 Yes, okay. All right, well, thanks very much for the invitation. So the very first math conference that I ever went to was the 1997 edition of PCMI, which was more than half my lifetime ago. I was in the undergraduate program at least. But I had an amazing time, so it's really nice to be able to come back and teach. So before I start, maybe let me just ask, did anybody read the preliminary reading that I sent out? Maybe hands up if you did. Okay, now let me, wait. Okay, so how was the level of the preliminary reading? Hands up for easy. Okay, hands up for not so easy. All right, good. Okay, so I'm not gonna talk about anything in the preliminary reading today. So all of those of you who were desperately looking it up on your phone, you can stop. So instead, I'm gonna talk about the Alexander polynomial. And so basically, my plan is that, all right, now I've lost it. How's my voice without this thing? Terrible. All right, so my plan is that I'll talk about the Alexander polynomial today, and tomorrow in lectures two and three, I'll talk about the Havana homology. And then we'll go on to talk about the Humphley polynomial and Humphley homology and colored homologies. But first, let's start basic. So what's a knot? So a knot, okay, is a smooth embedding of S1 into S3. And usually we draw pictures of knots by projecting them on the plane of the blackboard. So here's a knot called the unknot. There's another knot called the trefoil knot. So we'll say two knots are equivalent. So k knot is equivalent to k1. If there's an isotope, let's call it phi, mapping from S1 times zero one. Again, this is a smooth map to S3, such that if I look at kt, which is phi on the time slice t, this is a knot. So for example, this unknot is equivalent to this knot here. But our experience with shoelaces in electrical cables and things like that suggests that it's not equivalent to this trefoil knot, which is a little annoying knot you get in your shoes. OK. So how could we prove that? Well, one way we can think about this is to observe that if k knot and k1 are the same knot, then actually they're complements or homeomorphic. This isn't completely obvious. So if you've never thought about why this is true, you should. So it actually depends, for example, on the fact that I remembered to choose a smooth map here and not a continuous map. But once we know that's true, we can use all our arsenal of algebraic topology invariance to attack this guy. So for example, that means that pi one of the complement k knot is isomorphic to pi one of the complement of k1. OK. So maybe if we want to use this, we ought to know how to compute pi one of the non-complement. So let me tell you how to do that quickly. So a diagram D of k gives me what's called the Fierdinger presentation of pi one of the non-complement. And the way that this works is that for each arc of the diagram D, I get a generator in pi one. So here's my arc. I haven't written down my base point in pi one, which is always a dangerous thing to do, but I'll continue to do it. So really, the base point for this presentation is out at the back of the room. And so the generator associated to this bit of arc comes in from the back of the room. I'll throw an x here to indicate that it's coming in this way. Goes underneath the arc and then goes out by this O, back out to the back of the room. And really, I should have said, how do I know which way I was going here? I chose an orientation. So this was an oriented diagram D. And so for every crossing, I get a relation in this presentation. So for example, if I have a crossing that looks like this, there are three arcs to the diagram here. I'll call them A, B, and C. Then I get the relation that says that C is A inverse B, A, I think. And the way that I check that is, well, so I do A inverse. I come in from infinity. I go underneath here, come out. Then I do B. That's going to carry me around like this. I come out and go back to infinity. And then I go A. So it's going to take me like this. Go underneath here and come out. Now if I join these x's and o's together, you see that I can swing this orange loop underneath so it goes right under C. So that is the right relation. And similarly here, for this diagram, so let's say here, I have arcs A, B, and C. Again, I get a similar relation. Now C is A, B, A inverse. Obviously, I need to figure out how to fix this better. OK. So for example, if I look at this truffle oil, and I'll color the three arcs three different colors, red, white, and blue, so I can see what's going on. So let's call this blue arc A, the red arc B, and the white arc C, orient things this way. Then I get a presentation that looks like pi 1 of S3 minus T is generated by A, B, and C with relations that say that C is A inverse B, A. That's that crossing. So A is B inverse C, B. And B is C inverse A, C. Or if I just use this relation here to eliminate B, I'll find that I've got a presentation with two generators, A, B, A, B inverse, B inverse. And you might think that I should have another relation here, but actually, I find that I get the same relation again. And that's always true. I always have a redundant relation in this presentation. So now I know how to compute pi 1 of a non-compliment. What can I do with it? You see, the problem is that here I have a presentation of a non-Abelian group. And non-Abelian groups are complicated. If I give you two presentations, it's not