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Why is quaternion algebra 4D and not 3D? Complex algebra is 2D and what is known as quaternion algebra jumps to 4D.

$ i^2 = j^2 = k^2 = ijk = -1 $

Using $1, i, j,$ and $k$ as the base (where complex uses $1$ and $i$ (or $j$ if you are an EE)) which results in a 4-axis space. Why is there no 3D algebra?

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possible duplicate of Is there a third dimension of numbers? –  J. M. Oct 12 '11 at 17:10
    
This is related... –  J. M. Oct 12 '11 at 17:11
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1 Answer

There are only four normed division algebras (algebras where division by nonzero elements is possible) over the reals: the reals themselves, the complex numbers, quaternions, and a strange (alternative but nonassociative) algebra called octonions.

The reason that the dimensions are in geometric progression 1, 2, 4, 8 is that they can be derived from repeatedly applying the Cayley-Dickson construction, which doubles the dimension at each step. This explains the absence of dimension 3.

Generally, as you go up or down the Cayley-Dickson ladder you lose properties (as well as gaining some properties). From the reals to the complex numbers you lose order; going to the quaternions you lose commutativity; going to the octonions you lose associativity; going to the sedonions you're no longer alternative or a division algebra.

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But the question is deeper than this -- just because Cayley-Dickson does not lead to a 3-dimensional division algebra doesn't mean that there isn't another construction that does. I think it's a fairly deep result that there isn't; I've never seen it proved. –  Henning Makholm Oct 12 '11 at 15:20
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@HenningMakholm: I agree -- because of its depth I intentionally skirted the issue. If Christopher indicates that he is knowledgeable about the subject more details may be in order, but I suspect that my answer is already somewhat too technical (given that it was originally not posted on math.se). –  Charles Oct 12 '11 at 15:26
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@Henning It is not too difficult to show that a 3D-division algebra over the reals cannot exist. If $D$ were such a beast, and $x\in D, x\notin\mathbf{R}$, then consider the left regular representation of $D$ by 3x3 real matrices. The matrix $A$ representing multiplication by $x$ from the left cannot be scalar, because $x$ was not one. OTOH the eigenvalue polynomial of $A$ is cubic, and hence has a root $\lambda\in\mathbf{R}$. We have just shown that $x-\lambda$ is not invertible. –  Jyrki Lahtonen Oct 12 '11 at 15:46
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Henning: The simplicity of that argument (with hindsight) makes it all the more amusing that Hamilton spent years trying to multiply triples before he hit on the idea of using 4 coordinates instead of 3. –  KCd Oct 12 '11 at 17:59
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@KCd: we tend to consider linear algebra as a natural phenomenon, yet it is one of the truly great inventions of humanity! –  Mariano Suárez-Alvarez Oct 12 '11 at 19:14
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