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 $k[t_1, \dots, t_n]$ is a polynomial ring, then we can view $t_{i}$ as coordinate functions on $k^{n}$; i.e., $t_{i}(x)=x_{i}$ when $x=(x_{1},\dots ,x_{n}).$ This suggests the following: given a vector space V, let k[V] be the subring generated by the dual space $V^{*}$ of the ring of all functions $V\to k$. If we fix a basis for V and write $t_{i}$ for its dual basis, then k[V] consists of polynomials in $t_{i}$; 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 $S^{q}(V)$ denote the vector space of multilinear functionals $\textstyle \lambda :\prod _{1}^{q}V\to k$ that are symmetric; $\lambda (v_{1},\dots ,v_{q})$ is the same for all permutations of $v_{i}$'s.

Any λ in $S^{q}(V)$ gives rise to a homogeneous polynomial function f of degree q: we just let $f(v)=\lambda (v,\dots ,v).$ To see that f is a polynomial function, choose a basis $e_{i},\,1\leq i\leq n$ of V and $t_{i}$ its dual. Then

$\lambda (v_{1},\dots ,v_{q})=\sum _{{i_{1},\dots ,i_{q}=1}}^{n}\lambda (e_{{i_{1}}},\dots ,e_{{i_{q}}})t_{{i_{1}}}(v_{1})\cdots t_{{i_{q}}}(v_{q})$,

which implies f is a polynomial in t_i's.

Thus, there is a well-defined linear map:

$\phi :S^{q}(V)\to k[V]_{q},\,\phi (\lambda )(v)=\lambda (v,\cdots ,v).$

We show it is an isomorphism. Choosing a basis as before, any homogeneous polynomial function f of degree q can be written as:

$f = \sum_{i_1, \dots, i_q = 1}^n a_{i_1 \cdots i_q} t_{i_1} \cdots t_{i_q}$

where $a_{i_1 \cdots i_q}$ are symmetric in $i_1, \dots, i_q$. Let

$\psi(f)(v_1, \dots, v_q) = \sum_{i_1, \cdots, i_q = 1}^n a_{i_1 \cdots i_q} t_{i_1}(v_1) \cdots t_{i_q}(v_q).$

Clearly, φ ∘ ψ is the identity; in particular, φ is surjective. To see φ is injective, suppose φ(λ) = 0. Consider

${\displaystyle \phi (\lambda )(t_{1}v_{1}+\cdots +t_{q}v_{q})=\lambda (t_{1}v_{1}+\cdots +t_{q}v_{q},...,t_{1}v_{1}+\cdots +t_{q}v_{q})}$,

which is zero. The coefficient of t₁t₂ … t_q in the above expression is q! times λ(v₁, …, 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,

${\displaystyle f(x+y)=\sum _{n=0}^{\infty }g_{n}(x,y)}$

where g_n(x, y) are homogeneous of degree n in y and only finitely many of them are nonzero. We then let

${\displaystyle (P_{y}f)(x)=g_{1}(x,y),}$

resulting in the linear endomorphism P_y of k[V]. It is called the polarization operator. We then have, as promised:

Theorem — For each f in k[V] and x, y in V,

${\displaystyle f(x+y)=\sum _{n=0}^{\infty }{1 \over n!}P_{y}^{n}f(x)}$.

Proof: We first note that (P_y f) (x) is the coefficient of t in f(x + t y); in other words, since g₀(x, y) = g₀(x, 0) = f(x),

${\displaystyle P_{y}f(x)=\left.{d \over dt}\right|_{t=0}f(x+ty)}$

where the right-hand side is, by definition,

${\displaystyle \left.{f(x+ty)-f(x) \over t}\right|_{t=0}.}$

The theorem follows from this. For example, for n = 2, we have:

${\displaystyle P_{y}^{2}f(x)=\left.{\partial \over \partial t_{1}}\right|_{t_{1}=0}P_{y}f(x+t_{1}y)=\left.{\partial \over \partial t_{1}}\right|_{t_{1}=0}\left.{\partial \over \partial t_{2}}\right|_{t_{2}=0}f(x+(t_{1}+t_{2})y)=2!g_{2}(x,y).}$

The general case is similar. $\square$

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). In this case, one may impose an additional axiom.

The operator product algebra is an associative algebra of the form

$A^i(x)B^j(y) = \sum_k f^{ij}_k (x,y,z) C^k(z)$

The structure constants $f^{ij}_k (x,y,z)$ are required to be single-valued functions, rather than sections of some vector bundle. The fields (or operators) $A^i(x)$ 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 $|x-y|$. 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.

See also

• Algebraic geometry of projective spaces
• Polynomial ring
• Symmetric algebra
• Zariski tangent space

Notes

References

• Kobayashi, S.; Nomizu, K. (1963), Foundations of Differential Geometry, Vol. 2 (new ed.), Wiley-Interscience (published 2004).