necessarily easy to decide whether they represent the same group. Well, one thing that I could do, here I've got pi 1 of s3 minus k. I could say, why don't I Abelianize? That'll take me to h1 of s3 minus k. So I'll call this map Abelianization. So for example, here, I'll get an Abelian group. I'll write that presentation with brackets with generators A and B. And now I can commute these things to just get A, B inverse as 1. So in other words, this is, let's see, there should be, maybe I should erase this. So the convention is that if I don't write anything here, that just means that that's equal to 1. If I don't wear this, how is the sound back there? It's iffy, huh? Not loud enough. Not loud enough. All right. Obviously, I need a lesson on how to fix this self thing to myself. All right. So here, here I get z. So it's generated by, say, one element t. And A and B both go to t under this Abelianization map. OK, so now here's the disappointing thing. That was easy to compute. But it was too easy, right? Because actually, no matter what not I pick here, when I Abelianize, I'll get z. So that's a terrible invariant for distinguishing knots with. But all is not lost. So I could think about, here I have a surjective homomorphism. It has a kernel. The kernel is a subgroup. If I have a subgroup of pi 1, I get a covering space. Here I have a covering map corresponding to that subgroup, let's say, s3 minus k twiddle goes by p to s3 minus k. So pi 1 of this space is this group here. And the deck group is just this quotient group, z. So it's generated by some phi mapping from the covering space to itself. And now, well, I guess I could Abelianize this again, which would be like taking the first homology of this covering space. So I could look, let's call it ak. This is what's called the Alexander module, is h1 of s3 minus k twiddle. Let's just use rational coefficients. So this is a module over the ring r, which is the ring of Laurent polynomials. q would join t plus or minus 1. Well, it's obviously a module over q. So I just need to tell you how multiplication by t acts. And that acts by the deck transformation. So t times an element alpha is phi lower star of alpha. And phi was a homeomorphism. So I can also act by t inverse. So r is a PID. Everybody knows that the polynomial ring, q would join t as a PID. But it's also true for the Laurent polynomials. So that means we have a structure theorem for what this module looks like. k is some number of copies of r. So I should have said it's a finitely generated module. Plus some torsion stuff. So a thing that looks like r mod alpha 1 plus up through r mod alpha n. And now here's a little exercise. Actually, this free part is always trivial. This is always a torsion module. r is 0. And now we'll do the simplest thing that we could possibly do to get an invariant out of this module. Let's make a definition. Alexander polynomial, delta k of t, Alexander polynomial, is just the product of the alpha i's. And what that is, let's call it the order of this module, ak. So what do we call it the order? Well, if instead of using this PID, we thought about modules over everyone's favorite PID, which is the integers. OK? So a torsion module over the integers is just a finite abelian group. OK? And in that case, this product is just the order of the group. This is sort of the simplest invariant that we can take of this module. And so you may know a trick for computing the order of a group. So if I have a presentation, so if ak is the co-kernel of some map, say, from r to the m given by multiplication by an m by m matrix A, then what's the order? It's just the determinant of this matrix. OK. So let's think about some examples. So if I take k to be the nought. Another good exercise, if you've never done this before, is to see that the complement of the nought is s1 times d2. s3 minus k is s1 times d2, maybe the interior of that. So in this case, pi1 is equal to h1. The kernel of the abelianization map is 0. And our covering space is just the universal cover. So s3 minus k twiddle is r times d2. All the homology of that is easy to compute. h1 of s3 minus k twiddle is the trivial group, which is the same thing as taking the ring r modulo the ideal generated by 1. So that means that delta k of the nought is 1. And I should really be careful here. I'm going to write a twiddle. Why am I going to write a twiddle? Well, let's come back and think about this presentation that I took here. See, these alpha ones are really only well-defined up to units in this ring. So the way I've defined it here, this is well-defined up to multiplication by, say, c t to the k. All the units in this ring look like a rational number times t to some power. I'll write a twiddle to remind myself that I'm working at the moment with this equivalence relation. So we'll fix that ambiguity later. That ambiguity can be completely fixed, not so much bioframing, but it's definitely fixable. OK, let's keep this presentation. Actually, let me keep this whole board and move over to here. How about we try the trefoil knot, since that's the only other knot that we've computed pi 1 of the complement of? How am I going to think about this covering space? Well, s 3 minus t, pi 1 has this presentation. And in fact, you can see that the knot complement is homotopy equivalent to a two-dimensional cell complex. So this is homotopy equivalent to a cell complex x with 1,0 cell. Maybe I'll call that E. E is just a point. 