Natural transformations, dual vector spaces, Yoneda lemma

0:00 But slowness is stupidity. They exemplify it. And it didn't seem to make any difference. So, did you tell them that last week, we looked up the answer in the book, we didn't believe the answer that we looked up in the book, and then we carried on struggling. Well, we can't put it there, because I guess in the way it will. How much time? How are we over there on that edge, on that thing? Do you think it will pick up? and then we are absolved of any responsibility. No, we were just pointing, we were talking about, I mean, Richard was just pointing out that if you have products and equalizers, then you've got pullbacks. Right. So if you, by duality, if you've got co-products and co-equalizers, you've got push-outs. Yeah. What the blazes of push-out is, is another matter. But, I mean, I also think that, um, there's a lot of this stuff that's, um, you almost have to see how you can use it before, because it's otherwise, it's a load of abstract nonsense, really. I mean, you know, we should probably get into this. well-known to myself sections on the process you know the young when you're which one so i don't know if i think that we don't want to get a war right yeah so it says it's not but it's a book recruiting hands he's got something about interesting So there's a nice, so discussing how Bulbaki tried to define structure, and he said, and he discusses this a bit, and he says, in sum, Bulbaki's structure theory follows category theory and using morphisms to handle structures. It was developed by largely the same people who developed category theory. It failed. Well, Bugatti stipulates what morphisms are. They're the suitable functions. The category axioms nearly say how morphisms relate to each other. They compose associatively with identity elements. And if we suppose everything is a set, categorical morphisms may not be

2:30 functions. I thought this was interesting relating to the John Dayes' analysis of the Yoneda Lama that says this is a living example of relationalism where all you need to know about objects and how they relate to other objects in Yoneda Lama is the food factor to the best of the standard. And in this passage to pull that out nicely that from one point of your structure is given by saying what everything is categorical you'd find the structure just by saying how everything relates to each other well i mean this if that's a thought by a set is right unique and not just a category just is a set of objects and arrows What we're talking about is we're trying to peer inside the object as a category. That's the point Colin is making. Okay, so Bourbaki gives a definition of a category where you say it's this particular species of sets and these particular maps between them. yeah okay and that and the species of sets is specified exactly for example in the category groups the sets in question are ordered pairs consisting of sets and their and the binary operations and so on and the morphisms of the volume sure so what what what i think what color is saying here is that you want to do i mean you're you're looking at the examples of groups and rings and topological spaces and manifolds and all the sort of familiar things. And in each case, you get a net, you give a a conventional structuralist definition of what a, whatever it is, it's a manifold or a group. And then you specify what a morphism of manifolds or groups is. Yeah. Okay. And then you're often right. Okay. So now you can talk about the category of those things.

5:00 We shall not appreciate that. No, he's not around. He won't be back until Monday. Oh, really? Yeah. Oh, good. Thank you. But, you know, that's sort of... I see that from the point of view, you know, for a particular category. Neither one doesn't say anything about a particular category. It says about any category. It doesn't say that you have to specify the structure. It doesn't have to say that you have to put conditions on the objects in advance. It just says something about all categories, and all categories are just classes of objects and classes of arrows. But the point is... So I don't see how it can say anything about these cases. But look, the point is, what Vorgat Bakke was trying to do, we've got all these particular examples, and they've got odd features. For example, the morphism condition on a topological space is sort of ass-backwards. Because you don't, it isn't, you know, you can't say you're preserving structure. With all the algebraic examples, you can say the morphisms are preserving structure. This is sort of anti-preserving, higher-level structure. So, he wants to say, but there's something in common in all this stuff. you've got this thing called an object, which is an object of the appropriate sort, like a manifold or whatever, and then you've got the morphisms. Now, Bobacki wanted to give a general definition of what you could consider as a candidate object for a structure type. Yeah. So it's got to be, he's got to take in all possible higher order things. I mean, And the point about the category theorists is they just said, well, forget about what the internal structure of the categories is, of the objects in the category. Okay, but that's, this business of things, getting their existence by taking in other people's washing, which is, you know, which is what this idea is, it's only the relations that matter. that can work on all different levels and in the category case it's working on the level of when we're talking about concrete categories of looking at a whole system a whole class of objects of a given type like groups and considering

7:30 that class as being determined by the interrelation the morphic relation Morphosis between objects and a category. Okay. Now, as I got excited about what Stuart said, quoted, um, Bias is saying, is that Bias is saying, take the category theory perspective, agree with McLean that this Bobaki thing is nonsense. That's not the way to go about it. Nevertheless, curiously, by virtue of your need is there not? You get something like a Bulbaki picture back, right? Yeah, you can recover the internal structure. You can recover the internal structure of the objects as, if they're, you can, you can think of, by innate is there not, you can replace the given category by another category in which the objects are, these concrete structures. Yes, so you can take it, so you can take, you can take an arbitrary category, take a functer from that intercept, set, and that sort of acts like a concrete category where everything really, so rather than just having these abstract, you know, rather than the objects just being these anonymous dots or arrows, you can actually have a set that plays the role of that object category and a set of functions between sets that play the role of that arrow. So it's a curious kind of reversal where you go from, it's this sort of anti-abstraction, right? Because to get the category, you say, forget about what groups are made of, just look at the morphisms and so on in the category of groups. Now we're saying, here I've got this abstract picture with just morphisms and objects, and now I'm going to tell you what structures these things correspond to. Well, what happens if you do UNEDA on group? Do you get back something like Kaley representations? I don't know. Um... no, it's going to... So I forget how the UNEDA... Well, we ought to really work that out. Yeah. That would be a good... But it doesn't say anything about what... It doesn't say anything about the sort of, it just says any functor from this category into the set, into the category of sets.

10:00 And indeed, it gives the example of the representable functor, which isn't anything like the underlying set or something like that. It's just a bunch of morphisms, in fact, that go from the object into everything else, which looks nothing like the Rebac-E structure. structure. So it doesn't say anything about recovering from the structure. You get a Burbank-y type structure. I'm not even sure if it's like that. The only example we have of a sort of uniform way of doing a functor intersect is the representable functor. But But that just takes it to arrows. It's just a sense of arrows relating to other things. So that's nothing like Burbank's structure. That's still just concentrating on the arrows. Sorry, you... how does it work? You take a category and... You take a category and you have a functor. Doesn't nobody tell you what it is. An arbitrary functor from your category into set. So this is the first question. That's... Yeah. So imagine you had a very simple category with... uh... it didn't take me and uh... there are a lot of people that is our community margaret needs to say but most of the country uh... and now uh... from this category will take any will not any into just the fact of our race um, the arrow AB and the arrow AC, and that's, that's, that's, the arrow AA, yeah, um, it'll take B into the arrow BC and the arrow BB, the identity arrow, yeah, no, there is no arrow, but there's no arrow BA, and then it'll take C just into, into its identity. Since everything has an identity, we know that there's some target for it, but something... Yeah, so what is this? This is the HOM target. Yeah, exactly. It's the... I think John Byers writes it as the HOM A blank. Yeah, exactly. That's what I think he writes it as just the first...

