Okay, so I want to talk about this vanishing theorem that means that, you know, now we've discussed all these 1, 1, 1, 1, 1 components, nested Hilbert schemes. I want to talk about this vanishing theorem which says nothing else contributes in prime rank. So Gail already talked about this. I'm going to do it slightly different way, but it's more or less the same. So I'm going to exhibit a vector field on the modular space. And I'm going to use this duality that has kept coming up in the course. On Calabi, our three-folds, obstructions are dual to deformations. At any point, the obstruction space is the cotangent space. And the same is true on this SUR Vaffawitton space with its perp obstruction theory, that the obstructions are dual to tangents. And really it's tensed with a, this isn't entirely invariant under the C star action, it's equivariant. It's tensed with a one-dimensional character of the weight-1 character of the C star action. But I won't get into that. It's important, but I won't get into it. So a vector field, what it's going to give me is a map from the obstruction space to the structure sheath. And this is what's called a co-section. And I'm not sure why I said it's a surjection, but in many cases it's a surjection. Okay, so when it is surjective, this forces the virtual cycle to vanish. So we've seen that the virtual cycle you could think of roughly as the Euler class of the obstruction bundle. So if the obstruction bundle has a section, a nowhere vanishing section, or its dual has a nowhere vanishing section, you would expect this Euler class to be zero. And that is what turns out to be true. Okay, so if you have a perfect obstruction theory, so the obstruction, you know, this is sort of... In the fantasy situation, this would be the... It's the dual of the obstruction theory, which is what I'm using here. So the virtual tangent bundle would be the thing sort of tangent bundle to the ambient space, mapping to the vector bundle cutting out, the vector bundle whose section cuts out the modular space. Okay, the co-kernel of that map is what's called the obstruction space. So that's the general definition. Okay, so a co-section, a map from the obstruction space to the structure sheaf, so in our case a vector field by the duality. What that gives you is it gives you a map from E1 by composition, right? E1 goes to the co-kernel, that's the obstruction space, goes to the structure sheaf. So we get a map from E1 to the structure sheaf. And when that's onto, it means the virtual cycle vanishes. And I'm going to discuss that in a second, and why that's true. And what Keeman-Lee do is they show that actually the cone, if you remember, the graph made vertical of the section lies in the kernel of this co-section. Okay, so just the existence of a co-section forces the cone to lie in its kernel. And that's more or less sort of clear. If you remember, because the obstruction space is the co-kernel of the derivative of the function here, what that means is that to first, to leading order as you move off the modular space, the section is taking values not in E1 but in the kernel of E1 goes to ob. So ob is the bit that the section is essentially not seeing to first order as you move off the modular space. And so the cone lies in the kernel of the map from E1 goes to ob. And so this is why you get this thing here. So this is just saying that to leading order, to first order as you move off the modular space, where do the equations lie? They lie in the kernel of this co-section. And in fact they lie in the kernel of the map from E1 to the obstruction chief. And so exercise, we did a similar exercise to this before, or I set a similar exercise to this before where you had a section lying in a sub-bundle of a bundle and it was transverse to the zero section of the sub-bundle. And then you found that the obstruction bundle was the quotient of those two bundles and the virtual cycle was the Euler class of that. And this is just that exercise repeated a bit more abstractly and it uses the fact that the Euler class of the structure sheaf is zero. Okay, so in the class maybe they're two to count. So it does mean that the virtual cycle vanishes. All right, so that's the background. I'm going to discuss co-section localization briefly again later. So more generally when the co-section has zeros, or it isn't onto, then you can instead localize to its zeros. So that's something more general. But here we're going to produce a vector field which is nowhere zero. So this co-section is subjective and therefore the virtual cycle vanishes. Sorry, why are you happy when you say that the virtual cycle vanishes? Because it means that a whole bunch of components don't contribute to the variance. So we've in some sense computed the contribution of these nested Hilbert schemes but there's other components with profile which doesn't look like one, one, one, one. It looks more complicated where the young diagram is not constant and I want to show all those vanish because they're harder to compute. Well, they're not, they're zero. They're quite easy to compute. All right, so on these various components we're going to produce this vector field and show it's non-zero. So basically we're going to find it's non-zero except in the situation that we've already been computing where the Jordan blocks are all of constant size. So that's when your chief on the three-fold is supported on some thickening of the surface and it's got constant rank over that thickening and because of the prime rank assumption we've been making that constant ranks can have to be one. All right, so I've probably made a bit of a pig-zero of this but the upshot is we need a vector field on these components in the modular space which doesn't vanish anyway. So let's do that. Okay, so here's the vector field. It originated in earlier work with Rahul Pandrapanda on stable pairs but it works perfectly for any kind of sheaf theory. So here's how you adapt it to our situation. So we're going to discuss things not in terms of Higgs bundles at the moment but in terms of sheaves on the three-fold. It all depends on the surface having holomorphic two-forms so it only works on general type surfaces and what we do is we pick a holomorphic two-form and we use it to start moving our sheaf, our torsion sheaf on the three-fold. And more generally in the semi-stable case you can also do that with these Joyce song pairs or mochizuki pairs but I'll concentrate on sheaves. Okay, so we have some kind of torsion sheaf. There's a beautiful picture you can't argue with this. So we have a torsion sheaf on a thickening of the surface and we're going to flow to other points of moduli space and then take the derivative that gives us a vector field on moduli space. And the flow is, what you do is you pick off the last layer. So you're thickened to sort of length D, D times S. You pick off the last layer, so the bit annihilated by z if z is the coordinates of the fibre, so that the maximum ideal. And you send that up the fibres by taking the graph of this sigma or T times sigma as T runs through the complex numbers. Okay, so you send that up the fibre and then the rest you'll have to flow down a little bit slower because you want the centre of mass zero condition. All right, so that's the rough idea. So I'm going to describe it just on one fibre. All right, so we're just on one canonical fibre, so just one copy of the complex numbers. I've picked a coordinate, so it's really the complex numbers with parameter y. Okay, and my holomorphic two form at this point I'm going to say is lambda in this trivialization. It's just some complex number lambda. Okay, so I'm going to move at sort of speed lambda up and then, you know, lambda over D minus one or something down. So we have a question that I guess you just answered Should we think of this as moving the top layer away from the zero section to sigma section? Yeah, that's great. Yeah, all right, and we take our C star fixed sheaf and, you know, by the classification of principle modules over a principle ideal domain or whatever, it just looks like a sum of structure sheaves of fat points. And what am I going to do to such fat points? Okay, ignore the mu for now. What I'm going to do is I'm going to leave them alone unless they're really fat. So the thin guys are kind of discriminating here based on their slimness.  