2026-10-13
Random Choice in Idempotent Splitting
Probability Distributions and the Square-Root Law
Imagine that humanity has set out to search for extraterrestrial life. Upon reaching another planet, the explorers randomly choose the direction of their next flight. For simplicity, assume that all flights have the same length, all directions are equally likely, and each choice is independent of the previous ones. At every stage, the set of possible travel histories splits into new branches. The expedition realizes one of them, while the full tree contains every possible continuation of the search.
A civilization living a thousand such flights away from Earth is trying to estimate the scale of the search. If humanity traveled directly toward it, a thousand steps would suffice. However, random directions partially cancel one another: after a million flights, the expedition's root-mean-square distance from Earth will be only a thousand flight lengths. Knowing the duration of a single step, the aliens can estimate the corresponding timescale, although this calculation does not determine the exact date of an encounter.
Why does the set of possible histories grow rapidly, while the characteristic distance grows only as the square root of the number of steps? Below, we construct a tree of idempotent channels, specify a rule for random choice, and derive the distribution of the accumulated outcome. We first consider two possible outcomes at each step, then move on to spatial directions. Counting channels yields the binomial distribution, averaging yields the square-root law, and a large number of splittings yields the normal limit.
This is a mathematical study. The idempotent algebra specifies the structure of alternatives, while the probability measure and accumulation rule are introduced separately. The resulting distributions are familiar from probability theory; here they are derived using a representation in terms of splitting channels.
1. Binary Splitting and the Channel Tree
The algebraic origin of two complementary idempotents is discussed in “Origins of the i-Basis: From the Hyperbolic Unit to Idempotent Geometry.” Here we repeat the necessary construction and extend it to a sequence of splittings. Consider the free commutative real algebra generated by the units \(\j_1,\ldots,\j_n\), subject to the relations
\[\tag{1} \j_k^2=1,\qquad \j_k\j_l=\j_l\j_k. \] The free construction means that no additional relations between distinct generators are imposed. In particular, distinct \(\j_k\) are not identified with one another. Each step corresponds to two projectors, \(\pmp_k\) and \(\pme_k\):
\[\tag{2} \pmp_k=\frac{1+\j_k}{2},\qquad \pme_k=\frac{1-\j_k}{2}. \] \[\tag{3} \pmp_k^2=\pmp_k,\qquad \pme_k^2=\pme_k,\qquad \pmp_k\pme_k=0,\qquad \pmp_k+\pme_k=1. \] Any channel \(P\) constructed at earlier steps splits into two subchannels:
\[\tag{4} P=P\pmp_k+P\pme_k, \qquad (P\pmp_k)(P\pme_k)=0. \] Here, “orthogonality” means that the product of the projectors is zero. It does not yet imply perpendicular directions in physical space. Splitting does not create two copies of the whole: together, the subchannels constitute the original channel.
Denote a history of \(n\) choices by \(\boldsymbol\varepsilon=(\varepsilon_1,\ldots,\varepsilon_n)\), where \(\varepsilon_k=\pm1\). The corresponding terminal channel is
\[\tag{5} P_{\boldsymbol\varepsilon} =\prod_{k=1}^{n}B_k(\varepsilon_k), \qquad B_k(+1):=\pmp_k,\quad B_k(-1):=\pme_k. \] The notation \(B_k(\varepsilon_k)\) selects one of the two idempotents: \(\pmp_k\) when \(\varepsilon_k=+1\), and \(\pme_k\) when \(\varepsilon_k=-1\). If two histories differ in at least one sign, their product contains the factor \(\pmp_k\pme_k=0\). Therefore,
\[\tag{6} P_{\boldsymbol\varepsilon}P_{\boldsymbol\eta} =\delta_{\boldsymbol\varepsilon,\boldsymbol\eta}P_{\boldsymbol\varepsilon}, \qquad \sum_{\boldsymbol\varepsilon\in\{-1,+1\}^n} P_{\boldsymbol\varepsilon} =\prod_{k=1}^{n}(\pmp_k+\pme_k)=1. \] After \(n\) steps, there are \(2^n\) nonzero terminal channels. Each retains the complete history, so two channels can have the same outcome while remaining distinct histories.
