It's a core mission, an ongoing core mission of universities to work with young people who are exploring and discovering their talents, who become aware of who they want to be, where they want to focus on in life, and that it's still our core mission to engage with them, also in questions of sense and meaning, and that in this regard education via Wissenschaft remains a tenet. In this sense, it is not student-centered versus teacher-centered, but it continues to be both. I fully agree with Joe that the future is neither fully digital nor fully back-to-business as usual. But nevertheless, we have a certain danger maybe of falling back to our default mode, and this is why it's crucial to rethink our teaching mission and to balance these tensions between disruption and acceleration, but also growth and development and thus deceleration and contemplation, so that we need to carefully consider which formats we employ in which circumstances.


All right, so over here we have some detergent, liquid detergent. This is lemon juice. This is vinegar. This is some oil. This is normal water and this is salty water. And we have red and blue litmus paper for each spoon. So what we will do is we will let each litmus paper come in contact with the substances over here and we will observe the color change if there is any to the litmus paper. All right, so let's begin. So let's start with the red litmus paper and bring it in contact with the liquid detergent. Red litmus paper turns blue like this. So this means that liquid detergent is a base. A base can turn a red litmus paper blue. Okay, now let's bring the blue litmus paper in contact with the detergent. So we see there is no color change really. That means when blue litmus papers when they come in contact with bases, they do not show any color change. Okay, now let's move on to the lemon juice. So I have the red litmus paper with me. Again, there is no, there is no color change. Which, which tells us that lemon juice might be, might be acidic. Let's see what happens when, when we take in the blue litmus paper. So did you see that? The blue litmus paper turned red. Which tells us that lemon juice is slightly acidic. It is acidic in nature. Okay, now this is vinegar, two papers. But as you can see, nothing's happening to the red litmus paper, right? When it comes in contact with the vinegar. Let's take the blue one now. And there you go, it turns red. Which tells us that even vinegar is slightly acidic. Okay, now this is oil. Again, no color change for red litmus paper. And also no color change for blue litmus paper. No change in the color of the red litmus paper tells us it could either be acidic or neutral. But there was no change also in the blue litmus paper. So that means this, this substance is neutral. It is neither acidic nor basic. A substance will be neutral if there is no change in color for both red and blue litmus papers. Okay, take a prediction of what do you expect to see when litmus papers come in contact with normal water and salty water. All right, let's see. So no color change with red and no color change with blue litmus paper. That means that water is neutral. Again, no color change with red litmus paper for salty water. And still for blue litmus paper, we do not see any, any changing color whatsoever. Even for salty water. This shows that even salty water is neutral in nature. All right, so litmus papers act as indicators and it can tell whether a substance is acidic or basic by changing its color. So if a blue litmus paper turns red when it comes in contact with a substance, then that substance must be acidic. If a red litmus paper turns blue when it comes in contact with a substance, then that is base.


Welcome back to Tennis Talk, my name's Ken Williams and Simone Hallop and Darren Cahill's coaching relationship has come to an end after six years of being together as a player and a coach, and with great success as well, they've come to an end. This is what Simone Hallop had to write on Twitter only about an hour ago. So over on Twitter, this is what Hallop had to say. After six wonderful years working together, Darren Cahill and I decided that it's time to end our working relationship. Thank you D for everything, for making me a better tennis player and a better person. So a short and sweet message there. There's no real word on why they're split up as a coach and a player. Darren Cahill obviously has a lot of media obligations. Maybe that had something to do with it. And actually Simone Hallop just got married as well.  Who knows? We can always just speculate until we hear word from Cahill or hear word from Simone Hallop on a little bit more information. But during those six years together, they won two grand slams, one at Wimbledon and one at the French Open. Of course, making the world number one ranking as well over that time. So Cahill did a lot for Simone Hallop's career and where to from now? Because remember Simone Hallop has dropped down out of the top 10. She was in the top 10 at the start of the year. She's out of the top 10 at the moment. And it's kind of strange that they would part ways midway through the season. I know we're sort of on the back end of the second half of the season. But still, most of these coaching changes happen at the end of a season. So we don't know why this has happened. But let me know down in the comments below. Do you think this is gonna affect Hallop's game? What do you think? She's fine without a coach. She's got enough experience. She doesn't need a coach to get her to another Grand Slam because she has got two at the moment. She has played well at the Australian Open in the past and maybe looking to get a third one in January if we do get the Australian Open. Let me know down in the comments below. How's this gonna affect our matches in the future? So Simone Hallop, Darren Cahill, they've split as player and coach.


Yeah, it's like nookin' a cow. Dirty, dirty. Enjoy. Okay. That one ain't going nowhere. I didn't check back in yet, I'm just here. Technically.  Well, I have a pretty robust first lieutenant team here that works very hard in setting up the base for any type of emergency we might have. I have about 20 folks that are going around to all the tenant commands that include fleet forces, sub-land and surf-land, to ensure that their facilities are maintained during this tropical cyclone. Mainly our concern is, since we're in a low-lying level here, we're concerned about the storm surge, so a majority of my folks are setting up sandbags to protect most of the buildings from flooding and ensure that power is maintained during the storm. Navy Region Mid-Atlantic has set cyclone condition of readiness three. So, we went from four to three earlier today, and these guys will probably be working a good 12, 14 hours getting ready for cyclone condition three. Us, in addition with my neighbors at Joint Expeditionary Base Little Creek, Naval Air Station Oceana, and Naval Station Norfolk, are working hard to ensure all our sailors, military personnel, and other civilian personnel that work here on our bases are safe, cared for, and that all our military assets are protected. We're seeing the weather phenomenon coming up the coast, and either breaking to the left about where we are up north toward Washington or New York, or heading out to sea. Obviously, we're preparing for the Norfolk arrival, and with that, we're preparing for winds in excess of 50 knots, but expect them to be between 30 to 40. I'm confident the base will survive. We survived Irene. I think we'll survive this. I think I may have to swim to work Monday morning, but other than that, I think we'll be fine.


तो हमने रन किया तूवे रेपीटिट मेयर आनोवा अपूट के लिए में सारी जो अपूट के में टेबस दे वहां से कोपी पेस करके इस पावर पोईंट के अंदा में पेस की हैं अई बल एक स्पलेईन वंबाई वं. तो हमने रन किया अनालसेज उसका पहला जो तेबल हमारे पास आईगा वो मारे पास शुरू में बताता है कि विदिन सबज्ट हमारे प्टरस कितने हैं और यो बतारा है कि अपका पहला जो वेरिएबल द्रिंग हैं जिसके डो लेवल से हैं अप एक वाड आप की इमज्री है थो हर एक लेवल के अंदर उसका मीन. यहनि कितना हैं और उसका बाज उसकी च्टन्त्र डीवेयषन अप फिर असके हम आवाड पास उसके नहीं ञमे देद गया है. वेड़ of the data, add it at the level of each predictor, हमारे बताता वेरिएबल और थो थोस तेबल बताता है कि वो, उनका मीन अस्टान्र दिवीश्यन अख्राँ सिक्स कंदिश्यन्स क्या है सेकिन जो मारे पस टेबल आती है मैंने जिसे बता आता के हमारी एनोवा की होमवोजनाइति अफ वेरियन्स यन देपन्न्नट एनोवा और मार्ष्ट्रले टेस्त हम युस करते हैं फो रेपीटिड मैंगर एनोवा और मार्ष्ट्रले टेस्त का मतलब होता है के इक्वालति अफ देपन्ट वेरियन्स बत्वीन इच पेर अप देपन्ट लेवल बजाए ज़सा हम होमवोजनाइति अफ वेरियन्स छोग लिवाए लिविन्स टेस्त करते थे के हमारे दोनो गुरुट्स में जो इन्टिपन्टन्ट ते के हमारा वेरियन्स इख्वल है अप चुके मारे पस अलरी देपीटिड मैंझर है अगी पार्टिस्पन्ट से हर लेवल पे तो आब हम यह कते है के वेरियन्स एख्वल अक्रोस इच पेर अप देपीटिड देपीटिड मेंच्ड़ाए ग़िए खोग और आप अपना आपना आपनी और देपीटिड but as you can see, the values are same, so we have reported the value of green house geyser and above it is one and same thing. So we have to report three types of values, first for the main effect of drink, main effect of imagery and then the interaction of the two. For a report, we mainly report the value of F and SIG, this is P value. And we know that P value should be smaller than 0.05. So if you see, the main effect of the drink is significant, yes, because 0.013 is smaller than 0.05, so there is a significant effect of the drink on the attitudes towards drinks, and then our imagery is also significant effect, because the value is less than 0.05. It is 0.00, when we report, usually we report this 0.00 to 0.001, P is greater than P is equal to 0.001. So the main effect of imagery are significant, main effect for the drink are significant, now how imagery and drink are interacting with each other, again the results are significant. So we have three test values for the F and all three are significant. The value tells us that there is a significant main effect of the type of the drink and type of the imagery as well as significant interaction between these two variables. And what I have told you that the main effect of the interaction effect is from this plot, which is called interaction plot or profile plot. Actually graph literacy again is a separate kind of skill like our numeracy skill, our mathematical quantitative ability skill, graph literacy again is a separate skill. Many people are very good in math, in graph readings, they are not very good. So here you can see that this is the line of drink 1 and this is for drink 2, i.e. these are the ratings about coffee. The ratings are our attitude towards it. So now whether you have used positive imagery, negative or neutral for it, our attitude towards coffee has not changed much, almost it is like this. But for the alcohol, when we have used its positive imagery, our attitude has become very positive and high. And when we have used negative imagery about it, negative imagery means that you associate it with a negative thing or with the name of alcohol, link it with a very negative thing. So you will start developing negative attitude towards that stimuli. So now when you have used your negative imagery, your attitude towards it has gone down a lot, i.e. it has become low or negative and it has become neutral in our neutral. So you can see that the main effect of imagery is significant. The main effect of alcohol is significant. Whether it is any imagery for coffee, your ratings are positive or more or less. But the ratings for your alcohol are different. So there is a main effect between these two lines. Yes, there is main effect for the imagery. Yes, 1, 2, 3. And then there is an interaction significant that when you have used imagery, then he did not do so much work for the positive or negative coffee. But for this particular variable, which is alcohol and which was the purpose of the scientist to develop negative attitude towards alcohol, that was really, really significant. And that's the beauty of the interaction effect that sometimes our main effect does not have a difference. But we know that imagery will work. Maybe coffee does not work for imagery. But when we use this imagery for the other type of drink, then it works. So you can see that graph shows that there is a significant interaction effect between the two. And this is the same which I have told you that our significant interaction effect is there between the two variables and the significant main effect for drink as well as for the type of imagery. Again, this is the interpretation which I have already explained. I can read the significant interaction effect tells us that the type of imagery used and a different effect depending on which type of drink was being rated. The graph shows that the pattern of response across drinks both for alcohol and coffee was higher for positive imagery when compared to negative and neutral imagery i.e., ratings were positive for coffee and alcohol with a slight mean difference. I have just explained everything to you. Negative imagery had the desired effect on attitude toward alcohol but much less impact on the ratings of the coffee. So negative imagery which we have used did not change the attitude of coffee but our attitude towards alcohol has become very negative. Therefore, the interaction is likely to reflect the fact that imagery has the expected effect for alcohol that is rating highest for positive and lowest for the negative and neutral somewhere between but not for the coffee. So the interaction effect is significant.


