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End of text for video More notes... I have been trying to argue that the only solutions to Schroedinger's equation (and related quantum field theory equations) which are physical are those which are in terms of orthogonal Sheffer polynomials. I now share my idea of what that means physically. Sheffer polynomials are those that build space incrementally Combinatorially, Sheffer sequences are those which have the generating function {$\sum_{n=0}^{\infty} P_n(x)s^n = C(s) e^{x u(s)}$}. This means that the terms in the polynomial {$P_n(x)$} are assembled by pairing a component in {$C(s)$} with one or more components from {$x u(s)$}. On Monday I told John that we have the equation {$n! P_n(x)$} = the {$n$}th derivative with respect to {$s$} of {$C(s) e^{x u(s)}$} evaluated at {$0$}. So I became interested to calculate that {$n$}th derivative with respect to {$s$}. The number of terms in the derivatives grew as {$1, 2, 5, 15 \cdots$} So I looked it up at the online encyclopedia of integer sequences and I got the Bell numbers. The Bell numbers count the number of ways of partitioning a set of size n+1. Namely: for {$P_0(x)$}: {$ [0]$} for {$P_1(x)$}: {$ [01]$} & {$ [0][1]$} for {$P_2(x)$}: {$ [012]$} & {$ [01][2]$} & {$ [02][1]$} & {$ [0][12]$} & {$ [0][1][2]$} The sequence grows as {$1, 2, 5, 15, 52 \dots$} It also counts the number of ways of setting up equivalence classes on n elements. I realized that I can prove by induction straightforwardly that Sheffer sequences are, combinatorially, those that build up space in terms of partitions. Differentation of {$C(s) e^{x u(s)}$} generates two kinds of components. On the one hand, we can differentiate {$C = C(s)$} to get {$C'$} and then {$C''$} and {$C'''$} and so on. On the other hand, the derivative of {$E = e^{x u(s)}= e^W$} is {$W' E$}. And then we simply use the product rule, and if we have a product of many factors, then we simply differentiate one of those factors and sum up the possibilities. What this means, spatially, is that we start of with a space {$C E$}. I won't write the {$E$} because it is omnipresent. {$C = [0]$} is the initial (empty) space. Then we can add a new cell 1 in two ways. We can add it in a new compartment: {$C W' = [0][1]$} Or we can add it to an existing compartment by lengthening that compartment, which means by differentiating that compartment. {$C' = [01]$}. Consider now adding another new cell 2. We can add it as a new compartment: {$C W' W' = [0][1][2]$} & {$C' W' = [01][2]$}. Or we can add it to an existing comparment by differentiating: {$C' W' = [02][1]$} & {$C W'' = [0][12]$} & {$C'' = [012]$}. The way to prove the claim is simply by induction to suppose that the terms in the nth derivative match the partitions of {$n$}, and then to show that there is a bijection between the growth of the partitions and the effects of differentiation. Adding the {$n+1$}st cell as a new compartment {$[n]$} matches the differentiation of E. Inserting the n+1st cell into an existing compartment matches the differentiation of some factor in the existing derivative. So the two ways of thinking, in terms of partitions and in terms of differentiation, match up. So it is easy to get all of the terms of the nth derivative with regard to {$s$}. Simply write out all of the partitions of {$\{0,1,..., n1\}$}. Then a part which contains {$0$} and has size {$k$} will be given weight {$C^(k)$}, which is to say, the {$k$}th derivative of {$C$}, with regard to {$s$}. And any part which does not contain {$0$} and has size {$k$} will be given weight {$W^(k+1)$}, the {$k$}th derivative of {$W$}, with regard to {$s$}. Multiply together the weights for the parts in a partition and then add up the resulting weights for the partitions. Furthermore, each part (which does not contain {$0$}) gets a weight {$x$}. So the term {$x^d$} in {$P_n(x)$} gives the partitions with {$d+1$} parts. Thus the power of {$x$} is the measure of compartments. As the number {$n$} of cells grows, as given by the degree of {$s$}, and the index of {$P_n(x)$}, then so does the possible number of compartments, as given by the degree of {$x$}. I have to remember to divide by {$n!