Ring of polynomial functions

In mathematics, the ring of polynomial functions on a vector space "V" over a field "k" gives a coordinate-free analog of a polynomial ring. It is denoted by "k"["V"]. If "V" is finite dimensional and is viewed as an algebraic variety, then "k"["V"] is precisely the coordinate ring of "V".
The explicit definition of the ring can be given as follows. If formula_1 is a polynomial ring, then we can view formula_2 as coordinate functions on formula_3; i.e., formula_4 when formula_5 This suggests the following: given a vector space "V", let "k"["V"] be the commutative "k"-algebra generated by the dual space formula_6, which is a subring of the ring of all functions formula_7. If we fix a basis for "V" and write formula_2 for its dual basis, then "k"["V"] consists of polynomials in formula_2.
If "k" is infinite, then "k"["V"] is the symmetric algebra of the dual space formula_6.
In applications, one also defines "k"["V"] when "V" is defined over some subfield of "k" (e.g., "k" is the complex field and "V" is a real vector space.) The same definition still applies.
Throughout the article, for simplicity, the base field "k" is assumed to be infinite.
Relation with polynomial ring.
Let formula_11 be the set of all polynomials over a field "K" and "B" be the set of all polynomial functions in one variable over "K". Both "A" and "B" are algebras over "K" given by the standard multiplication and addition of polynomials and functions. We can map each formula_12 in "A" to formula_13 in "B" by the rule formula_14. A routine check shows that the mapping formula_15 is a homomorphism of the algebras "A" and "B". This homomorphism is an isomorphism if and only if "K" is an infinite field. For example, if "K" is a finite field then let formula_16. "p" is a nonzero polynomial in "K"["x"], however formula_17 for all "t" in "K", so formula_18 is the zero function and our homomorphism is not an isomorphism (and, actually, the algebras are not isomorphic, since the algebra of polynomials is infinite while that of polynomial functions is finite).
If "K" is infinite then choose a polynomial "f" such that formula_19. We want to show this implies that formula_20. Let formula_21 and let formula_22 be "n"+1 distinct elements of "K". Then formula_23 for formula_24 and by Lagrange interpolation we have formula_20. Hence the mapping formula_15 is injective. Since this mapping is clearly surjective, it is bijective and thus an algebra isomorphism of "A" and "B".
Symmetric multilinear maps.
Let "k" be an infinite field of characteristic zero (or at least very large) and "V" a finite-dimensional vector space.
Let formula_27 denote the vector space of multilinear functionals formula_28 that are symmetric; formula_29 is the same for all permutations of formula_30's.
Any λ in formula_27 gives rise to a homogeneous polynomial function "f" of degree "q": we just let formula_32 To see that "f" is a polynomial function, choose a basis formula_33 of "V" and formula_2 its dual. Then
which implies "f" is a polynomial in the "t""i"'s.
Thus, there is a well-defined linear map:
We show it is an isomorphism. Choosing a basis as before, any homogeneous polynomial function "f" of degree "q" can be written as:
where formula_38 are symmetric in formula_39. Let
Clearly, formula_41 is the identity; in particular, φ is surjective. To see φ is injective, suppose φ(λ) = 0. Consider
which is zero. The coefficient of "t"1"t"2 … "t""q" in the above expression is "q"! times λ("v"1, …, "v""q"); it follows that λ = 0.
Note: φ is independent of a choice of basis; so the above proof shows that ψ is also independent of a basis, the fact not "a priori" obvious.
Example: A bilinear functional gives rise to a quadratic form in a unique way and any quadratic form arises in this way.
Taylor series expansion.
Given a smooth function, locally, one can get a partial derivative of the function from its Taylor series expansion and, conversely, one can recover the function from the series expansion. This fact continues to hold for polynomials functions on a vector space. If "f" is in "k"["V"], then we write: for "x", "y" in "V",
where "g""n"(x, y) are homogeneous of degree "n" in "y", and only finitely many of them are nonzero. We then let
resulting in the linear endomorphism "P""y" of "k"["V"]. It is called the polarization operator. We then have, as promised:
Proof: We first note that ("P""y" "f") ("x") is the coefficient of "t" in "f"("x" + "t" "y"); in other words, since "g"0("x", "y") = "g"0("x", 0) = "f"("x"),
where the right-hand side is, by definition,
The theorem follows from this. For example, for "n" = 2, we have:
The general case is similar. formula_48
Operator product algebra.
When the polynomials are valued not over a field "k", but over some algebra, then one may define additional structure. Thus, for example, one may consider the ring of functions over "GL(n,m)", instead of for "k = GL(1,m)". In this case, one may impose an additional axiom.
The operator product algebra is an associative algebra of the form
The structure constants formula_50 are required to be single-valued functions, rather than sections of some vector bundle. The fields (or operators) formula_51 are required to span the ring of functions. In practical calculations, it is usually required that the sums be analytic within some radius of convergence; typically with a radius of convergence of formula_52. Thus, the ring of functions can be taken to be the ring of polynomial functions.
The above can be considered to be an additional requirement imposed on the ring; it is sometimes called the "bootstrap". In physics, a special case of the operator product algebra is known as the operator product expansion.