What does $Y(1,z)$ = id for vertex algebras mean?

Question

I'm adding an update to this post here with my current understanding of the situation for context. I read some Wikipedia articles and two texts. I am having some trouble so I figure I would attempt to reverse engineer this stuff and ask questions. For the original post please see below the line.

It seems that the formal power series is something of a tool for identifying numbers. For example it seems $3x^0 + 2x + 4x^2 + 5x^3$, holds the numbers $[3,2,4,5]$. So $[x](3x^0 + 2x + 4x^2 + 5x^3) = 2$ and $[x^2](3x^0 + 2x + 4x^2 + 5x^3) = 4$ for example.

Guesses:

From formal series stuff.

There is a zero element [. . . .0,0,0. . . ]

There is a multiplicative identity [1,0,0. . . .]

Ok

From the above, I want to believe that Y(x,z) is a notation for a formal series with some additional structure.

So $Y(x,z) = \sum_{ n \in \mathbb{Z}} x_n z^{-n-1}$

then $Y(x,z) = . . . x_{-1} z^0 + x_{0} z^{-1} + x_{1}z^{-2} ...$

From formal power series I want to guess that we are holding $[x_{-1}, x_{0}, x_{1}, ...]$

If one takes "1" as a polynomial to be the vacuum

then it seems $Y(1,z) = 1z^0 + 0z^1 +0z^2 . . . . .$

so [...0,0,0,1,0,0,0... ]

If any of this is true then "1" has subscripts then $1_n$

For example $1_{-1} 1_{-2}$ = 0 Is this true?

From $Y(1,z) = 1z^0 + 0z^1 +0z^2 . . . . .$, It would seem that $Y(1,z)a =1z^0 a = a

These are more or less guesses as the material was a bit abstract so I am constructing fleshed-out guesses as to what the authors were trying to say

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I'm about to embarrass myself, get popcorn, and text your friends!

Ok, so I have been looking into vertex operator algebras in my spare time.

I tried to convince myself that $Y(1,z)$ is indeed the identity in which case $Y(1,z)x_b$ I want to believe should give us back $x_b$. This could be true and sounds like a simple proposition.

$V = \mathbb{C}[x_{-1}, x_{-2}, . . . .]$ . I think V is a good place to start. It is the space of polynomials in those variables, so it should necessarily include $1$ it seems.

$Y(x,z) = \sum_{ n \in \mathbb{Z}} x_n z^{-n-1}$

setting the below as people do:

$a_{-} = \sum_{n\geq 0} x_n z^{-n -1}$

$a_{+} = \sum_{n < 0} x_n z^{-n -1}$

and then interpreted the definition of $Y(x,z)$ to mean:

$((x_0 z^{-1} + x_1 z^{-2} + x_2 z^{-3} . . . . )+ (x_{-1} z^0 + x_{-2}z^1 + x_{-3}z^{2}. . . . ))x_b$

It seems from a reference I am reading that.

$Y(x_{-1},z) = \sum x_n z^{-n -1}$

$Y(x_{-2},z) = \sum x_n (-n-1) z^{-n -2}$

and so on

I am having trouble directly showing that a $Y(x_i,z) = id$ exists.

I can't write down/ pick $x_i$ such that a $Y(x_i,z) = 1$.

Does fixing the vacuum state literally mean just picking an $x_i$ in V and setting that to 1, then saying that T(1) = 0? where T here is the translation of 1.

Page 35 of this reference in my understanding seems to say we pick an element of V and make it the vacuum, then say the translation which they call D is 0.

Answer 1

**This is an attempt at an answer. I hope it gets edited by someone.

This first part hopefully should just work for the commutative case.

Let there be a formal series in variable z with coefficients in a vector space V.

The notation goes something like $v[[z,z^-1]] = \Sigma_{n \in Z} v_n z^{-n-1}| v \in V$.

Let $A(z)$ be a field $A(z) := \Sigma_{n \in \mathbb{Z}} A_n z^{-n-1}$ with $A_n \in \mathbb{C}$

Let $B(z)$ be another field $B(z):= \Sigma_{n \in \mathbb{Z}} B_n z^{-n-1}$ with $B_n \in \mathbb{C}$

It would seem that the Cauchy product of A and B can be taken since the fields are defined by formal series.

If $I(z)$ is a special field called the identity field, it would seem that the Cauchy product of this field with another field should be that field.

$A(z)I(z) = A(z)$

$B(z)I(z) = B(z)$

In Mathematica

 G:=Sum[A[n],{n,4}]

 F:=Sum[i[n],{n,4}]

 G*F // Expand 

 output:  A[1] i[1]+A[2] i[1]+A[3] i[1]+A[4] i[1]+A[1] i[2]+A[2] 
  i[2]+A[3] i[2]+A[4] i[2]+A[1] i[3]+A[2] i[3]+A[3] i[3]+A[4] 
  i[3]+A[1] 
  i[4]+A[2] i[4]+A[3] i[4]+A[4] i[4] 

A[1] i[1]+A[2] i[1]+A[3] i[1]+A[4] i[1]+A[1] i[2]+A[2] i[2]+A[3] 
i[2]+A[4] i[2]+A[1] i[3]+A[2] i[3]+A[3] i[3]+A[4] i[3]+A[1] i[4]+A[2] 
i[4]+A[3] i[4]+A[4] i[4] /. i[1] -> 1 /. i[2] -> 0 /. i[3] -> 0 /. i[4] - 
> 0

The output is A[1]+A[2]+A[3]+A[4] 

What this tells me is that $I(z) := I_0 z^{-1} + I_1 z^{-2} + I_2 z^{-3} . . $ must be of the form $I(z) := 1 z^{-1} + 0 z^{-2} + 0 z^{-3} . .$

Basically the pattern is something like $I_n = [1, 0, 0, 0, . . . . .]$ where $I_n$ are coefficients in $I(z)$

This should define the identity.

It is left to guess 'a' such that Y(a,z) = I(z)

'a' is the vacuum

'a' seems to be 1

For the second part, I want to try the normally ordered product to deduce the identity field. This should technique should presumably also work for non-commutative cases.

The normally ordered product goes like $A(z)_{-1}B(z)= :A(z)B(z): = A(z)_{-} B(z) + B(z)A(z)_{+}$

Mathematica code to test show this is below:

 Aplus:=Sum[A[n]*z^(-n-1),{n,-3,-1}]

 Aminus:=Sum[A[n]*z^(-n-1),{n,0,2}]

 Bplus:=Sum[B[n]*z^(-n-1),{n,-3,-1}]

 Bminus:=Sum[B[n]*z^(-n-1),{n,0,2}]

 M:= Sum[B[n]*z^(-n-1),{n,-3,2}]

 Aminus*M + M*Aplus //Expand
  
 Then I did this to the result 

This means

/. B[0] -> 0    /. B[1] -> 0/. B[2] -> 0 /.B[-1] -> 1 /. B[-2]->0 
/.B[-3] -> 0 specifies the identity field.