Session: FW Lawvere & Anders Kock

0:00 Absolute co-equalizer for commutativity. You have an honest drawing, right? I'd like to see it. Yeah, I have a synthetic one. Oh, you heard it. I got sick. I'm really glad to see you. I missed being in the hospital. I saw a few by the way. Oh, hi. Hi Anders. No, don't let me stop you talking to Bill. Cheers. Great to see you though. I did already say hello. Okay. All right. We live on the same floor. Let's hope you're also on the same page. Singing from the same hymn sheet, I think, is the version, the usual one. Ah, okay, yes. Thank you for interpreting, Michael. Sorry, okay. Coming back here after Cambridge? But only in order to leave again? Well, I'll be here for a day. I suppose we're going back from Cambridge on. I'm going back from Cambridge on. I will be leaving on Wednesday, Wednesday evening. Well that means that you could have two days here, two days, great, at least one more day for full participation in the workshop. I don't imagine you're going to do anything on the Monday, you know, travelling back from Cambridge. Depends on what time people are going to be able to get back. How are you actually traveling over to Cambridge? Do you have arrangements all made, or how are you doing? Well, I was planning to take the train. Okay, all right. There isn't a direct train. I think you'd have to go to London and change. No, it's a paid train. I looked it up on the internet. Okay. So there's a three-stage journey.

2:30 Yeah, it's quite a... And the center stage goes from... The first stage ends at 9.45 and the third one begins at 10.45. So there's this 50 minute thing that you have to arrive at instantly and leave instantly in order to make the connection. What it means really, as Richard explained, is of course the tube drive is only five minutes, and you just have to do it sometime between that time and the other time, but... He's ambiguous, isn't he? It's a three-stage train journey, you say. Yeah, what the middle one is just a tube drive of five minutes. Ah, okay. Well, it's a bit more than that. Why longer? Yeah. Yeah, because I sort of would prefer... Hi, Davidi. I think I would... Hi, Davidi. Hi, Davidi. Even the four states were avoiding London. Yeah. Yeah. And that's a pretty good change for Birmingham. Oh, God, you don't want to try to go to Cambridge via Birmingham. That would be a disastrous idea. Would it? Yes. Definitely. It would take you much longer. No, no, I'm not holding myself out as a great authority, but on the subject of getting to Cambridge from here, I definitely will be quicker to go via London. Yeah, but the connections are far faster. Oh yeah, yeah, yes, but the subway and all that, you have to get... Yeah, but then you'd also have to mess around in Birmingham. True, you have got to get across London from Paddington. There's only one time station, right? True, but you'd still have an awful lot longer to wait, I think you'd find. Well, you're the native guy, so... No, it would be a lot quicker to go via London. It would. I'm actually going on the coach which of course is the cheapest way. There's a coach but it's again it's not direct you have to go to London and change but you only change you don't have to change stations in London you go to Victoria coach station and then from Victoria there's a coach to Cambridge so you don't you don't you don't have to go anywhere when you go to London you just change coaches at the coach station so it's a lot cheaper but of course rather more uncomfortable. The train from here to London is very fast, but as you say, you do have to go across London from Paddington to King's Cross to get the train to Cambridge. It's not difficult, it's only a...

5:00 You have to go from King's Cross to Liverpool Street Station, right? No, no, you come in at Paddington. You come in at Paddington from Bristol, and you have to get to King's Cross. Actually, there are trains from Liverpool Street, but the service to Cambridge from King's Cross is much better and more frequent. It's about 8 or 9 stations. I'd allow an hour to be on the safe side, really on the safe side, but you don't have to change on the tube. You only have to go directly from Paddington to King's Cross to get the train to Cambridge. It's not too bad. I was going to say there is an alternative, which is what I'm doing, which is to get the bus. ...to Victoria and then you just change and get a bus to Cambridge. You don't have to go anywhere while you're in London, but... But that, of course, no, from here, from Bristol. In fact, there's even a stop right here at the university, which takes you to Victoria, which takes you to Victoria coach station. I've got my ticket in London, in London. Yes. And then at Victoria, there's a, from, from the same place is a coach to Cambridge. But, but of course that is a longer, because it's coach journey, it's a longer journey, not so very comfortable. So I'm only doing it because it's a lot cheaper. But the train would be much quicker, but also hellishly expensive. Oh, of course. But you're going a day early, aren't you? You're going a day early on Thursday to talk to Roger Astley, to the people at CUP. No, it will be easiest for you to get the train, I would say. It will be easiest for you to get the train, I would say. It's a very easy journey from Paddington to King's Cross. It's on the circle line. You don't have to change it. How much is the bus ticket? Actually, I've got my ticket here. No, I only got it a few days ago. Ah, famous last words. I went and left the thing with all my tickets on my, on the desk in your room. I think it was something like, um, I think it was something like 17 pounds, 50. That was a round trip.

