 Welcome back. All right, so for part two today, I'm going to talk about chordism maps in Hager flow homology. And on the course description on the website that Paul shared with you, there's some links to some references. OK, so let's remember what we did yesterday, right? So yesterday we described our free manifold via Hager diagram. And then from that Hager diagram, we constructed a chain complex whose homology is invariant of the free manifold. In particular, the homology is invariant under Hager moves. Great, and so let me just remind you of some of the calculations from yesterday. So we computed Hf minus of S3. Great, and we also computed this from this Junus 1 Hager diagram. We also computed Hf minus of S2 cross S1. We computed this to be two copies that I have for joining you. OK, can people hear me in the back? Great, and we computed this for some technical reasons that I kind of didn't say too much about. We had to make sure that there were intersection points here. And then if you want to label our generators, let's say this is generated by, I'll call it theta plus and theta minus. Theta plus will be the one that's in higher degree. So remember the differential lowers degree by 1. So this one here is theta plus, and then this one here is theta minus. So I'm going to erase Paul's announcement about the photo. I'm going to erase his wonderful little person running away, so don't run away after this. All right, and maybe this will give you one more example. Hf minus of negative 1 half surgery on the right-hand Hf well. This looks like after joining you plus two copies of F. This is in grading 0, and these are both in grading 0. OK, maybe you're looking at these here saying, oh, well, what are the options for what Hf minus over three manifolds could look like? And then you stop and think for a minute, and you say, ah, OK, well, this is going to be a finitely generated module over the PID after joining you. And then you remember, well, finitely generated modules over PID are classified right there, a direct sum of three parts plus torsion parts. So, well, it's a PID. The degree of U is minus 2, right? So, well, yes. The question was, shouldn't the second example be the whole Hager flow homology and not for just one of the spin C structures? You would think so based on what I've said, but I haven't said everything. So there's some, like, I sort of waved my hands and said, you had to isotope this to get intersection points to make it admissible. And then for Hf minus, it has to be admissible relative to, like, the spin C structure. So for the others, this is admissible with respect to the spin C structure, but it's not admissible with respect to the other spin C structures. So you actually have to, like, isotope it more. Yeah, so, great question. Other questions, it's graded. The degree of U is minus 2. And Hager flow homology gives you out a graded module. So that means, well, OK, what are the only homogeneous, what are the only homogeneously graded polynomials? Well, since U is degree minus 2, the only homogeneously graded polynomials are monomials. So in particular, right, so any finally generated module looks like a direct sum of three parts plus a direct sum of torsion parts, where the torsion parts, well, it has to look like f of join U mod U to the ni for some natural number ni. And so f is just f of join U mod U. That's how that fits into this picture. What else? Well, we know that if y is a rational homology sphere, as you all know, tell us a little more. They, in fact, show that you have exactly one free sum and so if y is a rational homology sphere, hf minus looks like a single free sum and plus some torsion parts. And so what do I want to tell you about today? So today, I want to tell you about cobaltism maps. Great, so let's say I have a four-manifold w. That's a cobaltism from y0 to y1. So that means the boundary of w is minus y0, this joint union, y1. Let's also pick a spin C structure on w. So these are in one-to-one correspondence with h upper 2. And then this will give us a map from hf minus of y0. And see, if you have a spin C structure in the four-manifold, you can restrict it to the boundary, did to y0 to hf minus of y1 with the spin C structure, t restricted to y1. And so I want to tell you how these maps are defined. Well, again, for the three-manifold invariant, we described our three-manifolds in a certain way. We described it with a Hager diagram. And then from that Hager diagram, we built our invariant. So for the cobaltism, well, we'll describe our cobaltism in some certain way. And then from that description, we'll build our maps. So if you've seen some three-manifold topology, you know that, well, a Hager diagram was secretly it was built out of a single zero handle. G1 handles, G2 handles, and then a single three handle. And so the main idea is, well, w can also be built. So we're going to build a cobaltism from y0 to y1. We'll start off with y0 cross i. And then we can attach some 1, 2, and 3 handles to get to y1. So, well, OK, we have 1 handles, which are b1. We're going to thicken it up so that it's four-dimensional. And then this is attached along a boundary of b1 cross b3. Then we have some attached along boundary of b2 cross b2. And then some three handles attached along a boundary of b3. And so the way I'll define this