# Squared norm in Clifford Algebras

• mnb96
In summary, the squared norm in Clifford algebras is a mathematical concept used to measure the length of a vector or quaternion in a higher-dimensional space. It is calculated by multiplying the vector or quaternion by its conjugate and then summing the squares of the resulting components. This squared norm is especially useful in geometric algebra, which utilizes Clifford algebras to represent and manipulate objects in multi-dimensional space. It allows for the calculation of distances, angles, and other geometric properties in a concise and elegant manner, making it a powerful tool in mathematical and scientific applications.
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!

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" , Proposition 2.

Last edited by a moderator:
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:
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 remaining cases?

mnb96 said:
is it always possible to define a positive-definite non-degenerate symmetric scalar product for Clifford algebras CL(p,q)?

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?

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!

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

Last edited by a moderator:
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.

You are welcome.

## 1. What is the squared norm in Clifford Algebras?

The squared norm in Clifford Algebras is a mathematical concept used to measure the magnitude of a vector or element in the algebra. It is similar to the concept of squared magnitude in standard vector spaces. However, in Clifford Algebras, the squared norm can take on negative, zero, or positive values, making it a more versatile and powerful tool for representing and manipulating geometric objects.

## 2. How is the squared norm calculated in Clifford Algebras?

The squared norm in Clifford Algebras is calculated by multiplying an element by its conjugate. This operation results in a scalar value that represents the squared magnitude of the element. In other words, it is the sum of the squares of the coefficients of the element's basis vectors.

## 3. What is the significance of the squared norm in Clifford Algebras?

The squared norm in Clifford Algebras has several important applications in mathematics and physics. It is used to define distance and angles, as well as to measure the similarity between geometric objects. The squared norm is also closely related to the concept of inner product, which is used in many calculations and proofs involving Clifford Algebras.

## 4. Can the squared norm be negative in Clifford Algebras?

Yes, the squared norm in Clifford Algebras can take on negative values. This is due to the presence of multivectors, which have both scalar and vector components. In traditional vector spaces, the squared magnitude is always positive, but in Clifford Algebras, it can be positive, negative, or zero, depending on the elements involved.

## 5. How is the squared norm used in geometric algebra?

The squared norm is a fundamental concept in geometric algebra, as it allows for the representation and manipulation of geometric objects in a more compact and efficient way. It is used to define the magnitude and direction of vectors and to perform operations such as projection, reflection, and rotation. The squared norm is also used in the construction of other important algebraic concepts, such as the inverse and reciprocal of elements in Clifford Algebras.

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