Categorical structures & cohomology (lecture 2)

0:00 In the integral sense, let's say, it's xn, yn, so it's component by component, with obviously the arrows, so plus or minus 1, towards x-1, 3, y, n, plus or minus 1, more generally, which are the component arrows, for example, here I have the arrow ti, there it's going to be ti, so the phase operators and degenerations, These are those of each component. There are those of x and those of y, which defines those of x cross y, and the same for the other operators. And also, what I did not say in the same way, is that a morphism of definition 2, so that's 1, and 2, a morphism of simplicio,

2:30 In the simple category, there are morphisms with the usual laws of composition, and we have a product in the category as well. So now I want to talk about geometric realization. We saw that from a topological space, we could associate a simple complex by the singular complex. And in the other direction, here, now, either x, an integral set, the geometric realization, generally x with two bars, 2x, well, in the following way, it is another thing that I wanted to remind you of the last time, we saw, for any integral set, it is complicated to write it, we said at the end of the last time that x itself, As the reunion is a tautology, on all the x's of the n-simplex type delta n, so with xn, so it's the reunion on an application, it's like an application like that, an element of xn is an application like that, it's like this reunion there, and we can also write it under the form of delta n cross xn.

5:00 And if I take into account the morphisms here that send this to delta m, with the y's here, I can identify things here. In other words, here at this level, this is the union of these, and quotient T by a relation of equivalence, the following. I'm going to make a mistake here, it's delta n-1 3x here, the following, which is that for T in delta n-1, And the same goes for degenerative operators. I would like to remind you that every time I have xn here, in xn-1, the operator di corresponds to the operator delta n-1 in delta n. We have di, which is the corresponding operator, and the same goes for degenerative operators here. This is a functorial construction that will be implicitly used to bring back statements on x to statements on space. The simplex x is a simplexial set, and then it's just the meeting of things like that, glued together, and so statements on it, to demonstrate them, are done on it. So that's exactly it. In the case of a topological space, because for a topological space, we already have, as here, we have a type-synthetic set, we have a type-topological space, we have defined the type-synthetic set, and so we will pose, we pose now, here is the definition, finally, we define a thing, the reunion, this time, not of delta-n-synthetic sets, but of topological spaces.

7:30 The notation prepared this notation here, but I defined this set, it is the space in Rn plus 1 generated by the n plus 1 point linearly independent and so there I continue this cross xn modulo the same relationship of equivalence. The geometric realization of delta n is the space bar delta n that I defined the last time. And so it's justified now. We now have a word. Are there questions in English or am I going too fast? All right. We defined last time the nr in plus one, n plus one space, Euclidean space, a space which we call this way. And now for every simple set we're defining such a notion. And I'm saying that there's no contradiction between... If there are people who are one hour late, I will delete this list, these are people whose emails do not work when they are communicated with me, so if there are people who have arrived who are on this list, come see me later to give me your external address. So now, the two factors, topological spaces are the affichio ensembles, and so we saw the last time that we had the singular complex application,

10:00 And now we have the geometric realization vector that goes in the other direction. So, the proposition is that the vector in the infinitive sense, x in s of x1, b in the basic of the proof, gives us the proof that we go back to delta n. The total space of delta n with a value in y is equal to the nth part. So I say, but this is the definition.

12:30 So if we have this, then it's just a question of limits. We write then that, let's say I'm going to start the rest of the demonstration, The delta-n. We write this as a limit on the part of B.S. and on this we know this statement on delta-n and the commutes as a limit for applications, well, after all, we have to find them one by one, we don't have to find them, it's easy. And make a little quick description of the topology of a space, what kind of spaces are these geometric realizations. It is necessary to not give false statements, however, it is really a bit orthogonal to the spirit of the course, there are questions of general topology, and so I will just give a little guide to say things quite quickly, I will eventually send those who are interested to references. And even in English, in the middle of the road, I'll say roughly what it means, but in English, it's for all of us, thinking about the realization, its geometrical realization, complex, but at least the letters, that's for Roger Feinheis, which is practically forgotten, but in the end, that's the reason, and so, let's just say, so I'm not defining that at the moment, by the way, I'm not going to define it completely, we'll see, but let's say, since the vector is a joint,

15:00 This is the diagram, the Cocartesian diagram, the following reality, so you will know the diagram, the realization of the K-1 skeleton of X, this space here I will call it YK-1.

