Hey guys welcome to SS Minute Tech, so see in this side and today we are going to see one of the real time interview question and answer. So this question was asked in PwC interview. So recently one of my friend has attended interview in PwC so this question was asked over there. So let me quickly go inside the YouTube so I have already uploaded few of the videos which is based on that interview only. So you can go on the YouTube and you can search for PwC Power BI interview questions and answers and after that my YouTube channel name. Then you will see the first video which is scenario based question. So this question was asked on the same interview. So I would strongly recommend to watch this video as well. So let's start with today's video. So in this video what we need to do? So here as we can see we have this slicer. So this slicer is having the values which is one two and three and here as well the columns are one two and three. So once we select any one of this then we can see the color of this column has been changed. Once we select two then the color of top two will be changed. Once we select three then top three color will be changed. So accordingly this will be changing. So first we have to create this slicer because this is the dynamic slicer. If the column will be having four here then the value that should be four instead of three. So instead of static it should be dynamic the first case. The second case we will be implementing the row numbers that we can see here in this table and this row number is totally based on the total sale which is the value here. So if the value is having the highest for the UK so rank will be one. For the US the second highest so it is two and then India it is three. So once we will be implementing this and at last we will be implementing for this coloring option. So let's start how we can do that. So I am having a dummy file so where I have imported this data model and in this data model we have this table. So the first thing that we need to create is slicer here. So that slicer should be the dynamic and the column values we are having here which is a country. So if the country will be having five distinct values then the slicer value should be starting from one to five. So for that we have to create a new table by using the dash. So we can go into the data modeling side here modeling tab and we can see this. So we can click on this. So we need to write a dash expression to generate that sequence from one to whatever the number of countries. So let me call this as country list. Now here we have to use the generate series. So this generate series will be going to start whatever the starting value we are passing whatever the end value we are passing and what will be the incremented value. So it will be generating the sequence. So it should be starting from one and the end value. So end value we cannot specify any specific value. So this should be dynamic. So we can use the distinct count function of the country column. So we can select that and after that we can incremented one by one. Let me put it in. So this will be generating a new table and the sequence will be one, two and three because total three countries we have here. So in the country list that should be having total three values that we can verify. So this is the first part of our question. So let me try to use a slicer now. Let me format the slicer a little bit and after that let me use the value column here and after that let me make this as list. So the first part of the question is done. Now the second part of the question that we need to generate a rank here and here let me try to create a new measure for creating the rank. So as we can see this measure, so we need to write the formula for creating the rank. So let me call this as rank value and here for creating the rank we need to use the rank x function and under the rank x function first as we can see it is asking the table. So we need to specify the table here. So the table which is the sales table so we can directly specify the table name then it is asking the expression. So the expression will be having the summarize of the value column. So if we are going to use the sum directly here so that will not work because it will be going to check row by row. So we don't want to check row by row. We want to check on the entire table. So for checking the entire table first like in this table we have to use the all. So it is not going to indicate for any particular row the expression that we are applying on total column. So second we need to use the calculate function then the expression will be the sum of the value. So we can select like this then we can see the value. So that is optional. Let me skip that order we need to specify descending order and after that we can close the bracket. So let me put enter. So it is generating the rank but the problem is that we will see here let me try to drag and drop this rank value in this table. So here we will be going to see the rank the new column will be added but the rank we can see 1, 2 and 5. So why this is pretending like that because if we can go in the data model then the rank which is generating on this table. So whenever we are generating the rank on this table so mainly if we can drag and drop the city value here as well then it will more clear. So now we can show you that. Let me sort this. So now we can see it is sorted according to the unique combination of the country and city. So that is why we are returning 1, 2, 3, 4, 5 like that. If we are removing this city column here the rank which is not updated. So the main concern how we can do that for this ranking. So for that we need to go on the rank value measure and here we can see we are supplying all the table. So instead of supplying whole table we can supply only the column by which we want to generate the rank. So we want to generate the rank on country column so we can specify the country here and put enter on this. So now we will see this rank will be updated and it will be having values like 1, 2, 3 as we can see. So now it has the proper rank value here. Now the last requirement is so we want to change the color of this column by selecting the values from the slicer. So how we can do that? So for that we have to create one more measure and under that measure we will be writing the expression. So go to on the new measure and let me call this measure as color code. Now here we need to check if then the logical test. So what is the logical test? We need to check the rank and the selected value from the slicer. So first let me use the selected value. So selected value from the country list value. So this is the slicer value. If this value is greater than or equals to the rank which we have generated so that is the rank value last column that we have generated. If the rank value is greater or equals to of this country value that is selected one then we want to make the color as green. So let me use the green here otherwise we want to make this as blue something like this. Let me put it in here. So we have generated this expression and we can use this expression on the dynamic coloring option. So we can select this visualization that is the column chart go to the format and here we can see the data color. So default color we can see this. So instead of the default color let me click on this expression and under this expression we can go on the format by and need to select the field value. So the field value that we have generated the color code measure. So we need to select the color code measure here. So this is under the sales table and we need to select this color code. Once we select this measure everything has gone. We can click on OK directly. So now we can see this is all in blue color that is OK. Once we click on one then it should be filtered and only top sale will be having the green color. Once we select two top two will be having the green color. Once we select three then top three will be having the green. So this is the question was asked and this is very important question and we can say this is the complex as well. So let me recall what we have done in this video. So first we have generated this series as per the distinct countries and after that we have generated this rank and this rank is based on the country. So let me recall the rank function again. So we have used the rank X and after that all along with the country. So this will be checking for all these countries and then calculate with the sum function. So I will record another video where we will explain why we have used the calculate over the sum. So directly we cannot use the sum in this case. So we will make a separate video on that. So thank you so much for watching this video. See you in the next video with new scenario.


