Talk:Outermorphism

Notation
The previous LaTeX

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showed errors, I changed them to

and

Seems like brackets are needed. M&and;Ŝc2ħεИτlk 05:40, 26 May 2013 (UTC)


 * Yes, MathJax does not appear to parse the expressions correctly. I am still getting some MathJax processing errors of a different nature.  — Quondum 18:30, 26 May 2013 (UTC)

Definition incomplete?
The description/definition of an outermorphism is not adequately covered by the lead. In particular, the lead is silent about the extension to scalars, its statement in terms of direct devolvement to vectors is overly complex. It also doesn't define the outermorphisms when applied to non-blades, even though this is assumed in the "thus we take...". I think that it would make more sense to define an outermorphism as a nonzero extension that preserves the wedge product (the nonzero requirement excludes ambiguity on when we are extending the zero map). Dorst defines an outermorphism as the extension of a linear function, constrained by the properties:
 * $$ \underline{\mathsf{f}}(\alpha) = \alpha $$
 * $$ \underline{\mathsf{f}}(A \wedge B) = \underline{\mathsf{f}}(A) \wedge \underline{\mathsf{f}}(B)$$
 * $$ \underline{\mathsf{f}}(A + B) = \underline{\mathsf{f}}(A) + \underline{\mathsf{f}}(B)$$

I think Dorst's requirement of additivity is redundant, and his requirement on scalars is nearly redundant (thus can be weakened to my nonzero requirement). Whatever approach is used, it should be complete. — Quondum 21:06, 26 May 2013 (UTC)
 * I think you technically need the constraint for scalars because otherwise the outermorphism of the zero map is ambiguous, and there's no reason for the definition to exclude the zero map. Basically you want this to work for anything you can represent with an n x n matrix, even the rank deficient ones.  I think the addition rule is redundant for some blades (like the derivation I showed), but you still need a rule to know how to deal with multivectors of mixed grade and such. Teply (talk) 22:23, 26 May 2013 (UTC)
 * Yes, once you have assumed $f(x) = f(x)$, $f(A ∧ B) = f(A) ∧ f(B)$ and $f(A + B) = f(A) + f(B)$, you need a constraint $f(1) ≠ 0$. AFAICT, this deals with all cases (n × m linear maps, where the domain and co-domain need not even be the same vector space, including all levels of map degeneracy).  I see you're right (I missed the point about the mixed grades), hence the axiom of distributivity over addition.  My mention of excluding the zero map refers to $f$, not to $f$, which is to say that the zero map (on the geometric algebra as a whole) is not an outermorphism, but we are allowing the zero map $f(x) = 0$ on the 1-vector space(s), which extends to $f(A) = ⟨A⟩$, not to the zero map.  The last inequality is purely to deal with when $f(x) = 0$, because except for this case we can deduce that $f(α) = α$, and in this case we can deduce that $f(α) = α$ or $f(α) = 0$.  But which scalar constraint is used is a matter of taste.  — Quondum 23:06, 26 May 2013 (UTC)
 * I've hopefully captured the result of discussion in the article. I ended up using a compromise for the scalar constraint, one which I seem to remember from one of the references.  — Quondum 00:22, 27 May 2013 (UTC)

A comment on GA versus exterior algebra as the context for outermorphisms
It is curious that this concept (outermorphism) appears to find itself within the study of geometric algebra rather than exterior algebra (which would also place it in geometric algebra since a GA always contains an exterior algebra). Its natural setting is exterior algebra, since its definition and applicability is entirely independent of the GA metric. Exterior algebra is also the more natural setting for statements about linear algebra. Is it not possible that this concept occurs in exterior algebra under another name, but due to some bright spark's proclivity for renaming existing concepts, visibility of the connection was lost? It seems to me to be the case – see Exterior algebra. — Quondum 00:55, 27 May 2013 (UTC)