Mother, why did daddy switch to Postum? Your father says there's no caffeine in Postum. Nothing to spoil your sleep. And your father knows best. Yes, it's Father Knows Best, transcribed in Hollywood, starring Robert Young as father. A half-hour visit with your neighbors, the Anderson, brought to you by Instant Postum, the good-tasting drink that's entirely caffeine-free, and by Post's 40% brand place, America's largest-selling brand place. A field who said, if wrinkles must be written upon our brows, let them not be written upon the heart. The spirit should not grow old. What does this have to do with the James Anderson family of Springfield? Well, really nothing, because here is a family wherein the spirit burns warm and bright. But sometimes a man's mind plays little tricks on him, and on such occasions he needs a little nudging to return to a belief in his own basic opinions. For example, the other evening when the White Frame house on Maple Street was bustling with the activities of Bud and Betty as they made preparations to attend a wedding, Jim received a phone call which was to send him on a mental roller coaster ride. Like this. Well, that sounds great, George. We'd like to very much. Well, that's fine with us. Now, we'll see you around eight then. Fine. Goodbye. Margaret, better get out your tennis shoes and your gym bloomers. I'd better do what? Take a light workout and a quick rub down. Pack plenty of liniment and a few splints. What in the world are you talking about?  Badminton? Oh, dear, are you out of your mind? Not at all. That was George Phillips on the phone, and they've invited us over for a few games this evening. Oh, no. Most be over there around eight. Dear, you're not serious about this, are you? Well, certainly. Why shouldn't I be? Well, after all, we're not exactly teenagers. So what?  For heaven's sake, Jim, we can't do this. Why not? They invited us. Well, it's a very flattering invitation, but it came a few years too late. Now, where did you get an idea like that? You talk us over a couple of doddering old octogenarians. Well, you look as young as the day I married you. Thank you, dear, but I really... And you know as well as I do that old saying that you're only as old as you feel. I don't feel any older than I ever did. As a matter of fact, I think I feel better than I have in my whole life. Dear, even if I felt up to this, I don't know how to play badminton. Neither do the Phillips. We're just going to get some fun out of it. A little much needed exercise. Well... And this is a good night to get out. Didn't you say the kids were all going to be gone this evening? Yes, Betty and I are going to Evelyn Brickard's wedding. And Kathy's going to spend the evening with Patty Davis. All right, then. What are we waiting for? Well, I feel we both should have our heads examined for this, but... all right. Good. Where do you think those old tennis shoes of mine are? Probably in your closet where they've always been. That's a good place to start looking. Are you coming up? In a minute. I've got to put away a few things in the kitchen first. Okay, better hurry. You sound sharp tonight, Father. Thank you, Princess. I feel great. I wish I felt that good. I'm completely but completely worn out getting ready for this wedding. Well, you can feel good if you want to.  Let's have an awful tired attitude, then. Ow! Ouch! What's that? Oh, that's Bud. He's in the bathroom shaving. Shaving? Bud? He doesn't shave yet. He just started tonight. I didn't think this historic moment would arrive for a few years yet. Well, it hasn't really. All he's got is a little bit of fuzz. But he says if a boy is old enough to wear a tuxedo, he's old enough to shave. Ow! Doesn't sound like he's exactly enjoying it. I've been trying to get him out of there so I can take a shower. How long has he been at this operation? For hours. We're both going to be late if he doesn't hurry. Bud, hurry up, Bud. Get away. I'm shaving. I need to get in there and shave myself. Bud, are you about finished in there? Bet you, Dad.  I'll be out in a second. In the length of time it's taking him. You'd think he had a beard. Okay, next. Well, what was going on in there? All those low moans. Oh, nothing. Just combing my hair.  What's all the adhesive plaster on your face? Oh, nothing.  All right, Gabby. Who asked you anything? Who weren't you? I'm going to take a shower while you have a chance, Betty. All right. But go on and tell him what you were doing, Bud. Go on. Oh, get lost. Bud, you don't need to be so secretive about shaving. That's something all of us men have to put up with. Us men? Sure. As a matter of fact, if you told me what you were doing, I could have given you a few pointers. How can you stand to go through that every day, Dad? Well, there's one way of avoiding it. Don't cut yourself. You mean you can shave without cutting yourself? That's possible. Mommy! She's downstairs, kid, and she'll be right up. Bud, you better start getting into that tuxedo you rented. The tie alone will take a good hour and a quarter. Yeah, I better. Where did you say Mommy was, Daddy? Downstairs. She'll be up. Anything I can help you with? I don't know. What do you think I ought to take over the patties to play with? Oh, better take some dolls. Dolls? What do you think I am, a child? Well, I was under that impression. I'm pretty sure you're not a dog or a kangaroo.  I see. What are you, an old lady?  Well, so now I'm getting that from you, too. Dear, I forgot to mention you got a letter today. Mommy, what should I take over the patties tonight? Well, you'd better take some pajamas and a toothbrush. I mean to play with. Oh, well, why don't you take your sewing set? Yeah, I forgot about that.  She's given up dolls and she's discovered that I'm corny. Oh? Those are two sure signs right there. I'm afraid they're all growing up. I was just thinking today about this wedding they're going to. That's the third one of Betty's friends. It's gotten married lately. Next thing we know, it'll be Betty. Yeah, I guess we are going to have to face that sooner or later. Bud's the only one who doesn't seem to grow any older. You think not? You should have seen what he was doing about 15 minutes ago. Shaving. Shaving? Oh, no. He's got the scars to prove it. Oh, dear. The years are certainly slipping by fast. Yeah, fella doesn't realize. Oh, here's that letter I was telling you about. What letter is this? It came today. I forgot to give it to you. Well, what do you know? Look here, it's from old Eddie Gilbert. I thought you'd be surprised. I haven't even thought about Eddie in years. I wonder what he's up to these days. Well, I'd better see if Betty's got some rubber-soled shoes I can wear tonight. She's been taking a shower. Old Eddie Gilbert. I think I've seen him since we graduated from high school. What a guy. I remember he had more energy than sense. Oh, say, Betty, do you have any rubber-soled shoes of any kind? Rubber-soled shoes? Mother, I hardly think that would be the thing for the bridesmaid to wear. I don't mean you. I want to borrow them. Well, Eddie works for the gas company. What do you want rubber-soled shoes for? No, your father and I are going to play badminton tonight. What? Oh, you're kidding. Ask your father. Well, what did you say I was reading here? Father, you're not really going to play badminton, are you? Well, certainly. Why shouldn't we? Well, it just seems that you're too... Well, I guess it's all right. It seems that we're too what? Nothing. Gee, I've got to get dressed or I'll never make the wedding. We'd better get ready too, dear, if we're supposed to be over there by eight. Margaret, are you sure you feel up to this? Well, I don't know. I feel all right now. I don't want to urge you into this. No use taking chances. Well, I doubt if they'll be up to much of a game either. What does Eddie have to say? Oh, I haven't finished his letter yet. Says he might have to make a trip over here to Springfield in a day or so. Says he'll drop in if he does. Yeah, that'll be nice. Yeah, that'll sure be good to see, old Eddie again. Old Fireball Eddie. Do you think this skirt and a sweater would be the thing to wear? Yeah. Eddie says he's been having a little trouble with rheumatism lately. I can't imagine anything like that catching up with an active guy like Eddie. Well, of course he's not as young as he used to be. Well, you mean he's no older than I am. In fact, he's a year younger. Oh. Well, even so. Oh, no, listen to this. He's a grandfather. Eddie is? He's got three grandchildren. Well, they just can't be. Well, let's see. Why don't you say once that he had one daughter a year or two older than Betty and then another one just a little younger? He says the kids call him Gramps. This I just can't believe. Mother. Gramps. Margaret, he's a year younger than I am. Yes, that's what you said. Mother, will you zip me up? Certainly, dear. Gramps. Mother, do you think there's anything to that catching the bride's bouquet? I mean about being the next one to get married? Well, it's an old tradition. I don't suppose it always works out. Gramps. Well, Evelyn's going to sort of throw her bouquet in my direction. Oh. Who's going to referee? Well, look at our handsome young man here. Yeah. How do I look, Mom? Well, you look fine, bud. Real nice. Better fix that tie.  Yeah, I couldn't get a tie dried. Somebody will have to do that for me. Don't hunch up your shoulders like that. Just because you've got a tuxedo on, you don't have to go out of your way to look uncomfortable. Well, it does make you feel kind of stiff-like. Well, relax. I want my children to have an uncle they can be proud of. Your children? I'm going to have two boys and a girl. Marcia says she's going to have three girls, but I think two boys and a girl would be nicer. Don't you, Mother? Well, that only car would be there, Uncle, wouldn't I? Uncle Bud. I wish you had a more dignified name. Hey, Cathy will be their aunt. Aunt Cathy. Imagine having that shrimp for an aunt. Father, what do you want them to call you? What's that? What do you want them to call you? Your grandchildren. My grandchildren?  