Ring of polynomial functions
In mathematics, the ring of polynomial functions on a vector space V over an infinite field k gives a coordinate-free analog of a polynomial ring. It is denoted by k[V]. If V has finite dimension 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 is a polynomial ring, then we can view as coordinate functions on ; i.e., when This suggests the following: given a vector space V, let k[V] be the subring generated by the dual space of the ring of all functions . If we fix a basis for V and write for its dual basis, then k[V] consists of polynomials in ; it is a polynomial ring.
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.
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 denote the vector space of multilinear functionals that are symmetric; is the same for all permutations of 's.
Any λ in gives rise to a homogeneous polynomial function f of degree q: we just let To see that f is a polynomial function, choose a basis of V and its dual. Then
which implies f is a polynomial in ti'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 are symmetric in . Let
Clearly, φ ∘ ψ is the identity; in particular, φ is surjective. To see φ is injective, suppose φ(λ) = 0. Consider
which is zero. The coefficient of t1t2 … tq in the above expression is q! times λ(v1, …, vq); 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 gn(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 Py of k[V]. It is called the polarization operator. We then have, as promised:
Proof: We first note that (Py f) (x) is the coefficient of t in f(x + t y); in other words, since g0(x, y) = g0(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.
Operator product algebra
When the polynomials are valued not over a field k, but instead are valued 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).[clarification needed] In this case, one may impose an additional axiom.
The operator product algebra is an associative algebra of the form
The structure constants are required to be single-valued functions, rather than sections of some vector bundle. The fields (or operators) 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 . 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.
- Kobayashi, S.; Nomizu, K. (1963), Foundations of Differential Geometry, Vol. 2 (new ed.), Wiley-Interscience (published 2004).