2. Probability Measure and Random Choice
Introduce a symmetric rule: at each splitting, the probability of the parent channel is divided equally between its two subchannels. For any channel at the preceding level, assume
\[\tag{7} \mu(1)=1,\qquad \mu(P\pmp_k)=\mu(P\pme_k)=\frac{\mu(P)}{2}. \] Applying this rule successively gives
\[\tag{8} \mu(P_{\boldsymbol\varepsilon})=2^{-n}, \qquad \sum_{\boldsymbol\varepsilon}\mu(P_{\boldsymbol\varepsilon})=1. \] Thus, a random realization selects one history, and the measure specifies its probability. Since the conditional probability of each subsequent sign is \(1/2\) for every preceding history, the choices are independent. This is a property of the probability measure, rather than a consequence of the zero product of the projectors alone.
Every real element of the algebra has a unique representation \(F=\sum_{\boldsymbol\varepsilon}f(\boldsymbol\varepsilon)P_{\boldsymbol\varepsilon}\). Extend the measure linearly and define the mean and norm:
\[\tag{9} \mu(F)=2^{-n}\sum_{\boldsymbol\varepsilon}f(\boldsymbol\varepsilon), \qquad \lVert F\rVert_\mu^2:=\mu(F^2) =2^{-n}\sum_{\boldsymbol\varepsilon}f(\boldsymbol\varepsilon)^2. \] The last equality follows from channel orthogonality. All weights are positive, so this defines a positive-definite quadratic norm on the real algebra.
3. Accumulated Outcome and Grouping Histories
Assign the value \(+1\) or \(-1\) to the choice of each branch. Define the accumulated outcome of a history as
\[\tag{10} S_n(\boldsymbol\varepsilon)=\sum_{k=1}^{n}\varepsilon_k. \] This is a separate rule mapping channels to numbers. It requires no spatial interpretation. To express it algebraically, introduce
\[\tag{11} A_n=\sum_{k=1}^{n}\j_k. \] From \(\j_k\pmp_k=\pmp_k\) and \(\j_k\pme_k=-\pme_k\), it follows that
\[\tag{12} A_nP_{\boldsymbol\varepsilon} =S_n(\boldsymbol\varepsilon)P_{\boldsymbol\varepsilon}, \qquad A_n=\sum_{\boldsymbol\varepsilon} S_n(\boldsymbol\varepsilon)P_{\boldsymbol\varepsilon}. \] If a history contains \(m\) plus signs and \(n-m\) minus signs, then \(S_n=2m-n\). Group all such histories:
\[\tag{13} Q_m=\sum_{\boldsymbol\varepsilon:\,N_+(\boldsymbol\varepsilon)=m} P_{\boldsymbol\varepsilon}, \qquad m=0,\ldots,n. \] \[\tag{14} Q_mQ_r=\delta_{mr}Q_m,\qquad \sum_{m=0}^{n}Q_m=1,\qquad A_n=\sum_{m=0}^{n}(2m-n)Q_m. \] The projector \(Q_m\) describes their common outcome, while its components \(P_{\boldsymbol\varepsilon}\) describe different ways of obtaining that outcome. This grouping preserves completeness and orthogonality.
