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 $S^q(V)$ 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: let $f(v) = \lambda(v, \dots, v).$ To see that f is a polynomial function, choose a basis $e_i, \, 1 \le i \le n$ of V and $t_i$ its dual. Then

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

Thus, there is a well-defined linear map:

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

It is an isomorphism:^[1] 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

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

Then ψ is the inverse of φ. (Note: φ is still independent of a choice of basis; so ψ is also independent of a basis.)

Example: A bilinear functional gives rise to a quadratic form in a unique way and any quadratic form arises in this way.

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

1. There is also a more abstract way to see this: to give a multilinear functional on the product of q copies of V is the same as to give a linear functional on the q-th tensor power of V. The requirement that the multilinear functional to be symmetric translates to the one that the linear functional on the tensor power factors through the q-th symmetric power of V, which is isomorphic to k[V]_q.

References

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