Yes, it says we continue with long waits, actually, and from a little bit different perspectives since I'm not an expert in textile but in archaeologists that work with textiles. And this is also not a paper about big transformations, but about the macroscopes of transformation in craft production because it can be seen as I think it can be seen in the archaeological material. And I have to make some very quick introduction because there is no time to talk about the background. We might discuss it later, but let's say to follow my line of thoughts you should give for granted, believe me, that one of the big bronze age innovation, we have been working with it in Gothenburg for some time now, is supposed to be an introduction of wool around in the Latte in the already in the third millennium. In the area of the Near East and Anatolia you have clear evidence of intense wool production, the first traces of wool already from the fourth millennium. And they are documented in many different ways including archive documents, so we have numbers, we have names of people working with wool and so on and so forth. In the second millennium we started having fragments of wool in the continent including the big collection of woolen clothes from Denmark that has been recently analysed with Strontium isotopes by Karin Fry in Copenhagen and the surprising result was that the wool of these clothes doesn't look to be done with sheep that have been grazing in actual Danish territory, so they really strongly suggest that there was wool trade in Europe by the second millennium, at least mid second millennium BC. And we know that wool cannot be really easily produced with bronze age sheep because bronze age sheep don't look like modern sheep, they have very little wool so you need a lot of them in order to produce enough wool to have an economically sustainable production. So it's very possible that you had centres of production and then wool trade in the style that you could see already a thousand years before in the Eastern Mediterranean. And one very good candidate for this wool production I believe is located here in the population, not in Italy where I'm working with, where you have evidence of high numbers of sheep among the animal population and a lot of textile tools. As I said, the whole chain of wool is a complicated one, it's a year round activity as I said, bronze age sheep are different from modern sheep, they require different techniques, technology they're not, you can share them, you can put the wool from them and then as you can see there are several different steps, they keep different people busy and what we are going to focus today is the weaving part, which is the last one actually. In the population, which is the middle bronze age, Italian middle and recent bronze age culture so-called the Terra Maria culture, you have evidence from all these steps from the sheep herding to the weaving, and actually if you look at the archive documents you understand of course those who practice they also know, but the weaving part it is the final part and it takes, if you think about time, maybe the least time it's the quickest part of the whole process, but it has a very important, it's a key moment because if the weave, it's a poor quality, actually this wool, chain of wool doesn't really have any economic sustainability, the product it's not sellable, it's not usable so I think it's a pretty fundamental step, nevertheless this is the rock in Malcomonica where if you have good eyes, maybe you know you can see a lot of looms, like this one, with a long white there is some kind of them if you want to exercise and weaving is also, I would say, a sort of social craft because it's something that requires teaching, tutoring, you need to practice and you need a dialogue between the one that teaches you and the one that is learning, there is a lot of ethnographical evidence of large looms where you have one or two or three people working together, so weavers so I think there is a lot of professional interaction around weaving and this is why in this paper I wanted to talk about, introduce what I believe it fits very well the material, the concept of community of practice that I have been getting familiar with recently and in order for archaeologists to have a different approach to the material culture not only look at the tools but look at the tools as the results of this professional interaction between people, so they are the results of the practice that was going on and a community of practice, if you want a general definition so it's a group of people who share a concern or a passion it's very used in modern studies about a lot of different internet communities so for something they do and learn how to do better, they interact regularly so I think it fits very well with the weaving process and more in detail, in order to have a working community of practice you need to have, it's called domain, so it's a shared competence people do the same thing, they interact about this domain that they are concerned with so there is a community, some members that engage in activities and discussions and help each other share information as I believe for example between masters and tutors, between wevers and there is the practice, and the practice is the repertoire of resources experience, stories, tools that you use and this is distinguished the group, the community from other people doing other things and I think that in archaeology we have the possibility to find to discuss all three products because we could see the practice through the long waits that we have, the nature of the weaving this social interaction for the weaving process helps to construct the domain and the analysis of the tools which I'll try to demonstrate to show now briefly, I think it can help us to find key to interpret the domain so how this craft production develops and transforms this is the area where I have been working so the populate in north Italy and as you can see it's a densely populated area it starts around 1600, it is basically a force before 1600 and it's rapidly growing culture, the settlements increase in numbers and with time there are changes some settlements become much bigger in general in the beginning there are around one actor, two actors and at the end of the reason is that you have 20 actors two and some stop existing so we cannot go in any detail some of these places have been excavated and there are several textile tools and I've been looking at the published and well-published textiles and these are these three sites that I will mention I particularly am working with Montale which is our typical Trumare settlements look like with the beach and an embankment this is the Montale site the apartment and the large beach it was 40 meters wide so it's quite impressive the sediment itself is only one hectare as you can see here the modern excavation has been just working here all the rest has been shoveled away in the 18th, 19th century because it was used as compost it was very rich dark soil and it was spread all around to make a fertile agriculture but these excavations are very accurate and there is a lot of also long-waited that were collected by the local museum people during the compost query they were done and they were taking all the archaeological objects that were chewed away so we have nowadays there are 144 long-waited from this site and 17 from the modern excavation in this little area and this is the excavation area very small and in phase 2 they could see two houses here is the 3D reconstruction how they could look like it's a very densely-inhabited sediment and this phase 2 long-waited out of the 17 from the whole 11 phase the 10 of them come from this house and this is where I started thinking about a group community of practice and specific sets of tools because all of them are of cylindrical form there is many different kinds in the whole collection from Montal but all of those from this house are cylindrical and as you can see here they are the one in the red square if you see this one is the weight and this one is the diameter of the face with the hole these are all the long-waited from Montal they all have relatively similar diameter but the weight spread from 400 grams up to 1.5 kilo and the one from phase 2 they are all also similar in shape but different in weight and I was thinking if this is a physical way to work with and in the neighbor site of Santa Rosa you have a similar situation with two houses with the two lines of long-waited found on the floor which have been interpreted as one standing loom and the situation is rather similar they have many different weights which you can see here and somehow at least a group of them are relatively similar diameter and what's more interesting they all have very similar shapes and these two houses and the house from Montalé are contemporary so my idea is that actually there is some sort of groups, community that work with textile in a similar fashion in different villages at the same time using relatively similar long-wait with different weights so there are light and heavy ones used at the same time in Santa Rosa here you can see that the light ones are in the middle of the loom and the heavy ones at the two sides of the loom which is something that in some it is recorded it's no logical we have another in the next phase in the middle of the age three there is another site where the possible loom have been found and here you can see that in the next phase something has changed because looms are really similar they have I don't know the picture but they look the same they also have similar decoration but they are not very different in weight anymore there is a relatively limited range of weight so it feels like there is a different set, maybe a different way of weaving or maybe the technology has developed all over we don't have unfortunately many well-dated loom weights from Montalé to Saïda but we have from the 19th century excavation group of looms that look very much the same they have similar decoration and some of them like this one they have almost exactly the same weight so they really look very much like that something similar to what could be seen in the middle of the age three it's happening at Montalé too then Montalé stops existing in reason from age one and in reason from age two instead in another site that becomes much larger you have a house with many loom weights one that the group that will be interpreted as a loom and one that the area where they were just kept and all the loom weights from this late period are very heavy so maybe it's also changing they are all about one kilo so something has happened and you remember Montalé was very similar for Ville this one was very similar to Montalé in the middle of age two but now it has changed completely and this is again Montalé all the that stops before and during the reason from age one so you can see that until the reason from age one everyone is using loom weights below one kilo and in the next phase like in Pauville they become above one kilo so there is a lot of dynamic going on in the in the plane the weights of the loom the loom weights it changed very much during the different phases and I try to I will not go in any detail I try to compare with the Trentino Garda area which is very close to this area and other archaeological evidence show a lot of interaction between these two areas and here you have a completely different situation all the loom weights that have been found from the early Bronze Age until the final Bronze Age seem to have always a relatively limited range of weights below one kilo and so it really talk about a completely different way of weaving or of tools to win the results I don't know so the conclusion is that I think that there is reasons to see to look for transformation in the archaeological material even in this that could be considered like minor transformation because the possibility that community practice have to develop or to adapt to external for example market possibility it actually gives to the entire ability the possibility to have a sustainable economy and when the loom weights become the rest of the play Montaille actually stops existing so it might be also something very big has happened there from the economical economical point of view that made it possible to live there anymore and yeah