cobaltism map w is, well, I'll define it. I'll say what it is for the one handle, so the two handles, and three handles. And then since any cobaltism can be built out of a composition of those, well, we just compose each of those maps. All right, maybe I'll draw a cartoon picture. So we can start by attaching our 1 handles. Let's say that I have l of them. And then it's an exercise to check that if you attach l1 handles, can someone tell me what three manifold I'll end up with here? Each one handle gives you a copy of s2 cross s1. So if I have l of them, well, I get l copies of s1 cross s2. And now I'll attach m2 handles. And then let's say next letter is n, n three handles. That's numbers three. Great. OK, could someone tell me what three manifold should appear here? Yeah, because the point is, well, this is just the dual argument to here is one way to see that. And if you don't see this, then that's a good exercise for you to think about ahead of the TA session this afternoon. Well, we computed what we did a partial computation of hf minus of s2 cross s1. This is these two copies of f of join u. Moreover, yesterday I stated the kernel formula for you. So you could compute hf minus of the connected sum of a bunch of copies of s2 cross s1 in the connected sum of all these spin c structures. That's also a nice exercise to see what you get. Great. OK, so let me tell you what the one-handle map looks like. This should be a map from hf minus of y naught to hf minus of this thing. So I'll do it for a single one-handle. To do multiple one-handles, you can either iterate or you can also do it all at once. But let's do it for a single one-handle. So this is going to be a map. This is a co-bordism from y naught. Let's say a single one-handle. It's a co-bordism map from y naught to y naught connect sum s2 cross s1. And while this map makes the fw1, that's just the map induced by this co-bordism. And while this just sends, there's some class x here. Can people see right here? Just get sent. OK, well remember, we have this kernel formula that just says, you take this tensor product. Maybe I'll put some spin c structures in here. Well, this is just going to get sent to x tensored with the top generator here. So I'm just telling you that this is what the map is. And then Ajrat and Zaba do a lot of work to tell you that everything works the way you'd want it to. This is just like a declaration. So that's the map associated to a one-handle. The question is, do they actually define this co-bordism map this way? Yes. Yeah, it doesn't depend on all the choices you made. And we're not going to do any of that. But it doesn't depend on any of those choices. Not that I know of maybe someone else knows of a way. This map is going to count how much the curves. The question was, this map, I define doesn't seem to count how much the curves. That is correct, but the two-handle map does. So the question was, no matter what spin c structure you choose on, why not you always land in here? And yes, that sounds right. More questions. Skip to three handles, because that's also easy to do and doesn't count how much the curves. And since three handles, it's similar to the one-handle story. Great. OK, so w3. Let's just do a single three-handle for simplicity. This is going to go from y1 connect some s2 cross s1, sorry, to y1. And it's defined in much the same way. So hf minus of y1 connect some s2 cross s1, s connect some s0 to hf minus of y1, s. OK. So in here, again, using the current formula, well, all of these here are going to have x tensor theta minus, where theta minus is this guy, this here. And we'll also have things that look like x tensor theta plus. You could have linear combinations of such things. We'll define it on these. And this map, OK, so here it's basically, this map is just projection onto the theta minus factor. So I'll write it like this. So we did one handle, some three handles. Any question about that before we go on to two handles? For simplicity, let's do the case where we have a single two handle. Yeah. It needs to be a spin C structure on the whole three manifold. The question was, what if it's trivial on s2 cross s1? Yeah, maybe we can talk about that later. I don't want to get too deep into spin C structures. Yeah, so in case it wasn't clear, my mini course is meant to paint a broad picture of how Hager flow homology works. In three and a half hours, I didn't think I could sort of get into the details of everything. Yeah. Yeah, OK, yeah. So the question was, this map doesn't seem so well defined because of how I've sort of chosen this identification via the couldn't formula. Yeah. And again, there's a way to do it. I'm stating it this way just sort of in hopes that it's sort of clear that the idea is to project onto one of the factors coming from s2 cross s1. More questions? On to two handles. Yeah, we'll do a single two handle. I have a two handle. And let's say this is a two handle taking me from y1 to y2. Great, OK. And so this counts some holomorphic curves. This counts holomorphic triangles. OK, so how does it do this? So OK, so we'll have a Hager diagram that describes y1. So particularly well adapted to sort of how this two handle is being attached. So it's a surface, a g tuple of alpha curves, a g tuple of beta curves, and a base point. So this is a Hager