$}. Investigation: What does it mean to differentiate with regard to {$s$} ? So I need to think why differentiating a compartment is the same as lengthening it. Or why integrating has to do with shortening. I can also think of what differentation (with regard to {$s$}) means for a space {$s^n$}. Compare with multiplying out We can also simply multiply out the terms of the generating functions {$$\sum_{n=0}^{\infty}P_n(x)t^n=(1+a_1t+a_2t^2+\cdots)(1+xh_1t+\frac{(xh_1t)^2}{2!}+\cdots)(1+xh_2t^2+\frac{(xh_2t^2)^2}{2!}+\cdots)(1+xh_3t^3+\frac{(xh_3t^3)^2}{2!}+...)\cdots$$} We likewise get terms that correspond to configurations of parts. Here we ignore the initial element [0]. Each term contains a factor of the form {$a_i$} where {$i+1$} is the length of the compartment that contains [0]. (Note that {$a_0=1$}, in which case this compartment simply consists of [0].) The number of additional compartments is given by the degree of {$x$}. And then each term is a combination of compartments, with the additional compartments having weight {$h_j$} with length {$j$}. This gives all of the terms generated. But to match them up with the space builders above, I need to figure out how to account for the frequencies. This is where the {$n!$} is important because it arises on the left hand side in differentiating {$n$} times. For example, for {$n=1$} we have two configurations: {$a_1 \leftrightarrow [01], h_1x\leftrightarrow [0][1]$}. And for {$n=2$} we have four configurations: {$a_2 \leftrightarrow [012], a_1h_1x \leftrightarrow [01][2], [02][1], h_2x [0][12], h_1^2 [0][1][2]$}. So here we need to have that {$a_1 h_1 = 2C'W'$}. The roles of the Lie group and Lie algebra From the combinatorics we see that the Lie algebra {$xh(t))$} contributes the additional compartments whereas {$A(t)$}, which expresses the real form (the compactness or noncompactness), gives the original compartment, thus the nature of the spacetime. Is there a sense in which they constitute a pair of conjugate variables? The compartmentalization is given by {$x$} and so that expresses the amount of energy but also any observable for which the powers in {$x$} are meaningful. And the growth of the combinatorial objects suggests that it should be expressible as representations of a group. Idea: Orthogonality of Sheffer polynomials makes space threedimensional Furthermore, if the Sheffer polynomials are orthogonal, then their recursion relation is even more limited. The generating functions for {$C(s)$} and {$u(s)$} depend only on the first few terms, and all the remaining terms can be thought of as "echo terms", depending only on the first few. Indeed, for x u(s) only the first three terms matter. The {$x$} terms only arise from differentiating {$e^{x u(s)}$}. Each of the first three terms is qualitatively different, the first term is constant, the second term is {$α + β$}, the third term is {$α^2+ αβ + β^2$}, and the "echo terms" are similar but with higher powers, {$α^n +... + β^n$}. So I am intrigued that this might yield an explanation why space is threedimensional. Time is compatibility with space Stone's theorem says that a wave function {$f(x,t)$} evolves as {$\sum_{n=0}^{\infty} a_n P_n(x) e^{i λ_n t} \sqrt{w(x)}$}. Substituting in for {$P_n(x)$} as above this means that {$f(x,t) = \sum_{n=0}^{\infty} \frac{a_n}{n!