First question is from ZW Spivey. What is the best way to increase grip strength? Hmm grip strength we get a lot of questions on grip strength and I do and I made this comment before but I do think it's important Did you do a YouTube video on this? I did and I put some like forearm developing Exercises on there, but here's here's something. This is something that's very important to consider. Okay Your hands literally connect you to everything that you're doing and they're we evolved to have very You know intricate fingers like we could place them well, but also strong hands Literally if your hands aren't strong enough to support the weight that you could lift with your shoulders and your back and your legs And you can't lift that weight. What's happened is Art we barely ever use our hands. In fact, there was a study done With college aged males they tested their grip strength and it was recent was like within the last five years And they found that their grip strength was as good as a 60 year olds in the 1980s Like we just don't do anything with our hands and now this has far-reaching effects on the body It affects the way that your shoulders move of course affects your workouts It affects wrist health and so it's definitely something that's important to focus on so if you're working out and you can't hold on to the barbell or the The dumbbells you find your hands fatigue and you're not like some champion powerlifter who's lifting ungodly amounts of weight This is something you should focus on one of the best things you could do to improve your grip strength Is to do a little bit of grip work every single day So this is an old exercise device. It's probably one of the first pieces of exercise equipment that was ever sold It's those old school hand grippers that you can like a spring loaded one a spring loaded one And and now careful because you can overdo this but literally at your desk have it there and every hour You know mess around with it for five minutes and then put it down. That's it.  Well, they'll have to use way less weight because that makes it harder initially. Yes, I think um They I mean they could just focus on the regular bar to get their hands stronger If it's progress if it's stopping them from progressing I mean you're gonna have to do some stuff to get your grip stronger or just be super patient because eventually it'll catch up You know, I like farmer carries, you know doing things like that to do it I mean, I mean I like the the squeezing and contracting and I think I think that's gonna help But I feel like something that is More closely related to what they are trying to try it like it's for barbell rows, right?  Um, this is back when newspapers existed So you might be able to use something else, but you take a big sheet of newspaper I guess you could do this with butcher paper And with one hand you start at the corner and you crumple it up little by little until you get the whole thing into a ball It's actually a very good Um hand exercise and then in the gym Of course, you can hold on to things for time. That's more of an isometric thing But don't forget that because it's isometric the strength tends to be I mean there's definitely It definitely radiates out, but a lot most of strength is in that position, right?  Okay. So the rest of your body is not but your hands are and so doing things in the eye So hangs, you know hanging with your body weight on a pull-up bar. I think would help I think the fat grips on on the barbell would work and then of course just getting getting good Frequency and volume of holding heavy things. I mean it's like to your point of a farmer walks That was a big game changer for me just holding weight for longer periods of time because it and two You don't really realize that you're holding it for a longer period of time You're just trying to get to you know, so many yards of carrying these objects So at least it takes kind of the focus away from it But yeah, like it just exposing, you know, your hands to more different types of You know textures and like even like the like we have the rice bucket stuff like in in our ocr program It's like things like that are you know things you don't normally do and if that's really a focus of yours, you know expose Like just more dexterity and more function out of your fingers Which then helps with the overall hand strength, which then you know goes up to your grip and your your wrist strength and everything else It's just a matter of like Contracting and moving your fingers and your hands and picking things up Yeah, and of course the forearm, you know flexors and extenders are kind of connected to that, right? So there's exercises that'll flex Uh extend but then don't don't forget lateral a lot of people forget that this also is important Yeah, so I'll I'll I used to get a Like a dumbbell Like you know the ones that you can load with weight the old school ones So you can load weight or whatever and I would hold one end and I would load it with a tiny bit of weight on one end So it was offset and then I do this exercise here or Hold a bit here and go in this direction. So I'm doing for that with Indian clubs I'll do that sometimes too just because yeah, you have like that long lever To account for yeah that that being said all these exercises we're talking about right now like for forms and trying to develop for your your grip strength The greatest gains I ever had on my grip strength came way later when I started just heavy ass deadlifting and farmer carries Like I and and I've done I've like I remember potatoes wrist curls and reverse wrist curls and doing all this You know all the different moves to develop my forms to try and work on my grip strength Nothing gave me better grip strength than actually just getting Stronger at deadlifting stronger at farmer carries that brought it up more than anything else And in a stronger grip for people listening or watching who are like, I'm okay, right? Like what's the benefit? Your presses your rows your curls your extensions when your hands feel strong You're so much more connected to the exercise and you get better Activation up the kinetic chain so they find for example that wearing wrist straps, which tends to make us use a weaker grip Changes the activation up in the neck and in the shoulder So get having a stronger grip even on presses even exercise You don't think you need to have a strong grip when your grip is like try this next time you bench press Try putting wrist support on and wrist wraps on around the bar and all of a sudden you can lift more weight Like how is that possible? It's simulating a stronger grip So strengthen your grip and don't be surprised if you don't see all your lifts improve as a result


Welcome back to Kids Fun Science. My name is Ken. Today's experiment is how to make a stem popsicle rope climber. What you need for this experiment is five to seven popsicles, seven toothpicks, one and a half meters of string, about three feet, hot glue gun, pliers, and a cutter, rubber band, and a drill. All right, so here's the experiment. I'm doing it a little bit backwards this time. I'm going to do the experiment first and then show you how to set it up and make this experiment. It is a little bit more complex. It's not as easy as normal experiments I do, but it's well worth at the end. So this is the popsicle stick climber and so I got a string here. I'm just going to hold the top end, hold it, but on the bottom I'm going to give poles and it's going to climb to the top. So here's his maiden voyage and I'll start here and he's starting to go up. There he goes. Take a little while. He's getting a little crazy there and success. So that's your popsicle stick climber. To reset him, it's very easy. Just hold the top of the string, hold his hips and slide, pull the string on the top, keep pulling the string. So it's almost to the end. You don't want to have it all the way go out and then you're ready to climb again. So that's the popsicle stick climber and next I will show you how to create this or make this and thanks for watching. Here are the measurements for the legs. As you can see, you have two different popsicle sticks that you're going to drill four different holes and the measurements are there and then we go onto the arms. There's three holes you're going to drill on two different popsicle sticks. So the back, this is the back and it's just one popsicle stick with two different drill holes and there's the measurements and then here's the face with the measurements and there's two different popsicle sticks. You could use one popsicle stick. Depending what you use to cut the popsicle sticks, be very careful. I'm using wire cutters and they clip pretty easy for me but once again, be very careful that then to put it together, you're going to have to drill holes and you're going to have one on each side of the back of the popsicle stick like this and then the face is optional just kind of just to see what it's a climber and then take your drill and you're going to drill. I just did both at the same time and drill through and be careful with that. Then you're going to take your popsicle stick.  