How about gramps? I think you two had better get going or you'll be late. Oh, creepers, yes. Look what time it's getting to be. Come on, Bud. OK. Oh, Betty, you were going to find some rubber-soled shoes for me. Look around my closet, Mother. You'll find some. All right. Jim, you'll never get ready just sitting there on the bed staring at Eddie's letter. Don't you have to shave? Margaret, maybe you'd better call the Phillips and tell them something came up and we can't make it. Can't make it? I feel a little tired, I think. Why, Jim? Have we got any vitamin pills around the house? I guess plenty of us feel a little like Jim does. Kind of old before our time. Well, here's a man with a few cheering words for us, Ed Prentice. Say, if you feel kind of loggy and dull, sort of droopy, fresh out of energy, well, usually there's nothing wrong with you that a good night's sleep won't cure. Trouble is, how do you get a good night's sleep? Is that your problem? Well, friend, could be the caffeine in your coffee. That caffeine could be keeping you up nights and then letting you down days. If you're a victim of Mr. Coffee Nurse, here's your answer in one word. Postum. That's right. Postum. Instant postum. You just switch to instant postum and see what a difference postum makes in your sleep, in your nerves. That's because postum is caffeine-free. Contains not one iota of caffeine. Of course, caffeine doesn't bother everybody. Lots can handle it. But if you can't, you can switch to postum and love it. Yes, sir, good old instant postum. Get a jar tomorrow and drink postum for just 10 days. Now, I mean drink postum exclusively. And then you'll be the judge.  There have been a few changes in the Anderson household since last night. Jim Anderson, the sprightly young father, seems to have a slightly different outlook on life this Saturday morning as he comes carefully downstairs, makes his way through the house and slowly seats himself at the breakfast table. Like this. Well, good morning, dear. Good morning, Margaret. Where are the children? Well, they slept later than usual this morning. But I'll call them. They'll be down in a minute, dear. Why the dark blue suit? Oh, well, I think it's time I started wearing more conservative clothes.  Well, I don't recall that you ever dress like a schoolboy. Those tweed suits and things, they're all right for younger men. As a man gets on in years, he has to dress accordingly. Time takes its toll, Margaret. The fire is burning low.   What happened to jumping Jim the badminton boy? Well, that was yesterday, honey. I'm considerably older this morning. Well, it's been a long time since yesterday. Look, honey, let's face the facts. We're getting on in years. We're not kids anymore. The old body wears out, you know. Whose old body are you talking about? We can't go on pretending we're youngsters. We're middle-aged. We're past middle-aged. Oh, for goodness' sake. Look at the way the children are growing up any day now. Betty will be getting married, flying away. Bud will be married before we know it, and he'll fly away. Next it'll be Kathy. She'll be flying away. And who's left? Just you and me and our old flying school. Something we have to think about seriously. Our time is running out. Dear, just yesterday you were insisting that it's all in our mental attitude. That we have to keep young in spirit. Well, I see a lot of things. You know what's wrong with you, don't you? Time, that's all. The minutes, the hours, the days. Getting older all the time. You can't stop it. It's that letter from your friend Eddie Gilbert. That's what started the whole thing. The minute you read that he was a grandfather. He's a year younger than you. He's a year younger than I am. Well, what of it? He's an old man with rheumatism. His children both married and gone, grandchildren that call him Gramps. Eddie Gilbert. He's younger than I am. You said that, dear. I had kind of a twinge in my hip when I got out of bed this morning. You suppose it's rheumatism? Oh, Jim. Kind of a sharp sticking pain. I know, I know. You know? I forgot to take the laundry tag out of your pajamas. Well, it felt like rheumatism. What would you like for breakfast, dear? Bacon and eggs? No, I think I'll just have some dry toast and prunes. Hi, Dad. You at the table? No, I'm running a hot dog concession at Coney Island. Am I at the table? I just asked. Treat your father gently this morning, Bud. He's not long for this world. You going someplace, Ed? No, Bud. I'm just sitting here waiting for my breakfast. I'm gonna get the morning paper. Come on, dear. Smile. Well, what's funny? You know, Margaret, it's the kids growing up that makes us feel old. I've come to that conclusion. But you're just as young as you feel. Well, you said that yourself. Well, that was... Hey, take a look at this, Dad. Story here in the paper about a guy shaving. Got tired of fighting with his whiskers every day, so he put weed killer on his face. Oh, no. I guess it didn't work too good. They got him all bandaged up. I can imagine. I don't think I have to shave for a while yet, Dad. Oh? I was pretty sure I could see some whiskers on my face last night, but I can't find him this morning. Here's your post-em. Mm, does smell pretty good. Hi, Daddy. Good morning. Kitten, what do you have on your head? This is my Captain Video Space Helmet. I'm gonna wear it when I go to the moon. Oh. When are you planning to leave? I don't know exactly, but they say we're going to the moon anything now. And I gotta be ready. Hey, let me see that. Don't take it off. I'll lose my oxygen pressure. Let him try it, Kitten.  I will not.  It's my space helmet. I'm not gonna hurt it. You will, too. Quit pulling. Just let me see it, will ya? Well, dear, are those the grown-up children you were talking about? Daddy! I'm not even touching her. All right, you two. Break it up and come to the table. Okay. Tell Bud to stop it. Bud, stop it. I'm starved. Let's eat. Bud, we don't read at the table. Put the comics away. Okay, okay. I don't know, honey. Maybe I will have just one egg and a little bacon. All right, dear. And that's to me. Probably Joe. I'll get it. I'll get it. Look out. Don't knock the table over. I'll get it. It's probably Ralph. I've got it. I've got it. Quit pulling. I've got it. Why don't you do something, dear? I think I'll take off this coat and vest, Margaret. Make it two eggs instead of one. Thank you, Father. Me? Who is it? He didn't say. It's a man. Hello? Yes? Eddie! Eddie Gilbert. Sure, I remember you. Got your letter yesterday. How are you feeling? Good. Good. Sure, we're going to be home. We'll be glad to see you. Can I come down to pick you up? All right. 607 Maple Street. That's right. Fine, Eddie. Goodbye. That was old Eddie Gilbert. He's on his way out. Who's Eddie Gilbert? He's an old friend of mine, Princess. We went to school together. Haven't seen him in years. He's a grandfather now. Well, gee, Dad, if you want to school together, how come he's so much older than you are? Oh, he's not older, Bud. In fact, he's a year younger than your father. Bud, you must remember I'm not young anymore. Are you kidding? My father, there are boys at college who look older than you do. Betty. Well, isn't it true, Mother? Isn't Father just as young as he's always been? Well, that's what I keep telling him, but he won't believe it. That's what I keep telling him, too. Well, now, kids... The last time Father and I were over at school together, the girls were simply swooning. Wanted to know where I caught that gorgeous man. Oh, Betty, stop it. I mean it, Father. So do I. They thought you were my new boyfriend. That's what they thought. What are you talking about, Shrimp? I don't know. Well, can we all get back to the table now? Breakfast is on. Yeah, come on, kids. Get it while it's hot. First come, first served. Let's get with it. Good portion. Well, the wedding was divine last night, Father. Who caught the bride's bouquet? Oh, that Janie Ligget. She jumped right up in front of me. I had it right in my hand. You should have seen it. It looked like a basketball game. Everybody jumping for the ball. The guy could have got killed in there. I had it right in my hand. That Janie Ligget. What a center she'd make. Hey, wonderful breakfast, honey. Thank you, dear. I don't know what I put on this darn blue suit for. Oh, if you're friends coming, Father, I've got to get dressed. Let me know. Look out, you're stepping all over me. Well, move your big feet. Where's my yellow blouse, Mother? In your drawer. I better put another shirt on, huh, Dad? In your drawer. You better go change your dress, Angel. Okay, I'll put on my blue one. It's in the drawer. The kids, I tell you, Margaret, have a person young. I really feel sorry for old Eddie Gilbert. Why, dear? Well, his children have grown and gone away. He and his wife left alone. That's probably why he's turned into an old man before his time. Nothing left to live for. Well, maybe he's like you, dear. Old one day and young the next. No, when the children have grown, that's the end of the line. I could tell by the tone of Eddie's letter. Rheumatism, you know. Oh, there he is, dear. Let me get my coat and vest on. I kind of dread meeting old Eddie face to face. He was such a youthful, bubbling sort of fellow when he was young. Well, hurry, dear. Let's not keep him waiting. Come on, honey. Try not to notice his age, you know. I'll try, dear. Look through the window. Is that his car out in front? That English sport job? No, not Eddie. Well, Anderson, I'm Gilbert. Not Eddie, Gilbert. Sure. Don't you remember me, Jim? Yes, but where's the...  Neither of you.  