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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 do not understand what you mean by "elements of the algebra that can be written as the *outer product* of $k$ vectors" : the Clifford algebra can be defined as a quotient of the *tensor* algebra of $V$ but I've never heard its elements described as being generated by *outer* products (unless you are working with the zero quadratic form)... If you mean the product of some vectors $v_1\cdots v_k$ then the answer will be (up to sign, there are varying conventions for Clifford algebras) $Q(v_1)\cdots Q(v_k)$.2012-05-03
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    @OlivierBégassat Those geometric algebra folks like to use outer product representations of blades sometimes. They showed (for real non-degenerate forms anyhow) that the exterior product of $k$ linearly independent vectors is a $k$-blade.2012-05-03
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    Olivier, what do you mean by your answer? It that the proof? What is Q? Also, I'm not generating anything. It is a fact that some elements of the algebra are blades, or simple k-vectors. See [http://en.wikipedia.org/wiki/Clifford_algebra#Relation_to_the_exterior_algebra] for details.2012-05-03
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    $Q$ is the quadratic form on $V$. The wikipedia page you link to contains my notations and doesn't mention "blades" (maybeI missed it). How do you define Clifford algebras? Why do you use the wedge product notation instead of the more customary dot notation?2012-05-04
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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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