2026-10-09 · chapter Yang–Mills and the mass gapPart 10 Towards a proof of the mass gap — building the first step of the staircasedetour plans R1–R24, the line that killed it, types of breakage · Kiichi, Shiori and Rin

The list of detour plans — types of breakage

The one remaining proposition of route 5 was that the logarithmic Sobolev constant $A_W$ of the site window lies within the needed value $A_{\max}$ (10-09 §03). A detour plan moves one side of this inequality, or the form of the inequality itself.

• The results of the twenty-four. The twenty-two listed first split into 13 dead, 2 on hold, 3 conditional and 4 alive. Of the 4 alive, 2 (R1 and R2) enter every later form as parts, and the other 2 (R8 and R13), together with R24, which was born along the way, became the three strands of 10-09.
• The side of tools and the side of structure. What moved the needed value a lot were the plans that corrected the choice of tools (R1 and R2: $0.20\to90.3$ at $10^{-3}$, about 460 times), while the plans that moved factors coming from the structure (the depth of defects $b$, the rarity of the defect set $\ell$) all stopped under a ceiling of 1.15 times.
• Seven types of breakage. One source (the curvature at one site at the top); a ceiling for structural factors that enter as ratios; paying for the same place twice; mixing up currencies; the address of $1/\hbar$; a mismatch between the direction of fluctuation and the soft direction; transfer. Of the 13 dead plans, 12 fall under one of these.

Copied from 10-09. The sector framework, which controls the gap $\lambda$ of the core $Q_k$ uniformly in the core length $k$, combines the three inequalities of Barta, componentwise localisation and Bochner, and closes under the following condition (10-09 (7)).

Of the six constants, $\kappa=3/1000$, $b=493/1000$ and $\ell_{\min}$ are quantities of the structure, determined by the model and by how the defect set is defined, while $c_H=\tfrac14$ (the negative curvature paid uniformly over the whole core) and $\ln2$ (the Gibbs tilt fixed at $t=\ell$) are quantities of the choice of tools (10-09 Table 1). In the original form the needed value was negative / $0.20$ / $7.13$, while the upper bound on $A_W$ was $1857／2794／9106$ Reading (including estimates) and the lower bound $1.71／1.74／2.59$ Reading (values from a family of test functions).

What a detour plan is. To close (1), one raises the needed value, lowers the upper bound, or changes the form of the condition. In this chapter we split the plans into three sides by that direction.

The other axis is the character of the factor moved. Factors of the structure are quantities determined by the model, and moving them requires changing the definitions of the defect set or the weights. Factors of the tools are choices of steps in the proof, and can be changed with the same sets. The "Structure / tools" column of Table 1 is this axis, and "tools → structural floor" marks plans that changed tools but stopped on hitting a structural floor.

The words of the verdicts in Table 1 mean the following. Alive = in that form it came within the needed value (still not a proof of (1); remaining assumptions exist). Conditional = the remaining assumptions can be named. On hold = not decided on paper. Dead = the killing test put it outside the needed value, or no form came out. The line that killed it shrinks to one line the fact that stopped the plan, and all numbers are readings from floating-point grids or transfer matrices Reading (lower bounds are values within a family of test functions).

The count. R1–R22 are 4 alive (R1 and R2 absorbed as parts; the Poincaré forms of R8 and R13 conditional), 3 conditional (R3, R12, R21), 2 on hold (R11 absorbed into R8, R20) and 13 dead. Of the 13 dead, 8 are on the floor side (R7, R14–R19, R22), 4 on the needed side (R4, R5, R6, R9) and 1 on form (R10). Of the two born along the way, R23 is on its own dominated by R3, R24 entered strand 1 as conditional, and the form combining it with R23 became strand 3.

Contrasting the side of tools with the side of structure. Putting the "Structure / tools" column of Table 1 next to the results column, every row that moved the needed value a lot is "tools". R1 and R2 kept the same defect set and only corrected how the negative curvature is paid and how the Gibbs tilt is chosen, and moved the needed value from negative / $0.20$ / $7.13$ to $25.1／90.3／312.3$ (about 460 times at $10^{-3}$, about 44 times at $3\cdot10^{-4}$, from negative to positive at $3\cdot10^{-3}$) Reading. The "structure" rows (R4, R5, R6, R7) and the rows that hit a structural floor (R9, R10), on the other hand, did not exceed a factor 1.15, or no form came out. Most of the loss on the needed side lay not in the structure of the model but in the choice of tools. Conversely, the gap that remains after the tools have been fully corrected (5–13 times in the form of R3) goes back to the structural floor of type 1 in §04.