} (\textrm{d}/\textrm{ds})^n (C(s) e^{u(s)x  i λ_n t}) _{s=0} \sqrt w(x)$} I think this means that the time evolution is simply compatible with all that I have written above. The key term is {$e^{u(s) x  i λ_n t}$ and differentiating it with regards to s brings down {$u'(s) x$} but does not bring down {$i λ_n t$}. I think this means that the time dimension is what is compatible with space. Note that what goes down is {$u'(s) x$} and the first three terms matter. So time is like the constant term in {$u(s)$} whose derivative with regard to {$s$} is simply zero. There is a question of how to think about {$s$}. The powers of {$s$} are what link together the spatial dimensions. But only the first three dimensions are independent. The five kinds of Sheffer polynomials are five interpretations of space building The possible values for {$α$} and {$β$} are limited. The general case is {$α, β$} and the four special cases are {$α$}, {$0$} and {$α$}, {$α$} and {$0$}, {$0$} and {$α$}, {$\bar{α}$}. These further interpret and specialize the meaning of the space building. Investigation: Relate to the moments of the orthogonal Sheffer polynomials The moments are the numbers {$μ_k = \int x^k \textrm{dw}$} where {$\int \textrm{dw }$} is the norm with regard to which the polynomials are orthogonal. Sheffer polynomials are space builders, as dictated by the Bell numbers, and furthermore, if they are orthogonal, then their coefficients, considered as a vector, are orthogonal to their moments, considered as a vector. {$\int P_n(x) \textrm{dw} = \int \sum p_{n,k} x^k \textrm{dw} = \sum p_{n,k} \int x^k \textrm{dw} = \sum_{n=0}^{\infty} p_{n,k} \mu_k$} And note that {$P_0(x)$} is constant and orthogonal to {$P_n(x)$} for all {$n\neq 0$}, which means that {$\int P_0(x) P_n(x) \textrm{dw} = \int P_n(x) \textrm{dw} = 0$} and consequently {$0=\sum_{n=0}^{\infty} p_{n,k} \mu_k$} for all {$n\neq 0$}. There are precisely five sequences of moments that are orthogonal in this way. And each of these has a combinatorial meaning. The numbers {$μ_k$}, in the general case (the Meixner polynomials) {$α$}, {$β$}, are the ordered Bell numbers, which count the ordered set partitions of {$n$}. In the special case {$α, 0$}, which is the Charlier polynomials, they are the Bell numbers. In the special case {$α, α$}, the Laguerre polynomials, they are the number of permutations {$n!$} In the special case {$0, 0$}, the Hermite polynomials, they are the number of involutions without fixed points, which is to say, permutations where each cycle is disjoint and has length {$2$}. (As with the links in Wick's theorem.) In the special case {$α, \bar{α}$}, the MeixnerPollaczek polynomials, they are zigzag permutations of {$2n$}, which resemble "kinks". So I need to understand how these moments relate to the Bell numbers as derived above and how we get specializations of space building. Investigation: Consider Bell polynomials with coefficients T(n,m) OEIS A060400 Count of set partitions of n with block lengths given by the mth partition of n. Note that the conditions {$j_1+j_2+\cdots+j_{nk+1}=k$} {$j_1+2j_2+\cdots+(nk+1)j_{nk+1}$} are precisely those which Galiffa and Riston use to calculate the leading terms. Investigation: Why are these Lorentz invariant? As I learn more I can try again to understand why the orthogonal Sheffer polynomials might be Lorentz invariant. I will be reading Mattuck's book about Feynman diagrams because he is explicit in doing calculations. I think that the invariance should relate to the effect of translating {$x$} to {$xf$} which combinatorially has a nice interpretation in terms of objects going from having double roots to single roots. Math 4 Wisdom Show Would you, could you, should you eat a grape on a plate? find a place in empty space ? chain rule  bicycle chain product rule one grape two grape one plate two plate use your table if you're able first derivative is applicative big space small grape give your grape to an ape with a cape flying grapes and flying plates in backdrop of space Grapes on plates. Grapes look good in empty space. Sound good. Taste god. Origin stories  Hercules and the hydra but worse  the hydrangea, a monster  