Philosophy of Mathematics Discussion Group — Marco Panza's Paper

0:00 In the future, we will have the habit of writing R equals n plus i, and so really look at the application of H, n plus i, k, a, n, to H, n plus 1, k, a, n plus 1. If the value of n-1 is the same as the value of 2p, the difference between n-1 and r remains constant, stable, stable, stable, stable, stable, stable, stable, stable, stable, stable, stable, stable, stable, stable, stable, stable, stable, stable, stable, stable, stable, stable, stable, stable, stable, stable, stable. These are the main principles in the Sainte-Pitiale case, which are used in ordinary topology, for the cohomology of the Sainte-Pitiale cases, and so it is not Sainte-Pitiale in Sainte-Pitiale, as I said. So, importantly, the spaces, as I said at the end of the recalls, the spaces of the Sainte-Pitiale universe, I have given two answers.

2:30 I explained the last time that they have a very simple homotopy and a calculable homology, while spheres are the opposite. Spheres have a very simple homology and a very complicated homotopy. So the first reason is that we can decompose. I'm not going to give you a demonstration of the verses, I'm just going to finish the discussion. Space, what we call the decomposition of Kozmikov, which is done quite simply, we just don't have the time to talk about it. It is the following, supposing, in an iterated way, that x is equal to and reduced to a point less than or equal to n. Not only is it less than or equal to n, but we killed the motor j in degrees n by putting it aside.

5:00 In a situation like this, we have a number of proches in proches. We have x, we can think of this by saying that k with the fiber is a k, plus 2 in the degree of 2, plus 3 with the fiber is a k, plus 3 in the degree of 3. If you want the limit, it's square root of x. We have a linear system, a problem of generalization there. For example, we have x2 times x3, etc. In this way, we can decompose a space in the space of a magnetic field, and in this way, we can understand the spaces, especially in this case, in this case, in this case, In any case, this decomposition is better if it is a space that is a finite number of groups of entropy, because, as I said, we can reposition everything from there. If we have groups of entropy in each degree, which is the case for the spheres in general, well, in this case, of course, we will not really have to proceed infinitely to go up to x.

7:30 There is a bit of a filmation, but we know that in these cases, we have B, that this K plus B is equal to B, where above, we saw it with a space that we called E, of course, but we now call it the entire space, where the fiber is K, in this case. So, in fact, if I have this filmation, which you will not see, I will show you what it is.

10:00 In this case, we have a situation like this. For a term like lambda, we saw that it was classified because it is something that happens in a space that leads to a fraction, b plus 1, which reminds us that not only a is different from b, but really b is larger than a. That's why I take things in the form of a and b in the form of a. So b is strictly larger than a. So hb plus 1 is the square of pi a squared a squared. I'm going to show you here the post-Mikhovs, it's called the post-Mikhovs course, on two floors. I'm going to show you here the post-Mikhovs course, on two floors.

12:30 I'm going to show you here the post-Mikhovs course, on two floors. I'm going to show you here the post-Mikhovs course, on two floors. I'm going to show you here the post-Mikhovs course, on two floors. I'm going to show you here the post-Mikhovs course, on two floors. I'm going to show you here the post-Mikhovs course, on two floors. I'm going to show you here the post-Mikhovs course, on two floors. I'm going to show you here the post-Mikhovs course, on two floors. For these two-storied post-school tours, it is entirely characterized by geometry. We did not see the base in the notes, but for a second, we used in the calculations the groups of the motor and the spheres at the beginning. If I find a vibration like that, well, it used the groups of the motor and the spheres, but very quickly we switched to the calculations. It was the beginning of the calculations, it was with methods like that. I'm not going to go into too much detail here to show you that this notion of space in HM, which we are now building, is something important. And the second and last application, the second application, after the first one, will be the next one.

