## Summer School: Day 1

As I mentioned, I’m participating in a summer school on the Geometry of Quantum Fields at Penn.  I’m in Katrin Wendland’s mentoring session this week, which means conformal fields and vertex algebras.

First off, we talked about formal distributions.  The idea being to algebraicize quantum field theory, which runs on operator valued Schwartz distributions.  Now, to me, that sounds sufficiently awful that I want it algebraicized so I never need to think those words again.  (Well, actually, I’m ok with them at this point, due to a REALLY hardcore first year real analysis course.)

Let $R$ be a $\mathbb{C}$-algebra, and set $A(z_1,\ldots,z_n)=\sum_{i_1,\ldots,i_n\in\mathbb{Z}}A_{i_1,\ldots,i_n}z_1^{i_1}\ldots z_n^{i_n}$, where each of the $A_{i_1,\ldots,i_n}\in R$.  So $A\in R[[z_1^{\pm 1},\ldots,z_n^{\pm 1}]]$.

Now, this space is merely a vector space.  The product turns out to not be well defined unless we start thinking about convergence, and it’s best that we don’t.  However, we can multiply things in this by Laurent polynomials.  So each element of this vector space gives us a linear functional on $\mathbb{C}[z_1^{\pm 1},\ldots, z_n^{\pm 1}]$.

Now, fix $R=\mathbb{C}$ and look at $\mathbb{C}[z,z^{-1}]$.  We define the functional by $f\in \mathbb{C}[[z,z^{-1}]]$ acts on $g\in \mathbb{C}[z,z^{-1}]$ by taking $g$ to $\mathrm{Res}_{z=0} f(z)g(z)dz$.

With this in mind, we can give an algebraic version of the Dirac delta.  Define $\delta(z-w)=\sum_{m\in\mathbb{Z}}z^mw^{-m-1}$.  To see that this is the $\delta$ that we know and love from physics (where we claim it’s a function) and analysis (where we know better), have it act on $\sum_k A_kw^k$.  Then we take the product, which is $\sum_k A_kw^k\sum_m z^mw^{-m-1}$, and we take the coefficient of $w^{-1}$.  This gives us $\sum_k A_k z^k$, so $\delta(z-w)$ evaluates series in $z$ at $w$ and vice versa.

Now, taking $z$ as our polynomial, we get that $w\delta(z-w)=z\delta(z-w)$.  And so, we get the equation $(z-w)\delta(z-w)=0$.  So the delta function is supported on the diagonal, as we would want.

Now, we can break $\delta(z-w)$ into $\delta_-(z-w)=\frac{1}{z}\sum_{n\geq 0} \left(\frac{w}{z}\right)^n$ and $\delta_+=\frac{1}{z}\sum_{n>0}\left(\frac{z}{w}\right)^n$, and have $\delta(z-w)=\delta_-(z-w)+\delta_+(z-w)$.  So, analytically, when $|z|>|w|$, we have $\delta_-=\frac{1}{z-w}$, and when $|w|<|z|$, we have $-\frac{1}{z-w}$.  So pretending these representations are valid in both domains, we can see that $\delta$ is supported on the diagonal, because they cancel everywhere else.

Now, these give us some nice maps, $\mathbb{C}((z,w))$ to $\mathbb{C}((z))((w))$ and $\mathbb{C}((w))((z))$, the first by $\delta_-$ and the second by $\delta_+$.  So we want to be able to compare the images, so we can take the difference, and the things we get this way are our formal distributions.

So next up, we try to define what a field is.  Let $V$ be any $\mathbb{C}$-vector space.  Take $A(z)=\sum_j A_j z^{-j}\in \mathrm{End}(V)[[z,z^{-1}]]$.  We’ll call it a field if for every vector $v\in V$, there exists an $N$ that for all $j>N$, we have $A_j(v)=0$.  The idea here is that the $A_j$ are creation and annihilation operators, with the annihilation operators being of higher index.  So every “state” is killed by sufficiently annihilating operators.

So now, for any $v$, we have $A(z)v\in V((z))$, the formal Laurent series with vector coefficients.  For future use, we define $\mathscr{F}(V)$ to be the space of all fields on $V$.

If $V$ is a $\mathbb{Z}$-graded vector space, and each $A_j$ is a degree $\Delta-j$ map, for some $\Delta$, we say that our field is a homogeneous field of conformal dimension $\Delta$.  Then differentiation in $z$ increases the dimension by one.

Now, we want to figure out what it means for two fields to be mutually local.  For that, we need to look at $v\in V$ and $\phi\in V^*$.  Then take two fields, $A(z)$ and $B(w)$.  Then we look at $\langle \phi, A(z)B(w)v\rangle$ and $\phi,B(w)A(z)v\rangle$.  Their difference is denoted by $\langle \phi, [A(z),B(w)]v\rangle$.  We say that a pair of fields are mutually local if there is some $N$ such that the previous object is annihilated by $(z-w)^N$.  So we’re insisting that the commutator be supported on the diagonal.  Equivalently, the mutually local distributions have their difference in the space $\mathbb{C}[[z,w]][z^{-1},w^{-1},(z-w)^{-1}]$.

So finally, we finish up the day with a definition:

Definition: A vertex algebra consists of the following data:

1. A “State Space” $V$ which is a $\mathbb{Z}_{\geq 0}$-graded vector space.
2. Translation Operator: a map $T:V\to V$ of degree 1.
3. A distinguished vacuum vector: $|0\rangle$.
4. A State-Field correspondence: $Y(-,z):V\to \mathrm{End}(V)[[z^{\pm 1}]]$

The data must satisfy the following axioms:

1. The Vacuum Axioms: $Y(|0\rangle, z)=\mathrm{id}_V$, and $Y(A,z)|0\rangle|_{z=0}=A$.
2. Translation Axiom: $[T, Y(A,z)]=\partial_z Y(A,z)$ and $T|0\rangle=0$.
3. Locality Axiom: For all $A,B$, we have that $Y(A,z)$ and $Y(B,w)$ are mutually local fields.
3. No, it’s an action. Multiply a formal series by a Laurent polynomial and get back another formal series, not an element of $\mathbb{C}$.