They move in this direction. So we have a, you know, we start up. So the guys which look like C of y over y to the D. What I do is I take that bracket, that y to the D and I factorize it as y to the D minus one times y. Okay, the y to the D minus one, I kind of leave alone. And the y, I move off at speed lambda. So I move it upwards, up the fiber at speed lambda. Okay? And then actually I put this sort of drifted mu to make everything gently move down just to balance out so that the center of mass remains zero. Okay? So I'm sending one bit one way and then I'm pulling everything back a little bit to keep the center of mass zero. Yeah. Any questions about that? It's kind of important. This is the key thing. And it's easiest to write down and make rigorous in the Higgs language, in fact. So the exercises check that that that I just wrote down amounts to this in Higgs language. So what are we doing in Higgs language? We never change the bundle in Higgs language because we're only ever moving vertically so when you push down it doesn't change anything. All right? So the bundle always remains the same. We're just changing the Higgs field and in terms of this weight space decomposition which amounts to, you know, how the sheaf looks at various orders of thickening. All we're doing is the following. This right-hand term here is the drift. So this is the mu. You can ignore this. This is just what keeps trace equal to zero. That's very easy to write down because you just work out the trace and subtract it. All right? So you can ignore that. That's just the drift. This is the important piece and you see it's zero on every, every sum and except the last one. And on the last one it's just sigma times the identity. So I'm just increasing the Higgs field by sigma on that last piece and that's what sends me upwards. Okay? So, and then that makes perfect sense of the whole manifold, in fact, in a family over modular space. That's really the definition of the vector field. Any questions? So can you also move like 2 s up and then keep your p minus 2 copies kind of constant? Ah. Yeah, so now you'd have another vector field. Yeah, you could certainly do that. I wonder if you would just find the zeros more or less the same. Because I'm going to find the zeros are the uniformly thickened guys. So then they'd be zero on that as well. Yeah, maybe it's not useful, I'm not sure. Yeah. Why have I never thought of that? Because it's not useful. Okay. Good effort, but no banana. Okay. Here's the miracle. This is beautiful, okay? If you don't follow this lecture, try and follow this bit. This is cool. This isn't the miracle, sorry. I'm just leading up to the miracle. Okay. So I've defined a flow. So now I can just take it infinitesimally. So instead of letting t vary over the complex numbers, I just let them vary over the dual numbers and that's the definition of a vector field. Okay? So I'm getting this vector field. It doesn't preserve the fixed locus. The fixed locus sits inside the modular space and it's a vector field just on the fixed locus but not tangent to it. It's moving the fixed locus out of the fixed locus. That turns out to be important to get the weights right, but I won't go into that. So what is the rotation ct? Yeah, that just means the affine line, the complex numbers with parameter t as the coordinate, okay? So it looks like, at first sight, that vector field doesn't look like it's going to vanish anywhere, right? On my component, which would be bad because it would mean that all invariants vanish and we'd have computed all Vaffa-Whitney invariants except those on the really scheme theoretically on the zero locus, the instanton branch. But amazingly, it does vanish this vector field and this is the miracle. I never fail to enjoy this miracle, okay? The miracle is that if the if the sheaf is uniformly thickened so it's just identical as you go up the it's just a pullback as you go up vertically and then chopped off at some thickness, then it vanishes on those guys. So what that amounts to in the Higgs language is that all those weight spaces are identical up to tensoring with powers of the canonical bundle. In that case, this vector field is zero. Okay, and here's why. Here's the local computation. Check that if you take y to the d, the default thickened point in the affine line and you move it apart in this way so you just move it, there I've put the drift in. But it's exactly the same as on the previous slide. I've worked out what the mu is, right? So you move up at some speed. You take the top layer, move it off at some speed. Then move the bottom, what's remaining at sort of much slower, at one over d minus one times the same speed downwards to keep the center of mass zero. Then what you get to first order is the trivial ideal. It doesn't move anywhere. That is craziness, right?  You have to be careful. For d equals two, it's very easy to see algebraically. So when you have two, I send one layer up and I send the other layer down at the same speed because they're the same size. That's just this ideal. y plus t, y minus t. One's at plus t, one's at minus t. Other way around. So this is a first order deformation of the fat point and it's the zero first order deformation. That's extraordinary, right? But if you multiply it out, you can see it's y minus t squared and t squared is zero. You can interpret this in terms of the non-zero first order deformation. So what is the non-zero first order deformation? It's y squared minus t. So the point is that there's a first order deformation where you pull apart at sort of twice the speed, where you pull apart, you go to t to the half and t to the minus half. That's a first order deformation that's non-zero and this one is t times that one. And that's why it's zero, okay? You can u and r if you like, all right? So what you find is this vector field and on the components that we've already computed, these nested Hilbert schemes, it vanishes. It doesn't tell us anything. On all the other components, it's non-zero and it says that the invariance vanish, okay? Okay, this is very brief. I'm just going to mention at the end, Keemley's localization. Even when you take one of these components of the fixed locus where the profile is constant all the ranks are constant, this vector field or this co-section can be non-zero. You need more than just the ranks are constant. You need the sheaves to be uniformly thickened. So you need everything about them to be constant. So what you find is on these components, you do actually get a non-zero vector field, but it has zeros. And it turns out that essentially what it proves is that these sheaves have to be uniformly thickened. There's nothing interesting happening away from the zeros of this holomorphic two-form.  But what it means is you can localize the computation to this canonical curve in the surface where the holomorphic two-form vanishes. So in Higgs language, what you find is that away from the holomorphic two-form, all these guys, all these weight spaces have to be the same. And if you're doing nested Hilbert schemes, there's no non-trivial nesting. The sub-schemes have to be identically the same away from this holomorphic two-form. And so I've referred, and so has Lota, to T's Laracus computations. What he does is he localizes all these nested Hilbert scheme computations. He doesn't need to really compute away from this canonical curve because the nested Hilbert scheme is trivial there. There's no non-trivial nesting. Whatever the sub-scheme is, they're all the same. All the z-eyes are the same. But near the canonical curve, interesting things happen. And so you can localize there. And how do you localize? This is the technical thing, and this is really a bit much. I think this is more for experts who already know this. So yeah, never mind. So this is the abstract setup. If you have one of these co-sections, then they show how to localize the virtual cycle to its zeros. So the case we've been dealing with is where it had no zeros, so you just get zero. But when it does have zeros, how do you localize to this z here, which is the zero locus of this co-section? They produce a class which is supported on z and pushes forward on the modularized space to the virtual cycle. And how do you do that?  No, in our case, they're all positive. And we're going to work one irreducible subvariety at a time, so we can assume this c without loss of generality is an irreducible subvariety. That's important. If it's entirely supported over z, then we're done. When we take the Fulton-McPherson refined intersection over here, this intersection with the zero locus, then we get something supported on z and we're done. So what that means is we can assume it's not supported over z and it's irreducible. So let's do that. And now what we can do is we can blow up so that we may as well assume that z is a divisor. So we can actually assume that this co-section is zeroes over a Cartier divisor. And then you replace the c by its proper transform. That's why it's important that it was irreducible and not entirely supported over z because otherwise it wouldn't make sense to take the proper transform. So we're going to work upstairs. Later we'll push back down. Everything will be independent of choices. There's no problem with that. It just means you can replace c by its proper transform. It just means you can assume in everything I've done that the zeroes of the co-section are a divisor. And that makes things really easy because what that means is this co-section, so it mapped from our e1 to the structure sheath, because it vanishes on a divisor, it factors through this. And now this is onto. So now our co-section has given us an exact sequence of vector bundles. There's no sheaves here because of the divisor condition. The ideal sheath of z has become a line bundle of vector bundle. So what we end up with is that we've managed to split, more or less, our vector bundle into two pieces. The piece where the co-section is taking its values and the curl. And the point is that Keeman-Lee's show, as we discussed before, that the cone that we're interested in lies in this kernel, Ke. So now it's clear how to intersect with the zeroes of e while restricting to, while localizing to d, the zeroes of the co-section. What you do is you intersect the cone, not with the zeroes of e1, you just intersect it with the zeroes of k.  