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Bott periodicity, Bott Periodicity Chat GPT, Spin representation combinatorics

The Combinatorics of Bott Periodicity

A perspective is a distinction between what is known and what is not known. A symbolic expression of a perspective (such as {$1_k+0_k$}) is a syntactic distinction regarding the symbol (what is gets expressed whereas what is not does not get expressed) but also a semantic distinction regarding the referent (what is referred to as meaningful). This double distinction grounds the possibility that the opposing syntax (what is not known) is distinguished semantically as relevant.

When a perspective aligns syntax and semantics, then it expresses what is known, and I call it a concept of the answering mind. When a perspective opposes syntax and semantics, then it expresses what is not known, and I call it a context of the questioning mind.

Algebraically, the (unconscious, direct) concept is given positive sign and the (conscious, indirect) context is given negative sign.

Real Bott periodicity comes from the constraint of pairing each concept (expressed as {1_{k_1}$}) with a context (expressed as {$0_{k_0}$}). This is to say that a concept is not paired with a concept, and a context is not paired with a context. Combinatorially, this means that 00 and 11 have opposite signs and are paired and cancelled away. What survives has the form 10 or 01.

This means that subsets of {$1_1,\dots,1_n$}, each {$1_k$} contributes multiplicative weight {$i$}. Thus there are two tracks, off by one, odd numbers of {$1$} and even numbers of {$1$}, which alternate in sign. This means that the signs are {$+,-,-,+,+,-,-,+,\dots$} or {$+,+,-,-,+,+,-,-,\dots$} or their negatives.

Furthermore, when there is an extra perspective, adding either {$0_k$} or {$1_k$}, then the number of perspectives is odd. Since we can also choose no perspectives, the number of choices {$n+1$} is even, thus divisible by two. The row in Pascal's triangle is symmetric and the number of choices for {$\frac{n+1}{2}$} and {$\frac{n+1}{2}-1=\frac{n-1}{2}$} have the same value. They have either the same sign in the sequence (doubling their individual contributions with constructive interference) or opposite sign (cancelling out their individual contributions, yielding zero through destructive interference).

Thus suppose we have the sign sequence {$+,-,-,+,+,-,-,+,\dots$} where we start with zero.

• The zeroth perspective is the empty string and has positive sign by the sign sequence.
• The first perspective is either {$1$} or {$0$}. The options {$\binom{1}{0}$} and {$\binom{1}{1}$} have opposite sign because {$0$} has positive sign and {$1$} has negative sign. So they cancel away and the total is zero.
• The second perspective has a negative sign. We have {$01$} or {$10$} and they have the same negative sign. Whereas {$00$} and {$11$} cancel away.
• The third perspective yields the options {$\binom{3}{1}$} and {$\binom{3}{2}$}, both with positive sign given by {$1$} and {$2$} in the sequence.
• The fourth perspective has a positive sign.
• The fifth perspective yields {$\binom{5}{2}$} and {$\binom{5}{3}$}, with positive sign given by {$2$} and negative sign by {$3$}, thus cancelling away.
• The sixth perspective has a negative sign.
• The seventh perspective yields {$\binom{7}{3}$} and {$\binom{7}{4}$}, both with negative sign given by {$3$} and {$4$}.
• The eighth perspective has a positive sign.

Thus eightfold Bott periodicity actually arises from the fourfold periodicity of the odd perspectives. The fourfold periodicity comes from considering the two approximate halves of {$k$}, namely {$\frac{k-1}{2}$} and {$\frac{k+1}{2}$}, which are adjacent integers, and considering whether they have the same sign in the sequence (interfering constructively, yielding positive or negative) or opposite signs (cancelling destructively). Which is to say, the fourfold periodicity comes from the ambiguity of whether the extra perspective is attributed to concepts or to contexts, and whether that changes the sign or not, and whether the sign is positive or negative. So it is a fourfold periodicity in the analysis of that ambiguity.

Minimal pattern

The basic pattern is given by {$\binom{3}{0},\binom{2}{1},\binom{1}{2},\binom{0}{3}$} which is {$1,3,3,1$} as {$+,-,-,+$}. So it is rooted in the three minds but also the foursome. But this pattern regards the axes, which is to say, pairs of perspectives.

Clifford algebra recursions

The logic seems to be as follows.

A concept expresses compatiblity (of syntax and semantics). It squares to {$+1$}. A context expresses incompatibility (of syntax and semantics). It squares to {$-1$}.

A concept should be paired with a context. They are compatible. A concept should not be paired with a concept. They are incompatible. Likewise, a context should not be paired with a context.

Two minimal algebras arise.

If we have two contexts, they square to {$-1$}, and then they are incompatible and thus their product squares to {$-1$}, also a context. This gives a three-cycle of contexts, thus the quaternions.

If we have two concepts {$A^2=+1, B^2=+1$}, then they are incompatible and their product {$AB$} yields a context {$(AB)^2-1$}. But this context is compatible with the concepts, and its product with either squares to {$+1$} and is a concept, namely, the other concept. This algebra is the matrices {$M_2(\mathbb{R})$}. The matrices can be written:

{$A= \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}, B= \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}, AB = \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix} $}

Compare and interpret {$(1\pm 1)^n$} and {$(1\pm i)^n$}.

{$1\pm i$} expresses the twosome, the division of everything into two perspectives: opposites coexist {$\pm i$} and all is the same {$1$}.

Bott periodicity expresses the combinatorics of the eighth root of unity {$e^{\pm\frac{\pi}{4}i}=\frac{\sqrt{2}}{2}(1\pm i)$}.

{$e^{\pm n\frac{\pi}{4}i}=\frac{\sqrt{2}}{2}(1 \pm i)^n$}

{$$D_n=\underset{k\equiv 0,3(mod 4)}{\sum_{k}}\binom{n}{k}-\underset{k\equiv 1,2(mod 4)}{\sum_{k}}\binom{n}{k} = \sum_{k=0}^n\binom{n}{k}(-1)^{\frac{k(k+1)}{2}}=\sqrt{2}^{n+1} \cos ((n+1)\frac{\pi}{4})$$}

{$$D_n= \sqrt{2}^n( \cos \frac{\pi}{4}n - \sin \frac{\pi}{4}n)$$}

Explain the cos formula.

{$D_n$} sums the {$n$}-th row of Pascal's triangle weighting the entries with signs {$+,-,-,+,+,-,-,+\cdots$}.

The first values are

We have {$D_{n+4}=-4D_n$} and {$D_{n+8}=16D_n$}.

We also have that {$|D_{2n}|=\sqrt {2^n}$}. The combinatorics swaps sequences with pairs {$00$} and {$11$} leaving {$01$} or {$10$}. This suggests that a concept must come with a context and vice versa. Two concepts would get swapped with two contexts.

I need to explain the sign of {$D_{2n}$}.

Additionally, I need to show that {$D_{4n+1}=0$} and that {$D_{4n+3}=D_{4n+4}$}.