 Užnjen sem zelo vsezaj, da je odličila prejz, in je bilo vsezaj dobro vsem in da bom prejzvalo to v pomečno vsezaj, ali je bilo vsezaj, da je bilo vsezaj, da je bilo vsezaj, prist je, ki so videli, super-symmetrično teori na prikazku In se videli, da je počkaj 4 superčajče, vse začnega vradi in vsega SC2. Tajzometrije s tajsvega, ali ne začnega vsega SC2. In sem različil toga komutacijna vradi, ali pa izgledaš, da se videli, da sem mislila začnega. Zvuk sem, da je toga začnega, ali je toga plus. In to da je nekaj uvaj, da je odpravljava s minusem, nekaj je tako dobro tebe, da je očeljava, nekaj je nekaj uvaj, ki jaz je nekaj uvaj, ali tudi tudi tudi nekaj reloveni nekaj. Čak je tudi tudi koment, kako je to vsak 2,1 x 1? In jasno, da je Brunov je zelo vse, da se je zelo vse operejtor P vzelo, kaj je hl plus r, danas pa je bilo srepljati komutator vzelo v r in v q, vzelo vzelo v p in q, ker h vzelo v q. Unbezvičaj, to je ta operator, ki pa je v vsečenju V2 s 1 superalgebra. Takoga vsečenja V2 s 1 je vsečenje V2 s 1, večenja V2 s 1 vrstvite v H. Vsečenja V2 s 1 vs 1 različje v ešelji. Vsečenja V2 s 1 je in NEkaj odypovali na superargi. Zdaj smo izgledali, če je tudi teori prezentacij, tudi tudi superalgebra. Zdaj je tudi dve pravne multipletje. Zdaj je jedno, kako je spina nekaj, in tudi je komprizena, in tudi spolj, kaj je kaj bila ampak na 2j plus 2 minus r. ... Nači načo, zelo je bilo, da je to, če je to, nazaj, zelo je, da je tako, čekaj je spolj, nekaj spolj, da je spolj, da je spolj, da je spolj, tukaj čas. Čas je to zdaj z tukaj srčanj Markovč, in kaj je jaz vzelo. Čas je, ki je, da je vzelo. Čas je vzelo. Čas je vzelo. Čas je vzelo, ki je vzelo. Čas je vzelo. Čas je vzelo. Čas je vzelo. Čas je vzelo. N technologies at level 2, another multiple with spin j, and r-charge r-minus 2. Tis is fine for j greater or equal to 1 1 3. When j is equal to zero, you lose the multiplet, which would have j minus 1 1 3. And you are left with this. Then we looked at what happens when the multiplets are short, so the short multiplets happen for h equals 2j plus 2r. So again, yesterday I had the sign of r wrong because I had the sign of that r they're wrong. And also there is another thing that I should say that here I am suppressing l. Zato sem zelo l-1, zato vzelo l v vseh, da je, da je. Zelo bomo dve štje vseh multiplets, v kajim je, da je jah, s rčakom r, taj zelo h, in taj dveh multipletov, z vseh jah plus 1.5. cuisine the J minus a half and the level two guy. And then there is very short multiplite which is the singleton which is comprised of something which has h equals minus r and j equals 0 and nothing else. So then, given the structure, you can study how this multiplite can, boiled the clock multiple. it's put into short multiplets and indeed let's look at what happens when this multiple is the saturation bound, then you can sprite it in a short Kir minded of this form. So you take the Level Zero one, so that would be j r and then h is going to be 2 minus r, OK? And so you want this kind of short multiplet, so you will have another j plus a half of 2j plus 2 minus r and r minus 1. But this only takes care of two parts of the long multiplet. You still have two multiplets to take care of. And those have to become another short multiplet, which should still be of this kind. So let's check that that is indeed the case. So now the top component of the short multiplet will be j minus a half with r charge r minus 1 and the same h, 2j plus 2 minus r. And the relation between the charges here should be the same as the relation between the charges over there. So in particular, indeed, you can write this as 2j minus a half plus 2 minus r minus 1, because this minus 1 cancels with this plus 1. So indeed, the relation of the charges is the same. And then you are left with j r minus 2 and 2j plus 2 minus r. So indeed, this long multiplet here splits into two short multiplets of this kind when j is greater or equal to a half. Then you have to figure out what happens to this long multiplet. And there is the same story, except that this long multiplet will split into a short multiplet of this kind and one singleton. So this you can check by yourself. And then you can try to write down the most general index that will count this multiplet up to recombination. And also here I have messed things up. So this index should be a sum over all the spins and all the h's in your theory of some constant alpha j that, in principle, can depend on h. But let me put the h-dependent state. So we have constant alpha j times the number of short multiplets of spin j and charge h. And then there would be, so by this, I mean the number of these short multiplets. And then I'll have sum over h. That's sum over j of this. Then I have some coefficient beta times the number of singletons with charge h. And now here I can sum over h. And I could put some fugacity for h. So again, as you see, in these recombinations, because h commutes with everything, the value of h for the multiplets doesn't change. So this is sum index. And then, as I was discussing last time, you can figure out what alpha j is in bet as f to b in order for this thing to be an index under those recombinations. And you find that alpha j is to be equal to minus alpha j plus 1 half, or j minus 1 half. And then that alpha half should be equal to minus beta. And therefore, you can write down what the answer should be. So what you get is the trace of minus 1 to the f times e to the minus beta h. And is it a multiplicity? Yeah, I'll see if there is still. I cannot write things straight. So this is indeed the form of some width and index. And actually, if we had been even more, well, we could have put even more labels, because the multiplets can also have spin under the other S2, the S2 right. So I could, in principle, add also fugacities for the S2 right. And if I have conserved flavor charges, then I can put even more fugacities here. As long as I have operators which commute with all of the supercharges, I can always add fugacities for them. And h over l. What do you mean it's not really