12:30 A goes to HOM A blank. yeah yeah exactly so and then you obviously do this thing for um for the arrows for the arrows well what no what do what is the obvious thing for the arrows you just compose you you just uh you just compose yes no what is it now i have a problem with this actually because i i always The first figure of central functions is to think that arrow is going into an object, and then you can just do a composition, but I couldn't see how you could do it without things going out of the objects. That's why I was a bit puzzled by... Maybe you have to put the compositions in the other direction, by the other order. I don't remember that. Well, what happens if you do it the other way? Then A becomes the sparse one and C becomes the rich one. Yeah. Yes, exactly. But then how does that help us with figuring out the arrows? Right. Well, take an arrow in the first one, A to C. Yeah. Then let's hang on. Which punks are we doing? Let's do the one that Stuart's written down. Okay, so we take the arrow, A to C. Yeah. And what do we want to say? So now we want to... We want to map. So in set, we want to map from the set of three arrows, A to G, A to C, A to A, into the set C to C. And we've taken what's an arrow from A to C? Yeah, so... Yeah, okay, so from an arrow from A to C. Given that, you want to map that, so an arrow between those two bubbles there. Exactly. We want to take the arrow from A to C in the original category and turn that into some function between this set of three elements and this set of one element. Well, there's only one function that works because C to C is the only target. Exactly. So what about the arrow from A to B? What does that do? Well, before we do that, we also know that the arrow from b to c in the original category has to be a function from this set of two things into just so it's all got to go to there that's also just okay so that's no

15:00 problem okay so the if we don't understand what's going on if we could understand what a that arrow from a to b goes to well one thing we know is that um in the original category we've got of course we've got composition for functions. So we know that if we follow the arrow from A to B, and then we follow the arrow from B to C, we get the same as the arrow from A to C. But I think that's automatic, isn't it? Because whatever, yes, because whatever we do here in set, from A, from the representative of A to the representative of B, and then to the representative of C, inevitably we've carried this thing down to just the identity So whatever we fill in here, composition is going to follow, in this special case, where C is a sink. Well, I mean, what is the arrow from A to B then? What is its image under this functor? he does it differently I wondered about this when I read Bias as well you take an arbitrary object not an arbitrary, just take an object and that's what defines your home function there's not a different home functor there's not a different And the functor doesn't take each one, so all the maps going out of it, it takes, just take an element of the initial category at the start, and then the functor's value at, say, B, say you take A as your thing, then the functor's category at B is all the arrows from A into B. I mean, in this case, you've only got one, but of course, in general, you may have many so that's what he thinks that i don't know that that was the sponsor uh... objects so so you want to move from the front of the head of the head of the head of the front of the head of the head of the head of the head of the head of the you know the other is that the other is home yes exactly and what it would be

17:30 Anything here, x, is hom A, x. Exactly, yeah. Yeah. Okay. Now, okay, so now what's the function? So now what's the function of the lower x? So, okay, so if you go up an arrow from B to C, so this is, this is hom, this goes over to hom A, B, So that's all arrows from A to B, and then this one. Well, obviously, every one of these things determines a unique arrow from A. Yes, yes. Because we, for each arrow from A to B, we compose that with... That arrow. And that gives us an arrow from A to C. Oh, that's nice. Okay. I mean, it really does, it is really reminiscent of Cayley, right? I mean, well, this is what he calls, or maybe someone... Well, Veer calls it that. Someone said... Oh, Veer calls it the Cayley, the nato-limber. Yes, really. It's a big generalization of the Cayley. Okay. Now, so now it's begun in clear. And I can, okay, so I can always, I don't know whether this is the unate of them or not, but let me state what the nice thing that we might be able to say. Okay, take any category and any object in that category. Then we can do this trick. Okay. And then that identifies every object in the category with a collection of morphisms. Okay, so it's, and then morphisms are then, the morphisms in that new category are the things that are obtained by combining the morphisms. See, what do we want to say about it? Okay, let's... Okay, so here's... Oh! Here's my original... I want to see why it says what that is. What? I think this version he's got here is too general. Yeah. You're meant to just, it's just the idea...

20:00 No, it's not the identity. What is it? Well, look, what you're saying is that you've got always, you've got an arbitrary category, C, you take an element in it, And now, here's set, okay, and now you've got a functor going into set, let's call it F sub A, hom sub A. Yeah, I'll call it A dash. A dash, okay. Okay. That functure, that takes every element here, that takes a typical element, a typical object B, into the set of all morphisms A to B. And now, what do we do then with a typical morphism F from A to C? Well, that's the morphism that simply... This is a particular map from B to C. so this is a map that just composes each that takes each of these F's down here let's call this G so it takes each of those F's to G following F and if you wanted to do the thing the into version of it you'd do this backwards okay so now that means that these homesets The function composition are the objects in this concrete category. Okay, so, it's like somebody says, but I know what a group is. It's a bunch of transformations caused under identity, inverses, and composition. And somebody else says, no, groups are really, they're all kind of, they're matric groups, all this kind of stuff, and so on. You say, bang, I insist that this is what a group is, right? And then Cayley's theorem tells you that what I say and what you say are essentially the same, modulo this map. Right, right. Because you have this sort of abstract idea of the group in the Sunkat, so a group is just a category of one object, in that sense?

