For your amusement, here is a little glimpse into a horribly undisciplined mind.
From time to time, my brain gets tackled by some monstrous mathematical idea that dominates all my thinking for hours, days, weeks, even longer.
Early in 2005, I was innocently walking one evening (pushing my son in his stroller, trying to lull him to sleep), and thinking about something thoroughly un-mathematical, when ... suddenly, I found myself suddenly convinced that I had a counterexample to Poincaré's Conjecture. I didn't ask for this to happen. I didn't intend to ponder topology for the next two weeks. It just happened.
(Like most of the world, I was not aware at the time of Grigori Perelman's proof of the conjecture, since reviewers were still busy checking it.)
The (alleged) counterexample was "obvious". Only the fact that better minds than mine had failed to prove or disprove the conjecture for 100 years or so made me mull it over, again and again, to find out what was wrong with my idea. I figured it out eventually. (The beast in question failed, in a rather subtle (to me) way, to qualify as a manifold in the first place.)
And you wonder why I never accomplish anything...
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11 comments:
First oops:
The square isn't like the cube in that twirling around a square vertex isn't a 2d rigid motion, but twirling around a cube vertex is. There is only 1 (trivial) solid motion of the square that leaves a (any) particular vertex fixed. Rotating around, say, a diagonal would take the figure out of the plane. Perhaps putting a cube through the fourth dimension might in the same way get to it's mirror image?
For the cube, for any vertex there are 3 solid motions of the cube that fix that vertex. I've got a nasty feeling that in a tesseract there are going to be more than 4 solid motions that fix a particular vertex.
So for the tesseract, it might not be obvious how to form stereo numbers, and the configuration symmetry group operators might not use simple scalars to describe what happens to the extension after it lands on a new vertex. Or I might be pleasantly surprised.
Suppose you rotate a tesseract around the line [(1,1,1,1),(-1,-1,-1,-1)]. (Does that even make sense?) will the point (2,1,1,1) describe a circle that hits (1,2,1,1),(1,1,2,1),(1,1,1,2) and is there something like a consistent order around-the-circle for all the sets of extensions?
(I needed to make an adjustment to my comment dated August 13, 2008 5:23 PM)
It's very fortunate you brought up the subject, because I have an urgent need to ask for your assistance with a new (to me) monster that is rapidly getting out of hand. And I know that this is related to work you've done before.
Q (short version): For each of the 4 vertices of a square, you can (so to speak) break off a bit of the two edges adjacent to the vertex, and twirl the broken bit 180 degrees so that the broken bits of edge reattach (to the other edges). For each of the 8 vertices of a cube, you can break off a bit of the three adjacent edges, twirl them 120 degrees clockwise (say) and they reattach to other edges. Repeat for 240 degrees, and after doing it a third time you're back where you started from. This is sort of what happens when you twist the corner of a Rubik's Magic Cube. ... OK, so here's the question -- for each of the 16 vertices of a tesseract, is there an analogous way to break off the 4 adjacent edges, and twirl the construct 90 degrees to reattach the broken bits?
I did a Google search on |Rubik's tesseract corner twist|. One of the hits resulted in this hint:
"When pieces or cubies rotate in a cycle, their facelets undergo
n seperate cycles if they have n facelets. The important thing
is that there are always n disjoint cycles. In the tesseract,
every corner rotation boils down to four 4-cycles of facelets
for each cycle of four cubies."
Obvious supplementary Q: does this generalize to 5,6,... dimensional hypercube.
OK, now here comes the motivation. I'm supposed to be preparing a talk on stereo numbers for the Math guys at St Mary's. I'm only a bit over a year late ;-) and about seven weeks ago had an idea to introduce a simple proxy for a chemical molecule that would make the concepts easier to visualize. Yeah, right.
The "twist cube" is a 2x2x2 Rubik's Magic Cube in two colors. At START, all the U (up) and D (down) facelets are colored "X", while all the other facelets are colored "blank". This (rather boring) cube consists of just the 8 corners of an ordinary Rubik's Cube, except that all the corners are identically colored, and the only information you've got is the state of the cube twist at each corner.
Each of the 8 corners can be twisted independently.
Problem: to name, classify, and count the different patterns you can have.
NOTE: If you can get from one pattern to another by a solid rotation of the cube, they are not different patterns.
