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I've not been working on the geometric algebra article for the time being. That does not mean I think it is in good shape; I think it still needs lots of work. I'll probably look at it again in time. I'm guessing that you may have some interest in learning about GR. If you want to exercise your (newly acquired?) geometric algebra skills, there might be something to interest you: a development of the math of relativistic gravitation in a Minkowski space background. Variations on this theme have been tried before, with some success. There are related articles around (Linearized gravity, Gravitoelectromagnetism). It might be fun to get an accurate feel of the maths behind gravity. Aside from a few presumptious leaps, the maths shows initial promise of being simple. Interested? Quondum^talk_contr 17:47, 8 December 2011 (UTC)[reply]

I've been having a good time reading Macdonald's survey on relativity. I finally feel I'm getting some small footholds. There are many equations in physics I would like to understand, both in the classical way and in the GA way. If you think this might be the "easiest" place to start, I would be happy to start here :) Rschwieb (talk) 18:30, 8 December 2011 (UTC)[reply]

I'm guessing you are referring to his recent "General Relativity in a Nutshell" (also called "Elementary General Relativity")? I've still to read it, but it looks like it might be very readible. It might be instructive to develop the ideas from postulates, especially if it is developed from scratch. It'll be helpful if you first get a feel for the EM (electromagnetic) formulae. I'm also on a learning curve. I'll create the page User:Quondum/sandbox/Relativistic fields in the next few hours, and we can take it from there. I'm certainly aiming at it being "easy". Quondum^talk_contr 20:27, 8 December 2011 (UTC)[reply]

I learned elementary classical mechanics and the vector calculus to do mechanics, but I never have learned Maxwell's equations. Some people's plight is that they grasp the physical situation, but not the advanced mathematics, but I have the opposite problem. For me, the gap between abstract mathematics and application is broad. You're right, I'm reading the "nutshell" article. It's been quite a boon, and I think it will be an excellent stepping stone. Rschwieb (talk) 22:46, 8 December 2011 (UTC)[reply]

I am somewhere in between. I love to work from a more "mathematical" (abstract) perspective, but really am only at an undergraduate level. I am fascinated by the elegant synergy between maths and physics, e.g. that Maxwell's equations may be expressed in GA as ∇F=J. The math also gives one exceptional insights into physics (e.g. the symmetry of the time and space coordinates). Physics seems to be a good way to focus one's math. See what you think of the page I've started in my sandbox. The only physics needed is identification of the model. Quondum^talk_contr 05:25, 9 December 2011 (UTC)[reply]

One question on the title you chose for this subsection. Can gravity exist in flat space? My (amateur) impression was that gravity is a result of curviture caused by the presence of mass. Let me know if I'm off, or maybe you mean something else in the title. Rschwieb (talk) 15:45, 9 December 2011 (UTC)[reply]

Very insightful of you. In GR, the postulate is that gravity is the curvature of spacetime, and in that context, the answer would be "No". What I am proposing is to replace the GR postulate with another (two others: flatness and that gravity is a field in that space). GR has some problems of its own, such as that the question of global energy conservation is extremely problematic (according to Penrose). The flat spacetime model almost certainly will not suffer from that problem, but is also presumably not equivalent to GR. Quondum^talk_contr 16:30, 9 December 2011 (UTC)[reply]

OK. This again reminds me that I don't know where to begin :) I guess any start is good. I'd like to understand whatever important models of electromagnetism and gravitation that exist. Rschwieb (talk) 17:15, 9 December 2011 (UTC)[reply]

But before we move on, could we tackle the difference between the (1,3) and (3,1) geometric algebras? It's strange because the +--- and -+++ sign conventions seem to have been on an equal footing before, and yet if the geometric algebras are different, that would seem to indicate something important. What leads do you have? Of course we can begin with the fact that Cℓ_3,1(R)≅M₄(R) and Cℓ_1,3(R)≅M₂(H). Let me know if you feel this warrants a talk page of its own. Rschwieb (talk) 15:30, 10 December 2011 (UTC)[reply]