2,1 cells, I'll call them A and B. So that gives me two. I have a wedge of two circles. This is giving me the generators in my fundamental group. And one two-cell that I attach to get the relator. Maybe I won't try to draw that here. But it goes along this path. Let's say this is attached w is A, B, A, B inverse, A inverse, B inverse. So that's the word, the circular boundary of this two-cell. Trace is out. So now I can draw a nice picture of what the corresponding covering space of x looks like. So these cells are all contractable. So they're going to lift up to an integer's worth of cells. These cells are going to lift up to an integer's worth of cells in the covering space. So for example, maybe here's a lift of E. Here's another lift. I'll call it t times E. Here's another lift, t squared times E, so forth and so on. I won't label everything. t inverse times E is over here. And now I have to think about how I lift the one-cells up. What determines how the one-cells lift is exactly their image under this map. And they both go to the generator of z. So what that means is that they lift up something that looks like this and like this. And maybe here's ta and tb and t squared a and t squared b. And you can imagine drawing the rest of them, but I won't. All right, so that's pretty easy. So now all I need to do is I need to lift f up. So what is f supposed to do? So let's say it this way. Maybe I'll have to whittle for the lift. This is attached, the curve that I'll draw in blue. So what does it do? It's supposed to trace out A. And then it's supposed to go along B. And then it goes along A again. And then it goes along B inverse. And then it goes along A inverse. And then it goes along B inverse. And it closes up. If it didn't close up, I would have done something wrong. Because this disk has to lift up the disk up in the covering space. OK. And then again, I get a sort of integer's worth of translates of this two-cell. I won't try to draw any more of them. But now I can just look at this picture and write down the cellular chain complex. So c star cell of x twiddle. Well, in degree 0, there's going to be a one copy of r, which is generated by this E. In degree 1, there are going to be two copies generated by A and B. And in degree 2, there's going to be one generated by this f. And now I just need to write down the boundary map in the complex. So this is a linear map. It sends the boundary of A as Te minus E. So this is given by a matrix that looks like T minus 1. And the other entry is also T minus 1. Over here, these six cells with their orientations like this tell me the boundary of f. So I see that the boundary of f looks like 1 minus T plus T squared times A plus minus 1 plus T minus T squared times B. Notice that D squared equals 0. That's good. So here's the chain complex. Now I can compute the homology. So the kernel of this map D1 is just generated by an element that looks like 1 comma minus 1. And the kernel, whoops, the image of D2 is generated by this column vector. So it's, let's say, T squared minus T plus 1 times the same vector 1 comma minus 1. So H1 is kernel modulo image is R. This is R modulo T squared minus T plus 1. So in other words, delta of the truffle is T squared minus T plus 1, which is not some invertible multiple of 1. So the truffle is not the unline. I'm assuming you believe this sort of claim that I made about this x here. OK. So the process that we've just done here can be turned into a machine. The machine is called Fox Calculus. So say that pi 1 of s3 minus k is D, which has generators a1 up through ag. And relations, maybe we got it from a very good presentation, has relations w1 up through wg minus 1. We've got one more generator than relation. We'll define the free differential, Fox derivative. So say D with respect to this generator, ai, that's a map from the free group to, oops, maybe here I'm going to abelianize. So let's take g, abelianized here. And again, I'll write the abelianization map with these vertical lines. OK. So I have this di, it goes to fg of z of the group ring. Of, and you know. OK, so let's notice that in the case where we have a knot, which I just said we were in, this gab is just isomorphic to z. And the group ring of z is the Laurent polynomials, right? So this gives me a Laurent polynomial ring. OK. And it satisfies and is determined. So fg, oops, sorry. Yes, OK. So this here, this g, fg is the free group generated by these letters. And g is the quotient by these relations are. OK, so it's determined by two relations. One is that if I take di of aj, this is delta ij. It's supposed to make you think of a partial derivative with respect to ai. OK, but now I'm going to write down the Leibniz rule, which you have to be very careful not to teach your calculus students. OK. So the Leibniz rule says that if I take di of a product, ww prime, this is di of w plus the abelianization of w times di of w prime. Maybe just to see what this means, let's compute the di of an inverse. So I know that 0 is di of 1. Ask yourself why that's true. You can prove it. That's the same thing as di of w inverse w, which is di of w inverse plus the abelianization of w inverse di of w. In other words, this rule says that di of inverse is minus the abelianization of w inverse di of w. I think there's enough room on this board. OK, and this Fox derivative is fun and easy to do.