 Okay, so the first take and move is isotope, right? So you can, yeah, you can sort of wiggle around your curves. So we're gonna be interested in isotope through this joint simple closed curves. And then we also wanna make sure that we miss the base point. So for example, I suppose this was your set of attaching circles, well, and I should have a base point somewhere. Well, you're allowed to sort of wiggle them around. And so if you notice, well, these curves bounded discs are a handle body. Well, these ones will also still bound discs. The second move is a handle slide. Great, so this is probably easiest to describe the picture. Great, so let's say these are my attaching circles. So let's say we had gamma one, gamma two and gamma three is with a new set of attaching circles. Guess I'll go over here. Great, so what, oh, there's a base point somewhere. Great, and so what do I wanna replace them by? Okay, so let's say that we're gonna slide alpha one over gamma one over gamma two. So what should you do? So you can think of this as, well, choose an arc from gamma one to gamma two that misses your base point. So say here's an arc. Okay, so throughout this gamma three years is gonna stay over here. And gamma two is also just gonna stay like this. And we're gonna replace gamma one with a new curve gamma one time. And what are we gonna do? Okay, so here's, great. Okay, so first, I like to think of this as first, we sort of isotope gamma one along this arc. So it gets really close to gamma two. And then once you get really close to gamma two, well, take the connect sum with gamma one and a parallel copy of gamma two. So now this is gamma one time. Great, so this is a handle slide that we started with gamma one through gamma G and then we ended up with gamma one time, gamma two. And maybe you can convince yourself that, well, if these are a set of attaching circles to the handle body, well, these are also a set of attaching circles. So I guess the work is to convince yourself that gamma one time is gonna bounded disc. Oh, right, so, right. So we're starting with a set of attaching circles for some handle body. So it's saying that you have the handle body. So it's saying that this surface here bound some handle body and these three curves bound discs in that handle body. And then the claim that I'm making is that, well, if these three curves bounded a disc in that handle body, well, these three curves also do. Oh, right, so this handle body is sort of hard for you to see, but so this surface here is the boundary of some handle body and then in that handle body, this curve bounds a disc. Other questions? I'll say that again. Oh, yeah, the orientation of the curves doesn't matter, right? Nothing we've done sort of depends on the orientation of curves. We're just using the curves to sort of, yeah, attach discs along them. Okay, so we have these two moves, isotope and handle slide. Okay, and then the third move is called stabilization. So let's let sigma alpha beta be a Hager diagram. Great, so stabilization. Well, we're just gonna take the connected sum with something that's homeomorphic to this picture. So here is a torus and then here is a beta and here is an alpha, so an example. So maybe this is the Hager diagram that we're starting with. Well, then if we stabilize it, we'll just take the connected sum with that picture over there and we end up with, so this is maybe beta one, this is beta two. Here's alpha one and alpha two, great. And then there's also the opposite of stabilization, destabilization, and that's just the reverse. So going this way, stabilization, and going this way, it's destabilization. And so, great, so this is just, right, this is S3 and if you take the connected sum of two Hager diagrams, well, you can convince yourself that gives the connected sum of the two three manifolds, so connect summing with S3, so we're not really changing our three manifolds. And that's good because of the next theorem that I wanna state for you. So I'll attribute this theorem to Reitermeister and Singer, even though maybe the way that I'm stating it is not the way they originally stated, the way that I'm stating it is, I guess, due to Ajvat and Sabo. Suppose I have two Hager diagrams for the same three manifolds. So I have H and H prime, so let these be two Hager diagrams for the same three manifolds. Well, and the statement is that they're related by a sequence of these three Hager moves. So isotope handle slides and stabilizations and destabilizations, so then H and H prime, are related by a sequence of isotopes, handle slides, and stabilizations, destabilizations. Right, so you should think of this as analogous to Reitermeister moves, right? If you have two projections of the same knot, then there's a sequence of Reitermeister moves taking you from one projection to the other. And this, well, if you have two Hager diagrams representing the same three manifold, there's a sequence of Hager moves taking you from one diagram to the other. That's not quite true. You might have to stabilize both of them, but there's a sequence of moves taking you from one Hager diagram to the other, right? And so in not very why are Reitermeister moves great? Well, they're great because if you can define, you know, maybe define a knot invariant in terms of a projection, and you want to prove that it didn't depend on your choice of projection, will you just show that your invariant is unchanged under Reitermeister moves? So in the same way, well, you can define a three manifold invariant in terms of a Hager diagram, and if you show that our invariant is unchanged under Hager moves, then we know that it's actually invariant of the three manifold and doesn't depend on our choice of Hager diagram. That is true. I just am confused with what you mean by saying alpha goes to, or is that determined? Isotopia classes? Yeah, I guess I'm not sure. It may probably does, so you don't want to take the alpha curves to the beta curves because, but the alpha curves bound discs on one side and the beta curves bound discs on the other side, to take the alpha curves to the beta curves, I think you're just gonna always get a connect sum of S1 across S2s. Yeah, but maybe we can talk about this later. So we talked about Hager diagrams for three manifolds. We also want to be able to define a knot invariant, and we want a knot invariant to sort of be similar to the three manifold invariants, and maybe we want to describe it via some sort of souped up Hager diagram, so we want a way to be able to describe a knot in a three manifold. And for simplicity, we're gonna stick with knots in S3, even though the setup