h? So this one is the one that tells what is the BPS condition. No, that's h. So this is h, which is the operator which commutes with all of the q's. So it's not the one which sets the BPS condition. The operator which sets the BPS conditions, you could, for instance, select one particular supercharge, say q1. And then you could sum only over states which have q bar 1, q1 equal to 0. And that basically gives rise to this kind of condition. So but this h is not the operator that sets the BPS condition. It's just like an operator which commutes with all of the supercharges. OK, so we have this object. Now we can. Sir, I think you said it, but the condition of alpha j is so that minus alpha j is minus 1 half. Yeah, and alpha half equal minus. These conditions are in turn so that when two short, short moments can combine. Yes. But in two long, they should not contribute. Exactly. So we can indeed write down the more general index. Supposing that we have some other conserved charges, then we can write trace of minus 1 to the f. We can introduce two complex numbers, p and q. Then we can write p to the j3 left plus j3 right minus r over 2 plus 1 times q times j3 left minus j3 right minus r over 2 plus 1. And we can also insert some fugacity for some flavor charges that I have in my system. And I am supposed to take the trace over the illber space on s3. And you can see that if p is equal to q e to the minus beta over l, then this just becomes this guy over here. That's, well, you have to use the fact that h is equal to j plus 2 minus r. OK, so now we can interpret this index also as the partition function of our theory on s3 times s1, where the radius of s3 is l, and the radius of s1 is beta. So this should also be the partition function of our theory on s3 times s1, and this is radius l and radius beta. And from this discussion, we see that the answer can only depend on the ratio between the two radii. So in particular, it does not depend on the size of the manifold, but only on the ratio between the two radii. So p, q, and u. u is some other fugacity if you have some other flavor symmetries. But are they related to alpha j and beta? Yeah, beta, yes. So beta over r. So if you take p, so this is more general, because I've turned on also the fugacity for the j3 write, which is the other rotation. Is that expression more general than? Yeah, so this expression is more general than that, and it reduces to that expression if you take q equal 1, and then you take p equal q equals e to the minus b over l. But can you motivate that expression using the same logic? Yes, I said, because you have other charges that commute with all the q's, you can introduce fugacity for those. So you can choose a carton in the s2 write, and you can add it. So that will be p over q, and then you can introduce flavor charges. So I will have more to say about the interpretation of p and q later, but the interpretation of u, we can also give a different interpretation for u. Instead of saying, well, we can add fugacity into the trace, we can also say, well, we could add a background gauge field, which couples to the conserved current corresponding to the flavor symmetry, and we can put it along the s1. And then we can compute this partition function, where besides having the field, the background fields that we talked about last time turned on, we also have this other background gauge field, which wraps around the s1. And this is a background gauge field for the would-be conserved flavor symmetry. So this gives a geometrical interpretation for this u, and for the geometrical interpretation of p and q, you have to wait a little bit. But there is going to be a geometrical interpretation also for those. And another comment is that, again, these fugacities correspond to adding background gauge fields, and these background gauge fields can be complex. And indeed, this u can be some complex number. OK. Then an important fact is that you can see from here that the values of h of the states that do contribute at the index is fixed by their angular momentum and their r-charge. So h is fixed by l and r. So in particular, if I have some rg flow from the uv to the ir of some theory, and along the rg flow, I can conserve a u1r symmetry. So there is a u1r symmetry, which is conserved all along. Then the values of h of the states that contributes are fixed in terms of charges in angular momentum, so they don't change. And the index is invariant along the rg flow. It's another way of saying that indeed it does not depend on the size, but only on the ratio of l and beta. So you might worry about all sorts of bad things happening to this index, but you can actually check that, for instance, the spectrum of the theory on the cylinder provided that the r-charges of the fields are in certain bounds is discrete. So it actually is a pretty well-behaved object. Oh, sorry, this is j. They are fixed in terms of the spin and the r-charges. And so in particular, they do not depend on the scale. So as an example, we can just write down what the, so that, OK, so this in particular means that we can just compute this index in the UV where we can start from some asymptotically free theory. So we can actually do the computation for the index. It's a computation that you can done in free field theory. And then it tells you something about the IR. So that's an important comment. The other interesting fact is that you can actually check. So then you can ask, what does this index do, compute in the IR? So you can actually check that this s2 slash 1 times u1, et cetera, times s2r, is a subalgebra of the conformal algebra in 4D. So it's a subalgebra of what is it, su4 slash 1. And actually, you can do this for exercise. It's a maximal subalgebra, which does not involve on the right-hand side any actual conformal transformation of the cylinder. So if you take the n equal 1 super conformal algebra and interpret it as the super conformal algebra on the cylinder, then you can ask, what is the maximum subalgebra that on the right-hand side involves only isometries and not conformal