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

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.
 A: **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.

