This is a blog based on Topological Robotics. Everything here will be posted by an academic student researcher at Wilbur Wright College on topics related to Topological Robotics and most specifically Topological Complexity, the term first coined by the mathematician Michael Farber. The posts on the blog will be published every week on various topics.
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Cartesian Product of a number 8 to a circle
$S^1\times 8$, if 8 intersects, then it is going to give us the shape bellow
if 8 is like two circles in top of each others, then it is going to give us this shape
Where are, in your product drawing, the original $8$ and the original $S^1$ that you are multiplying?
It looks to me that in this last drawing, each point of the original $8$ got multiplied just by and interval, not by the whole circle. Start by drawing the $8$ and try to follow each of its points in a trip around the circle.
And let's just consider the figure $8$ with an intersection point in the middle, otherwise it is just a figure $zero$, i.e. a circle. And we already know that a circle times a circle is a torus.
As we discussed in our last meeting, this will not be our Cartesian product. I think the Cartesian product of $8 \times S^1$ will be a Genus-two surface(double torus or two holes torus). As I was looking for this double torus, I encounter Genus-three surface and Riemann surfaces. They are some good topological surfaces although I didn't understood Riemann surfaces.
So I found out that the Cartesian product of $8 \times S^1$ or $S^1 \times 8$ will not be a Genus-two surface. It will be a pretty interesting shape. You will get two torus glued together and the gluing will be by the whole circle, not just by a point.
Yes, I agree that is an interesting space as you described it, and it is exactly what Watheq showed in the new drawing. Now, are we sure that this space is not actually a doble torus?
On the other hand, you can also think of what spaces have the same homotopy type than the figure 8. You know that a cylinder can be deformed into a circle. What space can be deformed into a figure 8?
Where are, in your product drawing, the original $8$ and the original $S^1$ that you are multiplying?
ReplyDeleteIt looks to me that in this last drawing, each point of the original $8$ got multiplied just by and interval, not by the whole circle. Start by drawing the $8$ and try to follow each of its points in a trip around the circle.
And let's just consider the figure $8$ with an intersection point in the middle, otherwise it is just a figure $zero$, i.e. a circle. And we already know that a circle times a circle is a torus.
Please everybody comment on this problem.
please take a look at my modified drawing of moving the figure 8 around the circle.
DeleteAs we discussed in our last meeting, this will not be our Cartesian product. I think the Cartesian product of $8 \times S^1$ will be a Genus-two surface(double torus or two holes torus). As I was looking for this double torus, I encounter Genus-three surface and Riemann surfaces. They are some good topological surfaces although I didn't understood Riemann surfaces.
ReplyDeleteSo I found out that the Cartesian product of $8 \times S^1$ or $S^1 \times 8$ will not be a Genus-two surface. It will be a pretty interesting shape. You will get two torus glued together and the gluing will be by the whole circle, not just by a point.
ReplyDeleteYes, I agree that is an interesting space as you described it, and it is exactly what Watheq showed in the new drawing. Now, are we sure that this space is not actually a doble torus?
ReplyDeleteYes. It is not a double torus. But, we can think about if we can find the Cartesian Product that gives us double torus.
ReplyDeleteGood idea.
ReplyDeleteOn the other hand, you can also think of what spaces have the same homotopy type than the figure 8. You know that a cylinder can be deformed into a circle. What space can be deformed into a figure 8?
Watheq, can you think of a flat diagram (with rectangles) representing the surface in the figure of the two tori in your modified drawing?
ReplyDeleteTo answer Dr. Colman's question, the letter $B$ would be homotopic to the number $8$.
ReplyDeleteYes, and also, after our discussion yesterday, we can also say that the torus minus a point is homotopic to the figure $8$.
ReplyDelete