isometries. And so this is the answer. So then, by doing a little bit more work, you can check that, indeed, this index would flow in the IR to what is called the superconformal index of the whatever superconformal filter you should leave in the deep IR in this RG flow. So let me give an example. So the examples are going to be a little bit trivial, but you can find more complicated ones in the literature. Say, the index will flow in the IR? No, the index is always the same, but in the IR, it equal. In the IR, it has the interpretation of these other index. It doesn't flow because it's actually invariant along the flow. So let's consider a chiral field of R charge R. So then you can compute this index. And what you find is the following. It's a product over j and k. Maybe I should put m and n, because they are not related to of 1 minus pq to the minus r over 2 times p to the m plus 1 q to the n plus 1 divided by 1 minus p times q to the r over 2 times p to the m q to the n. And here I've suppressed the possible fugacities in case you want to charge the chiral field under some other symmetry. But so that's the expression for the index. And now we can check that these statements are right, at least in some very simple case. So, for instance, let's suppose that the R charge of our chiral field is 1. Well, then that means that we could, in principle, add a mass term to the superpotential, because the superpotential will have R charge 2. So we can have mass. And then we would expect the field theory to just go to the empty to some, I mean, it's gapped in the IR. So let's check what happens if R is equal to 1. And indeed, if R is equal to 1, you can see that these guys will become m minus m plus 1 half. And also here you will get m plus 1 half from this. And so the numerator and the denominator are the same. And well, if you believe that you can cancel them one by one, I mean, it's an infinite product, but you have to regularize it in some way, then indeed the index is equal to 1, which means that you only have the vacuum. So that's correct. So then we can also consider another maybe stupid case. What's the ratio for m and n in 0 to infinity? Yes, so that you should say, m greater than 0. Yeah, they're greater than 0. I think it's equal to 0. OK, so now we can do another example, which is R equal 2. So but R equal 2 means that we can just add to the superpotential linear piece. So we can take the superpotential to be F times phi. And then supersymmetry is broken. And therefore, we would expect to find no supersymmetric vacua. And so let's check what the index is for R equal 2. But for R equal 2, you can see that for m and n equal 0, and R equal 2, in the numerator there is a piece, which is 1 minus 1, that's 0. And therefore the index is 0. So it works also in this case. So, well, naturally, this is not maybe too satisfying. But indeed, by using this, so you can compute, you can take some gauge theory of your choice. Compute the index in the DPUV. So suppose your gauge theory has some number of vector multiplets, has some number of chiral multiplets. Then you will have factors like this for each of the chiral multiplets. But there will also be fugasti corresponding to the gauge charges. And then in order to impose the gauge symmetry, we will have to integrate over these fugacities. There will also be some factors for the fields in the vector multiplets. But in total, the index will be some integral over the gauge field fugacities of some expression like these. It will be infinite products of p's and q's. And indeed, these integrals can be, there is some theory of how to handle them. And you could check, for instance, that if you have two theories which are IR dual, they indeed have the same index. So that's a very non-trigger check. It's much, much more non-trivial than this very simple examples. So I just want to give you some reference. So the original paper on this index was the ones by Romersberger. So 0,5, 1, 0, 0, 6, 0, and 0, 7, 0, 7, 0.3, 7, 0, 2. So this is actually, the second one is actually the more readable paper. Any question? This expression looks like a particular case of some of these five-dimensional partition functions, the p and q being the omega deformation parameter. Is there a reason for that? So you would like the p and q to be the omega deformation parameter. So p and q are kind of two? Yeah, yeah, yeah. Are they the sphere? Well, yeah. Well, yes, yes. They do, I mean at least one certainly can be interpreted some rotationally. They do correspond to j left and j right, yes, with some combinations. But OK, I don't know of actual realization in 5D where this become the. The party has more privileges. So there is, for example, the gases for the internal particles that looks like if you specify. Yeah, no, yeah. It might be that there is some background which reduces to this. Yes, I don't know. That's an interesting. Is there actually a sphere inside R4? Inside R4, yeah. Yeah, so there should be some background which reduces to this. Maybe some kind of boundary. That's a very interesting comment. But I don't know of any precise relation. OK, so I'll just make some extra comment on what, well, in the spirit of this last comment by Nikita, we can think about using this background that we have on the S3 times S1, also to say something about theories in 3D. So in particular, if you have an n equal 1 theory in 4D with a u1r symmetry and reduce down to 3D, then you will get an n equal 2 3D theory with u1r symmetry. And so we can think about, well, we have this background with all these isometers. Maybe we can reduce it and get some interesting things in 3D. So one thing we can do is to take our S3. And the S3 is a vibration over S2. So we can reduce along the off fiber. Because nothing depends on S2, right, then we can take the off fiber aligned with S2, right, and reduce along that. All the supercharges are going to stay. And we are going to get a background, which is the 3D, the so-called 3D index. So that would be S2 times S1. So this is S3 times S1, S2 times S1. So it's usually called the