Editing Symmetric algebra (section)

==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 {{mvar|n}} is an element of {{math|''T''{{sup|''n''}}(''V'')}} that is invariant under the [[group action (mathematics)|action]] of the [[symmetric group]] <math>\mathcal S_n.</math> More precisely, given <math>\sigma\in \mathcal S_n,</math> the transformation <math>v_1\otimes \cdots \otimes v_n \mapsto v_{\sigma(1)}\otimes \cdots \otimes v_{\sigma(n)}</math> defines a linear [[endomorphism]] of {{math|''T''{{sup|''n''}}(''V'')}}. A symmetric tensor is a tensor that is invariant under all these endomorphisms. The symmetric tensors of degree {{mvar|n}} form a vector subspace (or module) {{math|Sym{{sup|''n''}}(''V'') ⊂ ''T''{{sup|''n''}}(''V'')}}. The ''symmetric tensors'' are the elements of the [[direct sum]] <math>\textstyle \bigoplus_{n=0}^\infty \operatorname{Sym}^n(V),</math> 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 <math>\pi_n</math> be the restriction to {{math|Sym{{sup|''n''}}(''V'')}} of the canonical surjection <math>T^n(V)\to S^n(V).</math> If {{math|''n''!}} is invertible in the ground field (or ring), then <math>\pi_n</math> is an [[isomorphism]]. This is always the case with a ground field of [[Characteristic (algebra)|characteristic]] zero. The [[inverse function|inverse]] isomorphism is the linear map defined (on products of {{mvar|n}} vectors) by the [[symmetrization]]
:<math>v_1\cdots v_n \mapsto \frac 1{n!} \sum_{\sigma \in S_n} v_{\sigma(1)}\otimes \cdots \otimes v_{\sigma(n)}.</math>

The map <math>\pi_n</math> is not injective if the characteristic is less than {{mvar|n}}+1; for example <math>\pi_n(x\otimes y+y\otimes x) = 2xy</math> is zero in characteristic two. Over a ring of characteristic zero, <math>\pi_n</math> can be non surjective; for example, over the integers, if {{mvar|x}} and {{mvar|y}} are two linearly independent elements of {{math|1=''V'' = ''S''{{sup|1}}(''V'')}} that are not in {{math|2''V''}}, then <math>xy\not\in \pi_n(\operatorname{Sym}^2(V)),</math> since <math>\frac 12 (x\otimes y +y\otimes x) \not\in \operatorname{Sym}^2(V).</math>

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 number]]s.