And then you've got to intersect with the zeroes of o-d. But there's an obvious way to do that. It's just to take the first churn class of that line bundle, which is minus d. So you do all the intersection on the left-hand side and then at the end you just impose this condition of taking the first, the Euler class of that o-d and that immediately restricts you to d because there's a canonical representative of its first churn class and it's minus d. So you just take this definition here where you restrict everything to d and then intersect with the zeroes of k. And that is, and then maybe you push that down under the blow-up map. And that's the co-section, localized. Each of them we have a question. I'll do it with any way. I'll do it with a finite cone of sin. Oh, this is this bare and fanteky cone from lecture one. Yeah, maybe I won't go into that. I refer you to lecture one. Is the localized virtual class can also be seen as some perfect obstruction theory over the zero set? No, there's this thing I keep referring to that the cone is not schemed theoretically in the zeroes. If it is, then you can. But actually it's only set theoretically in the zeroes and so then it doesn't work. Okay, this is some stuff that I don't understand but I felt I should include. Martin Kuhl helped me with this and Lothar already talked about it. So I'm just going to skim through this. You can't absorb this anyway unless you're an expert. This is just to give you an idea of I keep referring to estuality as swapping the two loci, the instanton locus that was heavily studied in the past and this monopole locus. I'm just going to say a little bit about it just to justify that. So estuality really is supposed to relate the SUR Vaffawitton Generating Series so the sums of all these invariants to the PUR Vaffawitton Generating Series which I haven't told you about. And that's done by generalizing everything I've said to PUR and Yun Feng Zhang and Kuhl have done that and they use twisted sheaves to do that. Okay, so it's possible to define a PUR theory. SUR is Langlin's dual to PUR. I don't really know what those words mean but there you go. And now this modularity, there's an SL2Z acting on the whole theory supposedly but what you can do is pass to a judicious sort of subgroup of it and when you do that you can see that it should just swap the SUR theory with parts of itself. So I don't really need to deal with this PUR theory if I pass to a subgroup of SL2. So let me do that. And then I can explain for a certain subgroup how it swaps the instanton and monopole loci. So Lothar already said some of this. I'm probably going to say it slightly wrong but that's entirely Martijn's fault because I just took it from an email. Okay, so for simplicity let's make these assumptions here. There's some standard modular form. This is a standard module form and then associated to the AR lattice there's also a standard modular form. This theta series. And this SUR generating series I keep referring to the normalizations are as follows. But it's just a sum of Waffa-Whitney invariance over all Cheung classes. And then by the vanishing theorem you can write that in terms of two contributions. The instanton contribution. That's this one. She's supported on the zero section and the nested Hilbert scheme contribution in prime rank of things supported on R times the zero section. And then one way of expressing Lyraka's results is that there are universal formulae functions such that this generating series looks like this. Okay, so sorry. When you calculate using the co-section localization the vanishing theorem and these virtual degeneracy loci so using all the techniques of the last lecture and this one. You can calculate and using this Elling's Rood-Gutcher-Leyn way of computing on Hilbert schemes it means you can actually calculate this second term this monopole contribution. Okay, in some sense. You get cyber-Whitney invariance. We discussed this last time. You get some topological things. You get some modular forms and you know, you're not supposed to absorb that. I've been looking at that for five years it didn't mean anything to me. And Götcher and Kool-Kerjecture and Lothar talked about this and they prove in low rank that to do this estuality transformation what you should do firstly is you should remove the grey piece here so just take the sum over the things above and then you should swap. If for each cyber-Whitney invariant you should put in a certain root of unity it's really not important. You should change one of the modular forms by a change of variables in this way. Very simple. That AR theta series you should switch to the dual lattice and that gives you a different theta series. Okay, they're all minor things and then the important things is that these sort of universal functions here these C0 and CABs what you should do is you should do the standard sort of modular transformation TOR goes to minus one over TOR on that. All right? Where TOR is really everything's function of Q and re-expressing in terms of TOR. TOR and Q are more or less there's some asymptotic re-expansion going on here but TOR and Q are more or less the same variable. Okay? So once you do that once you do this TOR goes to minus one over TOR modular transformation then you turn that monopole contribution into the instanton contribution. Is there a geometric motivation for making those changes? I have no idea. We certainly don't have enough of one. We need a better one. This is really physics, voodoo, magic. This is not. But more this is done by mathematicians doing computations and observing things by now. So here by proving low rank it means proven for certain coefficients up to certain coefficients for entire generating series. I refer you to Lota's lecture. Yeah, I don't know. I don't remember. Yeah, there's... Okay, great. Thanks Lota. And thank you for not pointing out all the mistakes on these slides. Okay. All right. This is an extremely powerful conjecture of theirs. For instance, it generalizes the famous cyber-witten conjecture expressing Donaldson theory in terms of cyber-witten invariance. Similarly for the higher rank versions of that due to Marino and Moore. It hides lots of integrality things in there. There's all kinds of predictions in there that various rational numbers, even complex numbers must actually be integers. It's hiding a ton of miracles. Okay.  And I'll give no further details. And all the modular forms seem to become Jacobi forms and it's all very wonderful. Okay. Any questions before I... All right. So this is more... I've only got, I think, 12 more slides. I'll go through rather quickly and we'll consider this to be some kind of just review of... just some kind of survey. I'm not going to try and get too much understanding out of this. Let's see if I can get rid of this. The point is that Lotta is answering the question. I will finish and then it will disappear automatically. Oh, I see. That's the issue. Got it. Yeah, yeah, yeah. Okay. Sorry. I should ask. So this was this question where the modern program is automatic to use in that case and then actually Thomas Richard already said it is automatic to use it. Yes. Okay, great. Thanks. Okay, so if you talk to physicists they'll more or less say they understand vaffer-witten theory and why are we all mucking about with it? Because they already, you know, conjecturally at least know how to compute it. Certainly in rank two. As I say, they thought maybe in higher rank but Lotta and Martijn have been correcting some of their conjectures. But nonetheless, they know that the theory can probably be expressed in terms of cyber-witten theory more or less and then some modular functions and so on.  