 Yeah, go ahead. Hello, everyone. So we are going to continue today. So if you recall, yesterday we introduced the notion of a Riemannian manifold connections. And we talked about symmetric spaces. And I said that we are going to classify Riemannian manifolds. And part of the classification is to first separate the symmetric or really actually locally symmetric spaces. So once we do that, then we can classify everything by their holonomic group. So we first said, OK, we also have to separate things into reducible things and irreducible things. And we said that Riemannian manifold is reducible if every point has a neighborhood isometric to a product. And it's called irreducible if it is not locally reducible. And if you have a manifold which is irreducible, then the tangent space at any point is an irreducible representation of the holonomic group. All right. And then we had the definition of a symmetric space here. We said that a Riemannian manifold is called symmetric. If for any point there exists an isometry from M to itself, which is actually an involution. And P is an isolated fixed point for this involution. And then we had the definition of locally symmetric. We said it's locally symmetric if every point has an open neighborhood, which is isometric to an open subset of a symmetric space. And we have this nice theorem which allows us to characterize being locally symmetric in terms of the curvature of the Riemannian metric of the Levitivita connection, the curvature of the Levitivita connection. We said that the Riemannian manifold Mg is locally symmetric if and only if the curvature tensor is actually constant for the Levitivita connection. So that means nabla R is 0. OK. And I promise you that I would give you a theorem today about symmetric spaces. So OK. But before we do that, we also need to talk a little bit very briefly about geodesics and completeness. That is another thing that we're going to need. So oops, sorry, wrong color. All right. So first of all, what's a geodesic? A geodesic is a parametrized smooth curve. And I will denote the map giving us the curve as gamma from an open interval AB to the manifold M such that for any point t of the open interval AB, you have that the covariant derivative given by the Levitivita connection in the direction of the velocity vector of the curve of the velocity vector itself is 0. So I mean, intuitively, you want to think of this as the curve having basically 0 acceleration. But it's 0 acceleration, again, with respect to the Levitivita connection. So in some sense, in physics, you have a particle moving at constant speed along your manifold. And by constant speed, that constant speed is relative to the manifold. It's determined by the Levitivita connection. What do we mean by constant speed? OK. So this is a geodesic. And it is you can prove that geodesics are at least locally length minimizing, meaning that if you have a point, there exists an open neighborhood such that inside that open neighborhood, the geodesics are exactly the shortest curves for the Riemannian distance, for the distance that you get from the Riemannian metric. OK. And we have a nice theorem about the existence and uniqueness of geodesics for any point of the Riemannian manifold m and any tangent vector 2m at the point p, there exists a unique geodesic. On some open interval, we don't know exactly what that open interval is, but there exists an open interval on which the geodesic is well defined, such that the value of this geodesic at 0 is the point p, and the velocity vector at the point 0 is the vector v. So OK. So this is nice because it assures you the existence of the geodesic. And it tells you that as long as you know the initial condition, the geodesic is actually unique locally speaking. OK. And what's important to us here is this notion of completeness. So a manifold, a Riemannian manifold mg is complete if every geodesic can be defined on all of R, where R, we think of R is the real numbers. It contains our open interval on which the geodesic a priori is well defined. And what we want is for the geodesic to be extendable to all of R, right? OK. So anyway, so this notion of completeness means that basically once you have a geodesic, you can continue the geodesic to infinity, right? So that's a notion of completeness. And most of the manifolds that you're familiar with from algebraic geometry are complete. So and now we have the characterization of symmetric spaces that I had promised you. So that's our next theorem. What does it say? Suppose that our Riemannian manifold is connected and simply connected. Sorry, let me say that a little bit differently. Is a connected, simply connected symmetric space, then