Editing Clifford algebra (section)

=== Relation to the exterior algebra ===
Given a vector space&nbsp;{{math|''V''}}, one can construct the [[exterior algebra]] {{math|⋀''V''}}, whose definition is independent of any quadratic form on {{math|''V''}}. It turns out that if {{math|''K''}} does not have characteristic {{math|2}} then there is a [[natural isomorphism]] between {{math|⋀''V''}} and {{math|Cl(''V'', ''Q'')}} considered as vector spaces (and there exists an isomorphism in characteristic two, which may not be natural). This is an algebra isomorphism if and only if {{math|1=''Q'' = 0}}. One can thus consider the Clifford algebra {{math|Cl(''V'', ''Q'')}} as an enrichment (or more precisely, a quantization, cf. the Introduction) of the exterior algebra on {{math|''V''}} with a multiplication that depends on&nbsp;{{math|''Q''}} (one can still define the exterior product independently of&nbsp;{{math|''Q''}}).

The easiest way to establish the isomorphism is to choose an ''orthogonal'' basis {{math|{{mset|''e''<sub>1</sub>, ..., ''e''<sub>''n''</sub>}}}} for {{math|''V''}} and extend it to a basis for {{math|Cl(''V'', ''Q'')}} as described [[#Basis and dimension|above]]. The map {{math|1=Cl(''V'', ''Q'') → ⋀''V''}} is determined by
<math display="block">e_{i_1}e_{i_2} \cdots e_{i_k} \mapsto e_{i_1}\wedge e_{i_2}\wedge \cdots \wedge e_{i_k}.</math>
Note that this works only if the basis {{math|{{mset|''e''<sub>1</sub>, ..., ''e''<sub>''n''</sub>}}}} is orthogonal. One can show that this map is independent of the choice of orthogonal basis and so gives a natural isomorphism.

If the [[characteristic (algebra)|characteristic]] of {{math|''K''}} is {{math|0}}, one can also establish the isomorphism by antisymmetrizing. Define functions {{math|1=''f<sub>k</sub>'' : ''V'' × ⋯ × ''V'' → Cl(''V'', ''Q'')}} by
<math display="block">f_k(v_1, \ldots, v_k) = \frac{1}{k!}\sum_{\sigma\in \mathrm{S}_k} \sgn(\sigma)\, v_{\sigma(1)}\cdots v_{\sigma(k)}</math>
where the sum is taken over the [[symmetric group]] on {{math|''k''}} elements, {{math|S<sub>''k''</sub>}}. Since {{math|''f''<sub>''k''</sub>}} is [[alternating form|alternating]], it induces a unique linear map {{math|1=⋀<sup>''k''</sup> ''V'' → Cl(''V'', ''Q'')}}. The [[Direct sum of modules|direct sum]] of these maps gives a linear map between {{math|⋀''V''}} and {{math|Cl(''V'', ''Q'')}}. This map can be shown to be a linear isomorphism, and it is natural.

A more sophisticated way to view the relationship is to construct a [[filtration (abstract algebra)|filtration]] on {{math|Cl(''V'', ''Q'')}}. Recall that the [[tensor algebra]] {{math|''T''(''V'')}} has a natural filtration: {{math|1=''F''<sup>0</sup> ⊂ ''F''<sup>1</sup> ⊂ ''F''<sup>2</sup> ⊂ ⋯}}, where {{math|''F''<sup>''k''</sup>}} contains sums of tensors with [[tensor order|order]] {{math|≤ ''k''}}. Projecting this down to the Clifford algebra gives a filtration on {{math|Cl(''V'', ''Q'')}}. The [[associated graded algebra]]
<math display="block">\operatorname{Gr}_F \operatorname{Cl}(V,Q) = \bigoplus_k F^k/F^{k-1}</math>
is naturally isomorphic to the exterior algebra {{math|⋀''V''}}.  Since the associated graded algebra of a filtered algebra is always isomorphic to the filtered algebra as filtered vector spaces (by choosing complements of {{math|''F<sup>k</sup>''}} in {{math|''F''<sup>''k''+1</sup>}} for all&nbsp;{{math|''k''}}), this provides an isomorphism (although not a natural one) in any characteristic, even two.