I can't believe it. Excuse me. I'm Margaret Anderson. Mr. Gilbert. Oh, how do you do? Delighted to meet you. Margaret, this is Eddie. Yes, dear. We've met. Well, come on in. Give me your coat, Eddie. Well, thanks, Jim. Man, you look great. Well, so do you. I thought, I mean, well, it's been so long. Well, I'll go up and help the children. How many do you have now, Jim? Three. They'll be down in a minute. Come on in the living room. You know, I just can't get over it. Can't get over what, Jim? You. From your letter, I expected, well, you'd be sort of older... Why? Because I'm a grandfather? Well, yes. I don't feel like a grandfather. I don't feel any older than I did when we were in school. It's all in your attitude of mine, Jim. A man is just as young as he feels. Well, I've always believed that. I had a touch of rheumatism. But I got that when Gladys and I were out ice skating. Fell through the ice. You know. But it's gone now. Never felt better. But Eddie, what do you do now that the children have grown up and gone? What do we do? Man, we're living. We told the kids, look, we love you both. We'll always be around if you need us. But we're not going to be babysitters for our grandchildren. Gladys and I are doing all the things we couldn't do before. We're free, man. We're having more fun than we ever had in our lives. I guess I never thought of it that way. You know, Jim, I'll confess I hesitated about coming out to see you. It's been a long time. Well, I was afraid maybe you'd, uh, well, change. I never felt younger, Eddie. You'll look great, Jim. Well, like I always say, it's all in your attitude of mine. We never grow old while the heart is young. For goodness sakes, eat post-brand flakes. Ladies, that's a fine little melody to keep in mind every time you go shopping. I say that because new post-40% brand flakes really are good and so good for you. You see, something wonderful has happened to brand. Today's post-brand flakes have a delicious new magic oven flavor that's fast-making at the favorite cereal of more and more people every day. And of course, post-brand flakes give you those important keep-regular benefits that brand is famous for. So next time you go shopping, look for new post-40% brand flakes. America's largest selling brand flakes. And remember that friendly melody. Post-brand flakes are so good. And so good for you. For a while, it looked as though Father Time had Jim Anderson by the coattails. But the old man with the hourglass apparently just couldn't stand the gaff at the White Frame House on Maple Street. Witness the scene taking place at Mount Hall. Margaret, are you ready? Your mother and I are going over to the Phillips for a while. What's in the suitcase? Just a few things we're taking along. You have a gleam in your eye, Father. What are you up to?  I'm ready, dear. What's in the suitcase? That's our equipment. We're going over to the Phillips to play some badminton. After that, we're going bowling and we're going horseback riding. The next week we'll be back with Father Knows Best starring Robert Young as Jim Anderson. Until then, good night and good luck from the makers of Post 40% brand flakes, America's largest selling brand flakes and Instant Postum, the drink that's entirely caffeine free. In our cast, we're Rhoda Williams as Betty, Dorothy Lovett, Ted Donaldson, Helen Strome and Ben Weldon. It's the best hot cereal you ever ate. Post wheat meal. The best hot cereal anybody ever ate. Rich and delicious with a nut-like flavor you'll never want to miss. And hot Post wheat meal is so good for you. Packed full of solid whole wheat nourishment, especially good for children. Post wheat meal takes just three minutes to cook. Get the big family economy size with a picture of Roy Rogers on the package. Post wheat meal. The best hot cereal you ever ate. This is Robert Young, ladies and gentlemen. I want to welcome all of you youngsters who have recently joined the Robert Young Good Drivers Club by signing the man-to-man or dad-to-daughter pledge for safe driving on our streets and highways. If you are not yet a member, please drop a line to me at this NBC station and your membership card and copy of the agreement pledge will be sent to you. Thank you. Father Knows Best was transcribed in Hollywood and written by Paul West and Roswell Rogers. This is Bill Foreman speaking. Tonight, play Truth or Consequences on NBC. Thank you.


So, jesrde, we were in the process of performing this localization computation of the partition function of the dimensional and equals to comma 2 gauge theories. And here, we are constantly into vector and还是 multiplets on the untwisted around 2. Pistim, da bi bilo pozorovati srednji, da smo vsega taj viče odvrli vsega. Zato smo odvrli vsega vsega vsega vsega. Nelam ne jazno izgleda, bo da je zelo narednico. To je vsega, da je vsega večo večanje, in je bilo komplexivno. kratične interakcije, ki je standard. In zelo smo zelo skupili materiakcije, zelo skupili kinetično for karan multiplič. In karan multipliči je v nekaj reprezentacij, zato v plizi je, da je kontrakcija z zeločenimi indice. Zeločen R je zeločen R, kako R je zeločen kratične. Vse je Vse. Vse je�o u punic Najlandski sama, kaj se vse začne, tač nemoramo z kavim aktivistim, da vse nas tačne vse je, začne je rank attribute dr. Tako, tako vidim, skoliko, tess, bila, nimi je Slatom vzačne. Tako, tako, tess, bila, tess, bila, tess, začnost. T, bo, bo, bo, to, Tukaj imamo 4 na vse predstavne predstavne, ali posledamo 1,2 tukaj, 1,2 tukaj tukaj. Tako, če so tukaj, v teori in interakciju? Zelo, da imamo super potenšel interakcij, in v sej super potenšel interakcij, tako, da imamo tako, tako, da imamo tako, da imamo nekaj olomorfički funkcij na karelj multipleti. Tako, to je karelj multiplet, tako, da imamo karelj multiplet in tako, da imamo f terma in tako, da imamo vse karelj multipleti v teori in to prijevajte, kar super potenšel interakcij in teori in zelo, tako, in vsej, in tako, in tako, da imamo karelj multiplet v teori in teori in tako, da imamo kaj je tukaj komponent tukaj karamultipleti, je to samo izgleda, da je tukaj v vsej spas, da je vsej derivati superpotenšel v svojimi karalfitih, da je tukaj, in tukaj da je vsej ukawačne interakcije. Tako je, da je tukaj izgleda, da je tukaj vsej tukaj vsej, in tako vsej, da je vsej, da je vsej, Ko ostajremo polinomi nekaj, to ognist愵 za košnje, ali ne pa ne bi tem, ki oblijemo patintečne nakoččne za košnje na superpotenčne. Izdrža in je nekaj, da teorič naredil independent z superpotenčnev. Zato, kako superpotenčni da bo vzelo R2, fpsestimo R-cimetrij, kaj ne kaj ma drugi del, je ni zgledaj, da bi njihovi postanje za vsega, ki je to samo vsega, ki je to dobro potenčenja. Zelo je bilo? Zelo smo tudi tudi tudi potenčenja, in tudi je semila, da je to je zelo vsega, da jelem tudi tudi kali grafik V, tudi tudi karal multiplet. In tudi je to, da so tudi malo morav vsega. Zelo, da smo priči vsega, da je to liniar, Zelo, da je linijačne, nekaj je konstant, nekaj je konstant nekaj je teori. Linijačne termi so kojefitični. Zelo, da sem se parametrizala z nekaj rejalnih part in nekaj imaginaričnih part. Zelo, da se počekajte, nekaj ga je v dela granžan, nekaj ga je komplekcij fajetiljopolis termi. If you wish this real part controls a linear term in proportional to the D term, that you put in the action, this is called the phytelopolis term and imaginary part controls the topological term. The field strength integrated in a two-spher, we are in two dimensions so we have such a topological term. What is it called, the teta angle, because the theory is invariant under shift of this angle by two pi. This is very similar to the term FJGF that you have in 4 dimensions. Of course, we could consider a higher, higher power of the sigma, but let's consider this very simple case. And the interesting thing now that this time is not Q exact. So in fact, our partition function will depend on the parameter, almost generally, on the various parameters so we haveittingijs with super cryptocurrency. Pomembno so bili vzdušnjih mase, in vzdušnjih mase, kaj je zelo svoje vzdušnje, in zelo počusti sva, kaj je vzdušnje mase se postavili na našlih smetri, včasno, ils tega vzdušnja v kartanju, subalge, in vzdušnjih smetrih. Kaj je, če ti je smetri, ali se vse vzdušnje nr. vse, in je šeljena v rovku vektormultipljete, zazve si ga prejsteši in pošliši, od nekaj posluši vzvečenji, a zelo je korisponsov, kaj plači se odvajne. To plače, in tako, je velice in zavečenje vzvečenji vzvečenji vzvečenji vzvečenji. Zato to pličenje vzvečenje v izku, vzvečenji vzvečenji vzvečenji. Na zelo je vzvečenji vzvečenji vzvečenji. In zelo sprem, da je izgled, da se da vse sveti. Zelo prejder, da vse vse vse vse vse, zelo izgleda spetetjče tudi vse vse vse vse vse vse vse. Kako iko tudi vse teori, posebno vsega, kako izgleda toga gravitino in tudi održiv. Vse vse vse vse vse vse vse vse vse. Zelo vse. In je ta vse obtah, da se dajimo, zelo se njimljeznje. Zelo se imamo zapraveno taj kundišen, kaj imamo opravboj do vse. Zato da sem začala povradi, da je vse, kaj je povradi. Ok. Zelo sem mužel njega premačnje v skupi vsej lokalizacij. Zato bi se nekaj dvjih mladov, zelo sem kaj povradi. Svet začak, da smo počkati vzahvali tudi iz veči dobro vzpečne termje, in da se zelo zagravimo hodnje. To bi smo počku, da smo počkati, da so jeli vzpečne nekaj vzpečne. To bi smo jeh tudi nekaj, da je vzpečne začak, da bi je tudi začak zelo nagrače, je to superimetrične, je to na veče, da so vzpečne in pomečite, da se počkati in vzpečne. To je tudi tudi vzpečne, Here comes the obvious contour which is just take real fields, which are complex conjugates, and this is manifestly positive. This is the sum of positive terms Of course let me stress once again if you don't impose a contour, there is no way that this is going to be positive, because we become complex fields, it is getting the generic values. But on the real contour this is a positive, so this is okay. T JOJSOKI Povedajte, da je tko, četeno o dvoreza povedajte bozonicje bozice, ker, zato, četno tako downloadede ili da je to vse pa je tko, tako, tko, četno tko, je to ovoreza povedajte četno da je tko, zamerak tko za tvoje 0 in 2 in tko so dodačimo komputation v tom, četno tko, Vspite sodak, da to neč koputka tega se odpeljezava. If you do the same thing, sum that you sum over the fermions of Q psi, the agru psi in you have to have Q on that. You get precision with this. This one is not in the kanonical form. If you prefer, and this is related to the fact that there is this imaginary part that is not positive, so if you prefer you could instead kronikl form, je veliko izgleda z toga, ali je bilo dobro. Zdaj, da je bilo dobro. Ok.  