 It's this that you're supposed to get right. What the co-bordism is that goes between these guys in, you notice that this picture has actually a three-fold symmetry. So if you take the tetrahedron and draw one picture, and you rotate it, the three clicks, you get each of the other pictures. So that kind of fixes, you know, the importance of the sense of the rotation, and then what's interesting is understanding the composites of these guys. So to say, you know, so let's define i-sharp of k for k and not in the three-sphere, i-sharp of k is i-sharp of s3 and k, and then the claim is that there's an exact triangle, and these guys are induced by these co-bordisms. This exact triangle is exactly analogous to the Havanoff exact triangle. Again, there's a difference in convention, so the relationship's not going to be between the Havanoff homology of the knot and the Havanoff homology of... an instant-on homology, but rather with the Havanoff homology. The mirror will be related to the Havanoff homology of the knot. Or, you know, or I could decide to take co-homology of something. Or is it Havanoff co-homology? Or is it Havanoff homology? Or Havanoff co-homology? Anyway, you know, okay, great. So the interesting thing is understanding how these guys compose, and you know, all I'm doing is putting a certain kind of twisted one handle that goes between these guys each time. And it turns out to be interesting. Let me try and draw. Schematically, what kind of describes the combinatorics are one of these guys. Actually, let me... This is the first one. It was red. And then that one's going to blue. And then when I go again... So this is supposed to describe how, you know, there's this pair of red edges, a pair of blue edges. And with that band, the twisting kind of means that there's this little extra twist in here. Right, this is a one-handle attached. And if I think of the knot times I, this is attaching a one-handle. This is where the core of the one-handle attaches, and this is kind of where the co-core attaches. And if I... So this is sort of a, you know... Yeah. So I can do this again. It's green. These guys go in the same sense. So this is what the composite of two of these guys looks like. And the thing to notice that we've constructed a mobius band when we do that. So that composite co-bordism, if I look at just thinking about the surfaces, if I look at sigma 21 composed with sigma 10, this guy, it has, you know, a mobius band has, of course, only one boundary. So I can think of taking it out. Then I have something that has boundary, just one circle boundary, and I replace it with a disc. So topologically, this thing turns out to be, actually, it turns out to be sigma 20 so backwards connect some RP2. Okay? I mean... You're going to check that. And... let me spend two seconds about RP2s. So this is, you know, I've done my first guy, this is sigma 21, sigma 10, I can't draw that much better than that, but then this composite, somehow there's an RP2 in here. And this is happening in R times S3. Now, you might complain for a moment, but are non-orientable surfaces okay? Well, that's the beauty of the particular holonomy that we're looking at. The I and minus I are conjugate inside SU2, so it's fine. I don't have to tell you the difference. I don't have to specify the direction in which I take the holonomy. We're looking at connections mod conjugation, so it doesn't matter. And so it's non-orientable surfaces are fine for this story. They wouldn't be fine if I had a different holonomy than I'd have to know which way I'm going. So non-orientable surfaces are okay. And RP2s, let's think about them in S4, they're already quite interesting. So the first observation is that non-orientable surfaces have a well-defined kind of self-intersection number. So you'd think, if you have an orientable surface inside S4, it's normal bundles necessarily trivial. The normal bundle tells you what the self-intersection number of the surface is, when it's orientable, the self-intersection number of the surface is necessarily zero because it's homologically trivial. But it's not true for non-orientable surfaces, and what you can... In fact, so what's the definition of the self-intersection number? If I have a non-orientable surface, non-orientable or not, a common definition is that I just... I make the surface transverse to itself, it meets at finally many points, then I pick an orientation, local orientation for the surface at each of the intersection points. I use the push-off to orient the other guy. Then I compute that self-intersection number. Count up with sign, that number. That's an integer. Of course, if it's orientable, I'm free to choose the same orientation everywhere I did. I want to. Everywhere I made a choice, I could just pick the orientation that the surface gives me. But if it's non-orientable, it turns out that's well-defined, it can be non-zero even if the surface is non-homologous. And for example... What am I doing? Very badly. Okay. Yeah, so... Let me say it. Yeah. So just to see that there... I'll show you... Here are two interesting non-orientable surfaces in S4. So it's a beautiful fact that it's observed by several mathematicians from the... several of the greats from the... from the 70s, Arnold, Kuiper, Massey, a few other people. It's observed independently that if you take a CP2 mod conjugation, that's actually, in a natural way, diffeomorphic to S4. You have to be a little bit careful about, you know... be a little careful about that. Okay. It's a model for the force sphere. Inside CP2, there are two very different interesting... Well, sorry. So there's an obvious interesting RP2, the fixed