index. There is another way of preserving supersymmetry on S2 times S1, so that correspond to doing a twist on the S2 with u1r symmetry. So that's a different one. And then we can also reduce just along. We can just place our theory on S3 times S1. And then we can reduce along this one and get a theory on S3. And, again, both these theories will have four supercharges. And so the four supercharges for the 3D theory on S3 are going to have the same properties of these. They are both going to transform only under S2 left and be inert under S2 right, which means that you can actually squash the S3 in various ways. And people have studied various possible squashings, which preserve even four or less of the supercharges. Can you take a linear combination of S1 and S3? Yeah, so what you can do is that you can glue the cylinder after a rotation. And then you can reduce on this, if you wanted to be reducing along some twisted thing, as you were saying. And then what you get instead of S3 is the squashed version. No, not rational. I mean, you can just decide to take this S3 here and identify it with the S3 here after a rotation of an arbitrary angle, and then you get an arbitrary squashing. You don't get the most general squashing in this way, but you get a good sum. So all the squashing will, in particular, preserve four supercharges, because they only involve the S2 right. OK, so there is still one extra comment on this, which is the interpretation in 3D of the background fields for the flavor symmetries. So as we said, in here we can add background gauge fields, which couple, let's call them, I don't know, little a mu flavor, which couple to the conserved flavor symmetries of your theory. And these can be in principle complex, so they have a real and imaginary part. So you can try to interpret what this real and imaginary part will be on S3. And, well, the real part of a mu is just something which, so you can imagine, it just shifts the arc current by something proportional to the flavor current. So when you write the theory on S3, there will be various couplings to the arc current, in the same way as we add couplings to the arc current in 4D, and the coefficients of these couplings can be shifted by turning on this flavor fugacities, and imaginary ones become real masses on the S3. So on the S3 you have the possibility of turning on real masses, and those are encoded in this dimensional reduction in the imaginary part of the flavor fugacities. OK, so I think this exhausts what I can say about this index, at least for now. So I will jump to a somewhat different subject unless there are questions. I was told that I'm pretty bad with these blackboards, so I should try to be better. Can you say something about generalization before the n equals 2? Yes, I can say something, but it's not going to. OK, this is a complicated topic. But let me just make some remarks. So first of all, it's obvious that if you take an n equal to theory, it's also n equal 1, and it has a u1 r symmetry. You can just take the Cartanian SU2, so you can put it on this background. Now, you would think that this would break the SU2 r symmetry to u1. But actually, if you do things carefully, you discover that that is not the case. So the corresponding background in n equal 2 will preserve eight supercharges. And unfortunately, it's not particularly useful. So one thing which makes this background useful is that if you look at the Lagrangian that I brought down, for instance, for the chiral field, it depends on the r charges of the fields. And in particular, if the r charges are between 0 and 2, then there is an actual potential for the chiral fields, which is some kind of master, which goes down like 1 over r squared. So when the manifold is very small, there is a large trap for the fields. So they are trapped near the origin in field space. So that makes the computation resilient to all sort of, there are no flat directions. But if you do the similar computation in n equal 2, then when you reduce from n equal 2 to n equal 1, the r symmetry of the resulting field turns out that there is some field of r charge 0. And then this field has a flat direction. And it makes, using this background, much less obvious. So that's one comment. There are many more comments that I can make about n equal 2, but we will leave it for later because it will take a long time. OK, so if there are no other questions on this, I will skip to a somewhat different topic. So I said at various points that this idea of coupling to background supergravity was useful because it would allow us to classify what are the possible manifolds on which we can place a certain theory preserving some amount of supersymmetry. So in particular, let's see how this works for the case of n equal 1 theories with a u1r symmetry, which as we said, coupled to a new minimal supergravity. So the generalized Killing-Spinov equation, in this case, we brought it last time, but I'll rewrite it again as the following form. And then there is another equivalent equation for the zeta-tildes. But because this equation only involves zetas and the other equation only involves zeta-tildes, we can just forget about the other one for now and look for what manifolds which have one solution, backgrounds which allow for one solution to this equation. OK, so some comments. Again, so we are working in Euclidean space. So that means that if I take a spinor zeta, alpha, and a complex conjugate, it's components. I don't get the components of zeta-tilde, but I get the components of another spinner. So let's call it zeta-dagger, which also transforms as a left-handed one, but with an upper index. So this is, if you want, by definition of zeta-dagger. And with this definition, then you can check that zeta-dagger zeta is just equal to the absolute value of zeta-1 squared plus the absolute value of zeta-2 squared. So one thing you have a little bit careful about is that if you take the complex conjugate of zeta with upper indices, this is going to give minus zeta-dagger with lower indices. So you can check this. OK, so this is