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• CommentRowNumber1.
• CommentAuthorEric
• CommentTimeApr 17th 2010
• (edited Apr 17th 2010)

Why vector spaces and not modules?

Why nondegenerate symmetric bilinear form?

Is there an inverse?

Is there an antisymmetric product

$u\wedge v = u v - \langle u,v\rangle e$

?

• CommentRowNumber2.
• CommentAuthorTodd_Trimble
• CommentTimeApr 17th 2010

You mean composition algebra. :-)

I guess you know that the classical application of all this is to composition algebras over $\mathbb{R}$, where there are only four: the reals, the complexes, the quaternions, and the octonions. This result is due to Adolph Hurwitz, over a century old now I think. It’s a gorgeous piece of pure algebra which works over any field whose characteristic is not 2, but I’m pretty darn sure the result will not hold in the absence of nondegeneracy. (Disclaimer: I’m not an expert on this stuff by any means, although I’ve been through the proof before.) I haven’t gotten to the point in the proofs where nondegeneracy is invoked, but it’s coming.

As you know, modules over fields are the same thing as vector spaces over that field, so I guess your question might really be: why fields and not more general commutative rings? I haven’t thought about it much. I think nondegeneracy of the bilinear form in that case would force the module to be finitely generated projective, which is close in some sense to its being free (“locally free”). I’ll need some time to consider whether the proofs to come are going to break down in this generality though. Interesting question.

I guess you mean multiplicative inverses of nonzero elements. “Yes”: $v^{-1} = \bar{v}/N(v)$. I’ll give a proof a bit later that this works.

There is an antisymmetric product that’s often used, but the formula for it is $u \wedge v = \frac1{2}(u v - vu)$.

• CommentRowNumber3.
• CommentAuthorEric
• CommentTimeApr 17th 2010
• (edited Apr 17th 2010)

Thanks Todd.

So if the antisymmetric product is not $u v - \langle u,v\rangle e$, I guess that means $\langle u,v\rangle e$ is not $\frac{1}{2}(u v + v u)$ so there is probably some other symmetric multiplication

$S(u,v) = \frac{1}{2} (u v+ v u) - \langle u,v \rangle e.$

Yeah, I know this stuff is ancient, but that doesn’t take away from the fun of thinking about it :)

• CommentRowNumber4.
• CommentAuthorEric
• CommentTimeApr 17th 2010
• (edited Apr 17th 2010)

I think nondegeneracy of the bilinear form in that case would force the module to be finitely generated projective, which is close in some sense to its being free (“locally free”).

Oh! That sounds interesting. This is related to a question I was going to ask after skimming one of Mike’s papers recently. What are finitely generated projectives? The relation to nongenerate symmetric bilinear forms on modules would be interesting. Is there some kind of relation between the two concepts? For example, given finitely generated projectives can you construct a nondegenerate symmetric bilinear form on the free module? Given a nondegenerate symmetric bilinear form on modules, does it imply finitely generate projectives (whatever those are)? Is there a relation between inner products and free modules?

These are all questions dear to my heart, but approached from an engineering perspective.

• CommentRowNumber5.
• CommentAuthorDavidRoberts
• CommentTimeApr 17th 2010

A finitely generated projective module is as close one can usually get to a module being a finite dimensional vector space (=fin. gen. free module over a field). A definition equivalent to the usual one is that $P$ is projective iff for every epimorphism $M \to P$ of modules (over the same ring) there is a section $P \to M$. This implies that $M \simeq P \oplus N$. For a finitely generated projective $R$-module ($R$ a commutative ring), we can take $M = R^n$, and so f.g. projective $R$-modules are direct summands of f.g. free modules.

• CommentRowNumber6.
• CommentAuthorEric
• CommentTimeApr 17th 2010

Thanks David. What is “projective” about it?

Is this related to algebra of projections (ericforgy) where we have a complete set of idempotents $\pi_i$ satisfying

$\sum_i \pi = 1$

and

$\pi_i\circ\pi_j = \delta_{i,j} \pi_i$

and the resulting free (differential graded) module here or is it totally unrelated?

• CommentRowNumber7.
• CommentAuthorDavidRoberts
• CommentTimeApr 17th 2010

Not sure where the name came from, but it is dual to injective. My guess is that the epimorphism $M \to P$ is, under the identification of $M$ with the direct sum, just the projection onto the $P$ factor. As far as I know, it is completely unrelated to an algebra of projections (and I don’t expect it to be)

• CommentRowNumber8.
• CommentAuthorTodd_Trimble
• CommentTimeApr 17th 2010

It may be at least a few days before I can get back to editing at the Lab, because I’ll be traveling with my kids. Just so you know if I seem silent.

What I was saying about nondegenerate symmetric bilinear form on $P$ ($P$ finitely generated) implying projectivity of $P$ might be wrong, but what I vaguely had in mind when I said that is that the pairing $\langle, \rangle: P \otimes P \to k$ would be the counit of an adjunction between $P$ and itself, with the unit being a map $k \to P \otimes P$ such that the triangular equations hold, and I know that such an adjunction forces finitely generated projective. This all has to do with enriched Cauchy completion; the Cauchy completion of a ring as one-object $Ab$-enriched category is the $Ab$-enriched category of its f.g. projective modules.

“Projective” means $\hom(P, -)$ is right exact (preserves colimits of finite diagrams), or that every epi mapping to $P$ splits. “Injective” means that $\hom(-, P)$ takes finite limits contravariantly to finite colimits, or that every mono mapping from $P$ splits.

When I said f.g. projectives over a ring $R$ are “locally free”, I meant that when you localize such things with respect to a prime ideal of $R$, you get a free module over the corresponding local ring, and I thought I remembered that the converse is true too (if all such localizations of a f.g. module are free, then the module is f.g. projective). But it’s late and I’m tired and I could well be wrong!

• CommentRowNumber9.
• CommentAuthorEric
• CommentTimeApr 17th 2010

Have a good trip! :)

• CommentRowNumber10.
• CommentAuthorTodd_Trimble
• CommentTimeApr 29th 2010

Am in the process of adding more material to composition algebra.

• CommentRowNumber11.
• CommentAuthorTodd_Trimble
• CommentTimeMay 1st 2010

Okay, I think I’m about done for now adding stuff to composition algebra. At some point material should be added which relates automorphism groups of these structures to exceptional Lie groups. John Baez is expert on that, but I don’t know that he reads the Forum much.