So you keep pushing in until it gets very very tight and then on the opposite end take your other leg and you're going to have to kind of twist it around back and forth because I got exactly really snug so it took a while to push all the way down in and be careful because the toothpick is sharp and so I just kept nudging it back and forth until it got snug against all three pieces and then from there you'll be able to use your cutters to cut the toothpick down so it's not extending out like it is now and so that's the start of the bottom of the legs there and then I take your cutter and I'm just going to cut each side you could use hot glue to hold it in place but I found it to be pretty snug where I didn't have to do any more so then on the leg you're going to continue the same process now but you only have two toothpicks to go through because the back isn't there. So on the rubber band I started off with a four centimeter rubber band but it was too loose so I had to switch to a smaller one and I went to a three centimeter I know it's not fully straight out but three centimeters is what worked for me you're going to do the rest of your legs there and you're going to take your third toothpick the closest to the leg you want to put the rubber band in first and of course I'm using the wrong rubber band in this video I had to push it in with the toothpick because it's a little bit tight in between depending on the size of your rubber band but you just want it to be hanging down before you put the toothpick into that final slot right there which is the second slot of the leg and then when you put the toothpick in now the rubber band will be trapped around it to help it hinge and decline so you can once again you like all toothpicks keeps wiggling it until it's really really snug and tight and then once it is and now you can see when you pull up with the rubber band that it's you know hooked in there and it can't come out and then you're going to take your cutters and cut all the pieces off from there now you've got the bottom part done with the leg now you're going to be working on the arms and we're going to do a similar technique with the arms so we're going to start off with the cut piece on the outside one side of the outer back and you're going to put that toothpick through just like this and you're going to have to keep twisting it if you drilled your holes the right size and then you're going to twist this one on twist it in a circle kind of get it nice and snug and then when you do then you're able to put the other side on and once again it's going to take a little while to screw that on and get it all the way on so I skip forward and I've got it all the way on you want to make sure it's snug and then face the arms outward and you're going to see that the second one this is easier this way to actually have the rubber band put in before and I probably should have done that about legs you know live and learn right and then you're going to put the toothpick on from one of the outsides and go through let the rubber band pass you don't want to go too far and then have the rubber band go inside it and then line it up with the outside arm and then you'll be able to push it all the way through and then you'll be secure the rubber band will be secure with the arms and the legs which will help it climb that rope so the toothpick broke off there I didn't even have to cut it and once again you can use hot glue on each end make sure that rubber band goes through the middle there you only want to through the second toothpick the middle toothpick there before you put the last toothpick in so this is where I found out it was a little bit too loose of a rubber band it wouldn't pull back so that's why I ended up redoing it and use the three inch rubber band and then so once again once you get the last toothpick in you're going to cut them all off and from there I'll go the next step so here it is see how it's when I'm pulling it up and down it doesn't have enough tension and so that's why this rubber band was too big and I had to use a smaller rubber band and so that's one technique if it doesn't work so now I've got my smaller rubber band and now it's very important here on the hands that you put some hot glue or a rubber band around the end I probably could have used a rubber band but you want it so the toothpick on the very end and this hot glue has a very small gap in between because we're going to put the string in between that and it's got to catch onto this glue or a rubber band if you hadn't wrapped around to be able to get that friction to make the experiment work and to be able to make the climber climb so you can see it's hard to see here I'm trying to get focused but it's a very small gap and there there it is just enough for the string to hit around the the toothpick and rub against the hot glue and so you want to make sure the hot glue dries before you do that and then what you're doing is you're coming around the bottom there you go over the leg and then you go underneath the third pin and then you come up in between the two from there so what I did is I took out the third pin I couldn't get the string to fit in between the actual third toothpick on the arm and hot glue so now I've got the toothpick I took out and then now I'm going to put it back in I'm just using a small little toothpick again but I'm using the pliers to kind of give it a good force through and hold that string back because you want to be able to make sure the strings in between there it is right there just like that in between the toothpick and the hot glue and therefore now the climber has the ability to move its legs and arms so here's the end results with a little bit of a climb here and there he goes and all I do is hold the top and pull on the bottom and it will go all the way up well I hope you enjoyed this experiment if you did please remember to click thumbs up and to subscribe and let me know in the comments below if you have any questions I'd be more than happy to help you get this to work thanks for watching


Next up I'd like to invite Peter Anderson from the ANU College of Engineering and Computer Science and the title of his three-minute thesis tonight is The Language of Sight.  Could you still find your way? Could you find an ATM or a table that was free at a cafe? Worldwide there's 285 million people with impaired vision that face challenges like this on a daily basis and despite all the advances in medical science many cannot be cured. So in my research I'm using artificial intelligence or AI to help. I'm building intelligent visual assistants that can help in ways the guide dogs and white canes cannot by describing what they see. Looking at this image my visual assistant describes it as a group of people walking down a street with several shops. Now that's helpful but of course there's much more to a scene than this. That's why you can also ask questions. Questions like is there an ATM here? Can you see a street sign or where's the coffee shop? And each question should receive in response a helpful correct answer. Now you and I can describe our surroundings and answer visual questions like these with these but this is still very difficult for an AI system but we're making rapid progress. To tackle this problem I'm using a deep artificial neural network a computer model made up of millions of connections inspired by the vast network of neurons in our brain. In my particular approach I train the network to first identify every single visible object then inspired by our own visual system the network learns to focus its attention. That means it looks only at those objects that are most relevant to either describing the overall scene or answering your specific question. Using this technique right now our work leads the world in terms of describing images and visual question answering. So it's my hope that in a future filled with self-driving cars and autonomous robots that we can also offer greater autonomy to millions of people with impaired vision.


So yesterday we wrote the localization formula which will be especially the hero of my next lectures. And so the formula says that the partition function reduces to an integral over a subset of fill configurations. Let me put this label BPS and I will explain in a minute what I mean. And then we have the classical action evaluated on these fill configurations. And then there is a remnant of the fact that we reduced on this sub manifold, which comes from small quadratic oscillations around this sub manifold. And this is because these small quadratic fluctuations or infinitesimal fluctuations are not suppressed. But in fact since they are quadratic this is a set of harmonic oscillators. And so this just gives us Gaussian factors, a ratio of determinants. I think this was the notation I used. OK, so something that I said at the end of the lecture, but let me stress it again. So this is still, I mean in general, this is still some infinite dimensional space. So it's still a hard problem. But in good situation, well the good situations are the ones in which this sub manifold becomes finite dimensional. Because then this reduces to a standard integral and we can have some hope to solve it. OK, the other comment is that this in fact is an exact formula, so we computed it with a sort of saddle point approximation, but because of the fact that there was no dependence on this deformation parameter, the result is exact. And then finally I would like to explain this label. So at this point how this was derived was just the zero locus of this functional that we chose with some nice properties. In fact, this most examples, it turns out that this zero locus corresponds, or it can be chosen to corresponds to BPS configurations. So configurations that solve the BPS equations. And the reason is the following. One can try to make a canonical choice for this B. And this canonical choice, so if you wish this is similar to what we did in the finite dimensional case where we started with the form beta, but then we say let's choose beta to be actually the one form which is due to the vector field. So here also one can try to make a canonical choice. So this is a sum of all the fermions in the theory. And then we take q psi, double dagger psi. So this double dagger, what is this double dagger? Well essentially, as we say, we go to Euclidean signature, so we complexify all the fields. And so our fields that are complex conjugate in Lorentzian signature they become independent in Euclidean signature. And so here we can just try to choose some anti-linear operator, which is like a dagger, but it doesn't need to be the dagger in Lorentzian signature. So I make this standard choice. And then what happens? So OK, we have to check whether this v satisfies the property that delta bv is equal to zero. So, but then the nice thing is that when this happens if we compute qv, so q can act on this. And then so what we get here is q psi, double dagger q psi and then it can act on this. And then we get q of q psi, double dagger psi. Now this is a fermionic term, because this is a fermion and this will also contain at least another fermion. Well, this is the bosonic term, but then you see that this is automatically positive. And the zero locus precisely corresponds to q psi equal to zero. So for the zero locus we find the BPS equations. And then OK, if you have chosen a contour, we have to restrict. So in general these BPS equations can contain lots of solutions, because in general we have complexified fields, but then if you have chosen a contour, we should restrict to these equations to the contour. And then OK, one gets what they get. OK. So, are there questions? OK, so now let's start with another topic that in my notes is lecture two. And so now I would like to discuss a concrete example. In fact, everything was very general. And in fact, this is the main theme of other lectures where other cases are discussed, and probably you will hear, or maybe you will hear this story again, explain with different words, which is good. But now I would like to go to a specific example, or examples. And so I would like to discuss two-dimensional theories with a certain amount of supersymmetry, which is 2,2, which corresponds to four supercharges. And in fact, this is the dimensional reduction of four-dimensional and equal one. So essentially you should already be familiar with this type of supersymmetry. Many things are similar to four dimensions. And OK, this is a nice example for various reasons. Well, for me, this