Editing Clifford algebra (section)

== Examples: constructing quaternions and dual quaternions ==
=== Quaternions ===
In this section, Hamilton's [[quaternion]]s are constructed as the even subalgebra of the Clifford algebra {{math|1=Cl{{sub|3,0}}('''R''')}}.

Let the vector space {{math|1=''V''}} be real three-dimensional space&nbsp;{{math|1='''R'''<sup>3</sup>}}, and the quadratic form be the usual quadratic form.  Then, for {{math|1=''v'', ''w''}} in {{math|1='''R'''<sup>3</sup>}} we have the bilinear form (or scalar product)
<math display="block">v \cdot w = v_1 w_1 + v_2 w_2 + v_3 w_3.</math>
Now introduce the Clifford product of vectors {{math|1=''v''}} and {{math|1=''w''}} given by
<math display="block"> v w + w v = 2 (v \cdot w) .</math>

Denote a set of orthogonal unit vectors of {{math|1='''R'''<sup>3</sup>}} as {{math|1={{mset|''e''<sub>1</sub>, ''e''<sub>2</sub>, ''e''<sub>3</sub>}}}}, then the Clifford product yields the relations
<math display="block"> e_2 e_3 = -e_3 e_2, \,\,\,  e_1 e_3 = -e_3 e_1,\,\,\,  e_1 e_2 = -e_2 e_1,</math>
and
<math display="block"> e_1 ^2 = e_2^2 = e_3^2 = 1. </math>
The general element of the Clifford algebra {{math|1=Cl{{sub|3,0}}('''R''')}} is given by
<math display="block"> A = a_0 + a_1 e_1 + a_2 e_2 + a_3 e_3 + a_4 e_2 e_3 + a_5 e_1 e_3 + a_6 e_1 e_2 + a_7 e_1 e_2 e_3.</math>

The linear combination of the even degree elements of {{math|1=Cl{{sub|3,0}}('''R''')}} defines the even subalgebra {{math|1=Cl{{su|lh=0.9em|p=[0]|b=3,0}}('''R''')}} with the general element
<math display="block"> q = q_0 + q_1 e_2 e_3 + q_2 e_1 e_3 + q_3 e_1 e_2. </math>
The basis elements can be identified with the quaternion basis elements {{math|1=''i'', ''j'', ''k''}} as
<math display="block"> i=  e_2 e_3, j =  e_1 e_3, k =  e_1 e_2,</math>
which shows that the even subalgebra {{math|1=Cl{{su|lh=0.9em|p=[0]|b=3,0}}('''R''')}} is Hamilton's real [[quaternion]] algebra.

To see this, compute
<math display="block"> i^2 = (e_2 e_3)^2 = e_2 e_3 e_2 e_3 = - e_2 e_2 e_3 e_3 = -1,</math>
and
<math display="block"> ij = e_2 e_3 e_1 e_3 = -e_2 e_3 e_3 e_1 = -e_2 e_1 = e_1 e_2 = k.</math>
Finally,
<math display="block"> ijk = e_2 e_3 e_1 e_3 e_1 e_2 = -1.</math>

=== Dual quaternions ===
In this section, [[dual quaternion]]s are constructed as the even subalgebra of a Clifford algebra of real four-dimensional space with a degenerate quadratic form.{{sfn|McCarthy|1990|pp=62–65|ps=none}}{{sfn|Bottema|Roth|2012|ps=none}}

Let the vector space {{math|1=''V''}} be real four-dimensional space {{math|1='''R'''<sup>4</sup>,}} and let the quadratic form {{math|1=''Q''}} be a degenerate form derived from the Euclidean metric on {{math|1='''R'''<sup>3</sup>.}}  For {{math|1=''v'', ''w''}} in {{math|1='''R'''<sup>4</sup>}} introduce the degenerate bilinear form
<math display="block">d(v, w) = v_1 w_1 + v_2 w_2 + v_3 w_3 .</math>
This degenerate scalar product projects distance measurements in {{math|1='''R'''<sup>4</sup>}} onto the {{math|1='''R'''<sup>3</sup>}} hyperplane.

The Clifford product of vectors {{math|1=''v''}} and {{math|1=''w''}} is given by
<math display="block">v w + w v = -2 \,d(v, w).</math>
Note the negative sign is introduced to simplify the correspondence with quaternions.

Denote a set of mutually orthogonal unit vectors of {{math|1='''R'''<sup>4</sup>}} as {{math|1={{mset|''e''<sub>1</sub>, ''e''<sub>2</sub>, ''e''<sub>3</sub>, ''e''<sub>4</sub>}}}}, then the Clifford product yields the relations
<math display="block">e_m e_n = -e_n e_m, \,\,\,  m \ne n,</math>
and
<math display="block">e_1 ^2 = e_2^2 = e_3^2 = -1, \,\, e_4^2 = 0.</math>

The general element of the Clifford algebra {{math|Cl('''R'''{{sup|4}}, ''d'')}} has 16 components.  The linear combination of the even degree elements defines the even subalgebra {{math|Cl{{sup|[0]}}('''R'''<sup>4</sup>, ''d'')}} with the general element
<math display="block"> H = h_0 + h_1 e_2 e_3 + h_2 e_3 e_1 + h_3 e_1 e_2 + h_4 e_4 e_1 + h_5 e_4 e_2 + h_6 e_4 e_3 + h_7 e_1 e_2 e_3 e_4.</math>

The basis elements can be identified with the quaternion basis elements {{math|1=''i'', ''j'', ''k''}} and the dual unit {{math|1=''ε''}} as
<math display="block"> i = e_2 e_3, j = e_3 e_1, k = e_1 e_2, \,\, \varepsilon = e_1 e_2 e_3 e_4.</math>
This provides the correspondence of {{math|1=Cl{{su|lh=0.9em|p=[0]|b=0,3,1}}('''R''')}} with [[dual quaternion]] algebra.

To see this, compute
<math display="block"> \varepsilon ^2 = (e_1 e_2 e_3 e_4)^2 =  e_1 e_2 e_3 e_4 e_1 e_2 e_3 e_4 = -e_1 e_2 e_3 (e_4 e_4 ) e_1 e_2 e_3 = 0 ,</math>
and
<math display="block"> \varepsilon i = (e_1 e_2 e_3 e_4) e_2 e_3  = e_1 e_2 e_3 e_4 e_2 e_3 = e_2 e_3 (e_1 e_2 e_3 e_4) = i\varepsilon.</math>
The exchanges of {{math|1=''e''<sub>1</sub>}} and {{math|1=''e''<sub>4</sub>}} alternate signs an even number of times, and show the dual unit {{math|1=''ε''}} commutes with the quaternion basis elements {{math|1=''i'', ''j'', ''k''}}.