cette uniformisation sera définitivement définie parce que cette map est juste exponentielle donc c'est exponentielle 2 pi i dot à la n donc maintenant j'ai donné la picture pour Ziegelset en h donc Ziegelset en h est juste un produit donc let's sigma h comme d'habitude c'est le standard Ziegelset où Z est x plus iy donc maintenant j'ai choisi coordinates où x est entre 0 et 1 et y est plus grand que 1 et puis donc le point est que delta star à la n est couvert par p de sigma h à la n et p de translate de sigma h à la n et donc pour prouver la définitivité c'est suffisamment suffisant pour moi-même pour p de ce type et donc vous voyez que exponentielle 2 pi i dot comme fonction sur sigma h delta star est définitivité parce que c'est x plus iy est map à exponentielle minus 2 pi y et puis vous avez question de 2 pi x signes 2 pi x mais maintenant x est boundé, c'est le point crucial parce que de la choisi de Ziegelset j'ai tué la période donc c'est en rn c'est la restriction de la fonction pour un set compact x entre 0 et 1 et quand je le vois en anglais, i est y c'est mal et ici c'est juste la fonction donc je vois que c'est definable en rx en rx donc si je regarde mon diagramme maintenant je suis en business parce que nous sommes réduits pour prouver donc j'ai mon Ziegelset en h à z n j'ai mon map pi tilde d g mod m projection pi s gamma g m et je voulais prouver que cette map ici delta star p sigma h je voulais prouver que cette map est définie donc c'est suffisant pour prouver que cette map est définie parce que j'ai juste prouvé que cette map est donc c'est suffisant donc vous pouvez l'inforcer c'est suffisant pour prouver que cette projection composée pi tilde est définie donc qu'est-ce que ça veut dire ? c'est impliqué pour prouver que c'est un TORM A peut-être, je vais l'écrire comme ça donc vous êtes dans cette situation local let's phi delta star sigma g m, votre map local puis le claim est qu'il existe vous vous rappelez que la structure de l'air ici vient de la set de ziggler donc il existe finitiellement beaucoup sets ziggler sigma i en g mod m comme ça d'abord et c'est le point important l'image de phi sur sigma h à la n est contenu dans l'union des sets finitiellement ziggler et b si vous regardez phi tilde restricte pour ces sets ziggler dans le produit alors c'est r et x finitiellement pour prouver que cette map est r et x finitiellement c'est suffisant pour prouver que c'est ça mais aussi que ça arrive en finitiellement beaucoup de sets pour celui-ci et maintenant vous voyez que c'est un statement en théorie asymptotique vous voulez comprendre comment votre variation de structure génére quand vous allez à l'infinité et c'est une théorie très développée assez technique, malheureusement donc c'était vous pouvez le demander, je suis sûr qu'il avait beaucoup de plaisir de préparer le talk donc c'était développé dans les années 70 d'abord dans le cas où n est equal à 1 et c'était fait par schmitt et c'est un très bon papier ce qui est surprise c'est que passant au large n est extrêmement difficile donc il a pris schmitt et d'autres personnes plus que 10 ans pour faire ça et le couple et schmitt ont une théorie plus grande plus que 1 donc le premier remarque est que à l'aide du papier schmitt c'est explicitement écrit dans le papier vous avez déjà ce statement pour n est equal à 1 en autres mots, si vous avez une variation d'un single puncture disque alors vous avez ce surprise un bon statement et quand vous avez un fondamental en each c'est mappé vous pouvez trouver un single ziggle set pour lequel vous êtes mappé c'est juste que non, vous pouvez faire ça avant mais ici c'est juste que c'est plus facile si vous vous restez à 0 et 1 plus tard dans la blouse c'est pas donc, la structure du ziggle set sur le source n'est pas très importante vous pouvez choisir un ziggle set j'ai choisi celui-ci parce que c'est convainable mais le statement est vrai pour n et aussi ce que vous savez c'est couvert par un translate c'est pas incroyable oui ok et finitly ce qui est important c'est de finitly finitly many translates je pense que n est equal à 1 ah, je ne suis pas allowed d'arrêter ça donc, ce que je dis n est equal à 1 c'est de finitly et ce, aussi, est de finitly ce qui est difficile c'est ce statement pour n c'est classique c'est impliqué par schmidt n'est pas important pour t1 comment vous avez le temps ok donc, nous savons que nous avons cette monadromie t1 et tn donc, le groupe phanométrique de cette data start to the end c'est z to the end so let t1, tn in d of z be as a unipotent monadromies right, I have reduced myself to the case where the monadromie is unipotent not quasi unipotent and then, this means that I can write each ti as being exponential of ni where ni is anipotent element in g the d algebra and because the final of the group is commutative the ni nj commut ok then schmidt remarks what is anipotent orbit theorem schmidt noticed that in fact it was noticed