From Crossing Equations to Convex Optimization

A numerical bootstrap exclusion starts before any solver runs. One must first show that the assumed spectrum contributes to crossing with nonnegative scalar weights or positive-semidefinite quadratic forms. Only then is the crossing equation a cone-membership problem, and only then can a separating functional rule out the spectral hypothesis. This page builds that logic twice: first for an identical-scalar CFT, then for an exact rational fixture that will follow us through the next three pages.

Required background. Crossing equations and positivity fix the CFT sign conditions. Convex cones, separation, and conic duality supply the separation theorem. Helpful background. Conformal Hamiltonian truncation illustrates a different finite approximation whose errors must not be confused with functional truncation.

Crossing as a cone-membership question

Fix the external operators, block normalization, symmetry sectors, and a trial spectral hypothesis $H$. For an identical Hermitian scalar, crossing has the schematic form

$$\mathbf F_{\mathbf1} +\sum_{a\in H}p_a\mathbf F_a=0, \qquad p_a\geq0.$$

Here $a$ abbreviates the exchanged dimension, spin, and representation. In a single-correlator problem, $p_a$ is a squared OPE coefficient. In a mixed-correlator problem, the corresponding statement is matrix-valued: OPE vectors contract positive-semidefinite blocks. A gap, an operator-uniqueness assumption, or a spin-dependent unitarity threshold belongs to $H$; none of these assumptions is produced by the eventual bound.

Let

$$C_H=\operatorname{cone}\{\mathbf F_a:a\in H\}, \qquad \mathbf t=-\mathbf F_{\mathbf1}.$$

The primal question is whether $\mathbf t\in C_H$. A real linear functional $\alpha$ belongs to the dual cone when $\alpha(\mathbf F_a)\geq0$ for every allowed generator. If

$$\alpha(\mathbf F_{\mathbf1})=1, \qquad \alpha(\mathbf F_a)\geq0\quad(a\in H),$$

then

$$\alpha(\mathbf t)=-1.$$

Thus $\mathbf t$ lies strictly on the opposite side of the separating hyperplane. Applying $\alpha$ to crossing would give

$$1+\sum_{a\in H}p_a\alpha(\mathbf F_a)=0,$$

whose left side is strictly positive. The hypothesis $H$ is impossible. Finding such a functional proves exclusion on its stated domain. Failing to find one in a finite search space proves neither feasibility nor the existence of a CFT; a converse separation statement also needs the relevant closure or regularity hypotheses.

An identical-scalar gap exclusion

The one-dimensional crossing calculation supplies a concrete CFT example. For an identical scalar with $\Delta_\phi=1$, define

$$\alpha[f] =f'''\!\left(\frac12\right) -49f'\!\left(\frac12\right).$$

With the block convention used there,

$$\alpha(F_0)=32, \qquad \alpha(F_\Delta)>0 \quad\text{for every }\Delta\geq\frac92.$$

The identity is $F_0$, so a unitary solution whose only operator below $9/2$ is the identity would make the functional sum strictly positive. Crossing says it must vanish. Therefore at least one nonidentity primary has $\Delta<9/2$. This is an exact, deliberately nonoptimal exclusion; no block interpolation or solver tolerance enters it.

Now remove the positivity hypothesis. If some $p_a$ may be negative, a negative contribution can cancel the positive identity action. The same algebraic crossing equation may still be studied, but this functional no longer supplies a positive-cone contradiction. Nonunitarity is one common reason for the failure, but the precise stop rule is narrower: stop whenever the particular channel and pairing do not provide the scalar or PSD positivity the proposed cone requires.

Worked fixture, stage 1: exact cone separation

Before approximating conformal blocks, test the logic on a problem with no floating-point ambiguity. Let

$$v(x)=(1,x,x^2), \qquad x\geq0,$$

and let $C$ be the cone generated by all such $v(x)$. The control target

$$t_{\mathrm{in}}=v(1)=(1,1,1)$$

is feasible: it is one generator with weight one. Consider instead

$$t_{\mathrm{out}}=(1,1,0), \qquad \alpha(y_0,y_1,y_2)=y_0-2y_1+y_2.$$

For every $x\geq0$,

$$\alpha(v(x))=1-2x+x^2=(x-1)^2\geq0,$$

but $\alpha(t_{\mathrm{out}})=-1$. Therefore $t_{\mathrm{out}}\notin C$. The same functional separates the finite control cone generated by $v(0),v(1),v(2)$, where its generator actions are exactly $(1,0,1)$.

The stop rule becomes visible in the same numbers. If signed weights are permitted, then

$$t_{\mathrm{out}} =-\frac12v(0)+2v(1)-\frac12v(2).$$

The target has an exact signed decomposition even though the positive-cone certificate excludes it. Thus the separator rules out nonnegative weights, not arbitrary algebraic decompositions—the precise distinction needed when reflection positivity is absent.

This toy problem is not a CFT. Its value is diagnostic: it exercises target signs, ordered bases, continuous positivity, a boundary zero at $x=1$, and exact serialization. Linear Functionals and Positivity will treat $\alpha$ as a basis-dependent coefficient vector; Block Approximations and Semidefinite Programs will encode $(x-1)^2\geq0$ by a PSD Gram matrix; and Solver Certificates and Independent Verification will verify the saved record independently.

Finite projection and the claim it supports

An exact crossing equation lives in a function space. A derivative computation chooses finitely many probes $e_i$ and maps each crossing vector to

$$\mathbf f_a= \bigl(e_1[\mathbf F_a],\ldots,e_M[\mathbf F_a]\bigr) \in\mathbb R^M.$$

Separation in $\mathbb R^M$ is exact for the projected vectors when those entries and their positivity are exact. It is generally weaker than separation in the full function space: increasing $M$ can reveal additional separators, while nonexclusion at finite $M$ says only that this search did not find one.

