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The product of a quaternion multiplied by a real number is a quaternion, but the product of a quaternion multiplied by a complex number is not in general a quaternion. Why are the quaternions defined so narrowly? What was gained by banning zq=q' where z is complex and q and q' are quaternions?

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closed as not a real question by Alain Valette, Dan Petersen, Yemon Choi, Misha, Andreas Blass Oct 21 '12 at 12:07

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

What do you mean by "multiplying a quaternion by a complex number"? If your quaternions are the elements of usual, $\mathbb R$-algebra $\mathbb H$ then it does not make sense to multiply them by elements of $\mathbb C$. Now, there are ways in which one can make sense of your question (e.g., extend scalars to $\mathbb C\otimes_{\mathbb R}\mathbb H$, consider $\mathbb C$ as being contained in $\mathbb C$ (in one of the many ways you can do this)), but if we pick one of this interpretations, then it is no longer true that you do not get a quaternion when you multiply one by a complex number... –  Mariano Suárez-Alvarez Mar 6 '10 at 4:33
I'm afraid that I do not understand the question as stated either. On the one hand, Hamilton's quaternions are an R-algebra, not a C-algebra (by definition; also, there are no finite dimensional complex division algebras). On the other hand, there are numerous embeddings of C into the quaternions... Could you clarify your question please? –  Ben Linowitz Mar 6 '10 at 4:42
(I would imagine a quarternion's phase to be an element of $S^3$...) –  Mariano Suárez-Alvarez Mar 6 '10 at 4:45
That should read «as being contained in $\mathbb H$», not in $\mathbb C$. I hope editable comments are not long in their coming! –  Mariano Suárez-Alvarez Mar 6 '10 at 4:46

4 Answers 4

The question is a little ambiguous. However, maybe you are asking why don't people work with quaternions having complex coefficients as opposed to real coefficients.

The answer is that as an algebra, complex quaternions are isomorphic to the algebra of 2x2 matrices with complex entries, so there is nothing gained by using the quaternion notation, that isn't already available in a simpler form. You can find a proof in "Spin Geometry" by Lawson and Michelson.

There is still some fun to be had though. For instance, the group SU(2) is the set of real quaternions so that $q\overline{q}=1$ under quaternionic multiplication. We can define $\overline{q}$ for complex quaternions in analogy to the real quaternions so that it is complex linear, That is if $q=a+bi+cj+dk$ then $\overline{q}=a-bi-cj-dk$. The set of complex quaternions with $q\overline{q}=1$ under quaternionic multiplication is SL(2,C).

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Just a shout-out for Lawson and Michelson's book "Spin Geometry" it's great! –  Scott Morrison Mar 7 '10 at 3:42

There is a unique copy of $\mathbb R$ in $\mathbb H$ and it is the center. On the other hand, for every quaternion $q$ not in $\mathbb R$, the generated $\mathbb R$-algebra with unit is a copy of $\mathbb C$, but none is central.

I like to view quaternions as $q=(x,X)\in \mathbb R\times \mathbb R^3$ with multiplication $$ (x,X).(y,Y) = (x.y - \langle X,Y\rangle, X \times Y + x.Y + y.X) $$ This uses oriented Euclidean $\mathbb R^3$, and it turns into a definition. One can then take any oriented orthonormal basis $i,j,k$ for the basic imaginary quaternions.

Edit: Note that $X\times Y$ is a Lie bracket and $-\langle X,Y\rangle$ the corresponding Killing form ($\mathfrak s\mathfrak o(3,\mathbb R)$ up to a constant), one can repeat this construction for every real Lie algebra. Only for 3 of them one obtains an associative algebra.

Second edit: The reference for the first edit is:

Peter W. Michor, Wolfgang Ruppert, Klaus Wegenkittl: A connection between Lie algebras and general algebras. Rendiconti Circolo Matematico di Palermo, Serie II, Suppl. 21 265--274, (1989)(pdf)

Moreover, a similar construction as the above on $\mathbb C\times \mathbb C^3$ using also complex conjugation (see "Greub: Multilinear algebra, 2n ed. 1978" page 289) leads to octonians.

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This is sort of implicit in Ben Linowitz's answer, but if you allow complex scalars then you get zero divisors, so that $x \cdot y = 0$ does not imply that one of $x$ or $y$ be zero. To be explicit, look at the product $$(I + i J) \cdot (-i I + J) = i - i + K - K = 0$$ where $i$ is the complex scalar and $I, J, K$ are the quaternion generators. So these complexified quaternions don't have the same nice algebraic properties that the quaternions do.

That said, these things aren't "banned" --- they are studied under the name "biquaternions" and, while less useful than proper quaternions, still show up occasionally. I've had a brush with them while studying the differential geometry of plain old surfaces, for example.

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I do not understand Your question. Here You may find that quaternion may be concerned as pair of complex numbers: So If You proper define quaternion coefficients it may represent every complex number. As quaternions are algebra, so You may multiply one quaternion by other, so whole complex algebra is within quaternion algebra ( You may say it is isomorphic to subalgebra generated by c=0, d=0 in quaternion definition)! From the other side: if You see quaternions as abstract objects outside real numbers, then multiply quaternion by real number have to be defined! It is the same natural/unnatural embedding for real number within quaternions as for complex numbers within quaternions!

It is the same as with complex numbers. You may multiply complex number by real number and You obtain complex number, because You do not go out complex numbers indeed.

Please: explain what You are asking about?

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