A nonempty low space after vacuum puncture

Why the retained space needs a second direction

The simplest vacuum-window example has a rank-one low projection. Its correct Schur complement vanishes, and removing the transported vacuum removes the entire low space. That proves an obstruction to an incorrect mixing estimate, but supplies no remaining excitation on which to demonstrate strict positivity. The present example adds exactly one retained direction. It is the smallest real model in which vacuum cancellation and nontrivial punctured coercivity can be seen together.

Let P retain the first two coordinates and Q retain the third. A vacuum with nonzero first and third components crosses this split. The second coordinate is orthogonal to the vacuum and survives puncturing. This arrangement models a decomposition of Hilbert-space amplitudes, rather than three points of a configuration space. In particular, positivity of the transfer operator means positivity of its eigenvalues; its entries need not define transition probabilities.

The purpose is to check exactly what the source theorem consumes: a specified vacuum-corrected operator, an invertible high block, transport of the vacuum through that block, and a positive form on the remaining low space. None of those algebraic checks supplies the regulator-uniform estimates required in a gauge theory.

Source: Vacuum obstructions and form Feshbach reduction, §6.1, Proposition 6.3, printed/PDF p.16.

Constructing the vacuum and two excitations

Write q = 1 − p and define three orthonormal vectors. The angle α rotates the excitation plane while leaving the vacuum fixed:

The transfer operator is T = |Ω〉〈Ω| + t₁|u〉〈u| + t₂|z〉〈z|. Its spectral resolution is already visible: Ω has eigenvalue one, while u and z have eigenvalues t₁ and t₂. All three are positive, and the vacuum is unique because both excitation eigenvalues are strictly below one. Degeneracy t₁ = t₂ affects the excitation basis but does not create another vacuum.

For the threshold r, define a = r − t₁ and b = r − t₂. Both are positive. The vacuum-compatible operator takes the particularly simple form G = rI − T + (1 − r)|Ω〉〈Ω| = a|u〉〈u| + b|z〉〈z|. The added rank-one term cancels the negative vacuum contribution r − 1 exactly. Omitting it would make the threshold test fail on Ω before any block reduction began.

Source: Vacuum obstructions and form Feshbach reduction, §6, Proposition 6.1, printed/PDF p.15.

The block matrix before elimination

Introduce three scalar entries in the unrotated excitation basis:

Since this is an orthogonal rotation of diag(a,b), its determinant is hk − ℓ² = ab. The low and high blocks of G are

Here B is a two-by-one column and G = [[A,B],[B^T,D]]. Because p and h are positive, D is invertible at every admitted exact parameter value. The low vacuum is ω = PΩ = (√p,0), and the high component is ζ = √q. Substitution gives −D⁻¹B^Tω = √q, reconstructing the missing vacuum component.

This transport identity matters because PΩ is not the full vacuum. Treating it as an independently normalized ground state of A would discard its coupling to Q. The low block contains a positive expectation in the vacuum direction that the exact return through the high block must cancel.

Source: Vacuum obstructions and form Feshbach reduction, §6.1, equations (63)–(66), printed/PDF p.16.

The cancellation leaves a harmonic mean

Compute F = A − BD⁻¹B^T. Its first diagonal entry is qh − qh = 0. Its off-diagonal entry is √qℓ − √qℓ = 0. The second diagonal entry uses the determinant identity:

The last expression makes the positivity and bounds transparent. The number c is a weighted harmonic mean of b and a, so min(a,b) ≤ c ≤ max(a,b). At α = 0, c = b; at α = π/2, c = a. If a = b, c is constant at every angle. In general, increasing the smaller spectral margin improves the retained coefficient even though elimination mixes the directions.

The normalized transported vacuum line is the first low coordinate. Thus P° = P − |ω〉〈ω|/p = diag(0,1,0), and its rank is one. On its range, F is exactly c times the identity. On the full rank-two low space, F still has the required zero eigenvalue. Strict positivity is therefore a statement about a declared subspace, with the vacuum kernel accounted for separately.

Source: Vacuum obstructions and form Feshbach reduction, §6.3, Lemma 6.7 and Proposition 6.8, printed/PDF p.18.

