Editing Clifford algebra (section)

== Properties ==
=== 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.

=== Grading ===
In the following, assume that the characteristic is not&nbsp;{{math|2}}.{{efn|Thus the [[group ring|group algebra]] {{math|''K''['''Z'''{{px2}}/{{px2}}2'''Z''']}} is [[Semisimple algebra|semisimple]] and the Clifford algebra splits into eigenspaces of the main involution.}}

Clifford algebras are {{math|'''Z'''<sub>2</sub>}}-[[graded algebra]]s (also known as [[superalgebra]]s).  Indeed, the linear map on {{math|''V''}} defined by {{math|''v'' ↦ −''v''}} ([[reflection through the origin]]) preserves the quadratic form {{math|''Q''}} and so by the universal property of Clifford algebras extends to an algebra [[automorphism]]
<math display="block">\alpha: \operatorname{Cl}(V, Q) \to \operatorname{Cl}(V, Q).</math>

Since {{math|''α''}} is an [[Involution (mathematics)|involution]] (i.e. it squares to the [[identity function|identity]]) one can decompose {{math|Cl(''V'', ''Q'')}} into positive and negative eigenspaces of&nbsp;{{math|''α''}}
<math display="block">\operatorname{Cl}(V, Q) = \operatorname{Cl}^{[0]}(V, Q) \oplus \operatorname{Cl}^{[1]}(V, Q)</math>
where 
<math display="block">\operatorname{Cl}^{[i]}(V, Q) = \left\{ x \in \operatorname{Cl}(V, Q) \mid \alpha(x) = (-1)^i x \right\}.</math>

Since {{math|''α''}} is an automorphism it follows that:
<math display="block">\operatorname{Cl}^{[i]}(V, Q)\operatorname{Cl}^{[j]}(V, Q) = \operatorname{Cl}^{[i+j]}(V, Q)</math>
where the bracketed superscripts are read modulo&nbsp;2. This gives {{math|Cl(''V'', ''Q'')}} the structure of a {{math|'''Z'''<sub>2</sub>}}-[[graded algebra]].  The subspace {{math|Cl{{sup|[0]}}(''V'', ''Q'')}} forms a [[subalgebra]] of {{math|Cl(''V'', ''Q'')}}, called the ''even subalgebra''. The subspace {{math|Cl{{sup|[1]}}(''V'', ''Q'')}} is called the ''odd part'' of {{math|Cl(''V'', ''Q'')}} (it is not a subalgebra).  {{math|This '''Z'''<sub>2</sub>}}-grading plays an important role in the analysis and application of Clifford algebras. The automorphism {{math|''α''}} is called the ''main [[involution (mathematics)|involution]]'' or ''grade involution''. Elements that are pure in this {{math|'''Z'''<sub>2</sub>}}-grading are simply said to be even or odd.

''Remark''. The Clifford algebra is not a  {{math|'''Z'''}}-graded algebra, but is {{math|'''Z'''}}-[[filtered algebra|filtered]], where {{math|Cl{{sup|≤''i''}}(''V'', ''Q'')}} is the subspace spanned by all products of at most {{math|''i''}} elements of&nbsp;{{math|'''V'''}}.
<math display="block">\operatorname{Cl}^{\leqslant i}(V, Q) \cdot \operatorname{Cl}^{\leqslant j}(V, Q) \subset \operatorname{Cl}^{\leqslant i+j}(V, Q).</math>

The ''degree'' of a Clifford number usually refers to the degree in the {{math|'''Z'''}}-grading.

The even subalgebra {{math|Cl{{sup|[0]}}(''V'', ''Q'')}} of a Clifford algebra is itself isomorphic to a Clifford algebra.{{efn|Technically, it does not have the full structure of a Clifford algebra without a designated vector subspace, and so is isomorphic as an algebra, but not as a Clifford algebra.}}{{efn|We are still assuming that the characteristic is not {{math|2}}.}} If {{math|''V''}} is the [[orthogonal direct sum]] of a vector {{math|''a''}} of nonzero norm {{math|''Q''(''a'')}} and a subspace {{math|''U''}}, then {{math|Cl{{sup|[0]}}(''V'', ''Q'')}} is isomorphic to {{math|Cl(''U'', −''Q''(''a'')''Q''{{!}}{{smallsub|''U''}})}}, where {{math|''Q''{{!}}{{smallsub|''U''}}}} is the form {{math|''Q''}} restricted to {{math|''U''}}. In particular over the reals this implies that:
<math display="block">\operatorname{Cl}_{p,q}^{[0]}(\mathbf{R}) \cong \begin{cases}
  \operatorname{Cl}_{p,q-1}(\mathbf{R}) & q > 0 \\
  \operatorname{Cl}_{q,p-1}(\mathbf{R}) & p > 0
\end{cases}</math>

In the negative-definite case this gives an inclusion {{math|Cl{{sub|0,''n'' − 1}}('''R''') ⊂ Cl{{sub|0,''n''}}('''R''')}}, which extends the sequence
{{block indent|em=1.5|text={{math|'''R''' ⊂ '''C''' ⊂ '''H''' ⊂ '''H''' ⊕ '''H''' ⊂ ⋯}}}}

Likewise, in the complex case, one can show that the even subalgebra of {{math|Cl{{sub|''n''}}('''C''')}} is isomorphic to {{math|Cl{{sub|''n''−1}}('''C''')}}.