is a nice example because I've worked on this example, so I'm more familiar with this setup. But it's also, I mean, we are in low dimensions, so things are relatively simple, and it's easy to do all the computations on the blackboard, or almost all of them. But still one can get very interesting physics, as we will see. So, OK, so let me say a few things about this supersymmetry. So if we start in Lorentzian signature, so as these notations suggest, so there are left-moving and a right-moving spinors in two-dimensions, which in Lorentzian signatures are not related by charge conjugation. And so we have one complex left-moving supercharges, one complex right-moving supercharge. And the biggest asymmetry that we can have is u1 times u1. There is a u1 that acts on this complex supercharge, another one on this other complex supercharge. And, well, if you want, we can do a change of bases and talk about a vector like our symmetry and an axelar symmetry, just the diagonal, anti-diagonal combination. But, of course, as also Guido stressed various times. So this is not part of the supersymmetry algebra in flat space. This is an outer automorphism of the algebra. And so supersymmetric theory need to have this r-symmetry. It can have it, but it's not required. It becomes required if the theory is super conformal, because then the r-symmetry is part of the algebra. However, we restrict to theories in which these r-symmetries is present. So we insist, or we chose, class of theories, that the vector like is present. So you want the r-symmetry. OK, so now in general there can be two central charges for this algebra. But if we insist that these are symmetries present, then there is only one central charge, which is related to the possibly broken axel symmetry. And the reason is that these central charges are charged under the two symmetries. So if you have the central charge, you are breaking the symmetry in vice-versa. And so in this case we have one complex central charge. In fact, in the super conformal theory where we have both symmetries, these central charges are zero. And if you want to have both central charges, you need to break those symmetries. And so now if we go on Euclidean signature, but still on flat space, the supersymmetry algebra looks like the following. Hopefully I don't have notational mistakes. So now in Euclidean I will use a tilde to indicate what should be the complex conjugate in Lorentzian signature, but is independent in Euclidean signature. And so what we have here is of course we have translations. Or if you want this is like a lead derivative. But then there is also this complex central charge. So let me write in this term, in this way. OK, so what I mean by this notation is that really I want to use bispinor notation. So this P mu should be rewritten. Well, this is in fact rewritten by spinor notation with the gamma matrices. And these projectors are the chirality projectors. So spinors in two dimensions are two components, the complex Dirac spinors. So these are chirality projectors in positive chirality and negative chirality. We could call them 1 plus or minus gamma 3 over 2. OK, while this is z... Sorry, didn't write z. So this z and z tilde, so this is the complex central charge and I'm using a tilde again because in Euclidean becomes independent. OK, while the qq and q tilde, q tilde are anti-commutator as 0. And then we assign r charges to these supercharges.  OK, so... So now we want to put these theories on curved space. So we follow the root that Guido explained to us. So first you should discuss supercar and multiplets. But as I said, this amount of supersymmetry is dimensional reduction of for dimensional equal 1. And so in fact, everything is essentially dimensional reduction of that case. And so in particular since we have these r symmetry, there exist an r-multiplet, which is the dimensional reduction of the r-multiplet for dimensions. And it contains the following operators. So it contains the stress tensor. It contains the supersymmetry current. And then it contains the current, the conserved current for the r symmetry. And then there is also the central charge. And so it contains a complex conserved current for the central charge. Correspondingly to this r-multiplet there is a two-dimensional off-shell supergravity theory, which is the dimensional reduction of new minimal supergravity for dimensions in which the graviton-multiplet precisely contains fields which are going to be paired with these operators. Because at linearized level the fields in the graviton-multiplet couple linearly to the operators in the superkarem-multiplet. And so in this graviton-multiplet we have the metric, we have the gravitino. And then we have vector fields for these guys. So there is a vector field that couples to the r symmetry and there is a complex vector field that couples to the complex central charge. So, of course, we know how the metric appears in the in the theory. How do these vector fields appear in the theory? Well, first of all, they appear in covariant derivatives, obviously. So every time there is some covariant derivative, this will be a covariant derivative in respect to the curved metric, but then it contains terms like this. So where r is the r charge and this is the vector field that comes to the r charge. And then in this particular notation where I use complex currents and complex vectors, again, if I didn't make mistakes it should look like the following. So this vector field appears in the covariant derivative weighted by the central charge. So depending on which fields it acts will be coupled in a different way. It turns out that these fields also appear through their field strength. And so, but since we are in two dimensions, so field strength is two form, but we are in two dimensions, it's convenient to dualize it to a scalar. And so we can use some scalar h, which is up to some factor. It's just the dual to the field strength for c So these are the fields that appear in the supergravity in the supergravity multiplet in the graviton multiplet. And so if you want to study supersymmetry on curved manifolds we should look at why essentially we should set to zero the variation of the fermions in this multiplet. And in this case the only fermion is the gravitino, there is no gauge or something like that. And so setting those variations to zero is sometimes what we usually call the generalized killing-spinner equation. So this equation looks like the following. I call this v. And there is a similar equation for the tilde fields. OK, so these are the gravitino variations in this supergravity theory. Now here the dots corresponds to terms which are, I mean, appear in the funeral linear theory, but that are automatically zero if we set the gravitino to zero. So we don't care about them in this generalized, I mean, just if we want to set these variations to zero because we also set at the same time the psi to zero. So the equation that we have to solve is this. And yes, and so we have so we have this, so the derivative, so this is the parameter for supersymmetry variations. This parameter is not charged under the central charges, so only this r-symmetry vector field appears, but those fields appear through the field strength. And moreover, here I'm using a notation. So if you want, this is a spinor and here for clarity I've chosen a base in which chirality is diagonalized. So I've chosen a base in which chirality is one minus one. Otherwise you can write in a invariant way with projectors, but it's more messy. I mean, it's not messy, but it's longer to write. Ok, and this reflects the fact that two supercharges are opposite charges. Ok, so essentially if we want to study supersymmetry on core manifold we have to solve these equations. So before doing that let me just say something which is useful. So we are in two dimensions and so the group of rotation is SO2 which is U1. And so essentially the bundle in which any field with spin transforms is the same as a billion-gauge bundle. So there is no real difference between the spin and some standard abelian charge. And so we can use these to simplify our notation a little bit. So first of all we can take the spin connection and so we can take the spin connection which if these are some field-bind bases, these are anti-symmetric and this is the algebra of SO2 and we can just convert in a standard abelian-gauge field. So we just multiply by epsilon and we can define some omega mu. This is the spin connection abelian-gauge field. But then this is useful because now if we take the covariant derivative and we act on a field with some spin S and we look at a specific component. So of course each component will have some value for the spin, for the Z component if you want. So this is just going to be the same as so the spin appears as a charge and we have this abelian-gauge field. So the spin just appears as some abelian charge and if a field is actually charged under some other gauge field these other gauge fields are in the same way here. I mean this is trivial but it's useful. Do you have any questions so far? OK. So another observation that we want to make is that again before going to solve these equations it's about the form algebra. So as we said when we go on a curved manifold the number of generators is a subset the number of generators of the supersymmetry algebra is a subset of the generators that we have on flat space but the algebra can be deformed and we are restricted into relevant deformations so the formation that disappear when the manifold is scaled to have infinite volume and we can read off this the form algebra into the supergravity and so essentially even before solving the equations I mean you can write in general what the form algebra is in terms of those fields and the specific algebra will depends on what is the profile that those fields will take on a solution and so the result that you obtain is the following I rewrite it in terms of some delta epsilon for me delta epsilon is just the contraction so as I say is a contraction of the parameter supersymmetry parameter which is a solution of those equations with the supersymmetry charge that we have in supergravity so it turns out that here so you get a lead derivative so there is a translation and this lead derivative is along a vector field which is specified by infact is a biliner in these spinors so this K mu is a standard formula which is sometimes called a sandwich because you take these two spinors but then if you act on some gauge variant field this should also be the gauge covariant lead derivative but then there is also an extra piece which is again a sandwich which depends on a matrix q so this is a matrix in spinor space and this matrix q is given by the following so again it is diagonally in chirality and so this matrix q is controlled by the central charges it is controlled by the profile of this scalar H which is the field strength of that field and then there is also this sigma that I haven't introduced so this appears when you couple to some matter action and then you go in best domino gauge and essentially we will describe this in more details afterwards but essentially this is related to twisted masses so some masses you can turn on in the theory and there are background fields so if you want these are scalars which appear in the vector multiplet so if you want