15:00 We can, if I take M as a whole in HM, an element of N, and natural in the cohomology, and A as the cohomology of the application HM, And so, the answer, very simply, at least from a theoretical point of view, is the following question, to understand a space, the method of algebraic topology is to calculate invariants, groups of homotopies, groups of cohomologies, of homologies, we can say what are the natural transformations for... There is the following question, what are the natural transformations? This means, of course, that the whole HM of the XA...

17:30 In any case, we are in H, N, X, B, but only in X. So if I have H, M, Y, A, I can write A as a qualitative result for all these natural transformations. So there is an example, you will all tell me where there is an example, but no, this is the example. M equals N, and from A to B, an anamorphism, that implies an application H, M, X, A. We have an example of a natural transformation, but the question is to know if there are other more sophisticated natural transformations than this one, and that from a thermic point of view, not only can we calculate all these natural transformations, but we can do it instantly.

20:00 We have shown that HM is the same as XEA, which is the same thing as an association in this category of XEKM. So HM is a counter-variant of H with O, K, A, M. And we know what it is by definition of this speaker, otherwise it is an element of H, N, K. After, of course, to bring it is the real thing to do, but no, the conclusion is at another level.

22:30 The knowledge of cohomology is still very strong, it is a very general question, we have a perfectly concrete answer, and as I said earlier, it is a calculable answer, so we can really give the complete answer. So here is the second justification I gave you for the importance of these complexes. To calculate these... so it ends with this thing here. We have not at all calculated these topologies or cohomologies. We are just going to touch a little bit on the limits without any attention. To make us touch the law, it is the fact that, in three ways, the complexes are... We are going to come to the Chateau de Thiers, this series of spaces, and so this series that we are going to do is to make really explicit calculations and algebraic objects.

25:00 For Heineken 1, which is very well known and known by many of you, it is a familiar object. For Heineken 1, we will see that it is just an apartment. I remind you, we are trying to understand what it is, that Z, that K, A. We have previously seen that K1 itself is the classifying space of A. We're not going to go through them all necessarily, but that's fine. And so here, I'm going to take an element of 2. We take a to the degree of a. I'm going to write down my question. a3a, a3a. I'm going to write down the beginning of the equations here. Here, if a1, a2, I send it by t0 over a2, by t1 over a1, times the two components. I'm going to go to the end of my logic. I wrote the product multiplicatively from A1 to A2, not from A1 to A1.

27:30 There are always two arrows. It is the final object of the category. There is no other choice. The elements are not on the elements and all the others are on the elements. There is no other application of a, because an element is not an element. So now, the only thing in Paris, I remind you that if they, you have already seen, if they are together, Thank you for watching this video, see you in the next one!

30:00 Speakers include ZA, ZA, ZA, ZA, ZA, ZA, ZA, ZA, ZA, ZA, ZA, ZA, ZA, ZA, ZA, ZA, ZA, ZA, ZA, ZA, ZA, ZA, ZA, ZA. And now the answer is the associated complex, so if I take now Z, I do all the details, if I take Z of A, we will remember that the associated complex application says that I have a group of initial variables, we take the same objects and we take the sum of the differentials to have the differential in the complex. So that means now that it is Z here. ZA has nothing to change here, this is the trivial application, the one that sends A to the zero element. And for the others, here I can rewrite this because there is ZA, ZA force, with one arrow, and this, and more generally ZA. This is the answer to the question of the answer to the question of the answer to the question of the answer to the question of All of this will be equal to A plus 1 minus 1 power n, which I hope I did not forget, plus 1 minus 1 power n, which I hope I did not forget.

32:30 Even for people who have never done a simple theory, we recognize that this construction is an object of art and so I call it constantly art. And so in fact, more generally, and so we note that there is a P to replace the letter P here, or a P in P.

35:00 The algebra is made and this B has the complex in the same way. B of R is the function of R and therefore it is defined in the formula. So it was not specifically for the chains on the spectra, but for any algebra. So, in this case, the situation is very clear.

