 This is a homology three-sphere. And it's a beautiful exercise to see what its SE2 representation variety is. So a nice way to think about, we need to understand what's the fundamental group of this guy. Well, a nice way to access it is to observe there's nearly a vibration, if I think of S1 acting on, just as multiplying, sorry, S1 acts on here by multiplying, so e to the i theta z1, z2, z3, is e to the i qr theta z1, e to the i pr theta z2, e to the i pq theta z3. So that preserves the equations. We can take the quotient of this guy by that action. And we get it is really an orbifold two-sphere. So there's an orbifold angle of 2 pi over p, 2 pi over q, 2 pi over r here. And the fundamental group of this orbifold is quite easy to describe. It's a triangle group, so the orbifold fundamental group. Probably. Yeah, I will. I will write larger letters. The triangle group tpqr with generators tu, tu, and v, so that t to the p is equal to u to the q is equal to v to the r is equal to tuv is equal to 1. Now, on the other hand, the fundamental group of. So what this tells you is that the fundamental group of these briskorn spheres is an extension of the triangle group by z. So the fundamental group that we're interested in. So this is a central element. And if I look at representations of this into SU2, so I look at a rho, rho has to send, so this is central, it has to send that guy to plus or minus the identity. In other words, if I push this representation, the representation under study into SU2, if I push it down into SO3, it's just a representation of the triangle group. And what do I want to say about that? So these guys you can understand. So rho descends to rho tilde mapping t, p, q, r to SO3. And so in SO3, so what we're getting is tuv inside SO3 that satisfy these relations. Now, what you can check is that what you can check is that these guys, let me draw the picture here. So these are rotations. So here's the two sphere, not the orbital two sphere, but just the honest two sphere. So t is a rotation about some axis 2 pi k over r and et cetera. So s is a rotation about this axis. Sorry, tu is a rotations. And the constraint that you end up getting is that you get a spherical triangle. If tuv equals 1, you get a spherical triangle. So here that angle is pi k over p. The angle here is pi l over r. The angle here is pi m, q, r. So this product relations tells you, given this, that there's a spherical triangle with these angles. The spherical triangles are rigid. So there's a up to rigid rotation of SU2. They're a finite number of such guys. There's some constraints that come from the possible areas of the possible triangles that you can get. And here you see kind of an amazing coincidence between, I shouldn't have erased that. So the number of such spherical triangles up to rigid motion is equal to minus the signature of the Milner fiber of that singularity. So the Milner fiber, remember we had this picture. There's s5. Here's this equation. If we give this equation a little kick, then we get, well, if we get a little kick that doesn't change too much what happens on the boundary, but we get actually a smooth four-manifold. So if I take this thing, intersect it with b6, that's the Milner fiber. So that's a sort of canonical four-manifold that this three-manifold bounds. And there's a beautiful fact, which eventually has several different proofs. But you're already starting to see that something that kind of trying to get some kind of feeling for what these representations are doing. And already just on the level of a naive count, you see that the count of these representations remembers something interesting about a canonical four-manifold that that thing bounds. That was a link of a complex singularity. And somehow, the count of these representations tells you something subtle about that canonical four-manifold that it bounds. So I want to show you that there are other interesting connections just looking at representations to other interesting problems. So let me see where they are. OK. So let's take a look at representations of not groups. Where is my eraser? So take the same. So let's take, we're going to look at s3 minus k. Look at its fundamental group, k for brevity. And then I'm going to look at representations of the fundamental group of k into su2. But again, I'm not going to look at arbitrary representations. I'm going to look at, so here's my k. Again, we can take a base point and then take a meridian for k. Here's k. Here's a meridian. And what we want to look at is representations into su2 so that rho of meridian, well, I can write it several different ways, so the trace of rho of a meridian is 0, i.e., rho of meridian. Well, that means its square is minus the identity because it's an su2. So another way to say it, here's the identity. Here's minus the identity. This is su2, the 3-sphere. And then there's a 2-sphere, which I'll call s2i, which is the conjugate's, that's the right identity. The conjugates of the quaternion i in su2, so the unit imaginary quaternions, if you like. So what happens if we