Whereas we're fascinated by the geometric origin of these dualities. However, one thing that they were really emphasizing is that there should be a refined version of vaffer-witten theory predicted by one of their physics theories. And that they really don't know how to compute or they're very interested in. So there's an attempt to find this refined theory. So a few people have asked about bare-end weighted, you know, expressing the virtual cycle in terms of weighted Euler characteristics which you can do in the compact case. And, you know, I've always said, well, that's the wrong thing to do here in this non-compact case. It doesn't give the same answer. It turns out to be the wrong thing to do. And that's a shame because in the bare-end weighted case there's an obvious refinement introduced by Joyce in his school, and basically you upgrade Euler characteristics to core homologies. So numbers to vector space, graded vector spaces. And so you can't do that for virtual cycles. Okay, but I have this project with Davesh and partly also with Martin to use other core homology theories other than usual core homology. What used to be called extraordinary core homology theories but nowadays are just called core homology theories. In place of ordinary core homology. But you can't use just anyone. You need it to be something called oriented which means it has a theory of churn classes. And you also need a certain rigidity property that basically when you have a projective bundle, so this is how you deal with a semi-stable case, remember. What you need for the way you deal with a semi-stable case to be compatible with the way you deal with a stable case, you need this property that when you have a projective bundle, the core homology theory should not see the twisting of the bundle. It should have a cunith formula. And it seems that elliptic core homology is the best theory that's going to give the nicest answers, gives crazy formulae, Borcher's modular forms and so on. And we haven't written that up. I apologize to Davesh if he's in his pajamas now. So for two years I haven't written that up. Okay, but the k-theory version where our oriented core homology theory is k-theory is what I'm going to discuss now. So what's k-theory about? Remember how we defined the virtual cycle in core homology? We took a graph, we made it vertical, we got a cone and we intersected it with a zero section. Well, you can do that intersection in k-theory instead of core homology. What that means is you take this, if you want to intersect two sub-varieties, so here the horizontal one and the vertical one, what you do is you take their structure sheaves and you tensor them, and that should give you the structure sheaf of the intersection. And the way to do it properly and homotopylic invariantly and so on is of course to take the derived intersection, the derived tensor product. So really the correct way of doing it is to take the pullback to the zero section of E1 of the structure sheaf of the cone. So that's the way to do intersection theory in k-theory. And so that gives something called the virtual structure sheaf on the modular space. You should think of it as the structure sheaf of the virtual cycle in core homology, but it's not quite that, especially because the virtual cycle might have, it might be a sum of sub-varieties with possibly negative coefficients. And so then taking structure sheaf for that's rather problematic. But anyway, this is perfectly good definition. Gives you a k-theory class. Using derived geometry, you can probably make this an element of the derived category by now, but you don't need that for defining enumerative invariance. All right, so this is a virtual structure sheaf. Any questions? So this is the k-theoretic virtual cycle. And then we want to modify this. So in physics, oh, I beg your pardon. So this replaces the usual Euler character. Let's suppose the modular space is smooth. You have an obstruction bundle. The virtual cycle will be its Euler class. In k-theory, the virtual cycle will be this kind of causal resolution of a representative of that Euler class. It's this wedge of the dual of the obstruction bundle. So in good cases, when the obstruction bundle has a section which cuts out transversely by regular sequence, some z, then this is a causal resolution for the structure sheaf of z. So that's the sense in which this is the structure sheaf of the virtual cycle. But in physics, they're really doing real geometry. Ridiculous, I know. They're not actually algebraic geometers. So they're using something called KO theory. And in real geometry, there's real bundles. Orb and ob-dual are actually isomorphic because you can pick a reman... whatever, a metric. And this Euler class here does not have that invariance. Lambda of ob is not the same as lambda of ob-dual. There's this twisting here. And in KO theory, the right thing to take is something that's exactly halfway in between. You twist by a square root of this. And you have to worry about integrality properties and so on, but we don't care about that. So we just invert two or something and there's no problem with this. Okay, so this thing is called the Atia Bot Shapiro Euler class. Can I ask a question? I don't understand this line. If ob-dual are isomorphic, then... There are real bundles, but I am taking the complex wedge, right? I'm taking the alternating sum of all the complex wedge powers. So this cannot be a kosher operation in KO theory. But it turns out that this is a kosher operation in KO theory, and it can be expressed in terms of Clifford algebras and... Okay, there's another way of doing it. Using spinners. This existence of the square root means there's a spin structure, and then you can take a Clifford module, and then it turns out, ultimately, to be this. Okay. So, you know, whereas we're always interested in things like holomorphic Euler characteristic of the structure sheaf and Todd classes, in KO theory, you're not supposed to do that. You're supposed to take the index of the Dirac operator and Ahac classes, Ahac genera, okay? But what that amounts to in complex geometry, if you rephrase it, is the holomorphic Euler characteristic of the square root of the canonical bundle. Again, that's to do with this square root here of this line bundle. So what happens to the index of the Dirac operator? It's fine. All I'm saying is that that's what's of interest in physics, and you can express that in complex geometry in terms of the square root of the canonical bundle and taking its holomorphic Euler characteristic. And you know, an indication of all that is this special property that because everything gets twisted by this square root canonical bundle, what you find is that this Euler class here, this ABS Euler class, has this nice property now that when you have a transverse section and so on, it really is the structure sheaf of the zeros up to, again, this same twist by the square root of the canonical bundle, okay? So if you take the holomorphic Euler characteristic of this, you end up getting the index of the Dirac operator on z. So is it true that in the real case when you take the index of the Dirac operator, but if always in the counterpoint, in the complex case, you have the square root of this Km of, yeah. The holomorphic Euler characteristic of that square root when it exists is the index of the Dirac operator. That's correct, yeah. Okay. Oh, there you go. All right, so there's just motivation, okay? I'm just going to tell you that for various physics reasons and various duality reasons, what you should do in all this theory is not take the virtual structure sheaf, but you should twist it by a certain line bundle and it's the square root of the canonical bundle. But because we're in virtual geometry, it's the square root of this virtual canonical bundle, all right? So in other words, you should take the index of the Dirac operator. You should twist everything by a certain line bundle and take the holomorphic Euler characteristic, all right? And that's a completely, if you're uncomfortable about that, you shouldn't worry, it's completely harmless because remember in my case, the virtual dimension is zero. And so all I'm doing is taking some zero-dimensional sheaf, more or less, and twisting it by a line bundle, I'm doing nothing at all, all right? So it doesn't make any difference, all right? But this turns out to be the right thing to do. You'll see why in a minute. Sorry, what do you mean that it doesn't change at all? I could have twisted by any line bundle at all, I would get the same answer. Why are you doing this thing? Well, you give me this. Let me unravel the... The reason I'm doing this is that actually I'm in a non-compact setting and so actually I can't take this holomorphic Euler characteristic, I'll get infinity. So the virtual dimension is zero, but... Yeah, anyway, let me give you an example. So what I'm going to do is use the C star action and I'm going to take this holomorphic Euler characteristic equivalently, okay? So I give an example here. If you want to take the holomorphic Euler characteristic of the structure sheaf of the affine line, well, that's just the