A production calculation adds several logically independent layers. Conformal blocks are evaluated or approximated; dimensions remain continuous; spin sectors and high-spin behavior must be covered; numbers are rounded and rescaled; and a solver returns finite-precision variables. The numerical-method reduction is reviewed in Poland, Rychkov, and Vichi 2019, §§IV.A–IV.B, pp. 25–29, while the polynomial-matrix formulation is derived in Simmons-Duffin 2015, §§2.1–2.3, pp. 4–8.

Layer	Exact object	New finite choice	Passing evidence	Failure or nonresult means
Crossing	functions and an infinite spectrum	derivative or integral basis	algebra and normalization regenerated	wrong physical problem
Spectral domain	all allowed $\Delta$, spins, and sectors	interval and tail representation	positivity covers the full declared domain	sampled statement only
Blocks	exact conformal blocks	radial, pole, or polynomial approximation	independent block comparison and remainder control	finite surrogate only
Conic problem	cone membership or optimization	matrix scaling and serialization	canonical unscaled input with hashes	irreproducible finite problem
Solver	primal or dual equations	precision and stopping tolerances	candidate variables with margins	algorithmic status only
Verification	original unscaled equations	independent arithmetic or enclosures	residual, cone, interval, and tail checks pass	no certified claim

Primal, dual, and boundary cases

The primal search asks for nonnegative spectral weights reproducing $-\mathbf F_{\mathbf1}$. The dual search asks for a separator. Strict feasibility gives a margin that can dominate verified numerical error. Bootstrap optima, however, often lie on cone boundaries: the exact toy separator already has the zero $\alpha(v(1))=0$. Near such boundaries, conditioning deteriorates and a solver label can be ambiguous.

A verified dual separator supports exclusion for the declared finite problem, with any additional CFT statement limited by block and continuum control. A verified primal point supports feasibility of that finite conic problem only. It does not construct an exact crossing solution, much less prove that a local CFT exists.

The diagram shows the information that must survive this progression. Inspect both outcomes after independent checking: a verified dual separator and a verified primal point have different claim ceilings, while any failed check stops the inference.

Schematic certification path from an exact positive crossing system to a finite conic problem. Solver variables are candidates, not certificates. Independent reconstruction checks identity, residuals, cone or PSD membership, normalization, continuous intervals, and tails; only then can dual separation support a conditional exclusion or primal variables support finite-problem feasibility. The diagram is schematic and not to scale.

The same logic in structured form is:

Stage	Required declaration	Independent check	Passing output	Maximum conclusion
Exact crossing	external data, sectors, gaps, positivity	algebraic crossing and sign derivation	well-posed hypothesis	cone question exists
Finite representation	functional basis, blocks, all truncations	regenerate entries and cover approximation domain	canonical finite problem	no physics conclusion yet
Serialization	order, scaling, precision, hashes	byte identity and inverse scaling	solver-ready input	one named finite problem
Solver output	candidate primal or dual variables	none from status alone	candidate record	algorithmic run only
Independent verification	residual, cone, interval, tail criteria	separate evaluator on unscaled data	verified dual or primal record	finite exclusion or feasibility
Interpretation	all physical assumptions and error scope	refinement and exact-to-finite comparison	bounded statement	conditional exclusion, never automatic CFT existence

Common pitfalls

Losing the target sign. With $\mathbf t=-\mathbf F_{\mathbf1}$ and $\alpha(\mathbf F_{\mathbf1})=1$, one must have $\alpha(\mathbf t)=-1$. Changing only one convention reverses the conclusion.

Treating nonexclusion as existence. Failure to find a separator means the selected search space did not exclude the point. It does not supply positive OPE data for the full crossing equation.

Using positivity outside its domain. A signed OPE product, an indefinite pairing, or an unorganized mixed system cannot inherit a scalar positive-cone proof merely because its truncated equations look similar.

Exercises

Apply $\alpha$ to $\mathbf F_{\mathbf1}+\sum_a p_a\mathbf F_a=0$ and prove the exclusion statement. Which line fails if the $p_a$ are allowed to have either sign?

Solution

Linearity gives $0=\alpha(\mathbf F_{\mathbf1})+\sum_a p_a\alpha(\mathbf F_a)$. The first term is strictly positive and every other term is nonnegative, so equality is impossible. If a coefficient can be negative, then $p_a\alpha(\mathbf F_a)$ need not be nonnegative and can cancel the identity term; the cone-membership premise has been lost.

Show directly that $t_{\mathrm{in}}=(1,1,1)$ is in the cone generated by $v(0),v(1),v(2)$. Then prove that $t_{\mathrm{out}}=(1,1,0)$ is outside both that finite cone and the cone generated by all $v(x)$ for $x\geq0$.

Solution

The weights $(0,1,0)$ reproduce $t_{\mathrm{in}}=v(1)$. For $t_{\mathrm{out}}$, the functional $(1,-2,1)$ has actions $(1,0,1)$ on the three finite generators and action $-1$ on the target, so finite-cone membership is impossible. On the continuous family its action is $(x-1)^2\geq0$ for every $x\geq0$, so the same contradiction excludes membership in the full half-line cone.

References

• Poland, David, Slava Rychkov, and Alessandro Vichi. “The Conformal Bootstrap: Theory, Numerical Techniques, and Applications.” Reviews of Modern Physics 91 (2019): 015002. DOI. Open PDF
• Simmons-Duffin, David. “A Semidefinite Program Solver for the Conformal Bootstrap.” Journal of High Energy Physics 06 (2015): 174. DOI. Open PDF