37:30 And so we found that the chain of Witten and Hawking is the same, because it's not the same when I read it. And now, there is another thing. So let's suppose now that we have the hypothesis that we have in front of us. Let's suppose that we have a group A. And so, by classifying a group G as something, it is a symplecial ensemble. And as I have already said once, if the group A is binary, it is a group A-binary-symplecial, because the arrows that will be in front of us are homomorphisms as well. And so, in this case, ZBA is not just a unit, it's a Z-algebra. Well, it's because, indeed, we have B, A, look at B, A3, A, we only have Z of B, A,

40:00 force Z of B, A, type of commutator and two complex scales, so the two parts of the structure. This is a natural arrow and an equivalence of homotopy on the scale of B, A, 3A, B, A, 3, B, A. This is not the direct parallel between the two. The total complex of the product of the tensor, which we will see at the end, is the diagonal. The Z of B A is equal to the total vector Z B A times the total vector Z B A times the total vector Z B A times the total vector Z B A times the total vector The only new thing here is that we can easily see that this is the same thing, that B A cross B A, no, there is nothing else than B of A cross A, so this is the same thing as the chains on B of A cross A, the chains on B of A cross A, the chains on B of A cross A, the chains on B of A cross A, the chains on B of A cross A.

42:30 This is where I can associate an anamorphism. So, in reality, if I take Z, B, M, I send it to Z, B, A. And this is the structure of an algebra that I have built, an anamorphism produced in a complex cell, towards the same complex cell. This is what gives the structure of an algebra to be translated. And all of this seems very long, and I could have just said that you had here, at the A, the term of the Hilbert space, which is what we have here, and we have added it here, even though I want to demonstrate it to you here, since we have said that we want to normalize it, and it is true that it is the same thing as the normalization of P-A, taking the normalization of P-A, this kind of inverse normalization of the D'Holme-Hugues' counter, as we have already done, so it is the same thing. I have created here another statistical space where normality is acceptable and there is a regression between contexts and things, so that's enough to judge. I gave you the arrow explicitly. So here, I still have to remember here, there were mathematicians here in the late 30s and early 40s, and I said that there were more rigid arrows in both directions. There is an equivalent of topology, but there are arrows that go from one to the other, and this one here is the sharpness arrow. This arrow is not unique, it is only used in autofiltration, and in autofiltration itself, it is unique in autofiltration, etc., but there is still a representative that is good, and with this arrow, this one tells us that the arrow, in this case, this arrow, so finally I'm going to write at once the total arrow, the arrow, by using the candle arrow, I'm going to leave that on the control, in an explicit way.

45:00 I'm going to show you how to solve a problem with A, with A, P plus 1, sensor P, A, P plus Q, in the opposite direction, Epsilon, Sigma, or Symmetric, Parcours, where I had P, Q, Chardonnay. It's a little hard to see in French, but you have to think of the matrix of 4. And so here, the permutation corresponds to the exact formula, the good formula is not to take sigma, but sigma-1, the inverse of that, and for the questions of A, sigma-1, the inverse of that, and for the questions of A, sigma-1, the inverse of that, and for the questions of A, sigma-1, the inverse of that, and for the questions of A, sigma-1, the inverse of that.

47:30 I'm going to give you a few examples because we actually have two degrees, so it's going to be three times B and a C. They're going to be in the two sections where A and B are in degree 1. And then we're going to take A and B and we're going to take B and we're going to take C. Well, it's the batting of A across B and C. So the order is going to be B and C unchanged and A across B. It's A transpose B transpose C, minus B transpose A transpose C, plus B transpose C, A transpose A, minus B, A, you can easily use these examples in the same kind of things you do. So it's a completely different case, you don't know any of the theorems, you just have to verify that it's a little bit more complex, it's a little bit more different from the part of the solution. I'm not going to go into too much detail here, because we're looking at cases where the degree is higher, so we're...