dimension of the space of sections of the structure sheaf of the affine line, which is infinite. So you need to regularize it in some way and so the way you do that is you use the C star action on the affine line to say that equivalently what's the holomorphic Euler characteristic? Well, you know, this Cx now splits into weight spaces and they're all finite dimensional. So there's C that has weight zero, there's X which has weight one or minus one depending on your conventions, there's the weight two space and so on. Okay, so these T's are really just characters of C star and it's just saying that this infinite dimensional representation of C star splits as an infinite sum of finite dimensional representations. One, the weight one representation, the weight two representation and so on, okay? And so this I can make sense of as one over one minus T. All right, and that's what we're doing here. That's actually this definition here and now it is important which line bundle I put in here. But does this square root of a virtual canonical bundle involve some choices? Yeah, but I'm going to localize everything to the fixed locus and on there there's a canonical choice and in reality the invariance don't depend on any choices. Is that true, Noah? Yeah. So even though, yeah, in general the answer's yes but then the invariance, the answer's no. Okay, so in other words what does this definition mean? It means that I replace the standard definition of holomorphic Euler characteristic there by its character as a virtual C star representation. And to do that I need to localize by inverting powers of one minus powers of T and also introducing a T to the half because I need to keep track of the C star action of this virtual canonical bundle. But these are all technicalities. Okay, so in other words what I end up with is a function of T and actually it's a rational function of T and did I write this? I'm sure at some point I'll have written it. It's also got this invariance property that it's invariant under swapping T to the half with T to the minus half and that's because of serduality. Okay, so I end up with a function of T and I can localize by k-theoretic localization I can localize it to the fixed point locus. This formula shouldn't scare you, it should be obvious in some sense. What you're doing is you're saying when you want to take the functions or sections of this line bundle on the whole modular space what you can do is approximate the modular space by the fixed locus and then it's normal bundle in the normal directions and what are the functions on the normal bundle? They're the symmetric powers of the dual normal bundle. Okay, so if you have vector space, functions on it are the dual vector space and then all its powers, all the polynomials in them. Okay, so you end up with this symmetric power of the dual normal bundle and so you end up with this localization formula and you can use Riemann-Röcken express it cohomologically in terms of the original cohomological invariance what you're doing is changing the cohomological invariant that we've been using up until now to this one. So there's a minor change going on here. But as I say, don't worry about this. Yeah. Why do you need a pretence also by all? Because I mean it's like we're always working with virtual things. The original definition had a no-virtual in it. I guess it. Okay, it should have done. There you go. Yeah, I never want to take a holomorphic Euler characteristic of a sheaf over the modular space. I want to take it over the virtual cycle. So I want to tense it with the virtual structure sheaf before I do it. Yeah. Okay, I'll correct that. Thank you. So the dependence on T is in this trunk vector? Yeah, that's right. That's right. And in passing from K theory to cohomology the T becomes an E to the T because that's the Turing character. Honestly, you should ignore this. Okay, the bit at the bottom is the important thing. It's that what this does is it defines functions of T which specialize to the old numerical invariance at T is one. And that our invariant under T to the half goes to T to the minus half. And that invariance comes from serduality more or less from the fact that on a scheme if you take the holomorphic Euler characteristic of the square root of the canonical bundle by serduality it's the same as the holomorphic Euler characteristic of the dual of the square root of the canonical bundle tense to the canonical bundle. But that is again the square root of the canonical bundle. Okay, very briefly. I should get a move on. There's an exercise for you. This is just a geometric exercise, nothing virtual but there's a virtual version. Okay, that symmetric power of, you know, those functions on the normal bundle can also be written as one over the exterior algebra of the normal bundle, okay, by a standard causal resolution argument. In other words, the symmetric power tensor with the exterior power is one because it is just the causal resolution of the structure sheet. But sorry, I'm kind of running out of time, so I'll speed up. Okay, so you can rephrase the virtual localization formula in this form, this is perhaps more usual. So this is a more or less, this is our definition of the refined Vaffa-Whitney invariance. Okay, so there's a couple of things just to say. So one is if you remember the numerical Vaffa-Whitney invariance on the instant on locus gave the virtual Euler characteristic. Here what we get is the virtual kaiwai genus. So that's a refinement of the Euler characteristic given by this formula here. So in the smooth case, instead of taking the Euler characteristic you can do this, you can take the alternating sum of the holomorphic Euler characteristics of the i-forms on M and you can put this T to the i here. Okay, so this is sort of one-half of the Poincaré polynomial, Cephalanomial or E-Phalanomial. Okay, and then there's this just usual yoga that we saw in previous lectures that when you have this modularized space of sheaves on S but when you can consider it in Vaffa-Whitney and this series sheaves on X, the three-fold, then you pick up the dual of this virtual tangent bundle, you pick up an extra piece which is the dual of this virtual tangent bundle shifted. So you get new obstructions, dual to the old deformations and you get new deformations, dual to the old obstructions. And so roughly speaking, everything just as before, everything simplifies rather quickly and what you end up with is that you're taking the holomorphic Euler characteristic of roughly the Durham complex, the wedges of the dual of the virtual tangent bundle. So the virtual tangent bundle and when you do that, what you end up with is the virtual kai-wai genus. So I've run out of time, so I won't go into that. Is it just T dot T or anything with dot? Sorry, that's just homological notation. So wedge dot means the alternating sum in k theory of the wedge powers. And similarly you can do the same not just for counting sheaves on X but also for counting pairs on X. Again, you have a refined version of our conjecture as to how that defines how the invariants define counting joy song pairs depend on N. This is our conjecture for how they depend on N. This is the one you should really think about. And that defines refined VW invariants in the strictly semi-stable case, modulo of this conjecture. And all you're doing is everywhere you replace more or less numbers before this was just kai of alpha N, this was an Euler characteristic of a projective fiber, a projective bundled fiber. Now you replace it essentially by Poincaré polynomial of a projective bundled fiber. So that amounts to replacing numbers by quantum numbers. So it's all very pretty. So I have a question on this. Constraint H01 as well as H02, you wrote. Usually in this VW video because it's not the opposite, H20 is 0. Yeah, but they're the same numbers. Here it's equal, you're assuming. But it may not be equal. H20 and H02 are always the same. That's the symmetry of the Hodge diamond. Is there a word for it? Does anyone know? A seduality, Poincaré duality. Okay. Oh yeah. Okay, so we don't need to say this. Okay. We proved this conjecture in some cases. What's very nice is in the compact case, suddenly you can use these other refinements. In the compact case, suddenly you can replace numerical invariance by, or virtual cycle invariance by bare end weighted invariance and they're the same. And it turns out the same is true at the refined level. So this k-theoretic refinement becomes equal to Joyce's refinement using DT invariance. So this fits in very well with DT theory when the vanishing theorem holds, when the canonical bundle, when the curvature of the surface is positive. Okay. And then I just consider this maybe an announcement. Just go to, Lothar already explained a lot of this anyway. So that standard modular form that keeps cropping up gets refined in this way to a certain Jacobi