Although the following sequence is actually composed in \( \mathbb R^3 \), I'm trying to make it look 2-dimensional. The 'coordinates' correspond to (#arcs, #alternating dimensions over \( R^3 \)), where an 'arc' is a transition. The sequence is: {(0,0), (1,1),(2,1), (2,2),(3,2),(4,2), (3,3),(4,3),(5,3),(6,3), (4,4),(5,4),(6,4),(7,4),(8,4), (5,5),(6,5),(7,5),(8,5),(9,5), (6,6),(7,6),(8,6),(9,6),(10,6),(11,6), (7,7),(8,7,),(9,7),(10,7),(11,7),(12,7),(13,7), (8,8),(9,8),(10,8),(11,8),(12,8),(13,8), (9,9),(10,9),(11,9),(12,9),(13,9),(14,9) (10,10),(11,10),(12,10),(13,10),(14,10), (11,11),(12,11),(13,11),(14,11)} You can assume there are actually 10 transitions between the second and last rows which are important or "non-trivial", so the first and second rows can effectively be removed. I'm not that good at linear algebra, but a question I have about the chart is, is there a way to determine some kind of diagonalisation formula, given just the numbers?
I have no idea what you're talking about. Are you doing some homework problem, if so what precisely is the question. If not can you explain a little further what you're doing to get those numbers?
Perhaps I can explain what is meant by a transition a bit better. The numbers are explained as (#arcs, #alternating dimensions over \( R^3 \) ), so are 'coordinates' like (1,0) and (0,1) also valid? Well, (0,0) means 'no arcs, no chosen alternate dimensions'. So (1,0) means a single arc but no dimension, and (0,1) means no arc and a single dimension chosen. A way around this is to envisage an arc, which is "taken" in one of two directions, as a translation of the origin by a unit length along a chosen dimension followed by a construction of an arc with length \( \pi /2 \), from the point (1,1) or (1,-1)--the sign corresponds to a chosen direction--to the initial origin at (0,0). A double arc, such as in a (2,1) transition, has length 2 along the chosen axis, so its endpoint won't be above or below the axis, it "loses" the sign that single length arcs "preserve". I think, sticking to this idea of a set of coordinates ordered by the number of alternate choices of dimension (in 3-space), and with directional transitions between rows/columns, we can see that the transition (0,0) -> (0,1) is a case of choosing one of three spatial directions. Then the transition (0,0) -> (1,0) makes sense if it gets "added" to the first, so it's actually (0,0) -> (1,1); the (1,0), (0,1) 'coordinates' exist in some inner sense and get 'handled' by (1,0)+(0,1) operations--they have to be taken together. There is something about the construction of a semicircular arc along a chosen axis. This leaves a distance of 2 along R if the start and end points intersect R. However, if instead the startpoint is the endpoint of a quarter-circle arc, this leaves an endpoint which is effectively a change of sign for the existing (1/4) arc, and no distance added along R. The (2,1) kinds of transition either add a distance of two 1/4 arcs in the same direction, or invert the sign of a single 1/4 arc and add zero distance. In the first case they also make the direction vanish, but preserve the distance along R.
Actually that last paragraph is confusing. There is a difference between a single arc and a double arc on the same axis, because both (1,1) and (2,1) are the same distance from (0,0), since they are also on the same row. There are three real axes; there are two choices for direction for each axis, so (1,1) can reach 6 different places, or the coordinate 'covers' 6 points. But (2,1) only covers 3, because the arc-lengths are twice as large. In both cases the distance covered is equivalent. Likewise the next row: (2,2),(3,2),(4,2), are at equivalent distances from both (1,1) and (2,1), but in pairwise fashion, i.e. (2,2),(3,2) are a distance of "1" from (1,1), and (3,2),(4,2) are the same distance from (2,1). The pairwise arrangement only corresponds to adjacent columns in adjacent rows--the coordinates have to be next to each other, and arc-transitive.
'sigh' It looks deceptively simple. There must be a sequence from (1,1) to (11,11) that contains only single arcs. This sequence (a 1-dimensional curve in 3-space) can be seen as the 'application' of the constructive aspects of the "trivial" (1,1) as a repeating operator on a single alternating dimension, i.e. x,y or z. You get a real interval of 'glued together' sections Rx,Ry,Rz,... up to 10 units long--given that being 1 unit from the 'origin' is a redundancy. If the (2,1) construction is allowed, things get more complex--you get extra columns. Also, some 'coordinates' are missing or unreachable. Starting with the 5th row, there is no (10,5) at the end, but there is a (10,6) in the 6th row. Another clue: the table or chart in post #1 was generated by an algorithm that searched all the possible transitions exhaustively, and is presumably an accurate set of 'data points'.
