Linear Functionals and Positivity

A bootstrap functional is more than a list of coefficients. It is a linear map together with a domain, an ordered basis, a normalization, and a proof of scalar or matrix positivity on every allowed sector. This page constructs a low-order functional that excludes a one-dimensional scalar gap, then separates the exact functional from its numerical coordinates so that rescaling and serialization cannot silently change the proof.

Required background. From Crossing Equations to Convex Optimization supplies the cone and exclusion logic. Helpful background. Forms, adjoints, and isometries clarify dual pairings and basis transformations.

Derivative functionals at the crossing-symmetric point

For scalar crossing in two cross-ratios, a finite derivative functional has the form

$$\alpha[f] =\sum_{m+n\leq\Lambda}a_{mn} \left. \partial_z^m\partial_{\bar z}^{,n}f(z,\bar z) \right|_{z=\bar z=1/2}.$$

The cutoff $\Lambda$ defines the search space, not the validity domain. Positivity must still hold for every allowed $\Delta$ and spin in every included sector. Exchange symmetry removes redundant derivatives, and a real basis should be chosen before coefficients are optimized.

In one dimension the crossing vector obeys $F_\Delta(1-z)=-F_\Delta(z)$. Writing $z=1/2+y$ shows that $F_\Delta$ is odd in $y$, so all even derivatives vanish at the symmetric point. A two-derivative ansatz is therefore

$$\alpha_{u,v}[f] =u f'\!\left(\frac12\right) +v f'''\!\left(\frac12\right).$$

For external dimension $\Delta_\phi=1$, the identity crossing vector has

$$F_0'\!\left(\frac12\right)=-32, \qquad F_0'''\!\left(\frac12\right)=-1536.$$

Thus identity normalization $\alpha_{u,v}(F_0)=1$ requires

$$-32u-1536v=1.$$

The rational choice

$$u=-\frac{49}{32}, \qquad v=\frac1{32}$$

gives the normalized functional

$$\widehat\alpha[f] =\frac{f'''(1/2)-49f'(1/2)}{32}, \qquad \widehat\alpha(F_0)=1.$$

Proving the sign on a continuous spectrum

Let

$$g=g_\Delta\!\left(\frac12\right)>0, \qquad r_\Delta= \frac{g_\Delta'(1/2)}{g_\Delta(1/2)}, \qquad C_\Delta=\Delta(\Delta-1).$$

The one-dimensional Casimir equation reduces the functional action to

$$\widehat\alpha(F_\Delta) =2g\left[ \left(C_\Delta+\frac78\right)(r_\Delta-12)+11 \right].$$

This formula turns a derivative sign into two elementary inequalities. The Euler integral for the hypergeometric block gives a positive measure

$$d\mu_\Delta(t)\propto t^{\Delta-1}(1-t)^{\Delta-1} \left(1-\frac t2\right)^{-\Delta}dt, \qquad 0<t<1,$$

and

$$r_\Delta =2\Delta\left[ 1+\mathbb E_{\mu_\Delta}\!\left(\frac{t}{2-t}\right) \right].$$

To make the inequality transparent, first remove the increasing factor $w(t)=(1-t/2)^{-\Delta}$. The remaining beta measure $d\nu\propto t^{\Delta-1}(1-t)^{\Delta-1}dt$ is symmetric and has $\mathbb E_\nu t=1/2$. Since both $t$ and $w(t)$ increase,

$$\operatorname{Cov}_\nu(t,w) =\frac12\mathbb E_{\nu\times\nu} \bigl[(t-s)(w(t)-w(s))\bigr]\geq0.$$

Reweighting by $w$ therefore gives $\mathbb E_{\mu_\Delta}t\geq1/2$. The function $t/(2-t)$ is increasing and convex, so Jensen’s inequality yields

$$\mathbb E_{\mu_\Delta}\!\left(\frac{t}{2-t}\right) \geq\frac13, \qquad r_\Delta\geq\frac{8\Delta}{3}.$$

For $\Delta\geq9/2$, one has $r_\Delta\geq12$ and $C_\Delta+7/8>0$. Consequently

$$\widehat\alpha(F_\Delta)>0 \qquad\left(\Delta\geq\frac92\right).$$

Together with $\widehat\alpha(F_0)=1$, this excludes a nonidentity gap at or above $9/2$ in the declared reflection-positive one-dimensional problem. The sign is pointwise strict, but it is not a uniform absolute margin as $\Delta\to\infty$, because $g_\Delta(1/2)$ decays. A numerical verifier must not replace this proved sign by a fictitious constant lower bound.