I doubt it, but who knows. As entire algebras they are not isomorphic, and this is something that has intrigued some people before. Someone has even suggested that some double neutrino emission experiment could resolve which applies. My own feeling is that since we generally use only subspaces of GA and hardly ever the whole algebra, it may well be that there is no difference on the applicable subspaces. One of my targets (but I forget where I wrote a note to this effect ^{found: see first bullet on my trial ideas page}) was to determine whether the two algebras were totally isomorphic on the relevant subspaces. My feeling is that they probably are. I suggest deferring this. Will chat later; I'm on my way out. Quondum^talk_contr 16:14, 10 December 2011 (UTC)[reply]

Okay, a little bit more on the subject. Then you can decide whether you'd like to discuss it; I'd be happy to. It is something I'd like to look into at some point, and you might feel you want to explore it now. My idea is this: the metric on the space is a matter of convention anyway; whether a vector squares to the algebra's identity or its negation probably makes no difference. You just adapt the signs in your formulae. So what is "spacelike" is detemined by the majority sign in the metric. I think one will be able to show that show that the math is equivalent over vectors, versors acting on the algebra, and blades generally. The wedge product is identical (it is not affected by the metric). General elements probably never occur or interact. If you want to consider the detail, tell me and I'll create a page or a section in my sandbox for it. The following might be of interest: [1], and a related presentation: [2] Quondum^talk_contr 22:13, 10 December 2011 (UTC)[reply]

Great, this looks like interesting reading, if I can get around to it. It just seems like if the algebras are this different, then the geometry they describe might be different. On the other hand, maybe Clifford algebras are "overstructured" and even though they aren't identical as algebras, they might be identical for the purposes of geometric algebra. Rschwieb (talk) 04:10, 11 December 2011 (UTC)[reply]

Very nicely put. I would be very interested in finding out exactly where the two algebras are equivalent, and hence exactly where the boundary is on the math as a model for geometry. This would also be useful inasmuch as it would suggest a specific area of investigation for a possibly necessary change to the geometric interpretation; I imagine that this is effectively how the field of quantum mechanics arose. Spinors are a nice example of where a CA works but a straightforward geometric interpretation fails. Rotate an electron 360° on any axis, and you don't end up with the same thing, but with its negation. Geometrically there is no evident way to deduce thus, except that a 360° rotation is topologically distinct from 0°, whereas a 720° rotation is equivalent 0° (holds in any number of dimensions). This corresponds to an experimentally observable effect. Quondum^talk_contr 05:50, 11 December 2011 (UTC)[reply]

One thing Macdonald does not do a lot of is give you an idea of how to translate cross products into geometric algebra. Do you know any tricks/identities that are useful to translate between the two? Maybe there is something in that paper on products we were reading, but I don't remember anything in depth. Rschwieb (talk) 04:09, 15 December 2011 (UTC)[reply]

The two are directly related in ℝ³ (see here and here; cross product is chiral, wedge product is not). Given e₁×e₂=e₃, e₁²=e₂²=e₃²=+1, a and b both vectors, we get:

${\displaystyle a\times b=-e_{1}e_{2}e_{3}(a\wedge b)}$ and ${\displaystyle a\wedge b=e_{1}e_{2}e_{3}(a\times b)}$.

