Symmetric algebra

In mathematics, the symmetric algebra $S (V)$ (also denoted $Sym(V))$ on a vector space $V$ over a field $K$ is a commutative algebra over $K$ that contains $V$ , and is, in some sense, minimal for this property. Here, "minimal" means that $S (V)$ satisfies the following universal property: for every linear map $f$ from $V$ to a commutative algebra $A$ , there is a unique algebra homomorphism $g:S(V)\to A$ such that $f=g\circ i,$ where $i$ is the inclusion of $V$ in $S (V)$ .

If $B$ is a basis of $V$ , the symmetric algebra $S (V)$ can be identified, through a canonical isomorphism, to the polynomial ring $K [B]$ , where the elements of $B$ are considered as indeterminates. Therefore, the symmetric algebra over $V$ can be viewed as a "coordinate free" polynomial ring over $V$ .

The symmetric algebra $S (V)$ can be built as the quotient of the tensor algebra $T (V)$ by the two-sided ideal generated by the elements of the form $x\otimes y-y\otimes x.$

All these definitions and properties extend naturally to the case where $V$ is a module (not necessarily a free one) over a commutative ring.

Construction

From tensor algebra

It is possible to use the tensor algebra $T (V)$ to describe the symmetric algebra $S (V)$ . In fact, $S (V)$ can be defined as the quotient algebra of $T (V)$ by the two sided ideal generated by the commutators $v\otimes w-w\otimes v.$

It is straightforward, but rather boring, to verify that the resulting algebra satisfies the universal property stated in the introduction.

This results also directly from a general result of category theory, which asserts that the composition of two left adjoint functors is also a left adjoint functor. Here, the forgetful functor from commutative algebras to vector spaces or modules (forgetting the multipliation) is the composition of the forgetful functors from commutative algebras to associative algebras (forgetting commutativity), and from associative algebras to vectors or modules (forgetting the multiplication). As the tensor algebra and the quotient by commutators are left adjoint to these forgetful functors, their composition is left adjoint to the forgetful functor from commutative algebra to vectors or modules, and this proves the desired universal property.

From polynomial ring

The symmetric algebra $S (V)$ can also be built from polynomial rings.

If $V$ is a $K$ -vector space or a free $K$ -module, with a basis $B$ , let $K [B]$ be the polynomial ring that has the elements of $B$ as indeterminates. The homogeneous polynomials of degree one form a vector space or a free module that can be identified with $V$ . It is straightforward to verify that this makes $K [B]$ a solution to the universal problem stated in the introduction. This implies that $K [B]$ and $S (V)$ are canonically isomorphic, and can therefore be identified. This results also immediately from general considerations of category theory, since free modules and polynomial rings are free objects of their respetive categories.

If $V$ is a module that is not free, it can be written $V=L/M,$ where $L$ is a free module, and $M$ is a submodule of $L$ . In this case, one has

S(V)=S(L/M)=S(L)/\langle M\rangle ,

where $\langle M\rangle$ is the ideal generated by $M$ . (Here, equals signs mean equality up to a canonical isomorphism.) Again this can be proved by showing that one has a solution of the universal property, and this can been done either by a straightforward but boring computation, or by using category theory, and more specificaly, the fact that a quotient is the solution of the universal problem for morphisms that map to zero a given subset (Depending on the case, the kernel is a normal subgroup, a submodule or an ideal, and the usual definition of quotients can be viewed as a proof of the existence of a solution of the universal problem).

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 S⁰(V) = K and S¹(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 and denoted Sym²(V).) It has a universal property with respect to symmetric multilinear operators defined on V^k.

In terms of the tensor algebra grading, S^k(V) is the quotient space of T^k(V) by the subspace generated by all differences of products

v\otimes w-w\otimes v.

and products of these with other algebra elements.

Relationship with symmetric tensors

As the symmetric algebra of a vector space is a quotient of the tensor algebra, an element of the symmetric algebra is not a tensor, and, in particular, is not a symmetric tensor. However, symmetric tensors are strongly related to the symmetric algebra.

