Loop equations and lattice bootstrap

The bootstrap starts by asking which collections of expectation values could belong to a consistent theory. Positivity, symmetry and exact identities can rule out large regions of possibilities before any direct sampling is performed.

Prerequisites: the finite lattice measure and positivity inequalities.

After this page: build a small positive matrix, explain an outer bound, and distinguish exact algebra from a numerical solver’s output.

Scope: the worked examples are elementary moment problems and linear elimination. They illustrate the logic, not numerical bounds on Yang-Mills observables.

Why positivity creates a matrix

For observables $O_i$ and complex numbers $c_i$, probability positivity gives

$$\left\langle\left|\sum_i c_iO_i\right|^2\right\rangle =\sum_{i,j}\overline{c_i}M_{ij}c_j\geq0, \qquad M_{ij}=\langle O_i^\ast O_j\rangle.$$

Thus $M$ is positive semidefinite. On a lattice, one can form gauge-invariant entries from pairs of Wilson paths sharing endpoints. Guo, Li, Yang and Zhu give this construction in arXiv:2502.14421v2, section 3.1, equations (11)–(15). Their section 3.2 treats reflection positivity separately. The two forms must not be conflated.

Worked example: a feasible interval

Let a real random variable $X$ satisfy $|X|\leq1$. Set $u=\langle X\rangle$ and $v=\langle X^2\rangle$. For the pair $(1,X)$,

$$M=\begin{pmatrix}1&u\\u&v\end{pmatrix}\succeq0.$$

Its determinant gives $v-u^2\geq0$. The support condition also gives $v\leq1$. If another exact identity fixes $v=1/4$, then $-1/2\leq u\leq1/2$.

This is a rigorous deduction from the stated assumptions. It is not a determination of $u$. Many distributions meet those constraints. Adding further moments and identities can reduce the feasible set, but a finite list need not identify a unique distribution.

The same logic explains an outer relaxation: every true theory supplies feasible data, while some feasible data may fail omitted conditions. Maximizing a quantity over that larger set gives an upper bound on its value in the true theory, provided the inclusion and optimization bound are both sound.

Why eliminating larger objects matters

Suppose retained variables $p,q$ obey exact relations with an auxiliary variable $r$:

$$2p+r=1,\qquad q-3r=2.$$

Eliminating $r$ gives $q+6p=5$. Throwing away every equation that contains $r$ would lose that constraint. This example is original teaching algebra, not a loop equation.

Liu and Yang discuss the corresponding issue for truncated loop systems in arXiv:2601.04316v2, section 5.1, equation (29) and the following elimination discussion. The loop-equation dossier explains the source’s exact setting. Keeping a richer collection of intermediate loops can expose relations among shorter loops without treating those longer loops as known measurements.

Three levels of confidence

Layer	What supports it	What is still needed
Exact constraint	Algebra, measure positivity, or a stated theorem	Check that the model and conventions satisfy its hypotheses
Computed feasible or dual solution	Solver output at a stated precision	Residuals, conditioning and the direction of the claimed bound
Certified numerical bound	A validated certificate with controlled rounding	The omitted hierarchy and physical limiting arguments

A nearly positive floating-point matrix need not be mathematically positive. A solver reporting “optimal” does not certify all equations or a continuum extrapolation. Conversely, useful numerical evidence need not be dismissed because it is not a proof; label its actual basis.

What remains toward the target?

A finite hierarchy can bound local Wilson observables without determining long-distance decay. Reaching the mass gap requires the appropriate spectral information and control as the lattice regulator and volume restrictions are removed. The positivity-bootstrap dossier preserves its dimensional and group scope; this guide makes no claim that a finite bootstrap solves four-dimensional Yang-Mills.

Exercise: a false certificate

A solver returns $u=0.5001$, $v=0.25$ for the matrix above. Is it feasible in exact arithmetic? Could the discrepancy be harmless if the program’s tolerance is $10^{-3}$?

Solution: $v-u^2=0.25-0.25010001=-0.00010001$, so the exact matrix is not positive semidefinite. A tolerance can explain why the solver accepted it but cannot turn it into an exact feasible point. A validated result must account for this error or construct a certificate with the appropriate margin.

Try the positivity lab, then compare the different kinds of estimates in rigorous lattice methods.