So I rocked up to the counter at IGA and I noticed these bags. Where are we going? Excuse me, miss. You seem to have showbacks for sale at the IGA. Please explain. I think it's because the real show went happy this year. Well, take the lot. Well, you can. All right, all right. Surprise me. Which one should we choose? Oh. Are you more like a child? Oh, surprise me. No, you can't ask me a question. Surprise me. Can I give you a hug? Yeah. Bye. Best I ever.


It's going on everyone Sam here late in sports cards. We are ripping at 20 22 Panini absolute football hobby six box half-case break number 13 pick your team style. Let me change our break title I'm gonna read off the list of Names and teams we've got Allen T with the Packers Anthony P. Buccaneers Austin K. Chargers Calum F. Giants Charles T Patriots Chris S. Bears Doshmere K. St. Diego. I with the Niners Emily G Broncos Francis P. Cowboys Gail L Ravens Vikings Eagles Lions Gilbert D. Browns Gregory T. Chiefs Harry G Bill's Jeff N. Texans Falcons Jennifer V with the Colts John P. Seahawks Jordan S football team Joshua B Panthers. Thank you, Joshua Sold us out Mark S. Rams Raiders Michael V Jets Mitchell H. Jags Paul K. Bangles PF with the Cardinals that's Philip G. Dolphins Robert O. Steelers and Will L. Titans. Good luck here everybody. Let's do it It's up smack. Hope you're doing well too Welcome in Welcome smack daddy I like the name Smack daddy And guys we've got 14 spots to go and tops Chrome Star Wars black breaking late breaking later That is random auto or pack style $60 a spot is on sale from what it usually has been Yes, the fresh case Yeah, whenever we do this is a PYT. So the second half will be the same Case That's it. Well, whenever we do half cases generally they're always two at a time. So We always do that red John Metschy to a hundred for the Texans Got Davis Mills tools of the trade 67 out of 99 going to the Texans and Then going to the Jags Snoop Conner 18 out of 35 on the blue auto nice one there congrats Jacksonville Super Han Solo that'd be cool Paul. I'm in on a super Han Solo I Carry Fisher Harrison Ford dude. Yeah Magic one thing I don't like about absolute is that he's clear Got last call Mason Crosby to a hundred for the Packers Mack Jones Damien Harris Trey McBride rookie that's 75 out of 275 Go into the Cardinals, and then we've got Zemir White To 233 out of 399 rookie premiere triple Jersey ball Jersey for the Raiders. Congrats Raiders Orange Lamar Jackson To 70 or yeah to 75 for the Ravens Travis Kelsey Mike Williams Got Jersey nice jumbo Garrett Wilson 39 of 49 Nice one there for the Jets Love these jumbo patch guards But one of ones of those have to be disgusting and then Daniel Jones 23 of 25 Danny Dimes Jersey ball Jersey tools of the trade Going to the Giants. There's Sean's guy right there Danny half sent yeah He does Ryan Tannehill 11 out of 25 going to the Titans base of Aaron Jones and Joe Burrow Got tools of the trade Zeke Elliott on the quad 14 out of 99 nice one there for the Cowboys and Jelani Woods 65 out of 75 down to 13 to go now on Star Wars top throne black Thank you guys. Thank you there Justin took a spot Aaron Rogers for the Packers to 75 on the draft diamonds Darren Waller Elijah Moore, and there's Isaiah Spiller to 275 for the Chargers followed by Peter Strong jr. 41 out of 49 four-color ball three-color Patch auto congrats Patriots nice one there That's Bill and Daniel What up bill David Montgomery to a hundred two of a hundred there's Devo Marquise Brown Breeze Hall three-color patch 16 at 49 and Michael Pittman, Jr. 64 out of 99 tools of the trade that auto going to the Colts nice one there MPJ What's happening bill thumbs up there we go Oh Geez, that's brutal Taylor What now it's a goat no that is an eagle Yeah, I'm a big big Mets fan. No, I'm not no time on what that Illegal I got to be honest with you that might end up be the worst signing of the entire the entire off-season Taiwan Walker's horrible last year There's Lamar to the Ravens to a hundred Austin Eckler David Montgomery like when he's on he's on but when he's off he is off One of those kinds of things Jevon say Williams to a 49 nice one there for the Broncos and Cam Taylor Brit 42 out of 75 there you go bangles I'm no Sean's a Dolphins fan. I'm not a Dolphins fan. Yeah, are you like Miami or no University Miami? Do you know? Oh, yeah, cuz Sean went to Florida State. So now Sean is a SSU guy There's Chris Olava. It's a 50 Donald Michael Carter. There's Eric as you con ma for the Dolphins to 275 and George Pickens three of five Nice one there Steelers Rookie premier patch auto. He's got Pittsburgh Robert. Oh, congrats, man. Yeah, that's true. That's all that's a very good point No, you're probably right Yeah, see if you can get in There's Kaderios Tony to put a Giants to 75 Nick Bosa TJ Hawkins in 130 Yeah Bree's Hall, that's two out of 25 going to the Jets the absolute Jersey and Deandre Swift nine of 25 on the color match going to the Lions nice one there You get that spin right up start using spider tack like Heracle and Chapman Do they win? James Illinois end up beating Texas Wait for Lane to get in here go horns down This is back for a hundred years that last call Brashad Paramann. Oh To this card 88 out of 100. Why does he have a card my god? Good Lord start giving anybody one now Ryan Tannehill Lamar Jackson We've got Troy Anderson to 275 for the Falcons. No, no, no, absolutely not For Sam. Yeah, right. We got redemption is Sam Howell rookie premiere materials auto five So that'll be the like five-way patch auto That's going to the commanders Can't believe they Legitimately just put over shot for a new part into this set. It's also true illegal if you went to the Rockies Aaron Rogers to 75 Talking about Cody Belanger illegal goes it would look bad for him if you went to Colorado and still sucked Oh, yeah You're locked Josh Jacobs Maybe There's Rashad Penny McKeneth Walker six out of 49 What's it to you infamil is that what you're on? Milk at jacked up. Oh, yeah Logan Halt One that's 68 out of 199. There you go bucks Right And there's Patrick Sir Tanne to 100 is the scarecrow the Broncos. Yeah Yeah That's Matt Ryan and Brandon cooks cool one here going to the Ravens Lamar Jackson 56 out of 99 nice one there and Legends for the commanders Carson Wentz 39 out of 50 on the out of here legend She daddy Was that one year They got hurt yeah, I'd never been the same Yeah Orange James Connor. I think it's cuz it's only one year James. So it's like if They end up If he ends up sucking then it's only one year But if he ends up playing decently well and the Cubs aren't in it, then you could just trade him Because it's only one year. So he's a good rental option base as Brandon cooks to 49 to a 49 for the Texans and Garrett Wilson on the auto six out of 35 for the Jets once that guys All right guys a couple emails I have to respond to before I forget all right There's bill. Let's go bill Let's go bill Gonna wreck the dolphins hat again Yeah, Traylon Birx to a hundred Base of Derek Henry and Jonathan Taylor We've got DK Metcalf on the burners 29 out of 49 for the Seahawks And Absolute legends Dallas Goddard 30 out of a hundred for the Eagles Drop you guys this link again to a Star Wars That's done We've got 13 to go now. We're so you hate what? Yeah, I feel that I would always be like Can't do anything All right, last break of the night with spots still is our top chrome black That is it is it that is all oh is that a kaboom we got a kaboom We've got a kaboom By storm Kenneth Walker to 50. What's your all right? Everybody's guess. What are we guessing kaboom kaboom? Who is it? I'm saying like Danian Tomlinson. That was my guess on Sunday What are you saying? It's Joe Burrow. Oh, that'd be cool. All right kaboom everybody with the kaboom predictions What do we got predictions? What do we got got Andy Dalton already back? Herbert Kirby what else we got Hurst doesn't have one okay Illegal is officially disqualified waddle waddle. Let's go. I like the waddle.  