diagram for y1 such that two nice things are going to happen. So the first nice thing, a lot of it more in this point because the way I'm stating it is not super precise. Let's remove beta 1. And then this is in some way is going to describe, so let's say this is a single two handle. Let's say this is attached along a knot k in y1. Because a two handle gets attached along a knot. So this is going to describe the knot complement such that beta 1 represents a meridian of the knot. So let me give you an example. And I also need to tell you how I'm getting this three manifold here from this collection of data. Let me draw a Hager diagram for you. Yeah, so the question was, what about the piece in the middle? I described what one handle is doing. I described what three handles do. In the middle, right now I'm describing what two handles do. For simplicity, I'm describing it just for a single two handle. You could either iterate it, or there's also a way to do a bunch of two handles at once. And that's going to be this counter polymorphic triangle in what's called a Hager triple, which I'll describe momentarily. Oh, topologically, what's going on? So you have y not connect some bunch of copies of s2 cross s1. That's with the boundary you're seeing at that point. And then, while you're picking some knots in that three manifold, as for the attaching regions of your two handle. More questions. What I'm drawing right now is, OK, so this is meant to be, it looks very flat, but this is meant to be a genus four surface. So now I need to draw some alpha curves and some beta curves. These are the genera of my surface. I've drawn four alpha curves. And now I'll draw some beta curves. And then this is beta 1. So from this surface, how do I get a knot complement? So on the alpha side, do the usual thing. So to all the alpha i's, plus 0. And then you get a result in s2 boundary. And so fill that with b3. Along the betas, well here I said, so ignore beta 1. So we're missing a beta curve. We're short one beta curve. So don't attach a disk along beta 1. And also, don't fill in with a 3-ball. So attach disks to beta i cross 1, where i goes from 2 to g. Here, i goes from 1 to g. And then do not with anything. So it's an exercise to check that since we're ignoring one of the beta curves, the boundary that's going to be left is going to be a torus. So from this description, so basically you take your Hager diagram. On the alpha side, do everything you're used to doing. On the beta side, ignore beta 1. Glue in thickened disks along the other beta curves. And then stop. And so now this gives you a 3-manifold of torus boundary. And so, well, a 3-manifold of torus boundary. It's a not complement of a some not and some 3-manifold. So that's what I meant here. And then you want beta 1 to represent a meridian. So if you think about this beta 1, beta 1, if you think about it, is going to be a simple closed curve sitting in the boundary of the 3-manifold we just built. So you have a simple closed curve sitting on a torus. Well, that's going to describe a meridian for some not and some for whatever not you had. And for some not and some 3-manifold. Great. OK. And maybe an exercise. For this example here, so convince yourself diagram describes the complement of a cheff oil in S3. And then if you want to get a Goldstein exercise, even figure out which cheff oil it is, the right-handed and the left-handed. And so now we also, OK, so we also want to have a Hager diagram that's going to describe Y2, right. So I had this two-handle cubordism from Y1 to Y2. I started off with this Hager diagram that describes Y1 in this particularly nice way that's sort of well adapted to the not along which we're attaching our two-handle. OK. Well, the way I set things up, Y2 is obtained from Y1 by surgery along K. Well, we sort of can see the complement of K in this diagram in a very nice way. And so now if you want to do surgery on it, we just want to fill in that torus boundary according to the framing of this knot here. OK. So well, great. So the meridian no longer bounds the disc. Well, that's great that a meridian showed up in this diagram in this really nice way. So while a race, your meridian, torus boundary that you have, well, just you can draw. You can always find, the way things are set up, you can always find a curved gamma that's going to be your end-framed longitude that's telling you how to attach your two-handle. So let's define gamma. So it's basically going to look like the beta curves except they're going to replace beta 1 with some other curve gamma that's the same longitude. So we have gamma 1. And then we'll keep the betas. So technical reasons will isotope them slightly. That's what the primes mean. OK, so this is going to be where gamma 1 describes the framing of the two-handle. And beta i prime is a small translate of beta i. So now we have, maybe in this example here, I'll draw an example of gamma of this gamma 1. I'm drawing this in green. I don't know if it looks green to anyone in the audience. And then the other, so this is gamma 1. This is green. And then the other gammas are just basically the other blue curves here. But I'm not going to draw a transfer to them because the diagram will get terrible looking. So the point is we have a single surface. We have alpha curves, beta curves, and