Thanks for another interesting video! Have you considered making a video for Numberphile before? I'm not sure how difficult it is to get in contact with them, but it could potentially be a great way of spreading the word about e.g. rational trigonometry.
+Asbjørn Lystrup Numberphile is a bit different, although just perusing their previous videos they have one about Ford Circles (perhaps worth a look). But the perhaps most interesting thing about #phile at present is its subscriber count. It has just passed sqrt(2) millions (@ 1,414,842 on Nov 3rd 2015) (As you know, there is no such thing. We would say Numberphile has _approximately_ 2 trillion quadscribers)
There is a theorem that has always struck me as remarkable, to the effect that the combination of multiple rotations about any axis vectors by any amounts of turning is equivalent to a single rotation about a single axis vector by a single amount of turning. (I suspect that if one starts with all-rational quantities, the result might have some irrational parameters.) This theorem is of great practical value in space navigation, etc.
+Doug Gwyn That is a remarkable theorem, not at all obvious. So in terms of sending one directed edge to another, we must be able to find a single rotation that simultaneously makes the necessary change in both position and direction. I can sort of see why it must be true: Let A and B be points on the sphere. At each point we have a little directed line segment (lying in the spherical surface.) Measure the turn-angle these segments make to the segment AB, call these turn-angles tA and tB. Since we have to send A to B with a single rotation, the path followed by A will be a circular arc on the sphere. And since A and B are both on this circular arc, the axis vector for the rotation has to lie on the great circle that perpendicularly bisects AB. So to prove the theorem, we just need to be able to find an appropriate point X on this bisecting line so that in sending A to B we also make the required change from tA to tB. Consider the 2 circles which cross at A and B: the great circle thru A and B, and the locus of rotation of A about the axis thru X. If X is at a pole with respect to the equator AB, then the locus of rotation and the equator coincide, while if X is at the midpoint of AB, then the locus of rotation is the smallest circle thru A and B. If the turn-angle between these circles at A is t, then, when they cross again at B, the turn-angle will be -t. So the effect of sending our directed line segment from A to B via the locus of rotation rather than via the great circle is to rotate it by 2t. And since t varies in a continuous manner from 0 to 1/2 as we move X down the bisector from the north pole to the south pole ( with t = 1/4 when X crosses AB), then 2t varies continuously from 0 to 1. And therefore, no matter what the difference in direction between our line-segments, we can find a rotation that sends A to B which also sends tA to tB.
+Andrew McKinnon If A and B happen to be antipodal, then the above doesn't work. In such a case we don't have different size circles thru A and B, we only have great-circles. But now we have a whole family of great-circles to choose between. One will coincide with the line segment at A, another with the line segment at B. And if we choose the great-circle which lies midway between these as our locus of rotation, then again we manage to simultaneously send A to B an tA to tB.
+Andrew McKinnon The theorem guarantees that if you can find a sequence of rotations that do the job, then there must be a single rotation that does the same thing. (Translation can be analyzed separately.)
+Andrew McKinnon And we might remark that the argument is primarily a physical one, involving continuity. It requires that two great circles (meets of a central plane with the sphere) on the sphere meet on the sphere. Actually this is a subtle number theoretical issue that may or may not hold in a given situation. So Doug's intuition that irrationalities may be involved is correct. And that means that the true theorem is actually more subtle. One way of seeing this is to inquire what the theorem looks like over a finite field. It turns out that something in this direction is still valid, but one has to be more careful in saying precisely what is going on. As usual it is quadratic equations which are the stumbling block.
It serves as another example (the Fundamental Theorem of Algebra was mentioned in an earlier discussion) of why many, I daresay most, of us are happy to have the real number system, despite definitional issues in its foundations. The behavior is nice, simple, and useful. To continue to use it without irrational numbers, we'd have to resort to some scheme of approximation, which would not have those behavioral properties.
The directed edge argument indeed shows the maximum number of places where the red arrow can be. But does that argument imply that these places actually can be reached by rotations? Perhaps it does, but do we need a proof?
+C Fortgens If you think about the vertices as points on a sphere, then a directed edge is specified by choosing a vector P corresponding to the starting vertex and then choosing a vector N perpendicular to P so that the directed edge lies in the oriented plane normal to N. If the initial edge is given by vectors P1 and N1, and the target edge by vectors P2 and N2, then 2 rotations suffice to take edge1 to edge 2: - P1 and P2 are both perpendicular to (P1 X P2), so let rot1 be the rotation about (P1 X P2) that sends P1 to P2 - Let N1' = rot1(N1) - Then N1' and N2 are both perpendicular to P2, so let rot2 be the rotation about P2 that takes N1' to N2. - Then rot2 * rot1 takes edge1 to edge2
+Andrew McKinnon Thank you very much. Is it correct then that the proposed method only can work with regular polyhedrons that can be projected onto a sphere?
+C Fortgens No it is true in general. Imagine 2 arbitrary directed line segments in space; call them v1 and v2 with starting points P1 and P2 respectively. Then three rotations will suffice to send v1 to v2: - rotate v1 (using the line P1P2 as your axis of rotation) to get v1' which is coplanar with v2. - then rotate v1' (in the common plane, about the midpoint of P1P2) to get v1' ' which is both coplanar and co-located with v2. - then rotate v1' ' (in the common plane, about the point P2) to get v1' ' ' which agrees with v2 in both location and direction. Actually, since a single rotation is sufficient for directed line segments on a sphere, only 2 rotations should be required for the general case: - take a sphere thru P1 such that v1 is a tangent to the sphere. - then there exists a rotation of the sphere that sends v1 to v1', such that v1' is parallel to (and therefore coplanar with) v2 but with opposite direction. - then rotate v1' (in the common plane, by a half turn, about the midpoint of P1P2) to get v1' ' which agrees with v2 in both location and direction.
+Andrew McKinnon Thinking about it further, the answer to your question should be 'yes', in a sense. Even though we can always find a series of rotations to send any directed edge to any other directed edge, this isn't guaranteed to be a symmetry of the polyhedron itself. This is only guaranteed if all the vertices, edges and faces are identical ie the polyhedron is one of the platonic solids.
This kind of symmetry is somehow used to classify viruses under the electron microscope. But since virologists don't know too much group theory, they can only have icosahedral, helical and complex symmetries. For example, the influenza virus has an icosahedral symmetry, cf www.twiv.tv/virus-structure
You say that the sphere is the most symmetric of all possible solids, but if I understand your viewpoint on real/irrational numbers correctly, then you must admit that the sphere is not technically a POSSIBLE solid. If there is no truth value to whether 10^10^10^10...+23 is prime then there is no truth to the symmetry of a sphere, which would require effectively infinite facets.
+Joshua Huntington We should distinguish between a physical sphere, which we construct (a marble say) and a mathematical sphere, which has various possible definitions. What you say might well apply to some definitions of the sphere, for example those that require us to contemplate infinite sets of points. But it does not necessarily apply to more sensible, rational, approaches to the sphere, for example that it is an equation of the form (x-a)^2+(y-b)^2+(z-c)^2=R for some rational numbers a,b,c and R. One can pleasantly discuss the symmetry of such a sphere in completely finite, algebraic terms.
