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Thread: Quaternions

  1. #1 Quaternions 
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    My understanding of quaternions begins with the 'Brougham Bridge Equations" i.e.



    where i, j, and k are all 'independent' square roots of -1.

    I seem to be unable to follow the logic of this statement. Taking 'independent' to mean the three symbols i,j,k are not equal to each other, I then stated the following:



    Dividing both sides by i, I get



    From this, we can infer that 'i' is greater than j or k.

    However,



    Dividing both sides by j, we get



    This implies, then, that j is greater than i or k.

    Repeating the same, this time with k^2,



    But this then implies that all three values are greater than each other, i.e



    which is logically impossible.

    Have I grasped the wrong end of the stick somewhere? Where have I gone wrong?

    Thanks for any help!


    In control lies inordinate freedom; in freedom lies inordinate control.
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  3. #2 Re: Quaternions 
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    Quote Originally Posted by Liongold


    From this, we can infer that 'i' is greater than j or k.
    Why? Consider . What is ? Is it strictly greater that either x or y?


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  4. #3 Re: Quaternions 
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    Quote Originally Posted by Guitarist
    Quote Originally Posted by Liongold


    From this, we can infer that 'i' is greater than j or k.
    Why? Consider . What is ? Is it strictly greater that either x or y?
    I understand what you're saying, but will only be less than and if both and have different signs.

    will be greater than and provided both and have the same signs.

    The problem is that i, j, and k have the same signs. Or is that wrong?
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  5. #4  
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    There is no linear ordering on the quaternions. In fact, should be viewed as the unit basis vectors in .
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  6. #5  
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    Quote Originally Posted by salsaonline
    There is no linear ordering on the quaternions. In fact, should be viewed as the unit basis vectors in .
    I should confess, salsaonline, I am not exactly a mathematician. Suffice it to say I stumbled on this in Roger Penrose's A Road to Reality and was intrigued. I know very little mathematics apart from basic high-school level mathematics.

    That said, I would really appreciate if you could explain to me what you mean by the unit basis vectors in . Thank you.
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  7. #6  
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    The standard coordinate vectors .
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  8. #7  
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    and this happens with being a four dimensional 'hyper-space'. If you have problems visualizing it, I suggest as a starting point, think of it in terms of space-time, where you have the three spatial dimensions and an additional time dimension, if that helps at all.
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  9. #8 Re: Quaternions 
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    Quote Originally Posted by Liongold
    Quote Originally Posted by Guitarist
    Quote Originally Posted by Liongold


    From this, we can infer that 'i' is greater than j or k.
    Why? Consider . What is ? Is it strictly greater that either x or y?
    I understand what you're saying, but will only be less than and if both and have different signs.

    will be greater than and provided both and have the same signs.

    The problem is that i, j, and k have the same signs. Or is that wrong?
    You can think of the quaternions, as salsonline said, as defining a "multiplication" on . But in this case the multiplication is not commutative.

    This Wiki article gives a tutorial on the quaternions.

    http://en.wikipedia.org/wiki/Quaternion

    The quaternions are one of three associative real division algebras (a vector space over the real numbers with an additional operation of multiplication of vectors that permits division by non-zero elements.)

    http://en.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras)

    There is a non-associative extension that sometimes comes up, and I think I recall Penrose mentioning it in that book, called the Cayley numbers, or also sometimes called the Octonions.

    http://en.wikipedia.org/wiki/Cayley_numbers


    Salsaonline: Do you happen to recall the classification of real division algebras ? I was at one time under the impression that there was a proof based on algebraic topology that showed that the reals, the complex numbers, the quaternions and the Cayley numbers were all of them, but I am not certain, and I think that possibly that perhaps one needs to add the assumption that the algebras are normed to exclude other candidates.
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    DrR: I've never seen the proof. But if I recall correctly, yeah, the only real division algebras are the reals, complexes, quaternions, and octonions.

    With the complexes, we lose the ordering property of the reals.

    With the quaternions, we lose commutativity.