17:30 We can see in the geometric realization of delta N, which we have seen, this is the syntax we know, always it is for X, so NXN, here we have the Cocartesian diagram with the skeleton KM. As I said, each of these inflective spaces is homomorph bn of Guilbault, and so we can find out from, there is an error here, there are two notations, k is the k here again, because they are the same, from y by attaching bk along, what we just attached, were the parts of the ball that were not on the edge, that were not in the sphere.

20:00 So, the difficulty, why? Well, that's pretty clear. In any case, everyone understands what it means. They have something else. The topology, which is the co-limit of y, k and z, that is, it is some kind of an application of y in z and continues if its restriction to y, k continues. In any case, an open one, yes, that's it. The idea that we are in this situation where we have to be careful of closure finite is that the number of cells, no one said that there was only one finite number of non-degenerate elements in the n-simplex, so at each point we are perhaps attaching here an infinity of cells to pass. And so that creates the topological difficulties, on which I will go a little bit by saying just the following thing, which explains the words closure finite, the important fact, I do not know this course, of course, to bring back something simple.

22:30 From topology to algebra, and here I am obliged at some point to do topology briefly, so it's going to be completely different, that means the following thing, for all, an element, a part a in y, so a is closed, if and only if, a inter c is closed, all compact, c in y, that's the property. Key terms include open data such as U-bar is compact, U-bar adherence is compact, which we call locally compact, and compactly generated. This example is the most common, but we can't do that because here we have the following,

25:00 for which there are F's. In general topology, we call this a sub-base, sub-basis, if the base of the webinar is a finite intersection of these parts. So why are we looking at all this? We would like to have in the category of Fermi-Laguerre, with topological spaces, the exponential property, which is often written in the form of non-exponentials.

27:30 That is to say that the application of a product is believed to be this space here, and homeomorph is the application of x. In the classification space, YZ, with the topological contrast, with the topological contrast of Hilbert, and it is false in general. So, that is, ZXY, homomorph, to XYZ, which is XY, is noted. What we're used to doing is expecting the exponential property of z to the power of x, and z to the power of y to the power of x. It's not even in the right sense, but that's why we call it exponential. That's the property. There's a bad name for those of us who need it, so we call it... Yes, that's right, that's what I'm... If you want me to say it right away, I'll say Wudeberg. Conditional. So it's false. It's false. It's false in general. So we call... So I found in a book, Gabriel Zisman's book, Kelly's space. Which is not a very good name. People don't use it anymore because I don't think it's because of Kelly, actually. But, well, I'll leave it there because there's no... Kelly's space. Space, analysis of spaces. It's the same. House-dorff is the same, exactly the same.

30:00 The other possibility is to call it CGH, not C-generated House-dorff. It's not much better in my opinion. So, as I said before, the properties that we have, so that's true, maybe I should have said earlier that this is elementary, it's false in general, but it's true already, For example, Y is locally compact, but what I said earlier is not enough for us, because complex CWs are not locally compact in general, so we can't just apply that, that's why we need these consortia. The continuous application of yk to y is the same as the universal application of the continuous application in y with zk. Each time I have an application like this, yk here, and I can each time factorize this in this way, each continuous application and each zk, it is factored like this.

32:30 The topology is equal to y itself, but with the topology, the same space, but the topology on y is 10k, which is the following, the thinner, no, the noise is always between thinner and thinner, the topology where the closed parts of yk These are, by definition, parts of the Y in the intersection of all compacts. I take this as the definition of my new topology, this is the space YK with a canonic axis arrow. This is my definition in the set Y and the closed parts. This is the one in Y, the one that the intersection with any compact is closed. So beware, the troubles continue, but we are now going to try, we want to say it remains true, for that it is true, but it is still not true, even for this situation, so we are going to cheat in a way that we will always do, and which is after all quite comfortable, it is if x and y in this space are equal to e, it does not imply that x believes y, the product, as a topological space, is equal to e. But we can qualify it, of course, x cross y, topological product. Here, I take the topological product as a space.