this formula is already a little bit more generic because as we explained if in our theory we have some symmetries we can play a similar trick with vector multiplets if you have some symmetry in theory to a background gauge field this background gauge field appears in a super multiplet and so there will be other bosonic fields in this multiplet one of these fields is a scalar and one can turn it on in a way that preserves supersymmetry and then this appears in this algebra while the other ones are zero and similarly for the tilted version so so yes yes you see it here that depends on the r charge so ok my writing is not the best so this is a small r so it depends on the r charge and so yes and so this it reflects the fact that in this deformed superalgebra the r symmetry enters yes so I think that the term that you are referring to let's see so that should enter I mean if something is charged under the r symmetry that will appear in this in this le derivative so this superscript means that you have to put all the gauge fields that are relevant I mean if it depends on what you act on yeah yeah we are not very precise in my notes but I think this is what what happens ok so this a means is that so if you act on a operator which is gauge invariant and neutral under everything this is just the le derivative but if this field is charged either because it is I mean you are acting on the fields in the Lagrangian so they are gauge covariant they are not invariant then this a is the gauge field the dynamical gauge field but I think if you act on an operator which is charged under the r symmetry then you should also include the v here so if it is charged it would act by like lk minus a mu or something or minus i a mu it would be like taking a covariant derivative or replacing del mu with the le derivative yeah so you can take for this le derivative I mean this define as the anti commutator of d and the contraction with a vector field and so you have to make d into a gauge you just bracket covariant derivative yeah if you want yeah ok but we will see this more explicitly in the examples ok so now what I would like to do is to take those equations so what we would like to do is to take these equations and solve them and on very general terms you would like to find the most general set of solutions so in particular you would like to solve in very general terms for the metric and then for some metric for the vector fields v and for the scalar h and then in such a way there is at least one solution in epsilon so we will not study in full details the solutions to these equations this has been done by Clausset and Cremonesi in 2014 instead I would like just to present a few solutions which are interesting so a first simple solution that we can have on any manifold or probably I should say any orientable manifold is the following so so we choose a vector field that goes to the r symmetry to exactly cancel the contribution of the spin connection in the covariant derivative in one of the two equations let's say that we do it for the first one well sorry let me say it back so we set the vector field to be equal up to factor one half to the spin connection in such a way that there is some cancellation and now and on the other hand we set the scalar fields to zero so now what's happening here so epsilon and those epsilon tilde they are two component spinors so we are using this notation in which there is an epsilon plus and an epsilon minus because we are in the base in which we diagonalize chirality so these components have spin plus one half and minus one half so in the first equation now we could write the equations in components and for one component transforms as it was charged and there will be a gauge with charge plus one half and the other component will charge minus one half and so you see that if we make this choice for one of the two so in the first equation for one of the two components we can precisely cancel v against the spin connection of course we cannot do it for both components because the arc charge is the same but the spin is opposite so we can only cancel the contribution one equation on the other one but then something similar happens in the other equation just for the other chirality and so now the equation becomes since h is zero the equation just becomes that the spinor is constant and of course this is a solution on any manifold and so we can just set epsilon to be zero epsilon minus with some constant epsilon minus and epsilon tilde to be epsilon tilde plus zero where epsilon minus and epsilon plus tilde are constant ok so this is a very simple solution but it exists on any manifold and in fact this solution has been known for a long time this is called the A twist so this is a type of topological twist what does this terminology means well the twist is a way to preserve supersymmetry in which the only thing that you do you have a theory with some r symmetry and then you turn on a background for the vector field that couples to the r symmetry which is related either is equal to the spin connection or is equal to some component of the spin connection in such a way that at least for some components of the spinor there is a cancellation vector fields that couples to the r symmetry and the spin connection and so for this for some components the equation simplifies and just becomes that component is constant and then there are solutions and in particular in the A twist you do that using the vector like r symmetry and so as I said this solution has been known for a long time however it is reassuring because it is contained in this general formalism that is supposed to contain all ways of at least using the r multiplet all ways to preserve all ways to preserve supersymmetry on a core manifold we can also go on and yeah, no, okay so this is the solution and and well, yes, sorry let me write what that what this supersymmetry algebra is and in fact becomes something very simple so this operator, so we have two operators which are important and moreover the anti-commute so this super algebra is very simple and if you wish the r symmetry is not part of this of this super algebra and it turns out that for genus bigger than one this is essentially the only solution so what do you invite essentially what would be exceptional for you well, let's see let's see so essentially because so as you see from the equation so let's take the so let's look at epsilon so the upper component is zero so if you look in the equation so gamma mu inverts the chirality and so you can choose any h tilde because it still solves the equations because it multiplies by zero so h tilde can be an arbitrary function on the manifold but this is not gonna so when you compute the localization results so partition functions is not gonna affect the answer so it's not the only solution but in terms of partition function that you compute it's everything you want to get you get it from there this is more just to make sure I'm understanding things but so you're saying this putting it in this background super gravity is exactly the same on the nose as the sort of older papers of people like Whitton on topologically assisted theories where they don't mention any background super gravity they just take the energy of an M-sensor and they shift it so that the right things happen that's exactly equivalent at some point I start mentioning it but maybe not in the very first page I just wanted to make sure I wasn't confusing things no no no it's a very important thing to remark it's just a different perspective on the old topological twist so one way to put it is that essentially you do some improvement transformation and use a different stress tensor which is in this different stress tensor essentially the words that are used is that you change the spin of the fields according to their charge because you regard rotations as now a subgroup so you take you want spin you take you want V then you declare that the new spin group is you want which is some linear combination of these two right this is the old language yes but what does it mean it means that now this group couples to the curvature of the spin connection so when you turn on a spin connection for this is like you are turning on a spin connection both here and here dictated by the particular linear combination that appears here so effectively what you are doing is that depending on the spin connection you also turn on a background for the vector for this vector for the cups to the R charge so this should show that they are exactly the same thing does it I mean are you are you saying that the improvement of an energy momentum tensor is equivalent to changing the background it should be in this case I don't know maybe Guido can correct me if I'm wrong but I mean you are changing the stress tensor I mean and you still have stress tensor which is symmetric and conserved so that should be an improvement transformation but the thing is that you follow and coordinate transformation with some out transformation yeah but at the level of changing the stress tensor so one way to put is that you take the stress tensor and you change it so you add something in such a way that becomes q closed q exact I think it can be interpreted as an improvement transformation so we can't think about that ok so another solution that I will call untwisted S2 so we take the round metric on S2 so this solution will be specific for S2 will not work for bitrariginos and then ok, I call it untwisted because this time we don't turn on at all a background for the vector field that couples to the r symmetry so we set this to 0 but it's decided we can still solve the equations so now we have to turn on the scalar fields where r is the radius of the sphere and now if you ok, you just write those equations the equations take the following form and in fact these equations are be known also for a long time these are called f and not mistaken conformal kininspinor equations and and well on S2 I mean you don't need to open a book to solve this equation on S2 they are simple to solve and it turns out that there are two solutions but in fact there are two solutions for epsilon and two solutions for epsilon tilde so in total there are four solutions and for instance for epsilon should I leave this as an exercise? ok, I mean they are simple enough to write so we leave it as an exercise find this epsilon but essentially there are two integration complex integration constants so there are two solutions for epsilon and then two solutions for epsilon tilde and so in fact let me this is something I didn't stress before let me compare with this topological twist so you see there is one complex solution for epsilon and one complex solution for epsilon tilde so there are two supercharges but we started with four and so the topological twist breaks half of the supersymmetries it's half pps however this solution does better we have the same amount of I mean we preserve all supercharges we have four of course we have to use the round metric however did the form supersymmetry algebra looks like the following so first of all this is not the same as in flat space because you don't just have you don't just have the derivative which is a translation but you also have an r symmetry rotation even though this vector field is zero so this comes from the second term from that bold Q correctly