Lining up the lines that killed the plans in Table 1 and Table 1′, the same reasons appear again and again in changed forms. We group them into seven types. For each type we give the plans it caught, a paragraph on why plans break in that way, and a question to test your own plan. The types are not mutually exclusive, and one plan can fall under two types.

Type 1 One source — the curvature at one site at the top

Plans caught: R6, R9, R10, and the uniform road and the Lyapunov criterion without sectors in Table 1′.

In one row of the Hessian of the action of the core, at the top the diagonal $4-\tfrac14=\tfrac{15}{4}$ falls $\tfrac14$ short of the sum of the couplings to the neighbours $4$ (10-09 (11)). This shortfall appears with the same value in the Barta weight rows, in the contraction criterion for chains and in the Lyapunov-function criterion, on the top face $x\equiv0$ (every site at the top). For example, the Barta row of a one-site weight $\psi_j=\phi(z_j)$ has the form $F(z,n)=w(z)+(\text{log-derivative of the weight})\times(\text{drift})$, and at the top $z=0$ with neighbours $n=0$ the drift $\varepsilon V'(0)+8(0-0)$ is $0$, so however the weight is chosen Proved here

(a one-line computation on the same assumption row as the Barta rows of 10-09) Leanfloor_top (assumption: weight value $\ne0$). The Agmon type of R10 is a tool that gains where the curvature has a positive margin, and the defect set is exactly where the curvature is negative, so the coefficient of gain there is $\le0$ Leanagmon_no_gain_on_negative_curvature (assumptions: curvature $w\le0$, $\lambda\ge0$ and others; the framework of Agmon estimates is not placed in the book). In R6, defining defects by long lumps made the rows of short lumps of length 2 or more negative, and the same shortfall came out on the side of short lumps.

Question to test: what does your plan pay on the top face $x\equiv0$? If the quantity paid there is a constant independent of $\hbar$, the shortfall of $\tfrac14$ remains as it is. Only the rarity (entropy) of the plateau at the top can pay for this shortfall, and $\hbar$ enters that payment.

Type 2 Structural factors that enter as ratios have a ceiling

Plans caught: R4, R5, R23, and the ceiling of R3.

Raising the threshold, with remainder of localisation $\delta\ge0$, remaining coefficient $\gamma$, Gibbs tilt $t$ and $L(t)=\ln(1+e^{t-\ell})$, the closing condition is $b(A\delta+D_W+L(t))\lt\kappa(t-A\gamma)$ (the form of R1–R3).

Ceiling lemma. If $0\lt\kappa\lt b$ and $\delta,D_W\ge0$, any $A\ge0$ satisfying this condition satisfies $A\gamma\lt\ell$.

Proved here Proof. The left-hand side is at least $bL(t)\gt0$, so $t-A\gamma\gt0$. From $b\gt\kappa$, $b(t-A\gamma)\gt\kappa(t-A\gamma)\gt bL(t)\ge b(t-\ell)$ ($\ln(1+e^x)\ge x$). Dividing by $b$ gives $A\gamma\lt\ell$. ∎ Leanceiling_same_set (assumptions: $0\lt\kappa\lt b$, $t-A\gamma\gt0$, $t-\ell\le L$, the closing condition)

Plans that move only $b$ and $\kappa$ without changing the defect set (hence $\ell$) cannot push the needed value above $\ell/\gamma$ ($4\ell$ in the form of R3). The needed value in the form of R3 is already at $1/1.15／1/1.04／1/1.01$ of the ceiling Reading, and the $1.05／1.01／1.004$ times of R4 lies within it. In R5, which changes the set, the rarity on the Barta side ($\ell_B$ of the set where the rows fall below $\kappa$) enters the condition only multiplied by $\kappa/(\beta_0-\kappa)\approx1/47$ Assumption (the paper argument is not placed in the book) Leanceiling_split, and making the set 1.40 times rarer stopped at a factor of 1.11. R23 changed the tool of the Gibbs step, but $A$ enters in the same form, so the ceiling is the same.

Question to test: does your plan change the defect set? If not, the needed value does not exceed $\ell/\gamma$. If it does, first write down which coefficient of the condition the rarity of the new set is multiplied by when it enters.

Type 3 Paying for the same place twice

Plans caught (on the side that fixed it): R1, R2, and the construction of the window in Table 1′.

