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An interesting question, which has been discussed in many forms on this site, is how many results from the study of linear algebra over vector spaces carries over when we allow the scalars to form an algebraic structure more general than a field.

For example, if you stop requiring multiplicative commutativity and allow the scalars to form an arbitrary division ring, then a surprising amount of structure carries over unchanged, as discussed here: you lose the notion of the determinant and eigenvalues become more subtle, but you still have unique basis cardinality, Gaussian elimination, the Rouché-Capelli theorem, and matrix representations of arbitrarily linear maps between finite-dimensional modules. A similar story applies if you allow the scalars to form an arbitrary commutative ring. However, all hell can break loose if you drop commutativity and division: a module over an arbitrary ring does not necessary have an invariant basis number, so much of the structure of linear algebra over fields immediately falls apart.

What happens if you further generalize the scalars' division ring to be a division algebra that is not associative? How much structure do you lose if you give up associativity but preserve division?

To be concrete, consider generalized modules with the octonions as their "scalars". (A precise definition of what I mean is given in this question, together with the requirement $(rs)\cdot x = r \cdot (s \cdot x)$.)

  1. Do these modules have a notion of invariant basis number?

  2. If so, can arbitrary linear maps between finitely generated such generalized modules be represented by matrices? Are the matrix elements given by the usual formula $A(\hat{e}_j) = \sum_i A_{ij} \hat{e}_i$?

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    $\begingroup$ It's very unclear to me a priori how to make sense of the notion of a module over something that isn't associative. The reason the question you linked to does not require $(rs) x = r(s(x))$ is that it would rule out the octonions as a module over themselves. $\endgroup$ – Qiaochu Yuan Jul 29 '18 at 22:35
  • $\begingroup$ @QiaochuYuan Yes, according to the definition that I'm considering the octonions would not form a module over themselves. When you say you can't "make sense of" my definition, do you mean you don't understand it, or you don't have much intuition for such an algebraic structure? $\endgroup$ – tparker Jul 29 '18 at 22:44
  • $\begingroup$ Qiaochu's observation does raise the question whether there is any nontrivial module over the octonions that satisfies your rule. $\endgroup$ – Henning Makholm Jul 29 '18 at 22:47
  • $\begingroup$ Let M be such a module. The requirement that (rs).x = r.(s.x) implies that (rs)t and r(st) act identically on M. In other words, M is really a module over the octonions modulo their associator - whatever this is. (I suspect it's the quaternions or something.) $\endgroup$ – Billy Jul 29 '18 at 22:52
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    $\begingroup$ @Billy: If I'm understanding Wikipedia's description of octonion multiplication correctly, quotienting out the associators would force first all seven imaginary units to equal each other, and then their common value to be $0$, since $e_4=e_1e_2=e_2e_1=-e_4$ and similarly for other values of $4$. But then $-1=e_1e_1=0$ and everything collapses completely. $\endgroup$ – Henning Makholm Jul 29 '18 at 23:22
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As Billy points out, the rule $(rs)\cdot x=r\cdot(s\cdot x)$ implies that $(rs)t$ and $r(st)$ will act identically on your module.

According to Wikipedia's multiplication table we have $$ e_1(e_3e_5) = -e_1e_6 = e_7 \\ (e_1e_3)e_5 = -e_2e_5 = -e_7 $$ so we must have $$ e_7\cdot x = -e_7\cdot x $$ whence (by distributivity) $$ 2e_7\cdot x = 0 $$ and by dividing by $2e_7$, $$ 1\cdot x = 0 $$ Since the module axioms also require $1\cdot x$ to be $x$, this means that the only module that satisfies all of your rules is $\{0\}$ -- not very exciting. Thus:

  • 1. Yes, the basis number is invariant. Every basis contains exactly $0$ vectors.

  • 2. Yes, if you allow $0\times 0$ matrices.

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