Scaling Degree and Extension of Distributions

A distribution on $\mathbb R^d\setminus\{0\}$ with finite scaling degree always extends across the origin without increasing that degree. The extension is unique when the scaling degree is below $d$; at or above $d$, two extensions can differ by derivatives of the delta distribution up to the degree of divergence. Scaling degree therefore proves existence and bounds finite local freedom, but it does not choose renormalization coefficients.

Required background. Epstein–Glaser induction produces the off-diagonal distributions to extend, and support and regularity domains fix their distribution spaces. Helpful background. Microcausal functionals supply wavefront compatibility, and counterexamples and hypothesis stress tests clarify the uniqueness boundary.

Scaling degree and extension theorem

For $t_0\in\mathcal D'(\mathbb R^d\setminus\{0\})$, define the scaled distribution by $t_{0,\lambda}(f)=\lambda^{-d}t_0(f(\cdot/\lambda))$ and

$$\operatorname{sd}_0(t_0) =\inf\left\{\omega:\lambda^\omega t_{0,\lambda} \longrightarrow0\text{ in }\mathcal D' \text{ as }\lambda\downarrow0\right\}.$$

Put $\rho=\operatorname{sd}_0(t_0)-d$. If $\operatorname{sd}_0(t_0)<d$, there is a unique extension $t\in\mathcal D'(\mathbb R^d)$ with the same scaling degree. If $d\leq\operatorname{sd}_0(t_0)<\infty$, extensions with the same scaling degree exist and any two differ by

$$t'-t=\sum_{|\alpha|\leq\lfloor\rho\rfloor} c_\alpha\,\partial^\alpha\delta.$$

Brunetti and Fredenhagen prove the two cases in Brunetti and Fredenhagen 2000, §5.2, Theorems 5.2–5.3. The proof subtracts the Taylor jet of a test function at the origin through order $\lfloor\rho\rfloor$, applies $t_0$ to the remainder, and then restores the finite jet with arbitrary coefficients. Uniqueness below $d$ follows because every nonzero distribution supported at the origin has scaling degree at least $d$.

More explicitly, choose test functions $w_\alpha$ whose derivatives at the origin form a dual basis for jets through order $N=\lfloor\rho\rfloor$, and define

$$Wf=f-\sum_{|\alpha|\leq N} \frac{\partial^\alpha f(0)}{\alpha!}w_\alpha.$$

Then $Wf$ vanishes to order $N+1$. The scaling estimate makes the limit defining $t_0(Wf)$ finite, while the omitted finite-dimensional jet is assigned numbers $c_\alpha$. Every extension is obtained in this way. If two extensions are subtracted, the result vanishes on all test functions supported away from the origin, hence is supported at $\{0\}$; the structure theorem for point-supported distributions and the scaling bound give precisely the derivative-of-delta sum above.

On a manifold or near a partial diagonal, one uses tubular coordinates and the microlocal scaling degree transverse to the submanifold. Covariance constrains the coefficients to local tensorial combinations of the metric, curvature, masses, and couplings. It does not remove all coefficients automatically.

Extending the squared Feynman propagator

For a four-dimensional scalar field, the short-distance singularity of the Feynman propagator is

$$\Delta_F(x)\sim\frac{C}{x^2-i0},$$

so $\operatorname{sd}_0(\Delta_F)=2$. On $\mathbb R^4\setminus\{0\}$ the square is defined and

$$\operatorname{sd}_0(\Delta_F^2)=4.$$

The degree of divergence is $\rho=4-4=0$. Therefore every extension preserving scaling degree differs by exactly

$$c\,\delta^{(4)}(x).$$

No derivative of delta is allowed at this degree. Lorentz covariance is automatic for the scalar delta term, while a subtraction condition fixes $c$. One explicit construction chooses a smooth cutoff $w$ with $w(0)=1$ and sets

$$t_w(f)=t_0\bigl(f-f(0)w\bigr)+c_w f(0).$$

The bracketed test function vanishes at the origin and lies in the domain on which the singular distribution has a finite extension. Changing $w$ shifts only $c_w\delta$. This completes the low-order extension needed by local counterterm renormalization before a physical subtraction condition is chosen.

At the logarithmic threshold, an extension normally introduces a scale. Denote one covariant choice by $t_\mu$. Since both $t_{\mu'}$ and $t_\mu$ extend the same off-origin distribution with scaling degree four, their difference must be

$$t_{\mu'}-t_\mu=C\log(\mu'/\mu)\,\delta^{(4)}$$

for a convention-dependent constant $C$. The logarithm follows from composing two scale changes; the delta support follows from the extension theorem. This is a local scaling anomaly, not a failure of the extension to exist.

An independent dimensional check gives the same answer. $\Delta_F^2$ has mass dimension four, as does $\delta^{(4)}$; a derivative $\partial^\alpha\delta$ has larger dimension and would exceed the scaling bound. Multiplying by a mass could change dimensional bookkeeping in a massive theory, but it cannot lower the scaling degree of the derivative distribution.

Existence is not uniqueness

Adversarial test. Claim that equality $\operatorname{sd}_0(t_0)=d$ gives a unique extension. If $t$ is one extension, then $t+c\delta$ agrees with $t_0$ off the origin and has the same scaling degree $d$ for every $c$. This one-parameter family disproves uniqueness exactly at the boundary.

Conversely, the presence of delta freedom does not mean that every coefficient is physically admissible. Symmetry, field equations, Ward identities, and chosen normalization conditions can restrict it. Those are additional theorems, not consequences of scaling degree alone.

The theorem is also transverse. For a diagonal in $M^n$, only the $d(n-1)$ relative directions are scaled; the common base point is left fixed. Using the full dimension $dn$ would give the wrong uniqueness threshold and therefore the wrong counterterm count.

Exercises

1. Homogeneous examples. Compute the scaling degree of $|x|^{-p}$ on $\mathbb R^d\setminus\{0\}$.

Solution

Homogeneity gives $t_\lambda=\lambda^{-p}t$, so $\lambda^\omega t_\lambda\to0$ precisely for $\omega>p$. Hence the scaling degree is $p$. The extension is unique for $p<d$ and has local ambiguity through order $\lfloor p-d\rfloor$ for $p\geq d$.

2. Delta derivatives. Show $\operatorname{sd}(\partial^\alpha\delta)=d+|\alpha|$.

Solution

$\partial^\alpha\delta(f(\lambda\cdot))=(-1)^{|\alpha|}\lambda^{|\alpha|}\partial^\alpha f(0)$, while the distribution rescaling contributes $\lambda^{-d-|\alpha|}$. Thus the critical exponent is $d+|\alpha|$, which is exactly why only $|\alpha|\leq\rho$ can preserve the original scaling degree.

References

• Brunetti, Romeo, and Klaus Fredenhagen. “Microlocal Analysis and Interacting Quantum Field Theories: Renormalization on Physical Backgrounds.” Communications in Mathematical Physics 208 (2000): 623–661. DOI; Open preprint.
• Steinmann, Othmar. Perturbative Quantum Electrodynamics and Axiomatic Field Theory. Berlin: Springer, 2000. DOI.