# Squared norm in Clifford Algebras

1. Sep 23, 2010

### mnb96

Hello,
I know that the squared norm of a multivector M in a Clifford Algebra $$\mathcal{C}\ell_{n,0}$$ is given by:

$$<M \widetilde{M}>_0$$

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 $$\mathcal{C}\ell_{p,q}$$ or at least for $$\mathcal{C}\ell_{0,n}$$ ?

Thanks!

2. Sep 23, 2010

Take the scalar part of
$$a^\tau b$$,
where $$a\mapsto a^\tau$$ is the principal anti-automorphism. This is a nondegenerate, symmetric scalar product for any $$Cl(r,s)$$ - in general indefinite. See for instance http://arxiv.org/abs/quant-ph/0608117" [Broken], Proposition 2.

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3. Sep 23, 2010

### mnb96

Is the principal anti-automorphism equivalent to the Clifford conjugation?
That is, for a given n-vector $x=a_1\ldots a_n$,

$$x \mapsto (-1)^{n(n+1) / 2} x$$

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
4. Sep 23, 2010

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5. Sep 23, 2010

### mnb96

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

6. Sep 23, 2010

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?

7. Sep 23, 2010

### mnb96

Thanks!
so if understood correctly, for a general Clifford algebra CL(p,q) it should be sufficient to define a "conjugation" of the kind:

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

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!

8. Sep 23, 2010

Get this paper: "http://hal.inria.fr/hal-00119996_v1/" [Broken]", René Schott, Stacey Staples, Eqs. (1.7,1.8)

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9. Sep 23, 2010

### mnb96

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" $$(-1)^{q} \widetilde{x}$$ was used implicitly.

Thanks a lot! You were very helpful.

10. Sep 23, 2010