Hey, there's Paul. Oh, there we go Paul. Congrats, buddy Let's go dude. Yeah, Taylor for Sean Pairman and Joe Burrow same player. I like them both about the same Chad Mooma to 275 And Christian Watson 14 out of 99 going to the Packers Yeah, so yeah Sean called the card Paul Sean gets it. Unfortunately. Yeah, sorry, bro House rules Nice hit though, Paul Nice hit nice hit Congrats take a burrow kaboom Said to pull the burrow kaboom. Yeah, the other green I was gonna say Purple Jalani Woods to a five nice one for the Colts low numbered hit there The purples are so nice and absolute dude Echler and Montgomery Now James Robinson tools of the trade 56 out of 99 for the Jags and Desmond Ritter 136 out of 199 that's going to the Falcons in one case, right? Yeah. I know I want to pull another one dude Then we had the green. Yeah, which was I think a burrow the one-on-one burrow. Yeah Yeah, we have that one and then we had a base of But we had a base to all all in one shot Matt Corral 3 to 25. There you go, Joshua Dude, I hope not. I really need to complete that trifecta. I think have I told you that? So boy on Bogdanovich has three cards and pretty nice spectra I pulled two of the nebulas of his car. I pulled the he had like a Full spectrum of him in the jazz and the jets Then I pulled the base or the jazz in the nets Then I pulled the base nebula and now all I need is is an autograph nebula. I need all three But that on the commercial screen I Really hope it's not pulled already That's Taekwondo into 275 not even illegal. I've just pulled two of the three nebulas out of spectra boy on Bogdanovich Crystal Lave nice one there 87 out of 199 for the Saints That would be very Sam to have boy on Bogdanovich as a PC. Yes, it would in fact. I do PC Paul Millsap. So yes, I I love Paul Millsap. I gotta rock my Paul Millsap jersey My guy Paul Millsap's my guy. We've got Dalton Schultz to 75 for the Cowboys Yeah, Piston's legend boy on Bogdanovich. That's legend boy on Bogdanovich There's Michael Carter Deandre Hopkins John Dotson eight out of 99 for the commanders Gem that they had my landscape full and then the poor chip full. Oh The burrow green gotcha gotcha gotcha We've got nice Herman Moore 70 or 13 out of 75 going to the lions nice one there I'll do it for the break guys do a quick recap. I'm not from Ithaca bill. I went to college in Ithaca I'm not trust me. I'm not from there. That's John Dotson commanders James Robinson Jags DK Metcalfe Seahawks Brandon Cooks Texans Lamar Jackson Ravens Seahawks dual Breeze Hall Jets Givante Williams for the Broncos Breeze Hall Jets Zikila Elliott for the Cowboys Garrett Wilson Jets Davis Mills Texans messing up Herman Moore lions Chris Olave Saints Desmond Reader Falcons And that's Christian Lawson Packers. That's true Dallas Goddard for the Eagles Garrett Wilson Jets Carson Wentz commanders Logan Hall Bucks Deandre Swift lions Cam Taylor Britt Bangles Michael Pittman jr. Colts Pierce Strong jr. Patriots Jelani Woods Colts Daniel Jones Giants Zemir White for the Raiders that Snoop Conner Jets Sam Howell for the commanders We had George Pickens three of five Then we had the kaboom going to the bangles Joe Burrow nice one there That'll do it for the break guys. Thanks again for the fill next up. We get the second half case


Are Israel and Taiwan neighborhood bullies? Or examples of democracy and freedom in a region surrounded by those who seek to destroy it. We're here in Taiwan to explore five ways these two tiny countries are similar. Taiwan and Israel both have shared values as liberal democracies. Both countries have made tremendous strides in only a few decades to protect civil liberties and to ensure that freedom of speech and freedom of the press is a reality. Israel is the only Middle Eastern country ranked free by Freedom House. And Taiwan is ranked sixth in the world for political and civil freedoms. Both countries have free and fair multi-party elections. And both countries protect freedom of religion under the law. But also both countries face similar challenges because of this commitment to freedom and democracy. Taiwan faces a serious threat from China where the Chinese Communist Party has repeatedly threatened to overtake Taiwan as part of their one China policy. As they did with Hong Kong not long ago. Israel faces threats from its neighbors as well. From Lebanon, from Syria, as well as from Iran. All of whom, at different points, have threatened to wipe Israel off the map. Additionally, both countries struggle in the fight against disinformation for Israel that is manifest in disinformation about what's happening in Israel on social media. For Taiwan, the threat arises from Chinese propaganda spread through social media and media. Now we all know that the UN focuses disproportionately on bashing Israel. But Taiwan is also discriminated against in the UN as China makes every effort to exclude Taiwan. And even Taiwanese reporters from UN recognition. As a result of their similar threats, Taiwan and Israel both have a close alliance with the United States and other Western democracies to help bluster the defense if existential threats arise. Because of the tiny size of Israel. And the tiny size of Taiwan. Both countries have been forced to innovate in new ways to build their economies. Israel is known as the start-up nation, where there are more tech unicorns than anywhere else in the world other than Silicon Valley. Taiwan is known as the scale-up nation, making it a natural collaborator with Israel. If there's a technology that impacts everyone's life, it would be the SD card or the SIM card, the vast majority of which are produced in Taiwan.  Like Taiwan, Israel has no shortage of tech developments that have changed the world and continue to. From medical research to penny-um-chip processors to electric car batteries and beyond. The truth is, without Taiwanese and Israeli innovation, Emily, we would be living in a totally different reality today. Israel declared independence in 1948 after the UN partition plan paved the path for a Jewish state. And Taiwan split from China in 1947. Both countries have faced existential threats from day one of their independence. And both countries have succeeded, despite all the odds, as liberal democracies. Today, the two countries regularly collaborate across industries. More recently, Taiwan was a world leader in dealing with COVID-19 pandemic, much like Israel led the world in the vaccine rollout and providing accurate vaccine data. This year marks 30 years of friendship and relations between Taiwan and Israel. And the relations are only growing closer. Two-way trade in 2021 reached a record $2.4 billion, a 33% increase year on year. There are many more examples of Israeli-Taiwanese connection. But one thing is certain, these two tiny countries have a massive global impact for the better.