I think I've worked this out, it's because of the modular arithmetic being used. There are only four 1/4-circle arcs in a circle (duh!), and I've already explained why the construction (1,1)+(2,0) = (1,1), since it only inverts the direction of the first arc. So (3,1) is unreachable or formally (3,1) -> (1,1). Likewise (4,1) -> (0,1), which is the case of 'choosing' one of 3 dimensions. So for (n,1) type constructions, every odd n is a case of (1,1) and every n mod 4 is a case of (0,1). This leaves every odd number times 2 for the case or 'type' (2,1). This is why the first row has only two entries. Suppose the next row, {(2,2),(3,2),(4,2)}, has only three entries for a similar reason--the modularity of the 'construction types' in the previous row? Demonstrating or proving this seems a bit harder, since the 'choice' (0,2) means selecting two alternate dimensions, and the left-hand numbers have to be divided by the construction. For instance, (2,2) is a composition of (1,1)+(1,1), where the right-hand numbers are mapped to alternate dimensions as explained previously. The recursive composition of (arc, dimension) 'units' descends into the space, the range of 'coordinates' in the chart is limited by the modular arithmetic (that has to be a really general observation). Constructing the entire chart from first principles is probably a fairly large and mathematically difficult problem, which is why a computer was used instead.
Can anyone recommend a graph drawing package so I can construct a visual or two? The exercise of putting together the arc-transitive lattice of 'points' in the given sequence isn't too hard. As explained, the (1,1) type of data point actually means one unit of distance in one dimension, with a directed arc. Abstractly, you construct a sequence of points along the line (0,0),(n,n) by adding the (1,1) to itself recursively. The n is limited, for some reason, to a maximum value of 11 units, after which the sequence shows the line goes vertical (i.e. is the line 'x' = 11) up to (14,11) (i.e. 'y' = 11..14). Note I'm plotting the righthand term of each tuple along the abscissa, and the lefthand term along the ordinal, but that's another free choice--actually it's irrelevant to the structure of the lattice. The line from (0,0) to (8,4) has a slope of 2. From (8,4) the line 'bends' inward--the slope changes to < 2, then returns to the 'geometric' y = 2x + c shape, until (13,7). Then, as we can see, things go pear shaped.
More numbers If you take the time to draw the lattice (using a graphics package so you can post it online, meh) you can see some 'interesting' stuff. If you draw three lines after deciding the ordinal and abscissa correspond to the left and right terms in each (m,n) tuple--"m" and "n" are just positive integer values--one of the lines is y=14, there are two lines: y = x and y = 2x, intersecting the first horizontal line at the points (14,7) and (14,14); there are 6 points on y = 14 between the two points of intersection. The total number of points on the 'boundary' of the lattice is 28 and there are 36 points in the interior, for a total of 64 points including (0,0). Of the 36 interior data points one is unreachable, which is the point (13,12). Of the 28 boundary points, 9 are unreachable: (10,5),(12,6),(14,7),(14,8),(14,12),(14,13),(14,14),(13,13),(12,12). There are 10 points in the 'full' construction of the lattice that are never included in any traversal of it; traversals build an interval along R in 3 alternating dimensions ("dimension" is the alternating group \( A_3 \)). Note that the missing interior point means there are 35 not lying on a boundary. There are 28 - 9 = 19 points on the boundary (including the origin). Excluding the origin means there are actually 18 boundary points, but because of the missing 'data' the lattice has boundaries internal to the three lines constructed as above. One of these is the line x = 11, which defines the righthand boundary. Another observation is that the lattice structure is time-independent, but traversals and constructions of unit intervals are time-dependent. Perhaps if I mention that I believe the lattice is a covering space of the Mathieu group, in fact it is a divisor of \( M_{12} \). The intervals that get glued together along R belong in the domain of fractional linear transformations, which someone here probably knows something about. The questions are: what is important about the direction an arc takes? Why are there points missing and why is their number = 10, the same as the number of 'transition levels' between (1,1) and (11,11)? Is there a path through the lattice that visits each point exactly once, and what does it mean if there is or not? Sometimes, you know, I wish I had better things to do than stuff about with lattice theory.
Knowing the direction is the same as knowing the direction a clock's hands are turning. That is, from which 'side' of the clock you are looking at the hands, or if the 'clock' is an electron say, then it's spinning in a clockwise or anticlockwise direction depending on whether its momentum is toward or away from you. Inverting the direction like: (1,1) + (2,0), is then the same as reversing the direction an electron is moving. If the direction is 'lost' as in: (1,1) + (1,0), this is the same as bringing an electron to a state of rest, or stopping a clock.
Ok, I've been looking at Delaunay triangulation and Voronoi diagrams. There are only two kinds of Delaunay triangle in the lattice. A smaller circle around three points which are 'in position' like: (1,1),(2,1),(2,2) has a center which always lies in the interior of the lattice. Circles around larger triangles like: (0,0),(1,1),(2,1) can only be drawn along the upper 'y = 2x' diagonal, and have centers exterior to the lattice. You can draw four Delaunay triangles this way, along the upper diagonal, from (0,0) to (8,4). Then there is the 'jag' to the right and a smaller triangle, then two more like the first four, but with centers in the interior. The 'uppermost' point of the triangulation is (13,7), which is connected to only two other points--i.e. you have to 'go backwards' and reduce the number of unit intervals by one (in the total along R) at this point, to 'leave' (13,7). Time is reduced to the concept of 'clockwise' or 'anticlockwise' along a preferred direction. In other words, you decide that a clock's hands should be moving clockwise, and if they aren't, you flip the clock around (you 'do' a (2,0) transition in place). The lengths of time intervals aren't relevant in this frame of reference, only the direction of particle spin and momentum.