35:00 And if I qualify it, if I take this part here, it is a definition. And so, we define the product... Well, we will write it later, we will just write the product, it will be the product. I really don't like the term K-K-K-K-K-K-K-K-K-K-K-K-K-K-K-K-K-K-K-K-K-K-K-K-K- The functional space is KE of x, y, x, y, applied by, of course, KE of x, y equals the functional space for the open contact topology. We define for x and y in KE the functional space, KE by that equals the case in which we have qualified. And so we show that the functional space produced as a Kekele space with Z is homeomorph to the functional space of X at value in the functional space in the sense of Kekele of Y with Z.

37:30 And therefore, we will always use, without specifying, in fact we will drop these products, the functional spaces will always be those. Last words, because I gave some examples without justification of key spaces, so I give now that I justified why we have to look at this to have this application, and so now let's say that I said everything I repeat, all spaces, locally compact spaces, all compact spaces, so let's just say that we don't care about it, I think, but all measurable spaces. And finally, this is the point of what we have said here in all CW complexes. In other words, implicitly, we will forget it, but when we do geometric realization, we will allow ourselves, when we take products, functional spaces from CW complexes, when we move on to geometric realization, there is no problem. But the only problem there is, if we want to look at what it really is, it is not exactly what we believe. We have re-topologized spaces every time. And if we remember that, we don't have much to do. So, the last thing I wanted to say about this, in particular, because you can see that sometimes we don't need to do anything, if Y is a functional space, sorry, is a topological space,

40:00 then S, the singular infinitial ensemble associated, is actually equal to S , Delta n, the spaces, are compact. So an application of the same simplex with value in y, and y, k here, it is factored in a unique way there. So there is a bijection between the continuous applications of delta n in y, that is to say the things of degree n here, and the continuous applications of delta n in y, k, that is to say this. So at least there is no need for the essence of a simple CIO to be used. The S of Y is the one we are used to. Now, a proposition that we will demonstrate later, if we have the time. We will at least partially demonstrate it, but we can already know it. If I take the topological relation, it is the same thing. What we have to say is that we don't have a top, but we have a simple one, the category of inflexible ensembles, and this is the S-factor, and this is the S-factor, general realization, geometry, which is associated with an inflexible ensemble, the topological space of which it is associated, so it's really a 10K, and with this notation, we have this. It's the same in Flanders.

42:30 Obviously, to do a little algebraic topology, the problem with the quintessential theory, which is completely visible in algebra, is that even the things that in an ordinary algebraic topology course are done on the first day, take a little time, so now we are going to look at the first thing, which is the notion of homotopy between applications, and therefore definitions. Equals this, to make it look more like topology, the unit interval. So my definition is that of everyone, but we have to describe it. So, we say that the superficial application of the y-axis is a superficial application of x cross y of the y-axis, 0 is equal to f and xh is equal to x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1x1

45:00 And so, in relation to that, we can write again in the same way. So now, that's the definition, but now, more concretely, it leads us straight to some quite interesting things. Well, I'm going to do the same, I'm going to finish the same, I'm not even going to do it. The primitive homotopy of a sub-space, a sub-sympathetic ensemble A in X. So we want H... So H would obviously be A, 3, 1, which is constant, not A, 3, 1, which is factorized by A with the projection of the same sign, constant homotopy on the sub-assembly in question. So a more concrete but more algebraic description of H-homotopy is in the old manuals, it's not the one I'm going to say here that defined homotopy, but without any justification. Before we get to this justification, I would like to give you a more algebraic description of the pi-h-mode. Since x is the inductive limit, as we have seen on the delta-n in all applications, we return to the description of the applications of delta-n over i.

47:30 In other words, we return to the case where x is delta n, where we have applications of delta n cross i in the y-axis. And so, if necessary, it is a better description of delta, which I recall once again, than i is delta. So, afterwards, we will be able to write like that. And so, what is that? First, these are drawings. And so, in the case of n equals 1, which is very instructive, we are looking at I croix I as a complex simplicial, so this is the square, 0, 1 in here, and then I do after 0 prime and 1 prime in the other direction, there are these four segments, this is a delta 1 here, and then this is the other delta 1 here, this is the other delta 1. And so, there is a triangulation. If we want to look at this with an abstract sense, there is a way to do it, which is to cut it like this. We are not allowed to look at the applications of squares in a space, it is the triangle, the n-simplex that we look at. And so there, I have two. So, yi, yi is composed of two simplexes, c0 and c1, such that, if we look, we have to see what happens, It is more instructive to look at the case n equals 2, so we can do the theory for that, but if you want, it is the data of two elements in the singular space of x. We are talking about delta 2 cross I, and so this is a prism, so at the bottom I have my triangle, at the bottom I have the delta 2, here I have my I, so the drawing is this one, and it is necessary to fill the prism.