A symmetric tensor of degree $n$ is an element of $T n (V)$ that is invariant under the action of the symmetric group ${\mathcal {S}}_{n}.$ More precisely, given $\sigma \in {\mathcal {S}}_{n},$ the transformation $v_{1}\otimes \cdots \otimes v_{n}\mapsto v_{\sigma (1)}\otimes \cdots \otimes v_{\sigma (k)}$ defines a linear endomorphism of $T n (V)$ . A symmetric tensor is a tensor that is invariant under all these endomorphisms. The symmetric tensors of degree $n$ form a vector subspace (or module) $Sym n (V) \subset T n (V)$ . The symmetric tensors are the elements of the direct sum $\textstyle \bigoplus _{n=0}^{\infty }\operatorname {Sym} ^{n}(V),$ which is a graded vector space (or a graded module). It is not an algebra, as the tensor product of two symmetric tensors is not symmetric in general.

Let $\pi _{n}$ be the restriction to $Sym n (V)$ of the canonical surjection $T^{n}(V)\to S^{n}(V).$ If $n!$ is invertible in the ground field (or ring), then $\pi _{n}$ is an isomorphism. This is always the case with a ground field of characteristic zero. The inverse isomorphism is the linear map defined (on products of $n$ vectors) by the symmetrization

v_{1}\cdots v_{k}\mapsto {\frac {1}{n!}}\sum _{\sigma \in S_{k}}v_{\sigma (1)}\otimes \cdots \otimes v_{\sigma (k)}.

The map $\pi _{n}$ is not injective if $n$ divides the characteristic; for example $\pi _{n}(x\otimes y+y\otimes x)=2xy$ is zero in characteristic two. Over a ring of characteristic zero, $\pi _{n}$ can be non surjective; for example, over the integers, if $x$ and $y$ are two linearly independent elements of $V = S 1 (V)$ that are not in $2 V$ , then $xy\not \in \pi _{n}(\operatorname {Sym} ^{2}(V)),$ since ${\frac {1}{2}}(x\otimes y+y\otimes x)\not \in \operatorname {Sym} ^{2}(V).$

In summary, over a field of characteristic zero, the symmetric tensors and the symmetric algebra form two isomorphic graded vector spaces. They can thus be identified as far as only the vector space structure is concerned, but they cannot be identified as soon as products are involved. Moreover, this isomorphism does not extend to the cases of fields of positive characteristic and rings that do not contain the rational numbers.

Symmetric algebra of an affine space

One can analogously construct the symmetric algebra on an 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

Given a module $V$ over a commutative ring $K$ , the symmetric algebra $S (V)$ can be defined by the following universal property:

For every linear map  $f$  from  $V$  to a commutative algebra  $A$ , there is a unique algebra homomorphism  $g:S(V)\to A$  such that  $f=g\circ i,$  where  $i$  is the inclusion of  $V$  in  $S (V)$ .

As for every universal property, as soon as a solution exists, this defines uniquely the symmetric algebra, up to a canonical isomorphism. It follows that all properties of the symmetric algebra can be deduced from the universal property. This section is devoted to the main properties that belong to category theory.

The symmetric algebra is a functor from the category of $K$ -modules to the category of $K$ -commutative algebra, since the universal property implies that every module homomorphism $f:V\to W$ can be uniquely extended to an algebra homomorphism $S(f):S(V)\to S(W).$

The universal property can be reformulated by saying that the symmetric algebra is a left adjoint to the forgetful functor that sends a commutative algebra to its underlying module.

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

\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) ^{[citation needed]}

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, 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 Hopf algebra. The article on the tensor algebra provides highly detailed mechanics showing how this is done.

As a universal enveloping algebra

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

References

Bourbaki, Nicolas (1989), Elements of mathematics, Algebra I, Springer-Verlag, ISBN 3-540-64243-9