this deformation disappears if you go into flat space limit because there is one over the radius of the sphere so as promised the supersymmetry algebra should reduce to the flat space one when we go to flat space but you see the r charges of the fields appear explicitly in this supersymmetry algebra and so in particular this is related to the fact that the r symmetry now becomes part of the algebra so this is different from flat space as I said before in flat space r symmetry is an outer automorphism but here instead in this algebra the r symmetry is part of the algebra and because of this reason we have to make a choice for the r charges and this choice of the r charges will affect the theory this is different from flat space in flat space you write down a Lagrangian if you want you can assign r charges but of course the Lagrangian is not gonna depend on the r charges just an assignment that we make and once again this is due to the fact that the r symmetry is not part of the algebra but in this case it will be different so we will see the Lagrangians explicitly depend on the choice of r charges they will appear as some curvature couplings and once again this is due to the fact that the r symmetry now becomes part of the algebra ok so the other ones are 0 I already wrote it there and this so in fact this K mu which we already wrote so this is a killing vector which generates as you vary this epsilon and epsilon t that this generates rotations of the sphere so the SO3 that acts on the sphere and in fact the full superalgebra turns out to be SU2 slash 1 plus bosonic subgroup is at the level of algebra is SU2 times U1 where this SU2 in fact is this well at the level of the algebra this is the same as the SO3 which rotates the sphere and this U1 is precisely the r symmetry ok is there any question? if you preserve inside of the vector r symmetry axial then you should have a corresponding solution for a V-model and is there an unquisted solution? yes so so if you want to discuss the axial case so that is not immediately obvious how to reduce it comes from four dimensions because the axial are similar to the reduction is accidental but in fact it should be very similar so in two dimensions it should look very similar to the r-multiple somehow it looks like an r-multiple for the axial symmetry in which now you have the conserved current for the axial symmetry in fact it turns out that is the mention reduction for dimensions but then when you go into two you get this vector if I understand correctly and so in that case everything goes through more or less in a similar way where instead of having the vector like r symmetry you have the axial r symmetry and everything just change between chiral and twisted chiral and vector and twisted vector the only thing to be careful about is that so this vector like r symmetry is never anomalous with this amount of supersymmetry but the axial one in general can be anomalous so the other twist since as we said in order to make this construction we really need that this is not just classically a symmetry, it should be a symmetry in the quantum theory so you need to make sure that the axial symmetry is a symmetry of the theory but otherwise everything goes through and in fact so this is very useful and in fact in the literature somehow this is called because it's the one related to the A twist and then the other one is called B because it is related to the B twist In principle vector and u r symmetry should be completely symmetric because of mirror symmetry they are not completely symmetric because of the anomaly you mentioned I mean of course if you go in the but the mirror is a different theory so you assume that your theory has a chiral multiplets which only chiral multiplets are charged under gaze matrices so in principle you could have twisted chiral multiplets charges under twisted vector multiplets in that case you can get anomaly for u and b I think you are right yes yes what I said is for the theory of vector and chiral you can write everything so if you have a theory with just vector and chiral you can just rewrite it in terms of twisted vector and twisted chiral you are right one last comment on this is that if you look at this solution for the scalars you see that one is not the complex conjugate of the other one because there is no minus we respect the two and so in particular this means that this background is breaking a reflection positivity now if we study some theory that flows in the infrared to a fixed point as we said yesterday we should not be worried too much because then in the infrared this feels capital redundant operator and we recover a fraction positivity but if you are studying a massive theory this has some consequences ok ok, any other question before we go on yes but essentially so if you do 0,2 essentially you are doing the twist so so that is essentially the only solution I think so there are no scalars in the graviton multiplet you just have this vector field for the say right moving asymmetry it's called half twist the reason is that the theory does not become topological when you do that twist it becomes essentially a chiral CFT but still it does depend on the complex structure it's not topological that is also true ok so now that we have understood these examples so now I would like to start discussing a localization computation so I would like to show one example in details so at this point you probably already saw one example in Wolfgang's lectures but so this will be another example but so the example that they want to study is precisely this one here of the untwisted background on the round S2 so let's actually take sorry, before you could I ask one question so this is curious in G greater than 1 we have no isometries and we can find a background that preserves two supercharges in gena 0 we have lots of isometries and you can find a background that preserves all supercharges on a torus for which we have a U1 times U1 isometries and you find a background that preserves all supercharges yes, I mean you don't have to turn on any background on a torus so this is the twist of that story so you can turn on so if you don't turn on anything what your partition function is essentially computing the width and index which is a number you can make the story and we will discuss this probably in the last or starting lecture or starting tomorrow I don't know you can make the story more interesting if you turn on some background for vector fields in vector multiplets or also for the r symmetry so then you really gets a function this is called the elliptic genus but this will break half of the supercharges ok, so we want to do localization for untwisted gauge theories s2 and this will be around s2 in fact this composition can be extended to a squashed s2 but the answer is not going to change so we will be happy just to do around s2 ok, moreover it turns out so we are in 2d2,2 and it turns out that there are many multiplets that one can use to construct theories in particular there are chiral and twisted chiral vector and twisted vector but I will restrict to a simple subclass of this theory so I will just consider vector multiplets and chiral multiplets and so in particular these are dimension and reduction of vector multiplets in four dimensions and chiral multiplets in four dimensions so ok, we already know these multiplets so let me just say very simple things so in the chiral multiplets just to maybe set up the notation so what fields do we have so we will not use superspace, we are just working components so in this multiplet there is a complex scalar there is a complex fermion two component complex fermion and then there is a complex auxiliary field that we call f and I will also use f for the field strength but the field strength will have two indices so hopefully there will not be confusion and and then there is also the anti chiral multiplet in which we have all the tilted fields and then there is a vector multiplet so in this vector multiplet we have a vector we have a complex scalar so I will use sigma and sigma tilde and then there is a complex Dirac fermion and then there is a real auxiliary field and of course all of these fields are in the adjoint representation of the gauge group while this will be in some generic representation and also it is useful to decompose this sigma into sigma 1 plus i sigma 2 and then sigma tilde is sigma 1 minus sorry, I am using a minus here so minus i sigma 2 is sigma 1 plus i sigma 2 however in Euclidean signature sigma 1 and sigma 2 are complex so these are not these are independent before we specify a contour and then it turns out that out of the vector multiplet one can construct so one can repackage these fields into a twisted chiral multiplet that I will call sigma and so the fields in this sigma are so the bottom components start with this scalar and then the various components which are containing here are essentially the gauge genie these details are not particularly important I don't want to waste too much time in details that will not matter okay you only want the details not important okay so when you do the splitting so you get some components of the gauge genome go here and then the auxiliary fields so in the twisted chiral multiplet not the vector field appears but the field strength and so one component will be something like this and it turns out that when you are on a curved manifold there is a mixing with sigma it is the other combination there is probably some type in my notes so the twisted chiral is a representation of the n equals to 2 algebra the twisted chiral is a representation of the n equals to 2 algebra so that means the vector is decomposable no, it contains the same fields these are two fields on the twisted chiral the twisted chiral are like two representations not on a day or day let's see let's see yes, that is true but ok, but it does not contain the vector it contains the field strength so if you really want to contain the vector you have to use the vector multivet and yeah, I'm not sure about this time because it looks like this is the same combination so I might have a type in my notes maybe there is not this minus sign ok so the important thing is what so now we want to construct gauge theories out of these multiplets so what is the data that enters in the finding constructing these gauge theories so essentially what is the data that we have to specify we can specify and we can play with ok, so first of all we can specify the gauge group because this will be some compact gauge group then we will have the matter in chiral multiplets so the matter will transform in some possibly reducible representation of g so we will specify a matter representation r there are many things that I call r hopefully this does not create too much confusion so this is a representation r of g and in other words the matter fields this phi take value on some vector space and this vector space can be decomposed into the weights of this representation so I will use often this notation rho in r these are the weights of the representation now as I promised before so the difference with respect to flat space is that now the r symmetry is part of the supersymmetry algebra and so we need to specify r charges because these r charges will control some couplings in the Lagrange once again we don't have to do this in flat space