Worked fixture, stage 2: functionals and coordinates

On the half-line fixture from the preceding page, the function $v(x)=(1,x,x^2)$ is expressed in the ordered basis $(1,x,x^2)$. The coordinate vector

$$a=(1,-2,1)$$

represents the functional $\alpha(v(x))=(x-1)^2$. These numbers are not meaningful without the basis. If vector coordinates change according to

$$f'=Sf,$$

then invariance of the pairing $a^{\mathsf T}f=a'^{\mathsf T}f'$ requires

$$a'=S^{-\mathsf T}a.$$

The same rule applies to derivative rescalings. A coefficient file evaluated in a differently normalized derivative basis is a different functional, even if its entries look numerically well conditioned.

Zeros also require care. The exact toy has $\alpha(v(1))=0$, and optimized bootstrap functionals often vanish at candidate operator dimensions. Such zeros can guide later spectrum reconstruction, but they add no theorem to the exclusion: the proof is the normalized sign on the full declared domain.

Scalar positivity and matrix positivity

For one identical correlator the condition is scalar:

$$\alpha[\mathbf F_{\Delta,\ell}]\geq0 \quad\text{for every allowed }(\Delta,\ell).$$

In a mixed system, an exchanged primary carries an OPE vector $\boldsymbol\lambda$ and contributes

$$\boldsymbol\lambda^{\mathsf T} \mathbf F_{\Delta,\ell} \boldsymbol\lambda.$$

After applying the functional, positivity for every OPE vector means

$$\alpha[\mathbf F_{\Delta,\ell}]\succeq0.$$

Positive diagonal entries are not sufficient. The quadratic form must be nonnegative in every direction, equivalently all eigenvalues—or all principal minors in an exact finite matrix test—must be nonnegative. Representation projectors, tensor-structure order, and parity signs must be derived before matrices are assigned to PSD blocks; see Mixed Correlators and Symmetry Sectors. The mixed-correlator SDP construction is reviewed in Poland, Rychkov, and Vichi 2019, §IV.B.1, pp. 27–28.

Under an invertible change of OPE-vector coordinates $\boldsymbol\lambda=S\widetilde{\boldsymbol\lambda}$, the matrix changes by congruence,

$$M\longmapsto \widetilde M=S^{\mathsf T}MS.$$

Congruence preserves PSD, but only when the matrix and vector transformations are paired consistently. Row scaling performed for a solver must therefore be inverted before physical signs and residuals are reported.

Integral and analytic functionals

An integral functional replaces derivatives by a kernel, schematically

$$\omega[f]=\int_{\mathcal C}dz\,h(z)f(z).$$

Its domain must specify endpoint behavior, contour or sheet, and any Regge bound. Interchanging $\omega$ with an infinite OPE sum requires a domination or convergence argument. Analytic functionals can be much stronger than low-order derivatives, but a formal kernel with the desired sampled signs is not automatically a valid functional on the crossing equation.

Item	What must be fixed	Independent rejection test
Crossing basis	ordered components, sectors, tensor structures	permute components without transforming coefficients
Functional basis	derivative or kernel definitions and scaling	evaluate the saved vector in an unscaled basis
Normalization	exact identity or objective action and sign	flip only the target convention
Positivity domain	every spin and dimension interval, including tails	insert a negative point between samples
Matrix sectors	complete PSD blocks and OPE-vector order	test a negative eigenvector with positive diagonals
Numerical representation	precision, rounding, and serialization	truncate a coefficient before re-evaluation

The next page converts continuous functional signs into polynomial and polynomial-matrix conditions. Solver Certificates and Independent Verification later checks that the saved coefficients, basis, normalization, and PSD domains still describe the same functional.

Common pitfalls

Confusing a functional with its coefficient vector. Coordinates depend on the ordered and scaled basis. Preserve the basis map or the numbers cannot be interpreted.

Checking only a dimension grid. A continuous spectrum can pass through a narrow negative interval between nodes. Sampling is diagnostic evidence, not a global sign proof.

Checking only matrix diagonals. Off-diagonal entries can create a negative eigenvector even when every diagonal entry is positive.

Exercises

In the derivative basis $e=(f'(1/2),f'''(1/2))$, the normalized functional has coefficients $a=(-49/32,1/32)$. Let $e'=Se$ with $S=\operatorname{diag}(2,1/4)$. Find the coefficients in the new basis and verify the pairing rule.

Solution

Because $a'=S^{-\mathsf T}a$ and $S$ is diagonal,

$$a'=\left(-\frac{49}{64},\frac18\right).$$

Indeed $a'^{\mathsf T}e'=(-49/64)(2e_1)+(1/8)(e_2/4)=(-49/32)e_1+(1/32)e_2=a^{\mathsf T}e$. Reusing $a$ with $e'$ would change the functional.

Show that

$$M=\begin{pmatrix}1&2\\2&1\end{pmatrix}$$

is not positive semidefinite even though both diagonal entries are positive. Give a vector that detects the failure.

Solution

The determinant is $1-4=-3$, and the eigenvalues are $3$ and $-1$. The vector $u=(1,-1)^{\mathsf T}$ gives $u^{\mathsf T}Mu=-2<0$. Componentwise positivity therefore cannot replace a PSD test.

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