gamma curves. Oh, right. So gamma 1, great. The question was, how do we determine what gamma 1 is? So according to this recipe, you get a three-manifold whose boundary was a torus. And gamma 1 is determined by the framing of the two-handle. So I haven't really said the details. I guarantee that this curve is always going to live on the Hager surface and not go across some of the beta handles. But it's true. You can always draw this framing for the not on the Hager surface. And that's going to be your gamma 1. Other questions? So we have a surface. We have this alpha curve that stayed the same throughout. We have this beta curves where beta 1 was a meridian. And then we have these gamma curves that are basically beta. But we traded out the meridian of our not for this same longitude that's telling us how to attach our two-handle. So from this data, we can get out actually three different Hager diagrams. By H alpha beta, that's just going to be sigma worth using the alpha curves. We can get H alpha gamma that's using the alpha and gamma curves. And we can get out H what's left beta gamma. OK, well, this one, this is what we started with. This describes Y1. This one, since Y2 is obtained by surgery on this not K and Y1, well, this one describes Y2. Great. And I'll leave it as an exercise for you to figure out what I'll just tell you. This one is going to describe a connect sum of G minus 1 copies of S2 plus S1. So maybe an exercise is to convince yourself of this. Maybe as a hint for this S2 plus S1 bit, remember yesterday, someone asked, OK, well, if you have a Hager diagram where the beta curves are all just equal to the alpha curves, what is that going to describe? Now what happens? OK, so this data, we're going to call it a Hager triple because we have three sets of curves, alpha, beta, gamma. Which this one? Yeah, the question was, the last one doesn't depend on the framing that you had on the knot. And the answer is no. This framing is coming from a two-handle attachment. So it's always going to be integral. So in particular, if I had kept beta 1 here, I'll isotope it slightly. So in particular, since it was always an integrally framed two-handle, the beta 1 and gamma 1 are always going to intersect once. That sort of comes from the setup. And then all the other curves are just, all the other beta and gamma curves are isotopic to one another. And that's enough to determine that. Great question. Other questions? So we have this thing called a Hager triple. We kept our base point everywhere. So now we'll define this map F alpha, beta, gamma from, so it's going to go from Cf minus of H alpha, beta to Cf minus, oh, no. There's a tensor with Cf minus of H alpha, gamma. This should be beta, gamma to Cf minus of H alpha, gamma. Great. And this map counts holomorphic triangles. Great. So this map is defined. So it's going to take in the tensor product of two generators. And then, OK. So then you sum over all z that are generators here. So generators here are intersection points between t alpha and t gamma. OK. And then we're going to sum over psi in pi 2 of x, y, z. So this is pi 2 of x, y, z is homotopy class's triangles connecting x, y, and z. There's some index requirement on the triangles. They should have index 0. And then you're counting points in the modular space of such triangles. And then, as usual, you count intersection points at the base point with powers of u. And then you get a z here. So basically, it's kind of like the disks from yesterday, but now there's triangles. So they go from three things. So what do these triangles have to look like? I'm going to use colors that may or may not be visible to the audience. This is red. This is blue. This is y. Y is beta gamma. And z is alpha gamma. Alphas are always red, betas are always blue, and gammas are going to be green. So it's the same sort of deal as where you're like looking for certain triangles where this is the source and it has to map in such that the decorations match up so that they should go to an alpha curve. They should go to a beta curve. And they should go to a gamma curve. And maybe for technical reasons, you should also split this along spin C structures. So the small print is that for technical reasons, yeah, you should split along. You should split this along spin C structures. So let me give you an example. The question is, should it be a hat over my modular space? And the answer is no. So remember yesterday, we looked for index one disks and we modded out by this r action. Here we're looking for index zero things. So this is already going to be a zero dimensional modular space. So this really is counting points already. And it has to do with sort of that your disk had this translation by r, but the triangle doesn't. Let's draw an example. It's going to be a genus one example. So this is a torus. Opposite sides are identified. Here's my alpha curve. Here's my beta curve. And then here's my gamma curve. So great. So x is going to be an intersection point between alpha and beta. So there's only one choice for x. Y is an intersection point between beta and gamma. So there's only one choice for y. Great. OK. And then while we're looking for some triangles, where the colors match. So well, here's a triangle where the decorations all match. So let's call this z1. This is on a torus. So