# Spurious length-scale cutoff emerges in propagator defined in Costello's “Renormalization and EFT”

In page 9 of the introductory chaper of Renormalization and Effective Field Theory (the introductory chapter is available free here), Kevin Costello defines a propagator $P$ for the Laplace operator $D$ in a Riemannian spacetime $M$.

In order to define a "length-scale version of the renormalization group flow", he provides a formula for this propagator in terms of the heat kernel $K_\tau$ of the Laplace operator: $$P(x,y):=\int_0^\infty d\tau\; K_\tau(x,y).\tag{1}$$ Using the interpretation of the heat kernel as a transition probability, one may then rigorously write (expanding upon equation 1) $$P(x,y)=\int_0^\infty d\tau \int_{\gamma:[0,\tau]\to M\\\gamma(0)=x\\\gamma(\tau)=y}\exp\left(-\int_0^\tau \|d\gamma\|^2\right).\tag{2}$$ My question regards the interpretation of this "path integral" above involving $\gamma$, as something is confusing to me. Since Costello defines $\tau$ to be the proper time of the worldline $\gamma$ as measured by the Riemannian metric, it occurred to me that the integrand in equation (2) vanishes for $\tau < \text{dist}(x,y)$, as there will not exist paths with lengths smaller than that of the geodesic.

Therefore, we can write, more subtly, $$P(x,y)=\int_{\text{dist}(x,y)}^\infty d\tau \int_{\gamma:[0,\tau]\to M\\\gamma(0)=x\\\gamma(\tau)=y}\exp\left(-\int_0^\tau \|d\gamma\|^2\right).\tag{2}$$ Look how, naturally, a length-scale cutoff emerges. Clearly, I have made a serious conceptual error. Which is it?

• frankly, I don't understand why for $\tau<\mathrm{dist}(x, y)$ paths do not exist. Can't we always make a reparametrization of $\gamma$ (if we can't, you probably have some bounds on $|d\gamma|$ that should be written out explicitly)? – user74900 Oct 15 '17 at 10:32
• In the path integral there is no requirement that the path has unit speed, i.e. $\|\mathrm d\gamma\| = 1$. This means that one can go as far as one wants in a given time interval by going "fast enough". I think the confusion stems from the fact that the parameter $\tau$ is not defined in terms of the unparametrized path $\gamma$, whereas you seem to operate under the assumption that we have $\tau = \mathrm{len}(\gamma)$. – Bertram Arnold Oct 15 '17 at 11:28
• Very sensible comment. However, according to Abdelmalek's answer, the issue was a little more subtle than that - indeed, the assumption $\tau=\text{len}(\gamma)$ was valid. The problem is that we can make the cutoff disappear by taking $|x-y|\to 0$. A true length-scale cutoff $\epsilon$ in the propagator for a quantum field theory is independent of $x$ and $y$, and in particular remains non-zero. – David Roberts Oct 16 '17 at 2:54
• Just to be clear, your formula (2) is not correct (as can be easily seen for flat space). There are other expressions for the propagator which do use cutoffs depending on $x$ and $y$, as in Abdelmalek's answer. If you look at the discussion in Paragraph 6.2 (page 49 in my copy), Costello says that the parameter $\tau$ should be thought of as intrinsic to the worldline, which is equipped with an “internal clock”; as the particle moves, this clock ticks at a rate independent of the time parameter on space-time.'' – Bertram Arnold Oct 16 '17 at 21:53
• Ah yes, I am wrong. Thanks for pointing this out :) – David Roberts Oct 21 '17 at 20:35

There is no space cutoff because it is $d(x,y)$ which depends on the two points. If you put a constant $c$ in your integral $\int_c^\infty d\tau$ then yes you would have introduced a spurrious cutoff in position space. If you are in $\mathbb{R}^d$ you can write the free massless propagator as $$(-\Delta)^{-1}(x,y)=\int_0^{\infty}\frac{dl}{l} l^{-(d-2)} u\left(\frac{x-y}{l}\right)$$ for some nonnegative, positive semidefinite smooth function $u$ with compact support say in the unit Euclidean ball around the origin. See for example my article "A complete renormalization group trajectory between two fixed points". Comm. Math. Phys. 276 (2007), no. 3, 727-772. This can be very useful for a rigorous renormalization group analysis (see this MO answer).
Clearly, there is no cutoff above because the integral over length scales $l$ is from $0$ to $\infty$. However, because of the support property of the infinitesimal multiscale slice $u((x-y)/l)$, the previous expression is equal to $$\int_{|x-y|}^{\infty}\frac{dl}{l} l^{-(d-2)} u\left(\frac{x-y}{l}\right)\ .$$ There is a theory for this type of compactly supported multiscale decompositions developed by Brydges, Guadagni, Mitter, Bauerschmidt, Talarczyk, and others. See the JSP article by Mitter "On a finite range decomposition of the resolvent of a fractional power of the Laplacian" (and its erratum) for a recent account.
• I think this formula only gives you a parametrix and not an inverse for the Laplacian. To get the inverse, you would need $u = exp(-\|x\|^2/2)$ (up to some constants), which does not have constant support. Of course, this whole business is exactly what Costello develops in the rest of the book - replace the inverse of the Laplacian by a parametrix and use renormalization flow to relate the expressions for different parametrices. – Bertram Arnold Oct 16 '17 at 17:40
• @Bertram: No. By the inverse Laplacian I mean the distribution given by a multiple of the locally integrable function $|x-y|^{-(d-2)}$ on the complement of the diagonal. This function is equal to the integral representation I gave. I forgot to mention though that u has to be radial. All you said about parametrix vs inverse, in the present context of this MO question, is a red herring. – Abdelmalek Abdesselam Oct 16 '17 at 17:53