7:30 Well, I might have been a little bit more than that. I'll check this evening. I could check, yes, I could check quickly on the computer for you what the fares are. But certainly the train, it would be a lot cheaper if we hired a car between us rather than a train. We have four all told. Probably not cheaper than us. But I'm not sure how many people would be happy to go by bus. It's a pretty tedious journey. It's not the most. But the other two are coming to us in three or four days. The advantage is that you don't have to change when you get to London. You don't have to mess around going from one place to another. You just go straight back out of Victoria on the train, on the bus to Cambridge. Whereas you do have to change stations if you go by the train, but it's about two and a half hours to London on the bus, and it's probably about an hour and a half to Cambridge, so you've got about four hours travelling time altogether, which is a bit, no, but come to think of it, that's about the total that you do on the train as well, because it gives an hour to get across London. Which I think to be on the safe side, if you don't, if you're not familiar with the tube, is probably sensible. Yeah, exactly. It might be worth thinking about the bus, if you know people, provided people haven't got a lot of luggage which they won't have to go to in two or three days. It might be worth thinking about it. Why don't we check it out at lunchtime on Richard's computer, just check what the fares are. I did get one of those pre-booked, you know, fares you have to pre-book to get the discount. So if you just go to the bus station here, it might be a bit more expensive. It's probably worth figuring it out and then seeing what the best deal one could get on hiring a, no, hiring a... People carrying, hiring a car would be... ...celebrate the start of the 70th birthday. I mean, if Davidius is coming to Cambridge as well, is he? Well, if he is and he's bringing his family, then it really does begin to make sense to hire sort of like a minibus or a people carrier, doesn't it? So we split it between like seven or eight of us. It's going to, I would have thought that would probably have worked out quite a bit cheaper than, even than the bus. But let's check the bus at lunchtime. I mean, even if it was like sort of 200 pounds, it would still work out. They're more expensive than the bus. And you'd have much more flexibility, obviously.

10:00 It's called Diego. Diego Moreno. I was thinking that. But now she's got to prefer this New Mexico to the North Carolina job temporarily in the back end of the market. Oh, yes, sir. We have this... No, he already told us that. But I thought it was coming in this morning. The fact that the product of cubed is another cube means that there are atoms as well. So, the fact that having this additional writer joint, I consider it as crucial for another reason, namely one of your first papers was about the Hilbert-McLean complex. Yes, yes, that's one of the topics I am interested in. Aiming at or turning to. I claim pretty wrong. Yeah. Yes. Right. So you think it's a viable idea? At least one should look at it. Yes. Can you explain the background? Can you explain the idea briefly? Maybe. Yeah. That's good. That's where this theory becomes nice. But as a simple set, of course, you have the entities. That's a set, and that comes about by applying a function. It's actually a representable function to sets. And that has not only a left-hand joint, but also a right-hand joint. So this means that if you have an obedient group say, The set maps from this set, so this group I, are usually called N co-chains on the sequential complex, but now because there's a rider joint, oh, a rider joint, a rider L, this set is in bijective correspondence to R in the sequential sets.

12:30 This is a special set x, top l, I should really call it l pi, so if I apply it to pi I get an object l pi n, so in other words the notion of n co-chains with values in pi, pi would typically be the natural numbers or something like that. The notion of n-co-chain is representable. It can be represented by a simplicial map, i.e. by a map of spaces from x into l pi n. Now, you are particularly interested in those co-chains which are co-cycles. And it turns out that there is a sub-complex of l pi n, which is usually called k pi n, which classifies co-cycles in the same way as l pi n classifies co-chains, l pi n. And this is the Eilenberger plane. That was invented by Eilenberger plane. And the set-up in this simplicial case is by, I found it in Seminaire Cartan, and made my master's degree on this relationship, but expressing against this in terms of adjuncts. Now, as Phil points out, there is a point of regression in this. This, the cohomology of x with coefficients in pi is the homotopy of k. So, the cohomology is reinterpreted as... That was the point of our topic last day. Cohomology means homotopy of maths into k pi n. Cohomology, that's the first co-cycle, modulo, and equivalence relation. So, it's consistent.

15:00 The problem is pi is a discrete group, and you might want to do this for a continuous group or in particular a simplicial group, so with a simplicial object that says an abstract set, for simplicial sets you can't do the same thing. For cubical ones you can, because of the product. Oh yeah, because this rider joint here, you can really, you can simplify, and then this rider joint really becomes the rider joint of the map. This one then becomes exponential, exponentiating to delta A. No, it's the fractional one. Yeah, yeah, yeah. This middle wall is blank until we delta m. That has no write agent, it's depletion set, but if you go to cubical sets instead, so you make box here, then since cubes can be multiplied, the main cubes, I mean the index category here, has finite products. You have the slider joint, which is the one. It builds a mutation. It would be 1 raised to the power of 1 over qn. This is the fractional exponentiation? Yeah. The further joint to the exponentiation. L pi n can be simplistic. Simplicial. Simplified, cubicalified, cubification, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no. There must be some fruits that can be part from that tree.

17:30 Well, so the idea is that also for a smooth, with Dirac cohomology, of smooth spaces, there could also be an Eilenberg-McLean space in the token, even if it's not a classical manifold. I have some information about that. Very good. At least for n equals 1. And it's in an old paper by a group of myself on one-form classifiers, where I can use what is L-I-1 or L-R-1 I take, the group I take to be the reals, or the number line, R. Don't rub off what you've just written, Anders, please, because I want to make a note of that. But, so in the, one of the standard totals for STG, say the classifier of commutative ranks, for Antlion schemes, in other words, for representatives, you have a notion of, so A bar is represented by the commutative rank A, you have a notion of R-rank with one form, with ranks of one form omega. Well, it's really an idea in algebraic geometry that this actually can be described by a map into spaces, and the space that was needed is called omega 1. Nothing to do with truth values. Nothing to do with truth values. No, no, this is nothing to do with the class of... Caleb inventions. Yeah, yeah, yeah. Because... For any commutative rank A, you have a notion of the k-log adventures of A. And that's a co-rater and fountain. Well, first of all, it's an abelian group, or really a k-log group, or k is, well, let's just say abelian group. So it's a fountain from commutative ranks to abelian groups, which means that it's an object.

20:00 And a really good object in the topos, because the topos are just co-variable functions from a really good process. And by a major length of its functional response to an element in omega-1a. So, the construction is really due to K-lock, but was made quite explicit in some of the work of the... The notion of one form, which corresponds to the notion of one co-chain in simplicity, it's a derived one form, is classified by a space map from the atomic alarm, at least in Tobias Affleck's schemes. Though, if, for example, our translation would define a one form to be a mapping to x2. And this really is an instance related to this idea without looking at it. Ah, yeah, all right, okay. Let's say one can construct omega 1 by carving it out of r to the 1 over e. Right, right. Well, I'm just finding out that it exists as a function, and I don't really know how to describe this one, whereas omega-1 and the Kähler referentials are. So what is new in Eduardo's and in my paper is that we do it not just for commutative lengths, but for any algebra, for what we call a Fermat theory, which It's an algebraic theory containing commutes and vanes, but where you have partial differentiation as a property, not as a pluralist function. So this works for, say, the catchment of the fluid derivative.

22:30 Yeah, we make this construction also for the smooth case, smooth scalar equations, not just algebraic geometry kind of equations. This works in these C-infinity models of SDG as well. How do you gap the space of these? Yeah, but of course what is... You start with one, and I don't know really how to construct a two-form classic model. 25 years old, but it's called one-form classifiers because we could not go beyond one. Because you did homogeneity, each variable separately, and antigenitivity, and those have to be incorporated into it. I mean, the two-form is a map of a double tangent model of some sort. Is the question of singling out the conditions? Well, yes. However... So we would be part of the fraction. Just carving out the right part is the problem. Which is really not a problem, but it's not really... at least we cannot calculate. And anyway, the clear differences... We do not mention the tangent vulnerable to construct. So they actually define the tangents in terms of the cotangents kind of thing.

25:00 No, no, but I mean in the classical algebraic... Yes, yes, which is, of course, a big discussion. Yeah, I mean, it's probably not correct. Tangents are precisely familiar to all of us. It's rather the other way around. Anyway, whatever you define things via double dualization. It's usually not the best thing to do. That's right. That was one of our very strong principles, you see. Yeah. Down with unnecessary double human things. Yes, yes, yes. That together with every law is really the associative law in disguise. Sounds trivial, but really these are very profound observations that I've learned for us. It applies in this case. Well, the remark about every law being the associative law in disguise is incredibly deep. I mean, there's all the work on tens of categories and stuff. Yeah, well, a lot of people had to work on it, but I was thinking of a stash of, okay, yes. Couldn't read yours. No. No. On a complete, just a historical question, the... This work of Elluci and Co. on, as it were, redoing the, you know, the one forms in the Durham Simplicial Cohomology in this setting, this kind of scheme-theoretic setting, was one of the first applications of the Grotendieck-Topos theory. At least that was one of the substantial ones, the work of... Yes, people on the topostheorians who were first version. There is a lot of material in it that has not been understood except by themselves yet.