A complete numerical checkpoint

Take p = 1/2, t₁ = 1/5, t₂ = 1/2, r = 4/5 and α = π/4. Then a = 3/5, b = 3/10, h = k = 9/20 and ℓ = 3/20. The blocks become

The second diagonal mixing contribution is ℓ²/h = 1/20. Consequently c = 9/20 − 1/20 = 2/5. The unpunctured Schur spectrum is {0,2/5}; the punctured spectrum contains only 2/5. The whole corrected operator has spectrum {0,3/10,3/5}, so the Schur coefficient need not equal its smallest nonzero eigenvalue. Elimination minimizes over a high component for each fixed low coordinate; it does not preserve the ordinary norm of the reconstructed vector.

The punctured-window explorer displays these blocks and the angle dependence. Tiny nonzero entries in a computed Schur matrix represent floating-point cancellation. The exact displayed formula identifies its vacuum zero, while residual diagnostics measure agreement with that formula without silently assigning a numerical sign to roundoff.

What the vacuum weight changes

The coefficient c contains no p. This does not mean the decomposition is uniformly well conditioned. The high block is D = ph, so its inverse grows like 1/p. At the excluded endpoint p = 0, the high space contains the vacuum and D is singular. The smooth value obtained by continuing c to that endpoint is not permission to use an inverse that no longer exists.

For the same spectral parameters and angle but p = 0.01, D = 0.0045 while c remains 0.4. The raw high transfer compression is QTQ = 0.9935, which exceeds r = 0.8. Nevertheless the vacuum correction makes QGQ positive, and the exact punctured Schur form remains positive. A direct high-transfer contraction is thus a useful sufficient premise in the source theorem, not a necessary property of every successful vacuum-corrected split.

Numerically, forming B²/D from extremely small entries can lose representability before the underlying formula becomes singular. The explorer evaluates the Schur subtraction through B/√D using its spectral factors and evaluates c as a harmonic mean. It separately reports whether the raw high block, inverse and literal vacuum reconstruction are representable.

Source: Vacuum obstructions and form Feshbach reduction, Proposition 6.6, printed/PDF p.17, and Proposition 6.8, p.18.

Transfer margins and logarithmic energies

The Hamiltonian for one declared transfer step is H = −log T. Its excitation energies are −log t₁ and −log t₂, and the exact toy gap is Δ = −log max(t₁,t₂). The threshold energy is δ = −log r. Because r exceeds both excitation eigenvalues, Δ exceeds δ. For the worked numbers, Δ = log 2, δ = log(5/4), and their difference is log(8/5).

The coefficient c is dimensionless; Δ and δ have units of inverse declared transfer step. If a step length a_t is later supplied, the energy expressions become −a_t⁻¹log t. A positive dimensionless Schur margin alone does not determine a physical energy without that calibration.

For example, prescribing tⱼ = exp(−a_tEⱼ) and r = exp(−a_tδ) with fixed Eⱼ > δ makes the spectral margins a and b of order a_t. Their harmonic mean c also scales with that step length. The meaningful uniform statement concerns c/a_t together with the correct vacuum transport and form estimates.

Source: Vacuum obstructions and form Feshbach reduction, Proposition 6.8, equations (79)–(81), printed/PDF p.18.

What the example contributes to a lattice proof

The model is an exact checkpoint for an implementation or a proposed inequality. It tests a nonzero retained excitation, a vacuum that crosses the split, two unequal transfer eigenvalues, and a mixing angle. It also exposes three independent quantities: the physical toy threshold, the punctured form coefficient, and the conditioning of the eliminated block. Conflating them would hide a genuine analytic obligation.

A lattice application must identify the actual Hilbert space, retain the physical vacuum of its specified generator, and prove that its projection preserves the relevant form domain. Its cross form must be continuous in the high-form norm. Those domain requirements are automatic for finite matrices and substantive for differential operators. Uniformity must then be established over the declared spatial volumes, boundary data, holonomy strata and regulator sequence.

The source’s form Feshbach theorem provides the exact completion-of-square identity once those assumptions hold. It does not establish the model-specific cross estimate. The three-state calculation therefore supports a useful local milestone: verify the vacuum transport and the punctured form in a nontrivial test problem before attempting a uniform gauge-theory bound.

Source: Vacuum obstructions and form Feshbach reduction, Theorem 10.3, printed/PDF pp.24–25.

Continue reading

• Explore the nonempty punctured window
• Form Feshbach reduction
• Counterexamples to threshold shortcuts
• A concrete lattice proof roadmap