22:30 No, no, in this case, no, what we're talking about, we're going from the category groups to the category of... Okay. Okay. So we're picking a... is that what we're doing or not? I think... Okay, so this is the crucial question, what you're at, what you're suggesting. Either we have to categorize internally what the structure is. Think of each Bobakian structure as itself not giving rise to a category, but just being one, like a group is one. It must be that you go from the category of groups to this home representation. What happens if we make this construction on a group, being a category of one object, Then, the only object we have to pick as our A, as the base point for this formate blank, is just the object, the only object. You get the same thing back, essentially. Yeah, and then you have to take its, um, and its value is going to be all its, all the maps. All of its maps to itself. So, yeah, we get something that's sort of... No, I think it's just, you get the same back, it's essentially the same thing back, but, okay, but now take, but what happens if you take the category of groups? Yeah. Okay. So the question is, it might make a difference what A you picked, right? Um, it might take A. Or, you might want to take the morphisms into, because then you could take the trivial group, because you've got a terminal object and group, right? because the trivial group, there's a morphism, no, sorry, it's the other way around, isn't it? There's a, no, there's a morphism from any group to the trivial group, just, just send everything to the idea. Yeah. I'm going to guess, I'm going to hazard a guess, that this construction is such that you get something different for each, um, you get a different from the bridge and

25:00 uh... related well i don't know what we're going to be but i don't need to get them but you need to assess the for any point the customer the sense of natural transformations between that folks are on the whole The home functor of A is the same size as the set taken by your functor from A. So, you have two. And this is why we haven't really touched the unatal element yet, because the unatal element starts with an arbitrary functor, and that says something about the relation. Here. I guess there's not much. No, there's nothing here that's better, really. OK, so, say you have a category, and you have a category of set, and you've got some functor X, OK, taking things in there, and then you choose some A, I guess best to put that here, and then do the home functor from everything, and all of a batch. Then, the set of natural transformations between, uh, from this factor, from one factor to this factor, is the same size as this set. That's what the unated limb is said. It's about a set size. I mean, it doesn't say anything like what a bias was saying. Well, that's why I've been so puzzled about this whole business. I don't see what it says here. Well, let's see. Did I bring my copy of McLean along with me? Oh, I've got it. That says it. Do you want to check? Well, I think I probably got it. I think that's better than that. Because I wanted, if I didn't want to spend the days talking about your native thing, transmissions i think we have to get that's yeah i think i think that's right sort of but i think this is a lot of this the great but i think that but we'll get going to get this

27:30 not possible and i have to restart by doing great uh... you have to do that with the uh... here's the best There's another passage in this Colton McCarthy paper, in a section entitled Natural Transformation. Yes, yes. Unsurprisingly. All right, let's see if you hear what Colin has to say. Two constructions might start with a group G and give different results, but always isomorphic results, where the isomorphism is defined the same way for all groups G. then the isomorphism is considered natural because it furnishes for each G a unique isomorphism not dependent on any choice of how to describe G it's natural in the sense that it's given simultaneously for all groups okay, we'll come back to this because I've been busting my ass over the double jewel example yeah, can you go through that the functions I find, the natural function on the internet, were just wrong. They weren't really even functions. They weren't isomorphisms into the joules, double joules, as far as I can make that. They seem to. Okay, so this will be a good, let me take, let me take this whole thing, and I'll take this as far as I can get. I mean, I just ran out of time, actually. Okay. So, let's remind ourselves. Do I need to go through the definition of what a natural transformation is? No. I don't care. It helps me to remind myself. In that case, I'm willing to help you by helping us. Okay, so what you've got, you've always got two categories, A and B, and you've got two functions, functors, F and G, both, as it were, functors from A to B, right? And you want to say that, what do you mean by alpha is a natural transformation? F to G.

30:00 So what you want to say is that you want to say that alpha is a function I don't know how you're going to put this alpha is a function the subscripts the subscripts that's how it's often done so alpha sub a what do we want to say it's a morphism a, f of a. Isn't that it? Is the morphism true? Where a, as an object... Where a, yeah, where, where, for all a in a, such that, if I connect f of a... No, sorry. No, it's a fungomorphism. From F of A to G of A. Yeah, exactly. Okay. So, if I've got this picture, so I've got, so, so I say, given F, A, and B, A, F, B, right? Then, Then this is big F of little f to big F of b, and here we've got g of big A, going by g of f to g of b. Okay, and now, what we've got is, this is a morphism inside the category B, right? So this goes, this is a morphism in the category B from there, from F of A to G of A. and f of b, a, alpha, b, goes from f of b to g of b. Okay, and that commutes.

32:30 Okay, so now, I mean, it's actually instructive to verify that the that the identity functor is a functor, well it's not a functor, but it's naturally equivalent to any functor, sorry, any functor is naturally equivalent to that functor composed of the identity functor, whatever. Right, yeah, the functor or equivalent to the energy group. Yeah, yeah, yeah. Okay, and then the composition one as well, yes. Can I just draw a picture? Yeah, sure. Okay. So we've got, so we can get category A sitting up here. Yeah. And category B sitting down here. Yeah. And we've got both F and G going from there. So we, well, we've got some, the hour difference we have two different functions f of a would be and F and the other the other, thank you over here, this is G that's a G of A and G of B and G of the left and now our natural transformation is a morphism within B that takes sorry, the natural transformation is a function which is assigned to each object in A, a morphism from F-A to G-F-A.

35:00 Ah, right, right. So it's something that takes in objects and gives back... It takes in objects in category A and gives back morphisms in category B. Which make the square commute. Yeah. Ah, okay. Okay. So now, are we ready to go for the double dual? Well, you don't believe we can do it. No, no, no. I don't believe we can do it either. So, I would go to the end. I mean, everyone says that when I'm trying to write those, it's obvious and simple, fun. All right. I said all this, you know, right? I said all this. Okay, so our category is going to be... B, the category of n-dimensional real vector spaces. Okay? Yeah. Well, this should work, Ronnie. And I'm in finite-dimensional students, I don't have to. I mean, I should... The spaces are all finite-dimensional ones. possibly, but let's if we can do it in a simple case okay, and so we now shall we try and simplify it for ourselves by using dirac notation for this? that won't simplify it for me it won't? just where you have your vectors in the right pointing that's a vector The problem is that mathematically there's no such thing as intrinsically being a vector or intrinsically being a dual. Well, there is actually, but there's no such... I mean, it's a, quote, set theoretical discussion. Okay, so we've got a map, which we'll call star, mapping V to V, where V star is the dual space of V, It's a set of all V goes by F to R, such that F is linear.

37:30 So essentially it's a set of all linear maps from V to R. okay that's what you mean by the jewel of the vector space okay so what's the trick so what's the trick there's this map is perfectly straightforward. There's nothing unnatural about the map. You've got a perfectly natural definition of what you mean by the dual space. What you haven't got is a natural isomorphism because these two things as dimensional spaces are isomorphic. So the way you get an isomorphism, You take a dual, sorry, you take a basis, take a basis V1 to Vn of V, and now you take the chronic or delta function on this basis as your dual basis. yeah okay so so this is everything here is natural module picking a basis but picking a basis is not a natural operation it's just we know there is one but we don't and no okay and then but we still but nonetheless it's natural because we have this operation and that takes us from the isomorphism relative to one basis Yeah, but we're right. So if I put the bases in, everything is natural. Yeah. But that's saying a vector space is a vector space, together with its particular bases. It's a vector. Sure, okay. Okay, so that, so by making an unnatural definition of the structure, we get, we can make everything natural modulo of that unnatural definition. Okay, but now notice, we've also got to map double star, which maps V to V double star, right? So that's the dual of the dual.