4. The Binomial Distribution from Counting Channels
To obtain a history with \(m\) plus signs, choose \(m\) positions out of \(n\). The number of corresponding channels is
\[\tag{15} N_m=\binom{n}{m}=\frac{n!}{m!(n-m)!}. \] Each channel has weight \(2^{-n}\), so the measure of their union gives the exact distribution:
\[\tag{16} \boxed{\Pr(S_n=2m-n)=\mu(Q_m)=2^{-n}\binom{n}{m}.} \] Normalization follows from the binomial identity:
\[\tag{17} \sum_{m=0}^{n}\mu(Q_m) =2^{-n}\sum_{m=0}^{n}\binom{n}{m} =2^{-n}(1+1)^n=1. \] For four splittings, the tree contains sixteen histories:
| Number of Plus Signs | Outcome | Channel Count | Probability |
|---|---|---|---|
| 0 | −4 | 1 | 1/16 |
| 1 | −2 | 4 | 4/16 |
| 2 | 0 | 6 | 6/16 |
| 3 | 2 | 4 | 4/16 |
| 4 | 4 | 1 | 1/16 |
All histories are equally likely, but six channels correspond to the zero outcome, whereas only one corresponds to each extreme outcome. The higher probability of central outcomes arises from the number of histories, rather than a preference for particular branches.
5. Statistical Orthogonality and the Square-Root Law
When averaged over all histories, each generator's signs occur equally often. For two distinct generators, all four sign combinations are also equally likely. Hence,
\[\tag{18} \mu(\j_k)=0,\qquad \mu(\j_k^2)=1,\qquad \mu(\j_k\j_l)=0\quad(k\ne l). \] The generators are therefore orthogonal under the inner product \(\langle F,G\rangle_\mu:=\mu(FG)\). The algebraic square of their sum itself contains cross terms:
\[\tag{19} A_n^2=n\,1+2\sum_{1\le k\lt l\le n}\j_k\j_l. \] Their mean is zero, so
\[\tag{20} \mu(A_n)=0,\qquad \mu(A_n^2)=n. \] \[\tag{21} \boxed{ \sigma_n =\sqrt{\mathbb{E}[S_n^2]-\bigl(\mathbb{E}[S_n]\bigr)^2} =\lVert A_n\rVert_{\mu} =\sqrt{n} }. \] If the outcome of a single choice has scale \(\ell\), then for \(X_n=\ell S_n\) we obtain \(\sigma_{X_n}=\ell\sqrt n\). An individual history need not end at this distance: the formula characterizes the distribution of the entire ensemble.
The statement “the square equals the number of steps” refers to the statistical norm of the sum of generators. It does not mean \(A_n^2=n\,1\), nor does it change the idempotent property \(P^2=P\). The measure remains normalized for any number of splittings; the growth in deviation is due to the accumulated observable.
6. The Normal Limit from the Number of Channels
For large \(n\), the central part of the binomial distribution acquires a Gaussian shape. Let us derive it from formula (16). For an admissible value \(s\) with the same parity as \(n\), set \(m=(n+s)/2\):
\[\tag{22} \Pr(S_n=s) =2^{-n}\frac{n!} {\left(\frac{n+s}{2}\right)!\left(\frac{n-s}{2}\right)!}. \] Use Stirling's formula:
\[\tag{23} r!=\sqrt{2\pi r}\left(\frac{r}{e}\right)^r \left(1+O(r^{-1})\right). \] In the central region \(s=O(\sqrt n)\), introduce \(u=s/n\). Substituting (23) into (22) gives
\[\tag{24} \Pr(S_n=s) =\sqrt{\frac{2}{\pi n(1-u^2)}}\, \exp\!\left[-\frac n2 h(u)\right] \left(1+O(n^{-1})\right), \] \[\tag{25} h(u)=(1+u)\ln(1+u)+(1-u)\ln(1-u) =u^2+\frac{u^4}{6}+O(u^6). \] For \(s=O(\sqrt n)\), we have \(u=O(n^{-1/2})\). Thus, the leading term in the exponent is \(-s^2/(2n)\), while the remaining corrections tend to zero:
\[\tag{26} \Pr(S_n=s)\sim \sqrt{\frac{2}{\pi n}}\, \exp\!\left(-\frac{s^2}{2n}\right). \] The left-hand side here is the probability of a single discrete outcome. To obtain a density, normalize the deviation:
\[\tag{27} Z_n=\frac{S_n}{\sqrt n},\qquad \Delta z=\frac{2}{\sqrt n}. \] The factor of 2 is necessary because adjacent admissible sums differ by two. Dividing the asymptotic probability by the interval width gives
\[\tag{28} \boxed{\rho(z)=\frac{1}{\sqrt{2\pi}} \exp\!\left(-\frac{z^2}{2}\right).} \] For any finite interval \([a,b]\), the sum of probabilities becomes the integral of this density. For completeness, verify that no probability is lost at large deviations. From (20) and Chebyshev's inequality, \(\Pr(|Z_n|>M)\le M^{-2}\) uniformly in \(n\). Therefore, the normalized outcome converges in distribution to a standard normal random variable:
\[\tag{29} \lim_{n\to\infty}\Pr(a\le Z_n\le b) =\int_a^b\rho(z)\,dz,\qquad Z_\rightarrow{\d\ \athcal N(0,1). \] Thus, the Gaussian limit is obtained from the number of terminal channels. The quadratic term in expansion (25) determines the exponential shape and the scale \(\sqrt n\). For a finite number of splittings, the exact distribution remains binomial; the normal shape is a limiting approximation to its central region.