there's also a bigger triangle that if you kept going here and then came down here like this. So z2. So this is saying that in this example, well, f alpha beta gamma of x tensor y, it's z1 plus z2. Oh, there's a base point somewhere. I'll put my base point out here. I'm sort of not saying much about spin-sea structures. So it turns out that these two triangles actually live in different spin-sea structures. The two triangles live in different. And so can you speak up? Oh, but the question is about the spin-sea structures. Yeah, I sort of don't want to get into that. That's the theme. Sorry. Other questions? Yeah, great. Yeah, I mean, for three manifolds, you can think about them as non-vanishing vector fields up to a certain equivalence. So define this map, f alpha beta gamma. And so now that's basically the key ingredient needed to define this two-handle map. Are there other triangles, but those triangles don't have mu equal to 0, like here? Maybe I'll put my r. Yes. Yes, yes, yes. Oh, no. I didn't find all the triangles. Thanks. The question was, are there other triangles, but they don't have mu equal to 0? And the answer is there are other triangles that I was sloppy when I was comparing my notes. So for genus 1, you want triangles that look nice like this. Also, these two triangles live in different spin-sea structures. And plus, maybe the point of this example was to show you what a triangle looks like. But I was careless and didn't find them all. So the question, what would happen if I chose beta 2? Yeah, so the point was that this diagram, we started with this diagram for y1, where beta 1 was a meridian. If you chose some other curve, well, that's going to be some meridian of some probably different knot in y1. So you could do this with any of the beta curves. And it's going to be some two-handle attachment. So if you do it for a different beta, will the maps be related? In general, if you chose a different beta, in general, it's going to be a different knot in y. So in general, it's going to be a different map, because it's some different two-handle. So why would beta 1 to be just like this? No. So it was drawn in that way, because if you sort of, the diagram that I drew, you can see that as like a complement of a knot in S3 in a particularly nice way. So it was particularly clear that that's a meridian for that knot. Any beta curve in any Hager diagram is going to be the meridian of some knot in that B manifold. More questions? Yeah. Yes. The question was, is there sort of a nice way to count triangles in a genus 1 diagram? And I want to say just sort of if they just sort of look like. As Josh said, if you go to the universal cover, you just sort of look for like triangles. Just like look like this one. Like the corner should all be a keel. I think that should be it. So to find this map from here to here, but the two-handle map should go from y1 to y2. So we're almost there. So my two-handle board is going to be called w2, I think, way back when. So this two-handle map from hf minus of y1 to hf minus of y2, well, it's just defined by it sends to f alpha beta gamma of x tensored with, well, remember, beta gamma describes connect sum of s2 cross s1s. So if you worked out with the Cunis formula, what that looks like, there's going to be a unique top graded element in there. So take this tensored with theta plus. So if this is a single copy of s2 cross s1, then this is the exact same theta plus that we had earlier. And then if it's a bunch of copies, then it's just the unique top graded. So we've described the maps for one-handles, two-handles, and three-handles. So you have a keyboardism built out of those. And it's just going to be the composition. Ashraf and Sabra did a lot of work to show that this is well-defined, and in particular doesn't depend on any of the choices that we made. Maybe you're curious about the gradings, how this map is graded. Very roughly, this map splits along spin-C structures along your co-boardism. And within each spin-C structure, there's a well-defined grading shift that's determined by some topological data associated to your four-manifold and your choice of spin-C structure. The question was about doing this in practice. I actually don't usually do this in practice from the definition. Maybe there's some people who do. But there's lots of nice tricks to compute this without actually counting holomorphic triangles. It's nice to know the counts exist and that they make everything work, but I don't actually do it that way. Other questions? Great. So the goal of this whole discussion was to give you an idea of how the co-boardism maps are defined. But maybe from my point of view, my goal is to use this tool to study low-dimensional anthropology. For example, one thing I care about is something called homology-cobortism. And so how can these co-boardism maps be used to study homology-cobortism? And so the main idea, and so if I lost you with the counts of holomorphic triangles, maybe what you should take from the past 50 minutes of this discussion is that there's a way to define these co-boardism maps on Higgins for homology. And it's well-defined and it works. But now, for applications, well, maybe one main idea for me is that nice co-boardisms induce nice maps on Higgins for homology. So there's many particular nice co-boardisms