In an orthogonal vector space with k elements \( k_1, k_2, ..., k_8 \) and 3 dimensions (D) there are a total of N states for all k; K is the set of 8 elements k: \( N = \frac {K!(D^7)} {KD} = 3,674,160 \). All states are integers, corresponding to rotations in K over D dimensions; I can write these integers as an ordered set \( q_n \) for n = 1, 2, 3, ...,, 14; q generates a "k-function" of n = a discrete integer #states in K. The series I have goes: \( q_0 = 1 q_1 = q_0 + 5 q_2 = q_1 + 21 q_3 = q_2 + 93 q_4 = q_3 + 414 q_5 = q_4 + 1722 q_6 = q_5 + 6713 q_7 = q_6 + 24089 q_8 = q_7 + 81091 q_9 = q_8 + 246359 q_{10} = q_9 + 570080 q_{11} = q_{10} + 420264 q_{12} = q_{11} - 568316 q_{13} = q_{12} - 692256 q_{14} = q_{13} - 90004 q_{15} = q_{14} - 276 \) ..note the roll-off in the value of q at \( q_{11} \). (The last value for n = 15 closes the series - there is no q there, it has a value of 0). Any suggestions about how to parameterize it? There are factors of 3 = D in some of the addends, but I haven't seen a pattern yet? Am I barking up the right tree here, or should I look at another kind of algebra? p.s. is Tex busted..?
Really having trouble understanding your question. Is it simple linear algebra or is this some kind of graduate-level topic with its own specialized jargon? How badly did the TEX mess up? I don't seem to have any trouble with it, but you might need to play around a bit.
As far as the TeX goes, use double backslashes (\\) to start new lines. You can also optionally use \begin{align} ... \end{align} to align things according to where you place & symbols: Code: [noparse][tex] \begin{align} q_0 &= 1 \\ q_1 &= q_0 + 5 \\ q_2 &= q_1 + 21 \\ q_3 &= q_2 + 93 \\ \end{align} [/tex][/noparse] gets you: \( \begin{align} q_0 &= 1 \\ q_1 &= q_0 + 5 \\ q_2 &= q_1 + 21 \\ q_3 &= q_2 + 93 \\ \end{align} \)
Thanks for the tips. This is definitely an orthogonal space. The series of q values represent eigenstates in the space. Each q(n) is a subspace or subset of "k-states". You can envisage \( q_0 \) as equivalent to \( K^0 \). The series obviously is discontinuous; I'm interested in formulating something like 4q + 3, say, since the 3 is = D the #dimensions. Also, this "K-space" is one in which the notion of "distance" is equivalent to the notion of "area"; hence it's either Riemann or pseudo-Reimann and there's a Riemann sum. Bugger all those primes that show up at n > 3... The series reflects the fact there are 8 elements in K each with "dimensionality'. K might be viewed as a "component, with slots for each k"; the "q-operators" assign the \( k_i \) to the slots; each k is a sub-component. I could call the K-thing something else, but want to leave that for now, and analyze the q. The numbers represent "all states which are isometric in K" as far as "orientation" over a distance or equivalent area (i.e. in x,y,z). The states are actually products like x.x' and y.y'; when \( k_i(x) = k_j(x) \) this selects a particular orientation and places it in one of the q subsets. The n is an ordinal for each q (n orders q) and as you can see, the ordering is complex so I'd like to simplify it (at least up to some n). When n = 0, all the x.x' = x.x, y.y' = y.y, etc. or K = 8k(i) is "flat". It's something I'm "toying" with, and I want to see if I can derive some formula and extend it to another, related K-space which has a different "geometry".