40:00 So this equals the set of all linear maps, V star to R, mapping V star to R, such that F is linear. Okay, so far everything is perfectly straightforward, okay? Okay. But now notice, however, that here, there's automatically associated a natural isomorphism. Okay. So, namely, let's call it alpha sub v, okay, which has to map v to v double star, takes, So alpha sub v of little v is a function. It has got to be a function from v star. So it's lambda f. Okay. And what's its value? Sorry, lambda f It takes a function. of v. Yeah. linear function, I'll say. It's value. It takes an object here, so it's the evaluation function. Okay. I'm beginning to think we're on the brink of success, because now I've got a... It's just I don't know what the functions are, but I've got these maps. Okay. This map hasn't got any sort of natural isomorphism involved. this one does okay so i'm going to say that this is i'm going to say that the double jewel is naturally there's a natural transformation from the identity yes that's right to the double jewel oh so that's exactly what it is and that is it let's check does everybody agree with this? from that on. I got to this point, and then I was in the coffee shop, and I got to this point, and I looked down at my watch, and it's fine, and it's still, sorry. So, let's spell it out. Spelling out, like the diagram I drew before, what are the categories? What we want to say is... This is of vector spaces. Okay.

42:30 We have linear... We'll just... As a first approximation, let's say, n-dimensional real vector spaces. Yeah. Okay, well, I don't think it's going to be required. It may not. It probably isn't. Okay, so, now you've got two functions. The first one... It just takes one to solve, the second one is the double solve. So it takes things to, they're double joules. So, okay, now, first problem, suppose we've got two boxes. Is this, what is the, what is the function map on this? What is the morphism part of the map here? Double joules. So you've got to take a double joule, say. So, so, to get a function, you've got the object part of the function, what's the map part of the function? So how do we specify what we want? Um... I've got the awful feeling that this is going to have to be the application map as well, somehow. of application, right? So the magic is going to be built into the functor and to the natural transformation. See what I'm saying? So the things in here are things that eat linear functions on V and return. So these that take F into F of V. No, no, they take F's too. Hang on, things in here, take F's to... Things in here, take F's and return F of V1. And we want, over here, we want the result to be things that take F and return F of V2. So, we want this to be something like composition, don't we? Yeah, yeah, we want this to be composition with... Yeah, this is already, you see what this is doing.

45:00 We're calling everything F. By default. So this function is... That function is from V to R. That's just... V1 to R. Well, you've got it here. V2 to R. So this is... Sorry. So little V1 double star is something that does this. Yes, that's right. Take something in here. Take some vector in here. It maps in here. See, that's what I'm saying. You've got to build this application map Yeah, that's a, that's a, that's a, that's a kind of, I didn't tease those things. Well, I mean, but what is it about the application, Matt? It is that it's a, a natural, I mean, if you're looking at it from the start, so you, I mean, remember when McLean and, and Annenberg did this, they didn't know what a category was either, or a functor, right? I mean, it might be actually interesting to look at their original papers. what they said. Okay, so all you have to do, Richard, is just specify what this... you've got this... what is double star? Yeah, that's what I was saying. So, suppose you've got F... maps... maps... those maps V1 to R. No, it's V star, so it's double 2. So, suppose you've got some F in there, you've got to take it over here. So, it takes something... it takes a linear function on something V. 1 into R. That's what, if it's in the double jewel. Can we have a moratorium on calling things F in any form whatsoever? No, no, this is, I'm trying to say, take the number of this, what number of this corresponds to it. Ah, okay, I see what you said. Take some F that goes from the single jewel into R. Yeah. So F is not B1, then what are we going to call this? That's going to be called, no, that's going to be called double star of G. OK, G double star. OK. OK, so what is G double star of F?

47:30 Well, it's another function. Well, this one goes from V2 star into R. Okay, so what value does g double star at x of some function in v2, so, I don't know, h, what's its value in r? Yeah. I think you're right. You've built an atom between B1 and B2. So maybe you... I think you're right. You've got to go backwards and die. Yeah, what you've got to do is compose either... It's got to either be G circle, whatever that was, F, or F circle G. Right. So what you want to do, you want to go... So, so here's V star, V1 star to R, and here's V2 star, is it just to R, right? And now, sorry, I shouldn't have these go the other way around. Okay. But you've also got V1 going by G to V2. Okay. That's an isomorphic. Yeah, but we haven't got, that's a whole point, because we haven't got an isomorphism here. Oh, sorry, you're doing, oh, you're doing, sorry, B. I've got to do it without, I can't bring in, I can't bring in unspecified. No, I was just meaning that, given you want to find this G double star, you could do that. No, because there's no, what is the going back now? What is the going back now? What is the inverse of the evaluation map? That's what you're asking. You know, the star of the identity. So there's an identity. So here, the application map goes from, the application map, application goes from

50:00 application at V goes from F in V star F of V. So is there a way of thinking of that as a function of V? So lambda V Okay. That gives us a map from, what, V star. That gives us a map from V1 star to R. It gives us something in here. Okay. And now, but if I compose, ah, it's, I want this, I want to do this one backwards because I composed, I composed that one with G on the left. Okay, so this is G double star, F, the H is G circle H, which maps. What? What's happened to F? Well, I'm not confused with the letters, I mean it's different. We need small, we need v's for, and we need v's and things for in this. We need small f's for in v1 star, and we need capital f's for v. We need superscripts and subscripts. So, we need to use tensor rotation. the trouble is we've got functions of functions of functions yes right so we know what the alphas are we know what the object maps are what the hell is the is the

52:30 Okay, takes functionals, and so on. Now, this thing calls it to us. Let's do this. Oh, no, this one is just... Oh, yeah, okay, that's double stuff. This is the identity, and this is just... Plus G. Yeah, that's G. That's meant to be a g, so small letters around f are meant to be members of the g. OK, so let's make this one... It's just meant to be a... ...morgulism, as they're all. That, and then, that. OK, so we want to know what... 5 double star of f is. That's right. OK. Well, 5 double star of f is going to be a function that goes from, it's going to be a number of this, it's going to be a function that goes from f to star of f. Oh, yeah. So, to specify, you have to specify, double star of f, you have to specify its value at some linear functional from f. What is big f? Well, big F is elements of V1 star. Yeah. And little F is? Little F is just an archery elements of V1 star. V2 star, sorry. Yeah. Okay, so if, but, but, can't we? Okay, so we've got to go from here to here, so why can't we reverse f here, follow it by phi, and then down? Well, no, that's what I was saying. You could do that, but what is reverse f? That's going from real to which? So you're thinking of reverse f being something that goes from the functional back to the vector? Well, we know that F inverse exists, right? Yeah, because it's massive. So we just have to calculate what it is. I mean, that shouldn't be that difficult, is it? I mean, if you've got something, if you've got a... It goes from V1 double star, right?