Oops, I made a mistake with the upper diagonals. The triangles in the interior actually have centers outside of the lattice. And the lower diagonal has 10 of the smaller ones with each center on the boundary of the lattice. So there are 10 exterior centers for the larger triangles--three have been removed from the uppermost diagonal because of the missing points leaving the four mentioned and six more along the next diagonal down; all the smaller triangles have neighbours. All the interior points except for those next to missing boundary points have 6 other points connected, which is a hexagonal tiling. Dammit, wrong again; I forgot a Delaunay triangle has a circumcircle where all three vertices are lying on it and no others, so that there are only 6 (bugger) that have exterior centers. I think it has to do with the same 6-fold symmetry the interior points have.
Apart from the bloopers, and even what Delaunay triangulation is about and the dual Voronoi lattice, it should be obvious that the fractional part of the fractional linear transformations is the unit interval along R, 'filtered' by A3. Note that after choosing the dimension of a preceding interval (including of course, the first interval) there are two left for the next, successor interval. Thus A3 alternates 'via' A2 and this is because of the nature of choosing from three dimensions, nominally x,y,z. Any transition, such as (1,1) -> (2,2) for example, transforms the origin along R in a single dimension. The directed arc for the above example can be given a sign to label the direction, i.e. t -> +t, -t. So that from the origin in 4 dimensions (the usual suspects), you get the transition \( (0,0,0,0) \to (\pm t,x,y,z) \) by choosing the first dimension and generating the first unit distance. Only one of x,y,z is = 1 and there is no reason to alternate over all three, since A2 can 'stay' in two of the three dimensions. Then you have the successive \( (t,x,y,z) \to (t',x',y',z') \), and the choice to include or exclude one dimension. One of three dimensions is excluded from any choice by being the predecessor. The 10 missing data points presumably can be 'predicted' from the existing data--there is a system of equations that does this and furthermore that 'explains' why they aren't in the set of points. I suppose there's a need to consider random walks and entropy and stuff.
Does anyone else see any correspondence here between the lattice model and anything else? Another thing that occurred to me about the fractional transformations, and since I threw the phrase "covering space" out there, is that the sequence in my #1 post is the set of all points where the Mathieu (sub)group, M12, intersects with the covering space. Since you can construct the M12 group as 12 spheres of equal size arranged around a central 13th equal sized sphere at 12 places--the outer spheres don't touch each other, the points of intersection [with the lattice] are like a further layer of spheres touching the 12 at three places each (but how many? [my "guess" is 8, which has to do with arranging 12 spheres on a plane surface]), something like Brownian motion then ensues and the outer spheres jostle around, but according to discrete rather than continuous motion, which is 'pairwise orthogonal' in respect of dimension. That is to say, once the 'system' chooses a direction, it moves perpendicularly to the remaining two dimensions along R. You don't 'get' the M12 subgroup until you label the 12 spheres, giving them coordinates, and allow certain kinds of tranformation of the coordinates around the central sphere.
I think the quaternion group is what gives the lattice the unreachable points. The point (3,1), which is 3 arcs in one dimension, isn't in the lattice because the vectors are 'pure imaginary'. Likewise the other points are missing (but not from the M12 subgroup) for the same reason. And it turns out the lattice is the simplest 'quaternion space' as a 3d face of a hypercube, sliced into 8 equal parts.
Wow you are looking like a broken water pipe Arf. Deceptively simple looking . I am looking at My lattice painting for the Music Professor. Yeah It looks like a lattice for sure and the next set is just more spread out, but the lattice is the same. One click at a time my man. Information Silos will fall. It is inevitable.
Time to wipe the blackboard, and come clean. I've been looking at a certain problem for a while and more or less asking questions about what the problem really is (it's a boundary problem), and so I suppose, since a painting is worth doing, here is what the problem looks like: Please Register or Log in to view the hidden image! Bwa-ha-haaa! The sequence in #1, determined by a computer program, tells me that you can only 'fold' (braid?, wreath?) the thing 11 times. The next boundary problem, which is what you are looking at when it gets the 12 extra labeled 'spheres' (note, the points of contact for packed spheres correspond to the inner flat surfaces of the packed 'cubelets', each external facet is labeled, but doesn't make any contact), is unresolved, but it's known to be at least 18-fold and less than 23-fold. The 8 'spheres' with 3 points of contact each, are a fixed lattice in each case, with a boundary which is the last row of numbers. If you remove the first row it looks "10-dimensional", which is interesting, given the fundamental group is Z3.
