matt grime

So, in reading up on the quaternions you didn't bother to look up the definition? You have confused them, Johnny precisely because you don't know what either of them is. 1. An algebra is a ring that is a vector space (over a field) It is the definition of what an algebra is. 2. The quaternions are R^4 equipped with a product that satisfies the correct properties to be a ring and is thus an algebra. We declare the basis to be given as 1,i,j,k with the rules you know so well: ij=k jk=i ki=j and ji=-k kj=-i ki=-j Thus H is the set of all symbols a+bi+cj+dk where a,b,c,d are in R and mutliplication extended linearly. I've told you at least once before if not twice. Q_8, the quarternion group is the set of quaternions with exactly one of a,b,c or d equal to plus or minus 1. It is a group with 8 elements. You recall that you implied that the quaternionic people had made a mistake with the naming? Well, if you don't even know what they are how can you say such things?

And what is your point? Note it states "two directed lines", not necessarily vectors in the sense we mean, though we'd have to guess what the old terminology meant. Of course as Hamilton invented vectors after inventing quarternions (yes, this surprised me too, but is apparently true) we cannot state he invented quarternions in order to "divide vectors".

They (the quarternions as a division algebra as you called them yourself in this thread) are not a group under multiplication, just as the real numbers are not a group under multiplication. What is the mutliplicative inverse of 0? The non-zero elements are a group undermultiplication. The link you posted to is about Q_8 the quarternion group (of integral unit length quarternions) comes around from the fact that the elements +/-1, =/-i, +/-j and +/-k are a finite group of order 8 under multiplication. You are confusing the Quarternionic group with the Quarternionic algebra (strictly speaking there are an infinitenumber of quarternionic algebras, this being the one we commonly talk about), this isn't that surprising, since they are both commonly called just quarternions. They are different. Only the second is a division algebra. The first is not an algebra. Why do I suspect you do not know what an algebra is?

A group with respect to what operation Johnny. Trivially they are a group under addition since they are a vector space. They are not a group under multiplication.

I have no idea, nor do I care, what is in this manual that you have and that I do not, and you are being obfuscatory to say the least when talking about this mythical manual. The Quarternions are famously a divisoin algebra, they are the 'it' I refer to, and I do understand them very well, as will most mathematicians: the unit integral quarternions and D_8 are the smallest non-isomorphic groups with the same character table for instance. For someone who is using "vector" without bothering to state which vector space he is referring to that is a very strange post you just made. I know how to divide quarternions, but you are just asking how to "divide vectors" what vectors; you are being very unclear as to what vectors you mean to divide. Do you mean the vectors in H, the quarternions, or not? I suspect you mean, from you original post, that you want to divide two vectors in R^3 in some sense, though you haven't been clear on this. It is you that needs to start explaining yourself clearly.

It is trivial Johnny since in a division algebra one can divide by definition. The quartnions possess a multiplicative norm.

Eh? What on earth? You wanted to call it a 4 element real vector space, no one else. It's a divisoin algebra, Johnny..... Only you seem to want to divide elements of R^3, so why don't you decide what you mean.

So what? No disrespect, but I do not trust your judgement on matters mathematical. A binary relation requires sets A, B, AxB, and a subset of AxB. I see nothing about a set of all sets being mentioned (which is Cantor's Paradox - Russell is the set of sets that do not contain themselves)

"4 element divison algebra" would indicate that it had 4 elements. There are an infinite number of quarternions (as there must be in any non-zero real vector space), so I really advise you to stop inventing notation that makes no sense.

Right, let's clear up a few things straight away. 1. S is not a rotation, but it can be *interpreted* that way if you so choose I imagine. They are *just* a 4-dimensional real division algebra. Saying the first is an angle is like saying the real numbers are lengths. If it helps to visualize them in a certain way feel free to do so but don't confuse the visuallization with them as abstract objects. The unit quartrnions aer iso to the 4-sphere, and thus SU(2), and various things like the hopf fibration can be described using them. 2. The quarternions (denoted H for hamilton) *are* real 4-dimensional vector space, they are also a 2-dimensional complex vector space. I could have used the word algebra too instead of vector space. 3. The elements i,j,k *are not* a priori orthogonal, since there is no such thing as a canonincal inner product, though there is an obvious one with respect to which we can delcare them orthogonal. 4. Hamilton was attempting to define a 3-d space over the reals that was a field, liek C is a 2-d real vector space that is also a field. He realized that in fact it needed to be 4-d, and that it couldn't be commutative, hence only a division algebra. we also have the 8-d octonions.