one comment. So the other comment is that, as we said last time, this spinor zeta is charged under the u1r symmetry. So that it actually lives into the unitary line bundle corresponding to the u1r symmetry times s2 plus, this left-handed spin bundle. So for a generic manifold, the spin bundle does not exist, but any orientable manifold in 4D is spinC. And in spinC, this unitary line bundle times s2 can be well-defined. So in principle, we can work on any orientable manifold, but there will be constraints on the kind of theory we can put in. We will comment on that later. So then there is another somewhat trivial point that one can make, is that, well, this equation has at most two non-trivial solutions. So how do we see that? Well, the first thing that it's clear is that this is like homogeneous in zeta, and it has first derivatives only. So if zeta is equal to 0 at some point, then the derivative of zeta is going to be 0. And then that means that zeta is equal to 0 everywhere, well, at least in some open set. So that means that if you have more than two solutions, you can always take linear combinations of them, which are going to be 0. And then that means that they are trivial. So in particular, this same point tells you that if there is a non-trivial solution, then zeta must be different than 0 everywhere. And as we'll see, just this fact has some very important consequences. So it's up to and out, so it's unique. So what you just said means that if it exists, it's unique up to and out. The solution? There can be two solutions. Yeah, because the? What was the argument that you take linear combinations in zeta at some point, when it vanishes everywhere? No, but you have two components of zeta. If they don't even align by moment. OK, so. I suspect it's useful if you say what spin C structure is. OK, this is going to take me a while. Let's say if people are interested, I can comment on that separately, because I won't be finishing this. So let's see. So now, OK, I'm going to mess with the blackboards, maybe. Let's see. I can put this up, and I can write on this one. OK, so now we can, in order to continue to study this equation, it is useful to just make some statements, which do not even use them. They just use the fact that the solution is nowhere vanishing. So for instance, we can construct the following object, for which I've chosen an appropriate name. By using, suppose we have a solution, then we can build up this object. So let's make some general comments. So first of all, you can notice that there is zeta at the numerator, but there is also zeta at the denominator. There is a zeta dagger at the numerator and the zeta dagger at the denominator. So that means that this is actually uncharged under the complexified U1r. So it is a good tensor on the manifold. And it is a good tensor because zeta is nowhere vanishing, so I can divide by the norm squared. This object has no zeros. So this is a good tensor. And we can try to figure out some properties of this tensor by just using Fierce identities. So you are welcome to check these in your spare time. So you can multiply two of these. And what you find is minus the identity. And you can also check that if I take g mu nu, that's basic nu lambda. What did I write here? Yeah, there in. Oh, yeah, yeah, I have some. OK, so what does this mean? So it means that this thing, j, is an almost complex structure. This is more or less the definition of an almost complex structure. And also that the metric is compatible with this almost complex structure. So fine. So from this we can already draw an interesting conclusion, which is if you have a solution, then there is an almost complex structure. Therefore, for instance, we can exclude that we're ever able to find the force sphere as a solution to our equation. Because the force sphere does not allow for an almost complex structure. So that's out. So as we saw, we can realize the force sphere, but for theories which couple to old minimo supergravity and not to nu minimo supergravity. OK, so now we have an almost complex structure. It's natural to ask, does this object actually a complex structure or not? So for that, you have to check that this tensor with a name that they cannot pronounce. Can I answer something? It's confusing. When we did with Volcker yesterday, the other day before, the S4 localization for n equal to 2, we had the killing spinors that they can be 0, the north pole, all the south pole. Yes, let me answer that question. So the n equal to killing spinor equations or the n equal 1 killing spinor equations in old minimo, they are different in such that they also have zeta tilde on the other side. So that means that it is possible to find solutions where either zeta or zeta tilde vanishes at some point. Not, however, both zeta and zeta tilde. Because the same comments goes through that if both zeta and zeta tilde are equal to 0, then you will find that the solution is everywhere 0. So, indeed, for instance, on Peston's theory on S4, like the left-ended part of the spinor, which gives the supercharge vanishes at the north pole and the right-ended vanishes at the other pole or vice versa. So as I said, it's natural when you have an almost complex structure to check if it is actually complex. So for this, you can do two things. Either you can check that this tensor, the noise tensor vanishes. So I can write down what this is. And hopefully I get the indices right. So I'm writing this just so that you see that it just involves j and the derivatives of j. So if this is a complex structure, then this object better be 0. So then you know what j is. You know what equations zeta satisfies. You can find what equations zeta dagger satisfies by taking the complex conjugate. Then you just can plug in and, well, and you can compute. Or maybe if you're good with some computer, you can have the computer compute for you and you find that it is indeed 0. So this is a pretty horrible computation. But there is actually a simpler computation that maybe you can try to do, which is the following. So you can check that if