    With the octonions, we lose associativity.

    edit: forgot we were talking about real division algebras, and said something dumb, which I corrected.
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  11. #10  
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    Quote Originally Posted by Arcane_Mathematician
    and this happens with \mathbb{R}^4 being a four dimensional 'hyper-space'. If you have problems visualizing it, I suggest as a starting point, think of it in terms of space-time, where you have the three spatial dimensions and an additional time dimension, if that helps at all.
    How do I visualize R^4? I think of R^3 or R^2, depending on the circumstances.
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  12. #11  
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    I meant my post more for liongold than you, salsa, as he inhabits the physics forum more than here, and I figured the analogy could be useful for him to use, not so much yourself.

    also, I made a tex error which I have since corrected, sorry about that, forgot the tags.
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  13. #12  
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    So what makes the sedenions not a real division algebra? (note that I don't exactly know what a real division algebra is and don't have time to look it up right now)
    "There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges
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  14. #13  
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    Quote Originally Posted by Chemboy
    So what makes the sedenions not a real division algebra? (note that I don't exactly know what a real division algebra is and don't have time to look it up right now)
    Well, when you have time, go look it up.
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    I mean, to be fair, imagine if I asked a chess player, "what's a strong opening for white," and then added "btw, I don't know how to play chess, and I don't have time to learn right now. But please answer my question."
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  16. #15  
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    Quote Originally Posted by Chemboy
    So what makes the sedenions not a real division algebra? (note that I don't exactly know what a real division algebra is and don't have time to look it up right now)
    I looked it up, only because I had never heard of the sedenions (and with any luck won't run across them again). They are not a real divisioin algebra becaue they are not a division algebra at all -- they contain divisors of zero.

    http://en.wikipedia.org/wiki/Sedenion

    Question, since you had obviously heard of sedenions before and somehow know that they were not a real division algebra, it must have occurred to you to type "sedenions" into the little Google box. Why didn't you?
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  17. #16  
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    wow... apparently that came across badly. there was discussion of real division algebras, I had read about the sedenions, wondered why they weren't one, and so I asked to add to the discussion a little. I would certainly look up info on sedenions and real division algebras later to figure it out on my own but there's no reason I can't ask a simple question to add a little to the thread. I do understand how that could be read the wrong way though, I should have worded it better or omitted that part entirely. I'm sorry you took it the wrong way. From everything I've been learning in this forum I'd hope you know that I really love math and don't want to take the easy way out.
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  18. #17  
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    Well, if you'd heard me in person, you would have heard me say it with a slight chuckle.
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  19. #18  
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    Quote Originally Posted by Chemboy
    wow... apparently that came across badly. there was discussion of real division algebras, I had read about the sedenions, wondered why they weren't one, and so I asked to add to the discussion a little. I would certainly look up info on sedenions and real division algebras later to figure it out on my own but there's no reason I can't ask a simple question to add a little to the thread. I do understand how that could be read the wrong way though, I should have worded it better or omitted that part entirely. I'm sorry you took it the wrong way. From everything I've been learning in this forum I'd hope you know that I really love math and don't want to take the easy way out.
    I did seem a little out of character.

    As noted earler ( and I edited it to call attention to the division aspect) an algebra is just a vector space with and additional operation, multiplication of vectors. A division algebra is an algebra in which everything except zero has a multiplicative inverse (which of course requires an identity for multiplication).

    An easy example of an algebra is the set of all nxn matrices, for some fixed n. The matrices can be added and subtracted, and multiplied by scalars, so they form a vector space. Now throw in matrix multiplication and you have an algebra. It is not a division algebra because not all non-zero matrices have an inverse.

    Divisoin algebras are not, at least to me, all that interesting. But algebras are very interesting.

    Here is another example. Take the set of absoluely integrable functions on the real line (you need Lebesque integrable here to make things work properly but if you don't know about Lebesque integrals just think of Riemann integrals since they are the same thing whenever a function is Riemann integrable). That is look at the set of all complex valued functions f defined on such hat . Call this set of functions

    For f and g in define a "multiplication" called convolution by





    Then is an complex algebra which has some more structure making it what is called a Banach algebra. This operation of convolution and the Banach algebra turn out to be very iimportant in the theory of differential equations and especially in the theory of the Fourier transform. In fact, in the slightly more general setting of locally compact abelian groups, one finds that the theory of the Fourier transform is deeply connected to the theory of the Banach algebra of where is the group. In fact as far as the theory goes, the Fourier transform is the same thing as the Gelfand transform which is an animal that is defined in terms of an abstract Banach algebra. This is the "right way" to study Fourier transforms, which includes the Fourier series (just the Fourier transform on a compact abelian group).

    Warning. Chemboy take my word for this, at least for the moment, as the background needed to make sense of what I have said in a rigorous way requires some knowledge of measure and integration and a bit of functional analysis. I have omitted some important details. Once you have more background, the book Fourier Analysis on Groups by Walter Rudin expalins this stuff in detail. Beyond that there is a wealth of interesting mathematics, but it is pretty specialized. It is also cool stuff.
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