Length of a Clifford Poly-vector?

The length of a vector is defined as:

$$||\mathbf{v}||^2=\mathbf{v}\cdot \mathbf{v}$$

In the case that $$\mathbf{v}:=a\hat{x}+b\hat{y}$$ is expressed in an orthogonal basis using $$\hat{x}$$ and $$\hat{y}$$ as the generators of a Clifford algebra $$Cl_{0,2}(\mathbb{R})$$, then $$||\mathbf{v}||^2=a^2+b^2$$

For a poly-vector (say $$\mathbf{p}:=c+a\hat{x}$$), one can also define an inner product. Then if one takes the inner product, one gets

$$||\mathbf{p}||^2=\mathbf{p}\cdot \mathbf{p}=c^2+a^2$$

However, I am skeptical of the inner product defines the length of the poly-vector primarily become the scalar $$1$$ and the 1-vector basis $$\hat{x}$$ are not orthogonal. Intuitively I would think the square of the geometric product of $$\mathbf{p}$$ with itself would be the length.

$$||\mathbf{p}||=\sqrt{\mathbf{p}\mathbf{p}}$$

In the case where the vectors are orthogonal k-vectors, the result is the same. For example

$$\sqrt{\mathbf{v}\mathbf{v}}=\sqrt{(a\hat{x}+b\hat{y})(a\hat{x}+b\hat{y})}\\ =aa\hat{x}\hat{x}+2ab(\hat{x}\hat{y}+\hat{y}\hat{x})+ bb\hat{y}\hat{y}\\ =\sqrt{a^2+b^2}$$

But, in the case where the poly-vector is not a k-vector, the definition differs:

$$\sqrt{\mathbf{p}\mathbf{p}}=\sqrt{(c+a \hat{x})(c+a \hat{x})}\\ \sqrt{c^2+2ac\hat{x}+ aa\hat{x}\hat{x}}\\ \sqrt{c^2+a^2+2ac\hat{x}}$$

Using this definition, we conclude that in the case of a poly-vector, a scalar length cannot be defined. Thus, define such a line as the inner product ought to erase some important geometric information about the length of the poly-vector.

Is this correct? Is there a standard definition for the length of a poly-vector?

• Your definition doesn't make any sense, since there is no well-defined square root function on the Clifford algebra. – Hans Lundmark Aug 8 at 13:59
• @Hans there might not be in the general case, but that just means that not all poly-vectors represent the length of another poly-vector. In the case of the geoemtric product of a poly-vector with itself, the resulting vector definitely has an inverse, which I define as the square root. $\mathbf{p}^{-1}\mathbf{p}\mathbf{p}=\mathbf{p}\equiv \sqrt{\mathbf{p}\mathbf{p}}$. – Alexandre H. Tremblay Aug 8 at 14:10
• But in general there are many elements in the Clifford algebra which have the same square! If $\mathbf{A} = \mathbf{p} \mathbf{p} = \mathbf{q} \mathbf{q}$, then should you take $\sqrt{\mathbf{A}}$ to mean $\mathbf{p}$ or $\mathbf{q}$? There's no way of telling, and that's why your definition is no real definition. – Hans Lundmark Aug 8 at 14:26
• By the way, contrary to what you're saying, the scalars are orthogonal to the vectors wich respect to the usual inner product that one defines on the whole Clifford algebra. – Hans Lundmark Aug 8 at 14:29
• @Hans If $4=2*2=(-2)*(-2)$, do you take $\sqrt{4}$ to mean $2$ or $-2$? – Alexandre H. Tremblay Aug 8 at 14:30

