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Clifford's algebras have basis elements squaring either to $1$, $0$ or $-1$. Why not $i$?

Is that already covered by $1$, $0$ or $-1$?

If there is an algebra squaring to $i$, what is his name? I have interest on the geometric interpretation of the geometric product of basis vectors squaring to $i$.

I mean something like $Cl_{p,q,r,s}$ where s is the number of basis vectors $e_j$ such that $e_j*e_j=i$

Colim
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    If you're talking about complex Clifford algebras, the scalar $i$ itself has a scalar square root, so anything squaring to $i$ or $-1$ (or anything nonzero, really) can be rescaled so it squares to $+1$. If you're talking about real Clifford algebras, there is no scalar $i$ to square to. – anon Aug 29 '22 at 00:50

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The situation of real Clifford algebras having bases which "square" to $\pm 1$ or $0$ is just a consequence of Sylvester's law of inertia: every quadratic form on a real vector space can be diagonalized, by a change of basis, with diagonal entries $\pm 1$ or $0$, with a uniquely-determined number of each.

But there is no compulsion to choose such a basis. So we could have a vector space basis whose "squares" were $\pi$, $\sqrt{2}$, and lots of other things.

For real vector spaces, no, we can't get $i$ as a "square", because it's not a real number. But, for complex vector spaces, with quadratic (not hermition) forms, we can certainly have "square" values $i$, or $\sqrt{-19}$, and so on.

Likewise, to make a Clifford algebra with "exotic" values, we just need to have vector spaces over a field that contains those exotic values. :)

paul garrett
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  • Why not simply define $e_j$ such that $e_j*e_j=i$? like $i$ was used originally? – Colim Aug 29 '22 at 01:06
  • @Colim Why is the word "simply" there in your question? The word "simply" to me implies that it is more simple or natural than defining $e^2=\pm1$, which is not true. And what do you mean by "like $i$ was used originally"? – anon Aug 29 '22 at 01:12
  • When $i$ was discovered, $i$ was originally used as "something" that squared to $-1$, without any understanding of how it worked, or what was his geometric interpretation. It was "simply" defined, and worked with. – Colim Aug 29 '22 at 01:15
  • It's not to "define $e_j$ so that...", but to give a quadratic form on a vector space: let the v.s. be $\mathbb C^2$, and quadratic form $Q(z,w)=iz^2+5w^2$... – paul garrett Aug 29 '22 at 01:22
  • Yeah, $i$ was used to square to $-1$. It was not used to square to itself (which would have been circular), so I'm not sure why you're suggesting $e^2=i$ is analogous to $i^2=-1$. The analogy with complex numbers is imperfect anyway: they were created to solve equations, and ended up modeling 2D rotations as a byproduct; Clifford algebras are created to model nD rotations (which is why there are multiple basis elements squaring to $-1$, e.g.). – anon Aug 29 '22 at 01:29
  • @runway44, I'm not saying that squaring to $i$ is the same as squaring to $-1$, but, only, that Clifford algebras can include all these things... – paul garrett Aug 29 '22 at 01:57
  • @runway44 Which algebra models $rD$ rotations ($r \in \mathbb{R}$)? For example, if $e_i, e_j$ are $0,5D$ blades, such that $e_ie_j=e_1$ ($e_1^2=1$, is an 1D blade), then $(e_ie_j)^2=e_1^2=1$. To preserve anticomutativity $e_ie_j=-e_je_i$, is necessary that $e_i^2=e_j^2=i$ – Colim Aug 29 '22 at 02:03
  • paul: Wasn't responding to you :P @Colim It's not clear to me how to interpret fractional-dimension rotations or why you're using the same basis notation for "0.5D blades" as 1D blades. But I see now you're trying to construct a "square root" of vectors in Clifford algebras by extending to a bigger algebra. But that would be something different from a Clifford algebra, so it still doesn't make sense to ask why we don't have $e^2=i$ in Clifford algebras to my mind. – anon Aug 29 '22 at 02:27