mg is actually complete. But that's not all. Actually, you use the completeness to prove the rest of it, actually. And if we put g, if we call g the following group, it's the group of compositions, sp composed with sq, contained in the group of isometries of the manifold m. Now remember that sp is the isometry, which is also an involution and has p as an isolated fixed point. So and the same for sq. So it's enough in some sense to just take competitions of two of these symmetries, the disinvolutions that have p and q as their isolated fixed points. And that's already a group. Normally you can take a bunch of isometries of the manifold and you can take the group that they generate inside the group of isometries. But here we're saying that just compositions of two of these is already a group. So then g is a connected lead group. If we choose now, let's choose a point and let h be the stabilizer of that point. Then h is a closed lease of group of g. And then you have an induced map. You can map g. You can first of all map g to m by sending an element g of g to the image of p by that element. But then because h is the stabilizer subgroup, naturally this map will factor through the stabilizer group of h. So I get an induced map like this is a diffeomorphism. So this pretty much tells you then, what are all the symmetric spaces? And what are all the locally symmetric spaces? OK. Any questions so far? And I think I should now maybe turn off my camera to avoid what happened yesterday when I couldn't connect to Zoom anymore from the iPad. Because I think Zoom gets a little bit overloaded. I think this is the time when I'm given a cue that I should turn off the camera. Sorry about that. All right. So now let me give you Duram's theorem about the decomposition of Riemannian manifolds. And then we will talk about the Burgess classification of polynomial groups. So what is Duram's theorem? So again, suppose mg is Riemannian and complete and simply connected. Then mg is isomorphic, sorry, not isomorphic, is isometric to a product as follows. So you have m0 times m1 times, et cetera, up to, for instance, mk, where m0 now is a Euclidean space. And m1 up to mk are irreducible. And OK, so this is the product decomposition of Riemannian manifold. So it's a Euclidean space times a bunch of irreducible things. And the decomposition is unique up to reordering m1 through mk. And the holonomic group, of course, is the product of the holonomic groups. Hollow of mg is the product of the holonomies. Holonomic groups of m1 through mk. Because when Euclidean space, Euclidean space is just the usual space you're used to, just Rn with the usual Euclidean metric. It has no curvature. And the holonomic group is just 0. So there's no holonomic group there. The only holonomic group that you're going to get through your manifold is the one that comes from the other factors m1 through mk. And it's the product of those guys. And then, and remember that we already know that the holonomic group, if you have a connected manifold, we already saw that the holonomic group only depends on the connected component of m that your point belongs to. So if your manifold is connected, then you have a well-defined holonomic group of the conjugation. So recall that when m is connected, then up to conjugation, we can just call it whole m of g, which is, by definition, whole x of g for any x is a well-defined subgroup, least subgroup of gl and r. OK? And I just need one more definition before I give you Berger's classification. The restricted holonomic group, g sub 0, is the connected component of 0 of the identity of whole of g, which is I will drop the subscript m whole m of g inside gl and r. So we're going to take connected manifolds from now on. So then the holonomy is well-defined up to conjugation as a subgroup of gl and r. And the restricted one is the connected component of the identity. OK. And now here's Berger's theorem. Suppose that mg is, well, Riemannian, of course, complete, connected, non-symmetric, irreducible. So if you allow reducible things, of course, I mean, I'm going to give you a list of groups. OK, so then the restricted holonomy group, whole g 0, is one of the following. OK, so I'm going to give you a list of these restricted holonomy groups. Now we assumed that, as I told you most of the time, the manifolds we're interested in are complete. And then if they're not connected, you can just take a connected component, right? If they are symmetric, if you have symmetries, I mean, we already saw that we can basically take any lig group and then mod it out by some least subgroup. And that gives you a complete Riemannian space, right? So basically, we already know what the