Symmetric algebra

In, the symmetric algebra S(V) (also denoted Sym(V)) on a V over a  K is the     over K containing V.

It corresponds to polynomials with indeterminates in V, without choosing coordinates. The dual, S(V∗) corresponds to polynomials on V.

A whose  is  is also called a symmetric algebra, but is not discussed here.

Construction
It is possible to use the T(V) to describe the symmetric algebra S(V). In fact we pass from the tensor algebra to the symmetric algebra by forcing it to be commutative; if elements of V commute, then tensors in them must, so that we construct the symmetric algebra from the tensor algebra by taking the of T(V) by the  generated by the differences of products


 * $$v\otimes w - w\otimes v.$$

for all v and w in V.

In effect, S(V) is the same as the over K in indeterminates that are a  for V.

Grading
Just as with a polynomial ring, there is a decomposition of S(V) as a, into summands


 * Sk(V)

which consist of the of the s in vectors of V of degree k, for k = 0, 1, 2, ... (with S0(V) = K and S1(V) = V). The K-vector space Sk(V) is the k-th  of V. (The case k = 2, for example, is the symmetric square and denoted Sym2(V).) It has a universal property with respect to symmetric operators defined on Vk.

In terms of the tensor algebra grading, Sk(V) is the quotient space of Tk(V) by the subspace generated by all differences of products


 * $$v\otimes w - w\otimes v.$$

and products of these with other algebra elements.

Distinction from symmetric tensors
The symmetric algebra and s are easily confused: the symmetric algebra is a quotient of the tensor algebra, while the symmetric tensors are a subspace of the tensor algebra.

By expressing the symmetric algebra as a quotient, it inherits the already present in the ; that is, quotienting preserves the universal property. That is, the symmetric algebra is the "most general" symmetric algebra, in that any other symmetric algebra (defined by some symmetric product $$F(V,W)$$) is homomorphic to the symmetric algebra.

Symmetric tensors are defined as invariants: given the natural action of the on the tensor algebra, the symmetric tensors are the subspace on which the symmetric group acts trivially. Note that under the tensor product, symmetric tensors are not a subalgebra: given linearly independent vectors v and w, they are trivially symmetric 1-tensors, but v ⊗ w is not a symmetric 2-tensor.

The grade 2 part of this distinction is the difference between s (symmetric 2-tensors) and s (elements of the symmetric square), as described in s.

In characteristic 0, symmetric tensors and the symmetric algebra can be identified. In any characteristic, there is a map from the symmetric algebra to the symmetric tensors, given by:
 * $$v_1\cdots v_k \mapsto \sum_{\sigma \in S_k} v_{\sigma(1)}\otimes \cdots \otimes v_{\sigma(k)}.$$

The composition of this map with the inclusion of the symmetric tensors in the tensor algebra and the quotient to the symmetric algebra is multiplication by k! on the kth graded component of the symmetric tensors.

Thus in characteristic 0, the symmetrization map is an isomorphism of graded vector spaces, and one can identify symmetric tensors with elements of the symmetric algebra. One divides by k! to make this a of the quotient map:
 * $$v_1\cdots v_k \mapsto \frac{1}{k!} \sum_{\sigma \in S_k} v_{\sigma(1)}\otimes \cdots \otimes v_{\sigma(k)}.$$

For instance, $$vw \mapsto \frac{1}{2}(v\otimes w + w \otimes v)$$.

This is related to the of the symmetric group: in characteristic 0, over an algebraically closed field, the  is, so every representation splits into a direct sum of irreducible representations, and if T = S ⊕ V, one can identify S as either a subspace of T or as the quotient T/V.

Interpretation as polynomials
Given a vector space V, the polynomials on this space are S(V∗), the symmetric algebra of the dual space: a polynomial on a space evaluates vectors on the space, via the pairing $$S(V^*) \times V \to K$$.

For instance, given the plane with a basis {(1,0), (0,1)}, the (homogeneous) linear polynomials on K2 are generated by the coordinate s x and y. These coordinates are s: given a vector, they evaluate to their coordinate, for instance:
 * $$x(2,3) = 2, \text{ and } y(2,3)=3.$$

Given monomials of higher degree, these are elements of various symmetric powers, and a general polynomial is an element of the symmetric algebra. Without a choice of basis for the vector space, the same holds, but one has a polynomial algebra without choice of basis.

Conversely, the symmetric algebra of the vector space itself can be interpreted, not as polynomials on the vector space (since one cannot evaluate an element of the symmetric algebra of a vector space against a vector in that space: there is no pairing between S(V) and V), but polynomials in the vectors, such as v2 − vw + uv.

Symmetric algebra of an affine space
One can analogously construct the symmetric algebra on an (or its dual, which corresponds to polynomials on that affine space). The key difference is that the symmetric algebra of an affine space is not a graded algebra, but a : one can determine the degree of a polynomial on an affine space, but not its homogeneous parts.

For instance, given a linear polynomial on a vector space, one can determine its constant part by evaluating at 0. On an affine space, there is no distinguished point, so one cannot do this (choosing a point turns an affine space into a vector space).

Universal properties
The symmetric algebra on a vector space is a in the category of commutative unital associative algebras (in the sequel, "commutative algebras").

Formally, the map that sends a vector space to its symmetric algebra is a from vector spaces over K to commutative algebras over K, and is a free object, meaning that it is  to the  that sends a commutative algebra to its underlying vector space.

The unit of the adjunction is the map V → S(V) that embeds a vector space in its symmetric algebra.

Commutative algebras are a of algebras; given an algebra A, one can quotient out by its commutator ideal generated by ab – ba, obtaining a commutative algebra, analogously to  of a group. The construction of the symmetric algebra as a quotient of the tensor algebra is, as functors, a composition of the free algebra functor with this reflection. The universal property can thus be seen as being inherited from the tensor algebra.

Analogy with exterior algebra
The Sk are s comparable to the s; here, though, the grows with k; it is given by
 * $$\operatorname{dim}(S^k(V)) = \binom{n+k-1}{k}$$

where n is the dimension of V. This binomial coefficient is the number of n-variable monomials of degree k. In fact, the symmetric algebra and the exterior algebra appear as the isotypical components of the trivial and sign representation of the action of $$S_n$$ acting on the tensor product $$V^{\otimes n}$$ (for example over the complex field)

Module analog
The construction of the symmetric algebra generalizes to the symmetric algebra S(M) of a M over a. If M is a over the ring R, then its symmetric algebra is isomorphic to the polynomial algebra over R whose indeterminates are a basis of M, just like the symmetric algebra of a vector space. However, if M is not free then S(M) is more complicated.

As a Hopf algebra
The symmetric algebra can be given the structure of a. The article on the provides highly detailed mechanics showing how this is done.

As a universal enveloping algebra
The symmetric algebra S(V) is the of an, i.e. one in which the Lie bracket is identically 0.