Ok, I'll repost the table for the quotients: \( \begin{align} q_1 = q_0 + 5 \\ q_2 = q_1 + 21 \\ q_3 = q_2 + 93 \\ q_4 = q_3 + 414 \\ q_5 = q_4 + 1722 \\ q_6 = q_5 + 6713 \\ q_7 = q_6 + 24089 \\ q_8 = q_7 + 81091 \\ q_9 = q_8 + 246359 \\ q_{10} = q_9 + 570080 \\ q_{11} = q_{10} + 420264 \\ q_{12} = q_{11} - 568316 \\ q_{13} = q_{12} - 692256 \\ q_{14} = q_{13} - 90004 \\ q_{15} = q_{14} - 276 \\ \) \end{align} I noticed after more playing around that the "remainder" approach gives me this, if I subtract q(n) from N, in the case of q(0) I get: \( 3^6(7!) - 1 = 17(216127) \); maybe I should look at the Li(x) Gauss fn, and count primes instead? Can I find the largest prime factor in any "remainder" (is it 216127?), and what does 8969 have to do with n or N? There are "too many groups to list" in the space, these quotient spaces are all external derivatives of the "inner" symmetry groups... something to do with prime numbers (maybe). I think I'm looking for a formulation that "folds" each q into the next, by factoring primes or something. The q with greatest magnitude might be the point of inflection in the series; perhaps I should try 'descending' both ways - back to n = 0, and forward to n = 14? Look for some sort of bijection, maybe think about how n = 0 is "joined" to q(15) -> zero additional states for N?
I think I got this: It's like being on the edge of a wheel, with 14 inner wheels that can rotate around q_0. the centroid. After 15, you're at infinity since no states are generated by n = 15 and q_0 = 1. It's the open disk, with 14 "layers", you have to find a path across - the cycle H is an alignment of s_1...s14, you need 14 "tickets" to make it to 1 = Z; Alternately infinity is 15 tickets away - the tickets are folded and the one you need to get from q_1 to q_0 is the trefoil: Hello Mr Jones... But look at those primes in q_1; 17 is a division of the unit circle - Gauss constructed this. What of p*q, if p is on the circle?
Hmm, I may as well show my hand since I have 14 cards, actually I will have to exclude the center from my hand since it's really "the deck", so I can't really get to the centroid of Q; that's ok, I can live on the trefoil instead - it has three circular measures I can rotate in both ways, it feels more comfortable. My ticket can't get me to q_2 unless it has s*27 on it, I need a 27 state Turing machine perhaps, to turn my trefoil into the next ticket, which has 27 ways it can be punched, to get me there. The void space will be the inside of the trefoil, at a crossing X in q_2
Ho k. So the path \( s_1, s2, ..., s_n \) is 14 spaces wide; in a sense it's also like having to find a span with the right components, to get from q to q. I conjecture that I have a presentation of K isomorphic to a \( K_6 \) which is incompleted; a single vertex at E has degree 4, the 5th is at infinity. Please Register or Log in to view the hidden image!
I'd like to propose that I can post a proof that a function f which composes q exists that is complete up to n = 3, the first 4 "numbers" including 0. I haven't finished it yet, it uses Turing numbers, and f(x) = r.x where r is a rational; f'(x) = x = f(x)/r and {f o g} the set of generators for rationals is in f*p. I also conjecture that the large prime factors in N - 1 = pq are the domain of f;p there are two more "gosh numbers" one a large prime given by \( N\;= \;8! . 3^7 . 12! . 2^{10} \), This N, with -1 added is a large prime (p = N -1); the other has one more power of 3 and 2 more of 2, with -1 added it factors into smaller primes. The number of digits is important in the Turing domain because a real number exists, r, that represents the prime (or any integer) as a number of digits to the right of a decimal point.
It's a doubling problem; in 2(f), you have a quotient space that has s1...s54 the same # as the next volume "form" which is 6(9) colored. 6 and 9 are magic numbers in G, with g_K the generator of the 1 + (N-1) in G. The f is least, and the generator inside it for the inner 14 (rings) the posted list is the image or pullback, that's the most recursive descent in T the #. the doubling problem is in P, the image is in #P. P and #P are not NP. Squares double cubes, that's neat.
Can someone please correct me if I have the wrong end of the string here, but are the homotopy groups on the n-sphere like a table of orbits around Z? If I have this right, there are 2x2 subsquares that appear in the table along the Z diagonal; the 0 or trivial groups must live in the group of infinite rotations, i.e. U(1): that is, this rotation group is a leaf of the table? For instance, where a diagonal for Z in a 2x2 subspace of the table first appears is in the first two homotopies of the sphere, for \( S^1, \;S^2 \). Any hints out there?
Aha; let's say we put 8 points on the n-sphere, somewhere near the equator, with each connected to either pole = 16 connections, one of each pair is i*r. Project \( K8;8 \) onto the sphere, find an inner volume in (n)|Q that maps the n-1 sphere to the edges in connections > 8 in the K. Repeat with next projection - the K will resolve into G.