of course but we have to do it here so if you want the fact that su2 slash 1 contains vector like r symmetry implies that we have to specify r charges then we have interactions and interactions are controlled by two functions one is superpotential and the superpotential is allomorphic and gauge invariant function of the chiral multiplets and should also be invariant it should transform homogeniously and r charge in particular in my conventions have r charge 2 if you wish in other words you can first specify the superpotential and then the r but you have to make sure because we need this r symmetry so the superpotential should transform homogeniously under this r symmetry so if you want this is some allomorphic function of the chiral multiplets so this is exactly as in for dimension n equal 1 but moreover we have a twisted superpotential and this will also be allomorphic function but now of the twisted chiral multiplets now after we have introduced all interactions in general there will be some flavor symmetry which is left over we call it gf so this will be a global symmetry that commutes with the supercharges so by flavor I mean that we don't include r symmetry and then we can play this game that already Guido mentioned yesterday so we have this global symmetry so in particular in the theory there is a current for this global symmetry and the useful thing to do in general is to couple this global current to an external vector multiplet sorry to an external vector field a background vector field since the theory is supersymmetric this vector field is in a super multiplet and so we will have other bosonic fields in this vector multiplet and we can try to turn them on of course we have to do that in a supersymmetric way so Guido wrote this equation that the variation of the background gaugino let me call this lambda favor so this is a background fermionic fields that is in this background vector multiplet this variation we impose it to zero this impose some constraint but model of this we can turn on these extra parameters and what they do they control some couplings in the theory and in particular what they turn on in particular there is a scalar in the vector multiplet so this scalar here if we turn on this scalar in a background vector multiplet which is a vector multiplet for this flavor symmetry what it does and then you see the effect in the Lagrangian what it does is turn on some masses so these are called twisted masses so you can think of them as expectation values for these flavor fields ok so this is the data that we control in the class of gauge theories that we want to study so as you see there is quite some some data we can play with ok when you wrote delta lab of the flavor equals zero are you saying that's automatic because it's non-dynamical no no that has to be imposed if we want to preserve supersimule what kind of constraint does that lead to on the twisted mass separation how does it tell you how do certain things so let's see so in let's see I think in flat space it imposes that sigma and sigma tillers to commute so because sigma is complex it tells you that it's diagonalizable what's in the carton? yes and then you can reduce to the carton oh so that's why for a billion theories you take the twisted masses it at least makes them constant incredible it's also true it's also true for eight ways omega deformation but twisted mass can actually depend on the position but in a specific way I guess in arbitrary way and then you can satisfy the equation delta lambda equals zero by turning on feel strength so of course these equations depend on the setup that you have the manifold the background for the fields in the graviton multiplet in fact there is a I mean there is a hierarchy if you wish there is a hierarchy of equations so we are working off shell so so the variation of each multiplet should stay within should only involve the fields in that multiplet and this was as we see in the beginning this approach that made this approach very interesting this approach of Guido and Cyberg because we can just discuss we don't have to discuss the specific theory we just discuss the variation in the graviton multiplet however something that we do is that we go in best zoom in or gauge we do it for the vector multiplet but in fact we also do it for the graviton multiplet so because you go in best zoom in or gauge what happens is that in this hierarchy of multiplets start with the graviton, the vector and the chiral the variations of the multiplets are below in the hierarchy they do contain the fields that appear above that you would see if I do everything off shell this should not happen but this happens because we are in best zoom in or gauge and so somehow we can first discuss the equations in the graviton multiplet they are only depends on the fields in the graviton multiplet but then the equation in the vector multiplet includes the fields that we have set in the graviton multiplet and then if you go at the variation in the chiral multiplet or whatever other matter multiplet we have they includes the fields of the vector multiplet and the graviton multiplet so in this sense the equations that we get for that somehow depend on what we have chosen for the auxiliary for the bosonic fields in the graviton multiplet ok so ok, so now that we have discussed the class of theories that we have decided the class of theories that we want to discuss we should construct actions and as we said this action can be constructed with the methods that Guido is teaching us so we will only give you the result so what action can we write so first of all we can write young mills kinetic action so the kinetic action for gauge fields and so let me write it and then if you want the details are not important but I just would like to point out some features so this is the bosonic part that involves the bosonic fields and then this is the fermionic part and probably in front of everything we want a gauge coupling ok, so this is our bosonic action so what do we notice in this action so first of all as it should if we send the radius of the manifold to infinity sometimes disappear and this action reduces to the flat space action so there is the standard young mills term there are kinetic terms for the scalars there is this quartic coupling that just comes from the dimension reduction from four dimensions because when you reduce from four dimensions two of the components of the vector become scalar and so just from the young mills term you get this commutator but the square of this commutator and there is the d term and ok and this is also the standard thing however when we put the theory on the sphere we have some deformations which are controlled by the curvature of the manifold so these terms here ok and let me also write so let me also write the meta action so we can write similarly a kinetic term for the meta action so here there is a similar story so when we send the radius to infinity we get rid of some of the terms and once again this action is just the standard dimension reduction from four dimensions however now here another feature that I wanted to show you is that as you see so these small r are the r charges for the current multiple that we chose and as prom is the action now explicitly depends on the choice of r charges and again this comes from the fact that now the r symmetry is part of the supersymmetry algebra and of course this feature disappears when we go to flat space now if we want to perform localization we need to choose a contour and in fact it's convenient to choose a real contour so a real contour is just the obvious thing to choose so fields that were real in Lorentzian signature we imposed also so now they are complexified become complex in Euclidean where we imposed that they are actually real and if we had a complex field in Lorentzian signature when we go to Euclidean the field and this tilted partner now becomes independent and we just imposed that a complex conjugate to each other so it's the obvious contour to choose and in fact if we choose this contour these actions the bosonic part of these actions becomes well the real bosonic part of these actions becomes positive definite and the integral the part integral becomes convergent so this is a good contour then we want to choose some super charge q that we want to use for the localization so a convenient choice is to choose some epsilon and some epsilon tilde so we know that you have two choices here so we made one choice here and we define this particular delta q and we will use this as the super charge that we use in our formal argument yesterday in this case no because first of all the final result should not depend on it in general because what we are computing is just the original part integral so localization is just a way to computing it of course the result should not depend on how we compute it but you might expect that the form of the result the way in which it is expressed depends on the choice of the super charge some choices might lead to some expression and some other choice might lead to some other expression these two expressions has to be equal but they might be equal in a very non-trivial way now in this case this does not happen essentially because of the SO3 rotation on the sphere yes thanks so essentially by rotations you can map any choice to any other choice but in general you might I mean if there are some inequivalent choices you might you might expect different looking answers and we will discuss tomorrow an example which is not exactly this but something similar so we will discuss how changing the localizing term can affect the form of the answer anyway we make this choice and then there is a nice feature something nice that happens which is the following that it turns out that both the superium mills action and the matter action are q exact so they can be written as q of something and so this is very nice for two reasons so first of all this is telling us that in fact the party integral that we are going to compute will not depend on the coefficients in front of this action so in particular there will not be dependence on this gauge coupling and well this is nice because in two dimensions the gauge coupling is dimension full and so if you wish the gauge coupling is setting for us a scale above this scale we have the strongly coupled gauge theory below this scale we have essentially the the theory in which we have quotiented by the gauge transformations now the gauge feeding to dimension is non dynamical but the other fields are dynamical so if you want below this scale we are in a strongly coupled phase in which we should use a different description now the fact that there is no dependence on this parameter means that there is no dependence in this sort of scale that we can think of as a cutoff and so in particular our result is independent of the RG flow at least for what regards of this parameter well the other property we can interpret it as the fact that there is no dependence on wave function renormalization so again it gives us some independence on the RG scale and so in particular if you have a theory that is closed to fixed point our computation will be directly sensitive to the infrared CFT so essentially doing computations using the UV description we will be able to access information about the CFT ok and I think that will stop