Everybody Matt here for late in sports cards ripping 2022 panini select football hobby six box half case break number eight pick your team We have Anthony P. Bucks Anthony G. Raiders mr. G Seahawks Ben C. Lyons Brooks v. Panthers Calum F. Giants Daniel C. Broncos Rams David K. Falcons Dennis F. Chargers the CQL M. Colts James a Cowboys Jeff M. Ravens Vikings Texans Jeffrey H. Steelers Jesus T. Niners and chiefs Joey F. Jets Laurie R. Eagles Matthew G. Bills Maximilian H. Jags dolphins Nicholas T. Saints Patrick K. Packers Rob V. Browns Robert R. Bears Seth. I'm sorry Seth B Bengals Steven W. Commanders Timothy S. Cardinals Todd W. Titans and Wayne S. Patriots. All right, here we go Yeah, we do have it tonight. I don't know if somebody's already grabbed the Seahawks tonight, but Later in the evening Sam has a couple pick your team style breaks of select six boxers My sports is ripping divisionals right now over on the Layton sports cards to channel Like some select football divisionals. He's ripping full cases Sam tonight has some six box random team styles some Three box random team styles some 21 22 Select two box mixers. There's Desmond Ritter along with full case divisionals and a couple six box pick your teams There's Dak Yeah, absolutely nipples. Let's see some monsters here Amon Ross St. Brown Joe burrow silver Here comes our Zebra rookie Duran Harman Zebra rookie Sweet level for the Raiders It's uh from Andre Stevenson Silver insert and Mari Cooper Damian Pierce rookie tricolor that one is to 249 and Raiders to 49 That is red auto Derek Carr There you go Raiders Almost said there you go Oakland Patrick Mahomes maroon That's to 149 rogers silver like white and this is Raiders to 10 sparks patch gold davante Adams Go Raiders with the gold patch uh Roger McCrary rookie silver rookie tricolor cave-on tibodeo Kenneth Walker rookie, um, sweet level. Yeah, sometimes they do Steven joku purple silver rookie drake Jackson Is that a redemption coming? Yeah Jalen waddle and Might be easy here. Let's trevor Lawrence prime selections prism signatures card number 40 Texans Damian Pierce There you go Houston rookie prime selections prism signatures Catered cahoe cahoo Uh tricolor there josh Allen So we had packers colts and chiefs three to go in the next break Jay brown breeze hall rookie silver channing tendle modsauce gardener Stingley and jefferson hurt warner silver new patrick bailey sitting 338 three Homer 17 RBI since he got culled up jeez Why somebody talk about that guy? Yeah Jalen tobert rookie tricolor josh allen breeze hall rookie insert a j brown patrick mahomes This is bill 75 jersey white staphon diggs Roger McCrary rookie silver That is tie-dye justin herbert 25 You go chargers It's premier level two of 25 You've done a recap we got Russell wilson tom brady silver That's evaniel rookie silver and clill mack He's so weird seeing clill mack and uh charges you That is rookie blue drake london This is to 199 concourse level trombone walker jack jones silver neon signs mixin and lions 49 that is rookie jersey auto purple adan hutchinson Nice hit for the lions Kevin harris A sweet level silver and breeze hall hexon hill Tyler lockett maroon and eagles 75 auto blue dallas goddard gerald alexander Kurt warner a j brown and rogers silver k von tibodeau travesy tn juniors Phenomenon silver moseley Box number three Packers cultson chiefs last three teams left for the day During our daytime picking team style breaks Nice falcons to 75 rookie auto blue Desmond ridder congrats falcons Lots of ridder today from andre stevenson tricolor pinyin green rookie silver Mark jackson christian watson snapshots rookie and sweet level rookie silver bailey zappy brian cook mod sauce gardener venerian mathis and giant 75 rookie jumbo jersey white wandale robinson quay walker rookie silver If in this box i pull i pick any rookie if i pull they're they're one of one They're black one of one you have to change your first name to wandale joe burrow We think Tricolor Deontay johnson silver I can change it right back to the state Trevor Lawrence ralando brown blue It's 149 justin herbert silver and vaughan miller What about shawndale shawnda pospery didder Yeah pospery do that is visionary insert devante adams Goaf i think that's the first visionary of the day we've seen Justin fields breeze hall randy moss field level blue And cowboys franchise numbers auto drew piercen You go nice on-card drew piercen auto If i pull a black in this box you'll get wandale tattooed on your arm Jamar chase silver. What if i pull? Let's see i'm trying to figure out how to get the tattoo from your arm up to your neck And red rookie derrick forest jr 49 A conco rookie silver If it's a wandale one of one in this box, i've only got a pack and a half to go here Chair elam rookie silver and justin herbert. Thank you, man. I don't see one Neon icons derrick henry Vaughan miller No watch i'm going to do it first pack next box one day a lot of one As we have packers and colts two to go on the next break.  Can you make this one, please? A nap when i get home. Gee whiz Raiders 75 jersey white max crossbeam Rashad white rookie silver net ingram Najee girls chakil inner Yeah, ebony hula graham slammer here kairan williams And davante white rookie light blue Tritofi rookie silver Adherian mathis marlin humphrey purple John runyon david ojabo rookie silver Uh-oh sam's sam's here. Oh, yeah, dude. I have set the bar quite high today Jalen hurts snapshots to a silver joe burrow neon signs garret wilson rookie and Packers to 199 rookie auto romeo dobs friar move patrick's sorry patrick sirtan blue And one of 75 chargers. That is jersey auto antonio gates. That's a cool one. Here we go chargers That's newton targan davis price um malik willis sweet level rookie Sean watson That is martin emerson rookie white 35 Baker mayfield silver it looks good chanel Kenneth walker sensations hufanga silver and bailey zappy rookie field level Two boxes to go in this break And we have one break left after that And we have just the colts left at 199 Now good timing length appreciate it and for us is live over on the latin two channel and he is ripping full-case divisionals right now Of select football tom brady sensations Jets to 75 rookie auto blue brice hall Here you go. Nice one jets Dawson knocks I have a half case in here for you slam. Oh, do you? Yeah, okay. Chad mumma Kenneth walker Phenomenon rookie kevin harris rookie silver diran bland light blue rookie and This is patriots super ball auto chris long Go new england rather conchlin bomb tibodeau christian watson rookie insert Sam williams rookie silver jordan and Patriots to 99 rookie jerseys pierre strong boston field silver axon hill mike white chiquil innard and Malik willis field level rookie rookie blue josh pascale 299 john ronion rookie daniel bellinger rookie silver jerry judy Allguyer tri-color pierre strong jr. Rookie silver and marcus jones holy quillis Trevor lorns and christian watson Williams and kenny picket red or not red uh maroon rookie that is to 149 nice one stealers k york rookie silver and tom brady two packs go in box five Monro same brown kenneth walker rookie silver This is tri-color trelon berks. Yeah, he did Isaiah likely your rookie silver right last box Somebody wants to close us out for the day. We have the colts Left in our next break and that is it for the day also grab up your spots for breaks on the two channel with forest And then for tonight's breaks on the main channel of sam and then kyle will be live on the two channel tonight ripping some uh personals josh jacob's and Cardinals to 199 rookie jersey auto tray mcbryde Smile scare it Here comes a gold That is seiquan barkley gold 10 for the giants One of 10 You go a nice seiquan rookie silver trelon berks Kenny picket neon icons rock party field level rookie silver This is lions to 25 tie-dye rookie patch jameson williams nice one for the lions seiquan Drake london rookie insert nick bonito rookie silver and rake london field level rookie This is rookie neon green the lime green kingsley enig enig bear Tom dotson Andrew booth rookie silver color murray judy rookie silver sweep level desmond ridder robert woods tray lance this corald daryl paterson tricolor and Colts award winners auto dairies lennard indianapolis untay taylor romeo dobbs rookie silver rookie blue trelon berks to one Sorry to 99 scowler thompson and bailey zappy That was to 99 trelon berks neon icons erin rogers Cordero paterson last pack rake jackson rookie white 35 derrick forest Jerome fort rookie silver and dalvin cook on the back So we had visionary davante adams for short print diran harman rookie zebra Both of those went to the raiders uh, justin herbert tie-dye to 25 And then relics we had lions rookie tie-dye to 25 jamus and williams Patriots rookie jersey pierre strong jr. It's 99 and then White parallels to 75. We had raiders max crossby white parallel relics. That is uh bill staphon diggs and giants rookie on dale robinson uh raiders gold patch to 10 davante adams Cardinals rookie jerseado to 199 trey mcbride Chargers blue auto or jersey auto antonio gates. That was to 75 lions rookie auto purple adan hutchinson.  That was to five That'll do it for the break. Thank you everybody. We have number nine closing us out next


The last lecture. OK, so in lecture 1 to 3, we have described how the algebraic, so we've introduced the category of stable maps, a bundle category over it, Cauchy-Riemann section functor. We have described the algebraic relationships, and we have put smooth structures on this. And sort of the packaging is essentially of when you want to resolve the transversality questions for SFT. So you have this diagrams of covering of, actually, now smooth covering, SC smooth covering functors. Some other relationships formalized by having some functors which behave in some compatible way is the Cauchy-Riemann operator. So now the goal then is, so now we have a complete setup to perturb the subcategory of sort of J-holomorphic objects into some other sub-weighted subcategory which actually will be smooth, SC smooth, so that will be enough to extract data from it. And we would perturb it into a functor which has certain properties. That is the algebraic thing, so that we discussed in length. And of course, geometrically, you want that this is a pair in general position as far as the Fritam series. So there's a Fritam functor as far as the Fritam series is concerned. Now, if you think of this here locally as a finite number of sections, so you get a finite number of equations, and you want that each of them is in general position, which can be achieved for this. It cannot be achieved when you homotope from one perturbation to the other. You cannot in general achieve general position because of all the structures which you have here. But for not doing homotopes, but when you just want to perturb, that's still possible. So now, so we have to perturb. And of course, we make choices to do so. So if you have two different perturbations, we want to make sure that in each connected component of the orbit space of our categories, the solution spaces are still compact. And we want to make sure that we can guarantee that we can connect this sort of as generic as possible, these two perturbations. And we also have compactness doing that way. So now, in order to actually make sure that this is a case, you ideally distinguish something like a convex set of perturbations that you can go around where it's also clear that, a priori clear that you would have compactness. So let me explain here. So we will actually be able to make an infinite construction. And there's some kind of an inductive procedure to produce it. So it starts off with low energy things, which cannot split further off, which only contain buildings with height one. Then you go to the next one where you have buildings of maximal height two. But then the boundary phase is just by the algebraic structure, which you have have to be explainable in terms of the data you already dealt with. So you pull up these perturbations on the boundary, and then you extend it. Then you have to distinguish between sort of connected components alone together and others with added cylinders and so on. So there's a certain number of things which you have to do on this level. So then, of course, when you just see this procedure, I go on and on and on and on. So it's not surprising that my perturbation gets larger and larger and larger and larger. So now if I only can control things in an a priori given neighborhood of the original solution set, you can imagine that at some point you fall out of it. That actually happens in every other approach. So what do you do? Well, you use homological algebra. So you perturb up to some level, then you encode this data, then you take a smaller perturbation to begin with and you can run higher up, and so on.  