Obviously, if you hold the cube still, there are 3 ^ 8 = 6561 apparent patterns. After an arbitrary ordering of the 8 corners, all of these patterns are coded (named) using 8 digit ternary stereo numbers. As you come to each vertex, the corresponding digit is 0 if the X facelet is U or D, 1 if it's rotated 120 degrees clockwise from that, 2 if 240 degrees.
The Configuration Symmetry Group for this mess consists of the 24 permutations of the 8 corners corresponding to the solid rotations of the cube, in cyclic form, with the additional information as to how much of a twist is added to each destination corner. Since ceil [6561 / 24] = 274, and there are a fair number of symmetric patterns, there should somewhere in the ballpark of 300 unique patterns.
So that was clear as mud, I've got a geometric interpretation. The idea arose out of the realization that the twist cube is analogous to the (very theoretical and probably impossible) chemical molecule (CoH2F)8. Cobalt can form bonds with 6 neighboring atoms in the geometry of an octahedron. So the 8 Co atoms join to each other in the form of a cube. Each vertex of the cube has 3 more bonds that extend the edges of the cube outward. For each Co, these extra 3 bonds attach 2 Hs and an F. The F corresponds to the X facelet of the twist cube, but in fact, it is the extension of one of the three edges of the vertex perpendicular to the X facelet.
So, here's a concrete construction you might actually be able to grok:
Let C be a unit cube consisting of the 8 vertices v = (x,y,z) where x,y,z = 1 or -1; along with the cube edges e = [u,v] where u and v are adjacent in the cube (they differ in just 1 of their coordinates).
Now for each vertex v = (x,y,z) choose a point t(v) which is one of the points (2x,y,z), (x,2y,z), or (x,y,2z). Then the extension T(v) for v is the line segment [v,T(v)]. That is, at each vertex of the unit cube, we choose to extend one of its adjacent edges for one unit beyond the cube.
So, once again, if we consider two of these beasts that can be isometrically mapped one onto the other using a solid rotation to be not different, how many different such beasts are there, and how may we name and classify them?
The extension to other dimensions in this formulation is obvious. So my original question comes down to whether we might expect work in the 3 dimensional case to generalize to other dimensions.
I'm confident that the 2nd theorem in my 1986 paper will be of help in the 3d case. Furthermore, I've been staring at the Wikipedia article on Wreath Products for some time, and have some faint hope that the concept of "blocks" they talk about there might be apropos.
Any multi-dimensional counseling you could spare is eagerly anticipated.
Hi John.
I haven’t solved your twist-cube question, but I do have a couple of comments on the multidimensional business.
(1)
Suppose we start with an n-dimensional hypercube, and we break off a corner piece, twist it around, and reattach it, so that the original vertex ends up where it started. How many different ways can we twist the piece?
The key here is to think about the “vertex figure” of the hypercube – in other words, the (n-1)-dimensional shape that gets revealed when you chop off the corner. More precisely, if O is the centre of the hypercube and V is a vertex, consider the (n-1)-dimensional hyperplane that is orthogonal to the line OV, and intersects the line-segment OV somewhere near V. The “vertex figure” is the intersection of this hyperplane with the hypercube.
A vertex figure of an n-dimensional hypercube is an (n-1)-dimensional simplex. Chop the corner off an ordinary 3D cube, and you get an equilateral triangle. Chop the corner off a tesseract, and you get a regular tetrahedron. And so on.
The number of ways that you can put the piece back depends on whether you are allowing rotations only or both rotations and reflections. (And yes, adding another dimension to the space you’re working in always makes your “reflections” achievable by a rotation.)
If you allow reflections, there are n! ways to put the piece back. Only n!/2 if you forbid reflections.
(2)
Re your question about rotating a tesseract around the line [(1,1,1,1), (-1,-1,-1,-1)]: The group of allowed rotations is spherical rather than circular. The locus of points that (2,1,1,1) could map onto, for such a rotation, is a sphere – and yes, it includes (1,2,1,1), (1,1,2,1), and (1,1,1,2).
In four dimensions you can rotate “around a plane”. More generally, in an n-dimensional space, you can pick any (n-2)-dimensional subspace and define a circle’s worth of rotations “around” that subspace – i.e., rotations that don’t budge any point in that subspace. If you only specify an (n-3)-dimensional subspace (such as the line in the 4D space of your question), you have a sphere’s worth of rotations around that subspace.
