 Right, so if I want to take, for example, dA of this word, A, B, A, B inverse, A inverse, B inverse, OK? Well, I just walk along the word and I pick off every time I see an A. If I see an A, I take the abelianization of everything that comes before. So here, there's nothing before, so I get a 1. Here, these both come before. They both abelianize to T. This goes to T squared. Here, this is an A inverse, so I need a minus sign. And I abelianize everything before it, including itself. This gives me a T if you do it. OK, and lo and behold, that's the same polynomial we had before. It's not an accident. Actually, what this Fox derivative is doing is it's just computing. So the derivative with respect to B, good. Let's do the derivative with respect to B. dB, AB, AB inverse, A inverse, B inverse. Well, so that's going to be, I abelianize what comes before T. That's this A that gives me a T. Here, I have a B inverse, so I abelianize everything up through here. That's minus T squared, minus because that's an inverse. Here, I abelianize everything up to here, but this is the whole word, which is 1 in the group, so this is minus 1. Really does work. OK, so let's say the following, maybe, C star cell of X twiddle, where X is something homotopy equivalent to this is going to be something that looks like r to the g minus 1. I get 1, 2 cell for each relation goes to r to the g. I get 1, 1 cell for each generator goes to r. I have 1, 0 cell. In this matrix here, I'll call A. And in this matrix here, I'll call B. OK, great. So excellent question. So the only thing that was really special about non-groups was that there was no torsion in here. So when I want to do anything, so it's easiest to think about rings, which are, say, polynomial rings. So instead I took this abelianization and divided out by the torsion part of the abelianization. So I got a free abelian group. Then everything would work exactly the same. And doing this, I get what's called a multivariable Alexander polynomial. So what's this A? So A is just a matrix, say, dAi of wj. And B is the matrix of abelianization. So it has one row and a bunch of columns. And the columns look like the abelianization of a1 minus 1 up through the abelianization of ag minus 1. And now you have to do a little bit of algebra. Not a whole lot, but a little. And at the end of the day, you'll see that Alexander polynomial of t is, one way that I could write it is it's the determinant of A i times the quotient t minus 1 over the abelianization of A i minus 1, where A i is, what I get by the i row. And this is also equal to the GCD of all of the data of A i's. So actually, I can define the Alexander polynomial just as an invariant of a group. And one way that, according to this definition here, really, one way that I can see that the Alexander polynomial is a polynomial over the integers, not over the rational numbers, is by taking this definition. So this actually lets me get an Alexander polynomial that's well-defined up to multiplication by plus or minus t to the k. But we still haven't gotten rid of that. OK, so the Alexander polynomial is a great invariant. The oldest and best not invariant there is, probably. Best not in the sense that it's the strongest, but somehow that connects to so many things. And so one very nice geometric thing that it connects to is the geometry of embedded surfaces. Maybe this is what I want to talk about for the rest of the hour. But before I do, let me comment. So this Fox calculus definition, if you're familiar with Heggart-Floor homology, Fox calculus definition is sort of very closely related to the definition of Heggart-Floor homology. If you're not familiar with Heggart-Floor homology, you should sort of think about that during Gen-Hom's lectures. So now let's talk about embedded surfaces. And first, let me talk about fibered manifolds. Let's give a definition. So we'll say y3 fibers over S1 if there's a submersion p mapping from y to S1. And then if we have that, then we can talk about the fiber of the fibration. So it's a preimage of a point. And since it's a submersion, it doesn't matter which point I picked. All these preimages are homeomorphic. And moreover, I can see that, say, y is really just f times 0, 1, modular to an equivalence relation, where I glue this end to that one. So let's say x0 is equivalent to phi of x comma 1, where phi napping f to f by a homeomorphism is what's called the monodromy. So now let's say that, say, k is fibered, S3 minus k fibers over S1. So if so, then I have S3 minus k. So that's this thing over here, f times 0, 1, modular twiddle. And the group Z on f times r. So the generator of the map, so Z, let's say, is generated by phi, what we'll call that big phi, big phi of x comma t is little phi of x t plus 1. So this space here, f times 0, 1, is a fundamental domain for the action. So the quotient is really none other than S3 minus k. So all modular of the action of phi is S3 minus k. In other words, we have a covering space, f times r goes to S3 minus k. This is a covering space with that group Z. If you think about it, since H1 of S3 minus k is Z, there's only one such covering space. So this is S3 minus k twiddle. So H1, let's call it, we call it ak. So that's really just H1 of f times r, which is H1 of f. And how does t act? t times alpha is big phi star of alpha. But that's really just little phi star of alpha. So another way that I could say that is that ak has a presentation. You know, let's go to the other board. ak has a presentation that looks like H1 of f tends to r goes to H1 of f tends to r. This is really just r to the 2g, or g is the genus of the fiber. This goes to r to the 2g. And what are the relations? Well, the relations say that t times anything, so t, say, tx is phi minus phi lower star of x is 0. So this matrix here is just t times the identity minus phi lower star. This is a 2g by 2g matrix of integers. So that means that the Alexander polynomial is the characteristic polynomial of the monodroming debt of ti minus phi star. And in particular, the highest power of t that I see is coming from the determinant of this i. It's t to the 2g is monoc of degree 2g. It's fiber delta k of t is monoc. And this is actually a really surprisingly effective way to tell whether or not it's fiber. So for example, a trefoil knot, we saw it had a monoc Alexander polynomial. It's a fibric knot. Every knot with 10 crossings or fewer that has a monoc Alexander polynomial is a fibric knot. No, it's not