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Symmetric algebra

From Wikipedia, the free encyclopedia

"Smallest" commutative algebra that contains a vector space
Not to be confused withSymmetric Frobenius algebra.

Inmathematics, thesymmetric algebraS(V) (also denotedSym(V)) on avector spaceV over afieldK is acommutative algebra overK that containsV, and is, in some sense, minimal for this property. Here, "minimal" means thatS(V) satisfies the followinguniversal property: for everylinear mapf fromV to a commutative algebraA, there is a uniquealgebra homomorphismg :S(V) →A such thatf =gi, wherei is theinclusion map ofV inS(V).

IfB is a basis ofV, the symmetric algebraS(V) can be identified, through acanonical isomorphism, to thepolynomial ringK[B], where the elements ofB are considered as indeterminates. Therefore, the symmetric algebra overV can be viewed as a "coordinate free" polynomial ring overV.

The symmetric algebraS(V) can be built as thequotient of thetensor algebraT(V) by thetwo-sided ideal generated by the elements of the formxyyx.

All these definitions and properties extend naturally to the case whereV is amodule (not necessarily a free one) over acommutative ring.

Construction

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From tensor algebra

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It is possible to use thetensor algebraT(V) to describe the symmetric algebraS(V). In fact,S(V) can be defined as thequotient algebra ofT(V) by the two-sided ideal generated by thecommutatorsvwwv.{\displaystyle v\otimes w-w\otimes v.}

It is straightforward to verify that the resulting algebra satisfies the universal property stated in the introduction. Because of the universal property of the tensor algebra, a linear mapf fromV to a commutative algebraA extends to an algebra homomorphismT(V)A{\displaystyle T(V)\rightarrow A}, which factors throughS(V) becauseA is commutative. The extension offto an algebra homomorphismS(V)A{\displaystyle S(V)\rightarrow A} is unique becauseV generatesS(V) as aK-algebra.

This results also directly from a general result ofcategory theory, which asserts that the composition of twoleft adjoint functors is also a left adjoint functor. Here, theforgetful functor from commutative algebras to vector spaces or modules (forgetting the multiplication) 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

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The symmetric algebraS(V) can also be built frompolynomial rings.

IfV is aK-vector space or afreeK-module, with a basisB, letK[B] be the polynomial ring that has the elements ofB as indeterminates. Thehomogeneous polynomials of degree one form a vector space or a free module that can be identified withV. It is straightforward to verify that this makesK[B] a solution to the universal problem stated in the introduction. This implies thatK[B] andS(V) are canonically isomorphic, and can therefore be identified. This results also immediately from general considerations ofcategory theory, since free modules and polynomial rings arefree objects of their respective categories.

IfV is a module that is not free, it can be writtenV=L/M,{\displaystyle V=L/M,} whereL is a free module, andM is asubmodule ofL. In this case, one has

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

whereM{\displaystyle \langle M\rangle } is the ideal generated byM. (Here, equals signs mean equalityup to a canonical isomorphism.) Again this can be proved by showing that one has a solution of the universal property, and this can be done either by a straightforward but boring computation, or by using category theory, and more specifically, 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, thekernel is anormal 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

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The symmetric algebra is agraded algebra. That is, it is adirect sum

S(V)=n=0Sn(V),{\displaystyle S(V)=\bigoplus _{n=0}^{\infty }S^{n}(V),}

whereSn(V),{\displaystyle S^{n}(V),} called thenthsymmetric power ofV, is the vector subspace or submodule generated by the products ofn elements ofV. (The second symmetric powerS2(V){\displaystyle S^{2}(V)} is sometimes called thesymmetric square ofV).

This can be proved by various means. One follows from the tensor-algebra construction: since the tensor algebra is graded, and the symmetric algebra is its quotient by ahomogeneous ideal: the ideal generated by allxyyx,{\displaystyle x\otimes y-y\otimes x,} wherex andy are inV, that is, homogeneous of degree one.

In the case of a vector space or a free module, the gradation is the gradation of the polynomials by thetotal degree. A non-free module can be written asL /M, whereL is a free module of baseB; its symmetric algebra is the quotient of the (graded) symmetric algebra ofL (a polynomial ring) by the homogeneous ideal generated by the elements ofM, which are homogeneous of degree one.

One can also defineSn(V){\displaystyle S^{n}(V)} as the solution of the universal problem forn-linear symmetric functions fromV into a vector space or a module, and then verify that thedirect sum of allSn(V){\displaystyle S^{n}(V)} satisfies the universal problem for the symmetric algebra.