Also easily rewritten in terms of components: a mapping between components of a vector (×) and bivector (∧) — Quondum^talk_contr 12:48, 15 December 2011 (UTC)[reply]

The article section Idempotence#Idempotent ring elements defines an abelian ring as "A ring in which all idempotents are central". Corroboration can be found, e.g. here. Since this is a variation on the other meanings associated with abelian, it seems this should be defined, either in a section in Ring (mathematics) or in its own article Abelian ring. It seems more common to define an abelian ring as a ring in which the multiplication operation is commutative, so this competing definition should be mentioned too. Your feeling? — Quondum^☏✎ 09:14, 21 January 2012 (UTC)[reply]

Yes, I ran into this problem when writing that part. Here are my chief considerations:

1. I found the usage in Googlebooks to be split about evenly
2. I found the usage among abstract algebraists overwhelmingly the "central idempotents" version
3. The other version seemed more common in engineering/information disciplines and some C* algebra and analysis texts by foreign authors.
4. To say "xy=yx for all x,y in R", "commutative ring" is far and away the most established and standard term. (In grad school we always thought that using Abelian for commutative here was a naive mistake, but then I found the texts which mentioned both.)
5. I don't oppose an Abelian ring article, but it would be a bit short and might be challenging to cite. (I can think of several things to say, only one citation comes to mind.) Mentioning the two usages would definitely a top priority in the intro. If you think this article is worth creating I'll be happy to help. Rschwieb (talk) 13:39, 21 January 2012 (UTC)[reply]

• If I understand you, the term "abelian ring" will mean one or the other with similar probability, but when a commutative ring is meant, it will only rarely be called an abelian ring. Ergo: rings with only central idempotents are not discussed in the literature nearly as often as commutative rings are.
• Since there is already an article Commutative ring, a new article would have as its subject rings in which every idempotent is central (with a disambiguating hatnote as appropriate). Two questions arise:

  (a) is this worth

  (i) its own article,

  (ii) a section under Ring (mathematics) (that someone may later branch out as its own article),

  (iii) a mere mention in Ring (mathematics), or

  (iv) nothing at all?

  (b) if we were to create such an article, would "Abelian ring" be an appropriate title, or is there some other term that gets used more, is less ambiguous or might be more suitable for some reason? Why should the absence of non-central idempotents be of any interest (we must answer this if we write anything)? AFAICT central idempotents split a ring into direct sums (and every ring is split by the trivial idempotents into the trivial ring and the ring itself; I love trivial cases), e.g. split-complex numbers, which is about all I would be able to say about them. Perhaps something about how the ordering of idempotents in such a ring works, but that may be WP:OR. There might be something about central idempotents forming an integral domain or something like that (I'm guessing wildly here; I do not even know what the term means).

• My preference for (a) leans towards (i) or (ii), maybe the latter, but this depends upon its noteworthiness in ots own right.
• On (b), if there is no rival term to describe a "ring in which every idempotent is central", then "abelian ring" must be it. Where are such rings treated as a class, except in the study of idempotents? — Quondum^☏✎ 11:04, 24 January 2012 (UTC)[reply]

Yes the use of abelian to mean central idempotents is mostly in research papers, and I've never seen a competing adjective for this usage. These rings are definitely not notable enough for (ii) or (iii). They could have their own article but I think it might be pretty stubby. The first place I found "abelian ring" in a text was in Goodearl's von Neumann Regular Rings. He establishes that a VNR ring is reduced iff Abelian.

Central idempotents correspond to ring direct summands of the ring, yes. It is a very strong condition: the full matrix ring of square matrices over a field is abelian iff it is the 1 by 1 matrix ring.

The only idempotents in domains and local rings are 0 and 1, so they are trivially abelian. Nontrivial idempotents are zero divisors, so they can't exist in domains. Rschwieb (talk) 18:49, 24 January 2012 (UTC)[reply]

Err... – what am I missing? Surely if abelian ring is not notable enough for (ii) or (iii), then it would not be notable enough for (i)? — Quondum^☏✎ 19:19, 24 January 2012 (UTC)[reply]

I'm not sure what your reasoning is, but the way I see it: even obscure topics can be worthy of wiki pages, but obscure topics do not deserve to waste space in their parent article. If one obscure notion got in, then 10,000 more obscure notions would go in too.

If "adjoint morphic transversalism" was an obscure notion in category theory, maybe it deserves a page, but it would be nonsense to put it in Category theory.