50:00 The reason why we're leaving the classifier is because of things that have nothing to do with quantum mechanics, and which are less known already. This product is less known than everything we know about quantum mechanics. And so, the most important point, the way to make it the most effective, is to notice that it's an iterative process. You see... I would like to remind you that if I now take an abelian simplicial, if you will, the vibration from A to B A is defined in the form of an abelian bisimplicial B A point by putting B A point in degree N, it will be a classifier. We will need to look at this, we will have to do this, so now this is an integral ensemble. So that means that B A point in pi degrees P Q will be equal to B 2.

52:30 In other words, B here is a neutral element with also above that is an integral ensemble and this is a loop here. We have a very simple set of equations like this one, where Q is equal to 0, and Q is equal to 0, Q is equal to n. So we have a symbol. I want the P to be here. I want this to be the P axis and this the Q axis, I think. So we're going to do this. This is going to be Q equal to 0. So here, we're going to have Va. B, A, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B, A, B The arrows here are the arrows given by the group law on B.A. in the previous lecture, but now it says that there is a problem.

55:00 We have to do something about it. And so now, we have the same with the construction that I introduced previously from B.A. I think there may be an implicit implication. The P of A is the fiber. The fiber will just be reflected, as in this case. The fiber is just A prime A itself. It's just A prime A. In fact, it's constant in one direction, so I don't know if it's true or not. An implicit implication is just one direction. So we can now move on to an artificial object, a diagonal object, with an A point, A point, A point, A point, A point, A point, A point, A point, A point, A point, A point, A point,

57:30 There are two types of constructions. For this one too, the homotopy is not contractive and the vibration, by taking the exact sequence of homotopies, implies a.e.k.a. a.n.ab.a.n. There is a model for this. The homotopy equation, as we have seen, is the three terms we have just described. This is a bit abstract, but it is to justify the following thing. I just want to base here, KON equals 2. Since we have built the KN equals 1, it is the classifying space, we will look a little more in detail at KN equals 2. The purpose of my lecture, because you may wonder why I insist on this space, is to show you that it is something concrete, which is not very well known, and which has a lot of importance for the classification of certain categories. So, there is a motivation to do this, isn't there? And other than simply to understand the academic spaces in which they exist. So, we will look at two or three dimensions at a time, so it's worth it to look at them one by one. So, I would also like to give you the data of the library. This is the original part, in the mathematics analysis, 154, I think. It's a bit hard to read, but it's the result of A1. So, A1 will be an element of Adenium.

1:00:00 Note that, for reasons you will understand now, A1 seems to be a bit lost. Now, let's look at... I need space to do this. We're going to try to decode this. I'm going to write it down here. But now we're going to move on to the second part of the B.A. lecture, and I'm going to let the B.A. fall now. I'm not going to distinguish between the group with the five pixels and the one where we do the sum of the differentials. I don't always want to carry this tilde with me everywhere in the lecture. So here, brutally, this is a model that we have here. It's a model that I just explained. In this case, if we look at the A, which is a square, we will have the effect of classifying the A itself, like this, if it is like this, then it implies that this thing is a model of this A, and already we will be able to write this distance. In this case, if we now want a model for a chain on K2, we can use Z everywhere. I'm going to put some z's in front of all this and I'm going to modify the part of the second column so that this is the same thing as z of dA, tensor z of dA.

1:02:30 I'm going to replace a column like this with a column like this. So now here I have z's everywhere. So here I have z, I have zA. And so here I'm going to replace these, this is the cross. So here I'm going to, so here I'm going to, in fact I already have it, so I'm going to, I'm going to go back to it later, but I'm going to replace it. Here I just put it in cells. And here I'm going to do something else. I'm going to use, systematically, this construction. The scale goes from 1 to 2. So here I wrote the previous equations, Z, Z. The most interesting problem for you here, I found that this is ZA, tensor ZA. So here I have Z, here I have the B-round construction bar of ZA, the big B of the construction bar of ZA.