But it turns out that in this problem, actually, and that has to do something really, it's basically an abstract factor. You don't have to know much. You have to know a little bit. You have to exploit a little bit more about the Cauchy-Riemann operator than people usually do. So there's more compactness in the background. And actually, this infinite procedure can be done. So geometrically, you can run up.  So that's a nontrivial effect. So you can make that infinite construction and get that infinite, smooth object. So that, of course, could be used in other approaches. So this finer compactness could be used in the other approaches as well, so there wouldn't be the need to do this. OK, so in order to deal with this aspect of the story, so you have to measure the size of perturbations and you have to come up with some idea how that controls compactness. OK, so we have heard what an auxiliary norm is in the case of an M polyfoil setup. So here, it's basically a functor. So it's actually defined on the things which have a little bit better quality in the fiber. But if they're not lying in this space, they'll just put it plus infinity. So a functor there restricted to this E is on each fiber norm. So if you go to orbit space, it should be continuous and should have the following properties. So if I take a sequence of objects so that n goes to 0, so that's the sequence of numbers, and if the isomorphism class of the base point below converges, then the isomorphism class of this vector converges to the isomorphism class of 0 over that point. So that's the requirement. It's some kind of local uniformity. So you take a sequence here. So if this is finite here, that means you're actually here. So then the size goes to 0. So this is just a sequence of objects, wherever. But now you look at the base of this object. So it lies over an alpha k takes the isomorphism class, assume this converges. Then the isomorphism class of this should converge to the isomorphism class of 0 over that object. So then there's something which I'm not going to explain. That's called reflexive auxiliary norm. That's an important refinement actually for this to utilize this more general compactness. So one would have to explain a little bit about that. But basically, this you can construct always if the fiber in your bundle category is actually reflexive, which is likely about space in our case. Then you can always get this. So one can presumably talk about an hour about the compactness issues. So then it should be compatible. So first of all, you define, if you have an auxiliary norm, you can define this on the fiber product. You just look at the individual factors, maximum. It's important that you take this and not the sum. That is what you can control here. If the maximum is controlled and you can control, and that means that you can control compactness here and here, you control compactness of the product on the boundary.  So if you're not too far away, you still have the contractive parts. So it still essentially looks finite dimensional and you can control this. So then you want compatibility with this structures which you have. Then you want to have something like this. It factors. So pi is this projection coming from this Whitney type decomposition, where you put the 0, 1 form on the non-trivial cylinder, on the trivial cylinder component either 0 or the other way around. Remember, that was one of our structural things. You want to have this. So it measures the size of a 0, 1 form on a non-trivial components or, in this case, over the cylindrical part. And you want to have that behaves like this. The reason is that you later perturb anyway where there's nothing on that side here, because over the cylinders, the perturbation will be 0. What is this E subscript 0, 1? Oh, this is E? This? No, cappy. Oh, yeah. So remember, for a strong bundle, K over O, has this bi-filtration, Mk where K is less than or equals M plus 1. And if you have the strong bundle structure on this and this E over C, it also gets a bi-filtration. So on the stable maps, you can talk about the regularity of the object. And then if you have something in the category of firing over it, you can talk about the regularity of the 0, 1 form over the underlying object. And that has an addition. So why do you allow the thing to actually go there infinitely on E? Yeah, because it's precisely finite if it's here, which is a sub-fiber. Can you say something about why is that? Yeah, so remember that, as Katrin pointed this out, that for a strong bundle, it's not just a bundle, but there's an additional bi-filtration.  And the SC plus sections are sections of the one level. So that is where you know that things are compact. So in the fiber, you have 0 level, 1 level, and so on. But if you are on the one level, you go down by compactness on the 0 level and you perturb by sections which lie on the one level. And that is precisely there are sections of this, the multi-section functors. A priori are given by sections of this, but the images of this section actually are in that better part of the space. So that is how a compact perturbation is built into the whole thing. And then it should have this property. So if you forget about the stuff over trivial cylinders and take the norm, or you move, or you even throw the component away, which, of course, is more or less, you don't need the trivial cylinder component, then you see the same measurement. So then, of course, you want, if you have disconnected non-trivial configuration and move them against each other, then you would see this. The measurement of those guys, they are maximum. Doesn't that contradict the thing you have to bother about n equals n equals to pi plus n equals to the Iw1 time? That's a plus right there. What a plus, no, no, why? So if, do you mean this? And I mean that one. Oh, no, no, no, no, no, no, no. So these are non-trivial components. So then you take this, and then there would be some cylinders here, which you discard, and some cylinders here you could discard in order to compute this. And if you look what happens over the cylinder, you take that. So this bit only refers to the things which are sort of non-trivial. Yeah, these are non-trivial. Non-trivial components, yeah. So once here in this compatible reflexive auxiliary norm exists, actually construction is rather easy to take a local model. And then locally you have a retract in the Banach space. It just takes the norm and take a partition of unity times the norm. And then you restrict it to the retract. And if you add it up, it has the size of this property. So if you have to construct such thing for a Banach manifold, for a Banach-Bandor-Banach manifold, it's precisely what you would do there. Is that, might we as well just use the thing you just said all the time instead of the general auxiliary norm? Yeah, so yeah, that is what I mean. Whenever I would construct an auxiliary norm, I would construct it like this. And it would have the property required from a general auxiliary norm. Then if I do the same procedure and have reflexive fiber, this procedure would precisely produce the norms which has the property which I want to have. But it turns out that reflexive auxiliary norms have some additional features which I haven't described. So there's some more properties you can. Because when you have reflexive fiber, you can start talking about reconvergence in the fiber. And that is actually a useful thing. Because the opposite freedom operators have a closed graph. Even if you don't have full convergence, you can see that in some weak sense, certain things would satisfy the equation. And this means you get a better organization of compactness just by that. But as I said, one needs some time to explain it. So once we have such an auxiliary norm, then we can measure the size of a multi-section function as follows. So first of all, since these are functors, I can pass to the orbit spaces. And then you get a function on the orbit space of stable maps just by looking at the maximum of n absolute value where I look at all u for which is non-zero. So basically, I measure the maximum norm of the different branches which I have locally. Can you remind us the SC plus in this? So SC plus meant that this is a section functor, but the local picture is that it's SC plus section. So they go into the 0, 1 fiber. So OK, so here's a local. So let me see if there's any. Hold on. Stay back. There was a dictionary. You should just think of compact perturbation. Yes, but nevertheless, thank you, Katrin. But let me just say that here, so this is the polyfoil bond, the strong bond. So then here you had this filtration where k is less than or equals m plus 1. So for example, the Kuschelman operator goes from the m level to the mm level. But there is one better fiber, m plus 1. So you could look at sections which go here. And these are sort of the compact perturbations because they have the better regularity. And you can go down to the mm level, and then you get compactness. And the SC plus multi sections locally are given by sections of that kind. So then I have locally the sections. And then I precisely, this norm takes real values on this. So I look when I have this picture here. This is O. And this is the local section structure. I'll just look at the maximum norm I see. So this function of z is over. Since isomorphic objects have the same norm, it just is a function telling me what is the maximum norm I see over the isomorphism class. Michael, happy?  So we fix such a norm. And now we also have to exhibit a compatible neighborhood u of the cos-modelized space of geomorphic curves so that we can guarantee certain compactness properties for the perturb problem. So what I want to say, for example, there is a neighborhood of this modelized space. And if I have a perturbation which is supported in this neighborhood, and the norm measurement is small, then the solution space is automatically compact of the perturb thing.  Like if I construct something here and it is in the neighborhood, then I bring it to the boundary. And then this boundary part doesn't belong to the neighborhood of the next level, then I'm stuck. So the neighborhood, so if I have neighborhood of the solution space, say, on the level which never degenerates, then it gives a product on the boundary of the next level. But that should be, in some sense, the trace of the neighborhood which I have there. So that is it. You guys with me? Just a bell of definition. MJ is the one that is actually geomorphic, or the one that is a stator? Yeah, so these are just the origin through the holomorphic objects, yeah? So this is the one for which stator is the 0, 1, 1. Right. And now, so this is a subspace of that, close subspace of this, which on each connected component is compact. So you require as input this geomorphic object? Yes, yeah, OK. So now I want, so now compatibility. So the compatibility is best explained by the hierarchy I use for inductive constructions. So this is a set of connected components of z. Then this splits into a disjoint union of those which never further degenerates. So they only contain building of height 1. They contain building of maximal height 2. So they contain components where the maximal is 2, but there is a building which of height 2, or top floor number 1, and so on. So that is also how the inductions go. Here come objects which you actually can count. And then you start some schemes. So there are some intermediate steps, I think, which would be sort of interesting to see if something at that level works. So this is sort of the mother of all schemes. But I think for certain specific things and subcategories, you might be able to get more structure, but you have to adapt the perturbation thing to this. So this is not written in stone, but that always works. But if you want to have more fine structure, you should on certain subcategories modify this. So then we had this forgetful functor which forgets unstable, forgets trivial cylinder buildings. If you pass, and this actually maps connected components to connected components, actually preserving this decomposition. So I go from a component which I have and some trivial cylinders. I just forget the trivial cylinders. And then in the image, I get another component. And let me write, if u is a subset of that, that is the orbit space of s. Let me write it like this as an index. A is an element here. So a lot of the stuff, I mean, the connected components actually appear as indices in some way. So an open subset is compatible provided. So if after forgetting trivial cylinder components, I get this class, then the open set over a prime of the