55:00 So it goes from a functional on V2 to the F... Okay. And it takes an arbitrary functional on B, real functional on B1, and sends it to a real. Okay. That's just what it is by virtue of being it, right? Yeah. Okay. Shite. As long as it was, I guess you'd prove it in Jacksonville. It was in Jacksonville. Well, I mean, how do you... Let's prove it in Jacksonville. We ought to be able just to construct a functional, that is, a member of V1, a V1 star, Look, if I pick an F, so, so, let's pick... Just kind of understand that. That's the word Prover. Jesus. Jesus. It can't be this hard. It looks like it's staring us in the face. I mean, you could just say, well, it's got an inverse, so just do it. Well, that's a little bit fishy.

57:30 It's a little bit unnatural, let's put it that way. And when you go to infinite dimensional spaces, this doesn't always work. No, so you got larger. You don't get back to where you were. The dual space has got the image of V under the application function is not the whole of the dual space. So we've got an object f that takes functions in v star to reals, by evaluating them at some vector that's built in there. But what we have to do is find the vector. We have to define a vector such that f, that aft. I can, I bet I can do it if I had a basis. But that doesn't help. Right, I wonder if this is as the f of the components that... Is that right? So, f is just some member of v2 star, which is this side. So, you want to know what this function is. Yeah. It takes a dual space into a dual space. Yeah. So, you have to say, what it's, what, give it a function f, which goes from the dual of v1 into r. Yeah. And give it a function, little f, that goes from v2 into r, so a member of v2 star. Yeah. What's its value? number. Sorry. No. Oh, fuck, man. So we put... Yeah, no, so, no, yeah, that's right, because

1:00:00 this is meant to take this, so it's meant to say what the real number is. So this, this thing is the, is the member of this. So you evaluate it as a function. So, okay, so we We know F goes from V2 into R. So first of all, take something V1, take this cross to V2 by Y, and then take it to R by F. So that's this bit. That gives you a linear function all from V1 into R. Yeah? And then just evaluate so that's a number of V1 star so then just evaluate take F of that linear function. So that gives you a number for R at least. And it includes large f, small f, and 5. The question is whether it is. So that's what we're trying to do. Oh yeah, go to the function. The functor. The arrows So, what do we have to prove to show that that's a function factor? That everything composes, but I'm still not sure what exactly... You're supposed to start out with a function. You're supposed to start out with a real valued function from v1 star to r a real valued function from V2 star to R? Yes, yes, exactly. Okay, so now how do you do it? Take a real valued function from V1 star to R, okay, and then you're supposed to get a real valued function from V2 star to R. Okay, so give me a member of V2 star. Yeah. F. Okay. So it goes from B2 into R. Okay, so it goes from B2 into R, F does. So F circle phi then goes from B1 to R, so it's only a functional. And it's in B1. So it's in B1 star, so the composition of F and X is a number of you want to stop over there for which is in the

1:02:30 one star star must take a real value so evaluate it at that. So this is our member of R. The question is whether that gives the right function. I'm sure it does. I'm absolutely... and how could you write... how could you write down the simple expression of all being composition and and the main components and have it come out fucking wrong. I mean, there's no way it's going to happen. It's like, once you work out the units, the value units, they're not ready to last. Yeah, exactly. I'm going to say, oh, I can't. Okay, well, what else could it be? Well, okay, let's suppose for a moment, I'm getting a little bit punctual, but let's suppose for a moment that this is correct for some varying of it is. so the last thing that comes out of this is that well of course the other piece blind faith we're operating under is that this functor now properly defined where the object part of the functor goes to the double jewel and the morphism part of the functor is this mess here that that functor that there's a natural transformation from the identity functor to this functor and that is witnessed by the map which is the application map So, if I say... But, hold on, before we demonstrate this... So, the identity function is from D to V, and the double dual is from Z to V double star. Oh, no, these are counts of the cascade of all vector spaces. Ah, okay. No, it's the cascade of both. That's why... Yeah, I mean, you don't try it, but you could just do an dimension because the dual is double dual. same dimension but finite. I think it's better to do it double on the same dimension first because the multiple dimensions might booger out this. I mean, because of this thing. Yeah,

1:05:00 yeah. They get that right in the simplest case and then I might do it. When the simplest case is this I don't want to jump straight into the less simple case. OK, so, here's two things, and here's maybe... So you have the identity map, obviously just takes it down to V1, and then each is going to take this one to V2, And then you have the double dual function, which takes this one to be two star, and this one to be two star, and this, okay, so suppose you have an amorphism here, well, yeah, sort of that, so you have it, well, that's, and this is the final star, and I'll just call that final star. So, it says that the application map here... Alphas of v1 and alphas of v2. Because alpha is the general map. Yeah, that's right, from the whole vector space into this. No, no, alpha is the map on the whole category. Yeah, alpha 1 is just the one from v1 into... of v1 so we can show that it's got v1 as its argument. Okay. Okay. Okay. So that's not the whole community. Okay, so now the lesson that comes out of this is that in order to make this, that's the way to state it in terms of natural transformations between functors, but in order to use this example you not only have to do the application thing for the um natural transformation map which is alpha but you also have to do it to give you because you have to specify the functor to which the identity is given is is uh naturally transformed naturally equivalent that's the dual functor yeah but the dual functor the whole problem is to say it's obvious

1:07:30 what the dual functor is on objects the whole problem is to say what the dual functor is you know that's what that's more business you don't have to invest you don't have to use the application function if that's right right if that's right it doesn't make any appeal to the application 5 double star of f applied to f, little f, r. So it really says that 5 double star of f is a function that equals g, which is an f, 2 double star, where G of F, which goes from B to R, equals, um, uh, what was this? Um, F of, big F of little f composed with phi. Yeah, exactly. Okay, so little f composed with phi takes you from v2 star to v1 star, right? Not from v1 star. No, it takes you from v2 star, so you've got v2 into r, and now you've got v1 into v2, so that's phi. this is, what do we call it, little f? Okay. Yeah. So when I compose that, so f is the following phi. Is a linear function. Is a linear function. So that's a number. You won't stop. Right. And now we can apply f to that. Yeah. There you go. That's it. Yeah. Okay, so we have to, the point is, we have to do the trick with the evaluation map. No, we don't. The point is that you don't. Yeah, exactly, exactly. It looks like using the backwards application map would work, I guess. But it also looks like you don't have to, because of that. Maybe it wouldn't get the right function if you go backwards on the navigation. So what's this? Does it commute?

1:10:00 Well, backwards application map would surely commute because the backward backwards application map the inverse application map would work well but does this does this diagram commute with this definition of that of a five star we just need to chase things around right well we want somebody can some great soul so we want to show that going this way around, which is alpha V2 following phi, equals phi double star after alpha V1. We want to do that by taking some arbitrary element in V1, and chasing it in two different directions. Okay. Okay. So... So it's A-B. Right. So we've got our two different journeys started out, and hopefully they're all in the same place. So, phi on V is just some V prime in V2? Yeah, well. What do you want to call it? I don't know if I'm calling it phi of V. Yeah, that will be fine because this thing will come out with phi of V to me. Okay. So that's alpha V2 of phi of V. So the application, which is the application map on V, alpha V. Which is, so it's some, I'm sorry, on phi V. So it takes, so you want to know what its value is at a linear functional. Some H in V2 double star? It's actually H maps. H maps K in V2 star.

1:12:30 To K evaluated at 5V. Okay. This, on the other hand, is the evaluation map of V. What do we define it to be? It's F. It's findable. So, first of all, take the evaluation map. The evaluation map of v. This is what we'll just run here. It'll be double star. Okay. How does the evaluation map work? It takes something like v double star and turns it into... It takes something like v star. So it's something in the dual space of V1. So this little V double star is playing the role of capital F over here, yeah? Because this is the thing that carries double VLs into couple VLs. Yes, exactly right. So what this is, we've got it like that, it's here, or whatever, yeah, there's some G in, I mean, two double star, such that G maps, what are we calling it there, F, or we call it K, just the same, I guess. Yeah, take so. G, that's K, into... Well, the valuation... Into, sorry, yeah, into... So, the valuation over a double star. So, into the double star. Evaluate to that. Value over that. Little F. Yeah, little F. Circle. So, in this case, K... Circle. Our original morphosome. Right, now we want to show that that function, that function are the same. Right, so take a K. Yes, that's right.

1:15:00 a cause takes on K the first one, we know it's this first one goes to this, the second one is the evaluation map of this which just takes a map like this and evaluates it to that yeah, which is exactly exactly right what are the chances of that try it year nineteen forty-four eighty uh... over sixty three years buying up to date uh... this is not this is what I was saying to John, for you kidding me, that something funny about all this stuff that when I figured out that, I didn't understand anything about why I'm out of it just because of the same dimensional analysis you just sort of chug through what it should be and it turns out to be well we're always talking about calculating he's always saying let us calculate so if you're a whiz bang algebrist you just I mean it's like you look at a polynomial and you think oh shit that's just these two you just factor it by side right and this is what you're doing there so you're factoring these things by side that's what factorization problem and you factored it by sight yeah it's the only thing it could be right that's what you did right so so so it is turning this stuff into algebra so so what this is like chasing the arrows inside the cat inside the objects of the category is set theory and therefore bad. But doing calculations just like that, that's doing algebra, and it's good. Okay. Okay. I may say, so what the problem here, there's a kind of general problem in the foundations is it's like trying to go from conventional algebra back into the theory of ratio and proportion, and as it were, figure out exactly what's going on everywhere, right? I mean, we've done a little bit.

1:17:30 It's a mess. I mean, God, that's hard enough. And this is... Yeah. So the problem, the foundational problem, to really understand it, maybe uh take you into uh regions where nobody really wants to go like who wants to figure out all the stuff about rational arithmetic starting from uh eudox's definition of proportionality and coming up to the sure huh but i mean somebody ought to because otherwise you don't understand what's going on sure sure the genius of the of mathematics is that you can as it were scoring the base degrees by which you did ascend you can always just you know you have this messy theory yeah you know and you move you just move up to some more abstract version of it and then the original problems all disappear because you can because the new thing is so simple I think the analogy is really good because there is something very similar about doing tricks doing just you know the calculus of fractions or even decimal addition and multiplication you don't you don't see what you're doing or why you're doing it you just do your thinking for you it's very necessary mathematically unsatisfactory so he wanted but it was natural about it is you can do it you know i mean that is what is the natural transformation uh... why is it natural what what so what is a natural what is what is natural here is that there's no we know that all these things are isomorphic because they're all in dimensional spaces that incidentally by the way requires a lot of work or it may be routine I can't remember I've done it, you've done it when you learned it linear algebra okay so what's natural here is that these are all

1:20:00 real spaces so they're all isomorphic okay so the dual and the double dual are both isomorphic to the original space but in the case of the dual you have to have in order to get in order to define you have to start with them I mean you have to you have to come up you have to specify a basis for the underlying for the sure domain space and then that gives you a dual basis in the in the other thing but you have to use the basis so it's not as it were coordinate free as the phrase goes right because putting in a basis is introducing coordinates yeah i just see coordinates so what this is a coordinate free definition of the isomorphism okay you've got a coordinate free definition of the isomorphism star from b to b double star but there's but but there's no coordinate free definition of in the case of the dual yeah although if you put the basis in there's well i don't know whether but if you put the basis in it the the map from you know from the vector from the basis to the Kronecker delta on that or the projection onto the particular so what is the what is the dual object if you've got if you've got if you've got A1B1 plus and so on plus A and BN so the dual just gives you AN which just projects onto the N coordinate so it's a projection on so those things, the VN stars now are the or the dual basis okay, so I mean that's natural so you've got to be careful about what's not natural but what's not natural is having to tie the definition No, I see that, but I don't see how the natural transformations say that. You know, you've said all of that, and I understood all of that before I knew what natural transformation was.