symmetric ones are. So we can assume that it's non-symmetric. And if you want to assume reducible, what you're going to have to do to get all the holonomy group is just take the products of the list of holonomy groups that I'm going to give you now. So these are the building blocks. These holonomy groups are the building blocks of the holonomy groups of Riemannian manifolds, of the non-symmetric ones, OK? All right, so here's the list. Number one, this is, if you like, the generic possibility. It can be isomorphic to SON. So what is SON? Well, these are the, basically, inside, remember the holonomy group lives inside GLN of R, as a subgroup of the group of isometries of Rn. And so it's always contained inside the orthogonal group of size n, right? And here, taking, I mean, it's actually usually always contained in SON, the special orthogonal group, the determinant is one. So these are basically automorphisms of Rn, right? And this is the generic metric. So these metrics, so these are most of the metrics are like this, right? Most Riemannian manifolds are going to look like this. This is a generic one, right? Your holonomy group is rather big. Number two, now we're going to, the holonomy is going to be a little bit restricted. If the dimension is even, the dimension of your manifold M is even, right? Then another possibility is that this restricted holonomy group is equal to U of M. So here, and U of M sitting inside SON, right? You want to think of this one, these are automorphisms of Cn. So what's going on in this case, Cm, sorry, automorphisms of Cm. So what's going on in this case is that you have a complex structure option. So we will come back to this in a little bit more detail later, but this is what we would, this is the Kealuk case. Number three, the third possibility, again, in the case of even dimension, oh, sorry, it has to be greater than or equal to four here. This one also has to be greater than or equal to four. Whole G zero, this time is SUM. So now these are the automorphisms of Cm that preserve the determinant. So, and this is, but not the determinant of Rn, the determinant of Cm, which is a different kind of beast. And so this is the Kalabiyaal case. These are, again, automorphisms of Cm. These are Kalabiyaals and they're also, again, all Kehler. Okay. The fourth possibility, this time your dimension is a multiple of four. And actually, sorry, I need to add one more thing. It's Kehler, the Kalabiyaal case is Kehler, but it's also a Ricci flat. We introduced the Ricci tensor, right? But being Ricci flat means that the Ricci tensor is the, the homology class of the Ricci tensor is zero. Okay. So, okay, so that's also true in this case. So number four, your dimension now is a multiple of four and equals four R is, and it has to be at least eight now. Your holonomic group, oh no, no, sorry. I'm getting confused here a little bit. Oh no, this is, again, just where we are and equal to four. Yeah. So number equal to four and the holonode restricted holonome now is SPR. Now SPR is the group of automorphisms of the quaternions now, okay? But we, we again think of this as a subgroup of the, of our SON, right? And as I said, we want, these are automorphisms of the quaternions, HR. And these guys are the hypercalors. And they're also, again, Ricci-flat, meaning the Ricci form is, has comology class zero. And they are also of course, I mean, not of course, sorry, I shouldn't say that, but they're also caler. Then the fifth possibility, again, the dimension is a multiple of four, but this time it has to be at least eight. This is whole of, whole of G zero is equal to now SP of R times SP one. Okay. And this is again a subgroup of SON. And we can again think of these guys. These are, these guys are again, automorphisms of the quaternions, HR. Can I ask a stupid question? What do you mean by SPR times SP one? Okay. What do I mean by SPR times SP one? Well, as a, well, you embed both of them in SON and you take the product. Okay, okay. Yeah, yeah. So, Okay, so elements of the type. Yeah, it's the product group. So this is actually, yeah, this is actually bigger than the other type. Is that? Yeah. And it's, and these guys are actually not caler. Okay. Yeah. This, this is, yeah. This is, yeah, it's true. I don't know why it's listed in this order. Maybe because people are less interested in these ones. So, yeah. Right. Okay. So these guys are, yeah, quite what we call quaternionic caler. The terminology is also not, not very good. Quaternionic caler. And they are not caler actually. Okay. And then, let me see. And I can say a little bit more about this. These are also actually, these are also not even reaching flat. And they are what we call