Zelo, da je bilo površen, bozonički površen je površen, zvrst received to zero each of the squares. So the equations are just that F is related to sigma two, d is related to sigma one. These derivatives of sigma one and sigma two are zero, that's valid in commutator. And then once you have fixed this, again, if we are in this region for the charges, In se je zelo, da je to vse odložite. Zelo je zelo, da je vse odložite. Vse je zelo, da je to zelo. Vsih je zelo, da je v sektor. Zelo je zelo, da je vse odložite. Prejde, tako, ta rekojst. Prejde. Takaj rekojst je zelo, da vse zelo, da je skelo. In zelo, da je skelo. In zelo, da je skelo. Tako, da je skelo. Bhejte nekaj dolič, bo se vse zelo, izgleda se, da je tukaj držav, izgleda se, da sigma 1 in sigma 2 je konstant, zelo všeč je konstant. Tako, zelo, da vse bps lokus, kaj smo lokalizati, je ekstremljno zelo. Zelo, da je tukaj, zelo, da je tukaj, zelo, da je tukaj. Zelo, da je tukaj, da je tukaj, da je tukaj, da je tukaj, da je tukaj, da je tukaj, da je tukaj, da je tukaj, doko jo, colluščemski soj, je mentalia, tako, da je tukaj, da je tukaj, cinem. Zelo, da je tukaj, da je tukaj, Tukaj modija začeljne objeljno bi bil, da je zelo, seveda, da je zelo v tukaj začeljno, ko je objeljno zelo pliz.  Tukaj zelo je, da je tukaj držav. Tukaj je tukaj kvantizacijen konditon na kompak spas, da je in tegral na tukaj, da je kvantizacija. Tukaj je tukaj gno kvantizacijen konditon, da nekaj kod tukaj k vsega se zoprati, nekaj kod tukaj trkaj. Tukaj be gno kvantizacija. Konditon je, da je tukaj vsega, zelo je tukaj, da je tukaj vsega. Konditon je, da je tukaj kvantizacija. Tako, zelo je, da je tukaj vsega, Tako, da je tukaj početil, da je ta m ozirčil. Načo je, da je izgleda, da je to izgleda o druh, se boš tamo vsega, da je to izgleda. Zelo je to kaj naša vsega, da je to brožem vsega, kaj ne zadači, da je to vsega, da je to vsega, kaj je to, da je to, da je to prist. Mi je ta začel, da je to, z vsega, of magnetic charges, and then what we have to compute, well, we have to compute the classical action and the one loop determinant, so the classical action is very simple. As we said, the only term which is not q exact is this twisted superpotential. So essentially you have to evaluate this very simple action on these configurations, and you get something very simple. And then the next task is to compute one loop determinants. Sorry, small a is diving on magic? Yes. So how do we compute this one loop determinants? So essentially what we have to do, we take for each of these points of these configurations, we take the localization term, we expand that quadratic order around that point. So the feature is that we have the space of the configurations, we have the special set, special configurations for each point, we should expand the quadratic action and quadratic level and multiply all the eigenvalues. And so we could lower the middle part a bit, the shape next to the impossible to see the last thing that was the SFI? Yes, but it's not very important. I kept that one will be more useful. So OK, I can do that, but OK. So how do we do that? Well, the most pedestrian way is that, so it's a ratio of determinants for the bosonic sector and the fermionic sector, and we do it. So we've compute the full spectrum in each sector. Now let me stress one of the advantages of working of shell is that for each multiplet, we have essentially a separate problem, and so we can compute for different multiplets the contribution. Now, in general, this is a very hard problem to compute the spectrum, because these operators, morally speaking, are a Laplacian or a Dirac operator. OK, this is a generalization thereof, but this is what they are. In general, we don't know how to compute the spectrum of these operators, this is a very hard problem. We are able to do that if you have a special situation with a lot of symmetry, like here you have the round S2. So as I was sure, this is doable, but if I already have some squashed sphere or some more general space, in general, we don't know how to compute the spectrum. It's a very hard problem. Luckily, as I said, there are many cancellations between the spectrum of these two operators, and so one is really interesting just in computing the eigenvalues that do not cancel. And there are more sophisticated methods to compute just those eigenvalues that do not cancel, somehow because the second-order equations reduce to first-order equations, morally speaking, because of supersymmetry. But as I said various times, I will not go into that, because I don't have time. So hopefully, someone else will do that. So I will just take the pedestrian approach of computing the full spectrum, which in this case we can do. And in general, so one just has to decompose the fields in harmonics in the space, and since this is S2, we know what the harmonics are. Now, of course, our fields, first of all, they have spin in general. So some of them are scalars, but some of them are not. So we should use harmonics for fields with spin, so sections of either a spin bundle or a vector bundle, sorry, tangent bundle and so on. And they're also charged. So this should also be a bundle and a gauge bundle. But luckily, in two dimensions, the spin group is abelian, and moreover, everything is diagonalized. So all the background is abelian at the end of the day. And so essentially, we only need harmonics for fields, which are in line bundle, essentially. And so these objects are called, in the physics literature, spin-spherical harmonics. So these are called YSJJ3. So S is the spin, but more generally, it's just the charge under the line bundle, and this J and J3 are the angular momentum on the sphere. So these J and J3 are half-integer, depending on S. Moreover, J3 and S are smaller or equal to J. And so essentially, if you want this S, it's a sort of a hectic spin, which contains the z-component of the spin, but also charges under gauge bundle. So this gets a contribution from the magnetic field. And so these are just the eigenfunctions of the Laplacian. So if you take the Laplacian, where you put both the spin connection and the various gauge connections, you have some eigenvalues of this operator. So this is just the standard scalar spherical harmonics. If you set S equal to 0, this is precisely that. OK, and so with this, we can decompose all our fields in these harmonics. And we just have to plug in and take the product of all the eigenvalues. So it's very standard, it's straightforward computation. So let me just give you an example. So let's look what type of thing that one get for the Karan multiplet. So what do we have to do for the Karan multiplet, where we take this matter action. So first of all, we look at, first, the bosonic sector and the fermionic sector. So the bosonic sector I wrote. And so essentially, you have to do the quadratic expansion around the background. And the operator that you get, let me call it O5, is the following, where what I mean here is that you have to plug in the values of the background. This is the quadratic expansion. So in S1 and S2 and D, you plug in what the background is. So in particular, the result will depend on the point, where we are. And you just plug in all the spherical harmonics and you collect the factors. And what you obtain is something like the following. So the details are not important. Just want to give you a flavor of the type of computations and results that one obtain, just that you see it once. So in particular, you don't have to copy these expressions. They're not particularly important. And you can find them in the papers. But the point is that you get some infinite product of eigenvalues like this. And then, OK, one repeats the same thing for the fermion operator, which I'm not going to write because I didn't write it. I wrote it yesterday. Now, in this case, these are two-component spinor. So really, you have to decompose both components in terms of these spherical harmonics. And they have spin 1,5 and minus 1,5. Yes, in this particular computation, phi is a scalar. And so my value for s that I'm using here is really just the magnetic flux. But here, I have two components, 1,5 minus 1,5. And so once again, I plug in and I compute the eigenvalues. And you get some similar expression, which there is nearly no point for me to write. Well, OK, let me just write it to show the feature that I was talking about. So the expression that one obtains is something like the following. So a very similar state. There is some product of this similar j. So as you see, it's very similar, but it's not quite the same. So most of the terms will cancel, but not all of them. And so again, one finds, so we are interested in this super determinant, which is the ratio of the two, the fermionic over the bosonic determinant. And this is, so this will be some simplified expression, but it will still be some, I mean, this is not completely trivial, it's still some infinite product. OK, so this is the expression. Now, well, this expression is not well defined because this product is not convergent. So we need to regularize it. So let me put this in quotation mark. So this is telling us what the function should be, but this per se is not a good, what is not convergent. So we should regularize it. And for instance, we can use zeta function regularization. And so, OK, in this particular case, what do we do where we can use the hord width zeta function? This zeta function depends on two parameters, z and q. And it is defined in some infinite sum. And this is convergent if the real part of z is large enough, but then you can do analytic continuation and you can define it in the whole complex plane up to some poles. And now, OK, so now if you take this expression and you compute a certain derivative of this expression, so you take the derivative respect to z and then you set z to zero. So formally, you obtain the following expression, which, OK, up to taking the