=== Antiautomorphisms ===
In addition to the automorphism {{math|''α''}}, there are two [[antiautomorphism]]s that play an important role in the analysis of Clifford algebras. Recall that the [[tensor algebra]] {{math|''T''(''V'')}} comes with an antiautomorphism that reverses the order in all products of vectors:
<math display="block">v_1\otimes v_2\otimes \cdots \otimes v_k \mapsto v_k\otimes \cdots \otimes v_2\otimes v_1.</math>
Since the ideal {{math|''I''<sub>''Q''</sub>}} is invariant under this reversal, this operation descends to an antiautomorphism of {{math|Cl(''V'', ''Q'')}} called the ''transpose'' or ''reversal'' operation, denoted by {{math|''x''<sup>t</sup>}}. The transpose is an antiautomorphism: {{math|1=(''xy'')<sup>t</sup> = ''y''<sup>t</sup> ''x''<sup>t</sup>}}. The transpose operation makes no use of the {{math|'''Z'''<sub>2</sub>}}-grading so we define a second antiautomorphism by composing {{math|''α''}} and the transpose. We call this operation ''Clifford conjugation'' denoted <math>\bar x</math>
<math display="block">\bar x = \alpha(x^\mathrm{t}) = \alpha(x)^\mathrm{t}.</math>
Of the two antiautomorphisms, the transpose is the more fundamental.{{efn|The opposite is true when using the alternate (−) sign convention for Clifford algebras: it is the conjugate that is more important. In general, the meanings of conjugation and transpose are interchanged when passing from one sign convention to the other. For example, in the convention used here the inverse of a vector is given by {{math|1=''v''<sup>−1</sup> = ''v''<sup>t</sup> / ''Q''(''v'')}} while in the (−) convention it is given by {{math|1=''v''<sup>−1</sup> = {{overline|''v''}} / ''Q''(''v'')}}.}}

Note that all of these operations are [[involution (mathematics)|involutions]]. One can show that they act as {{math|±1}} on elements that are pure in the {{math|'''Z'''}}-grading. In fact, all three operations depend on only the degree modulo&nbsp;{{math|4}}. That is, if {{math|''x''}} is pure with degree {{math|''k''}} then
<math display="block">\alpha(x) = \pm x \qquad x^\mathrm{t} = \pm x \qquad \bar x = \pm x</math>
where the signs are given by the following table:
: {| class=wikitable
 ! {{math|''k'' mod 4}}
 | {{math|0}} || {{math|1}} || {{math|2}} || {{math|3}} || …
|-
 ! <math>\alpha(x)\,</math>
 | {{math|+}} || {{math|−}} || {{math|+}} || {{math|−}} || {{math|(−1)<sup>''k''</sup>}}
|-
 ! <math>x^\mathrm{t}\,</math>
 | {{math|+}} || {{math|+}} || {{math|−}} || {{math|−}} || {{math|(−1)<sup>''k''(''k'' − 1)/2</sup>}}
|-
 ! <math>\bar x</math>
 | {{math|+}} || {{math|−}} || {{math|−}} || {{math|+}} || {{math|(−1)<sup>''k''(''k'' + 1)/2</sup>}}
|}

=== Clifford scalar product ===
When the characteristic is not {{math|2}}, the quadratic form {{math|''Q''}} on {{math|''V''}} can be extended to a quadratic form on all of {{math|Cl(''V'', ''Q'')}} (which we also denoted by {{math|''Q''}}). A basis-independent definition of one such extension is
<math display="block">Q(x) = \left\langle x^\mathrm{t} x\right\rangle_0</math>
where {{math|⟨''a''⟩{{sub|0}}}} denotes the scalar part of {{math|''a''}} (the degree-{{math|0}} part in the {{math|'''Z'''}}-grading). One can show that
<math display="block">Q(v_1v_2 \cdots v_k) = Q(v_1)Q(v_2) \cdots Q(v_k)</math>
where the {{math|''v<sub>i</sub>''}} are elements of {{math|''V''}} – this identity is ''not'' true for arbitrary elements of {{math|Cl(''V'', ''Q'')}}.

The associated symmetric bilinear form on {{math|Cl(''V'', ''Q'')}} is given by
<math display="block">\langle x, y\rangle = \left\langle x^\mathrm{t} y\right\rangle_0.</math>
One can check that this reduces to the original bilinear form when restricted to {{math|''V''}}. The bilinear form on all of {{math|Cl(''V'', ''Q'')}} is [[nondegenerate form|nondegenerate]] if and only if it is nondegenerate on {{math|''V''}}.

The operator of left (respectively right) Clifford multiplication by the transpose {{math|''a''{{i sup|t}}}} of an element {{math|''a''}} is the [[adjoint of an operator|adjoint]] of left (respectively right) Clifford multiplication by {{math|''a''}} with respect to this inner product. That is,
<math display="block">\langle ax, y\rangle = \left\langle x, a^\mathrm{t} y\right\rangle,</math>
and
<math display="block">\langle xa, y\rangle = \left\langle x, y a^\mathrm{t}\right\rangle.</math>