Einstein. I know I didn't define that. Sorry. Maybe I will get back to this one a little bit later and maybe explain a little bit more. Yeah. Sorry about that. Okay. So then number six, n equals seven, whole G zero is the exceptional group G two, inside S07 this time. Now this is the dimension, the dimension of the Riemannian manifold is now just seven. And these are the automorphisms of the octonians. Actually of the imaginary octonians, which is isomorphic to R7. And this is, of course, as I said, this is the exceptional group G two. And these are again, reaching flat. And then number seven, n is equal to eight, whole G zero is spin seven, inside now SO eight. This is a manifold of dimension eight. These are again, automorphisms of the octonians. Not the imaginary ones, just the octonians. And exceptional, reaching flat, okay. Yeah, sorry about this. I guess I'm not gonna get into a lot of these, these different cases, but it's a nice classification theorem. So I thought it would be nice to mention it. And then, okay, so is that okay? I don't know if my response was satisfactory there. There's something in the chat. Oh, oh, that's what you mean. Right. Whether they commute SPR times SC one. Oh, I see. You mean the product would not be, let me see, I would have to think about this. So SP one would be just the automorphisms of the quaternions themselves, right? So, and so you're kind of just multiplying by, yeah. Yeah. Okay, let me get back to this if you don't mind. I can get back to this tomorrow and we can discuss it a little bit more. All right. Okay, so, okay, so time-wise I've got another 15 minutes right now, Dwight, or? Yes. Okay, all right. Okay, so let me talk a little bit about the, so let me start by talking about the killer case. So killer manifold. So if you have a, so now we're going to go to the complex case. So now M is a complex manifold, okay? And we have multiplication by I, which defines map from an R linear map from the tangent bundle of M to itself and it satisfies I squared is minus the identity. Okay, so this is the complex structure, right? And a metric is called Hermitian if G of VW is equal to G of IV times IW for all vector fields, V and W. And we, if we have a Hermitian metric, then we have a one one form associated to it, associated to G, to the Riemannian metric and the complex structure. And that is omega, I'm going to define the one one form by its action on vector fields again. So in terms of its action on vector fields, you define it as G of I times V and W. For all V and W vector fields again. Okay. And the fact that it's a one one form means that omega is a one one form. What that means is that omega of IV IW is equal to omega of VW for all V and W. And so this is not so hard to check. So you can also check that omega is anti-symmetric. So then it's really a differential two form. It is really a differential two form, right? Which is of type one one. And it's also easy to check that any two of these determines the third. So that any two of the set I, G and omega determines the third. And so now we can say what Kailer means, right? So definition and proposition. The metric G is Kailer with respect to I if or sometimes I will just say that I is Kailer with respect to G. That's a little bit unorthodox in terms of terminology, but sometimes it's just faster if you talk about the complex structure being Kailer rather than having always to talk about the metric being Kailer anyway. So if one of the following equivalent conditions hold. So number one is that your differential one one from omega is a closed form. And the D here is the differentiation for differential forms. The other one is that when I apply the two levity Vita connection nabla to omega, I get zero. So omega is a constant covalently constant tensor, right? In terms of the levity Vita connection. And the third one is that the complex structure is constant for the levity Vita connection. So the proposition part is that these three conditions are equivalent. And the definition part is that if these guys hold, then we call the metric Kailer. Okay, so what do we get? So we get that G is Kailer if the omega, which we actually then call the Kailer form is a constant. So omega is then called the Kailer form of G. Okay, and what we saw about the Holognomy is that if a tensor is constant, if and only if it is invariant under the action of the Holognomy. So equivalently, the Holognomy group of G preserves I and omega. So now we have the generic Holognomy group as we saw is S1, right? So what, so we have now a subgroup, our Holognomy group, which is a subgroup of S1, right? And it preserves I and omega. So, and now we get the we get the unitary part. So if