exponential is precisely the thing that we want. It's an infinite product over this n for some shift. This shift will be this expression here, this expression here. But in fact, if you do it more properly, what you get is a log of a gamma function. And so this is the regularized expression that we associate to this regular function. We associate to this expression. And so because of this, so let me also say that so here, OK, here it was, OK, and this expression is like we have a single mode. But of course, this current multiple is in some representation of the gauge group. So we have as many components as the dimension of the representation. Each component is associated to a weight of the representation. And so really here there is also a product over the weights of the representation. And so the full thing that I can call z1 loop chiral, because for the chiral multiplet, is first of all a product over all the weights in the representation. And then there is a ratio of gamma functions. So once again, the specific function that we get here is not important in different examples and in different dimensions. We will get different functions. The important thing is that is explicit and relatively simple function. And so in general, it's not a big problem to compute in this one loop determinants. So this is not a hard part in localization. It might be technical, it might take some time to compute it, but usually it's not a big obstacle. OK, so this is the result for the chiral multiplet. Now we want to do something similar for the vector multiplet. Are there questions? OK. So the computation is similar. The only point that I would like to stress, which is different, is that we need to fix the gauge. We are dealing with a gauge theory. And we can use the standard Fadevko-Pov method. However, as this drawing, so let me make this drawing again, so we are expanding around a certain point, a certain background configuration. And so we have to do gauge fixing on a background, not around the zero configuration. And so what we do, so as we said in the general discussion, so we take this gauge field and we separate it into some background, which is essentially this constant magnetic flux that we have on the sphere, plus an oscillatory part. And we should perform gauge fixing on this background. Let me also say that we don't need to do gauge fixing for the background, because essentially we have already done it. So when we wrote the set of BPS configurations, we said that, OK, we have to sum over magnetic fluxes. So we already were using a gauge invariant description of that. So we don't need to do gauge fixing for this, but we need to do gauge fixing for these oscillations. And so we do the standard thing. We add a gauge fixing action. So we introduce ghosts, C and C tilde.  These are fermionic scalar in the adjoint representation. And now this parameter xa, so maybe I should use a different name. Let me call it, OK, let me use xai gauge fixing. So this is now the Fajeti-Lopoulos term. This is the parameter that we are using in arc C gauge. And so in particular, as a check, there should not be a dependence on this parameter at the end of the computation. And this covariant derivative, these are computed on a background. So in this covariant derivative, you only have the background. And so in particular, this action is quadratic in the oscillating fields, in this a hat and the ghosts. And so the only difference is that now we have to do so when you expand a quadratic order and we compute one loop determinant, we also have to include this and the ghosts in the computation. But otherwise, this is the very same thing. So in particular, we can use a superior maze plus this gauge fixing action. This is already quadratic and we do the same thing. And the one loop determinant will have a similar form. So there will be a contribution from the ghosts. Once again, we expand all the fields in harmonics.  Well, here we get a square root because these are really real fields, while all the other ones are complex. And there are some primes here. So we get some zero modes. So we get some zero modes in this operator and these zero modes will be associated to sigma. So there will be some flat direction from sigma. So we remove the zero modes and we'll have to integrate over them. You also get some zero modes from these ghosts. And in particular, zero modes corresponds to constant gauge transformations. These constant gauge transformations, you don't have this problem in flat space because in flat space usually you set to zero your gauge transformation at infinity. But now we are in a compact space, so we do have constant gauge transformations. And we know that we have to divide by them so we can just remove these zero modes. So we know that these zero modes are just related to gauge transformations. So we can either remove them by hand. If you want to be more formal, we could introduce ghost for ghost, but the result will be the same. So since we understand where these zero modes come from, I will just remove them. Okay. And so the expression that we get is the following. So there is a product over the roots, which annihilate the magnetic flux. So of course this answer depends on where we are. And actually probably a more better picture is the following. So if you want each of these slices, it's parameterized by A, but we have many slices because we have this M, this discrete parameter. And the answer depends where we are. So this will be product over the roots that are invariant under this magnetic flux, that commute with this magnetic flux. And then you find some product over the positive roots instead of finding a product over the weights, but these are the weights of the joint representation. And here the function that you get is even simpler than before because it's just a quadratic polynomial. Okay, so we have this very simple expression. Is there any question so far? What's the argument of the first alpha of the input for it? Well, it's not an argument. In the 1 over... Here? So this is 1 over absolute value, let me write it, 1 over absolute value of alpha of A, sorry. Yes, alpha M is 0. There's no argument. You just do the computation and this is what you get. And then the other alphas are also all alpha of A? No, this is alpha positive. So you divide the roots into positive and negative. Like, after that, the exponent minus 1 to the alpha of... Alpha of M. And then it's alpha of M squared plus... Because so this is a product over alpha M equal to 0, this is a product... These are two different products. So this is alpha of M and this is alpha of M squared and alpha of A. Once again, the details are not particularly important. I mean, they are important for this computation, not for the general idea. And so as we say, we have these zero modes and so we should integrate over these zero modes. In particular, these zero modes span the cartanto-balgebra, which commutes with the magnetic flux. So in particular, if the magnetic flux is... The magnetic flux is diagonal. If the magnetic flux is totally generic, all the again values are different, then it's just broken to, if you want... I don't know, U n is broken to U1 to the n and so the zero modes are only corresponds to these n directions. But if the magnetic flux is zero, then there is no breaking at all and so the zero modes span the full U n algebra. Essentially, because these magnetic flux lift the zero modes. So if this is generic, it lifts almost all of them, it cannot lift the ones in the cartan because they are neutral. But if it is special, then there are more zero modes. In particular, if it is zero, it doesn't lift anything. So we should integrate over these zero modes. And so what does it mean? Well, it means that we integrate over these a. So these are many as the rank of g, this a n. And you see essentially, so as I said in the general discussion, these zero modes are nothing else that the continuous direction that you have in this space. So the pat integral reduces to a lower dimensional space and the coordinates around that are precisely the zero modes. So we have this, however, for special values of m, in fact these zero modes are not just the diagonal, as I say, the span, the unbroken algebra, and so we should include a Van der Monde determinant. And moreover, we should also divide by the residual vile group. So when we break un, for any group to the cartan, we have a residual gauge transformation that corresponds to exchange in the cartan, but there is a gauge symmetry, so you should divide by that. So this is the vile group, which leaves the magnetic flux invariant. And so we have this integral to do. In particular, notice that this factor will cancel with this factor. And then finally, we have to sum over all these magnetic fluxes. So we sum over these lattice of magnetic fluxes, which depends on the group, not just on the algebra, as I said. But once again, when we sum, we have to divide by gauge transformations, sorry, by the residual gauge transformations, the residual vile transformations. And these are, so this is the total vile group divided by the vile group that leaves M invariant. So this is just the transformation that actually act on M. And once again, there is a cancellation between this factor and this factor. So we are ready to write the final formula. Is there any question on the, on the general procedure? Sir, from this question, but you only fixed flux of your connection. You only fixed flux, and no flux, yes? But why don't integrate the wrong integration which provides flux, or they are all gauging, they are either grimo, and what what. They are gauge equivalent, because so what are the gauge invariants? So the gauge invariant and the integral of the field strength, and then there are the Wilson lines. But we are on S2, there are no Wilson lines. So the only gauge invariant is the field strength. I mean, in general, the field strength is not the only gauge invariant, you should care about Wilson lines, but you need some cycle. But, okay, if we use the same argument for S4, for example. Yes. And we spent on S4. Yes. For