Which positivity are you testing?

Prerequisites: covariance of two random variables and the reflection-positivity lesson. No measure-theory course is assumed.

After this lab: you can distinguish two quadratic forms, construct a counterexample to reflection positivity, and explain why a few successful tests cannot establish a reconstruction theorem.

A covariance tells you how fluctuations at two times are related. Its ordinary positivity follows from the fact that a squared random variable has a nonnegative expectation. Reflection positivity changes which times are paired. It asks for another positive form, the one needed to turn suitable Euclidean observables into physical state vectors. The first condition does not imply the second.

TWO DIFFERENT FORMS

A covariance can pass the wrong test

Reset

Compare the same real linear probe before and after reflecting its first time argument. The default is the exact one-time cosine counterexample: ordinary form 1, reflected form −1.

Covariance kernelCosine: C(u) = cos(u)Massive exponential: C(u) = exp(−|u|)/2Frequency ω = 1 or mass m = 1 in reference inverse-time units.

First positive time t₁ / π

0.50

t₁ = π/2 at the default

Second positive time t₂ / π

1.00

Time in the same reference units

Real coefficient c₁

1.00

Real coefficient c₂

0.00

Ordinary form Q1.000000Uses C(tᵢ − tⱼ)

Reflected form Qθ−1.000000Uses C(tᵢ + tⱼ)

Vary the first time while holding the other controls fixedHorizontal axis: t₁/π. Vertical axis: the quadratic form in the kernel's reference covariance units.

Dashed line: ordinary form. Solid line and marker: reflected form. Values above are the text alternative to the plot.

Reflection positivity fails for this probe.

The ordinary cosine covariance is a positive Gram matrix, but its reflected form can be negative.

Exact cosine counterexampleDownload these test matrices (CSV)

Exact Gaussian toy models. The displayed tests use real coefficients at two positive times. Signs within 10⁻¹⁰ of zero are reported as numerical zero. A negative test disproves the condition for this kernel; finitely many nonnegative tests do not establish all reflection-positivity or reconstruction requirements. The exponential identity is proved on the page below.

A positive covariance that fails the reflected test

Let $X$ and $Y$ be independent real Gaussian variables, each with mean zero and variance one. Define an exact random process, with frequency $\omega>0$,

$$\phi(t)=X\cos(\omega t)+Y\sin(\omega t).$$

Expanding the product and using $\mathbb E[XY]=0$ gives

$$C(t-s)=\mathbb E[\phi(t)\phi(s)]=\cos\bigl(\omega(t-s)\bigr).$$

For real coefficients $c_i$ at times $t_i$, the ordinary form is

$$Q=\sum_{i,j}c_i c_j C(t_i-t_j) =\left(\sum_i c_i\cos(\omega t_i)\right)^2 +\left(\sum_i c_i\sin(\omega t_i)\right)^2\geq0.$$

This is a Gram-matrix identity: each time labels a two-component unit vector. Positivity is true for every number of samples, not just the two displayed. Complex coefficients replace the two real squares by squared absolute values.

Now reflect the first time through zero. For strictly positive $t_i$, the reflected form uses $C(t_i+t_j)$, giving

$$Q_\theta= \left(\sum_i c_i\cos(\omega t_i)\right)^2 -\left(\sum_i c_i\sin(\omega t_i)\right)^2.$$

The minus sign matters. Set just one coefficient to one and choose $t_* = \pi/(2\omega)$. Then $Q=1$ but $Q_\theta=\cos(\pi)=-1$. This is a valid ordinary Gaussian covariance with a failed reflected linear test. No numerical eigenvalue calculation is needed.

A comparison with a nonnegative reflected form

For the massive exponential kernel, with $m>0$,

$$C(t-s)=\frac{e^{-m|t-s|}}{2m},\qquad Q_\theta=\frac{1}{2m}\left|\sum_j c_j e^{-m t_j}\right|^2\geq0.$$

Positive times allow $|t_i+t_j|=t_i+t_j$, so the exponential factorizes. The identity holds for arbitrary complex coefficients and every finite set of positive times. The plotted real tests are examples of that identity. For $m=1$, $c_1=1$ and $t_1=\pi/2$, the ordinary form is $1/2$ while the reflected form is $e^{-\pi}/2\approx0.021607$.

The exponential also is an ordinary covariance: its Fourier transform is the positive function $1/(p^2+m^2)$ with the convention $C(u)=\int e^{ipu}\,dp/[2\pi(p^2+m^2)]$. Integrating a squared Fourier sum against this positive weight gives ordinary positivity. This calculation concerns the stated one-dimensional Gaussian model.

Diagnose the inference

Exercise 1. For the cosine kernel, choose two equal times and coefficients $(1,-1)$. What are both forms? Does their value show that the kernel is reflection positive?

Solution. The two random variables cancel, so both forms vanish. A zero test is compatible with failure elsewhere; the single-time counterexample above still disproves reflection positivity.

Exercise 2. Why does changing the minus sign in the definition of $Q_\theta$ to a plus sign not “repair” the same physical test?

Solution. That would restore the ordinary covariance form by changing the time pairing. It tests a different condition, so it cannot justify the physical inner product associated with reflection.

What this model does not establish

A negative linear test is enough to disprove the required positivity condition for that kernel. Successful finite tests alone prove no universal condition. The displayed exponential identity settles the linear-kernel form; extending it to a complete Gaussian field construction and then satisfying the other reconstruction hypotheses are additional arguments. Neither toy process constructs four-dimensional Yang–Mills theory.

For the reconstruction framework, use Osterwalder and Schrader, Axioms for Euclidean Green’s functions (1973), together with Axioms for Euclidean Green’s functions II (1975), whose publisher abstract explicitly describes it as extending and correcting the earlier formulation. These are background references; this lab makes no new assessment of their full reconstruction proofs. The elementary identities above are derived here. Continue to positivity and correlation inequalities and physical reconstruction.