you have an holomorphic vector, then x mu sigma tilde mu zeta is equal to 0. And then an equivalent characterization of a complex structure is something for which the commutator of two holomorphic vector is also called holomorphic. So if you take x mu and y mu holomorphic, which means with only such that 1 minus ij on them is 0, then the commutator should also be holomorphic. And by using that an holomorphic vector satisfies this condition, it's actually much easier to check this. You just take a derivative of this and then you dot it with another holomorphic vector and then you just antisymmetrize some indices and you get the result pretty easily. So as a conclusion, we have that if there is a supercharge, so zeta supercharging satisfying those equations implies that the manifold is complex and that the metric is compatible with the complex structure, which means that if we choose appropriate coordinates, we can write it as ds squared is going to be g i i bar d zi dz bar i bar. So in particular, the metric does not have any g ij or g i bar j bar components. So are there any questions so far? So is a statement that if there is a solution? So if there is a solution to those equations, then from that solution you can build a complex structure, therefore your manifold is complex, and the metric is a mission. So just a characterization of a complex manifold is that I can cover it in patches and on each patch I can choose complex coordinates zi's, in which the metric, for instance, will get that form. And in going from one patch to another, so this is the important part, the transition functions between one patch and another are allomorphic functions of the coordinates. So the new coordinates zi prime are allomorphic functions of the disease. So you can cover your manifold in patches which look like situ, and then in going from patch to patch, you just change the coordinates with allomorphic functions. OK, so this is showing that if there is a solution, then the manifold is complex. Now we can try to go in reverse and show that if the manifold is complex, then we have a solution, OK? So that is what we will try to do in the following. So now if the manifold is complex, and you are using species structure, then your zetas will become, now, what, 1-0 forms? Or 0-0-0-2? Well, the zetas will become a scalar. The supercharge will correspond to a scalar. Will be scalar? Yes, and so with the supercharge I can also build an allomorphic two-form. Give me a second. Can you write back the most complex in the module? Yes, then one can twist various, I mean, are you talking about trying to write down some twisting? You always have to do the same thing. If you can write the larger module. Yes, yes, that can be done. So that's one thing I could try to show tomorrow, but I don't know, depending on what people are interested in hearing about. OK, so... Let's see if I wanted to say something else. I think so. Oh, OK, one thing which is important is that you notice that the complex structure that I have written down. Is self-dual, just because sigma mu nu is self-dual. So that will be somewhat important. That's just a manifestation that it can be built out of zeta, which is a left-handed spinor. If instead you add a solution to the other, killing spinor equation, the one for the right-handed spinors, then again you would be able to write down a complex structure, maybe anti-self-dual. But the same comments will go through. OK, so let me make a comment first about... Let's look at this equation that I wrote there. And let's just set v to zero. So then it just looks like d mu minus i mu of z equals zero. So you might want to ask, when can I solve such equation? And, well, the answer is that you basically want the olonomy of the Levy-Civita connection. Well, so the olonomy of the Levy-Civita connection in general would be in s2 plus, I mean s2 plus, then s2 minus. But you would like the s2 plus part, the one under which the zeta transforms, to just be u1. So if that's true, then you can twist away the... So if the olonomy of the Levy-Civita connection is u1 times s2 minus instead of s2 plus times s2 minus, then you would be able to twist away the u1 part using the u1r symmetry. And, indeed, this can be done on scalar manifolds. So this happens on scalar manifolds. Again, you want your complex structure to be self-dual so that the u1 will be in the right place. If it's anti-self-dual, then you will do it with the zeta tilde. OK, so that's a comment, but it gives in spirit what you would want to accomplish. So this can be accomplished on scalar manifolds, but we would like to show that if we have the extra freedom of adding this other background field v, then we can do this on any complex manifold. Does it really imply scalar or just permission of the complex? Well, I think it's true for scalar. I'm not completely certain, yes. I'm not completely sure about the reverse. OK, so in order to proceed, OK, so the one thing that we can look at are covariant derivatives of the complex structure. So actually I think yes because of what I'm going to say. So it's a fact that the covariant and if the manifold is complex, then the covariant derivatives of the complex structure, they are not all independent, but they are actually all encoded into the divergence. OK, so in order to check this, you have to use the horrible tensor being zero. OK, so then if this object is zero, then the manifold is scalar and then you can just set v equals zero and use this trick to solve your Killing-Spinov equation. If this object is non-zero, what we want to show is that we can make use of v to accomplish the same. OK, so sorry. Yeah, yeah, this is the divergence. Maybe I should write it more carefully. So the statement is that if the manifold is complex, then all the derivatives of the complex structure are encoded into the divergence. That's a statement that in order to check it, you have to use the fact that j is an actual complex structure and not the almost complex structure. OK, so let's compute this object by using the Killing-Spinov equation. So you have