Suppose one has a geometric algebra $\mathbb{G}^{p,q} \equiv \mathbb{G}(\mathbb{R}^{p,q})$. And suppose $M$ is a general multivector in $\mathbb{G}^{p,q}$. One can define a real quantity $\Vert M \Vert^2 = \langle M^\dagger M \rangle$ where $M^\dagger$ is $M$'s reverse and $\langle \rangle$ is the 0-vector (scalar) part of the multivector it encloses. But that quantity is not always nonnegative when $q > 0$; as an example, take $M = \mathbf{e}_{p+1}$ from a standard orthonormal basis $ \left\{ \mathbf{e}_{1}, \ldots, \mathbf{e}_{p+q} \right\}$. So one can't in general take the square root of $\Vert M \Vert^2$ to form a real quantity $\Vert M \Vert$. (Complex numbers per se are not allowed in geometric algebra.)
(So defined, $\Vert M \Vert^2$ seems rather like the idea of special relativity's spacetime interval, $ds^2 = (c\,dt)^2 - dx^2 - dy^2 - dz^2$. When $ds^2 > 0$, one can regard it as the proper time squared; and when $ds^2 < 0$, one can regard $-ds^2$ as the proper length squared.)
Of course one could modify the above definition to read $\Vert M \Vert^2= \left| \langle M^\dagger M \rangle \right|$ and then take the square root to form the nonnegative $\Vert M \Vert = \sqrt{\left| \langle M^\dagger M \rangle \right|}$. But the mapping $M \mapsto \Vert M \Vert$ thus obtained is not a norm in the usual sense of that word. If nothing else, it's not positive definite when $q>0$, for then $\Vert \mathbf{e}_1 + \mathbf{e}_{p+1} \Vert = 0$ even though $\mathbf{e}_1 + \mathbf{e}_{p+1} \ne 0$.
So is anyone aware of a natural norm on $\mathbb{G}^{p,q}$, i.e. a norm that arises from the operations of geometric algebra and is not basis dependent? To what uses might one put such such a mapping?
To clarify my question: I am interested in whether there is a natural norm which works for any multivector $M$ in any geometric algebra $\mathbb{G}^{p,q}$. I already know about the Euclidean case ($q = 0$), for which $\langle M^\dagger M \rangle$ is a positive definite quantity and so may have its square root taken. And I already know how to handle blades. It's the general case I'm wondering about.
The motivation behind my question is improvement of the GAlgebra extension to the SymPy computer algebra system.
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I'm not sure where you would use an arbitrary multivector norm.
However, your $\Vert M \Vert = \sqrt{\left| \langle M^\dagger M \rangle \right|}$ is a useful quantity for vectors, bivectors, and trivectors in $\mathbb{G}^{3,0}$, as it compactly encodes the (unoriented) length, area, and volume of vectors, parallelograms and parallopipeds (i.e. of $\mathbf{a}$, $\mathbf{a} \wedge \mathbf{b}$ and $\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}$).
For Euclidean applications like this, you can drop either the absolute value or the reversal, as well as the scalar selection and use one of:
$\Vert M \Vert = \sqrt{\left| M^2 \right|} \qquad (1)$
or
$\Vert M \Vert = \sqrt{ M^\dagger M } \qquad (2)$
That is, provided you are interested in a norm for a blade and not a general multivector.
For example, if you want the unit bivector representing the plane spanned by vectors $\mathbf{a}$ and $\mathbf{b}$, you could write that as just:
$(\mathbf{a} \wedge \mathbf{b})/\Vert \mathbf{a} \wedge \mathbf{b} \Vert \qquad (3)$.
I feel that would be a pretty intuitive notation, but am not sure whether it is widely used (I've used it in my own chicken scratching, and perhaps also in more formal writing.)
The norm (1) above also works nicely for 2-blades in $\mathbb{G}^{3,1}$. Using the same example (3) above, in a relativistic context, that unit bivector could represent a spatial unit-bivector for a plane of rotation or spacetime plane that represents a boost "direction".