1:22:30 Ah, so you want to know what explanatory power of the natural transformation? Well, apparently it's not even explanatory power, but from what I could read, you can't even say certain things without the natural transformation. natural what they claim is the natural property of us can't even be sad okay so what's unnatural about this example of a natural transformation what's unnatural about it is the natural transformation naturally connects two filters the obvious we obviously want the identity functor here on the domain, I mean at the top. So we're connecting the identity functor with this double star functor. What's unnatural is that, what's unnatural is figuring out how to do the double star functor on Morphous, what it does on maps, not what it does on all this business here is is in lieu of doing a business about picking bases and so on yeah but the construction of the application the idea of the application you get from functions on the vector space just by evaluating at some point that's the straightforward part that's the bit that takes us from v1 to v1 level star that's the natural part well that's the straightforward part in order for this to in order to demonstrate that there's really something powerful it's not just a simple notational trick we also have to demonstrate that this construction this application of construction gets along with kinds of operations we'd want to do on vector spaces. And having this square commute, as the calculation we did shows, means that this isn't just something that works in one particular vector space and there's a different construction that works in a different vector space. So if it were the case that for each vector space we could make a construction like this, then that would be unnatural in that sense.

1:25:00 for this. It's localized. But what this, what the community of the square, the community of the square demonstrates that for each construction that you do in one vector space, you can port, you can find the exact counterpart of that construction in the other, and indeed it's the construction you would have made in the other vector space in the first class. I'm sorry to see that. I guess I don't totally see what the big deal about commuting squares is. What's so desirable about that square of commuting? Is it what you just said, that it's the thing you would do in one way? It's not the commuting of the square, it's the existence of the alpha that's significant. Yeah, but the existence of the alpha is such that square commutes. Yeah. You have to have such, you know, it's no good just having an alpha. Yeah, it's the existence. It does those two maps. It has to then commute, like Stuart says, get on. I see the metaphor, but I don't see exactly what it's significant. So, you know, what's so good if you have a vector space, you go into another one. So you're able to do it on your map into another back to space and then you can just do this map into its dual as the same it's just the same function as if you were to do the dual with the first one and then take this dual function back. This is supposed to be saying that there's a natural connection between spaces and their double duals which doesn't exist between spaces and they're duels even though we got isomorphism in both cases and in this case the isomorphism is kind of uniform but how does that how does this and i agree with richard how does this commuting square say that so for each of the v1s and it was over to that puzzle by what a big deal about commutative so we've got we know that we can make the construction so this So the straightforward part is this little trick, the evaluation function, yeah, the evaluation trick. It's alpha sub v1. That's the point. Alpha is the uniform part of it. So for each v1, v2, v3, there's some trick that gives us a double dual for that.

1:27:30 and but just given that fact just knowing that you can you can do something like then you know this is doing the straightforward thing means that you have one of each one of these for each v1 v2 v3 etc well there's nothing to tell you that there that there's anything related about them or there's anything uniform okay but okay but let's put it this way in the category thing in category theory I mean the insight that category theory gives us is it's not just algies it's algies and their morphisms yeah right so the this business it forces us to define the the functor because that forces us to say what happens to morphisms exactly so it forces us to do all this shit here, right? That forces us to calculate, right? See? Yeah, I see that. To see what it is about this alpha function. The definition of natural transformation doesn't say anything about uniformity. It just says that there's a class of morphisms. You know, to actually say the thing about uniformity would be to talk about this category of functors and this is what's getting me interested in this because the way I look at set theory, the only way you can do these big things is by means of rules you can't do species and you can't do global functions except in the context of rules I take that point but that's not how the category theory is what the category theory is saying is it so natural here given they're not saying I can be curious to say, the vector spaces, V1, V2, etc., are just objects in the category of all vector spaces, and any construction you've got that just treats the objects as disjoint things and does one thing, one operation for each of those objects, but that doesn't get along with the morphisms, isn't doing anything to do with what these objects are. really are because all of the content of what the objects really are is encoded in what's

1:30:00 preserved by morphisms. So just a set of all vector spaces isn't interesting, but the category of vector spaces in which the arrows are just exactly the thing that preserve what's interesting about these individual elements of the set. But in an arbitrary category, you can't think of arrows as preserving anything. They're just arrows. This is true, yeah. But this is in this case. We're interested in this. But in an arbitrary category, what we're thinking, one way of thinking of what the arrows do over and above nearly the set of objects us some kind of that's sort of a generalization or an analogy with the contents of the object it's a kind of like mitzian vision of what these structures are they're just monads and they got windows on the world namely arrows i mean so they got contact with the rest of the world so you you the whole thing is somehow together is it's their relationships that are important between the objects yeah and the relationships are given as these morphisms sure sure so so as we have here if we have just some construction that works for each vector space then that's not then that's just treating the vector spaces as elements of a set of but there's no guarantee that it's really doing anything that's relevant to the vectorial structure, to the linear structure of the vector space if this construction didn't get on with the morphisms between vector spaces in a natural, rule-governed way which is brought up by this algebraic calculation which is the same if it didn't commute with impotence it's the commuting definitions of the alphas and the phi double star. And what this way of looking at it calls our attention to is that this double dual thing is not just associating the object V with the object V double star. It's associating the object V with the object V double star

1:32:30 and any morphism out of V by the morphism Phi out of V by the morphism Phi double star so the whole message is that you've got to relate it's not enough to say it's not enough to make the make the transition between objects you've got to make the transition between morphisms but what expresses what the what the transition between morphisms is is this commuting square i guess okay it's just because you what you're saying about the linear you're preserving some of the linearity or the the linear structure but that doesn't there's no need that these two things should be equal to preserve that only that, say, taking two vectors in there the result of taking each of them down to here and then mapping across here is the same as taking their sum here and then going down to here and then going across here. It's all in here that the linearity is preserved. It's why do these have to be the same map? That's what I... have to be the same, then. This and this. What does that tell you? What does it preserve? So what we need, one way of thinking of it, perhaps, I don't know if this is going to be an answer to your question, is we have one of these alphas for each vector space. Now, when we have a map between the vector spaces, what we need is for there to be something corresponding does the same thing to a morpho so that if we transform the vector space into its double view and we transform the result of it I take that because that's generally what these commutative diagrams are supposed to do that somehow it's this part is sort of mirroring course it might be forgetful or like whatever but you somehow it's a parallel structure yeah exactly