connected components should be just the preimage of that. So that means the neighborhood of a trivial cylinder is not restricted at all. But I don't have to, because I'm not going to perturb to in the de-bar where this is 0, then I will always find my holomorphic cylinder. So it can be generous on that side. If the objects in A have more than one non-trivial component, then ua is invariant under this local family structure when I move things around against each other. And if A, so for any A, it holds, for the phases of this A, it holds that if I look at the category with objects in U, fiber product is a category object in U, take its orbit space and pull this back by, of course, the image of this is in a subspace of that, and pull back by the covering functor, then I get that. Which basically means if I have an open set here, an open set here, takes a fiber product and view it as a boundary phase, then the neighborhood which I had in the full component restricted to this boundary phase is precisely what I already had, but I lifted up from below. Yeah, is that clear? So in this inductive thing, so you start on level 0, so you have two open neighborhoods of, let's say, of your modelized space. Then you go to the next level, there's a phase. Now the phase pulls back this product of the two neighborhoods, so on the boundary have a product of two neighborhoods. And then this should be the restriction of the neighborhood which I have on that component. OK, so given our reflexive, so this is important, but you can ignore that for most. So I would just say that the statements later would be wrong without that word. So I want to at least make sightable statements. It's a subtle thing you wouldn't realize. I mean, I could tell you, I mean, if I wouldn't have told you, I wouldn't have figured it out. So but I want to say that. So so on the level of lies which I heard in this course here, so I think that's I could have just forgot saying, not mentioned it, but no, I was referring to other speakers.  So that is what I meant by that. So I tried, I didn't want to use an approximation here. So so then an open subset. So if you have an open subset of the orbit space, define, so the auxiliary norm is n, define u n to consist of all isomorphism classes of objects, which are belonging to you where the auxiliary norm on the Cauchy Riemann is less than or equal to 1. And the trivial cylinders occurring on any floor are j-halomorphic. So this j-halomorphic cylinders is black stuff. And this thing here has some data on it so that if I apply the Cauchy Riemann operator and evaluate its norm, it's less or equal to 1. So now, is that clear? So you have an open subset of isomorphism classes of objects and auxiliary norm. So now associated to this open set, you just take all objects in this open set for which the trivial cylinders occurring every floor are j-halomorphic. That means if you apply the debar, there's no contribution of a trivial cylinder. Then you apply n, and it should be less or equal to 1. So what is the condition of the trivial cylinders? Can you take a trivial cylinder? What is an open condition? Well, I mean, you'll say it's an open set. No, no, no. u is an open. I give you an open set. I don't say u n is open. u is open. u n is not open. It's definitely not open because of that condition here. And if I put less than not, then I would have to allow basically every cylinder here, or a neighborhood of those guys. Yes, yes, yes, so the thing is that the perturbations I know will precisely produce that, since I will never perturb over this. Yes, yes. So why do you ever then actually put trivial cylinders into the polyphonic if they're not j-trivial or morphic? Because if I have a building, suppose three floors, and then I have a cylinder here, and I do the gluing construction, I need the variation of those guys here, otherwise it's not gluing. So think about this picture here. And then maybe something non-trivial here. Then here is something non-trivial. But here is a trivial cylinder. And then something happens here. When I glue this part here, obviously, in the moment I glue, this thing will be close to this one, but it starts off deviating. So I need function space there. If you glue this in, I mean in the setup. And I need to vary the function here. No, it's glued. So I need this varying things in between here and the function space in the broken one in order that I actually get a frittongue problem. Otherwise you don't have too much enough input. The middle level, if you look at just the middle level, that is a trivial cylinder building on the far right. Oh, and you're saying the boundary stratum of your polyfoam to be a product of the polyfoam that you used before? Yes, yes, of course. Otherwise, OK. So I don't remember what and where can you set up? Well, just as implicit in the setup. No, not simple. They're just when I have, so I always distinguished between trivial cylinder components and j-holomorphic cylinders. The trivial cylinder component is just top and bottom. I have the same thing as homotopic to the j-holomorphic cylinder. I think there's a condition that the map from S to the breaking of it should be a finite problem. And that means you just. Sojective. Yeah, sojective and finite. OK, so now here. So you have the following strong compactness property given the reflexive compatible auxiliary norm. And there exists an open-neighbored U of the coarse-modelized space. So the unperturbed j-holomorphic curved-modelized space such that the closure of this UN, which I just described, has the following property. For every connected component A, the intersection is compact. So you can choose neighborhoods precisely with the properties which I described to you, which live under this inductive process. And you have a guarantee of compactness of the closure of this thing here, which satisfies this. So if you look at all those things, the closure of this associated UN is compact. And you can choose to be a neighborhood which is compatible so it satisfies these conditions. So what does it tell you? That tells you that in your inductive argument, the neighborhoods you are working in then appear as a product are the restrictions of neighborhoods you had already there. And then if you extend then in the perturbation by the construction before, the pullback of the perturbation by the covering transformation because of the definition of the norm maximum of the two things has the same norm on the boundary. So it's less than 1. So suppose the first we take a sequence of epsilon i's less than 1. So starting with epsilon 0, you perturb smaller than epsilon 0, then you get something on the boundary smaller than epsilon 0. Then extend smaller than epsilon 1. Then on the next level you get something smaller than epsilon 1. Extend smaller than epsilon 2, always controlled by this neighborhood so compactness never grows away. So let's see. OK, but I explained that in more detail. So then there is one issue. So let me first say if I have such a neighborhood u which has this property, I say n u as in the theorem that n u controls compactness. Yeah, so I think that's a good word. It controls compactness of the situation. Yeah? Can you say something about why the theorem is true? Yeah, because of reflexive auxiliary norms. It's almost true, but there's a certain problem in the transition when you lift up by the covering and then one extended to the boundary. So it's a little bit of a subtle argument. It's not true. So I wouldn't bet my life on it, neither your life, that such a property holds for arbitrary auxiliary norms. It's possible. Actually, nobody is alive in this room. OK, everybody happy now? OK, so. So now here is one issue which one should point out when one easily forgets about it.  If you take a product of two compactly supported perturbations, the product is generally not compactly supported. The solution set is still the same, because you could be here in the compact support and here outside. It's still non-trivial, the product. Can you draw a picture with your set? Yeah, so if you have a line and this is a support, cross a line where the support is here, then in the plane it looks rather like this. That's the support. But the solution set, of course, for the product still lies in here. So OK, so that's. I mean, if I take a function on one and a function on the other. Yeah, so if the support is here and here, then for the product the support lies in that. If f times f prime is non-trivial. No, no, no, it's a function now from. So you have a function from r into r, cross a function from r into r. So you get something from r2 into r2. Yeah, it's a Cartesian product. So that's what I have. But you're adding the function, right? No, no, no, no, no. I mean, you have a threat on problem here with the compactly supported perturbation, say infinite dimension in here. Then the product threat on problem would not be compactly supported. I mean, everybody likes this fact, so you find it in a lot of papers. But it thinks that compact support is compact support. So everybody is a generally accepted fact. But you have to be a little bit careful. What do you mean by saying that this solution set is convenient? No, the solution set is, of course, a product of the two solution sets. So if the perturbation was compactly supported and forces you to have a solution. So if you know that for each of the freedom problem, the solution set is compact, then the product set is compact. You want to know what that means? OK, suppose you have a map from the real line to the real line and the support is here. Suppose I have another map of this kind with the support last year. Takes the Cartesian product of the two things, you get a map. This means ff of x comma g of y. So it's a map from R2 to R2. So then the support is here, this. And that is precisely what we have here. So now, however, if you explain the notation that's in there, what is n sub lambda? What is t dot lambda? Yeah, I did, actually. There was a definition somewhere. t dot lambda was a rescaling by lambda when you were sort of multiplying by lambda. Oh, it was there. t dot lambda. So here, u n consists of all the morphism classes in u, such that this is true and this thing only has 12 cylinders. Little n sub lambda. Oh, a little n sub lambda. Here, it's just the maximal norm you see for your section. So you have a multi-section functor that goes down to orbit space. And over an object, you look at the maximal norm of one of the branches. And that's invariant under isomorphisms. So you get a function on the orbit space. So now, however, if you are in this situation and you know that this space, so let me give this in the simple case. So suppose your freedom problem is this. And then you take a perturbation. And you know that for this perturbation, the solution set where you allow any t between 0 and 1 is compact. So if you can switch it off, then so there's a little bit stronger assumption than just assuming that this is compact. But of course, it's true in these kind of sets of all ways. Then if you take the product which would not have of the perturbations, you take two such things. Suppose S of x has bounded support. Then of course, this was happened here. But it would be true with two parameters t and s. But because you can homotop away, you can put a bump function front of it, which actually is supported, which is just identically one here. And you cut everything off around it. And it would not influence the solution set. Why does two parameters lie in through that? No, no, that is even unimportant. So what is important is that it means if you put so, if you replace, so now replace, if you take the product, take just put then in front of this perturbation a cutoff function. And suppose the second problem was this, times S prime. So here's x times y. So because of that property here and the control compactness, you can, if you take the set of solutions of this problem for minus 1 to 1, then you can actually find a cutoff function which just on the solution set has to be 1, you can cut it off. So there's a different perturbation which actually would have bounded support. It would have the same solution set and the same linearizations as a solution. So there is this marginal difference. However, if I would require this, then this nice multiplicative property goes away. So if my perturbations are multi, I have a definition that on the fiber product and so on, then it's precisely the one which has this support. But it has the same function, then there's another perturbation inside which would have bounded support. So therefore, that is a little point one has to observe. So without loss of generality, you may assume that as far as compactness is concerned, the support is in you. And then we have