50:00 The basic of delta 2 is divided by 3 simplexes of delta 3. We have to triangulate this, so it's not one of the triangles anymore, but it's the case above. And so I just drew them, so we can see that it's clear if we look at the drawings. We can cut pieces in there, here, there, a stop and a stop there. And so, when I'm deciding here, I have, in fact, there's 0, 1, 0 prime, 0, 1, 2, that's this one with this one, and then there's 0, 1, 1 prime, 2 prime, there's this one, there's this one, and finally, there's, so that's C0, this one, that's C2, that's C1, and in fact, there's C0, finally. This one will be zero, zero prime, this one. And if you see, it's... And so, there are faces... The only one that is a bit hard to visualize is the one in the middle, because the others, you can see them well on the edge, we have three sides on the edge, and... And so, the relations are... They are... They are attached, glued along...

52:30 This is a common phase, so it means the following relations, d1 of C0 equals d1 C1, once we check what it is, that is to say, and on the other hand, d2 of C1 equals d2 of C2. That's it, and that describes everything to us. Now, perhaps I have erased too much, no, it's okay. So now, for everything, to explain this drawing in the general case, now what we do is the presentation. In general, there are not many references in the book, I'm sorry, so the place where this kind of thing really took place is the book of Gabriel Wiesman. Otherwise, there is a relatively old book, and not so much in the spirit of things, but for this kind of thing, it's very good. It's in the series of the Archimedes, I think, the first. And so, the name of the following is that we have a representation of this type there, if I take delta n cross i. I have here the union of a certain number of n plus one copies of delta n plus one with zero minus zero equal to i minus zero equal to n, or as before, delta n i is just a copy of delta n, but it is to check the copies, with here the arrows c1, c0, up to cn. And here I have a relation between these CIs, the ones you see appearing here. This tells me exactly that we can fill the delta-n-3i by these n plus 1 simplexes. And the relations are exactly the extensions of this one. So here, there are delta-n here, index i again, so it will be zero less than or equal to i less than or equal to n, let's say with... Alpha and beta here, alpha and beta, both values, both, di which goes from delta n, i-th component, to delta n plus 1, i minus 1-th component, and that's, for alpha, that's this, that's the union, and beta, that's almost the same thing.

55:00 I call this n plus 1, but this time 2i, in this way, equal to d i. So what we are saying is that the relation d i c is equal to d i plus 1 c i plus 1. So we have relations like this, with 0, 0 equal to i, 0 equal to n-1. So we have a presentation, not very simple, with a relation, and this relation is exactly the one we did here. If it's not this one, it's because I made a mistake here in the indexes, but it's the generalization exactly of what we can do. And so, now, we can understand a little, we could just, that would be enough to allow us to say what a homotopy is in the quintessential case, but I want to go a little further on that because I find this part very interesting. And so, how do we justify that? Let's say we can say a little more. What were they? They were expressions of the form a0 is equal to m, the whole is equal to n, with possible repetitions. We saw at the beginning of the course that this was it, an n-simplex in delta n.

57:30 This is the data of an invariant of a term here, 0, is equal to this. And for delta n, it's the same thing. If I now look at βn cross β1, that's i, in the second dimension n, it's the pairs a, i, b, x, with 0 equal to 0 equal to n, and 0 equal to b0 equal to bn equal to 1. So, in fact, the B0, there is not much. It goes either 0 or 1. And so, what we call an expression such as... So, there is a dictionary here, we can say that A0, it will be equal to A0.0 Ai. So here, it will be Ai.0, while Ai prime, it will be Ai.1. So this is equal to exactly what we add here, the elements 0 are equal to a0, ai, and then at some point ai prime, an. At one point, b0 ceases to be 0, it becomes 1, and the others continue to be increasing, so here with primes everywhere else, so it's sequels like that, it's the three simplexes that I've drawn, 0, 1, 1 prime, 2 prime, etc. It was summaries that were exactly things like this. As we have seen, 1 is equal to 2, 1 is equal to 1 and 2. So, with this presentation, with this example, it means that a homotopy H now has a value in, what did we say before, Y. Yes, that's it.