I think if "abelian ring" is not given its own article, there is no better place for it than where it currently is. Rschwieb (talk) 20:30, 24 January 2012 (UTC)[reply]

• Okay, I'm following you, and agree with the principle. What you mean is that an obscure subtopic should not be directly mentioned in its great-grandpappy article. However, this is not at variance with the idea that in a densely cross-linked medium, a chain of traceability should exist in both directions. Thus, it should be mentioned in the article of the slightly less obscure topic one level up: its immediate parent(s).
• The immediate parents of abelian rings (as per the meaning we seem to have settled on) would I guess be idempotence and perhaps representation/classification/decomposability of rings. Is the latter a topic independent of idempotence? If so, the traceability should exist in that direction too.
• My inclination, if some reference has covered the topic adequately (even if it is only a primary reference), is to consider a separate article, using that reference as basis. If it is merely mentioned in passing in texts, then I prefer that we should focus only on clarifying it where it is in Idempotence.
• Why I feel some clarification is needed is that its presence immediately suggests unanswered questions, such as why the absence of non-central idempotents from the ring should serve as a useful basis for classification (as is implied by the use of a name, "abelian ring").
• And as a side topic, the mention in Idempotence#Role in decompositions appears to be inconsistent at first glance, or at least I cannot connect the concept of "a unique idempotent ... such that ..." to "a central idempotent", the latter being what I would have expected to have the property given. But it is murky to me: modules seems to be a more general classification than rings (over fields). Should the mention of modules and rings be separated a bit more in this section? — Quondum^☏✎ 06:52, 25 January 2012 (UTC)[reply]

It seems like there isn't much to say about central idempotents. They correspond to ring decompositions of a ring. To say that all idempotents are central implies that any one-sided ideal generated by idempotents (any module direct summand, for instance) is actually two-sided. For von Neumann regular rings, this means that all right and left ideals are actually two-sided. I'm not aware of an exhaustive resource on the topic, but that doesn't mean they don't exist.

The problem with the "Abelian" condition is its strength. Very basic types of rings, like semisimple rings and von Neumann regular rings, are usually rife with noncentral idempotents.

I like the way modules and rings are presented together in the Role in Decomposition section. The idempotents flagging module direct summands do not have to be central.

I'll be waiting for feedback on specific points in the section from you. Maybe some things need clarification or citation. Rschwieb (talk) 14:31, 25 January 2012 (UTC)[reply]

I feel a bit like I'm being thrown in the deep end without my water wings. I can't think much beyond rings at this stage.

1. The five subsections under Idempotence#Idempotent ring elements do not strictly relate to rings. I suggest that this section should be renamed Idempotents in abstract algebras, and a subheading Rings below it should be inserted.
2. Do idempotents in rings always occur in orthogonal pairs?
3. The statement starting "A ring in which all idempotents are central..." is not directly relevant to the point in which it is stated – it is really just a follow-on. Should it not be added as a by-the-way type statement at the end of the list?
4. There seems to be a super-strong connection between idempotents, zero divisors and (when 2 is invertible) involutions. It somehow feels this has not been presented graphically enough. Although I am pretty sure there are zero divisors that are not idempotents. Do nilpotent elements tie in with idempotents?
5. I've seen involutions used to decompose Clifford algebras (involution → idempotent pair → decomposition) as though every involution leads to a decomposition as a direct sum. Is this the case? There seems to be something wrong with this, since the idempotents must be central. Can non-central idempotents usefully decompose a ring?
6. Should I be looking up endomorphisms? I think I just strained a meta-muscle.

