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) linear 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 $e_{i},\,1\leq i\leq n$ 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})

.

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_{0}\cdots i_{q}}t_{i_{1}}\cdots t_{i_{q}}

where $a_{i_{0}\cdots i_{q}}$ are symmetric in $i_{0},\dots ,i_{q}$ . Let

\psi (f)(v_{1},\dots ,v_{q})=\sum _{i_{1},\cdots ,i_{q}=1}^{n}a_{i_{0}\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.

Notes

^ 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).

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[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.

[1]

Symmetric multilinear maps

See also

Notes

References