Vertex algebras done right

Vertex algebras are a somewhat mysterious type of algebraic structure. Their foundational ideas come from conformal field theory in physics, while they also have seen deep connections to various areas of mathematics.

When I first started learning about vertex algebras for my undergraduate thesis, I found them quite baffling. Their definition was technical, and the standard references I looked at, such as [Frenkel, Ben-Zvi 2004] or [Kac 1998] (titled ‘for beginners’!), did not help much in the intuition aspect.

Now, having worked with these weird objects for some time, I have managed to understand them better. Still, I am very surprised that the simple intuitive ideas behind them, which are well-known to experts, are not easily accessible — in fact, I doubt that they exist in the literature (or anywhere) at all, in an explicit form.

Therefore, I decided to write this note to explain these ideas. I hope this will be helpful to others who are also learning the subject and feeling confused as I was years ago.

Idea§

In a nutshell, a vertex algebra is a two-dimensional algebraic structure.

We think of a usual associative algebra as a one-dimensional structure, as living on a line $\mathbb{R}$, in the sense that given several points on the line, if we assign an element of the algebra to each point, we can multiply them in the order given by the line. The product does not change when we move the points around, as long as they do not cross over each other. This is illustrated on the left in the picture below.

On the other hand, in a vertex algebra, we multiply elements on the complex plane. This means that given several points $z_1, \dotsc, z_n \in \mathbb{C}$, if we assign an element $a_i$ to each point $z_i$, we get a product $a_1 (z_1) \cdots a_n (z_n)$.

The main feature that characterizes vertex algebras is that the product $a_1 (z_1) \cdots a_n (z_n)$ depends meromorphically on the variables $z_i$. Moreover, the product can only have poles when $z_i = z_j$ for some $i \neq j$, that is, when two of the points collide.

The new definition§

Here is a new definition of a vertex algebra, which is, in my opinion, more enlightening than the standard one, which I will discuss later.

The associativity axiom might look complicated, but it is actually just the usual associativity of multiplication, with a bunch of $z$ and $w$ variables thrown in.

This more straightforward definition was hinted at in [Borcherds 1998], but not fully spelled out. A more rigorous treatment was done by [Kim 2011], who, strangely, did not state his result in terms of vertex algebras. I only learned about this from an unpublished paper by Dominic Joyce, and as of this writing, I am not aware of any published reference on this precise version of the definition.

Fields and translation§

If we look at the above definition more carefully, some useful information can be extracted. Here, I will focus on explaining the facts, while leaving out some details.

First, since the $1$-ary multiplication is defined, it means that every element $a \in V$ automatically comes with a power series $a (z) \in V [[z]]$, which some people would call the field associated to $a$ (while others wouldn’t!). The value at $z = 0$ is the element itself, $a (0) = a$.

It then makes sense to differentiate the field $a (z)$ with respect to $z$, which defines an element $\partial_z a (z) \in V [[z]]$. We denote this differential at $z = 0$ by

$$ T a = \partial_z a (z) \, \bigr|_{z = 0} \in V \ . $$

We also have that $(T a) (z) = \partial_z a (z)$, which justifies that it is reasonable to also call $\partial_z a (z)$ a field. We can also take higher derivatives of $a (z)$ similarly.

This operator $T \colon V \to V$ is called the translation operator, which I find a bit misleading. In fact, the actual translation operator is the operator

$$ \exp (z T) = \sum_{n = 0}^\infty \frac{z^n}{n!} T^n \colon V \longrightarrow V [[z]] \ . $$

Let me elaborate on this. There is the identity $\exp (z \partial_z) \, a (w) = a (z + w)$, which is a formal fact and does not involve anything about vertex algebras. This implies that

$$ \exp (z T) \, a (w) = a (z + w) \ , $$

which is the true reason for the name. In particular, taking $w = 0$ gives

$$ a (z) = \exp (z T) \, a \ , $$

which can be seen as an alternative definition of the field $a (z)$, and will be useful in understanding the old definition of vertex algebras later.

The old definition§

Let us now see how the standard definition of a vertex algebra fits into this new picture. The standard definitions differ slightly between different sources, but they all look something like the following, taken from [Frenkel, Ben-Zvi 2004].

To get this from the new picture, the translation operator $T$ is as in the previous section, and the product is defined as

$$ Y (a, z) \, b = a (z) \cdot b (0) \ . $$

The element $Y (a, z) \, 1 \in V [[z]]$ is what we called $a (z)$ before, and more generally, the product $a_1 (z_1) \cdots a_n (z_n)$ in the previous language corresponds to $Y (a_1, z_1) \cdots Y (a_n, z_n) \, 1$. More precisely, expanding $(z_i - z_j)^{-1}$ as a power series in the former product, in the limit $|z_j| \ll |z_i|$ when $i < j$, gives the latter product.

The translation axiom here is just another way to say that $T$ acts as the derivative operator on fields.

Finally, the locality axiom can seem very strange. In fact, in our new language, it is just the statement that $a (z) \cdot b (w) = b (w) \cdot a (z)$, but it is stated in a more convoluted way, because of the inconvenience that in this old definition, one has to choose an order to expand the power series. More precisely, in the products $Y (a, z) \, Y (b, w) \, c$ and $Y (b, w) \, Y (a, z) \, c$, the first is expanded in the limit $|w| \ll |z|$, while the second in the limit $|z| \ll |w|$, and hence they might not be equal, although they are both expansions of the same expression. The reason is that something like the following can happen:

$$ \sum_{n = 0}^\infty \frac{w^n}{z^{n + 1}} \sim \frac{1}{z - w} = -\frac{1}{w - z} \sim -\sum_{n = 0}^\infty \frac{z^n}{w^{n + 1}} \ , $$

where the leftmost and rightmost terms give different expansions of the same expression. However, in this particular instance, after multiplying by $(z - w)$, both sides become equal to $1$. In general, one might need a higher power $(z - w)^n$, but this is always possible, and the power $n$ we need is the order of the pole at $z = w$ of the original expression.

An example§

In my experience, the new definition is not only helpful for intuition, it is also often easier to work with in practice.

Just as an example, there is a property of vertex algebras, which is sometimes called skew-symmetry (for example, in [Frenkel, Ben-Zvi 2004: Proposition 3.2.5]), and refers to the identity

$$ Y (a, z) \, b = \exp (z T) \, Y (b, -z) \, a \ . $$

I find this hard to make sense of in the old language, but from the new perspective, this is nothing but the identity

$$ a (z) \cdot b (0) = b (0) \cdot a (z) = \bigl( b (-z) \cdot a (0) \bigr) (z) \ , $$

where we remember from the associativity axiom that when variables are combined, they get added together, which is what we did in the last step.

References§

• Borcherds, R. E. (1998). Vertex algebras.
  Topological field theory, primitive forms and related topics, 35–77. Birkhäuser.
  (zbMATH) (arXiv)

• Frenkel, E., and Ben-Zvi, D. (2004). Vertex algebras and algebraic curves, 2nd ed.
  Mathematical Surveys and Monographs 88. American Mathematical Society.
  (doi) (zbMATH)

• Kac, V. (1998). Vertex algebras for beginners, 2nd ed.
  University Lecture Series 10. American Mathematical Society.
  (doi) (zbMATH)

• Kim, N. (2011). Associativity of field algebras.
  Annales Henri Poincaré 12 (6), 1145–1168.
  (doi) (zbMATH)

See also:

• nLab: Vertex operator algebra.

• Wikipedia: Vertex operator algebra.