before by Griffith but the theorem was really proved by schmidt is that you can twist the period map so consider the following twist psi from each end so now we will not go to d but to the dual flag variety which consists in twisting or d twisting if you want your phi ok then now this function is monadromie invariant right, if you are adding one to each of those variables then you perturb by exponential minus zj which is tj minus 1 and here you have a tj getting out so this becomes monadromie invariant so this means that it descends to a map psi from delta star to the n to t-check so understanding the asymptotic of phi is the same thing as understanding the asymptotic of this simple function here and then what is schmidt theorem schmidt theorem tells you that this function extends so schmidt theorem psi extends homomorphically to delta n so the value f infinity which is by definition psi of 0 so this is an element in d-check usually this is not in d is called limiting hot filtration and it tells you that out of this limiting object you can construct a new more or less trivial period map which is an important orbit that approximates the original one moreover exponential sum of the zj nj applied to this f infinity lies in d for z large and is asymptotic is asymptotic to psi so the picture is the following you have d inside d hat so this is an open analytic you construct by this twisting you construct some element f infinity then you have a very homogeneous map this is just given so this is called a nilpotent orbit this is this nilpotent orbit and it's going there and you know that you have your original period map and you know that those things become very fast asymptotic so I will not write the asymptotic here but this is what happened so this is phi of z and this is exponential so here let's say you have a point here and very close you have a point of the form zj nj f infinity so they converge very fast so in some sense this is a simple algebraic model of what the period map is so in geometric terms this corresponds to the statement that I gave yesterday that the heart filtration on your vector bundle v extends to the delin canonical extension as a sub bundle ok so this tells you that phi tilde of z1 dn now becomes something of the form exponential sum of zj nj times psi of the projection of z1 dn in the polydisc so now psi is a lot more thick so and in rn because now this works really working on the polydisc so this is compact and I know that this map is definable rnx definable on sigma hn and this one looks bad because this is an exponential a complex exponential but the nj are nilpotent so this is a polynomial so this tells you that this map is definable and so this is your statement b ok so now we are restricted to prove a and as I said a for n is equal to 1 was proven by schmidt using a refinement but the very deep point of the nilpotentorbit theorem thinks that once you have the nilpotentorbit you can ever approximate it better by an sl2 orbit so you can totally embed totally geodesically embed some age that will also approximate in this picture there will be a third orbit even more regular because it will come from an sl2 coming from an algebraic morphine from sl2 to g ok but the point is that and this is a statement that took 10 years to be proven in multivariables case so alright so we don't want to do this we don't want to use the full power of the sl2 to the n orbit theorem in many variables so we will use only so the key word is that we use only part of what is in that paper and which was in fact proven before by kashiwa namely some norm estimates so we use only norm estimates which are implied by the usual sl2 orbit theorem in one variable so let me make a philosophical comment here I mean really I just erase a that's too bad so you want to prove that this of delta of sigma h to the n lies in finitely many Ziegels sets basically what schmidt tells you is that if you start with any delta star embedded in your delta star to the n then the image will for the corresponding Ziegels set will lie in finitely many and the idea is that of course you can cover this sigma star to the n by a lot of sigma star and you know that in restriction to each of them you are mapped to finitely many so in some sense what you are looking at is really a uniformity statement because it's not clear that all of them will lie in finitely a finite collection the same finite collection ok so and so this is where we die I don't think I will have time to prove everything today may I don't want to rush too much so maybe I will give more some details and try to finish finish this next time so this is really degeneration of hot structures so there is some mixed theory so I recall mixed hot structures on VZ is an increasing filtration on VQ which is called the weight filtration and such that the grue the grue pieces for this weight filtration is a rational pure hot structure of weight L so this is a standard