Relationship with symmetric tensors

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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 asymmetric tensor. However, symmetric tensors are strongly related to the symmetric algebra.

Asymmetric tensor of degreen is an element ofTn(V) that is invariant under theaction of thesymmetric groupSn.{\displaystyle {\mathcal {S}}_{n}.} More precisely, givenσSn,{\displaystyle \sigma \in {\mathcal {S}}_{n},} the transformationv1vnvσ(1)vσ(n){\displaystyle v_{1}\otimes \cdots \otimes v_{n}\mapsto v_{\sigma (1)}\otimes \cdots \otimes v_{\sigma (n)}} defines a linearendomorphism ofTn(V). A symmetric tensor is a tensor that is invariant under all these endomorphisms. The symmetric tensors of degreen form a vector subspace (or module)Symn(V) ⊂Tn(V). Thesymmetric tensors are the elements of thedirect sumn=0Symn(V),{\displaystyle \textstyle \bigoplus _{n=0}^{\infty }\operatorname {Sym} ^{n}(V),} which is agraded vector space (or agraded module). It is not an algebra, as the tensor product of two symmetric tensors is not symmetric in general.

Letπn{\displaystyle \pi _{n}} be the restriction toSymn(V) of the canonical surjectionTn(V)Sn(V).{\displaystyle T^{n}(V)\to S^{n}(V).} Ifn! is invertible in the ground field (or ring), thenπn{\displaystyle \pi _{n}} is anisomorphism. This is always the case with a ground field ofcharacteristic zero. Theinverse isomorphism is the linear map defined (on products ofn vectors) by thesymmetrization

v1vn1n!σSnvσ(1)vσ(n).{\displaystyle v_{1}\cdots v_{n}\mapsto {\frac {1}{n!}}\sum _{\sigma \in S_{n}}v_{\sigma (1)}\otimes \cdots \otimes v_{\sigma (n)}.}

The mapπn{\displaystyle \pi _{n}} is not injective if the characteristic is less thann+1; for exampleπn(xy+yx)=2xy{\displaystyle \pi _{n}(x\otimes y+y\otimes x)=2xy} is zero in characteristic two. Over a ring of characteristic zero,πn{\displaystyle \pi _{n}} can be non surjective; for example, over the integers, ifx andy are two linearly independent elements ofV =S1(V) that are not in2V, thenxyπn(Sym2(V)),{\displaystyle xy\not \in \pi _{n}(\operatorname {Sym} ^{2}(V)),} since12(xy+yx)Sym2(V).{\displaystyle {\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 therational numbers.

Categorical properties

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Given amoduleV over acommutative ringK, the symmetric algebraS(V) can be defined by the followinguniversal property:

For everyK-linear mapf fromV to a commutativeK-algebraA, there is a uniqueK-algebra homomorphismg:S(V)A{\displaystyle g:S(V)\to A} such thatf=gi,{\displaystyle f=g\circ i,} wherei is the inclusion ofV inS(V).

As for every universal property, as soon as a solution exists, this defines uniquely the symmetric algebra,up to acanonical 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 tocategory theory.

The symmetric algebra is afunctor from thecategory ofK-modules to the category ofK-commutative algebra, since the universal property implies that everymodule homomorphismf:VW{\displaystyle f:V\to W} can be uniquely extended to analgebra homomorphismS(f):S(V)S(W).{\displaystyle S(f):S(V)\to S(W).}

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

Symmetric algebra of an affine space

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One can analogously construct the symmetric algebra on anaffine space. The key difference is that the symmetric algebra of an affine space is not a graded algebra, but afiltered 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).

Analogy with exterior algebra

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TheSk arefunctors comparable to theexterior powers; here, though, thedimension grows withk; it is given by

dim(Sk(V))=(n+k1k){\displaystyle \operatorname {dim} (S^{k}(V))={\binom {n+k-1}{k}}}

wheren is the dimension ofV. Thisbinomial coefficient is the number ofn-variable monomials of degreek.In fact, the symmetric algebra and the exterior algebra appear as the isotypical components of the trivial and sign representation of the action ofSn{\displaystyle S_{n}} acting on the tensor productVn{\displaystyle V^{\otimes n}} (for example over the complex field)[citation needed]

As a Hopf algebra

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The symmetric algebra can be given the structure of aHopf algebra. SeeTensor algebra for details.

As a universal enveloping algebra

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The symmetric algebraS(V) is theuniversal enveloping algebra of anabelian Lie algebra, i.e. one in which the Lie bracket is identically 0.

See also

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References

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Basic concepts
Algebraic structures
Linear and
multilinear algebra
Algebraic constructions
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