Recent content by lavinia

It took me a while to understand what you are saying. Yes. You can't rotate one circle in the link so that it is flipped without banging into the forbidden circle. What I tried to do here was to show various ways to detect and define linkage. The definition which seems most intuitive, to me at...

Technically if sin(a) and sin(b) are both not zero. Without invoking a probability space, the set of pairs of angles with one or both zero has Lebesque measure zero. I think.

Can you make that probaility argument more precsise?

I think not all great circles. Some of them intersect with each other. The great circles in the Hopf fibration do not intersect with each other, this because they are orbits of the action of SO(2) on S^3. If two orbits intersected then where would the intersection point go under the action of...

Here is an image of a twisted cylinder that shows the linked boundary circles. This was taken from Wikipedia.

Clifford parallels and flat tori in the three sphere Start with a square in the xy-plane whose corners are (0,0),(2π,0),(0,2π) and (2π,2π). Consider the mapping of this square into R^4 given by the rule T:(x,y) -> (cos(x), sin(x),cos(y),sin(y)) The image of this mapping is a torus in R^4...

They are not homeomorphic. The three dimensional space deforms continuously onto a 2 dimensional torus. The idea of continuous deformation is important in topology. A space that can be continuously deformed onto another has many of the same topological properties. Here is the first example...

After continuous deformation it is a regular 2d torus. I had trouble describing the situation in a clear way even though the picture is clear, so I left it as a problem. If I knew how to post drawings this would be easier to see but I don't. My apologies. Here is another stab at it. Given a...

Another way to think about the linking of two fibers in the Hopf fibration is to look at the topology of their complement in S^3. Take for instance, the complement of the equator(the unit circle in the xy-plane) and the fiber that passes through the north pole of S^3. Since the north pole is...

yes I just wanted to emphasiize that by sliding the linked pair of fibers along the rays emanating from the north pole of the three sphere the linked fibers are continuously moved into R^3. Stereograaphic projection by itself without this sliding is just a mapping and is not a movement of the...

Here is a way of doing it using stereographic projection. Slide the two linked fibers along the rays emananting from the north pole of the 3 sphere until they reach R^3. Then raise one circle up into R^4 so that its fourth cooridinate is not zero. (This all occurs in R^4)

Right. BTW: Your answer proves that the Klein bottle can be embedded in R^4. So for two linked fibers in the 3 sphere, is the worry that they would have to cross over each other just to get out of the 3 sphere into R^4?

OK. So how would you move two linked loops in R^3 off of each other into R^4 without them crossing over each other? How would you move a point in the northern hemisphere of a sphere in R^3 to the southern hemisphere without crossing the equator if the point is allowed to move anywhere in R^3...

As an example, consider the equator of the three sphere, the circle in S^3 whose third and fourth coordinates are both zero. It divides the equatorial 2 sphere, the sphere whose fourth coordinate is zero, into two hemispheres and it is the boundary of both. Any fiber of the Hopf fibration...

What are your thoughts on this?