Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: varsigma on 25/01/2024 21:06:09
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In this Youtube vid, the author explains how to construct two graphs, and use them to confirm that both preserve a group structure. It isn't about the Rubik's cube, which is used as an example of symmetry (i.e. a group).
The structure of a group is determined by the group operation; if the operation is preserved the two representations (graphs) are homomorphisms.
It's in French, but with an English translation. The diagrams and graphs are what give the game away here.
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Hi.
As you've noticed in your previous post(s) and directly stated yourself, over half the people aren't going to watch videos. Thats fine, it's a forum where people may prefer to read. However, you may not get a lot of replies - that's ok as long as you're aware that you may not get a lot of discussion.
The structure of a group is determined by the group operation
Yes, that is a way of looking at it.
A group is (G, +) where G is the set of elements in the group and + denotes the binary operation. If you have a full description of the binary operation then this inevitably tells you what every element of the group must be. That's actually quite a lot of information even though it looks like just one thing. Every element of the group has been identified in the ordered set that is the Domain of the operation, so you have put a label on every element of the group.
An alternative way of describing or uniquely identifying a finite group is to specify the following:
(i) identify some elements of the Group as "generators".
(ii) Specify the important properties of these generators under the operation.
Here the attention or focus has been put on identifying some suitable elements that will generate the group rather than directly stating a full description of the operation. You'll end up with the same total amount of information and have a full description of the group upto isomorphism, so you could now provide a full description of the operation if you wanted it. It's just a shift in emphasis about where you start or what information you wanted.
More info: https://en.wikipedia.org/wiki/Generating_set_of_a_group and also this article about the presentation of a group: https://en.wikipedia.org/wiki/Presentation_of_a_group
For example, two groups are isomorphic if and only if you can find an equivalent presentation for both of them.
Best Wishes.
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Last night, I confirmed that the three 3x3 matrices I've chosen are a generating set for a larger group.
The three matrices fix a unit vector (one of say x,y or z) and rotate the other two through π/2, around the first. It's "quaternionic", but I multiply these rotations together, I can't add them together in any meaningful way. No ring algebra so far.
So I color each unit vector, and sure enough I can permute the position (one of eight) and orientation of this 3-vector.
Here the colors identify directions.
And sure enough, a rotation of the vector (x,y,z) "in place" is a trace 0 matrix (a cyclic permutation); this twists the vector to the left or right without changing its position. I therefore have a group which acts on positions and orientations; I'm pretty sure the three group elements I have are a generating set.
It's not hard to draw what I'm describing, for yourself. It's a group whose elements act on a single octant of the sphere.
It's a 2x2x2 Rubik's cube with one 'cubie'. I can at least conjecture that, if I add a second octant of S2, I'm going to need larger matrices.
But from my first "naive" algorithm with one octant, if I draw some x,y,z Cartesian axes and add the first octant, then with the group operations (rotations) I can explore the total space available; it's a graph with one vertex, so far. Adding vertices and edges in graphs is a well-understood algebraic operation.
I'm almost trembling with anticipation, here in the lab.
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About the lab.
This is the one (it's not really in a building) that Erno Rubik was in when he designed his prototypes.
Since the release of the original puzzle, several other versions of the 33 cube have appeared, some with different shapes than a cube, such as Bart Simpson's head, or a globe of the earth. Some have non-equal slices.
One is the void cube which removes the central frame; thus proving you don't require a fixed set of points; you can pack this problem without them.
All the central square pieces do, in the original puzzle, is assist with a packing problem. Removing them does not result in any "loss of generality". This solution--the void cube--will not work when there are only eight octants because the centres have been removed already. I haven't seen any examples of the void cube than 3x3x3.
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Since my last post I had a beam from somewhere that lets me sneak addition into my Cartesian system: the centre of the octant is identified by the sum of the unit vectors.
So I can think of a single vertex piece of a Rubik's cube puzzle as being this sum; I get the same thing if I add the three colors together on the vertex. But x,y,z is a familiar 3-tuple, and red,yellow,blue isn't.
This sum of the unit vectors is the centre of rotation for the orientation group action, a "slice function" isomorphic to the cyclic graph on three vertices. Or Z3. I can represent this as the 3rd roots of unity on that slice.
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Ok I'm going to wind this down for a bit because I've realised there's a few papers to look over.
But my detective's notebook has this:
Rotating an octant of the sphere rotates three points around a circle.
Assuming Mr Hopf's information is accurate, I can take three colored points on the sphere to a Mobius strip.
Since cutting the strip is only meaningful if the cut is parallel to an edge or is along the centre, the cutting itself suggests that removing a circle from the interior of the strip is part of the mapping (but where does it go?). Cutting along a circular path around the strip removes a circle, which cannot be twisted. Hence if the three points are placed across the strip they should lie on different linked "circles". I can draw the cutting path instead of using scissors and figure out if I'm right.
This idea brought to you by this publication: https://arxiv.org/pdf/2212.01642.pdf (https://arxiv.org/pdf/2212.01642.pdf)