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Squared norm in Clifford Algebras

  1. Sep 23, 2010 #1
    Hello,
    I know that the squared norm of a multivector M in a Clifford Algebra [tex]\mathcal{C}\ell_{n,0}[/tex] is given by:

    [tex]<M \widetilde{M}>_0[/tex]

    that is the 0-grade part of the product of M and its grade-reversal.


    Is there a more general definition of squared-norm (for multivectors) that works for any Clifford algebra [tex]\mathcal{C}\ell_{p,q}[/tex] or at least for [tex]\mathcal{C}\ell_{0,n}[/tex] ?

    Thanks!
     
  2. jcsd
  3. Sep 23, 2010 #2
    Take the scalar part of
    [tex]a^\tau b[/tex],
    where [tex]a\mapsto a^\tau[/tex] is the principal anti-automorphism. This is a nondegenerate, symmetric scalar product for any [tex]Cl(r,s)[/tex] - in general indefinite. See for instance http://arxiv.org/abs/quant-ph/0608117" [Broken], Proposition 2.
     
    Last edited by a moderator: May 4, 2017
  4. Sep 23, 2010 #3
    Is the principal anti-automorphism equivalent to the Clifford conjugation?
    That is, for a given n-vector [itex]x=a_1\ldots a_n[/itex],

    [tex]x \mapsto (-1)^{n(n+1) / 2} x[/tex]

    Is that what you meant?


    ***EDIT:***
    Probably not: it won't work for CL(2,0) because of [itex](e_1)^2 = -1[/tex]
     
    Last edited: Sep 23, 2010
  5. Sep 23, 2010 #4
    Last edited by a moderator: Apr 25, 2017
  6. Sep 23, 2010 #5
    Ah! I see.
    so I guess what is confusing me, is that the scalar product defined with Clifford conjugation is not necessarily positive, which is what I was looking for.

    So the question now becomes:

    is it always possible to define a positive-definite non-degenerate symmetric scalar product for Clifford algebras CL(p,q)?

    For CL(n,0) the answer is yes, if we use reversion
    For CL(0,1) the answer is yes, if we use Clifford conjugation
    For CL(0,2) the answer is yes, if we use Clifford conjugation

    What about the remaing cases?
     
  7. Sep 23, 2010 #6
    Sure, it is possible. Choose a basis and define your scalar product as the Euclidean one in this basis. Some people use it - sometimes it is handy. But the question is: what additional nice properties do you want to have for your scalar product?
     
  8. Sep 23, 2010 #7
    Thanks!
    so if understood correctly, for a general Clifford algebra CL(p,q) it should be sufficient to define a "conjugation" of the kind:

    [tex]
    x \mapsto (-1)^{q + n(n+1) / 2} x = (-1)^{q} \widetilde{x}
    [/tex]

    You said that some people use this, though I could not find it in the literature.
    Does this sort of "conjugation" have a name?
    If you have time, could you please point out one reference?

    I think the properties I was looking for were non-degeneracy, symmetry, positive-definiteness.

    Thanks a lot!
     
  9. Sep 23, 2010 #8
    Get this paper: "http://hal.inria.fr/hal-00119996_v1/" [Broken]", René Schott, Stacey Staples, Eqs. (1.7,1.8)
     
    Last edited by a moderator: May 4, 2017
  10. Sep 23, 2010 #9
    It seems that the authors of that paper do not introduce a "conjugate" at all. Instead, they simply define the Clifford inner-product in such a way that it yields the same result as if the "conjugate" [tex](-1)^{q} \widetilde{x}[/tex] was used implicitly.

    Thanks a lot! You were very helpful.
     
  11. Sep 23, 2010 #10
    You are welcome.
     
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