— Quondum^☏✎ 17:48, 25 January 2012 (UTC)[reply]

• Re1: When I look at the subsections, and they all look like they are concerned with ring idempotents. I don't think relabeling it with "algebras" makes any sense, because algebras are just special rings.
• Re2: Yes, if e is idempotent, then 1-e is idempotent and orthogonal to e.
• Re3: There might be some room for improvement here. Instead of listing things like "abelian ring" and "boolean ring" next to the "type of idempotents" entries, we could keep the current list strictly for idempotent types, then have a section for stuff like abelian rings and boolean rings later. I'll definitely take a whack at that.
• Re4: I am rather ignorant about involutions, other than the fact they are isomorphisms between R and the opposite ring R^op. This could be a place for improvement. The only nilpotent idempotent is 0! Another fact that I'm stating vaguely and incorrectly: idempotents will typically be outside of the Jacobson radical, nilpotents will typically be inside.
• Re5: Any idempotent decomposes the ring R as R=eR⊕(1-e)R and R=Re⊕R(1-e) into modules. If, in addition, e is central, then R=eR⊕(1-e)R=eRe⊕(1-e)R(1-e)=Re⊕R(1-e). You can check the decomposition in the middle is actually a ring decomposition (in addition to being a left and right module decomposition.) Good question about which involutions give rise to central idempotents: I don't immediately know but it seems like we can figure out the answer :)
• Re6: endomorphism of M just means homomorphism from M to M.
• Update: Your comments on involution helped me spot that it was unclear that module involutions were meant, and not ring involutions. I'll try to line this up with your question about Clifford algebras soon.
• Update: Any idempotent will give rise to a module decomposition, though not necessarily a ring decomposition. It looks like central elements that act as involutions correspond to central idempotents. Rschwieb (talk) 02:35, 26 January 2012 (UTC)[reply]

1. re Re1: Role in decompositions immediately starts talking about modules, which is a different class than rings, hence why it does not seem to fit under rings; I know it then talks about an endomorphism ring, but this is not an obvious connection to "ring elements" in the section heading. And I meant abstract algebra, not algebra (over a field). The first sentence in this section makes no sense to me. It mentions "idempotents of R" and then again ""idempotent e in E". There seems to be mixing of the meta-layers (E and R), whereas no motivation has even been given as to why they can be considered isomorphic, or what the map E → R is.
2. re Re2: So I'll assume that in rings, idempotents elements occur in orthogonal pairs of distinct elements, with the isolated case of the trivial ring in which the elements of the pair are not distinct. It may occur, I imagine, that elements of different pairs are also orthogonal. By the way, how is "orthogonal" defined here? I assume by ab=ba=0. This is confusing, because it is not the definition in, for example, a Clifford algebra, where it is ab+ba=0, and then only elements of the generating vector space.
3. re Re4 and your edit: I'm not entirely happy with the last change you have made. You seem to have lost the sense of an involution as an element, i.e. an element that squares to 1 under a binary operation (in a ring, by definition the multiplication operation). The connection between an involution and a pair of idempotents is as elements of a ring (as I see it), whether this be an endomorphism ring or otherwise. Think of an involution as any element x of a ring such that x²=1. From such an element (or its additive inverse), we can easily construct two idempotents (subject to the invertibility of 2:=1+1). No mention of endomorphisms required, I'd guess.
4. re Re5: Module decomposition versus ring decomposition – ouch, I'll have to figure out what this means. I am not familiar with modules.
5. In general, I'm way over my head here. Module vs. ring involutions, Jacobson radicals, endomorphism rings ... yikes. — Quondum^☏✎ 09:30, 26 January 2012 (UTC)[reply]