The Bochner step of the original construction paid the negative curvature $-\tfrac14$ over the whole core $X$. But negative curvature exists only for $z\gt-1/\sqrt3$, and that region is inside the defect set already paid by the Gibbs step as $Y$. Paying the same place again on the $X$ side made the coefficient of $A_W$ equal to $bc_H=0.123$, through the amplification $b/\kappa\approx164$. Localisation (R1) brings the coefficient down to $0.0038$ (10-09 (9)). Fixing the Gibbs tilt at $t=\ell$ (R2), and the $(1+\sqrt{C_w})^2$ in the construction of the propagation step (counting the same quantity twice), are of the same type. This is a type that has a large effect when fixed.

Question to test: is the quantity paid in that step on a set already paid for by an earlier step? Drawing the places where the two steps pay in the same figure makes it visible.

Type 4 Mixing up currencies — speed of exit and rarity

Plans caught: R14, the form that pays on the band by rarity and the double threshold in Table 1′, and the naive decomposition of the variance in 10-11 §04.

What a Poincaré inequality measures is the speed of exit (the ratio of the variance to the Dirichlet form), and what can pay for rarity $\mu(B)\le e^{-\ell}$ is the currency of entropy (10-09 (3): $\ell\int_Bg^2\le\mathrm{Ent}+\ln2\int g^2$). The ratio of the variance to the Dirichlet form does not change when the function is multiplied by a constant, so putting in a function supported on a rare set makes the rarity cancel between numerator and denominator. "The fluctuations are small because the path sticks to the band" is an intuition about the variance, and what the step that pays for rarity needs is an estimate of entropy. R14 is more direct: if the defect set is determined by the $\sigma$-algebra of the conditioning, the measure of that set under the conditional measure is $0$ or $1$ Proved here (the definition of conditional probability), and no rarity $\ell$ is left to pay with.

Question to test: does the step that pays for the rarity $\mu(B)$ pass through entropy (or an inequality of the same type)? If it uses only the variance and the Dirichlet form, rarity has no effect. Also check whether rarity is still left after conditioning.

Type 5 The address of $1/\hbar$

Plans caught: R7, R15, and in Table 1′ the linear collective coordinate, the sector-split chain, the multi-stage cut at the wall, the chain of wall marks, and the chain with the mean height added.

What strand 3 needs is the coefficient of $1/\hbar$ in the logarithmic Sobolev constant of the core. If one mistakes which quantity that $1/\hbar$ lives in, the plan only moves $1/\hbar$ to another place. Three facts came up again and again.

• $1/\hbar$ lives not in $\sum z$ but in the radius. On the plateau at the top, the action changes not in the direction of the linear collective coordinate $\sum_jz_j/\sqrt k$ but along the radius from the corner at the top (the direction of $\sum_jz_j^2$), and its length is the crossing length $\ell_\times\approx0.34/\hbar$ Reading. In the two-scale split into the collective coordinate and the transverse directions, the smallest curvature in the directions orthogonal to the collective coordinate on the top face, $4-4\cos(2\pi/(k+1))-\tfrac14$, becomes negative at $k=17$ Proved here (the closed formula for tridiagonal eigenvalues) Leantransverse_fails_k17 (no assumptions), so convexity holds only for $k\le16$.
• Cutting into sectors only allocates $1/\hbar$. The chain rule of entropy $\mathrm{Ent}=\mathrm{Ent}(\text{marginal of the marks})+\mathbb E\,\mathrm{Ent}(\text{inside the sector})$ is an identity Proved here Leanchain_rule_cells (assumptions: the chain rule, upper bounds on the two terms), and cutting the plateau at the position of the wall allocates $\approx0.8/\hbar$ between the side of the marks and the side of the sectors Reading. The sum does not decrease.
• A mark represents only the one step that crosses the face. In the chain of wall marks the constant of the chain itself becomes small, but a mark does not distinguish positions on the plateau and represents only the step across the face, so $1/\hbar$ moved into the constant $\ge13／26／66$ that compares the mark chain with the original measure Reading. Whether a variable added to the chain is good or bad is decided by whether its rate function is flat at the extremum (if flat, the cost of adding it is $O(1)$) LeanS_ext_O1_iff (algebra); the mean height is not flat and pays $\propto1/\hbar$, while the radius passed.

R7 (the rarity of the band and the tail through the band are both $\propto1/\hbar$ and cancel) and R15 (the cost of the Holley–Stroock perturbation $e^{\mathrm{osc}/\hbar}$) are also examples that only moved the place where $1/\hbar$ is paid.

Question to test: in your plan, which quantity is $1/\hbar$ attached to? If it has only moved to a sector, a mark, a comparison constant or the cost of a perturbation, it has not disappeared. Draw the rate function of the variable you add near its extremum.