study these guys? So, well, this is here. Now, let's look at the set of these guys. And the set of these guys is equal to r0 of k. So r0, if I look at the n0, the fundamental group is just z. And I'm just picking one point on this 2-sphere. That's all I'm doing. So it's just a copy of that central 2-sphere. What if I take a more interesting knot? Well, you can figure out what's your, if you think of the Verdinger presentation. So we consider Verdinger generators. So think of loops that come in from infinity. They go looping around a particular component of the projection. So here I have generators x1, x2, and x3 for the trefoil. And there are Verdinger relations, which say something like xi. So the Verdinger relations say that as I pass under this guy that the generator for x2 gets conjugated by x3. So this picture, if I have xi, xj, xk, then up to proper choice of which way these guys are going, xk is the conjugate by xj of xi. That's the Verdinger relation. And you can see, say in the case of the trefoil, so what does that mean? So remember, we're looking at representations where every meridional curve goes to this 2-sphere. So each of these guys labels a point on the 2-sphere. Let's call that the corresponding points capital X1, capital X2, capital X3. What this relation says, so if I look here, so maybe going this way, if I conjugate x1 by x3, let me draw them like this. I'm going to draw them like this so it'll be easier to say what I want to say. So that relation says at that circled crossing, it says that if I conjugate x1 by x3, I get x2. What that means geometrically in this case is that if I look at this great circle, they lie on a great circle, and these distances are the same. So this relation that x3 conjugates this guy, that implies this picture. And if you think about it, in the case of the trefoil, all of these relations are the same. They have the same consequence. So there's two possibilities. The representation either sends all three points to the same point, all three. One fair solution to this equation is that the images of x1, x2, x3 are all equal. Then they all commute, so there's nothing to check. Another solution is that there are three symmetrically distributed points on an equator on a great circle. That's another solution. You can check that the three relations imply the same thing. So the representation space of the trefoil has two components. There's a two sphere, all the same. Union, well, three ordered points on an equator. If you think about it, that's just you can pick where x1 goes, then pick where x1 and x2 go. That tells you where x3 has to be already. And x1 and x2 together, they're points on a great circle, ordered points. So that's a copy of the unit tangent bundle of the two sphere, a copy of SO3, or RP3, let's say. You can keep going with these examples a bit, and you can compute these for a lot of examples. But if you take a 2n torus knot, then this representation variety of a 2n torus knot. So there's two strands which get twisted around n times. Maybe I'll just put an n in here. That this representation variety is a copy of the two sphere. Union, n minus 1 over 2 copies of RP3. OK, so what? Well, in Jake's lecture on Havana homology, he told you what the Havana homology, I think, of these knots were. And as an ungraded gadget, I'm going to compare to Havana. So we notice that the homology of the representation variety of the unknot, the two spheres, so it's z plus c. Forget about the grading. If I look at the homology of r knot of the trefoil, there's a z plus c from here. And the z homology of RP3 is z, z2, 0z. So it's z to the fourth plus z2. And here, the homology of the representation variety of the 2n torus knot, well, it's z to the n plus 1 over 2 plus n minus 1 copies, n minus 1 over 2 copies of z2. And these are all isomorphic to the corresponding Havana homology, again, as an ungraded gadget. So this is a story that cries out for some kind of explanation why these groups coincident. It's far too naive, turns out. I mean, sometimes by being naive, it's great, and you're right. But it turns out to be far too naive that just looking at the homology of the representation variety gives you the Havana homology. But it turns out that once we look at the instanton homology of the term Simon's function, whose critical points are these representations, then we can access Havana homology. The story is a bit subtle. It's not that this actually agrees always with Havana homology, but what we can prove eventually is that there's a spectral sequence that starts with Havana homology and converges to the instanton for homology. And in this way, you can play off results in one theory to learn about the other. Should I stop at three sharp? Or? OK, a couple of minutes. I'll say there's one other interesting thing that you can think about. I mean, there are infinitely many interesting things that you can think about, but one that I might get a chance to talk about a little bit. Yeah, it's not two minutes' worth of stuff, though. Never mind. I'll just stop. Seems like a good moment.