Spinors, Wikipedia: Spin representation
Andrius investigates
Spinor combinatorics
Underlying eightfold pattern given by {$(1+i)^n$}
Note that {$\frac{\sqrt{2}}{2}(1+i)=e^\frac{\pi}{4}$} is the eighth-root of unity. The length of {$1+i$} is {$\sqrt{2}$}. Thus {$(1+i)^8=2^4=16$}.
| {$\begin{matrix} 1\;\;\;\;\; \\ 1+i \\ \;\;\;\;\; 2i \\ 2-2i \\ -4\;\;\;\;\;\; \\ -4-4i \\ \;\;\;\;\; -8i \\ 8-8i \end{matrix}$} |  |
Combinatorial interpretation of {$S\otimes S$}
Let complex vector space {$S=W\oplus U\oplus W'$} where {$U$} is one-dimensional and the dual of {$W$} is {$W'$}.
Define the exterior algebra {$S=\wedge^\bullet W$} of {$W$}. If {$W$} has dimension {$m$}, then {$S$} has dimension {$2^m$}.
{$S\otimes S\cong \bigoplus_{j=0}^m \wedge^{2j}V^*$}
{$2^m2^m=\binom{2m+1}{0}+\binom{2m+1}{2}+\binom{2m+1}{4}+\dots +\binom{2m+1}{2m}$}
And furthermore
{$\sum_{j=0}^m(-1)^j\dim (\wedge^{2j}\mathbb{C}^{2m+1}) = (-1)^{\frac{1}{2}m(m+1)}2^m = (-1)^{\frac{1}{2}m(m+1)}(\dim \textrm{S}^2 S - \dim\wedge^2 S)$}
{$(-1)^{\frac{1}{2}m(m+1)}2^m=\binom{2m+1}{0}-\binom{2m+1}{2}+\binom{2m+1}{4}-\dots +\epsilon_{2m}\binom{2m+1}{2m}$}
{$(-1)^{\frac{1}{2}m(m+1)}2^m=\binom{2m+1}{0} +\epsilon_1\binom{2m+1}{1} -\binom{2m+1}{2} +\epsilon_3 \binom{2m+1}{3} + \binom{2m+1}{4} \dots + \epsilon_M\binom{2m+1}{M}$}
{$(-1)^{\frac{1}{2}m(m+1)}2^m=\binom{2m+1}{0} + (-1)^m\binom{2m+1}{1} -\binom{2m+1}{2} + (-1)^{m+1}\binom{2m+1}{3} + \binom{2m+1}{4} \dots + \epsilon_M\binom{2m+1}{M}$}
What is the sign of {$\binom{2m+1}{1}=\binom{2m+1}{2m}$} ?
- If {$m$} is even, then it is positive.
- If {$m$} is odd, then it is negative.
What is the sign of the largest and final term {$\binom{2m+1}{M}=\binom{2m+1}{m}=\binom{2m+1}{m+1}$} ?
- If {$M$} is even then the term is {$\binom{2m+1}{m}$}
- If {$m$} is divisible by 4, then it is positive.
- Otherwise, it is negative.
- If {$M$} is odd then the term is {$\binom{2m+1}{m+1}$}.
- If {$m+1$} is divisible by 4 then it is positive.
- Otherwise, it is negative.
So if {$m=k$} or {$m=k+3$} it is positive, and if {$m=k+1$} or {$m=k+2$} it is negative.
The eightfold nature comes from the sign of the intial terms, either +,+,-,- or +,-,-,+ and the sign of the final terms, either +,+,-,- or +,-,-,+ or -,+,+,- or -,-,+,+. Yet I am dealing only with odd values {$2n+1$} and there are also even values {$2n$}. But Bott periodicity only deals with the even orthogonal groups. So I don't understand.