form. I think that's the right word, but I don't know much about it. And in the k-3 case, you can just calculate everything, you get formulae. But let me just say a tiny bit about that and where it comes from. So if you remember, we have this co-section. The localization works just as well in the refined case. You find that you localize just to these uniformly thickened things and then you can just compute. So everything comes down to these nested Hilbert schemes. And I've said all this before. What I claim is in these uniformly thickened cases it's rather easy to compute. So you have to use these pairs, but I will just here give you a calculation just for sheaves and the calculations are following. This is really a doable exercise, especially if you just take D equals 2 or something. Okay. What I'm saying is for these uniformly thickened sheaves, you can really compute the deformation obstruction theory. And it's given by the deformation obstruction theory of the restriction of your thickened sheaf to the zero section. Tenser with some sort of trivial pieces, but very important pieces when you keep track of the C star action. So these are all weight spaces for the C star action. Okay. That's really a rather simple exercise. And what that means is your modulized spaces are isomorphic to just the modulized space of sheaves on S. S is a K3 surface, sheaves on S, they're very simple things. They're governed by holomorphic symplectic geometry. We have tight control of them. And the obstruction theory is slightly different, but you can really compute it. It's the same up to some extra factors which you can really compute with. And from this means that you can really compute what's called a multiple cover formula for these refined invariants. And here's the multiple cover formula. It's that these uniformly thickened Vaffewittian invariants are equal to the usual invariants just of counting rank one sheaves on the surface. These are governed by just Hilbert's scheme calculations on the surface. Divided by this, so there's this one over D squared, but it's the quantum integer D, one over D squared multiple cover formula. Okay. And the reason I mentioned that is mainly for the experts. You should compare with the usual one over D squared multiple cover formula that we're familiar with, but you should contrast with the refined one over D squared multiple cover formula that appears in, for instance, work of Meinhardt and Davison. And what they get is D, the integer D times the quantum integer D. And for some reason we get the quantum integer squared. And I don't actually know the reason for that. Oh yeah. Yeah, you're the finite one. Is it here? Yeah. This is the quantum integer. Okay. It's the Poincaré polynomial of projective space. Just as integers are Euler characteristics of projective space as you're taught at primary school. Well, that's how I was taught. And matrices are torsion sheaves on the affine line. So quantum integers are Poincaré polynomials of projective space. I should finish. Okay. So everything that we've done in this course, I can't believe I've filled up five lectures and I still haven't finished on time. I apologize. Everything we've done in this course goes through in both the semi-stable case using pairs and the refined case using this definition. All right. So there's a refined version of the degeneracy loci, Tom Porsche's formula. You get pretty looking formulae and I've run out of time. All those calculations I told you that Laracca did, he didn't really do those. He did the much harder calculations. He did the pairs case and the refined case all at once. His work is extraordinary. He computed everything in as much generality as you could ever imagine. And so this is just a statement I gave you before, but with a T everywhere. This is just the refined statement. This is really what he did. And this is the simplest case of what he did, by the way. Okay. So last slide. Right. You can't hold me to any of this. You know, if in three years' time you come back and say I spent the last three years working on exercise three you gave me and the course of my career is now ruined, then you shouldn't have paid so much attention. Okay. This is just me mucking around. Okay. If you're interested in virtual psych, deformation theories, perfect obstruction theories, removing traces. Dennis, what's the word? You asked me a question about it. The book bites Flenner. Semi-regularity. Semi-regularity maps.  Okay. Then work out what rank zero Vaffa-Witton theory should mean. No one's ever done it. And passing to this perp theory is not the same there. The trace is zero, so it all gets more complicated. But there's surely a theory there. Okay. If you know some geometric representation theory, you know about Hilbert schemes, you can interpret these Tom Porsche's formulae I've been giving in terms of something called Carlson and Kuhnkopf operators. So that sounds an awful lot like vertex operators. And vertex operators sound an awful lot like modular forms. So I don't think really anyone's really exploited this properly. I mean, maybe if you understood what Tisse was doing and you understood modularity, so maybe Noah could tell me. Maybe you would just say, well, obviously he's getting modular forms because he's using a Carlson and Kuhnkopf operator. I don't really know whether there's something sensible in this exercise or not. But it seems to me this is a root to modularity, understanding modularity. Physicists who make great progress in Vaffa-Witton theory using TQFT and cut-and-paste operations. So I don't really know what they use as definitions. They don't really. I mean, they use properties they expect the Vaffa-Witton partition function to have. And then they cut and paste. And in this way, they make great progress in computing what they think Vaffa-Witton generating series should look like. And it would be nice if mathematicians could justify some of their calculations and emulate them in the complex case using degeneration and relative theories. So you degenerate varieties to normal crossing things and then you define relative invariance on both pieces and then you glue them together. So there's a well-defined relative theory and DT theory. And it would be interesting to explore to what extent that can recover these things that physicists are doing. There's a very recent paper of Manchur Mor, which is extremely exciting and it opens up the possibility of extending what I've been doing and discussing, sorry, to almost all four manifolds. In fact, four manifolds with just B plus odd. So that's kind of half of all four manifolds. And that's because they have an almost complex structure and where I exploited the complex structure to give a C star action on the modular space, they can exploit this almost complex structure to get a U1 action on a perturbed modular space, but it's good enough, okay? So now you can start localizing, you can get rid of these non-compactness problems and really it's possible you could really define Vaffel-Whitney invariance for actual proper four manifolds, which is really what you're supposed to be doing. Obviously I won't be taking part in that, there's no complex numbers there. Okay, another sort of approach to modularity is that everything comes down to Hilbert schemes and we seem happy now to accept that therefore everything's modular. I mean, I personally am never happy that the partition or the partition functions are modular for that makes no sense to me, but modular that miracle. If you express everything in terms of Hilbert schemes, then maybe you're not surprised that you get modularity. Okay, so I've shown you how to get monopole calculations in terms of Hilbert schemes. Gertcher and Kuhl get instant on calculations in terms of Hilbert schemes by using something called mochizuki wall crossing. So instead of just considering a sheaf on S, you consider a sheaf on S plus some field, some section or something like this with some stability condition, and then you change the stability condition. And at one end you're dealing with just stable sheaves on S. At the other end you're dealing with much more complicated objects, and the wall crossing in between, sorry, not more complicated, actually much simpler objects where essentially you get down to rank one. But more importantly, the wall crossing in between, all the contributions come from decomposing your bundle into sort of rank one sub bundles and extensions and so on. And so in that way mochizuki is able to produce a wall crossing which involves Hilbert schemes, really rank one sheaves. And so that's partly how they do their calculations. Okay, so then we end up with two sets of calculations on Hilbert schemes. Now what's a nice way of relating them? And is that what's giving modularity? Okay, another interesting thing you can look at is a blow-up formula when you replace the surface by its blow-up. What happens