Thanks! Obviously there's some work to do.
I don't know if this is important, but it might be a place for me to start getting a handle on how hypercubes move around. I read Harary's Graph Theory when I was trying to figure out stereo numbers in 1974-76.
"The Automorphism Group of a Hypercube", by Frank Harary (1925 - 2005), J.UCS, 2000.
I gave my computer the task of crunching through your 2x2x2 2-colour cube by brute force, it turns out that there are 306 distinct patterns.
3 of the patterns (one being the "at start" pattern) each correspond to 3 "apparent" patterns.
6 patterns each correspond to 6 "apparent" patterns.
51 patterns each correspond to 12 "apparent" patterns.
246 patterns have no symmetry to speak of, and therefore each correspond to 24 "apparent" patterns.
I haven't figured out an elegant way to get these results, nor a good way to name or classify the patterns.
Thanks for the info. At least my guess of "about 300" worked out :-)
The degenerate case of the one dimensional "2" cube (the X axis) yields 1 pattern with the one stereo number 00 (in base 1).
I stared at the 2x2 case for a while and concluded that the most aesthetic zero pattern was not the one with the extended bits going up / down (in the Y direction). My argument is that then the configuration symmetry group operations would then be things like (1,2',3,4') (numbering counter-clockwise from upper right, see below).
Instead the directions alternate so that the configuration symmetries aren't primed.
So the vertex order I'm using for calculating stereo numbers is (most significant digit vertex first) (1,1), (-1,1), (-1,-1), (1,-1) and the extra line segments for the zero pattern would end at (2,1), (-1,2), (-2,-1), (1,-2).
There appear to be 6 unique patterns:
2 correspond to 1 apparent pattern
1 corresponds to 2 apparent patterns
3 correspond to 4 apparent patterns
The 6 stereo number equivalence classes are:
{0}
{1,2,4,8}
{3,6,9,C}
{5,A}
{7,B,D,E}
{F}
Patterns 0 and F are a pair of mirror image opposites, as are patterns 1 and 7.
Patterns 3 and 5 are their own mirror images (i.e. you can find a line of symmetry in the pattern.
Searching for the sequence 1,6,306 did not result in a hit at the The On-Line Encyclopedia of Integer Sequences, so it is possible nobody else has bothered to explore these objects.
Guess: solid rotations of the cube (and therefore configuration symmetry group elements) exercise all the even permutations of the possible twist directions from one hyper-corner to the other. So the order of the directions at a corner should be chosen to respect these. That's why in Rubik's Cubology corner facelets are listed clockwise.
If I'm seeing how the 2x2 builds from the "2" cube, the natural numbering for the 2x2x2 corners might then be the 2x2 case on the top level, then the reverse of that at the bottom, sort of like the seams on a baseball.
I agree with your 2D results.
For the 3D case: I don't understand your "guess". But I do understand the image of the baseball. I like this way of ordering the digits, because it can be extended easily to any number of dimensions. (If you trace an n-dimensional baseball seam for the "top" n-dimensional layer (in (n+1)D), then trace the same thing backwards for the "bottom" layer, that pathway "defines" an (n+1)-dimensional baseball seam.)
So we seem to be agreed on the order of the digits. But I'm not convinced WRT the orientation of the digit values.
Let's go with your idea for the sake of argument. Assuming that I understand you right: Working in 3D, if we look at the cube from somewhere out on the +Z axis, the zero-position for the "near" layer should look like the zero-position that you described for the 2D case, while the zero-position for the "far" layer should look like the mirror-image of that 2D zero-position.
So the ends of the "hairs" in the 3D zero-position are at (in order):
(2,1,1),
(-1,2,1),
(-2,-1,1),
(1,-2,1),
(2,-1,-1),
(-1,-2,-1),
(-2,1,-1),
(1,2,-1).
Now, the "apparent patterns" are given by 8-digit ternary numbers, analogous to the 4-digit binary numbers that you collapsed into hexadecimal for the sake of presenting the 2D results.
So "00000000" (ternary) describes the arrangement with the "hairs" as described above. But how do we label a digit as "1" vs "2"?
