$8$x$8$ Diagonal

Analysis of $8$X$8$: Diagonal

On this analysis of the Cartesian product $8$X$8$, I will use an slightly different orientation for the visualization of $8$X$8$.

  

Figure 1: Solid of Cartesian product $8$X$8$

                                         

Figure 1b:$8$X$8$ in glass

Figure 1c: Top and Front view of $8$X$8$

We can confirm that this product is continuous (connected?) due to the cross section where all parts of the $8$X$8$ emerge (Figure 1c - Front View). By analyzing the Top and Front view we can also recognize a common path between figures. In the Figure 1c - Top view, for example, we can see the $8$ path that the $8$ figured followed. In fact, this helps us to understand the the Cartesian product of $8$X$8$ is the union of four torus sharing a common intersection. If we were to remove one point, in other words make a whole in each torus, we would get as result only their fundamental circles. For the torus, its fundamental circles are the following:

In this case, since the four torus intersect on a common point, they will overlap their path in the exact same way at this point. In fact, this guarantees us that this representation of $8$X$8$ is continuous (connected).

  

Figure 2: Fundamental Circles of $8$X$8$ (Front, General, and Top Views)

$8$X$8$ Diagram and its' Diagonal

In the same way that a torus have a diagram that facilitates it analysis, and the product $8$X$8$ is conformed by four torus intersecting on a common point, we can use the torus' basic diagram to figure out what would be the $8$X$8$ diagonal. 

In previous posts, we already analyzed what would be the diagonal of a torus. In fact, we used a similar process to find $S^1$X$S^1$ diagonal, than to find the diagonal of a circle. 

Figure 3: Torus ($S^1$X$S^1$) Diagram + Diagonal

In the case of $8\times 8$, we had to find a way to connect the four torus diagrams in such way that they share their fundamental circles as I showed in Figure 2. When constructing a torus from its diagram, we can see that the sides $>>$ and $>$ when joined, form a fundamental circle each. 

https://www.math.cornell.edu/~mec/Winter2009/Victor/part1(6).png

We can use this concept to create our own $8\times 8$ diagram. We have to take into account each of the circles that intersect themselves, that in this case are four. Also, in order to find  its diagonal from its diagram, we will have to take into account the number of elements that the whole figure has. I will count every torus as an element of the final Cartesian product. I did it in that way so it facilitates me to use previous facts that we conclude about the torus and have better chances to conclude fact about the $8\times 8$ diagonal. As result I obtained the next:  

Figure 4: $8\times 8$ Diagram + Diagonal

This is my principal objection: the diagonal should go only thru two tori, not four.

The reason that I oriented the diagonals in this ways is because if two robots are moving in the space $8\times 8$, it means that in every element of this space they could not follow the path where both robots have the same position (its diagonal). Therefore, four elements require four different paths where the robots would intersect.

I do not understand this. Think of the original $8$ figure. Two robots in it. You can think the first robot in one $8$ and the second robot in another $8$. These are the two figures $8$ in your cartesian product. Do you see that the only way that robot 1 and robot 2 are in the same place is in case that both are in the red circle or both in the blue circle? If one is in the blue and another in the red, no way that they can be in the same place!

Figure 5: $8\times8$ Diagram (this is not $8\times 8$, these are the circles that remain once that you take a point out from two of your tori in $8 \times 8$)

Since we already concluded that the diagonal for a torus is a Mobius Band (an annulus, not a Moebius band) with two twists (equivalent to a cylinder), in this case we would have four Mobius Bands intersecting on a common point. If we consider them as cylinders, and reduce them to a circle, we obtain a "flower kind of shape" that would represent our $8\times 8$ Diagonal.

I think you are confusing the diagonal with what remains after we cut the diagonal. Be sure you identify first clearly what the diagonal is, in the drawings. Then what remains once that you take out the diagonal. And then what that latter space is homotopic to.

(GRAPHS WILL BE POSTED LATER ON)