to the invariance? I think we know what the answer is supposed to be in the numerical case, but also Toda shows you what to do in the three-fold case when you use bare-end weighted Euler characteristics. So that work should be repeated in the virtual cycle setting. And then do elliptic calculations with the paper that one day I will write with Davesh. Okay, and then finally, when you set t to the half to be minus one instead of one, this is extremely speculative, but is there some real theory? There are some situations in nature where Gertrude in a different life with Schender finds some refined theories where when you set t to the half to be minus one, which is the wrong one, usually you would set it to one, but when you set it to minus one it comes down to some real theory, which is completely extraordinary. I don't think we understand it very well. So I don't know what happens here. Maybe something interesting. Maybe Vaffa-Witton theory for real groups or something, I have no idea. Okay, I'll stop that. Any questions? Sorry, could you summarize, why do you need the Refined Key Theoretic Conference? Why do you need it? You don't need it. What do people say? The mountain, it was there, so I had to climb it, I don't know. Also physicists sort of ask for it. They're really interested. Maybe not, yeah. And Gertrude and Kool were doing, they were already, I kept talking about them computing these virtual Euler characteristics, they weren't doing that at all. All along they were refining, they were only ever taking virtual Kai-Wai genera. So immediately that was crying out for an explanation for a theory which incorporated that refinement. So the theory was already refined to find the whole theory. Yeah. So when you refer to a botchup multiform, you mean a botchup product? I have no idea what I'm talking about. You shouldn't take me seriously. That's the thing we're missing in all of this, right? A vertex operator algebra, le algebra. Yeah, go for it, Lota. Yeah. Yeah, we computed some in the range of case, we computed some in the genus of the of the monopolar space, just on the on the incident side, don't do that for a week. And then you do have a botchup product. I mean, incomplete, there's one term missing, but basically it's just botchup products, yes. And the one-term missing should come from the monopole locus? No. No, you're kind of missing an elementary term. It's essentially just a botchup product with some incremental term.  In this case, you're more able to miss something slightly new. It's essentially miss out some in the Kofi form or something. And essentially it's just botchup product. Thanks. Yeah, sorry I overrun.  You've been a great audience. Thank you. Can I just thank the organizers for this huge amount of work? No, you may not speak. This is a huge amount of work. And the person who I've had to interact most with who's done just an amazing amount of work is Francesco.  I have no choice. Boris Johnson won't let me. And he would not let me cancel. And he proved to be right. It turned out. Yeah, exactly. There's Sicilian here. And at every stage it turns out when politicians say something, it means nothing. And actually you can do what you like. I haven't even broken any laws. It's been fantastic to be here, to be in a conference for the first time in two years. So yeah, I thank Francesco for that. But now you can ask a question. So what's about that point, too? Because you say that Tom Porscheu's Chinese class formula is related to Karstena Kuntur. And you show us that that formula depends also on the twisting by some line bundle, or some plus beta, right? Okay, that's when instead of nested Hilbert schemes of points you've also got curves and points. In the case in which I introduced that you have some modified version of... Yeah, we call it Carlson and Kunkoff operator, but you're right, that guy is not really in their paper. But we proved similar properties for it. We proved that k-theoretically it's really a bundle. It doesn't have any higher churn classes. We prove it, yeah, okay. So, yeah, someone should study that Carlson and Kunkoff operator in more detail. It's in your paper. Any more questions?  That's okay. Okay, Noah says it's fine. Okay, so let's thank each operator.


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Hi, friends. Welcome back to my channel and a special welcome if you're new here. My name is Jen. I'm a certified weight loss and nutrition coach, and I'm on the Weight Watchers or the WW Blue Plan. Today we're talking about Weight Watchers or better known as WW. We're going to go over some pros and cons of following the WW program. I get asked a lot as a weight loss and nutrition coach about pros and cons of WW versus calories or counting macros. So in a later video, we are going to compare WW and calorie counting. But today I want to focus on the pros and cons of being on the Weight Watchers or the WW program.  And if you're new and you're not part of my channel, I would love to have you here. Hit the subscribe button and click the bell right next to it so you never miss a single video. I do upload five days a week so you definitely don't want to miss out. Check out the description box down below for nutrition coaching. I offer personalized macros and calories as well as one-on-one coaching if you want to chat with me directly. Also, I just released two WW friendly, calorie friendly, all plan points included recipe ebooks, breakfast, and lunch. So make sure you're getting your hands on those for amazing, delicious recipes that fit into that healthy lifestyle. Links, discounts to all my favorite things and my Facebook group. Comment over, join us there. Are also down in that description box. So let's go ahead and jump in to talking about the pros and cons of WW. As we know, WW Weight Watchers is a very, very popular weight loss program that leans on a point system to track your food. There are endless zero-point foods on their list. And in order to track and stay on track, you track the WW points. Everyone has allotted a certain number of points every day based on several factors, weight, height, gender, and age. So not everybody receives the same number of points every single day. But in order to follow the WW program, every food is assigned a point value. When you consume that food, you track the points. And when you're out of points, you're out of food for the day. Now this sounds pretty easy and convenient. You don't have to track every single calorie that you're eating because WW does offer so many zero-point foods. Those foods do not have to be weighed, measured, or tracked. So it eliminates having to track every single thing that you're eating like you would have to do if you were following a macro or calorie counting approach. But that doesn't mean that Weight Watchers is the right program for everyone. So I want to share with you some pros of the program and some cons of the program so you can decide if, number one, you want to join the program if you're just looking into WW or maybe you're on WW and you're just not sure if it's the right program for you. Hopefully the pros and cons of today's video will help you make a better educated decision. First, let's start with the positive. Let's start with seven pros of the WW program. Pro number one is it is a balance and flexible program. In fact, WW is one of the most flexible diets on the market. By assigning fruits, vegetables, and lean proteins a point value of zero. WW encourages you to get the majority of your food throughout the day from this healthy zero point food list. With the points left over, you can pick adequate whole grains and dairy that fall within your daily smart point allocation.  