Symmetric algebra

In mathematics, the symmetric algebra "S"("V") (also denoted "Sym"("V")) on a vector space "V" over a field "K" is the free commutative unital associative "K"-algebra containing "V".

It corresponds to polynomials with indeterminates in "V", without choosing coordinates.The dual, $S(V^*)$ corresponds to polynomials "on" "V".

It should not be confused with symmetric tensors in "V". A Frobenius algebra whose bilinear form is symmetric is also called a symmetric algebra, but is not discussed here.

Construction

It turns out that "S"("V") is in effect the same as the polynomial ring, over "K", in indeterminates that are basis vectors for "V". Therefore this construction only brings something extra when the "naturality" of looking at polynomials this way has some advantage.

It is possible to use the tensor algebra "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 quotient algebra of "T"("V") by the ideal generated by all differences of products

:$votimes\; w\; -\; wotimes\; v.$

for "v" and "w" in "V".

Grading

Just as with a polynomial ring, there is a direct sum decomposition of "S"("V") as a graded algebra, into summands

:"S^k"("V")

which consist of the linear span of the monomials in vectors of "V" of degree "k", for "k" = 0, 1, 2, ... (with "S"⁰("V") = "K" and "S"¹("V")="V"). The "K"-vector space "S^k"("V") is the "k"-th symmetric power of "V". The case "k" = 2, for example, is the symmetric square. It has a universal property with respect to symmetric multilinear operators defined on "V"^"k".

Distinction with symmetric tensors

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

The symmetric algebra must be a quotient to satisfy its universal property (since every symmetric algebra is an algebra, the tensor algebra maps to the symmetric algebra).

Conversely, symmetric tensors are defined as invariants: given the natural action of the symmetric group 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 vectors "v" and "w", they are trivially symmetric 1-tensors, but $v\; otimes\; w$ is not a symmetric 2-tensor.

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

In characteristic 0 symmetric tensors and the symmetric algebra can be identified. In any characteristic, there is a symmetrization map from the symmetric algebra to the symmetric tensors, given by::$v\_1cdots\; v\_k\; mapstosum\_\{sigma\; in\; S\_n\}\; v\_\{sigma(1)\}otimes\; cdots\; otimes\; v\_\{sigma(k)\}.$The composition with the inclusion of the symmetric tensors in the tensor algebra and the quotient to the symmetric algebra is multiplication by $k!$ on the "k"th graded component.

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 $frac\{1\}\{k!\}$ to make this a section of the quotient
$v\_1cdots\; v\_k\; mapsto\; frac\{1\}\{k!\}\; sum\_\{sigma\; in\; S\_n\}\; v\_\{sigma(1)\}otimes\; cdots\; otimes\; v\_\{sigma(k)\}.$For instance, $vw\; mapsto\; frac\{1\}\{2\}(votimes\; w\; +\; w\; otimes\; v)$.

This is related to the representation theory of the symmetric group:in characteristic 0, over an algebraically closed field, the group algebra is semisimple, so every representation splits into a direct sum of irreducible representations, and if $T=Soplus\; V$, 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 $S(V^*)\; imes\; V\; o\; K$.

For instance, given the plane with a basis $K^2$, the (homogeneous) linear polynomials on $K^2$ are generated by the coordinate functionals "x" and "y". These coordinates are covectors: given a vector, they evaluate to their coordinate, for instance::$x(2,3)\; =\; 2,\; ext\{\; and\; \}\; y(2,3)=3.$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 $v^2-vw+uv$.

ymmetric algebra of an affine space

One can analogously construct the symmetric algebra on an affine space (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 filtered algebra: 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).

Categorical properties

The symmetric algebra on a vector space is a free object 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 functor from vector spaces over "K" to commutative algebras over "K",and is a "free object", meaning that it is left adjoint to the
forgetful functor that sends a commutative algebra to its underlying vector space.

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

Commutative algebras are a reflective subcategory of algebras;given an algebra "A", one can quotient out by its commutator ideal generated by $ab-ba$, obtaining a commutative algebra, analogously to abelianization 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.

Analogy with exterior algebra

The "S"^"k" are functors comparable to the exterior powers; here, though, the dimension grows with "k"; it is given by:where "n" is the dimension of "V".

Module analog

The construction of the symmetric algebra generalizes to the symmetric algebra "S"("M") of a module "M" over a commutative ring. If "M" is a free module 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, that is not true if "M" is not free; then "S"("M") is more complicated.

As a universal enveloping algebra

The symmetric algebra "S"("V") is the universal enveloping algebra of an abelian Lie algebra in which the Lie bracket is identically 0.

ee also

* exterior algebra, the anti-symmetric analog
* Weyl algebra, a of the symmetric algebra by a symplectic form