I am trying to clear my ideas about the relation between a Conformal Field Theory (CFT) and a Vertex Operator Algebra (VOA). For me a CFT based on a (complex) vector space $H$ is a projective monoidal functor from the Segal category $\mathcal{C}$ to the category of complex vector spaces. Objects in $\mathcal{C}$ are disjoint unions of $S^1\cong\mathbb{R}/\mathbb{Z}$ and morphisms are (class of conformal equivalence (which is the identity on each boundary component)) Riemann surfaces with boundary composed by disjoint union of (parametrized) circles (this is a very rough description of the Segal category, a precise one can be found in Segal). I.e. \begin{eqnarray} U:\,\mathcal{C} & \longrightarrow & \mbox{Vect}_{\mathbb{C}} \\ \; S^1 & \longmapsto & H \end{eqnarray} satisfying some properties (cf. Segal original paper, section 4). I have read in several papers and books that the chiral (holomorphic) part of a (rational?) CFT is a VOA. I have interpreted this by meaning that if $H$ is the base of a holomorphic CFT then $H$ is a VOA.

I will not recall here the definition of a VOA (but maybe Wikipedia can be useful).

Question: Where do field operators $Y(u,z)\in(\rm{End}\,H)[[z^{\pm 1}]]$ in the theory of VOA's come from?

Everyone says that such an operator is associated to the ''pair of pants'', i.e. a genus zero Riemann surface with three circle boundaries (just google it). Such a Riemann surface give me a morphism \begin{equation} U(\mbox{''pair of pants''}):H\otimes H\longrightarrow H. \end{equation}

More precise questions:

  1. Given such a morphism, we use it to define an endomorphism of $H$. What the parameter $z$ is? Why can we look at it as a formal parameter (defining the space $(\rm{End}\,H)[[z^{\pm 1}]]$)?
  2. It seems to me that we have a lot of conformal classes of pants: Why can we associate a 'unique' field operator?
  3. Am I completely misunderstanding everything?

Thank you very much!

P.S.: there are actually descriptions of such a $Y$ field operator in terms of a CFT, but all the references I have found use a physical language and notation, while I would like to understand it algebraically (or at least mathematically). Any reference is also much appreciated.

P.P.S.: sorry for the great amount of tags but I was not sure which one was the best one.


1 Answer 1



the most mathematical thing would be to view them just as part of the axioms. One of the interpretations of a physical field is an operator-valued distribution meaning that in more analytical settings a physical field $\phi$ acts on the Hilbert space only after smearing with a test function $f$: $$ \phi(f) = \int \phi(x) f(x) dx, \quad \phi(f) \in \mathrm{End}\, \mathcal{H}, $$ where $\mathcal{H}$ is a Hilbert space. Usual test function spaces on $\mathbb{R}^n$ are the Schwartz space of rapidly decreasing functions or the bump functions. Without smearing, most of the fields are very badly behaved and singular. If we were to abuse the notation and write $\phi(x)$ instead of $\phi(f)$ and moreover assume that $\phi(f)$ has some kind of a Fourier series, then $$ \phi(x)= \sum_{n\in\mathbb{Z}} a_{(n)} x^{-n-1},\quad a_{(n)}\in\mathrm{End}\,\mathcal{H} $$ and hence $$ \phi(x)\in \mathrm{End}\,\mathcal{H}[[x^{\pm 1}]], $$ if we also forget that our $x$ was initially in $\mathbb{R}^n$. Finally, we can also forget the Hilbert space structure to end up with a vector space $\mathrm{H}$ as you have correctly observed. Also, CFT is rather special and we have state-field correspondence meaning that to each state of the physical system (e.g. $|p\rangle\in\mathcal{H}$ is a state of particle having momentum exactly $p$) there exists a field and vice versa. This is included in the axioms as $$ Y(a, z)|0\rangle|_{z=0} = a \quad\text{(field to state)} \quad\text{and}\quad Y(\cdot,z)\quad \text{(state to field)}.$$

Note, however, that a Segal CFT can encode both: the full CFT and the chiral parts, whereas a VOA encodes just a chiral part and one has to patch-up two VOAs to produce a full field algebra.

More precise answers:

  1. Just add some formal variable $z$. Formal means that nothing about $z$ is implied, it's just there for bookkeeping purposes. You can of course say that $z\in \mathbb{C}$, but you don't have to. The extra $z$ as opposed to pair of pants, has to do with Schroedinger and Heisenberg approaches to evolution in quantum mechanics: whereas Segal CFT is a functorial QFT (FQFT), meaning that states of the system (vectors in the Hilbert/vector space) evolve and the operators acting on them don't, VOAs are in algebraic QFT (AQFT)- the states don't change, but the operators do.

    Moreover, according to nLab:

due to (Huang 91) it is now known that vertex operator algebras have equivalently an FQFT-type characterization which manifestly captures this relation to $n$-point functions in the 2d CFT:

There is a monoidal category or operad whose morphisms are conformal spheres with $n$-punctures marked as incoming and one puncture marked as outgoing (each puncture equipped with a conformally parametrized annular neighborhood). Composition of such spheres is by gluing along punctures. This can be regarded as a category $2Cob_0^{conf}$ of 2-dimensional genus-0 conformal cobordisms.

As shown by theorems by Yi-Zhi Huang and Liang Kong, a vertex operator algebra is precisely a holomorphic representation of this category, or algebra over an operad for this operad, i.e. a genus-0 conformal FQFT, hence a monoidal functor

$$ V:2Cob_0^{conf}\to Vect $$ such that its component $V_1$ is a holomorphic function on the moduli space of conformal punctured spheres.

  1. Due to the above nLab reference I think that we cannot.

  2. Certainly not :)


Simplest pair of pants explanation,

not all VOAs are rational, e.g. from Abe, Buhl, Dong "Rationality, regularity, and $C2$-cofiniteness":

Definition 2.6. A vertex operator algebra is called rational if every admissible module is a direct sum of simple admissible modules.

Intro to "On Axiomatic Approaches to Vertex Operator Algebras and Modules" of Frenkel, Huang, Lepowsky mentions VOA similarities to Lie algebras and commutative associative algebras. Also, see Schottenloher "A Mathematical Intro to CFT" pp. 191-192 for connections with commutative associative algebras or Section 1.4 of Kac "Vertex algebras for beginners".

  • 1
    $\begingroup$ Thank you very much for your interest in this old question! I was losing the hope for an answer :) It is very useful and also thank you for the great amount of references! $\endgroup$
    – User3773
    Commented Mar 29, 2016 at 13:05
  • $\begingroup$ You're welcome! VOAs are a big part of my (almost finished) master's thesis, their relationship to general QFTs too. I thought that I could be of help and that I would like to summarize some stuff for myself too while the material is still fresh in my head. $\endgroup$
    – Gytis
    Commented Mar 29, 2016 at 13:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .