Bill Cook

To understand how this stuff works you might want to get a copy of Li and Lepowsky's book: Introduction to Vertex Operator Algebras and Their Representations. It lays careful formal variable stuff out in its first chapter.

You are mixing symbols. Y(v,z) is a generating function where v_n (the coefficient of z^(-n-1)) is the nth mode of v. There isn't a vector "x" whose modes are the indeterminates x_1 etc.

Vertex algebras resemble Lie algebras in some ways and resemble unital commutative associative algebras in other ways. But these two kinds of structure are not really directly compatible - vertex algebras carefully weave the structures so they can live in a kind of uneasy balance together. This is why looking at them you might see Lie algebra stuff going on. If you look more you'll also see associative algebra stuff going on.

The choice of acting on the polynomial algebra comes from this family of constructions. Historically, certain representations of Virasoro, Heisenberg, and affine Lie algebras were built from "vertex operators" which were essentially generating functions encoding the actions. It was later realized that these vertex operators could be multiplied and these representations built from vertex operators had an amazing structure later dubbed a "vertex algebra".

For certain Lie algebras, some of their representations have the structure of a vertex algebra and so, yes, the representation has to be chosen carefully. Yes, infinite dimensionality is nearly non-negotiable requirement (except for a "trivial" class of examples).

If you are looking for a finite dimensional concrete vertex algebra, you're going to be disappointed. It is not too hard to show that any finite dimensional vertex algebra is just a finite dimensional commutative associative algebra equipped with a derivation. That's part of what makes learning about vertex algebras difficult - there are no "non-trivial" finite dimensional examples.

The Heisenberg vertex algebra is built from the infinite dimensional Heisenberg algebra. There are similar constructions associated with the Virasoro Lie algebra and affine Lie algebras. There is something very special about each of these Lie algebras that allow for the construction to work out. You can't just replace them with some finite dimensional Lie algebra and go from there.

To understand how this stuff works you might want to get a copy of Li and Lepowsky's book: Introduction to Vertex Operator Algebras and Their Representations. It lays careful formal variable stuff out in its first chapter.

You are mixing symbols. Y(v,z) is a generating function where v_n (the coefficient of z^(-n-1)) is the nth mode of v. There isn't a vector "x" whose modes are the indeterminates x_1 etc.

Vertex algebras resemble Lie algebras in some ways and resemble unital commutative associative algebras in other ways. But these two kinds of structure are not really directly compatible - vertex algebras carefully weave the structures so they can live in a kind of uneasy balance together. This is why looking at them you might see Lie algebra stuff going on. If you look more you'll also see associative algebra stuff going on.

The choice of acting on the polynomial algebra comes from this family of constructions. Historically, certain representations of Virasoro, Heisenberg, and affine Lie algebras were built from "vertex operators" which were essentially generating functions encoding the actions. It was later realized that these vertex operators could be multiplied and these representations built from vertex operators had an amazing structure later dubbed a "vertex algebra".

For certain Lie algebras, some of their representations have the structure of a vertex algebra and so, yes, the representation has to be chosen carefully. Yes, infinite dimensionality is nearly non-negotiable requirement (except for a "trivial" class of examples).

If you are looking for a finite dimensional concrete vertex algebra, you're going to be disappointed. It is not too hard to show that any finite dimensional vertex algebra is just a finite dimensional commutative associative algebra equipped with a derivation. That's part of what makes learning about vertex algebras difficult - there are no "non-trivial" finite dimensional examples.

The Heisenberg vertex algebra is built from the infinite dimensional Heisenberg algebra. There are similar constructions associated with the Virasoro Lie algebra and affine Lie algebras. There is something very special about each of these Lie algebras that allow for the construction to work out. You can't just replace them with some finite dimensional Lie algebra and go from there.