with this argument that we go to the, in the two gauging invariants, it's f and f squared, and we got no Wilson lines in S4. We have not? We don't have any Wilson lines on S4. Yes, there is no extra gauge invariant observable associated to. So the only configurations are enumerated by numbers also? Well, no, I mean, there is not just the integral of the field strength. I mean, the actual field strength is a profile. So the instantons have a modular space, right? You can move these instantons around. Yes. But you only have to tell me on S4, you only have to tell me what is the field strength. You don't have to tell me an extra parameter, which is some Wilson line somewhere, because it's completely fixed. So if you want on S4, so let's say that you tell me what is the field strength. Okay, you specify what is the field strength as f nu nu of x. And then, okay, I can start with a small Wilson line here. This Wilson line is zero, because it is extremely smooth. I am assuming it is smooth. And then I can make this Wilson line larger, and I know how it changes, because it is just the integral of the field strength, right? When I change the Wilson line, I know that the variation in the Wilson line is just the integral of the field strength. And so, if you tell me what is the field strength, I can tell you what is the value of any Wilson line on S4. So there is no extra information. I think the information that may be missing is that f nu nu is set to be constant by the gs equations on S2. Yes, no, absolutely. So, I mean, there are equations that tells us that f will be constant. In general, if you remove equations, I'm saying the only gauge invariant information is in the field strength, but which is a full function. No, any other question? OK, so, OK, we put everything together and we get the following expression. Again, I write it, the details is not super important, because they are specific to this, but then I will make some comments. Yes, the gauge fixing apprehend is qx. It's qx. So you want the result not to be... I'm actually not even using one which is supersymmetric.   I don't have to do that in a supersymmetric way. I could, then it will be more formal. As I said, you can also introduce gs for gs, but the result is not going to change. The sum of the charge to magnetic bias, is that supposed to be inside of that integral on the line above or is that something separate your writing edges? It's outside, because for each magnetic sector, if one of the zero modes are different. So you should have first... If you think first compute the above integral, then multiply it by that pre-factor and sum over the magnetic fluxes. Sorry, so this integral... So you first take the... OK, let me write the expression, it will be clear. OK, let me write it, and then let's see if this is clear. And you should not copy this. I mean, this is in the papers. There is no point in it. OK, so this is the expression. So let's see if it makes sense. So we have this... So let me repeat today the drawing. We have this modular space of special configuration where we localize. This is a set of submanifolds. So we choose a point on this submanifold for each point we have an integrand, which is this thing here. This is the classical action, and this was coming from this oscillation, so this was the one loop determinant. And then we have to integrate over these submanifolds and then to sub... Over the submanifolds, so we first integrate, and this is our integral over the submanifolds and in fact the zero modes that we integrate over precisely parametrajozati these manifolds. And then we have a sum over the values manifolds, which is a sum over magnetic fluxes. And of course here we are still using a description, which is gauge redundant, because there is the vile group that acts, so we divide by the dimension of the vile group. Does this make sense? Does this answer your question? OK. So let's comment on this. So the first comment is that... OK, I don't know what you think, but this expression is extremely simple. So after all we are computing, so this is an exact evaluation of an unperturbated evaluation of a pat integral. So the fact that we have such a simple expression is quite interesting. And of course this is a common theme in these localization computations. One is able to find exact expression, which is relatively simple. After all, it's just an integral that... OK, I mean, if you want, you can even do it numerically. And then there is a sum. You can do it numerically. So it's a very simple expression. In fact, as we say, so this is an exact unperturbative result. So in particular it should contain... I mean, there should be all the instanton corrections. So you might ask, OK, where are the instantons? And if you wish, this is... In fact, this expression is a bit different from what you get, for instance, on S4, as both can explain, because in that case you also have to include all nonperturbative corrections that you need to compute in some other way. But here we don't find extra nonperturbative corrections. So in some sense it might seem like this is just perturbative. At the end of the day we just put in this one loop determinants which are computed perturbatively. So what about the instantons? And the point is that, in fact, this expression does already contain the instantons. So they are not manifest from here, but it does contain them. And one way to see, but it will be more specific in a moment, is the following. So this is an integral over some real lines. So let me consider the rank one case like it was u1. So this is an integral over the real line. So this is the plane A. Of course there is also this sum over the magnetic fluxes. And we integrate along this real line. So how do we do this integral? One way is to do the Cauchy take. So we close the contour at infinity. Of course we have to make sure that there is no contribution if this circle is large enough. And you can check that this is the case. You will have to choose one of the two sides. That's OK. So then you have this contour integral. And so this contour integral reduces to a sum over residues. And if you... So here there are gamma functions. The gamma function doesn't have any zero, but it has poles. So you get poles from here. And so, OK, take into account onto this sum over the magnetic fluxes, you discover that if you wish at this point you can bring the sum in, because there is no... because this integration contour does not depend on M. So what we discovered, there is a sort of... rank to lattice. It's not really lattice, it's a wedge. But essentially there is a sort of... sort of wedge of poles. And so you can reduce the computation to computing residues around this poles and summing the residues. And, well, as we will see better in a moment, it turns out that, in fact, each of these residues is an instanton contribution. And so we can rewrite this expression as a sum of an instanton contribution. OK? Yes, so if you have pure gauge theory in two dimensions there are no instantons. But if you have a gauge theory with matter and you go on the X branch, there are vortices, and vortices are the instantons in two dimensions. Well, these are non-perturb... these are solutions to the Euclidean equation of motion with finite action. OK, so we will see this in a moment. Well, we will start seeing it. OK. Sorry, but is that... satisfying some... No, it's a vortex equation, which is a different equation. It's just a totally fine. Well, it's... in three dimensions it's a soliton. So if you add time, so this is the standard story. If you have an instanton in some... in dimension d, you can go in d plus one dimensions and becomes a soliton. So the vortex becomes a vortex particle in three dimensions, while it is an instanton in two dimensions. So these are the very same vortices that... I don't know if you are familiar with the particle vortex duality or... Sorry? Yeah. So there are... The vortices are solitons, so there are vortex particles. In four dimensions there are vortex strings and they play a role in superconductivity. They distinguish between type one and type two superconductivity. So they are very... the very same vortices. So let's continue with this comment. So here we computed just the pure partition function, but in fact one might be interested in insertions, operator insertions. And in fact it turns out that there are... So it's easy to make insertions of order operators. For instance we can insert twisted chiral fields at one point of the sphere and antitwisted chiral fields at the other point of the sphere. So here we can include some sigma or more generally some function of sigma gauge invariant and here some other function of sigma tilde at antipodal point and it's easy to include in the localization because essentially you just have these functions appearing in the integrand. And so this is already an example of what I said at the beginning. It's not a holomorphic correlator because this holomorphic is anti-holomorphic. So if you wish this is a type of correlator which you do not access with the topological twist. So this is different from what was done in the past. And in fact there is a lot of... these correlators are quite interesting and I think next week Komar Gotski will talk about this in four dimensions instead of two but essentially there is a story. It's also easy to include the Wilson line operators. So it turns out that you can insert the Wilson line operators along some parallel. So here you can have some Wilson line that depends on some arbitrary representation r. And once again this is very easy to include essentially you have just another exponential in your integrand and so you can compute them and once again this is very similar to the Wilson line. One can also include the other type of operators for instance these order operators of course these are more complicated because then you need to change your boundary condition integrate over a different set of fields because now the fields will be singular points with singular boundary conditions so you have to repeat the computation but this can be