How? I'd put money on it not doing.

Ok, don't buy the bridge thing (I don't think many people do), but Hamilton is still widely credited with inventing the quarternions. Are you going to post "that matrix"? I looked at the page (light font dark background) and didn't see anything like that.

SO, if you "divide" two vectors, presumably in R^3, you get something that is in a 4-dimensional real space, or a 2-d complex space, or the quarternions. Hmm. And you think that's a reasonable notion of division of two vectors, which Tom, as most poeple would, took to mean as some operation on R^3 that 'undoes' some multiplicative operation. Of course, since we can embed R^3 into the quarternions in any number of ways we can take inverses there - again the operation will not be closed with respect to the embedding.

Yes, but Hamilton is usually credited with creating the Quarternions. I've no idea who Gibbs is.

Take it to its (il)logical conclusion: if < denotes less than on the real numbers is duck<lion true? No, but neither is its negation: it is meaningless, since < isn't a relation on the animal kingdom.

Doesn't that strike you as being completely tautologous? It ought to, since it is. If x and y are not in T then xBy makes absolutely no sense. The domain is part of the definition of relation (and function). Moreover if x and y are not both in T then xBy is neither true nor false - it is out of the domain of B and is meaningless.

All manner of applied maths uses tensors in that sense (stress strain etc). And we in pure have a similar though stirctly speaking different thing too.

Tom said binary *operation* not binary *relation*, so what was the point of that I mean in what sense can a binary relation even be thought of as producing another element of the set? Anyway, it depends on how they define binary operation. Some leave the codomain implicit some explicit. Often they are are defined to satisfy closure since * will be referred to as a binary operation from GxG to G. Though quite why some 'fault' of binary relations means we ought to state something about a completely unrelated thing is a mystery. (nb no one should ever define functions as relations)

Indeed. SU(2) is the simply connected cover of SO(3). They are not isomorphic as Lie groups (there can be no homeomorphism between them, never mind one that respects the group structure). I doubt they are isomorphic as abstract groups either, though I can think of no compelling reason for that in all honesty. But that possibly reflects the fact I've been to the pub after work. Note it is possible for topological groups to be abstractly isomorphic without being isomorphic as topological groups, eg R with the normal topology and R with the Zariski, discrete or indiscrete topology.

Would people at least mind explaining in what ring they are finding the inverse of "infinity" or indeed in what ring "infinity" lies? Would it be better if I phrased it as: 1. What do you mean when you talk about dividing by things? 2. what are the domains of any binary relations you may wish to define? 3 what are the compatibility rules for these operations? 4. do you accept you aren't talking about real numbers?

I am not penalizing you I am asking you to clarify what you meant since the words you use have a strict mathematical meaning different from your usage of them. There may be some tricks, some short cuts, but if you really have to write it out in base 10 then you just have to get on and do it. Why should it be any other way? Have a big bit of paper handy though.

Here are the tools you need, I think, It may be overkill, but I don't think I've left anything out. If f is a map from R to Hom(M,M), then Im(f) is iso to R/Ann(M) where Ann(m) is the set of all x in M such that rx=0 for all r in R. If f is faithful, then Im(f)=R If M is simple then M=Rm for any m in M. Any simple module is iso to R/L for some maximal left ideal L. note, I'm assuming you mean left module. Is R unital? Commutative? Shuold we reallyl be talking about right modules? Answers to any of these?

What do you mean by solve? (nb there are no equations there; equations are characterized by having equals signs in them)

Johnny, I know what a Talyor series is, and how to find them (and I certianly wouldn't sue your method since you would need to prove that sum and integral were allowed to be intechanged, and that is difficult usually) and how to find its remainder in one of three forms, but I do not memorize them any more, nor do I have any need to, nor is it important what the Taylor series is. You have missed the point that I was trying to make to Sarah that as long as she has the correct idea as to how to solve it that is the main thing. The actual answer is pretty immaterial. Please don't hijack this thread.