7. Splitting Directions and Spatial Steps
Let us return to the flights. The binary construction extends to a finite number of directions: a parent channel is divided into orthogonal subchannels, each assigned a step vector. For directions \(\mathbf u_a\) and weights \(w_a\), assume
\[\tag{30} P=\sum_a P_a,\qquad P_aP_b=\delta_{ab}P_a,\qquad \mu(P_a)=w_a\mu(P),\qquad \sum_a w_a=1. \] Let all \(\mathbf u_a\) be unit vectors, and let the set of directions be symmetric:
\[\tag{31} \sum_a w_a\mathbf u_a=0,\qquad \lVert \mathbf u_a\rVert=1. \] At step \(k\), the vector \(\Delta\mathbf R_k=\ell\mathbf u_a\) is selected. If the choice rule does not depend on the preceding history, the steps are independent and
\[\tag{32} \mathbb E[\Delta\mathbf R_k]=0,\qquad \mathbb E[\lVert \Delta\mathbf R_k\rVert^2]=\ell^2,\qquad \mathbb E[\Delta\mathbf R_k\cdot\Delta\mathbf R_l]=0\quad(k\ne l). \] For the accumulated displacement \(\mathbf R_n=\sum_k\Delta\mathbf R_k\), we obtain
\[\tag{33} \mathbb E[\lVert \mathbf R_n\rVert^2] =\sum_k\mathbb E[\lVert \Delta\mathbf R_k\rVert^2] +2\sum_{1\le k\lt l\le n}\mathbb E[\Delta\mathbf R_k\cdot\Delta\mathbf R_l] =n\ell^2. \] \[\tag{34} \boxed{R_{\mathrm{rms}}=\sqrt{\mathbb E[\lVert \mathbf R_n\rVert^2]} =\ell\sqrt n.} \] The formula does not depend on the number of spatial dimensions. Uniform choice over the entire sphere can be described directly by a probability measure on directions; this is a continuous generalization of the finite tree. Law (34) requires only independent steps, zero mean, and the same mean squared length.
The distribution of distance in several dimensions differs from the distribution of a single coordinate. For example, for independent isotropic steps in three dimensions, the covariance matrix of the accumulated displacement is \((n\ell^2/3)I_3\). The multivariate central limit theorem yields a Gaussian limit for the coordinates; the corresponding continuous approximation to the vector density is
\[\tag{35} f_n(\mathbf r)\approx \left(\frac{3}{2\pi n\ell^2}\right)^{3/2} \exp\!\left(-\frac{3\lVert \mathbf r\rVert^2}{2n\ell^2}\right). \] The density of the radial distance \(r=\lVert \mathbf R_n\rVert\) includes the volume factor of a spherical shell:
\[\tag{36} g_n(r)\approx4\pi r^2 \left(\frac{3}{2\pi n\ell^2}\right)^{3/2} \exp\!\left(-\frac{3r^2}{2n\ell^2}\right),\qquad r\ge0. \] The last two formulas concern the scaling limit and a continuous approximation for large \(n\). For discrete directions, the distribution at finite step counts remains discrete. The detailed three-dimensional limit requires a multivariate probability result; the one-dimensional normal limit above has already been derived directly by counting binary channels.