to take a derivative of j and then you have to plug in the equation. And what you get is that this is v mu plus v bar mu plus i... No, these are mu minus v bar mu times j mu nu. OK, so it's not zero because the fact that it's non-zero is encoded in v. In particular, you can invert this relation and you can write v in terms of the divergence of the complex structure. But this does not completely determine v. There is still something that you can add, u mu. And remember that v has to be conserved. It's an auxiliary field in the supergravity multiplet which is a conserved vector. So that means that... OK, so this piece works. It's conserved. Then this piece also needs to be separately conserved. So grad mu... So the divergence of v mu is zero, of u mu is zero. And then in order for this to stall that, you also need to impose that the antiholomorphic components of u are zero. So this encodes some freedom of choosing v, which is not completely determined by the complex structure of your manifold. So you choose a manifold. It will have some complex structure. Then this complex structure is gonna... If it's... If it's a scalar, it's gonna tell you what v mu has to be up to changing the antiholomorphic components by a conserved part. OK. Now, to conclude, I have to use another fact about complex manifold, which is... So as you see here, the covariant derivative of the complex structure is not zero. It's encoded in v. But if you change... If instead of working with the Levy-Civita connection, you choose a different connection, you can choose a connection such that it is compatible with the metric. So I'm gonna claim that there exists a connection that is compatible with the metric, and it's also compatible with the complex structure. So... And J mu nu. So this connection goes... Well, actually, there is more than one. There is an entire one parameter family of such connections. But we are gonna choose one in particular, which is called the churn connection. So the definition of the churn connection is that... Like, see, we've write down the spin connection omega mu nu rho for the churn connection. That is gonna be the Levy-Civita one, the one that you all know about. But then you have to subtract a piece, which gives the contortion tensor. And the contortion tension is determined by the complex structure in the following way. Where this is just the Levy-Civita spin connection. So the claim is that if you use this spin connection, then it is compatible what with the metric and with the complex structure. So it makes sense to use this one instead of the churn connection. And why is that? Well, it's because this connection is compatible with J mu nu. And this solonomy is gonna be in u1 times s2. So what did I call it? Plus. So that's because if you take... When the menu for this complex, you can take frames which are holomorphic, and then, like, in going from patch to patch, you can just... You can... Two holomorphic frames are gonna be related by a u1 times s2 transformation. And because the connection is compatible with the complex structure, I can... An holomorphic frame is gonna stay holomorphic. OK. So now I can try to just take my equation over there and rewrite it in terms of the churn connection instead of the Levy-Civita one. So let's do it here. Let's see what this becomes. So we have to make some redefinition. We have to use, first of all, that V is now known in terms of the complex structure plus a piece which is undetermined. And then we have to use the fact that we want to change the connection to be the churn one. So after you do all this, you get the following. Where A mu C is just the old u1R gauge field, plus a well-defined one form which is some function of the complex structure. OK. So this means that you're just shifting the u1R connection by a well-defined one form. So that's legit. And then you can rewrite that equation in this way. And now it's clear because the claim is that the u1R connection is in u1 times s2 plus but we are just looking at left-handed spinors. So I can just twist away the u1R connection by using the u1R connection. What after the u? The ambiguous part drops from the equation. OK. So we can be a little bit more explicit and how much time do I have left? Like 15 minutes. OK. So in order to be a little bit more explicit, let's introduce another object. We call it p mu nu. And that's just going to be z sigma mu nu times z. OK. So this object, you can check in these two indices, it is holomorphic. But then it also has r-charge 2 because it has two spinors. So this guy lives into the square of the u1R line bundle times the bundle of holomorphic two forms which is also called the canonical line bundle. So you can check that indeed this thing is a line bundle in the same way as you check that the determinant line bundle is a line bundle. OK. But this guy is different than zero everywhere. Therefore, this tells you that this object is trivial. And therefore, we can identify L with minus a half of the canonical line bundle. So this does not make it... So then L is not going to be well defined, but L times k to the half will be. And that's basically what you use to construct the spinner. So very explicitly we can take some coordinate patch. So in some coordinate patch we can take the component 1,2 of p and we can call this little p. Then we can define another object s which is little p times g to the minus a quarter where g is the determinant of the metric. So we can construct this object. And so s is clearly different than zero and it has R charge 2. And now you can ask how does s change under changes of coordinates which are allomorphic. So now you do an allomorphic coordinate change. And what you discover is that the new s in the new coordinates is equal to the old s in the old coordinates times a phase. And this phase is the determinant of dZ prime over dZ divided by the determinant of dZ bar prime over dZ bar to the power a half. So it's important that this thing is a phase. Now s has R charge 2. So that means that in going from patch to patch in doing an allomorphic coordinate