out there. One particularly nice kind of co-boardism that I like is something called a homology-cobortism. So a definition. We say that a w. OK, so my four-manifold is going to be smooth and compact. Four-manifold w is homology-cobortism from y0 to y1. OK, so it should be a co-boardism from y0 to y1. So the boundary should be minus y0, district union y1. That's the co-boardism part. And then the homology part is that w should look like a product on the level of homology. So one way of saying that is that the map induced by inclusion from the homology of one of the boundary components to the homology of w, this should be an isomorphism. And then we say we write this. I'll just denote it with a tilde. So in particular, this is an equivalence relation on three-manifolds. That's a nice exercise to check. What are some examples of homology-cobortisms? Well, I said the homology-cobortism looks like a product on the level of homology. So one example is, well, w equals y cross i is a homology-cobortism from y to itself. And maybe another example is that if y is an integer homology sphere, w, if you take y minus a three ball and cross it with i and then remove a four ball, this is a homology-cobortism. Well, this part, you can check the boundary. It's going to be y connects some minus y. And this part, the boundary is going to be s3. So it's a homology-cobortism from y connects some minus y to s3. Yes, a picture of this. I can draw a cartoon of this. So here's y. Here's, I took out a three ball. Now I cross it with i. So this is going to be minus y. This part just realizes the connected sum. So this gives us y connects some minus y. That's coming from this part. And then you just remove a four ball from that. And then the boundary that's left is going to be s3. And so it's an exercise to check that since y was a homology sphere, this thing is going to be a homology. I'm going to take to be a nice-cobortism. And then Ashwath and Zaba tell us that this induces a nice map. So proposition one due to Ashwath and Zaba. So let w, a homology-cobortism, for simplicity, let, no, I won't say for simplicity. Great. And then let's choose a spin-c structure on w. The map induced by this cubortism from hf minus of y0, t restricted to y0 to hf minus y1, t restricted to y1. This induces a graded isomorphism on hf infinity, which I defined yesterday. And so the consequence of this proposition that I want to tell you about says something for rational homology spheres. So let's let y be a rational homology sphere. Well, remember, y is a rational homology sphere. At the beginning of today, I told you that hf minus always looks like, right, so there's going to be some sweep part and some grading. And then there's some torsion stuff. OK, and so now the exercise for you is to use proposition one over here to show this rational number here, right, the gradings take values in the rationals, that this grade in here, that d, I like this, it depends on y and our spin-c structure. So I'll write it d of yf is an invariant of homology-cobortism. So there's a few things that are wrapped up in this exercise. So I guess you should also convince yourself that this is actually well-defined, right, because I've chosen a split in here. So you should check this is actually this thing that I've called d, that I've claimed only depends on y and s. Your claims is actually well-defined. And then you should use this fact to conclude that, well, it's invariant under homology-cobortism. And so that relies on these cubitism maps, so u-equivariant. So that's the exercise. Great. This will commence with the question part, so you can clap and then I can answer more questions. Can I remind you what hf infinity is? Yeah, yeah, so hf infinity, this is just take hf minus and tensor it with invert u, right. So a good exercise in algebra would be, well, if you invert u, right, think about what happens to the free part and what happens to torsion stuff. Oh, yeah, so there's a question about this example, right. So when you delete this ball, it's not y anymore. So, okay, so ignore this part, but when you delete b3 from y and then you take this product, well, if you think about what the boundary of that product is going to be, right, well, you had y. You took out a free ball with it, but there's sort of this tube here that connects it to this other free ball in minus y. And so, but you should think about that, well, it's basically a connected sum, right. A connected sum is you're like, yeah, yeah. Oh, yeah, so the question was, can this, like, anything that's of this form, can that be realized as hf minus of sum three manifold? And that's, yeah, so the answer is no. So for example, oh yeah, so there's also a grating here. So if you just have like a single f here, this is just a single f. It actually has to be in a grating that's close to d. So if you put, I added, the question, yeah, the question, I added a grating to your question, so you can't get everything, but it's open sort of like exactly which things you can get. Yeah, the question was, is there a simplest example of a Vashley homology component three manifold so the d invariants differ? So the d invariants are actually invariants of spin C Vash, sorry, what am I saying? I don't think I can give you a good answer, I mean, I'll think about it and say something later.