compactness. So you need this property here, then this is well behaved. But that precisely we have because I have the statement about UN which precisely says, if this here is positive, which precisely says this. Let's see, where is this? Somewhere. Oh, here. This property. But this means if you perturb this by t times some section which has norm less 1 for t between 0 minus 1, you have that property here. So it's built in. So of course, I could have gotten the way without saying this because nobody really cares about that fact. But it's sort of good to know. But for the following argument, I will always say that the support is in you because I could modify it. So for the following, we assume we have compatible auxiliary norm, we have a compatible neighborhood of this, and you controls compactness. Now since the local models are built on SC Hilbert spaces, we have smooth partitions of unity and so that the supports are locally finite, of course, these things can be used to patch together and construct this SC plus multi-section functions that I explained to you last time, how to construct them. And you can construct them controlling derivatives and whatever you want. So precisely, so whatever you want to do in finite dimension. So the whole whatever you can dream up in finite dimension you can do with transversality you can do on this level. Because you have the same amount of freedom as far as perturbations are concerned. So on the previous slide, when you said N sub lambda, so this is the maximum norm? N sub lambda, yeah. So it means at every point, at every isomorphism class are less than 1. So again, over all of this? No, no, no. If this is less than 1, on what? Everywhere, everywhere, on the whole of S. But in addition, it has this property, which would be precisely, it would be an example here. If I just take the product, that would be a perturbation, which would be non-trivial on a rather large set, but it would have this property. So the end quality on evolving T in lambda, there's just an additional sign of T, isn't it? No, no, that's precisely this problem for each branch. So this is just the rescaling of the branches here. T in front of this. It's not T times lambda. T lambda is the rescaling of the local sections by multiplying them by T. So how does the point of lambda is supposed to be contained in U? Is that not true? No, no, no. This is still addressing the starting point of my discussion. So if you have this property here, then even if you arrive in this bad situation here, you can find a new perturbation, which has precisely the same solution set, with the same linearizations, which has support in U. So when I'm squaring back, so I now forget about this fact, and in my inductive procedure, I will always say the support is in U, which literally is not true, but would have to be true if it's compatible with the other structures. But since you can cut it off, and once you are a little bit inside this thing, then you can just go. So you have this big support here, then you go in, which is on the boundary, then you go inside, you just immediately decrease it to something that you line the set U. OK, so what I just said, transfers are like infinite dimensions in the case of sections. So now here's the perturbation algorithm. So the following result, given a compatible UN, so it's the things we just constructed, they exist for every epsilon between 0, 1, a compatibility SC plus multi-section function. So it's compatible with all the things we ever discussed, here controlled by N and U, satisfying the norm is less than epsilon, so that lambda and d bar are in general position. So that's the basic result. So here comes now the proof. So let's recall this. Then let me define pi 0 less or equals j, just the union of all things which have, so it consists of connected components, but the maximum occurrent flow number is j. So that is how we do the induction. Then of course I allow that, I don't know why I wrote this. Forget about this. So we say A has at least two non-trivial components if there exists an alpha, this d alpha equals 0, and alpha having two non-trivial components, so we know this already. We say two classes are related if the image in pi 0 is the same, that means this thing and this thing just differs by shedding a certain number of cylinders, triple cylinder buildings. Then E less or equals j is just E restricted to this, and this thing here just stands for the union of all. So it's S, the full subcategory associated to all objects which lie in some component, which have an isomorphism class lying in some component here. And lambda j is an SC plus multi-section function defined on this, and the inductive procedure starts, constructs a lambda 0. Then I go to the next step. It extends what I had before and goes to the next level, and so on and so on. OK, inductive statement. Given a strictly increasing sequence epsilon j in 0 epsilon, there exists for every j in SC plus multi-section function having the foreign properties. One level up, it's restricted to the previous level. It's the one which I constructed. The norm of the thing up to level j is less or equals epsilon j. The boundary of the thing on level j is the pullback of the data which I constructed up to level j minus 1, because the boundary faces are coming from stuff below. Then this restricted to Ej should be true. Then this has to be true, which precisely says that I don't perturb over the cylinder components, n is compatible with the local family structure. For some reason, that should be blue is black, I don't know. It's controlled by nu over the. So lambda j, the compactness is controlled over this set over the closed open subset of the orbit space associated to the connected components in j. And these things are in general position over this. So if you can go to the next step, then you have a perturbation which produces compactness and everything. A lot of statements here, but basically it says, OK, so let's start. So the induction first step is induction. So that's easy. So take first a component without trivial. Take first a connected component which does not represent object with trivial cylinder components. So these are the things which cannot decompose further. I have an open neighborhood around the solution set here, which lies in A. So of course I can have only one connected component because if I would have two, non-trivial could shift them against each other and I get boundary strata.  So I would have boundary strata. So actually, if I exclude trivial cylinder components, I only have one connected component. Then you find a small perturbation lambda A on this thing with norm less than epsilon 0 so that this here over SA is in general position and the support of these sections occurring there is in you, the neighborhood which I constructed. So this is basically the statement. If you have a freedom operator with compact solution set, you take neighborhood of this. You take a perturbation with support in this neighborhood and you wiggle a little bit and it seems becomes transversal. So that's the classical freedom result just like that. So then now take a component which is related to A. So this differs from this by having trivial cylinder components. So this is how you define it.  So now this is obviously transversal as well because the perturbation is transversal over the cylinders because you have automatic transversality and otherwise it's the previous space. So then the support of this lies in you because you intersect with A prime because there was no constraint whatsoever with respect to the cylinders by compatibility condition. So now the collection of these defines lambda 0, 1 and it satisfies ST0. That is straightforward. So now. I'm like, before you move on, can you briefly say what the theorem is saying? Because it was like. Oh, OK. So you are sure. So it says that in a consistent way I can perturb on each connected component. So there is a SC plus multi-section function globally which perturbs the freedom problem in general position and it has the additional compatibility conditions like the data before when it occurs as a fiber product of another component then the perturbation there is a pullback of the stuff which was constructed there. Then if I have a building which has two non-trivial components and I move it against each other and that was a solution, that was a solution, it's also a solution. Then if I have a solution, I add trivial cylinders to it which are geomorphic, it's also a solution. So this is precisely if you have this and now you compose, so if you have this thing and you compose this with the Cauchy-Riemann operator, you get a smooth weighted subcategory which locally can be represented by this file manifold, that is precisely what it says. And these are the things you can integrate over if you have orientations and so on. So that is precisely where you can sort of do what is in most papers when people normally are like we could do with transversality and then they end up at this point, then they say what they would do if things would have been generic. So there's a lot of documentation about what you have to do. So if you have this, you can basically actually take this paper about SFT and just start implementing the things rigorously. Okay, is that clear? Okay, so now. So assume this is constructed with this required property, so the norm is, for example, less or equal to epsilon n. So now we pick a component A, so one level higher for which the objects are connected. So which means there is, if you look at a building of height one, it looks like this, of course on the boundary it might have to compose this but it's still connected, yeah, in the boundary. So now consider the pullback via F theta. These are the covering functors of the previous perturbation. For all theta in the phase of A. So now for each, so the theta, that's, so your, so phase comes from the pullback, comes from a pullback of two problems which are in general position. And then, so for example, if you have two level one buildings, you have two of height one, and they are in general position on the boundary then, then you have still the normal direction to the boundary and you perturb it to make it in general position in the neighborhood with respect to the full space. Actually don't even, so near the boundary you wouldn't even need this because it's already onto there. And then you extend it. So, so they, so if you go, so there's one thing one should point out also that if you are, so there are certain things one should think about it, but they will not cause any difficulty. So, so I started in the induction with the components which don't compose further and obviously I can achieve general position there. Then I take a product of two of such guys. I'm on an honest boundary. So, so if I have a product of two things whose linearization is surjective, then if I just and pull it back, then on the boundary viewed as an operator with respect to the boundaries already surjective. And then you can extend it somewhat nearby would be surjective. And then inside you have to wiggle a little bit to make a surjective there. Then, so now what happens, so what happens if I'm at a corner? So when I have a corner I cannot argue like this. Then this part is restricted, so this is surjective and restricted to this is surjective, but if you linearize with respect to this direction, this direction, you know precisely what the linearization here is. You cannot say now my, I have an additional parameter to help you to make it surjective, so. But, but so you have to look what the linearization is and it's automatically surjective. So, so, so then the restriction to previous things are surjective, so it's in general position, so everything's fine near and you extend it, so.  So, I'm astonished how short the proof is.  