1:05:00 ZA, ZA, tensor 2, ZA, tensor 2, tensor 2, tensor 2, tensor 2, tensor 2, tensor 2, tensor 2, tensor 2, tensor 2, tensor 2, tensor 2, tensor 2, tensor 2, tensor 2, I'm sorry, it's not visible here, but I'm going to rewrite it differently at the moment. I have parentheses here. z to the power of 2 this time, plus z to the power of 2. And so, in fact, there are too many terms here, so there are terms that are useless, and so, in fact, this is ZA, Z plus A, I defined it last time, this is ZA divided by the generator, in fact, by the zero element here, and so, in this case, many of these terms will disappear, and even when I make tensorial products with terms of D greater than zero, they will disappear, and so there, I will rewrite a new...

1:07:30 In this context, this will be the last architecture. I have Z in degree 2. So here, I have Z in degree 1. On top of that, I have nothing at all. I have the first term non-null and in degree 1. On top of that, I have the other terms here. Now, if I have in degree 2, the novelty is that in the terms that I had here, I had three terms. There are two that have been transformed with Z, so they disappear. And I just have the term in the middle, which is here. So, what I have here, I have nothing here, and here, well, I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... I'm going to... A2 tensor, Z2A, plus Z2A, tensor Z2A2, with here the clashes that we will explain, but at least something like this.

1:10:00 And the last thing we will see is B of ZA, tensor B of ZA, tensor B of ZA, the tensor product, and the first term here is ZA, tensor ZA. These are the degrees, degrees 0, 1, 2, 3 and 6, in the other direction, it's 0, 1, 2, 3, 4. So this term is a degree 6. 10 degrees 3, 3 on the diagonal. The terms that start with the 0 degree are a, b, c, d, a, q. There are always degrees 0 and a. Excuse me for being a little technical with this. It's just to show that it's complicated. And so, now explicitly, so notation of Heisenberg-Karzenay, the case, we continue with that. We have a cell inside, which is the part of the restriction on zeta, tensor of the part of the restriction on zeta. We have already seen up to the a, p, tensor, p plus 1, up to the a. From this part, there is also the following part.

1:12:30 And more generally, an element that we factorize, we find the sequence in the same nucleus, in the same variable, and the same parenthesis or with the parentheses of two sorts of bars, and so on. And it corresponds with that. And so, an element like that, in a way, is summing up the data in that sense. And then, after that, there is the number of terms. Omega and S may be the total degrees of the elements. The next thing to see with this is the total degree. I talked about the degree of the complex theory.

1:15:00 I talked about the total degrees, so I want to see that the total degree, we have a zero degree here. Then here, we have two degrees. So finally I'm going to write in total degrees a version of this. Since it's the sum of the degrees, it works like this. So, what remains is the construction bar. The construction bar of ZA, we can say it like this. B of ZA, the construction bar is an algebra. So it's an algebra of the construction bar. The terms and the force on the negative side. We start with ZA, which is in degree 2, which is rationally a bar of 2. So above that, if you look here, the terms that work here are Z and this term here. So in fact we have Z to the 3 plus Z to the 2 with a arrow.

1:17:30 And now if I continue with the work, I have Z to the 3, Z to the 3 plus Z to the 3. So the last degree we're going to look at is Z to the 2, Z to the 3, Z to the 2, 4, 3, 3. And finally, we have 2z to the power of 5, 3z plus 3z to the power of 4, 3 times 2z to the power of 4, 3z to the power of 3. And when we write them like this, we see nothing at all. So we just have to write them as they are. Here, it's an element a. So it's an element a, b, a1, a2. What we did is we added a lot of things to that.

1:20:00 And here, I'm going to write it here. The degree of K corresponds to the inclusion.

1:22:30 The degree of 1 is written in degrees A. The terms are written in degrees A. That is to say that the Zena, which was in the construction part of degree 1 here, is the only one that has a arrow. To send this column, the first column, to the arrow of degree 1, the term here is in degree 1, in this complete degree. This is a general phenomenon. Thank you.