27:30 Yes, if even by them, I suppose. I'm just requesting it. Yeah, I'm able to understand it better. I think it goes way back to the idea of extended function. Extended function? Yeah. So in other words, a function on X from X to R. An extended function on X to R is really a function on X to the A, the kind of A. So it's a function of lines. Ah, okay. Yes. I mean, this is a function. If you have any three objects, you can think of it as a... It's a function, it's an algebraic operation on x of arity A, you could also say that, and so there's some sense in which it's like a function on x except it's extended, you know, it's a contrarian function of x and so on and so forth, so these were probably extended or so forth. Now, so taking A, it's like special figure shapes. By the way, Voltaire used the term element. And roughly the same way I use the word figure. Yeah, but you want figure. Figure. Figure. In other words, there were generalized elements. I mean, this is just . Yeah. But among the maps with the given co-domain, they actually alluded to the special domains and more general points in some restrictive sense. But figure seems to be . So functions of figures in x are like extended functions of x. And if x to the a is equal to d, then the linear ones are the non-linear ones are like Lagrangians. Lagrangian on a given space of configurations is a function on the tangent model. So Lagrangian is an extended function.

30:00 It doesn't only depend on Q, it also depends on C, in that sense, because it's going to diagnose the trivialization of the bundles of the separated B's and Q's. And another example, I like to think that Cailliau and his thermodynamics had some idea when he said that heat is a function. Later people realized it's really more like a differential form, like a function in the strict sense, but you see it is an extended function. If it's a differential form. So there was something, a big content in his idea, even though the narrow definition of what a function is came back to be wrong. And so in the case of cohomology, shapes of interest are triangles or spheres and balls, you know, basic, basic things. Balls and spheres, in some sense, they seem to be one and the other one. I'm not trying to go from the square to the total. The ball and the sphere. We're interested in the contrast between the sphere and the ball. The origin of functions defined on the sphere is extended to the ball. That's the good question, and that's great. That's sort of the thing. But if you use these various dimensions as your figure shapes, then these... Extended functions on X, those are the basic ingredient co-chains. So the idea that extended, certainly extended functions, are actually functions, but with values that are a bit more complicated. So if R is any kind of algebra, this is the same kind of algebra, because applying to the exponentiation is the product deserving. So if R is the range, so is this, so look at the algebra.

32:30 So, essentially, what Eilenberg and McClain were doing with cohomology was justifying Candeau and LeBron reinterpreting these extended functions as actual functions because of the very special nature of the figures being used here. So, the atomic nature of the figures doesn't differ from other figures in that this reduction of extended functions Of those kinds, the ordinary functions but with special values. That's what's involved in the algebraic plane space and in several other forms. It's also the key idea in Hisbel's proof that a category has a small etiquette subcategory, also has a small co-etiquette subcategory, provided there are no rule on cardinals. The new system consists of objects like that, where A and B are, you know, rather inaccurate. For example, it could be always taken as the ordinary wheels of the ring structure, from the, no matter what category you're in, to the ordinary wheels of the ring structure, as in the theory of real compact spaces. New functions can be generalized to functionals and then restricted back again to merely extended functions, which are merely functions of a mathematical system.

35:00 If we think A as an infinitesimal body and we take X equal to R, we would have that kind of paradox that extensive quantities... It would be represented as an intensive quantity over a different... Yes, a different body, so to speak. No, a different value. Yeah. Different space. Oh, yes, okay. Not about air value quantities. Extensive air value quantities would become intensive air 1 to the d value quantities. Yes, that's right. That's right. If you think of X as ordinary space and the placements of this body A in space, so measuring these placements can be seen as a function on the ordinary space, but it's sort of like the Lagrangian view of the fluid motion, right? In other words, isn't that right? It's sort of like at each point of space you have the density of the body. Instead of the body is embedded in space and its mass is pushed forward into the space according to the placement. Take the other view that, yeah, right? So we have ordinary space and we have a body, and now if you measure anything about the placements, If the volume of the field on space itself takes its values corresponding in R to the 1 over V, as I was saying, for example, if you had a mass measure on V and then you choose such a placement, the measure on V, the density of that, but somehow you measure things about the position of your space, the same number for each placement.

37:30 Can I make a suggestion about... Yes. I'm only speaking for myself. I don't know how Richard feels. I think it would be very helpful, perhaps when John is here, I don't know if he's going to be able to get here this afternoon or any time today, but at some point during these discussions, it would be very helpful to me if you could give a... Entirely general exposition on the whole issue of the distinction between intensive and extensive quantity and their behavior as respectively in a covariant and contravariant mappings into and out of domains and co-domains and a relationship to the understanding of the way you see quantity in space and structural domains of variation in general. But I think it is still Very little understood, I mean, by a broader audience. I know there are plenty of people in this room, apart from myself, who understand it a good deal better, but I think that the whole issue of intensive and extensive quantity and the way they operate would be just very, very helpful to have a general pedagogical exposition of that.

40:00 I don't know if you feel that would be a useful, that would be a worthwhile use of part of the workshop. It's absolutely up to you, of course, but I would find it very helpful, especially since we're going to be talking about Grassman at some stage. That might be the suitable point at which to develop it.