if and only if. So you can easily produce. Right, OK, so yeah, so that's not, OK, you're completely right. So that's not a proof, right? It's a suggestion that I should think that it's monoc, or that I should think that it's fibric. And then I go off and I check that it's fibric. But it's a theorem, actually, that if you can draw your knot as an alternating knot, whereas I walk along the knot, the crossings go over, under, over, under, OK? The knot is fibered if and only if the Alexander polynomial is monoc, OK? So I could also prove that theorem and then observe that the trefoil was alternate. The last thing I want to talk about is ciphered surfaces. Ciphered's theorem. So let's make a definition. So a ciphered surface for k is an embedded, let's say, sigma. And again, embedded into S3, whose boundary is the knot. So for example, say I take the trefoil, there's a surface here, OK, that has two sides, say one side is blue and one side is orange, OK? And here's the blue side and here's the orange side, OK? And if you've never seen that before, that's probably a little hard to see. So let me try and draw a better picture. So what did I do? I started with two disks that I took like this, one with the blue side up and one with the orange side up, OK? And then I took a strip of cloth, I should think about it as being like a one handle, OK? With the blue side on one side and the orange on the other and I gave it a half twist, OK? And I took this strip and I pasted this bit to here and this bit to there. And I got my surface. Well, I got part of it. I did it two more times. Which surface is this, right? It has some genus. What's the genus? How do I compute that? Well, one good way to compute it is to notice that the surface blue is not the right color. Red is. The surface retracts to this graph, OK? Where I had one vertex for each of these disks and one edge for each crossing. And the Euler characteristic of the red graph gamma is, well, there are two vertices and three edges. So that's 2 minus 3 is minus 1, OK? Now this surface, whatever it is, it has one puncture on boundary component. So actually, that means since its Euler characteristic is minus 1, that means that this surface sigma has genus 1. This is homeomorphic to a punctured torus, OK? Well, let's make a definition. Cypher genus, OK? An interesting geometrical invariant of k is the minimum of g of k, oops, g of sigma, such that sigma is the Cypher surface for k. So for example here, the truffle oil has Cypher surface of genus 1. So its genus is either 1 or 0. But in fact, this is a little exercise. So notice that the unknot bounds a disk. So its genus is definitely 0, OK? Exercise is that actually g of k is 0 if and only if k is the unknot. So we saw that the truffle oil is not the unknot. It's Alexander polynomial is non-trivial, so it must have genus 1. So here's a theorem about the Cypher genus. So this is a theorem of Cypher. Let's put it this way. So g of k is bigger than or equal to, oops, 2g of k is bigger than or equal to the degree of delta k of t. And so here I'm thinking about the difference between the lowest and the highest powers of t, right? So for example, for the Alexander polynomial, this number would be 2, and the genus is 1. So this is actually an equality. Excellent question, yes. So every knot has a Cypher surface. There's a kind of abstract way to prove that using, well, OK, I can tell you afterwards, there's also a nice algorithm due to Cypher, which constructs a surface from a diagram. And you can read about both of these methods in the notes, or I can tell you how they work afterwards. But I want to say a word or two about the proof of this and then conclude. OK, so, and actually, let me amuse myself. So just like the statement about fiber knots, this theorem is not always an equality. Maybe the simplest knot that you can have where it's not an equality is what's called the minus 3, 5, 7 press the knot, which is in your exercises. So let me draw a picture of it. Five crossings here, and now they're going to be seven here. Looks like kind of a complicated knot, but actually it's really pretty simple. So this knot has delta k of t is the same as the unknot. That's 1. But it's not the unknot. Actually, its genus is 1. OK, so maybe let me just say a few words about this. And then I will stop because I know we have lunch. OK, so what do I want to say about the proof of this? I'll say the following. So say that I look, say I take y. So I say that sigma is a ciphered surface. And then I take y as s3 minus a neighborhood of sigma. And then y sort of has two boundary components. Say sigma plus union 0, 1 times boundary sigma, union another copy of sigma. OK, and I have s3 minus k is y union sigma times i. I'll write this as minus 1, 1. So I get two gluing maps. Let me call it i minus and i plus mapping from sigma into boundary y. Those are the images of minus 1 times sigma and plus 1 times sigma. And so then I see that actually s3 minus k is sort of y, modulo of the equivalence relation that I set i plus of x equals i minus of x. So I glue, if I imagine, shrinking this interval. I get that. And then I can see that s3 minus k twiddle is y times z modulo the relation that if I take, let's say, i plus of x comma n is equivalent to i minus of x comma n plus 1. So that means that h star of s3 minus k twiddle has a presentation that looks like h lower 1 than sigma is isomorphic to 1 of y. It's a bit of algebraic topology. You have to go do here, h1 of sigma, this is h1 of y. And this is the isomorphism. And here I have, say, t yoda plus minus yoda minus. So lk of t is the determinant of t yoda plus star minus yoda minus star. This is 2g by 2g matrix. And so is this. Finish by saying one application. And then I'll stop. Let's say I can use this to show that delta k of t actually looks like the determinant of, let's say, t a minus a matrix at. These i and yoda minus transposes of each other, which means that I can normalize delta so that delta k of 1 is 1. That's an exercise. And delta k of t is delta k of t inverse. So for example, with this normalization, I'd write delta of t is t minus 1 plus t inverse. It's symmetric. And then this satisfies, so if you saw the stain relation for the Jones polynomial, stain relation, delta k of a negative crossing minus delta k, oops, delta of a positive crossing is t to the half minus t to the minus to the half delta of the resultant. OK. So I think that's a good place to stop. So let's go and get lunch.