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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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  • $\begingroup$ 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) $Q(v_1)\cdots Q(v_k)$. $\endgroup$ Commented May 3, 2012 at 11:39
  • $\begingroup$ @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. $\endgroup$
    – rschwieb
    Commented May 3, 2012 at 13:19
  • $\begingroup$ 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 [en.wikipedia.org/wiki/… for details. $\endgroup$ Commented May 3, 2012 at 22:54
  • $\begingroup$ $Q$ 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? $\endgroup$ Commented May 4, 2012 at 23:57
  • $\begingroup$ 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.) $\endgroup$
    – Ted
    Commented May 6, 2012 at 6:54

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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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    $\begingroup$ Since $\widetilde{AB}=\widetilde{B}\widetilde{A}$ for any elements of the Clifford algebra, including vectors, the proof follows trivially from your first assertion. However, it is not obvious to me that any blade can be written as you claim. Keep in mind that the metric is not necessarily Euclidean. $\endgroup$ Commented May 2, 2013 at 5:40
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    $\begingroup$ Your proof for a bivector works only if $a^2\ne0$ which is not the case in general. $\endgroup$ Commented May 3, 2013 at 2:59
  • $\begingroup$ if $a^2=0$ but not $b$ it can be done the other way around $a\wedge b=a'b$. If $a^2=b^2=0$ then for a bivector $(a\wedge b)(b\wedge a)=(ab-\lambda)(ba-\lambda)$ $=\lambda^2-2\lambda a\cdot b$ which is scalar. Proof could follow from induction although the inductive step might not be $n\rightarrow n+1$ i.e. $1\&\dots\&n\rightarrow n+1$. If I prove it will post afterwards. $\endgroup$
    – DFlores
    Commented May 7, 2013 at 15:39
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    $\begingroup$ 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. $\endgroup$
    – mr_e_man
    Commented May 24, 2018 at 3:35

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