Editing Clifford algebra (section)

=== 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>