Categorical structures & cohomology (lecture 3)

0:00 I have validated the product of two NNB and NN complex. We have a family of... I will not go into the details of each one, but I will also mention the general case. We have a right-hand presentation. Here, the families of NKJ. I have not specified exactly what the presentation will be useful for. Maybe... So here, I have said it, but I will repeat it. I'm going to show you a few examples. The first one is the most common. We have zero, two, three. We have... And here, we have... It's obvious. So, these are references. To classify a three-dimensional model, which corresponds to a partition. Plus one element, number of zeros, plus one, minus one. Thank you very much for your attention and see you next time.

2:30 In this way, it is a diagram that can represent points in it, and so when I have points in n plus 2, when I have the number of points exactly, which is the number here, so it becomes delta n plus n plus 1. And so we had already seen it, the only way we can do the math, it was the case where n equals 1, so it was in this case delta m. I'm going to explain this in great detail and remind you that I drew the prism corresponding to the last one, the two simplexes, which were, for example, three simplexes, which were like this one. And that is, if you want, in the drawing corresponding to this one, which is now a drawing, we saw it last time, this drawing here, so here is the path that goes from zero to two degrees, we're going to do it in point 2.1, that's it.

5:00 You can see that the points that are here at the bottom, but that are horizontally, correspond to a simplex, a two-simplex here. And now we go up a notch when we have to go to the other side. And so in general drawing, we have to remember that the lines are aligned here, at the top of the pentagram, and even here. So this is a diagram that tells us that we pass through the other factor from one point to another. In this sense, this is supposed to explain how we have, from a diagram like this, we have points of independent fineness in Rn plus 2, product of Rn plus 1, cross Rn plus 1. etc. etc. etc. etc. etc. etc. etc. etc. etc. etc. etc. etc. etc. etc. etc. etc.

7:30 If I had two, I would have two colors, that would be good. So, let's take the first path, the path like this, the one I have just described, the precipice, and then let's take the other path, excuse me, it's not going to be the precipice, it's going to be one step further, more administrative. So now we're going to have a tetrahedron, 3 delta 1, so imagine a triangle and a tetrahedron, and then in a quarter of a dimension, we're going to have a big one, and the other thing I'm going to do is I'm going to take this path here, so you can see here that the intersection of the segment is 10 to the 1st, it's 2 prime, 3 prime, but it's okay, the intersection of the two paths, which corresponds... So we said M plus M3 plus 1 to R6, and the point is that it's not exactly the same, but...