1:00:00 As a corollary, I will keep this in the table here. Corollary. Here we have the data of a family of applications, the initial application h i zeta n plus 1 with a value in y, which leads to the relations alpha and beta, which I do not rewrite here, C i equals D i plus 1 of C i, of H i maybe, it's D H i now. I'll come back next time, after, now in form. So that's it, we've demonstrated this presentation, we've studied microscopy, and it's really linked to the geometry of the prism. These applications may be inflexible, and you may recall that an inflexible application is just an element of y1n1. This justifies the definition in the following form, which is incompatible, f and g in an obvious sense. That is, if I restrict the delta n here, we can see, we can follow what is happening, the reflection must be what we want for the homotopy. So here we have a definition proposition, for us it is a proposition that is used in ancient literature, but also in us, it is the definition.

1:02:30 An homosapien of F by Rg for the terms of application x by y, the initial, to the data of the applications hn, hi, sorry, of xn. The data for all n of the application h i x n to y n plus 1 for zero equal to i equal to n, satisfying conditions 1, 2, 3, 4 to the following conditions. So I hope it's a bit repetitive because these are essential identities, but now I hope I have convinced you to justify it first, but I'm still going to ask you the following questions. First of all, we have to give the compatibility to F and G. It tells us that D0 of H0, of H0, so in the HIG we are, but yes, I don't know, no, maybe not, maybe. So the zero of H0 is equal to F, and on the other hand, the higher one, on the contrary, Tn plus 1 of the highest H is equal to G. So the zero phase of the first prism is F, the last phase of the last prism is Cg, and then in the first two, the flux has to stick together. So we had the following thing. The one we saw, the one in the middle, was J plus 1.

1:05:00 I could have concluded by elaborating on my drawing, that is to say that for the facial operators there are other things, there is also TI of HJ is equal to HJ minus 1 of T. When i is lower than j, it is also a simple identity. And then there is i equal to j, i equal to j or j plus 1, j equal to j and j plus 1. And then after the last one, it is di equal to hj equal to hjt i minus 1, when i on the contrary is greater than j plus 1. I'll let you prepare an entity that defines a centrician, and finally, the relationship with degenerations. SI of HJ equals HJ plus 1 of SI when I is lower than or equal to J, and is equal to HJ S I minus 1 when, on the contrary, I is greater than J. Well, personally, I find that when we define a homotopy like that without explanation, it's a bit hard, but there, well, I justified them. We could have justified absolutely each of these identities, the one I mentioned, which for you is the most important. It's the one that sticks together and is simpler. Well, then, I'm not quite done with... With this part here, because I want to go further with this, I put this where it fits, I put it to the left here. So, the diagram ending with a plus a simplex plus a time cross i, each plus a simplex was of the form, I have the axis A, the axis B here, with 0, 0 here.

1:07:30 At a certain point, we had j, here I had 0,1, and here I had the axis like this, and each simplex corresponded to 1, 0, j, that's 1, j, sorry, that's j, 0, and it goes all the way to the end, n, here, that's n, 1. And here it goes like this, corresponding to a path from 0,0 to n,1. Each n,1 plus a simplex, I should add non-degenerative. This is the only thing I looked at in my drawing. Naturally, if I look at the degenerate simplexes that come from below, there would be other relations, but those that are non-degenerate, at the end of the course, we have to get to n1 here, it really has to be of length n plus 1, we have to have a n plus 1 simplex, and so it's a path like that, so you see all the different paths. So I want to go further than that, since we did this work there, and look more generally, we can look at the official presentation as well, so more generally. Delta m cross delta m with n and m. So we did the case m equals 1 here, but we can look at the general case. So here it's harder to draw, of course, even the case, the first case, n equals 2, n equals 2, it's not easy to see.

1:10:00 So let's say, again, we know that it's zeros after zeros equals 1. The following points correspond to the A-I-B-I. A-I-B-I, in a sense, is an entirely ordered set. Note that n is totally ordained by the relationship of ordained orders. Similarly, m is totally ordained, but n cross m is not. For us, it is a totally ordained maximum set for non-degenerated things. We are now looking for, we will say here, it will be, it is a maximum set, totally ordained. When n plus m plus 1 simplex delta m cross delta m is non-degenerate. So this is the case that I want to look at. I mean, does it have to be degenerate? And so now the drawing is the following. These are drawings like this.