Type 6 Mismatch between the direction of fluctuation and the soft direction

Plans caught: in Table 1′, radius and angles on the sphere, the three steps, the variance version of two-scale, and the interior radius; and the paper upper bound of strand 3.

In a two-scale composition, taking the gradient outside the fibre with $\sup$ (the direction of fluctuation) and paying for it with the constant of the softest direction of the fibre (of the type $\kappa\times A_{\rm fib}$) makes the upper bound too large by orders of magnitude wherever the two directions are not aligned. Even in the version replaced by the variance, the lower bound on the loss, $0.66〜1.8$, was outside the needed $0.18〜0.50$ Reading. What must be paid is the gradient of the solution of the Poisson equation on the fibre ($|M^{-1}\nabla_\perp G|$ in the Helffer–Sjöstrand representation), and it must be estimated while keeping the direction. The gap in strand 3, where the Monte Carlo reading is 0.54 times the threshold and the paper upper bound $10^3$ times the threshold, also came from the step where the scalar Grönwall discards the direction of the gradient and reduces every direction at the softest rate (10-09 strand 3).

Question to test: somewhere in your upper bound, have you crushed a vector quantity into a scalar? If there is such a step, it pays at the rate of the softest direction there. Draw the direction of large fluctuation and the soft direction in the same figure.

Type 7 Transfer — the weight only moves to another column

Plans caught: the modified LSI of R13, R16, R17, R19, R21, R22, and the proxy series in Table 1′.

Known uniformity theorems (R16, R17) do not make the constant; they move the problem to the finite-volume constant of a box covering the correlation length (length $\gtrsim230〜810$). The window is finite, so existence was trivial from the start, and the reason they do not help is size. Plans that change how the window is taken (R21, R22) reduced the loss of the window, but the floor of the propagation step remained. In the cluster expansion (R19) the pin of the largest path stayed at the top, and in the modified LSI (R13) $\ln(1/\pi_*)$ moved to the energy side. In the proxy series of Proposition P5, when the assumption column became lighter by changing proxy or direction, the computed column became heavier by the same amount (an identity in which the sum of the two columns does not depend on the direction) Assumption (the definitions are not placed in the book) Leanthreshold_gap_invariant (assumption: the identity of the two sums).

Question to test: in your plan, has another column (the size of a box, the propagation step, a computed quantity) become heavier by as much as one column became lighter? If the sum is invariant, the margin has not moved.

Among the dead and on-hold plans, those whose type column in Table 1 is "—" are R18, which fixed a step that was not the limit, and R20, which is on hold (for the alive and conditional plans that column gives the chapter they went to). This table format, which keeps the line that killed each dead plan, can be used as it is to tidy up after sorting out detour plans for other problems too.

The alive and conditional plans were gathered into the table of needed values and the three strands. How far the three strands got and their remaining steps are in 10-09 Table 4, and here we list only the correspondence from the plans.

The place where all three strands stopped is the source of type 1, the same as the common wall of 10-09 §06. Another construction that controls the core constant from the side of the box rather than by the sector framework is treated in 10-11.

This chapter is a list, so its rows are limited to the lemmas supporting the types and the values of Table 1. None of the rows is a premise of the conclusion of route 5, and the dependence of the conclusion is in the graph of 10-09 §08. So this chapter has no dependency graph.

The "Proved here" rows are general facts that identify the types, and do not mean that any detour plan succeeded. None of the rows of Table 4 shows the closing condition (1).

The construction that controls the core constant in three steps from the side of reduction to the box (H3) and the gap inside the box (H4) rather than the sector framework, and the drawing of the boundary of the end sets, are in 10-11. This part places Assumption H at the axis of the study, and this chapter tidied up after the plans aimed at the core row of (H2) of its first step. Another viewpoint that handles the negative curvature at the top without relying on convexity is treated in Part 11 (11-06).

What this chapter can say

Sources and reproduction

Words used in this chapter — detour plan, the line that killed it, needed side, form, insufficient side (floor), structure and choice of tools, ceiling, types of breakage, paying twice, currency (speed of exit and entropy), the address of $1/\hbar$, crossing length, direction of fluctuation and soft direction, transfer. Their meanings are in A-02 Glossary.

Next chapter to read: 10-11 (H3)(H4) and the core constant — turns reduction to the box and the gap inside the box into propositions, and controls the core constant in three steps.

← 10-09 The sector framework and the three strands of route 5Contents10-11 (H3)(H4) and the constant of the core →