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In introductory textbooks & lecture notes on conformal field theory, it is usually stated that solving the highly nontrivial dimensional quantum field theory in 2 dimensions is possible due to the existence of the powerful infinite-dimensional symmetry. The symmetry under consideration is, of course, the Virasoro symmetry.

Now consider a completely different class of 2-dimensional field theories – those invariant under the group of area-preserving diffeomorphisms. Here the Lie algebra of the symmetry group consists of vector fields of vanishing divergence (on the flat $\mathbb{R}^2$, this means $\partial_{\mu} v^{\mu} = 0$). This is also an infinite-dimensional symmetry algebra, so naively, I expect it to be powerful enough to allow us to solve/classify 2-dimensional field theories invariant under area-preserving diffeomorphisms.

My questions is: to what extent can similar methods be applied to such theories? What are the most noteworthy results?

Finally, some motivation: at least classically, the 2-dimensional Yang-Mills theory is invariant under the area-preserving diffeomorphisms. We already know that there exists a non-abelian conformal gauge theory in 2 dimensions (the Wess-Zumino-Witten model) which is exactly solvable. Maybe the 2-dimensional Yang-Mills theory admits a similar definition, only its "blocks" are built by using "primary fields" of a different infinite-dimensional algebra rather than the Kac-Moody algebra?

I've heard that the 2-dimensional quantum Yang-Mills theory has been constructed rigorously by solving for certain heat kernels on the gauge group $G$. I wonder if there's an equivalent formulation which is more "Wightmanian" in its nature, specifically, one that utilizes explicitly specified $n$-point blocks that are used to build the correlation functions.

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To be concrete let spacetime be 2D with Minkowski signature.

  1. On one hand, the infinite-dimensional algebra $${\rm locconf}(1,1)~=~{\rm Vect}(\mathbb{S}^1)\oplus {\rm Vect}(\mathbb{S}^1)\tag{1a}$$ of locally defined conformal transformations in 2D is the algebra of conformal Killing vector fields (CKVF), which is 2 copies of the real Witt algebra, cf. e.g. this Phys.SE post. In light-cone coordinates, a CKVF satisfies $$ \partial_{\mp}X^{\pm}~=~0\qquad\Leftrightarrow\qquad X^{\pm}=X^{\pm}(x^{\pm}). \tag{1b}$$

  2. On the other hand, the infinite-dimensional algebra $${\rm sdiff}(1,1)\tag{2a}$$ of 2D area-preserving transformations is the algebra of symplectic vector fields (SVF), which includes the algebra of Hamiltonian vector fields (HVF). In light-cone coordinates, a HVF is of the form $$X^{\pm}~=~\pm\partial_{\pm}H, \tag{2b}$$ where $H=H(x^+,x^-)$ is an arbitrary function.

  3. The two infinite-dimensional algebras (1) & (2) behave naturally like oil & water, i.e. they almost don't mix: Their intersection is quite small, namely it is the 2D Poincare algebra $${\rm iso}(1,1),\tag{3}$$ which is generated by 1 Lorentz boost and 2 translations.

  4. However, the algebra (2) is much bigger than the algebra (1) in the sense that it is often possible to embed a Virasoro/$W$-algebra inside (2), cf. Refs. 1-2. So in that sense the conformal methods apply to theories invariant under area-preserving diffeomorphisms, cf. OP's question.

References:

  1. E. Sezgin, Area-Preserving Diffeomorphisms, $w_\infty$ Algebras and $w_\infty$ Gravity, arXiv:hep-th/9202086.

  2. D.V. Artamonov, Introduction to finite $W$-algebras, arXiv:1607.01697.

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  • $\begingroup$ Awesome, thanks for writing this up. So because $sdiff$ is bigger than $locconf \sim Vir$, there should be additional constraints on the Virasoro representations? How hard is it to solve these constraints e.g. for the minimal models? Has Yang-Mills in 2d been constructed in this way and if it has, what is the corresponding CFT? $\endgroup$ – Prof. Legolasov Aug 6 at 20:52
  • $\begingroup$ There is an error in the labelling of items in point 3, and possibly point 4. In particular, 3. seems like it should say "The infinite-dimensional algebras (1) and (2) behave..." and so I infer that 4. should say "However, the algebra (2) is much bigger than the algebra (1) in the sense..." I'm leaving this as a comment rather than editing, though, in case it is clear to someone with more expertise that 4., as written, is correct. $\endgroup$ – Matthew Titsworth Aug 7 at 15:48
  • $\begingroup$ Hi @MatthewTitsworth. Thanks for the feedback. I tried to relabel for clarity. $\endgroup$ – Qmechanic Aug 7 at 16:13

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