definition so an important fact due to the lean so you see that mixed hot structures are bifiltered object but the remark of the lean is that but bi-graduation are good and the remark is that the lean says this is equivalent to a bi-graduation so it looks like pure hot theory such that WL is the sum for R plus S smaller than L of I RS maybe I will continue there so this is the first condition the hot filtration in terms of this bi-graduation yeah I'm coming yes you mean ah here yeah sure ok so you have the filtration F on VC such that the induced filtration on those grue defines a rational pure structure thanks Fp is a sum for R larger than P of I RS and the condition now that means that this is not pure but mixed is that if you take the conjugate of that space then this is essentially the same ISR except that now this is a mod plus IAB for A smaller than R and B smaller than S I hope I'm not making a mistake about the indices here ok so when you don't have this condition this is pure but otherwise this is mixed ah and so now when so let me write this when you have really the equality then this is an equality now over R so we say that VC F is R split instead this is an important fact that when you study mixed structures usually you try first to reduce to the R split ones ah yeah this is uniquely so and in some sense this is a way you prove that this is a nabillion category ah so a remark of the lane is that if you start with any mixed structures then you can associate to it a canonically a split one a real split one so of course a mixed structure is not a direct sum of pure but if you extend the scalars to R then essentially it becomes in a can I mean you can deform it to a one which is where so you twist your hard filtration by some automorphism which is complex only where delta in fact you choose it in the sum a b negative ok so now this is a painful part where you have to make the degeneration or maybe I want to go you have no choice right you know what I will not make the degeneration now I will try I will try to give the heart of the argument because if I go to the degeneration anyway everybody will have forgotten next week so so forget about mixed structures and let's go directly even I so I don't want to do it 2 times so ok so so let me recall you what we want to prove we want to prove that if I start if I start from that guy then I can map it to finitely many seagull sets in gmod M so the first thing that I do is that I embed I use a fact that I have a canonical representation in GLVQ and this induces an embedding of D inside the associated Riemannian symmetric space and this map if you think about it is just to a point Z you associate the hodge norm at that point ok so the hodge norm where A is the hodge norm corresponding to the hodge structure Z ok so now what is the argument the argument is that you want to prove that to show that your sigma H to the N is mapped to finitely many seagull sets there I claim this is enough to prove the same statement in X and this comes from the second part of the TOM that tells you that the prime age of any seagull set is contained in finitely many seagull sets here so as yota-1 of seagull set is contained in finitely many in X finitely many seagull sets in D we are reduced to showing that to show that yota composed with phi tilde on sigma H to the N is contained in finitely many seagull sets of X so here you see something strange happen it's not really clean, I don't like this proof you get out of the pure domain you just look at the norm because we don't understand enough seagull sets in G mod M ok so now the trick is that asymptotic hodge theory does not tell you what is the situation that generate on this but only on very special sectors but those sectors will cover ultimately that space so let me define where is it let me define let sigma N so this is where you really use the coordinates you choose an order on the coordinates so let sigma N is a set of Z1 ZN in HN such that the XIs are 4N1 and Y1 is strictly larger than than YN larger than 1 ok so these are sectors and you will notice that if you use all the permutations of the coordinates then you will cover sigma H to the N ok so what you already use to prove so what we want to show you want to show the theorem is that there exist basis E of VQ and C larger than 0 such that for all Z in sigma N and then HZ is EC reduced so if you prove this and if you remember the first fact you exactly that YOTA composed with phi of Z which is this HZ is contained in a Ziegelset of X so this is the analytic statement that you already just to prove and then using all permutations of the viable you will get what you want ok alright so how do we do that so this theorem implies everything and you see this is just a statement about the Hodge norm so we don't use the full power of SL2 abit theorem and here this is kind of tricky so the idea is that we didn't know how to make work by restricting to