points at the real points, the real RP2 and CP2. Now, the real RP2 and CP2... Well, actually, that's already interesting. It has a non-zero self-intersection number. The normal bundle, that RP2 is Lagrangian. So the normal bundle is isomorphic to the cotangent bundle. The Euler class of the cotangent bundle with twisted coefficients is minus the Euler characteristic of... of RP2. So the Euler characteristic of RP2 is 1, so minus the Euler characteristic is minus 1. And now, if I push that down, I'm twisting the normal bundle, sort of squaring it. So the real RP2 inside here has self-intersection minus 2. Now, I'll give you another RP2. I take a real quadratic, real degree... homogeneous polynomial of degree 2 with real coefficients that has no real points. That's a two-sphere, a conic curve in CP2, which misses RP2 because it has real coefficients. It's invariant under conjugation. Conjugation is acting by the antipodal map on that RP2. Upstairs, it has self-intersection 4. It's a degree 2 curve. It has self-intersection 4. When I push it down, it has self-intersection plus 2. So there's an RP2 plus and an RP2 minus. So self-intersection plus 2 and minus 2 inside here. Okay. And there are a lot of fun ways to describe it. Some of them are described in the notes. And, yeah. So it turns out that this is RP2 plus. That's funny the real answer to your question. You got to make sure that it's RP2 plus. You can do this either way. One will give you RP2 minus, one will give you RP2 plus. You better get RP2 plus. And the beautiful thing is that now there's a new game that we can play. Where's my, never mind, eraser. So here's this, you know, this co-bordism. Sigma 2, 1, sigma 1, 0. And then I see this, there's this RP2 plus in here. Now what I want to think about doing is let's take this RP2 plus and kind of exhibit its connect sum geometrically. That is a connect sum. So what I'm going to do is think here is the co-bordism sigma 0, 2 backwards. So it goes from 2 to 0 as this one does inside here. That's the surface sigma. And then I want to, here's S4 containing RP2 plus and I tube them together. Now there's a natural one parameter family of metrics that goes from this picture to this picture. And one of the things that, you know, one of the bits of homological algebra that you teach yourself from thinking about modular spaces is that whenever you move things in a family, then the map, so these two will induce the same map on homology, but in fact the, if you look at the one parameter family of modular spaces that you look at, that you get from this family, look at its boundary, you're constructing it. Look at the one parameter family, count the number of points in it that gives you a chain homotopy between these guys. So these maps are actually chain homotopic. So, and it turns out that if you, when you study this a little more carefully, you realize that this guy, when you pull it off, it has to suck off some non-zero amount of energy. That's the difference between rp2 plus and minus. That's a computation that you have to do. So this guy sucks off some energy. What that means is that it doesn't leave behind enough energy to have a solution here. So this guy, when you stretch out infinitely, there's an empty modular space. So the chain map is zero when I use this model. It is whatever it is here. It's chain homotopic to this guy, which is zero. So this map is chain homotopic to zero. That sounds good. So what we've shown is that if you go around here, if you look on homology, each of these composites are zero. Now that one has a, so, let me say this. The fundamental group of the complement of rp2 minus, or either rp2 is z2. In one case, in the rp2 plus case, that guy is actually obstructed. In the rp2 minus case, it's not obstructed, and this map is, the composite in that case would be equal to this map. So if you change all the orientations, which are not the ones we want, this composite turns out to be equal to the composite with the opposite orientation. So it's not something that you see, the distinction is not something that you see on the level of representation spaces, it's something where you have to think of them as modular spaces, understand their deformation theory. As far as it goes, it's not hard stuff, but it's stuff. There they are. Okay, great. Now we want to see that there's an exact triangle. So what we just saw is that on the chain level, sigma12 star composed with sigma, sorry, chain homotopic to zero. In fact, let's remember the chain homotopy, give it a name, so that's a map from one to zero. Sorry, I'm going from two to zero. Okay, so we want to see if actually we're getting an exact triangle. So it's a nice little bit of homological algebra that you can figure out eventually. So abstract situation, sorry, to one, zero, and I have, I'm going to call it, well, sigma, let's call it sigma21, sigma, I'm just going to index it by where it starts. So I have chain maps, these guys, and what else do I have? Well, this composite is chain homotopic to zero. So there's a map this way, there's a K2 this way, this picture is completely symmetric, so there's a K0 this way and a K1 that way, which are the chain homotopies. So if I look at sigma i plus one composed with sigma i, sorry, sigma i composed with sigma i minus one in this convention, that's equal to di Ki plus di plus one Ki. Everything's mod two. I'm not worrying about signs here. So I have this guy, and I want to know if this is an exact triangle. The way you can tell is there are interesting maps at each of these vertices. So you consider, for example, let's look at this guy. So what can we do? We can do sigma zero and then K2 