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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?

Anon21
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  • Your definition doesn't make any sense, since there is no well-defined square root function on the Clifford algebra. – Hans Lundmark Aug 08 '19 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. $\mathbf{p}^{-1}\mathbf{p}\mathbf{p}=\mathbf{p}\equiv \sqrt{\mathbf{p}\mathbf{p}}$. – Anon21 Aug 08 '19 at 14:10
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    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 08 '19 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 08 '19 at 14:29
  • @Hans If $4=22=(-2)(-2)$, do you take $\sqrt{4}$ to mean $2$ or $-2$? – Anon21 Aug 08 '19 at 14:30
  • On the real numbers there is of course a well-defined (and well-known) square root function, but that's because the reals are ordered, so that you can decide to choice the nonnegative option. But already for complex number, square roots are much more complicated, and for Clifford algebras it's even worse. Not just the scalars $\pm 1$, but every unit vector, and other elements too, have the square $1$. So which one out of all these element do you choose as “the” square root of $1$? And how do you make this choice consistently and meaningfully for each element of the Clifford algebra? – Hans Lundmark Aug 08 '19 at 14:34
  • @Hans Hello, I get what you are saying but I do not understand why it is a problem to take the result of a square root on a poly-vector as the set of all possible solutions. As far as I see it, it just means that the length of a poly-vector contains less information about the poly-vector than the poly-vector itself (which is expected). In fact, multiple vectors have the same length. For instance, $v_1:=3\hat{x}+4\hat{y}$ and $v_2:=4\hat{x}+3\hat{y}$ have the same length, yet they are different vectors. One could ask for the set of all vectors whose length is $5$. – Anon21 Aug 08 '19 at 14:48
  • Well, I have to confess that I don't understand what it is that you want to accomplish with all this to begin with... – Hans Lundmark Aug 08 '19 at 15:35

1 Answers1

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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 $a$.

amnesiac
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  • 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? – Anon21 Aug 17 '19 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 '19 at 04:50
  • Moreover, I think you need a deeper look into the nature of the Clifford product: it does not make sense to take the square root of the product of an arbitrary poly-vector, since it may yield non-scalar parts. – amnesiac Aug 18 '19 at 04:53
  • 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. – Anon21 Aug 18 '19 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 $d^2=\mathbf{v}\mathbf{v}$) – Anon21 Aug 18 '19 at 13:53
  • A definition does not need to be aligned with our own interpretation of some structure. The scalar part in the Clifford algebra is not intrinsically connected to some value which should not be considered in the length of the poly-vector. In fact, one usually perceives the image of the $0,1$-projection $\langle;\cdot;\rangle_{0,1}(\mathcal{C}\ell(V))$ as the $(n+1)$-dimensional paravector space $\mathbb{R}\oplus\mathbb{R}^{n}$. Nonetheless, I will edit my answer and give you another suggestion, trying to encompass your vision of what the norm should be. – amnesiac Aug 18 '19 at 21:36
  • An example would be $p=\alpha+e_1$, where $\alpha\in\mathbb{R}\backslash{0}$ and $e_1^2=1$. Then $pp=(\alpha+e_1)(\alpha+e_1)=\alpha^2+1+2\alpha e_1\notin \mathbb{R}$. – amnesiac Aug 18 '19 at 21:39
  • Your example can be simplified further. One can rewrite $\alpha^2+2\alpha e_1+1$ to $2\alpha(a/2+e_1+a/2-a/2)+1$, then because $p=\alpha+e_1$, one can rewrite as $pp=2\alpha(p-a/2)+1$. Then finally, solve a quadratic equation $pp-2\alpha p=-a^2+1$ for $p$ and get $|p|=a\pm \sqrt{1}$, a scalar. – Anon21 Aug 18 '19 at 22:14
  • Let us continue this discussion in chat. – amnesiac Aug 18 '19 at 22:28