curve ultimately here it will work so it will be an argument arguing only on curves and using this so the proof is as follows so let theta be the ring and this is where you see that you need something more than pure hominimality you need a rate of growth at some point so and you take the ring of functions on sigma N obtained by pullback via P from HN to delta star to the N of real analytic function functions on delta N so what I'm saying is that you are allowed to use on sigma N any real analytic function coming from this this we know because we know that this psi will extend so then I will consider the following ring so this is the ring of polynomials in the variable X1 XN so now we do real analysis Y1 YN and then the inverse Y1-1 YN-1 with coefficient in O and now we call OXY its fraction field and then the claim is that I will restrict myself to very nice functions so which are the following so F in this fraction field is called roughly monomial F F is equivalent to Y1 to the S1 YN to the SN for some integers SI and it will be called roughly polynomial if it is of the form G mod H so this is a completely ad hoc definition but it works where G is in O is in the ring and H is roughly monomial so what is the use of those things why do we introduce those functions well the point is that those functions have the good idea that you can test the rough polynomiality on curves so the lemma is that ah, I should give you maybe the precise definition yes, the equivalence is where F is say to be smaller than G if there exist a constant such that there exist three positives of that F is smaller than CG and F is equivalent to G F is strictly smaller than G and G is strictly smaller than F well it's not strictly smaller much smaller and much smaller this is a usual kind of equivalence ah so the claim that makes a proof of work is that if F is roughly polynomial G is roughly monomial then the fact that F in absolute value is smaller than G on sigma N can be tested on algebraic curves can be tested it is enough for this to be true for each curves by restricting two curves tau of the form alpha 1 fixe un certain nombre de variables et t'as juste une bonne équalité t'as juste une bonne plane et puis pour les restes variables t'as juste une constante pour alpha i plus large que q positive et beta i en R donc c'est une claim que je ne veux pas prouver ok mais c'est bien c'est la stratégie qu'on ne sait pas comment faire le travail géométriquement c'est le travail purement en termes de fonctionner donc maintenant qu'on va réduire bien on veut comprendre quelle est la monoméalité ou la polynomialité de cette hz donc la proposition la proposition c'est que pour une uv en vc la fonction z donne hz de uv donc c'est la fonction en oxy où z est x plus iy et c'est et c'est quasiment un polynomial sur sigma n et c'est la première partie pour le diagonale donc si vous regardez à z uu j'ai juste écrit ceci ceci est quasiment la monoméalité pour quelques constants z n 0 plus 1 c'est zeta et 0 plus 1 c'est juste un constant et cette proposition qui s'accueille de la symptomatisation décrit par Kashiwara ok je vais prendre quelques minutes pour conclure la preuve comment conclurez-vous donc maintenant vous voulez finir la preuve de cette théorème qu'il existe sur la base et ici je ne peux pas le faire parce que j'ai besoin de la décomposition de la symptomatisation je pense que je vais stopper cette production parce que sinon ça va prendre toujours et puis la prochaine fois je vais expliquer comment vous prouvez cette proposition en utilisant la symptomatisation je ne vais pas prendre trop parce que j'ai beaucoup d'autres choses à dire mais toujours donc je pense que c'est mieux de stopper ici pour aujourd'hui et ceux qui sont intéressés dans les détails vous devez regarder la vidéo de la prochaine semaine ok ok questions j'ai une question, pourquoi c'est juste parce que j'ai compris les sets de ziggles mais pas vraiment en D mais vous devez dire qu'on a réduit les sets de ziggles oui, vous avez réduit pour montrer que c'est cette composition de Yota juste parce que si je sais que j'ai infinitely beaucoup de sets de ziggles ici je sais que la preuve de ziggles ici est contente en infinitely beaucoup de sets de ziggles alors j'ai prouvé la finabilité ici vous n'avez pas prouvé la finabilité vous devez prouvoir la finabilité de la poignée non, cette finabilité est définie c'est totalement géodésique c'est juste la finabilité d'une chose ici dans la direction originale ok c'est de ces statements que j'ai dit que si vous avez un morphisme de groupes algériques sur une map alors je ne sais pas si c'est un embêtement d'indiquer ces morphismes que c'est définitif c'est déjà évident c'est complètement semi-algebraque vous regardez la définition et vous regardez l'autonomie correspondant à ça c'est un statement le vrai statement c'est que le morphisme de Sgamma Gm c'est de l'espace localisé c'est définitif c'est plus fort