done. And for instance Takuja Okuda who is here did these sort of computations so you can ask him if you are interested. Ok. So I will not discuss this. Yes. So you say it's an untested theory and yet in the final expression for the partition function I don't see any explicit dependence on the form of superpotential. Yes. I don't know how you reconcile those two facts because I would imagine that an untested theory would also depend on the form that the superpotential dates. Yes so the full theory of course depends on everything. So if you take a generic operator and you compute correction function of generic operators those correction functions will depend on everything but we are either computing that we just computed the pat integral on the sphere with no insertion so just the partition function that particular computation does not depend on the superpotential. In fact if you and also as I said you can compute more generally these correlation functions for instance of twisted and untrained twisted but these are very special correlation functions so those objects do not depend on the superpotential but a generic correlator does and we don't know how to compute those generic correlators in the strongly coupled theory. Any other question? Similar, yes. If I remember correctly you have to integrate a and then there is also sigma and they have to combine there will be some conditions I don't remember the conditions but there is some conditions. Ok then let me just say that if one study 3D n equal to theory on gauge theory on S3 things are very similar of course the details are different the type of function that you get are different you don't get the sum of magnetic fluxes only the integral but otherwise the same philosophy and the expression is extremely similar so one could leave this as an exercise but you can find it on papers but let me stress once again instead the competition in four dimensions in these respects are different I mean there is a qualitative different that actually in the integrand you need to put all the non-perturbative and non-perturbative corrections but no you have to put the contribution from non-perturbative corrections that you have to compute in some other way so this makes the computations like on S4 or in five dimensions more complicated we can say much more complicated for D S4 requires instantons Ok and then let me stress once again something that we said before that this computation does not depend or my precise does not depend on the scale of the set by the gauge coupling and so in particular something which is a setup which is particularly useful is a setup in which we start with some gauge linear sigma model in the UV so gauge and sigma model is just a gauge theory and for instance we can take a gauge theory of vector and kernel multiplets and we have a formula for the partition function now this gauge linear sigma model may flow in the infrared to a non-linear sigma model and in particular this is particularly interesting if this non-linear sigma model is conformal because then there is no non-unitary deformation and what we compute really knows about the physical theory in the infrared and so this can become a very efficient way in the quantities in non-linear sigma models and but it is very useful in general but in particular for instance it is very useful for the string because the string is a two-dimensional theory and if you compactify the string for instance on some kalabijao there is a sector of the of the worksheet theory which is a non-linear sigma model on the kalabijao and what we compute thinks about the string on these manifolds ok so I'm gonna move to the next topic so if there are extra questions this is a good time so the next thing I would like to talk about is to clarify this story of how the instantons are included in this expression yes so how this expression knows about the instantons and how can we understand this in a more abstract way and so in particular this is related to I don't remember if this question was posed in this in this school but ok natural question one could have is so there are various choices that we have to do in this localization computation in particular we have to choose this localizing term what if we choose a different term are things gonna change or not of course the final result as a function must not depend on it otherwise it means that we are doing something wrong because localization is just a way to compute the original path integral however we might obtain some different expression for the same quantity so let's suppose that we add besides the two terms we were adding before we add a new localization term let me call this LH and I will not write this term with full details because once again the details are not so important but this term looks like the following here probably I should put a tilde in the notation that I am using here plus many other terms and this is a q-exact term so we want to add localization term so it is something q-exact I am not gonna write all the terms but the important thing is that in this term there is a constant so this k is an number and of course since this goes into a localization term there will not be a dependence on this constant and the end of the day it depends on whatever I want this is controlled by some constant here and then you see that this term looks a little bit like a D term sorry like a phyatelopoulos term because the phyatelopoulos term precisely is a D chi so there is a phyatelopoulos term but of course it is not just a phyatelopoulos term because as we commented before the final part integral the partition function does depend on the physical phyatelopoulos term explicit I mean it is written there so you see that the physical phyatelopoulos term was this chi and the final expression does depend on chi this one is a sort of phyatelopoulos term dressed with other stuff in such a way that there is no dependence on it but it looks like a phyatelopoulos term now action is not positive semi-definite and ok, it is not manifest from here so here this is the imaginary part but if you take the full expression that you can find in the papers and you see what is there this is not positive definite so it looks like this is not good however D appears in the action so there is no problem in doing the part integral over D we can do it by hand without localization and if you do the part integral by hand over D the expression that we find is positive definite so if you do this piece of the integral by hand then we can apply the localization argument afterwards and if we integrate out of this D once again since this is auxiliary field just means that we substitute D by the value that it gets by the equation of motion which is just algebraic and this is the reason why we can do the substitution and in fact one gets the following equation which, correctly, does not contain derivatives and D is just linear so this is why we can just substitute the equation of motion to the action and in fact if you look at this precisely it looks like the D term equation but with a phiatilopoulos term and the nice thing is that if we are in two dimensions and we turn on a phiatilopoulos term and we look what happens to the modular space the vacua are moved to the X branch and on the X branch there are vortices so we might expect that now somehow the localization locus is moved to a sort of something that looks like the X branch in flat space and so the configurations that matter are v vortices by contrast if you want the configuration that we are contributing there looks more like a Coulomb branch in flat space because essentially we integrate over this sigma the sigma was the scalar in the vector multiplet and when we turn it on we break the gauge group to the cartan and this is what happens on the Coulomb branch so in fact sometimes this formula is called Coulomb branch localization formula but here it looks like we might localize to something different so I don't have much time so I will not once again go into the full details if you are interested you can ask me or you can look in the papers but now that you impose this constraint now you go and solve the BPS equations and the solution to the BPS equations are different than before and you might ask why the BPS equations are different, the BPS equations do not depend on the action and of course you are correct the point here is that so when we go to Euclidean we have these BPS equations and they are for complexified fields and so there are lots of solutions which are general complex solutions and in general this is complicated however then we restrict to a real contour and then there is just a subset of the solution so in particular when I wrote those solutions those were the zeros of the real part of the actions which if you go to the BPS equations those are the restriction of complexified BPS equations to some real contour so it is a subset of the solutions and somehow imposing these constraints is changing the real contour but it is only changing it for the D term for the auxiliary field so after all it is not a big change because anyway it is non dynamical but still it changes the reality condition and then you get some other subset of the complexified BPS equations that I am going to tell you so this is how you can get different set of BPS solutions the first set of BPS solutions is I don't know what it is it is very large, very complicated or maybe it is not but I am not studying or I don't think anybody has studied it is there any question on this point? so what happens is the following that so we can imagine what would be the Coulomb branch on flat space the classical Coulomb branch so this is parameterized by this A and there are special points on this would be Coulomb branch in flat space where x branches open up and these points are points where one of the current multiplates become massless so these current multiplates have masses because there are twisted masses into the game but twisted masses can be offset by moving on the Coulomb branch and so special points one of the current multiplates becomes massless