9
$\begingroup$

Moshe's recent questions on formalizing quantum field theory and lattices as a definition of field theory remind me of something I occasionally idly wonder about, and maybe this site can tell me the answer. Are there any mathematicians working on defining quantum field theory by beginning with a rigorous definition of CFTs, and working from there?

The reason I ask is that I think this is how most of us in physics think about quantum field theory (that is, in a Wilsonian way): to define a QFT, you start with a UV fixed point, and deform it with some relevant operator. So if you had a general theory of CFTs, you would know how to understand how CFTs respond to external sources for operators, and getting a more general QFT would "just" mean turning on a spatially homogeneous source for some operator and seeing how it responds.

The other object we study is "effective field theory," which you might imagine in this language is a CFT that serves as the IR fixed point, together with some notion of equivalence class of irrelevant operators deforming "up" away from that point (being agnostic about whether you ever reach a UV fixed point).

Very (extremely) naively, I would suspect mathematicians might have better luck studying the space of CFTs rather than trying to begin with all QFTs. And you might imagine this approach would be well-suited for questions physicists might care about (like, say, whether there is an "a-theorem" or something similar, analogous to the c-theorem in 2 dimensions, characterizing RG flows as irreversible).

Axiomatic/algebraic/constructive field theory seems to worry about all kinds of field theory at once, and other mathematicians seem to be trying to dig up interesting structure in perturbation theory, which I'm not sure will ever lead to progress in nonperturbatively understanding field theory. I know there are some mathematicians who work on CFTs. (I found this MathOverflow question that has a lot of links to work by mathematicians on CFTs, for instance.) But I wonder if any have tried to work on CFTs as a route to understanding QFT more generally.

|cite|improve this question
$\endgroup$
  • 2
    $\begingroup$ This is not a full-fledged answer, so I will just add this small comment. CFT is of course a beautiful mathematical structure with its own new recipes to be formalized. One may see e.g. the "big yellow book" by di Francesco et al., amazon.com/Conformal-Theory-Graduate-Contemporary-Physics/dp/… ... One may obtain QFTs as deformations of CFTs in the UV, and running them down, but once the conformal symmetry is broken, the original advantages - new ways to define the axioms that were possible for CFTs - go away... $\endgroup$ – Luboš Motl Feb 11 '11 at 19:46
  • $\begingroup$ Just thinking about a specific context where this approach might have lead (or might lead ) to some progress: some relevant deformations of two dimensional CFTs lead to integrable theories. Seems a natural context to discuss whether some of the formal structure of CFT survives in the deformed theory. $\endgroup$ – user566 Feb 11 '11 at 20:35
  • $\begingroup$ I'm not quite sure how this would help them. They can't even get a single working example of an interacting relativistic 4dimensional field theory, much less something considerably more constraining. On the other hand, I believe their are some examples of rigorous theories (eg satisfying Wightman or Haag-Kastler axioms) in 2dimensions. So naively i'd guess they would have an easier time making string theory rigorous, and then deriving effective field theories from that. $\endgroup$ – Columbia Feb 12 '11 at 19:57
  • 1
    $\begingroup$ Kontsevich gave some lectures at UChicago a decade ago on this subject: defining QFTs as relevant deformations of UV fixed point CFTs. AFAIK, nothing has ever appeared in print. $\endgroup$ – user1504 May 4 '12 at 14:05
5
$\begingroup$

CFT consists for most mathematicians - who are interested in this topic - currently of the study of vertex operator algebras, see this question on math overflow:

You can find a little bit more about the topic and the work of several mathematicians here:

As you can see from the answers on mathoverflow, vertex algebras were not invented for the study of CFT, and that they form an axiomatic abstraction of operator algebra products was noted only later.

A personal and very subjective note: One should not underestimate the amount of theoretical physics that is necessary to understand what a QFT is to physicists. Most mathematicians that encounter physics for the first time since highschool through some QFT framework seem to be quite taken aback by the high intrance fee they'd have to pay to understand this. This is my own, personal explanation for the observtion that most mathematicians study the formal machinery only, in order to use it to prove some new mathematical theorems, but only very rarely in order to better understand what physicists do. Although you'll find quite a lot of work by quite a lot of rather famous mathematicians if you follow the links above, AFAIK there is none doing the work you describe.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Vertex operator algebras are only a formalization of 2d CFT, right? $\endgroup$ – Matt Reece Feb 11 '11 at 21:16
  • 2
    $\begingroup$ It's true that there's a lot of overhead to learning QFT, so I'm not that surprised that it's not an extremely active research area in math. On the other hand, there are a noticeable number of mathematicians working on things like the structure of perturbation theory, which suggests to me that they're taking the textbook QFT material too seriously at the expense of the more general lore they would get from talking to physicists. (Maybe they're doing important work and finding deep structures, of course; I don't have a clue what a "motive" is, so I'm out of my depth....) $\endgroup$ – Matt Reece Feb 11 '11 at 21:19
  • $\begingroup$ @Matt: Yes, vertex algebras formalize 2d CFTs. I assume you allude to the work of Connes-Kreimer on renormalization? That seems to be widely ignored by physicists - just like AQFT :-) $\endgroup$ – Tim van Beek Feb 12 '11 at 7:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.