1:12:30 Here I have 0, 0. Here I have the pair n, m. And what we are looking for this time, in the same way, In order to describe an application, we have to generate delta n plus m plus 1 times delta n times 3 delta m. All paths, this one in here, are paths with horizontal and vertical components. This was the case with m equals 1, the one we did earlier. So I note in the passage a drawing like this. Those of you who were at the workshop two weeks ago saw in the description of Capranoff the spaces of the assay, there was this approximation of the spaces of the path of the assay, it is these paths there, horizontal and vertical, which correspond to the elements of the engendered free group, in two variables, a non-communicative polynomial in two variables, I do not come here but I draw your attention to it. I think it's for the same reason. I would like to just... So, for each row, when we go from AI, BI to the next row, A, AI plus 1, BI plus 1, we have either AI plus 1 equals AI plus 1 and BI plus 1 equals BI, or the opposite, or AI equals AI plus 1. So, there are different ways to describe such a path. There is another way I like to describe such a path. Such a path can be described by a partition characterized by the following sequence,

1:15:00 So, we have a partition of the set M plus N, which is equal to the set of N, and the disjoint union with the set of N, so a partition in part totally retained, defined in the following way, where I in N plus N belongs to If and only if the segment i, i plus 1 is horizontal, so the i-th segment in this list is horizontal, and it belongs to the subtenu, if and only if the sector i, i plus 1, in its most obvious form, is vertical, where I have numbered my segments, this is the segment, I take 0, 1, 2, plus k, n plus m. This is the sequence of segments, and for each segment, it gives me a number. So, I'll finish before, maybe I'll have a good time to stop, but before that, I'll finish with two other descriptions of this combinatorial thing, one of which appears a lot in topological and algebraic things. So, it's such a part, we can also say that such a partition, partition, union, null, can also have a permutation,

1:17:30 The order of the set that preserves the order in the following partition is the partition defined by, we send i, well there is a question of i, but it is not towards the i, but since it is a growing sequence, we send i towards the i plus 1 for 1 less than or equal to i less than or equal to p. The point is that there are some things that start with zero and others that start with 1 in the notation, and then also it's equal to the opposite, so I only have sigma of n plus j equal to mu j, on the contrary, plus 1, with 1 is zero equal to j, interior equal, so P was n a n in the actual notation. So this is a rule to define this and so it is specified that this reconstruction are sigma permutations in the symmetric space S and N plus M which preserves an interior of each of the parts.

1:20:00 Sigma of i is less than or equal to i, less than or equal to n is less than or equal to j, less than or equal to n is less than or equal to sigma of i and less than or equal to sigma of j. Sigma of n plus j is less than or equal to sigma of n plus j prime, if I'm not forgetting in the notation. So within each of the two packages, the order relation is provided by the sigma permutation. And so that's what we call, if the English term is more common, we say that sigma, the sigma permutation, is a shuffle. Shuffle is the shuffling of cards. We take two packs of cards and we can mix them. We put them together without changing the order of the cards that were in the left hand and those that were in the right hand. It's just permutations that preserve their order. And so, as we rarely find, we find in certain fields a card swap, isn't it? And I'll finish with the last description. Finally, we can see such a shuffle. These shuffles play an important role afterwards, if you will. The deep reason is that it's short. This is the one we just saw because it describes these two compositions of theta1 cross theta1. So we can see a type shuffle in the latter way as a choice of a family of representatives in the group N plus M, the symmetric group S, N, N plus M, and the choice of the family of representatives of the classes.

1:22:30 We are looking at all the permutations and we do not distinguish them, but we want them to preserve order, so we do not allow permutations inside the null package or the null package. I notice that Sn is distinguished in Sn plus M, Sn also in Sn plus N, since it permits a sub-assembly, and finally that a commutator. Since these are partitions with permutations of disjoint elements, they commute between them, so, indeed, Sn is a subgroup distinguished from Sn plus M. And so what we're saying here is that it's really these classes that play a role here, the classes of SM plus N modulo and this group SM cross SM. And so that's another description, an objection, between the simplex scenes we're looking at and these classes of SM plus N modulo and SM cross SM. Well, I'll stop here for now and after I'll take a 10-minute break.