works more generally. So I wanna talk about doubly pointed Hager diagrams. Okay, so we're gonna have, let K be a knot in S3. Now I'm gonna define for you a doubly pointed Hager diagram for K. A doubly pointed Hager diagram for K is great, so it is a five tuple. We have a surface sigma, alpha curves, beta curves, and now two base points, which we'll call W and Z. So what are we gonna require on this five tuple? Well, alpha beta is gonna be a Hager diagram for S3. These base points are gonna be encoding or not. So K is the union of arcs A and B, where let me tell you what A and B are. So A is an arc in sigma minus alpha, connecting W to Z. But so right now, this is an arc in the surface. And now let's push this slightly into the handle body, H1. Remember, the alphas are a set of attaching circles for H1. And now beta is going to be an arc in sigma minus the beta curves connecting Z to W, and then we're gonna push it slightly into H2. So B is an arc in sigma minus beta, connecting Z to W, pushed slightly into H2. Remember, the betas are a set of attaching circles for H2. So let me draw an example. So I'm gonna draw for you a Hager diagram for S3. So this is S3, right? There's just an isotope taking us back to that standard genus one Hager diagram for S3. And now I'll put a base point W right here and a base point Z right here. And I'm claiming this is specifying a knot in S3. Well, we can get the knot by following this recipe. So maybe let's start here. So let's take an arc in sigma minus the beta circle connecting W to Z. Great, so if I wanna, maybe I'll go around here. So look something like this. And now I wanna push it a little bit into H2. So maybe I'll push it a little bit sort of up off of the surface. And now I wanna connect W to Z, missing alpha. And then I wanna push that slightly sort of inside. Great, so this is, I'm gonna push this slightly inside. So sort of this is gonna go through here and like this. Great, so this is a knot in S3. And I guess since I've pushed these, since I've pushed the arc A slightly into H1 and the arc B slightly into H2. Well, this circle that I got, it intersects our Hager surface in exactly two points, Z and W. And you can check, right, this, if you sort of forget the Hager diagram and just look at this orange knot, this should be the left-handed trough oil. So the point is that just by adding this extra base point Z, well now this diagram actually describes the knot in our three manifold. The knot you obtain exactly by this recipe on these two boards. Great, okay, and just like any two Hager diagrams for the same three manifold were related by a sequence of Hager moves. Well, any two doubly pointed Hager diagrams for the same knot are also related by a sequence of Hager moves. We're now, now we have two base points. So our isotopes have to miss both base points and the handle slides also have to be away from both base points. But so those are the extra additional restrictions and then any two doubly pointed Hager diagrams for the same knot are related by a sequence of these doubly pointed Hager moves. Great. So one thing you might be wondering, you might be wondering, well, can every knot in S3 be described by a doubly pointed Hager diagram? And so the answer to that is yes. And so I'll prove that through you by describing a recipe that takes a diagram of a knot and then spits out a doubly pointed Hager diagram for that knot in S3. So let's start with this knot here. So we'll start with this projection of the trough oil. Great. If you forget the crossings in this diagram, well, then you're gonna get a graph and then you can take a, let's, our surface is gonna be a boundary of a tubular neighborhood of that graph. So our surface will look like this, right? So if you forget the crossings, well, you get this graph and then just take the boundary of a tubular neighborhood of this graph. Your alphas and betas are going to come from. Well, your betas, you have these bounded regions and so around each of these bounded regions we'll put a beta circle. So we'll get beta circles here, here, and here. Alpha circles are going to come from. Well, each time we see a crossing like say this, we're gonna get a picture. We're gonna put in an alpha circle that looks like this, sort of locally, right? So at this crossing we want an alpha circle like this. At this crossing we want a circle like this. And then at this crossing we want one like this. And then, well, we have this white surface here, we have a genus four surface. I've described for you three alpha circles. And so now we'll just put a fourth one that's just a meridian. And then we'll put two base points on either side. So coming from the orientation. It's an exercise for you to check that this is indeed a doubly pointed Hager diagram for this non-S3, right? So maybe I'll sort of describe in words how one might think of it. So on the outside you have these beta circles that are gonna bound discs. And then when we fill in sort of the outside with a three ball, we'll sort of just left it sort of the inside of the surface, what's inside the surface. And now each crossing we're sort of gluing in this little, this little Pringle shaped disc. And now if you try to connect, if you try to draw the arc A described on that left hand board connecting W to Z, missing the alpha circles, well you find that sort of around each of here is sort of forced to wrap around in a way that's exactly compatible with the crossing and the diagram. And then I guess you should also check that this is indeed a Hager diagram for S3. But it is, right? So this gives you a recipe given not diagram and a way to build a doubly pointed Hager diagram compatible with that, not. So I'll stop there. That's right. So the arc B that was in sigma minus beta, right? So I guess I drew them both in the same color but that's this part, right? That was the arc B. So that should be pushed slightly into H2, which I sort of think of as like the outside handle body. And then the remainder part, this part winding around here, well that was the arc A and that should be pushed slightly into the inside handle body H1. No, so yeah, so both A and B should be embedded arcs. Nothing does. So this proof I gave you was specific to sort of a diagram for a not in S3. Yeah, so another way to prove that every three manifold has a Hager splitting is using Morse theory and then you can use sort of a Morse theoretic proof to prove that any not in a three manifold has a doubly pointed Hager diagram. You were asking about the point? Oh, that's right, yeah. I mean a single pointed Hager diagram is a base three manifold, point two three manifold.