you have a subgroup S1, which preserves a complex structure, then that is basically by definition, that is U of N, where N is equal to 2M. We have a complex manifold. So the real dimension is two times something, right? So basically, I mean, I mean, this is almost a proof, maybe not 100% but close. So basically what we get here is that your manifold is Kailer if and only if the Holognomy the Holognomy group is contained, the restricted Holognomy is contained in U of M, right? So, so M is Kailer and only if Holognomy is contained in U of M, all right? And again, so remember now the Ricci curvature, which was what did we call it? I think I don't remember what I called it. Let me just call it Ricci, which was, it was an isomorphism. It was a map from TM to TM dual, right? But this is how we introduced it. So just kind of like the way that, so we had the Riemannian metric itself is also a map from TM to TM star, right? And so then the Ricci curvature is a map from TM to TM star. So when we had the Riemannian metric for the Riemannian metric G, right? We introduced this one one form associated to G, right? This omega here. And we can, we're gonna do something similar with the Ricci tensor. So we define the Ricci form, which is a differential form, right? So the Ricci curvature is not a differential form, but this guy here that we're gonna define is a Ricci, is a differential form. So row, again, we define it by its action on pairs of vector fields. It's going to be Ricci applied to IV and W for all vector fields, the V and W. Or we can also express it so equivalently row is this composition here. So row is goes from TM to, so we can put I here and then from TM to TM star, we put the Ricci tensor. And it's the same with omega. You put I there and then you put now the metric. So they're defined in a very similar way. Okay. So the, so this, we have something similar to what we had for omega again. The Ricci form row is a one-one form and it's comology class in H2 of MR is comology class of row equal to two pi C1 of the canonical sheaf of M, which is also two pi C1, the class of the cotangent bundle. Okay. Any questions so far? So let's say I have two minutes. Okay. Let me just say something about Ricci flatness and then I will stop. So Ricci flatness. So we say that M is Ricci flat if this comology class is zero. Okay. And so if this comology class is zero, sorry, not the comology class. I'm sorry, if row itself is zero. So if another thing that we can check is that row is the curvature tensor of the connection induced by the Levy-Chavita connection, KM which is the top exterior power of the cotangent bundle. That would be wedge M of T star. Okay. So what we get is that, so if row is zero, then KM is a flat bundle, right? And I'm going to go a little bit faster here. Sorry about that. But so then in this case, in this case, you need to do a little bit of work. M is what we call Krabi-Yau, KM is actually trivial and the holonomic group is actually contained, not just in U of M, but in SU of M, right? So it requires a little bit of work, but I think maybe I'm running a little bit short on time, so I will stop here. So we have a break and the picture right now, do we? Yes. Okay. Yes. All right, very good. So I'll stop here and continue after the break. Thank you. Thank you. I actually have no idea what we're supposed to do with the photo. I think the people of ICTP are supposed to do it. Yes. We can identify who they are. Yes. Hi, it's Paco. Hello. Hi. Hi. So you're ready for the photo? Yes. Yes. Okay. Well, everybody turn your camera on. All right. Okay. Everybody smile. I'm going to take a couple. Okay. We still have some cameras that are off. Do you want me to log out of the iPad? That's my camera that's off. Couple more. Everybody smile. Great. Okay. Let me just check that everything is okay. Yes. Great. Looks good. Thank you. Oh, it's done already? I can take a couple more just to make sure that everybody has a nice smile. So there's 35 of you, right? Okay. I think we're good. We took enough. Okay. Great. Thank you. We can maybe still have a minute or two. Sorry, where will the photo be published on the webpage of our scope? Yes. That's what they said. Valentina, that's what Victoria said. So I guess. We will send Victoria and then they will decide where to publish them. But you will let us know. Yes. Victoria will definitely let you know. And Elham, there is another question in the chat. So maybe. Oh, okay. Well, whenever you start again, if you want to take your break. Oh, okay. The definition of Richie Carbacha. Right. Okay. Yeah, sure. I'll do that when I start again. Well, maybe you can restart connecting your iPad and doing all the preparation.