Let us assume that the quadratic space (V,g) over which the Clifford algebra $$\mathcal{C}\ell(V)$$ is being constructed is real. Just to fix the notation, we shall denote the Clifford product by juxtaposition. One can then define \begin{align} N:\mathcal{C}&\ell(V)\rightarrow\mathbb{R}\\ &a\mapsto \langle \tilde{a}a\rangle_0 \end{align} where $$\langle\;\cdot\;\rangle_0$$ is the projection onto the scalar part and $$\widetilde{ab}=\widetilde{b}\widetilde{a}$$ is the reversion anti-automorphism. In particular, for $$\alpha\in\mathbb{R}$$ and for a basis $$\{e_i\}$$ there holds $$\widetilde{\alpha}=\alpha$$ and $$\widetilde{e_i}=e_i$$. Then, you would have defined a (squared) seminorm in the Clifford algebra. If $$g$$ is positive definite, then $$N$$ is indeed a norm. Namely, in the case $$V=\mathbb{R}^2$$ with the canonical inner product, an arbitrary element $$\mathcal{C}\ell(V)\ni a=\alpha+a_1e_1+a_2e_2+b_{12}e_{12}$$ has $$N(a)=\alpha^2+a_1^2+a_2^2+b_{12}^2,$$ which is just the canonical inner product over the $$2^n$$-dimensional vector space structure of $$\mathcal{C}\ell(V)$$. Notice that for every $$v\in \mathbb{R}^2$$, there holds $$N(v)=g(v,v)$$, so $$N$$ is compatible to the original metric.
Trying to define a norm not considering the scalar part, I can only think of defining \begin{align} N':\mathcal{C}&\ell(V)\rightarrow\mathbb{R}\\ &a\mapsto N(a-\langle a \rangle_0) \end{align} In the $$V=\mathbb{R}^2$$ case, if you would even want to discard the bivector part then you can make $$N'(a)= N(a-\langle a \rangle_{0,2})$$. But then again, $$N'(a)=g(\langle a \rangle_1,\langle a \rangle_1),$$ which is just the norm in the vector part of $$a$$.
• Here are my impressions: By defining $N(a)$ as you have done, you are stating that the distance between a scalar and an element of the $\hat{x}$ axis is given by the generalization of the Pythagorean theorem. But this cannot be the case: The Pythagorean theorem relates two (or more) elements of an axis such as $\hat{x}$ and $\hat{y}$. Now, if you throw in a scalar, your definition gives $d^2=x^2+y^2+a^2$ and mine gives $d=\sqrt{x^2+y^2}+a$. In my definition, the contribution of the scalar to the distance is independent of the position on the graph (as it should be). Thoughts? – Alexandre H. Tremblay Aug 17 at 11:42
• @AlexandreH.Tremblay Why should the "contribution of the scalar to the distance" be independent? This is the most natural metric, since it is just the inner product in the vector space structure of $\mathcal{C}\ell(V)$. Even more: each homogenous space (of $k$-vectors) is orthogonal from each other with respect to $N$. In fact, the spin group is defined in terms of this metric. – amnesiac Aug 18 at 4:50
• Say you have a cartesian graph with axis $\hat{x}$ and $\hat{y}$. Now picture point (3,4) with distance 5 from the origin. Now picture a scalar unit that we call m (the amount of money in your pocket). I construct a poly-vector $\mathbf{v}=m+x\hat{x}+y\hat{y}$ and I inquire about the 'distance' from to origin to get there. If I travel along the x-axis for 3 units, then my distance to the point (3,4) on the cartesian plane is now shorter according to the Pythagorean theorem. However, my distance to having 10$in my pockets is not made shorter by having walked on the plane. – Alexandre H. Tremblay Aug 18 at 13:49 • As for your second point, would you have an example of such a multi-vector? My research led me to believe that the geometric product of a multi-vector with itself always produces a scalar. Using the matrix representation, I have tested it with a general multi-vector generated by the Pauli matrices and the Dirac matrices, and Mathematica does solve the geometric product to a scalar for these special cases. (Since the square root is such an unexpected issue, please consider that I define the square of the distance as the geometric product$d^2=\mathbf{v}\mathbf{v}\$) – Alexandre H. Tremblay Aug 18 at 13:53