change you can undo it by doing a U1R transformation. So this can be undone with U1R. And therefore under allomorphic coordinate changes followed by appropriate U1R transformations this s transforms like a scalar. Yes, indeed. So under coordinate changes plus U1R s is a scalar. And now we can just use s to construct our spinner as Nikita was saying. OK, so let's see. Put this up here. So again this is going to be very... I mean for people who are not interested in the explicit part you can just forget about it. The global existence of the spinner is already proved but you can actually write down expressions in each patch. So to write expression in this patch you have to make some choices. So you have to introduce some frame. So we choose some allomorphic frame which would be the following one. 2 square root of g11 bar dZ1 plus square root of 2 over g11 bar d21 bar dZ2 and dZ2 is going to be equal to square root of 2 over g11 bar g to the 1 quarter times dZ2. So then you can check that the metric is just d1 e1 bar plus e2 bar and that if you do... if you change... OK, that's... I think this is enough. And then you can just solve in this frame for the killing spinner equation when written in terms of the... of the churn connection, which I think is just erased. And what you get is that the solution is dZ equals square root of s over 2 times 1, 0 and you can also get what the components of the churn... of the u1r connection are. This is going to be minus i over 8 di of log g minus i over 2 di of log s and ai bar c is i over 8 di bar log g minus i over 2 di bar log s. OK, so in summary, like if you have a complex manifold, you can always find one supercharge at least on it and these supercharges correspond to... well, it's a complex manifold with a self-dual complex structure. Then, like you can find the left-ended spinner which solves the new minima equations and therefore you get at least one supercharge and these supercharges are charge 1 and it will square to 0. i over 2. Yeah, sorry. That's my horrible. OK, so that... Are there any questions on this? So in some sense this is just... once you realize that you can change connection so the churn connection and with that change the equation looks exactly like the one-keller manifolds. It's the same story. It's not very different. OK, so there are some more comments I can make about what if we want manifolds which allow more than one just supercharge. We could try to find manifolds that have two or more and now there are different choices. So you can ask for two independent zetas so two independent supercharges with the same r-charge. So I'm not gonna say much about this case here because it's complicated but also I don't know the answer completely. So what we found is that in the case the manifold is compact then this is a very restrictive requirement and in the case the manifold is non-compact there are some differential equations for the metric that you have to satisfy and it's not clear what the most general solution is. But another case which is interesting is to ask for so this is for two zetas for one z and one z tilde so one solution to one supercharge with r-charge one and one supercharge with r-charge minus one. Well then it's clear that your manifold will have a complex structure j mu nu another complex structure j tilde mu nu which is gonna be anti-self-dual instead of self-dual but as it turns out this is not the best way to describe this manifold it's much easier to work with bilinear of z and z tilde so you can just build out z sigma mu z tilde and as we said in the last lecture this is a killing vector and you can check various things about this killing vector so first of all you can check that k is oromorphic with respect to j and j tilde and you can also check that if you take k and you square it you get zero it's a complex killing vector so that's not unreasonable and finally you also find out that it's norm well ok, that's not too important it's norm is non-zero because both of the z and z tilde are everywhere non-zero different than zero and then there are actually expressions for j mu nu and j tilde mu nu in terms of k so j mu nu can be written as some object q plus a half of epsilon mu nu rho lambda q rho lambda where q is i over the absolute value of k squared times k mu k bar nu minus k mu k bar nu so now there are like two different cases that sorry so this is mu maybe I should just put mn nm is it better? jaj jaj tjeda will be with the opposite sign one is self-dual and the other one is anti-self-dual ok, so now there are two cases case one k commuted with k bar is different than zero ok, so then you have to do some work and you discover that this implies that the manifold is as three times as one or as three times r ok, we already know about that case number two is the case when they commute so this is more or less intuitive, because if it's non-zero then you start getting more than two killing vectors and they have to form an algebra there are not many options ok, so in this case if they are zero they commute and then you can actually write down an expression for the metric so what you discover is that your manifold, so these two vector fields are gonna generate a torus action on your manifold and the torus never shrinks to zero side because the normal of the killing vector is always never zero and then you can write down what the metric of the manifold will be, there will be some conformal factor dependent on some complex coordinate z which is never zero and then we will have some coordinate dw plus h zz bar dz times dw bar plus h bar dz bar dz bar plus some function c dz bar dz dz bar and this omega square is equal to q the norm of k squared so in this case I think k is just dw so I think I should stop but I'm gonna say a little bit more about this next time h and h bar and c are arbitrary functions so this tells you how the torus is fibred over the base and this is the metric over the base so in particular next time we'll see that as three times this one is indeed of this form any other question I think everybody is sweltering