So, this first, first without traversing the components, but, okay, so I was here. Extend to related as before, related means I add trivial cylinders. So, now take a component also on the same level without trivial cylinder buildings, but disconnected components. Then you study the phases, it takes a pullback of the perturbations from EP, blah, blah, blah, and it turns out that this thing here, by just, because it can decompose it into disconnected components, which I do, and it just takes a product. That relation was already true on the boundary. It just extended. So then, so then you have to verify that that is smooth, yeah, but sort of, that was smooth, so you just write down what that means. And so then, that's defined. And then, then this buildings here, I extend by adding trivial cylinders to it. So, now, so now here, when I take products and so on, you, so there is this point with what I discussed, this difficulty which discussed here. But if you believe me that I could sort of modify it and bring it into something which has support in you, then this procedure never, will always have multi-sections which have support in you and are small and a priori as a compactness result says that my solution spaces are compact. So then, this is a step and this is an extension of the previous one and so on. So, now you have transversality. So now, remarks, so I started 20 minutes late because of you. Six. Okay, no, no. Okay, so I had actually two endings for my talk and one ending was to show you the homotopy argument which is actually between two perturbations which is actually more interesting than that one. So that's rather straightforward. So you cannot have general position when you homotop from one to the other and the reason is that you have two T dependent. So if you take, if you go from, so on the zero level you can homotop nicely, but you also want to homotop for example like this that if you look at the T projection that you only see more singularities for example. So if I look at the modular space with the most singularities for example. So then you do the product and go to the next level, but then of course the perturbation is the fiber product with respect to T. That causes some constraints in general, so things are not, and that prevents you from putting things into general position because of all the other structures which you have. And the problem are precisely situations like this where you, so also if you have two different components below then of course when the modular space and have T you can talk about that you only see most type singularities and on different components which are really different you can make sure that the critical levels are different by generosity. But if you have a component and you add trivial cylinders to it the most function is the same. So this thing's already produced the same critical point, a critical point, so like this. Suppose you have this thing here, say beta, this is a periodic orbit, gamma. So then you can glue that thing on it and you have a beta here, gamma and you put a cylinder on this. Now you see that this thing plus the cylinder here is related to that object. This differs from this object only by adding a trivial cylinder and gluing it on. So the perturbation you see here is the same perturbation you see here and you build a fiber product that causes immediately difficulties. But the deficiency in the cocoon is one and you have a normal direction if you glue these two things you can kill this but you have to find out what a good perturbation is because when you iterate this process then precisely this kicks in, this problem. So the result of this is that you cannot achieve general position but you can achieve identifiable garbage plus good position and you only have to integrate about good stuff. The other is just, so identifiable garbage is something like this. You might have a solution here but where would the solution space be? It would be something like this, it lies outside. The stuff you see as too small dimension. Whereas other stuff looks like this.  I mean, if you're just interested in zero-dimensional components it's what happens is precisely you might have this picture where technically it looks like there would be something lying outside of the space so that doesn't play a role but then all you have something like this which is good. Which of course in general position would have been something like that. So you come out of a corner. Okay, so there you have to actually explain more in the transversality during the homotopy which was one ending but after my disaster's first lecture there only came through half of it. I had to cancel that plan and presumably now I have to cancel that plan here as well because our chairman is really tough. So remarks on, I have to stop? Okay, so here, okay, so let me just make some one minute. We do have a discussion. Right, but I have to prepare everybody for the discussion. So let me just say, so at that point you want to introduce invariance so what you have to do is work out orientation things and you run, you see that there are some issues and the best thing is to go to a covering which is precisely in the SFT paper where you number the punctures, you put asymptotic markers, you put random mark points on the periodic orbits and have some requirement, then you see that you're basically oriented except with the occurrence of bad orbits and then you can write down integrals and then you get the relations by this modular space is how data is related and then you can represent the data in whatever we said there in that paper. That's a remark. Great questions. Yes, so critical cylinder buildings, right? So you're saying you want an ampere 40-fold in which the constant cylinders sit. So in my world, what I'm hearing when I hear that is you want local slices to the R action on maps from R times S1 to a manifold M near the ones that happen.  So when I look at stable maps, then I would have maps from... Oh, the question is what is your ambient polyfold for just for the trivial cylinder? It doesn't. It never occurs by itself. It only occurs in combination with something else. But there is one, right? And it's sort of... No, but the construction you have is not a product construction. So you cannot have a trivial cylinder by itself because it's an unstable object. So moment, but a trivial cylinder occurs always together with a non-trivial component. And then when I construct a global slice, it's not a product of a construction on a cylinder and the other one. It takes data from both and mixes them up. It's not a product. So therefore, trivial cylinders never appear alone because they are unstable. They can only appear together with a non-trivial object. And then when I make a slice with respect, because when you make a slice with respect to the R action then you fix say the non-trivial part but you can slide the cylinder against it. Yes, yes, and that has to be built in but it's not a product situation. Because you also have to divide, you have to take a transversal constraint, you have to divide out the automorphism group for the cylinder as well. So there are several things which you have to do at the same time, which go into the construction of the uniformized. So let me get this right. You're saying there's not an ambient M-polypone in which the trivial cylinder sits. No. Okay, that's just what we need to break out of. I'm just going to push this over here.


Hi, I'm Dan and Baker. I work at Johns Hopkins as a member of the galaxy team and I contribute to the back-end working group that I'm representing here This is a short talk in our series covering recent efforts. We've made towards modernizing the galaxy back-end I'll cover Decomposing some of the work that the galaxy server does into smaller units for asynchronous consumption by a task queue We're doing this to both improve the end user experience and increase the scalability and robustness of the platform Okay, let's start with a good example of one of the primary problems. We're working to solve here Galaxy has traditionally performed many maybe even most activities other than actually running jobs within the synchronous request response web or API transaction To make this a little more concrete consider the diagram on the right here When a user in the galaxy web client requests the purging Which is actually deleting off the disk of a data set The server receives the request and then does the work all of it including the disk IO and updating the database and Then responds with a confirmation that the request succeeded and then that's what you see in the galaxy client This has worked pretty well until recently with advances in the number of data sets the galaxy can process and Now it can be a serious bottleneck for any functionality that performs file IO or expensive database queries Thinking about the data set purge task, but considering The tens of thousands of data sets you can have in a collection from a modern art Analysis pipeline the synchronous response when you click to purge a collection can take minutes or more I probably don't have to tell you that clicking on something in the interface and just waiting for something to happen for Tens of seconds or even minutes without any feedback is pretty bad Even worse for very long requests the galaxy proxy server Can think galaxies dead and drop the connection returning an error to the client even when it's actually still working on the back end Okay, so what can we actually do about this? The gist of it is that we need to refactor galaxy to encapsulate these expensive units of work as tasks that can be run asynchronously whenever appropriate In this diagram I've added two more components a queue and a worker process The client will now make the same data set purge request to the server But instead of actually doing any work the server just logs a task on the queue with the right parameters and then tells the client Yeah, got it. It'll get taken care of The worker is watching the same queue asking if there are any tasks it can handle when one comes in It does the work and the task is marked as completed however long it takes Without blocking the user if you're familiar with the galaxy code base You might be thinking that we already have a queue and a queue worker. We do kind of These older modules use the kombu library for messaging But those tasks are handled as a part of the existing galaxy process just in a separate thread It turns out it's not as flexible as what we need and what we implemented there ended up being mostly useful for inter-process communication Celery is a library that sits on top of the same kombu messaging layer providing some very nice abstractions for Formalizing chunks of code as tasks and distributing them and making sure they get taken care of Okay, let's take a quick look at the code for this focusing on the purge data set task again The left side is the original code for purging a data set taken from the HDA manager It gets a handle to the user figures out how much this data set purge should affect their quota and synchronously calls the method To actually delete the data set from disk and flesh out the results to the database The code on the right is kind of what it looks like after refactoring on the top. That's the purge method now It checks to see if celery tasks are enabled And then just dispatches a task by calling purge HDA dot delay The purge HDA method is below And that's the celery task wrapper itself It hopefully demonstrates how easy some of this refactoring can be the wrapper is just a function that calls the same exact code You see on the left just now as HDA underscore manager dot underscore purge with the the right parameters So looking at the purge HDA task again You probably notice the two decorators and additional typing in the function signature The first decorator celery app dot task just defines the function as a celery task and says that we don't need a return value from it Pretty standard celery stuff. The second decorator is where some magic happens and that's a custom decorator We've added uses dependency injection to bind an appropriate context to the method from wherever it gets called in this case from the celery worker I just wanted to point it out here because this is really cool new stuff in galaxy And if you'd like to learn more there's a much deeper dive in the architecture content So wrapping up you can try this right now You'll need to enable celery tasks in your galaxy config and then you can start galaxy up using circus to handle the Galaxy and celery process management for you Circus is the other new bit of the stack that we don't have enough time to go into but in short It's nice python process and socket manager kind of simpler supervisor That allows us to manage the seller worker processes and maybe more in the future The issue linked here is where we're collecting tasks that should be offloaded to celery if you'd like to take a look to help out or Recommend particular pain points you might have noticed and Lastly some specific feature work that we expect to be pretty impactful is to rewrite job Handling and workflow scheduling is tasks right now both of these processes are fairly rigid But indeed composing this work as tasks we can have much more flexible deployments as an example in a Kubernetes-based galaxy you could have robust Horizontal scalability pretty easily by linking the number of celery worker pods to the size of the task queue Your deployment would then autoscale up to meet an influx of requests or workflow submissions and then back down as the queue goes idle All while leaving the now thinner galaxy core web app running at high performance. Thank you