The key point is that when there are {$n$} bits there are {$n+1$} switches that can be placed around and between them. These switches then act on the default expectation, switching it (from {$0$} to {$1$} or from {$1$} to {$0$}). So this is an additional way of looking at information that takes the independent individual bits and considers them in a collective string with a default expectation. And that expands the symmetries to consider.
| {$2^2 = \binom{3}{0} + \binom{3}{2}$} |
| {$4 = 1 + 3$} |
| 00 | .1.0 0.1. .11. |
| {$-2 = \binom{3}{0} - \binom{3}{2}$} |
| {$-2 = 1 - 3$} |
| 00 {$\Leftrightarrow$} .1.0 |
| 0.1. .11. |
| {$2^4 = \binom{5}{0} + \binom{5}{2} + \binom{5}{4}$} |
| {$16 = 1 + 10 + 5$} |
| 0000 | .1.000 0.1.00 00.1.0 000.1. .11.00 0.11.0 00.11. .111.0 0.111. .1111. | .1.0.1.0 .1.0.11. .1.00.1. .11.0.1. 0.1.0.1. |
| {$-4 = 1 -10 + 5$} |
| {$-2^2 = \binom{5}{0} - \binom{5}{2} + \binom{5}{4}$} |
| {$-2^2 = \binom{5}{0} + \binom{5}{1} - \binom{5}{2} $} |
| 0000 <=> .1.000 |
| .1111 <=> .1111. |
| 0.111 <=> 0.111. |
| 00.11 <=> 00.11. |
| 000.1 <=> 000.1. |
| 0000. <=> 00.1.0 |
| 0.1.00 |
| 0.11.0 |
| .11.00 |
| .111.0 |
Note the sequence A002054 which is {$\frac{(2n+1)!}{(n+2)!(n-1)!}$}, namely {$\binom{3}{0},\binom{5}{1},\binom{7}{2},\binom{9}{3},\binom{11}{4},\binom{13}{5}\cdots$}
I solved the puzzle with the plus and minus signs. Briefly, in Pascal's triangle
| 1 |
1 1
1 2 1
| 1 | 3 | 3 | 1 | ||
| 1 | 4 | 6 | 4 | 1 | |
| 1 | 5 | 10 | 10 | 5 | 1 |
The first puzzle was to prove that 1 + 10 + 5 = 16 = 2^4.
An easy way to easy way to see that is to add 0 to the line 0 1 4 6 4 1 and then pair up:
0+1=1, 4+6=10, 4+1=5 and the sum of this fourth row is 2^4 = 16.
We can express this in terms of choices from 5 by considering the spaces around the digits and placing switches * there, as I showed you.
Thus **_** gives *1*0_0*1* for example.
The second puzzle is to show that 1 - 10 + 5 = -4 – -2^2. This is "5 choose 0" minus "5 choose 2" plus "5 choose 4".
Similarly, we pair up 0+1=1, 4+6=10, 4+1=5.
The "culling" that we introduce says that if we see a pair 00 then we replace it with 11, and if we see a pair 11, then we replace it with 00. This changes the number of 1's by 2, which changes the sign.
For example, we start this search from the middle of the four digits, which is to say, the second and third. If they are 00 or 11, then they cancel away. So the fixed points must have 10 or 01. Then work outwards and consider the first and fourth digits. If they are 00 or 11, then we switch them and they cancel away. So we are left with 10 or 01. So there are only the following four possibliites:
| 1010 | 0101 | 1100 | 0011 |
So the left side inverts the right side and vice versa. Curiously, this is like the two particles in a singleton, with opposite states.
Then we can write this out in terms of 5 switches. The middle switch will be left out. Then the second digit must equal the third digit as here:
| *_*_ | becomes *1_11*0_ |
| _*_* | becomes _0*11_1* |
| **_ _ | becomes *1*00_0_ |
| __** | becomes _0_00*1* |
Note that the middle digits (the second and third) are the same. But this brings to mind the root system understood as a duality between counting forwards and counting backwards where we fuse an internal zero so that we count
...-3, -2, -1=+1=zero, +2, +3...
This is when we are choosing from n=Odd.
(Although this is the pattern for the Dynkin diagram for D_n which is n=Even orthogonal)
When we choose from n=Even then it would be simply
...-3, -2, -1, +1, +2, +3
(Although this is the pattern for the Dynkin digarm for C_n which is for the symplectic group.)
So I have to think further about this and how it relates to the simple roots.