and the other thing that we can do is do K0 and sigma one. And if this is, so consider these guys plus cyclic permutations, thinking of the indices as being mod three. So if this is, so what you need, if these are all chain homotopy equivalences, then that implies that C0 is chain homotopy equivalent to the mapping cone of, C0 is to the mapping cone of sigma two. So plus cyclic. So if you look at everything that you get out of what you just learned, you get some interesting new maps to think about, and the thing you have to check is that those new maps are chain homotopy equivalences. And then you learn that each of these, that each corner is chain homotopy equivalent to the opposite, the mapping corner to the opposite face. Which, anyway, so great. Well, anyway, the thing that you, I'm going to draw the picture very schematically to try and indicate what's going on. So here's K2, K1, K0, K2. There's some... Okay, what have I drawn here? If I look at the composite going from K1 to K0, I observe that there's a way of writing it as a connect sum with an RP2 and S4. So this, it was also a knot a little bit earlier. Sorry about that. These are knots in the three sphere. This is the three sphere. It has a knot in it somewhere. Another three sphere with a knot in it. There's a surface inside here, surface inside, surface inside here. And what you see is that there's another three sphere that splits off an RP2. If I go from here to here, there's another three sphere that splits off an RP2. And then what you have to find is that if I go, if I stick these two together, there's yet a third three sphere that doesn't realize it's a little more complicated. It's not a connected sum picture, but what happens is that... So if I look at the composite... Sorry. Sigma 02 composed with sigma... I'm running out of steam. Sorry, let's go from here. Sorry, I look at K21 composed with K01 composed with K202. You know, whoever didn't use reverse polish notation should just be eliminated. Just, anyway, you know, the composite through the opposite direction, whatever. Anyway, you look at this composite, so there's this big surface in here. This thing turns out to look like it's a product with... This is a Klein bottle twice punctured, which is attached to a cylinder, which is R times K2. All right? So this big composite, it's a little more complicated than it connects some of surfaces. There are two boundary components. When I cut with the three sphere, this thing's a Klein bottle, and what you can... But this is starting to show you where the chain homotopy equivalents... I mean, in this case, it'll be... These things will be chain homotopic to the identity, actually. And the identity map is going to come out of this bit. It's a product. And, you know, there's some... Anyway, kind of out of time, there's a beautiful way of constructing a chain homotopy from these maps to the identity map that's induced by the product co-bordism. And that's eventually what allows you to prove that there's an exact triangle. And then, just to say quickly what happens, how you relate to Havana homology, well, once you have one exact triangle, you can take a knot projection, you look at the crossings, you number them, you do make an exact triangle for the first crossing, then you replace the original instanton complex for that guy by a mapping cone. Then each of the... You get two new knots, which are simpler in the Havana... I mean, with fewer crossings, you do pick up the next crossing, the same one in both of them, replace each of those complexes by mapping cones, etc. You get a cube. The cube looks just like the Havana cube, except that... So, I don't know, maybe there's a K11, K00, etc. There are maps along these edges. That's what Havana likes. But if you go through this construction and think about it carefully, you find that the instanton guy can give you other maps, which follow the descending diagonals in the cube. So, what you see is the instanton homology, there's a huge chain complex with lots and lots of maps in it that computes the instanton homology, but it's filtered by the sort of height inside the cube. I mean, starting from this is the lowest point, that's the highest point, there's a filtration. And what you can check is that if you take... So, that gives rise to a spectral sequence, and what you can check is that the E1 page, which only notices... which takes the homology at each vertex, that'll be the homology of the instanton homology or an unlink, which we compute, and then the edge maps are the Havana maps, but they're higher differentials, so eventually you get a spectral sequence, so the instanton homology necessarily has rank. Its rank is a lower bound for the Havana homology, instanton homology detects the Havana homology detects the Havana. There we go. Yes, Josh. Well, you know, I mean, it's a little tricky because it involves the mu map, the point map induced by the point class, so there's sort of a connect sum theorem, but it's the connect sum theorem tricky because it's not just... I mean, just like in Hagar for homology, I mean, the connect sum theorem's not trivial. There's some, you know, you have to take care of the fact that there are operators on each of them which become equal after you take the connected sum, but then you can do that. Yes, Josh. Josh Prime. No, there's a client... I mean, well, I mean, the triple composite still got a client... It's just the same... There's still a client bottle there that doesn't... The orientation doesn't... It's a different, you know... No, but then it's just the connect sum with... Well, if you get rid of the RP2, then you've gotten rid of, you know... No, but...