and those are the points where the x branch open up of course in the x branch you give wave to the sum of the current multiplates and so they better be massless wave and in fact in the BPS equation you find things like sigma 1 minus m phi equal to sigma 2 phi equal to 0 where this m I should use a different letter maybe I can use capital M so where these are the twisted masses and in fact now I realize I didn't tell you how the twisted masses enter in that formula so let me just say that so in this formula I set all the twisted masses to 0 for simplicity but you understand how the twisted masses appear because twisted masses are expectation values for the scalars in background vector multiplate so they are exactly on the same footing as this A so if you turn on the twisted masses you also have some other contribution which is the same as this A the only difference is that A is dynamical and so it's integrated over the twisted masses are not dynamical so they are not integrated over and there will be some parameters at the end of the day but otherwise they appear exactly in the same way so you find this type of equations and so this is what I described before so if you have a mass of course the field should be 0 but on special points of the Coulomb branch some of the components because of course this is a diagonal matrix in the cartano in the algebra while this is if you want this is a diagonal matrix in the flavor algebra but you can have a conspiracy in which some component cancel out and that component then can get a V because this equation allows for phi to get a V so these are precisely the special points where sigma 1 minus m has some vanishing eigenvalue and then you find some other equations now these equations involve so these are differential equations that involve this chi and we once again since there is no dependence on this chi we can take a limit in which chi goes to infinity we can take a limit in which chi goes to infinity and so I will just describe the type of solution that you obtain when chi is very large but I mean there is some differential equations for finite chi and one can study this differential equation so I take the limit and chi goes to plus or minus infinity and the type of solution that you get is the following so we have the s2 and and so outside so there are two special points which are the poles these two special points are selected by the particular q that you chose because this q is made of these epsilon and epsilon tilde of some components of them so there are two special points and outside of the points essentially this equation forces you on the x branch so you have the field strength is 0 and phi, phi tilde is equal to chi and this is precise what you have on the x branch so phi takes a wave and this is we can call it fake phi tilde opolost term in the sense that it is not the physical one however if we go close to the poles then we have some non-trivial differential equation so if you want this is similar to what happens in paston's case on s4 but with the difference that I will remark so we get this equation ok, so this plus or minus I think depends on whether we are at the north pole or the south pole so in one case we get a plus and this d plus minus maybe I should call them the z, z bar these are holomorphic derivatives so maybe I should call either z bar or z depending on the two and so these equations are equations which are known in physics and mathematics and these are called vortex equations so as we said before these equations are so the solutions are essentially instantons so you can study this equation in flat space and you can think of the instantons the difference is that the instantons in four dimensions are equations only for the field strength in two dimensions so if you just write the equation for the field strength you don't get something non-trivial you need the equation for the field strength and some scalar field so this is what you have in these vortex equations and so and so now what you localize on well ok, you will have some finite sum over these x branches but this is just a finite sum so this is a single x branch it's not an important point the important point is that now we have to include all configurations which are solved, yes oh, I was just wondering what's the relation between the off the side those vortex equations so those are the number of the configurations but the equations you wrote under x branch F1 is equal to 0, 5, 5 is equal to 5 is that just giving an ansatz to solve the vortex equations? Yeah, this is the behavior far from the core of the vortex so as we are taking the sphere to be big enough so that you can say ok, it looks like flat space at each pole? Yeah, so the point is that so if you take this equation of course the vortices have modules that corresponds to moving them the core of the vortex but really the vortices here are confined to the poles they are in a omega background essentially and are you saying that all solutions have the core mass topically, or that's kind of an easy class to construct? No, all of them has up to the, so if you go far from the core of the vortex, so the vortex solution looks like the following, so there will be a core the field strength is as a bump around the core so this is the field strength and then the field strength goes to 0 I think exponentially far from the vortex and so if the field strength goes to 0 you see this is constrained to be 0 so this is, so the field strength is 0, this just comes from solving the equation you see that it goes to 0 exponentially yeah and then this follows so if you want phi goes something opposite, phi is 0 at the core essentially because phi winds so it goes to the core so it goes like r essentially this goes like z to the n where n is the let me call it k, the vortex number so phi winds, the phase of phi winds as you go around set infinity if you count the number of windings this is the vortex number it's a topological concept charge and so since it winds it has to go to 0 at the origin to be smooth so this goes to 0 and then when you go far from the origin it goes to the value phi equal to square root of chi which is the so far from the vortex we are on the x branch but then in the core of the vortex I behave like this yes, in fact it's nice how this is related to superconductivity because so when you are in the x branch you know x mechanism is superconductivity essentially the photon is massive and the magnetic field cannot penetrate but when you have a vortex this breaking parameter goes to 0 so the mass of the photon goes to 0 and then the magnetic field can form a tube that goes through the vortex is a tube of magnetic flux that goes into the superconductor so there is this some actual real world physics in this of course also in instantons ok yes, the only difference so here I am a bit simplifying the equations in this limit because the only difference with paston case is that so somehow in paston case you have this point-like instantons at the pulse of the S4 if you would like to be completely let me just finish this sentence so if you would like to be completely precise you would need to do what Necrozov did in flat space you would need to do probably some non-commutative deformation and open up the instantons to smooth configurations still in paston set up these are not smooth these are really point-like and I don't think that this has been done on S4 I think everybody believes that it's gonna work but here there is a parameter that you can tune and when chi is finite you actually have a smooth solutions and if the guys go into infinity they become the standard vortex equations in general there are some smooth deformation of these equations so in some sense here you already have the deformation that makes all the configurations smooth but otherwise it's very similar ok so we will continue tomorrow


couple of people asking me for snacks. Let me show you my snacks. Sorry if I'm not reading the chat. I just cooked this up this morning because this is our food for today. I had a couple casserole dishes. I laid the bottom down with cabbage chopped up, nut dill, leek chopped up, and then put olive oil. In one casserole dish I laid it down with sweet potatoes, another one with potatoes. Some rosemary, salt, a little bit of balsamic vinegar, not too much. Then I put a metal thing on top, like grating thing. Then I put chicken thighs that had marinated last night in lemon and garlic.  So glazed it with the honey garlic and then broiled each side of the chicken, flipped them for about five minutes or so. This is what we got. This is what we got. This is the chicken and it's got a lot of juice sauce on the bottom. So this is the chicken. Both sides are, I flipped them, so both sides are glazed with honey garlic sauce. Honey garlic glaze, I guess. This is the sweet potatoes. We got potatoes as well, gotta love the potatoes. And we got tomatoes. And then we got leeks and cabbage. So it's quite yummy. I already had one. So let me bring a knife and have a little bit of salt. Well, I'll just take this down like this. I'm going to woof this guy down again, because I like the skin as well. Some people like to take the skin off and I do sometimes as well, but when I cook it like this, the skin is part of the amazing flavor. The skin is part of the amazing flavor they got. Very yummy. Very yummy. Very juicy. Very juicy. And it's super good. And I got some of the grapes from our garden. The unripened grapes, say that little green guy, round thing. And it gives it a nice tangy tart flavor. Super yummy. And this is what we're going to eat for later on today. And I cooked up 16 chicken thighs with the bone. Okay. When you cook chicken with the bone, you need to leave it on there in the oven for a while. So it cooks well. You got it with bone, it takes a little bit longer to cook.  And the potatoes are really nice. Nicely cooked up. Right. Super yummy. Our sweet potatoes are really good for you. Right. Super good for you. And it's like lukewarm right now because I just took it out of the oven like 45 minutes ago. Very nice.