1:35:00 you're saying that these two this thing plays the same role in this yeah well you know there's sort of a rule assignment thing yeah these sort of are mirrors or something and then you say well to say that it's genuinely a mirror is to say that um and some we're connecting that is is the same way as connecting that well that's what it's generally do i still don't see why that doesn't makes this no I mean the thing that one suspects I mean why is this an equivalence I don't see that it is an equivalence I don't see why it's an equivalence unless you have the thing going back the other way do you mean equivalence in a technical sense a natural what were what are these called natural equivalences? No, this is just a natural transformation. Natural transformations or equivalences there. Ah, okay. So now, if, now the question is, is, right. Okay, now, we know that there's a natural transformation from the identity functor to the double-dual functor. Is there a natural transformation from the double-dual functor to the identity functor? That would be what we would need, I think, in order to that they were equivalent naturally yeah yeah i think so so is there a natural transformation is there a natural transformation from the identity functor to the no from the dual function that's the identity so are there perhaps that do yeah well there's a well yeah presumably well So, if the space... The inverse... The inverse amplification, but we weren't able to say what that was. Yeah, but it does exist. So, that's small. Isn't that so? I know. Yeah. What's for that really nasty notation? Huh? Really ugly notation. Minus ones for inverse. Yes, I thought that could be. What's the mathematical notation about process of natural selection is made more and more elegant, and that one never has been. There was a nice lecture by Jerry Sussman, who wrote one of the standard textbooks in a computer program, talking about how to teach the way of thinking about how to solve a problem.

1:37:30 So, for example, he teaches in electrical engineering, and he drew out a big circuit on board, and he says, the way we teach our students how to analyse a circuit is the way that's easiest to teach, which is you teach them some very simple rules for junctions and voltages across the parallel, this kind of thing. but to actually use these rules to analyse this circuit, it was only like a three stage amplifier or filter or something and to actually use those rules is very contradictionally intensive and essentially impeasable. What do I do when I look at a circuit? Well I say well this is obviously just such and such a gate so I know that this thing is going to be equal to that thing and then because of that I know what a voltage is here and he's just using these sort of high level heuristics this isn't some kind of black magic this isn't merely the product of having had lots of experience and remembering finding analogies between things he's looking at and things he sort of remembers he's got an algorithm and in fact he's written out the algorithm precisely enough that he can hand it in five pages to his students and say if you learned this algorithm You can be as good at analyzing circuits as I am. And... So that's partly what we need here. That's a notation. Well, not so much a better notation, but... A pedagogical strategy for one thing. A pedagogical strategy. So all the struggling that ended up with is no longer struggling, but it's... Oh, I'm sorry. But it algorithmically emerges, so we say, well, we know what kind of thing we need, we know what the starting point and the ending point of this function has to be, so here's the obvious path that gets us from what we need. This is kind of what I was meaning earlier when I said that you can do all this stuff without thinking. No, that's an algorithm there. I'm already mad at studying what I'm saying. But I think that's one of the nice things about this, and possibly this is where some of the naturalness emerges, where it plays its role, is the fact that, so, the, the, um, the, like, uh, Dan the Pony

1:40:00 who wrote something else about the Unadolem, which I haven't yet translated into something comprehensible to one Haskell programmer. Right. But one of his other projects in Haskell is, um, writing things that, so Haskell is strongly typed, every variable of a type is that an integer or a list or something. And he's investigating what happens if you write functions that are type correct without any filters as to what they're meant to do, and then see if they're useful. And a lot of the examples... So that's what's dimensional analysis is what that is. Exactly. It's the type version of dimensional analysis. And he finds that, you just throw some, you take some already useful functions that are of certain, you know, of type A into B and type B into C. Plug them together and do some various obvious things with them, and you get things that are actually useful. And I think that's the same. Well, I just report by way of rousing your curiosity that, well, Veer, when I ask him if he'd ever worked out the stuff on dimensional analysis, he's in my course. where it says that was it he hasn't written it up you know but he's worked it out well i mean you know this business that's that's really deep stuff because it's it's getting it the idea of what a physical quantity is i mean that's deeply mysterious the simplest kind of physical quantity like a moment you know anybody that understands the principle of the yeah knows what these qualities are but what's a mass times the distance what kind of a quantity is that yeah right but but you can sort of see this picture the big mass on on the little end right yeah I know to go well that's a much actually I it's been quite useful yeah I think the thing is that if we're going to do what we set out to do we have to push off in directions that a category this might think we're crazy yeah huh yeah i think you're possibly right

1:42:30 we're trying to deconstruct this stuff not to but but we can hardly deconstruct it before we get the idea of what it's about yeah but i i still think if we get this you need of them i think somebody ought to write this up all right well this this example oh this one yeah maybe we should all do it in our privacy of our own bedrooms yeah piece might be the the cancer theory diagram thing it's very easy to use yeah I'd recommend that anybody how do you put we pull it down off the web well you probably have it in your thing already you just have to put use, I'll send you how you put it at the start, and you can try and compile, and if it says it doesn't recognize, then you'll have to take it from the web, but you'll probably have it. And then it's basically, so you just do it as arrays. So just as you would do an array, usually, just with your object names just in the places. And then beside each object name you want an arrow going away from, you just write backslash AR. you put either right or down and right or down down and right for the where you want the arrow to end up and then you just put a little subscript or superscript say where you want the name to be on that tower so that simple so you wanted to have something like that the next you would just write in your Oh, I could do that, but what about a triangular diagram? Slash, AR, right, superscript F, B, and then AND, B, so that's one part of your matrix. For a triangular diagram, you just space out the matrix a bit better. So if you want that, then you just make the matrix like that. So you have and, A, and, new line, and, B, and, and, C. So there's a space, A, space, B, space, C. Nobles, yeah, well, yes.

1:45:00 They'd be tripling space, huh? Well, because the ands, so the way it... Yeah, so the and, this can calculate the space here. So how do you get the arrows in? And then you put, if you wanted an arrow going down from here to here, and it starts at A, you put it in with the A in between the two hands that the A occupies, and you put backslash arrow. You want it going down and to the left, so down one and to the left. You put DL down left, and then put in the name that you want, superscript F, And that would give you that, or if your B was down two levels, you would put down, down, left. It's really simple. I thought it was going to be a nightmare. So it'll draw a straight line, but it'll allow you to cross straight lines, presumably, so you have to be careful. Oh, yeah, I should do all manner of fun, but you can always correct it. There are ways to do curved lines as well, ways to do dotted lines, dashed lines, double lines, so you can do funktos as well. It's really, really good. There are ways to do all this. There are ways to do boxes, big stacks, big arms between them. It's very powerful, but just to be simple, you can use sort of diagrams that sound... Right. Yes, that's... I really have got to...