In the spirit of this proposal for the zero-position, perhaps the little cubies should alternate CW/CCW in their 0-1-2 cycles.
That is, if you look "directly" at a cubie, along the line that connects the vertex to the centre of the cube, some cubies go clockwise as their (ternary digit) positions go from 0 to 1 to 2 to 0, while other cubies go CCW instead. And each cubie with a CW digit sequence should immediately follow (and immediately precede) a cubie with a CCW digit sequence, as you follow the cubies along the baseball-seam pathway.
The patterns with maximal symmetry (i.e. corresponding to only 3 apparent patterns each) are
{00000000,11111111,22222222}
{01010101,12121212,20202020}
{02020202,10101010,21212121}.
The first and third of these patterns are mirror-images of each other. The second pattern is the one with all eight "hairs" pointing along the same axis (all x, all y, or all z).
Now here's the punch line: If we labeled the ternary digits simply according to which axis the "hairs" are pointing along (say, 0 for x, 1 for y, 2 for z), then we'd get exactly the same results for the ternary numbers labeling the maximally-symmetric patterns. But this time, the 01010101 and 10101010 states would be mirror images of each other, while the 00000000 state would be the simplest one to describe.
And I think this is a powerful aesthetic argument for assigning digit values simply according to the axis along which the "hair" is pointing, notwithstanding the primes that appear in the 2D symmetry group operations.
By the way: Did you notice that the same analysis that gives you 274 as a lower bound for the number of (true) patterns in 3D gives you over 22 million (true) patterns in 4D? And this is just a 2x2x2x2 tesseract! No wonder it doesn't appear in the encyclopedia of integer sequences.
From your argument for the simple stereo numbers, I conclude that for the 2x2x2 case we don't quite have a precise concept of twist that lets us build configuration symmetry group elements.
That is, suppose a solid rotation of the cube maps c into d. In general what does it mean to say that c lands on d untwisted, versus twisted 1/3, versus twisted 2/3. Or in the decorated cyclical permutation notation: (...c,d...) versus (...c,d'...) versus (...c,d''...).
In Rubik's cubology, the U and D (i.e. +z -z direction) facelets of the corner cubies / cubicles are "prefered", so that a cubie is "untwisted" if its U or D face is U or D at its destination, twisted 1/3 if it's at the position 120 degrees clockwise from U or D as seen from outside (which is the x-axis direction half for half the corners, y-axis for the other half) and 2/3 if it's at 240 degrees clockwise.
So it looks to me that to define how a corner lands at its destination (how twisted), there needs to be a mapping that looks at the signs of the axes for the start and destination corners and defines the permutations that move the hair around depending on its being untwisted, 1/3 twisted, or 2/3 twisted.
Generalizing this to other dimensions would at least make the primes go away for the 2x2 case.
I'm guessing that corners can land in the 2x2x2x2 case with any of 12 "hypertwists" depending on the solid motion of the hypercube, with the shufflings again depending on the relative sign values for source and destination corner axes. And so on for higher dimensions.
My recommendation is that the base-n digits should simply be labeled according to which axis the corresponding "hair" points along, regardless of whether it's the positive or negative direction along that axis.
In the 2D case, your symmetry group is based on a rotation that applies the permutation (1 2 3 4) to the digit positions and (0 1) to the digit values.
(I always find the "primes" confusing. Let's just de-couple the permutation that applies to the digit values from the permutation that applies to the digit positions. This is always possible for solid rotations in any number of dimensions.)
So if we "square" this basic group element, we get (1 3) (2 4) to the digit positions and no change (identity element) to the digit values. Etc.
So the 2D patterns will be
{0, F}
{1, 4, 7, D}
{2, 8, B, E}
{3, 6, 9, C}
{5}
{A}
Now in the 3D case, following the "baseball seam" principle for the digit positions, we can make a 24-element symmetry group based on the following primitive elements:
(1 2 3 4)(5 8 7 6) to positions, (0 1)(2) to values;
(1 4 5 8)(2 3 6 7) to positions, (0)(1 2) to values;
(1 8 7 2)(3 4 5 6) to positions, (0 2)(1) to values.
If I bang my head into this for a day or two, I should be able to come up with the 6(?) primitive group elements for the 192-element symmetry group in 4D (i.e. the group of rotational symmetries of a tesseract).
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