Talk:Invariant decomposition

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	This article was accepted from this draft on 20 July 2022 by reviewer Femke (talk · contribs).

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Fundamental Issues

Latest comment: 4 days ago1 comment1 person in discussion

There are some fundamental issues with this article, which seems to be almost entirely based on the cited Roelfs & De Keninck "Graded Symmetry Groups" paper which has the same issues.

In the degenerate case, Cartan-Dieudonne and the composition of the Pin groups is absolutely not as described. (I also want to note that Roelfs & De Keninck's proof of Cartan-Dieudonne fails for *any* isotropic quadratic form, not just degenerate ones. Cartan-Dieudonne is still true for any nondegenerate form, however.) Several things happen:
1. It is false that all orthogonal transformations are products of reflections, though we can fix this be requiring the action on the radical to be the identity; this is the group that Pin covers. However, I believe it is unknown whether or not the number of reflections is bounded by $n=p+q+r$ . We still have bounds: a good uniform one is $4n/3$ , and a tighter one is $\min(2(p+q),p+q+2r)$ .
2. But this is separate from bounding products of vectors. The kernel of the Pin group into the orthogonal group contains not just scalars but also nontrivial invertible radical even multivectors. If we want to use the above bounds, all we can say is that every product of invertible vectors is e.g. a product of $\lfloor 4n/3\rfloor$ invertible vectors and some invertible radical even multivector. I do not know if its known how to isolate such a $4n/3$ product.
3. For the two cases that Roelfs & De Keninck are probably actually interested in, namely $r=0,1$ things happen to work out: when $r=1$ the only invertible radical multivectors are a sum of a nonzero scalar and a vector, which is not even and so not in the Pin group. It also known that we have a tight bound of $n$ vectors when $q=0$ .
Saying that Riesz was wrong about his counterexample to invariant decomposition is utterly disingenuous. Riesz is clearly not interested in the complexification; it's ridiculous to suggest he didn't understand that you could complexify. I also fail to see any geometric significance to considering the complexification. This seems to a be step backwards from the general geometric algebra tenet of removing "opaque" uses of complex numbers.

Loppy1243 (talk) 22:59, 13 July 2024 (UTC)Reply

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