Quote:Original post by Troll
I might be wrong, but I believe the most that can be said about quaternions is that they form a non-commutative ring with inverse.

Not really. Well, for one thing they form a non-abelian group. And while that is a property of quaternions, it isn't the most that can be said. As in, there are a number of algebraic object that satisify that property that are no isomorphic to the quaternions. For example, 3x3 matrices over the integers form a non-abelian group, but are not isomorphic with the quaternions. As I mentioned earliers, quaternions form a four dimensional normed division algebra over the real numbers, which is a sufficient description (all four dimensional normed division algebras are isomorphic to the quaternions).

Quote:Original post by SiCrane

Quote:Original post by Troll
I might be wrong, but I believe the most that can be said about quaternions is that they form a non-commutative ring with inverse.

Not really. Well, for one thing they form a non-abelian group. And while that is a property of quaternions, it isn't the most that can be said. As in, there are a number of algebraic object that satisify that property that are no isomorphic to the quaternions. For example, 3x3 matrices over the integers form a non-abelian group, but are not isomorphic with the quaternions. As I mentioned earliers, quaternions form a four dimensional normed division algebra over the real numbers, which is a sufficient description (all four dimensional normed division algebras are isomorphic to the quaternions).

I dont mean to intrude here but I think its only fair to point out that he said non-commutative ring with inverse. So there is a bit of a redundancy to note that "for one thing they form a non-abelian group". Also, he said ring and SO(3) does not fit the criteria of forming a [non-commutative] ring. He also says that there exists S^3 -> SO(3), which addresses your 3x3 matrices mention since SU(2), isomorphic to the set of unit quaternions S^3, are a double cover (any two quaternions can be used to represent a rotation in R^3) of SO(3).

It would seem that he did say everything that could be said save that the concept of division deos not exist natively for rings. To describe the quaternions most succintly would be to simply state that they form a skew field or a division ring, as dragongames states.

I will note that my mention of the existance of homorphisms and double covers of this and that is not to show that SO(3) is isomorphic to S^3 (which you are right,SiCrane it is not) but rather, to show that Troll is aware of this fact.

Quote:Original post by Anonymous Poster
I dont mean to intrude here but I think its only fair to point out that he said non-commutative ring with inverse. So there is a bit of a redundancy to note that "for one thing they form a non-abelian group".

Rings aren't necessarily associative. Groups are.

Quote:To describe the quaternions most succintly would be to simply state that they form a skew field or a division ring, as dragongames states.

Skew fields and division rings aren't necessarily normed division algebras.

Quote:Original post by SiCrane

Quote:Original post by Anonymous Poster
I dont mean to intrude here but I think its only fair to point out that he said non-commutative ring with inverse. So there is a bit of a redundancy to note that "for one thing they form a non-abelian group".

Rings aren't necessarily associative. Groups are.

Exactly, SO(3) would not appear to form a ring based on what I said, making your example unnecessary.

But I am guilty of rushing (in my haste i forgot that for multiplication, a ring only requires that it be associative), Matrices can in fact from a non-commutative ring so what you said applies. hehe. *abashed*.

Quote:To describe the quaternions most succintly would be to simply state that they form a skew field or a division ring, as dragongames states.

Skew fields and division rings aren't necessarily normed division algebras.

Here, I agree with your statement of [non-commutative] division rings not necessarily forming a normed division algebra but a skew field, inherently by definition, must. I argue here that the existance of a skew field is structure enough to allow for the definition of a norm. I could be wrong and am operating only on instinct but if I am wrong then I would appreciate you correcting my misconception.

And if we are going to be pedantic we would have to say quaternions form a non-commutative normed associative finite dimensional division algebra with identity over the field of reals. :)

hmm, i made an error with the quotes, sorry. I said:

Exactly, SO(3) would not appear to form a ring based on what I said, making your example unnecessary.

But I am guilty of rushing (in my haste i forgot that for multiplication, a ring only requires that it be associative), Matrices can in fact from a non-commutative ring so what you said applies. hehe. *abashed*.

There are only four normed division algebras: R, C, H and O. There exist skew fields that are not isopmorphic with R, C, H or O (such as all finite skew fields). Therefore skew field is not synonymous with normed division algebra.

Quote:
And if we are going to be pedantic we would have to say quaternions form a non-commutative normed associative finite dimensional division algebra with identity over the field of reals. :)

Again, descriptive but not definitive.

Quote:Original post by Anonymous Poster
Exactly, SO(3) would not appear to form a ring based on what I said, making your example unnecessary.

Wait, so first you complain that I'm not descriptive enough (using non-abelian vs non-commutative) and now you're complaining that I'm including unnecessary description?

Quote:Original post by SiCrane
There are only four normed division algebras: R, C, H and O. There exist skew fields that are not isopmorphic with R, C, H or O (such as all finite skew fields). Therefore skew field is not synonymous with normed division algebra.

Your reasoning is a bit confusing but I see your meaning. I am being too specific in my use of the word skew-field. Okay, thanks for helping me clear that up. Skew-field is not enough of a description to capture the properties of a quaternion. But that does not preclude the fact that quaternions are an example of a skew field no?

Quote:

Quote:
And if we are going to be pedantic we would have to say quaternions form a non-commutative normed associative finite dimensional division algebra with identity over the field of reals. :)

Again, descriptive but not definitive.

You said "As I mentioned earliers, quaternions form a four dimensional normed division algebra over the real numbers" is sufficient but what I stated is not? Can you give an example of something that forms "a non-commutative normed associative finite dimensional division algebra with identity over the field of reals" that is not a quaternion or at the very least isomorphic to it?

Lets see, it can proven with topological arguments that a division algebra can only exist for D = 1,2,4 and 8 or as R, C , H and O as you yourself mention. Both R and C are commutative. O is not associative. So what am I missing then? I believe I caught the essence of it. That is all that is needed to construct for yourself the quaternions with the basic background knowledge ofcourse.

Quote:Original post by SiCrane

Quote:Original post by Anonymous Poster
Exactly, SO(3) would not appear to form a ring based on what I said, making your example unnecessary.

Wait, so first you complain that I'm not descriptive enough (using non-abelian vs non-commutative) and now you're complaining that I'm including unnecessary description?

No, I said based on my initial reasoning that your example does not apply. But I later realized an error in the reasoing and recapitulated. No hard feelings intended.