• The last points 1 and 3 need to be tackled with a single answer. Stick with me here: the entire section revolves around idempotent elements in the endomorphism ring E=End_R(M) of a module M. Idempotent ring elements of E are precisely idempotent homomorphisms on M. Bottom line: idempotent elements of E correspond to decompositions of M. Now in the special case M=R, the ring R is naturally isomorphic to E=End_R(R): the endomorphisms are given by left (or right) multiplication by ring elements. In this case, then the endomorphisms we were discussing are exactly idempotent elements of R(which are hard at work decomposing the module R).
• Last point 2: Yes, ab=ba=0 is what orthogonal means here. It's possible that this does not jive with geometric orthogonality in the CA (do you have something I can read about what you are alluding to?).
• Last point 4: I think you will catch on quickly. I'll try to find time to summarize the situation clearly, but here is a sketch of internal direct sums: if you have seen M=A⊕B written for a module M, then A and B are submodules of M such that A+B=M and A∩B=0. These conditions amount to: Every element of m of M is uniquely expressible as a+b for some a in A and b in B. For rings, it is exactly analogous same: if you see R=S⊕T, then you know that S and T are rings, and that S+T=R and S∩T=0. Again, this amounts to elements of R being uniquely expressible as s+t for some s in S and t in T. Do note that the type of subobjects have changed. This is all that separates "module decompositions" from "ring decompositions" in our discussion. (I would like to tell you sometime how this is isomorphic to the external direct sum A⊕B, and explain that internal direct sums are just a reinterpretation external ones.)
• Last point 5: No worries, just enjoy the ride. You'll learn plenty, and there's no obligation to learn everything. Rschwieb (talk) 13:27, 26 January 2012 (UTC)[reply]
• Re My edits to idempotents page: I realize an idempotent element and an idempotent operation are two different things, but they both depend entirely on the binary operation. The unary-binary split seems most logical. I don't think anything is gained by singleing out elements. Rschwieb (talk) 17:45, 26 January 2012 (UTC)[reply]

1. I think you get the idea that in some aspects the article is not written to make sense to someone who has only the necessary knowledge to understand what it is saying. That is to say, it can be written to be mostly understood without already knowing about endomorphisms etc., but at the moment it requires such knowledge to make any sense of those same aspects. Maybe give a bash at fixing that rather than explaining it to me?
2. I'm surprised you want stuff on orthogonality. Orthogonality on a vector space seems to be defined in terms of a reflexive bilinear form, which seems to naturally suggest either ab–ba=0 or ab+ba=0 as a definition of orthogonality on, at least, the generating vector subspace of most K-algebras. Anyhow, it is interesting to note that orthogonality defined by ab=ba=0 ensures orthogonality in terms of any bilinear form defined in terms of a K-bilinear multiplication in a subspace of a ring, i.e. one implies the other over the latter's range of applicability, and might thus be naturally related. I'm surprised I haven't seen the concept of orthogonality in this sense ("centrally orthogonal"?) before; it does not appear at all in the article on orthogonality.
3. I like the idea of a decomposition defined in terms of "uniquely expressible as". Worth getting familiar with. But that is clearly not as powerful as "uniquely decomposable as", which seems to tie in with idempotents.
4. I disagree on your split on the types of idempotent. We are defining three distinct types. Of course they're related. Actually, the unary operation and the element are most closely related - they are in fact identical in some sense. The binary operation is the odd one out. In any event, it is disconcerting for the reader to refer to three types of idempotent, and only see two headings. From the reader's perspective, it is important to organize headings by concept, not by relatedness of explanatory detail. — Quondum^☏✎ 18:01, 26 January 2012 (UTC)[reply]

• Re 1+4: OK, please do your best but be judicious. We have a difference of opinion, but I don't really object to anything you've said that strongly. It would be nice if a third and fourth person would comment, eventually.
• Re 2: Since there are so many notions of orthogonality, I just wanted to know which one you were thinking of. I'm not really sure what to make of ab-ba=0 other than a and b commute, but I have some thoughts on the other one. I always associate orthogonality with a product being zero, such as the inner product. If we're in a real Clifford algebra and selecting vectors a and b, then ab+ba=0 if and only if a⋅b=0 (inner product), so that looks like ordinary "right angles orthogonal" to me. Secondly, in any ring R, ab+ba=[a,b] is a definiton for a Lie bracket on R, so [a,b]=0 could be thought of as orthogonality with respect to the Lie product on the ring.
• Re 3: To say that a module is uniquely decomposable (with nonzero submodules) would be to say there are precisely two nontrivial idempotents corresponding to the halves of the decomposition. I'm not sure when that happens, but I can give you two examples with integers: the integers modulo 6 and the integers modulo 12 have exactly four idempotents! Rschwieb (talk) 18:54, 26 January 2012 (UTC)[reply]
• Update: Hm I seem to have forgotten that ab-ba is the usual Lie bracket, and not ab+ba! This is getting more interesting all the time. I wonder how these two brackets interplay with the stuff we have been examining. Rschwieb (talk) 01:47, 27 January 2012 (UTC)[reply]

