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Is there a simple proof that the product of a simple k-vector with its reverse always yields a scalar? Assume real Clifford algebra of arbitrary but finite dimension with unspecified metric. By simple k-vectors I mean elements of the algebra that can be written as the outer product of k vectors, e.g. $a\wedge b\wedge c$, whose reverse is $c\wedge b\wedge a$.

The wikipedia page http://en.wikipedia.org/wiki/Clifford_algebra#Relation_to_the_exterior_algebra talks about natural isomorphism between Clifford and exterior algebra. Accordingly, some elements of Clifford algebra are isomorphic to elements of exterior algebra that have the form $a\wedge b$, $a\wedge b\wedge c$, etc. These elements are blades in exterior algebra, so I call them blades in Clifford algebra as well.

Feel free to use whatever definition of Clifford algebra is most convenient for the proof.

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    I suppose the identification of pure wedge products with Clifford algebra elements is $a \wedge b = a \cdot b - b \cdot a$, or in general, $a_1 \wedge a_2 \ldots \wedge a_n = \sum_\pi sgn(\pi) \prod a_{\pi(i)}$ (sum over permutations $\pi$ of $\{1,2,\ldots,n\}$). (Sometimes one adds a $1/n!$ constant factor, but this won't affect the answer to the question.)2012-05-06

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If the k-vector is simple it can be written as a (geometric)product of k vectors.

$A_r=a_1...a_r$

The proof is by induction. For a vector you know

$a^{\dagger}a=|a|^2$

which is scalar by definition, so you know this to be true for r=1 suppose it is true for r

$A_r^{\dagger}A_r=|A_r|^2$

it follows that

$A_{r+1}^{\dagger}A_{r+1}$

$=(A_ra_{r+1})^{\dagger}(A_ra_{r+1})$

$=a_{r+1}^{\dagger}A_r^{\dagger}A_ra_{r+1}$

$=a_{r+1}|A_r|^2a_{r+1}$

$=|A_r|^2a_{r+1}a_{r+1}$

$=|A_r|^2|a_{r+1}|^2$

$=|A_{r+1}|^2$

Which completes the proof

To see that $a\wedge b$ can be written as $ab'$ observe that:

$A=a\wedge b=a\wedge (b+\lambda a)$

so choose a $\lambda$ such that $a\cdot (b+\lambda a)$ is zero.

$a\cdot (b+\lambda a)=0$

$\frac{1}{2}(a(b+\lambda a)+(b+\lambda a)a)=0$

$ab+\lambda a^2+ba+\lambda a^2=0$

$\lambda=-\frac{a\cdot b}{a^2}$

call $b'=b+\lambda a$ with $\lambda$ as above and so

$A=a\wedge b=a\wedge b'=ab'$

Then by induction $A_{r+1}=A_r\wedge a_{r+1}=A_ra'_{r+1}$

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    If $a^2=b^2=0$, then $a\wedge b=\frac{a-b}{\sqrt2}\wedge\frac{a+b}{\sqrt2},$ and $\frac{a-b}{\sqrt2}\cdot\frac{a+b}{\sqrt2}=\frac{a^2+a\cdot b-b\cdot a-b^2}{2}=0.$ Any blade can always be orthogonalized.2018-05-24