No matter what diet plan you choose, it should be something that you want to follow for life. Now this doesn't mean that you can tweak the way you follow the plan as you reach your goals. But whatever you do to lose weight, you have to do to maintain your weight. So you want to make sure that the plan that you're choosing is something that can work for you long term. WW teaches essential lifelong skills such as weighing and measuring your food, watching your serving sizes, and in fact encourages you to prepare your meals at home. So you know exactly what goes into those meals. Pro number three is my favorite. And this is that no foods are forbidden, restricted, or eliminated. There is no list of foods that you will find in order to follow the WW program. No food is off limits. WW encourages you to eat healthy food choices. That's why they make those items zero points. But the leftover points that you have for the day, you can allocate those to your favorite foods. Sometimes it's pizza. Sometimes it's a big green salad. But no food is off limits. And that's what makes WW a more sustainable program and a sustainable lifestyle. Pro number four is it delivers slow and steady weight loss. And as I always say, slow and steady wins the race. You can expect to lose half of a pound to two pounds per week following the WW program. You're not going to lose five, 10, 15, 20 pounds a month, but you are going to lose weight at a healthy rate. And the slower we lose our weight, the more likely we are to keep it off. So don't fall into those fad diets that tell you that you're going to lose five pounds a week. Or even those fad diets I've seen on TV that say that you can lose a pound a day. That is not healthy. It's not sustainable. And chances are very, very, very good that you're not going to keep that weight that you lost off because you're not developing good, healthy, sustainable habits as you're losing that weight. The next pro is that WW delivers a ton of support and resources for you to utilize. There's many, many features such as the WW app that allows you to not only track your food, look up the point values of foods, build recipes, figure out what you can eat at your local restaurant. And it even contains a community called Connect, which is essentially Facebook for the WW app where you can connect with other WW members. In addition to the app, you can attend regular in-person support type of meetings where you go and you discuss a weekly topic, you weigh in, you ask questions, and you can utilize the support of like-minded individuals on the WW program. And there's even the option for a premium membership that includes one-on-one coaching that you can chat with a WW coach directly. And lastly, you can sync any type of Fitbit, Apple Watch, Garmin to the WW app so that you can rack up extra points when you're working out. These are known as fit points. So the fact that WW has so many resources and so many different options for support makes it a really good program if that is what you're looking for. If you need accountability and support, WW can be the way to go. The next pro of the program is it actually reduces your risk of diabetes. Because the healthy food options are zero points, people who follow the WW program generally consume a lot more healthy foods than they would if they didn't have those foods rated at a zero point. And because they're gravitating towards these healthier options, they have a reduced risk of type two diabetes quite substantially when following the WW program.  So WW not only helps prevent diabetes, but if you suffer from diabetes, it can really help curtail your symptoms and make it a little bit more manageable. And last but not least, when it comes to pros, is WW promotes and encourages exercise. WW encourages plenty of daily movement. And like I said, you can sync up a fitness watch to the app and earn additional points called fit points as you do exercise. So it's encouragement and it's motivation to get out there and move your body and exercise not only because you earn fit points, but because that is what also helps you tone up, become stronger and see more success on the scale. Now all of this sounds amazing. Sounds like the plan that you've been looking for, but before you jump in with both feet, let's chat a little bit about some WW cons. Number one and the one that I hear most often is it is a costly program. Now the cost varies from person to person depending on what services and resources you would like to utilize from WW. If you decide to be digital only or online only, this is going to be the most affordable option. Now the in-person workshops fall right in the middle for affordability and the most expensive option is going to be the one that utilizes one-on-one coaching. Now you can get discounted rates by paying for several months in advance. You can also utilize any promotions that WW is offering and down in the description box I will put a link for a free month of WW if you want to try it out and again see if it's the right program for you. But take advantage of discounts, free months, links for special promotions before signing up for the program and try to make it as affordable as it can be so that it's something that you can continue to stick with and that you can continue to work into your budget. Con number two is counting points can be pretty tedious. If you don't like counting calories there's a good possibility that you're not going to like counting points either. You still have to weigh, measure, and track all of the food that you're eating. Now depending on the WW plan you choose some involve more tracking, some involve less tracking but the reality is you're still going to have to track. Con number three is that weekly weigh-ins are necessary for the program. You need to weigh in once a week to track your progress. Your weigh-in is put into your WW app and that is how Weight Watchers tracks your overall weight loss that's how you earn charms and rewards is by tracking your weight every single week. You can either weigh in virtually or in person. Now if you're someone that does the digital only plan you would weigh in at home and track your weight in the app. If you're someone that utilizes the in-person workshop you actually weigh in with what is called a WW guide at the workshop and that weight is tracked right there with that individual. Now this can make some people uncomfortable some people don't want to step on a scale in front of someone else and track their weight and some people don't hold themselves accountable enough to weigh in every week at home and track it in the app but WW says in order for you to be successful you should weigh in one time per week. Another thing I want to address is sometimes for some people the scale can wreak havoc on how the rest of their day goes their week goes their month goes so if you are someone that struggles with seeing that number on the scale that maybe didn't do you any favors weighing in once a week could be a big con for the WW program. The next con is there is very limited evidence for cardiovascular benefits of being on the WW program. The program offered little results when it came to stabilizing blood pressure and cholesterol compared to other lifestyles or ways of eating that are out there for example if you're someone that wants to reap the most cardiovascular benefits you would want to gravitate towards something similar to the Mediterranean diet. Unfortunately WW hasn't been found to have a lot of cardiovascular benefits. The next con may actually be that the program gives you too much freedom. As silly as it sounds too much freedom can really be the Achilles heel for a lot of people.  It's completely possible and very common for people to use their points on less nutritious foods. Lots of processed foods or they're looking for lower point options fat free sugar free and these are much less nutritious full of chemicals and not foods that we should be putting in our body all of the time. So that extra freedom that is given to choose whatever foods you want to come up with low point foods or low point dupes can actually backfire and you may be consuming a lot less nutrient dense unhealthy food than you would be if you were given a specific food list to eat off of. And the last con is probably the most concerning con for me and this is that WW can lead to unhealthy eating habits and an unhealthy lifestyle. There's a lot of concern that counting points can actually lead to an unhealthy relationship with food. A lot of people save up their points to binge on unhealthy foods later they look for those WW dupes like fat free and sugar free and chemical filled foods and a lot of members gravitate towards processed foods that are generally low in calories but high in points so a lot of people will suffer from not eating enough on the WW program. So although they give you zero point foods that are healthy and nutritious if you don't lean on those foods for the majority of your meals chances are you're not eating enough calories maybe you're choosing unhealthy foods or restricting certain foods that can lead to binging so there's a big concern on whether or not following the WW program actually can lead to an unhealthy relationship with food. So now that we've discussed the pros and the cons I hope that you can walk away with a little bit more knowledge in the bank when it comes to the WW program and like I said if it's something you're considering you can weigh out the pros and cons and decide if this is truly the best program for you. We're all different some of us thrive on the program some of us struggle on the program and a lot of people under eat and develop an unhealthy relationship with food based on the program. Now if you've been on the program for a while it's time to maybe take a look at these pros and cons and reassess and decide if this continues to be the right program for you. No one can tell you what to do no one knows you and your body better than you do so deep dive into these pros and cons and make the best decision for you. 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