8. Unequal Weights and Dependent Choices
If the positive branch receives probability \(p\) and the negative branch receives \(1-p\), the choices remain independent provided these conditional weights remain unchanged. A history with \(m\) plus signs has weight \(p^m(1-p)^{n-m}\), so
\[\tag{37} \Pr(S_n=2m-n)=\binom nm p^m(1-p)^{n-m}. \] For \(0\lt p\lt 1\), the mean and variance are
\[\tag{38} \mathbb E[S_n]=n(2p-1),\qquad \operatorname{Var}(S_n)=4np(1-p),\qquad \sigma_n=2\sqrt{np(1-p)}. \] The square-root law still holds for the spread about the mean, but the mean itself grows linearly when \(p\ne1/2\). Consequently, the root-mean-square deviation from zero no longer coincides with the standard deviation:
\[\tag{39} \mathbb E[S_n^2]=4np(1-p)+n^2(2p-1)^2. \] If the probabilities of subsequent branches depend on earlier choices, covariance terms appear in the variance:
\[\tag{40} \operatorname{Var}(S_n) =\sum_{k=1}^{n}\operatorname{Var}(\varepsilon_k) +2\sum_{1\le k\lt l\le n}\operatorname{Cov}(\varepsilon_k,\varepsilon_l). \] For example, if the first sign is chosen randomly only once and retained at all subsequent steps, then \(S_n=\pm n\), so \(\sigma_n=n\). This behavior is specified by a measure concentrated on two histories of the tree. The idempotent channels remain algebraically orthogonal, but the corresponding choices become statistically dependent. Splitting alone does not guarantee the square-root law.
9. Search Scale and Conclusions
For a distance \(R\), equate the root-mean-square distance to this scale. If each step takes time \(\Delta t\), we obtain
\[\tag{41} n_{\mathrm{scale}}=\left(\frac{R}{\ell}\right)^2,\qquad t_{\mathrm{scale}}=\Delta t\left(\frac{R}{\ell}\right)^2. \] For \(R=1000\ell\), this is a million steps. With one flight per year, the corresponding scale is a million years; with one flight per day, it is a million days, or about 2,740 years. The idealized model considers a single expedition with a constant step length and duration. Branching into multiple actual expeditions, purposeful navigation, and a nonuniform distribution of planets require different accumulation and choice rules.
The quantity \(t_{\mathrm{scale}}\) is not the expected first arrival time at a particular planet. It indicates when the root-mean-square distance reaches \(R\). Estimating an encounter requires the geometry of available routes, the size of the target region, and current information about the journey. The remaining time cannot be determined simply by subtracting the steps already taken from a million.
The derivation follows a sequence: idempotent splitting creates a tree of alternatives, the measure assigns weights to terminal channels, and the accumulated observable groups histories with the same outcome. Counting these histories gives the binomial distribution. Averaging the square gives the \(\sqrt n\) law, while the asymptotic channel count yields the normal limit.
Thus, a probability distribution can be constructed within the algebra of splittings. Its form is determined jointly by the channel structure, the chosen measure, and the rule mapping a history to an outcome. In the symmetric independent case, this construction exactly reproduces the classical random walk.
This construction makes random choice within splitting channels a useful mathematical tool for developing physical hypotheses. It connects the structure of possible states with the distribution of observed outcomes and yields testable relationships: the growth of root-mean-square deviation, the shape of the distribution, and changes in statistics caused by correlations between choices. Physical applications require specifying what the channels represent, what determines their probability weights, and how the sequence of realizations maps to a measurable quantity. With these additions, random splitting can provide a basis for models of diffusion, fluctuations, and other processes involving the accumulation of random changes. The mathematical correspondence alone does not establish the existence of a physical splitting mechanism, but it provides a rigorous basis for formulating and experimentally testing one.