# How does locality decouple the UV and IR behaviour of a QFT?

I came a comment in this paper: Scattering Amplitudes and the positive grassmannian in the last paragraph of page 104 which says: "One of the most fundamental consequences of space-time locality is that the ultraviolet and infrared singularities are completely independent."

How do I understand this?

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On thing to keep in mind is that IR and UV divergences appear in different kinematical regimes: UV divergences are basically due to the fact that in loop integrals there are not sufficient propagators to make the integral fall off at infinity. E.g for a bubble integral

$\int d^4l \frac{1}{l^2(l-p)^2}$ will be logarithmically divergent. Do for instance a Taylor expansion of this expression for the loop momentum becoming large then this becomes obvious.

IR divergences however live in a completely different regime: they appear either when two particles becoming collinear $p_1\sim p_2$ or because some particles become soft $p_i\sim0$.

Or put a little more condensed:

UV: loop momentum becomes large

IR: external momenta become collinear/soft.

This is one way to see why these two kinds of divergence are not connected. Nima and company propably meant just this but in fancier terms.

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Isn't your first integral (UV divergent example) diverging logarithmically and not linearly. –  Learning is a mess Feb 19 '13 at 9:19
You're right. Corrected :) Cheers. –  A friendly helper Feb 19 '13 at 9:21
I think the description is right, but I'm still not sure I understand the connection to the locality of the theory. Is there an easy way to see that, if I have a non local theory, this separation doesn't occur? –  twistor59 Feb 19 '13 at 12:56
Sorry for not commenting earlier, but I've been waiting to se eif there are any other responses. I have 2 questions with regards to this answer: 1. As @twistor59 says, I don't understand what role locality plays –  Siva Mar 15 '13 at 4:30
Secondly, say we have a function $f(x)=\frac{1}{x^k}$. Then, in $n$ dimensions, the asymptotic behaviour ("divergence") of the integral $\int f(x) d^n x$ in the IR and UV is related (and complementary). So why should the divergences not be related? –  Siva Mar 15 '13 at 4:36

This quote shows a funny way how mathematicians perceive and describe what they see.

In fact, UV and IR singularities are to a great extent of different nature, but they both are results of a bad formulation of theory. They both are absent in a good formulation.

UV singularities are due to "self-action" contained in the interaction term. This interaction term part is wrong and nobody keeps the UV result intact. Rather, it is discarded (subtracted). Thus, the theory is modified. Is the theory local after renormalization - that's the first question to the authors.

After subtracting wrong self-action contributions, the reminder is "good", but, considered by the perturbation theory, it gives divergent results for low-frequency modes. This result is correct - push a low-frequency oscillator with a strong force, and you will obtain huge oscillations proportional to $1/\omega$.

When you have many modes in a superposition, each "soft oscillator" has large amplitude, but after summation (inclusive picture), they give a finite resulting amplitude. These modes are obligatorily all taken into account, not discarded. Only an inclusive picture is meaningful because the purely elastic cross section is zero and each particular mode diverges. Thus, one needs to sum up soft mode contributions to all orders, and that is equivalent to another initial approximation and another perturbative series, i.e., another formulation of theory.

You see, one has to work hard with the original "local" theory formulation in order to arrive at physical results. Is the resulting theory "local" after renormalization and summation of the soft modes - that's the second question to the authors. There are indications that a correct theory formulation is somewhat "non local" (see a popular explanation here or here).

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