re Re 1+4: I was hoping you'd give the editing a bash, with me acting as back-seat driver. I should not be working in main article space on stuff that I feel is wrong, but do not know what is right. However, when I have time, I could put together a sandbox suggestion and let you pull it to shreds. Remember that we are on your talk page: other opinions will be more likely forthcoming if we move it onto the article talk page.

re Re 2: You are suggesting a general concept of orthogonality with respect to a multiplication operator. All the examples listed other than the abstract algebra case may be characterised as reflexive: orthogonality is two-sided. The abstract algebra case does not fit into this picture: it requires two distinct identities to be satisfied simultaneously, and the multiplication is not assumed/required to be reflexive. Also, in general, the multiplication operator in most cases is frequently not the ring multiplication, but rather a related/derived operation that is reflexive. So the case that we are dealing with is atypical in many respects, and the term "orthogonal" is really a different meaning. This does however suggest the concepts "left-orthogonal" and "right-orthogonal", where the combination is what is what we are dealing with. I would not be surprised if these two new concepts are useful, especially in module decompositions.

re Update: Surely the Lie bracket is just another reflexive product? I've seen both ab+ba and ab−ba in physics contexts. They both fit the general picture.

re Re 3: Surely not quite. That's like saying 30 is not uniquely factorizable, because it has more than 2 prime factors. It has a unique prime factorisation, just like the external direct sum R⊕R⊕R has a unique complete decomposition (and by implication has four nontrivial central idempotents). A similar statement could be made for the integers mod 30, or for any squarefree modulus. See the Chinese remainder theorem. — Quondum^☏✎ 07:45, 27 January 2012 (UTC)[reply]

1. Orthogonality: Yeah, general ring multiplication can have ab=0 and ba nonzero, and as you said we want a minimum of symmetry if we want to call it orthogonality. The condition "ab=0 implies ba=0" is studied in a paper by P.M. Cohn under the term "reversible ring". Annihilation is studied a lot in ring and module theory, but I'm not sure how many authors use orthogonality as a motivating idea with annihilation.
2. Unique decomposability: I think I speculated what you meant and arrived at the wrong answer. I will try another answer: the Krull–Schmidt theorem is the main result on unique (in a sense) decomposability of certain modules.
3. I don't think I have any texts studying orthogonality with the commutator and anticommutator, can you recommend some sources? Rschwieb (talk) 14:22, 27 January 2012 (UTC)[reply]

• While I have run into commutator products and less often anticommutator products, I don't recall then being associated with the definition of orthogonality, so unfortunately I have no references. To me, orthogonality has always been associated with a bilinear form of some sort (generally symmetric, but that mention of reflexivity has thrown me). It was you that linked it to a more general sort of statement relating to multiplication. And the idea of idempotents being "orthogonal" may be a flash in the pan: an unfortunate term to mean a pair with a particular property. Perhaps we should regard it as an essentially unrelated homonym.
• Come to think of it, this "one-sided" orthogonality seems pretty useless. Every idempotent a trivially has its orthogonal twin b=1−a such that ab=ba=0. So it is centrality, and not reflexivity/"orthogonality" that is of interest in a true decomposition of rings.
• It would seem to me that all the idempotents only have to mutually commute (not be central) to produce unique left-decompositions and unique right-decompositions. I think my brain is going to mush on this. The Krull–Schmidt theorem is a little abstract for me at thos time of night. — Quondum^☏✎ 20:23, 27 January 2012 (UTC)[reply]